diff --git "a/stack-exchange/math_overflow/shard_23.txt" "b/stack-exchange/math_overflow/shard_23.txt" deleted file mode 100644--- "a/stack-exchange/math_overflow/shard_23.txt" +++ /dev/null @@ -1,23425 +0,0 @@ -TITLE: stacks that are not necessarily fibered in groupoids appearing in algebraic geometry and differential geometry -QUESTION [5 upvotes]: Question: - -What are (some of) the stacks (occurring in algebraic/differential geometry) that are fibered in arbitrary categories and not necessarily in groupoids? - -In the notes Notes on Grothendieck topologies,fibered categories and descent theory Angelo Vistoli introduce the notion of a stack over a site $(\mathcal{C},\mathcal{J})$ to be a fibered category (not necessarily fibered in groupoids) over $\mathcal{C}$ satisfying some "locally determined" condition. -But, examples of stacks of interest in algebraic geometry and differential geometry (a small set of examples I have seen) are always fibered in groupoids. So, what could be a justification or necessity for introducing the notion of stacks fibered over arbitrary categories, if "almost all" stacks that occur in Algebraic geometry (that I know) are fibered in groupoids. -There might be interesting examples of stacks outside algebraic geometry of differential geometry that are not necessarily fibered in groupoids. I would be happy to see such examples (please add them as answers if you wish) but as for this question, I would like to learn about situations in algebraic geometry or differential geometry. - -REPLY [3 votes]: Virtually any kind of algebraic structure (e.g., group, ring, module, vector space, affine space, etc.) leads to a stack in categories -whose objects are bundles of such structures and morphisms are fiberwise -homomorphisms of such structures. -For example, the stack Vect of (finite-dimensional, say) vector bundles -is a stack in categories over the site of cartesian smooth manifolds. -Likewise, the stack BGrb^n_A of bundle n-gerbes with structure group A -is a stack in (n+1)-categories. -As a practical application, one can immediately define the category -of vector bundles or bundle n-gerbes on a given stack or ∞-stack S as the category -of (derived) sections S→Vect or S→BGrb^n_A. -This also captures the symmetric monoidal structure, and in the case of -vector bundles, the bimonoidal structure.<|endoftext|> -TITLE: Explicit computation of spinor norm -QUESTION [6 upvotes]: I've asked this on math.stackexchange, unsuccessfully. I hope this question is appropriate for mathoverflow. - -Let $V$ be a finite-dimensional vector space over a field $K$ with $\operatorname{char}K\neq 2$, and $Q$ a non-degenerate quadratic form on $V$. The spinor norm is a homomophism -$$sn: O(V,Q) \rightarrow K^*/(K^*)^2$$ -defined as $Q(v)$ for reflections by a non-isotropic vector $v$. -Alternatively, for $g \in O(V,Q)$ let $a \in \Gamma(V,Q)$ be the element of the Clifford group that realizes $g$ via an inner graded automorphism. Then, $sn(g)$ is defined as $N(a)=a^t a$, which is a scalar if $a$ comes from the Clifford group. -I am interested in explicitly computing $sn(g)$ for a given $g\in O(V,Q)$. I know a bit in some special cases: - -For the Euclidean space and the corresponding $O(n,\mathbb R)$ group the spinor norm is trivial $sn(g)=1$, since the group is generated by reflections by vectors of unit norm -For an algebraically-closed field $K$, the spinor norm is always trivial since $K^*/(K^*)^2$ is trivial -For a metabolic space $V = W \oplus W^*$ with the form $Q(w,f) = f(w)$, any $g \in \operatorname{GL}(W)$ gives rise to an orthogonal transformation on $V$ by the formula $g \cdot (w,f) = \left(gw, \left(g^{-1}\right)^*f\right)$. The spinor norm of this transformation is equal to $\det g$ (this is half-anecdotal: I've heard it in a Russian video lecture on Clifford algebras, presented without complete proof). -In particular, for any quadratic space that has a metabolic subspace as a direct (orthogonal) summand, the spinor norm is surjective. -Clearly, the spinor norm of $\Omega(V,Q)$ (the commutator subgroup of $O(V,Q)$) is trivial, since $K^*/(K^*)^2$ is abelian. This article states that $\Omega$ is precisely the kernel of the spinor norm, providing an injective morphism $O/\Omega \rightarrow K^*/(K^*)^2$, though I don't see how it helps in actually computing the spinor norm of a given orthogonal transformation. -I've done some calculations with the real hyperbolic plane with orthogonal basis $\{e_1, e_2\}$ such that $Q(e_1)=1$ and $Q(e_2)=-1$ by explicitly computing the elements of the Clifford group that represent certain orthogonal transformations. It seems that the spinor norm of a matrix $A$ (which is $\pm 1$ in the real case) in this basis coincides with the sign of $A_{2,2}$. -Having in mind the connected components of an indefinite real orthogonal group $O(p,q)$ and using that the spinor norm is a continuous map to a discrete space $\{\pm 1\}$, it has to be constant on connected components, thus it is enough to compute it for a single representative from each component. This gives a generalization of the previous result, namely the spinor norm is $+1$ iff the transformation preserves orientation of the negative-definite subspace, and the spinor norm equals the determinant of the lower-right $q\times q$ submatrix (in a basis where positive-definite vectors come before negative-definite ones). This is basically my own findings, and I would appreciate a reference that supports/disproves this claim. - -In general, it feels that there should be some explicit (maybe polynomial?) formula $O(V,Q) \rightarrow K^*$ implementing the spinor norm, but I failed to find any references on this. In any way, I am happy with any explicit way of computing the spinor norm of an orthogonal matrix for a general quadratic form, or otherwise an explanation of why this isn't that straightforward or even possible. - -REPLY [2 votes]: Posted from the comments (1 2 3), by request. -Of course, the spinor formula itself arguably is an explicit formula, for some values of explicit, so I'll take explicit to mean polynomial in the entries, as you suggest—in which case I imagine one can prove rigorously that the answer is ‘no’. If some special formulæ are instead of interest, Jessica Fintzen, Tasho Kaletha, and I recently found ourselves having to do some such computations, and found that, at least for semisimple elements, there's a reasonably easy, mostly explicit (in terms of eigenvalues) answer. It's assembled from some facts described in §5.1 of Fintzen, Kaletha, and Spice - On certain sign characters … in the form that's of interest to us, but most of them come from Scharlau's book Quadratic and Hermitian forms. -Although it turned out not to be most useful for us, §2 of Zassenhaus - On the spinor norm, and specifically (2.1), might be closer to what you want. It's a nice paper, but not the last word on the subject; you may like to look at who cites it.<|endoftext|> -TITLE: Rigid monoidal and closed monoidal categories -QUESTION [5 upvotes]: I am trying to understand the relationship between rigid monoidal categories and closed monoidal -categories. First every rigid monoidal category is closed, with an adjoint to the functor $X \otimes -$ given by $X^* \otimes -$. -Let $\mathcal{C}$ be a closed monoidal category (i.e., with internal homs), such that for all $X \in \mathcal{C}$, the functor $X \otimes -$ and its adjoint forms an equivalence of the category $\mathcal{C}$ with itself. Does it follow that $\mathcal{C}$ is rigid? - -REPLY [3 votes]: Let $1$ be the unit of $C$. For every $X$, we define $X^* = Hom(X,1)$. I will assume $C$ is strict closed symmetric monoidal. Further assuming the condition the OP specified, we can show $C$ is rigid. -Let's unpack the additional condition the OP wants to assume. For every $X$, the functors $F(-) = X\otimes -$ and $G(-) = Hom(X,-)$ form an adjoint equivalence. In particular, the counit of the adjunction, $\epsilon: FG \to 1_C$ is a natural isomorphism. So, $\epsilon_1: X\otimes X^* = X\otimes Hom(X,1) \to 1$ is an isomorphism. Call this morphism $ev_X$. Define the coevaluation as its inverse. -Following Section 2.10 of Tensor Categories, we must show that the compositions: -$X \stackrel{coev_X\otimes id}{\longrightarrow} (X\otimes X^*) \otimes X \stackrel{\alpha}{\longrightarrow} X\otimes (X^*\otimes X) \stackrel{id\otimes ev_X}{\longrightarrow} X$, and -$X^*\stackrel{id \otimes coev_X}{\longrightarrow} X^*\otimes (X \otimes X^*) \stackrel{\alpha^{-1}}{\longrightarrow} (X^*\otimes X) \otimes X^* \stackrel{ev_X\otimes id}{\longrightarrow} X^*$ -are the identity morphisms. But the first is just $X\cong 1\otimes X$, followed by the associator, followed by $X\otimes 1 \cong X$, which is certainly the identity on $X$ (by strictness), and the second works the same way. So, indeed, $X^*$ is a left dual to $X$. -By symmetry, $X^*$ is also the right dual to $X$, and $X^*\otimes X \cong 1$, so the two morphisms in 2.10.2 of Tensor Categories are also identities. Hence, every $X$ has both a left and right dual, so $C$ is rigid.<|endoftext|> -TITLE: Should water at the scale of a cell feel more like tar? -QUESTION [64 upvotes]: The Navier-Stokes equations are as follows, -$$\dot{u}+(u\cdot \nabla ) u +\nu \nabla^2 u =\nabla p$$ -where $u$ is the velocity field, $\nu$ is the viscosity, and $p$ is the pressure. -Some elementary manipulations show that if you zoom in by a factor of $\lambda$, then you expect viscosity to scale as $\lambda^{\frac{3}{2}}$. So, for example, if you zoom in to the length scale of a cell, you expect viscosity to be around a million times larger than humans experience it. -This is not observed, however, which makes sense since we expect the components of a cell to move around extremely quickly. (EDIT: this is observed - see answer - my initial google searches were untrustworthy, damn google). Nonetheless, the calculation above suggests that they feel like they are moving through one of the most viscous fluids imaginable. -What then is the mechanism that prevents this? I have seen some explanations through the ideas of 'microviscosity' and 'macroviscosity' in the physics community, but I couldn't find much of a theoretic backing for them. -I'm wondering if there is a more mathematical explanation, perhaps directly from the Navier-Stokes equation itself (seems unlikely), or something from a kinetic theory point of view? For example some kind of statistical model of water molecules that reproduces the result? - -REPLY [16 votes]: You may be interested in Shapere, A., and F. Wilczek. 1987. Self-propulsion at low Reynolds number. Phys. Rev. Lett. 58: 2051–2054 where they use gauge theory to describe micro-swimming. Because the Stokes equation - the infinite viscosity limit of Navier-Stokes - is linear, it allows us to define a connection for the principal G bundle: -(located shapes) --> (unlocated shapes). Here G is the group of rigid motions of space, a located shape is (say) a volume-preserving embedding of the ball into usual 3-space, and the space of unlocated shapes is the quotient space of the space of located shapes by the action of G. Think of the ball as the cell (parmecium, E Coli, cyanobacterium, ..) which wants to move. A swimming stroke is then a loop in the space of unlocated shapes. The resulting holonomy for the Stokes connection is computed by solving the Stokes equation with zero boundary data at infinity. Shapere in his thesis estimates the curvature at the embedding which is a round ball, and thereby investigates ``infinitesimal swimming motions''. Some of this story can also be found in my book, A Tour of SubRiemannian Geometry.<|endoftext|> -TITLE: Suggestion for framing a course in Representation theory + Spectral graph theory -QUESTION [12 upvotes]: I am going to give a course in spectral graph theory to graduate students. I want to learn and teach the connection between the spectral graph theory and the representation theory of finite groups. I am good at both the areas but I am not sure where to start and what to include. It would be a great help to me if you can suggest what to add beyond the basics in both the areas. I haven't seen any book on the connection between these areas. -Some things in my mind are: - -Representation theory of $Aut(G)$ where $G$ is a finite simple graph. - -Reducible/Irreducible eigenspaces. - - -I am not aware of the literature where representation theory is used in spectral graph theory or vice versa. Kindly give your suggestions. -Thank you. - -REPLY [8 votes]: Update -I have since uploaded a preprint discussing this connection. This is probably not it‘s final form, but since I claimed writing on this some years ago, it is more than time to finally mention it here. Allow me to also point to my PhD thesis (follow the links in my profile) where I explore some of this in more detail. -Otherwise, the answer is unchanged and contains below some of the sources and main ideas I have used in the past. - -I alway put my focus on the idea of the graph realization, because it gives the subject a geometric touch. A graph realization is simply a map assigning to each vertex $i\in V$ a point $v_i$ in Euclidean space. -And such a realization can be highly symmetric (related to representation theory) or it can be some sort of balanced configuation (related to spectral graph theory). These ideas are not independent. -For example, suppose you have a realization that satisfies some kind of self-stress condition: -$$(*)\qquad \sum_{j\in N(i)} v_j = \theta v_i\quad\text{for all $i\in V$}.$$ -Let $M$ be the matrix in which the $v_i$ are the rows, then you can write $(*)$ as $AM=\theta M$ (where $A$ is the adjacency matrix of the graph). Immediately you see that $\theta$ must be an eigenvalue of $A$, and the columns of $M$ must be eigenvectors. -The columns need not span the whole eigenspace. -But if they do, then we call it a spectral realization (see also the link [1] below). -If you define the arrangement space $U:=\mathrm{span}(M)$ as the column span of $M$ (see also the link [3] below), then you have a handy way to define symmetric and spectral realizations: - -a realization is symmetric if its arrangement space is $\mathrm{Aut}(G)$-invariant. -a realization is spectral if its arrangement space is an eigenspace of $A$. - -And since eigenspaces are always invariant, we immediately find that spectral realizations are always as symmetric as the underlying graph. -In my opinion, it is this property of spectral realizations that tells us a lot about the structure of the graph (at least for highly symmetric graphs). -Others might use them on less symmetric graphs in graph drawing algorithms or optimization (but I feel this is less related to representation theory). -If you take the convex hull of the vertices in a spectral graph realization, you obtain the eigenpolytope of a graph. -The literature on these is quite scattered, but the initial source is probably "Graphs, groups and polytopes" by Godsil (I have since tried to organize the literature in this other (work in progress) preprint). -Godsil proved that the eigenpolytope is as symmetric as the initial graph. He also proves group theoretic properties of $\mathrm{Aut}(G)$ from these polytopes (which are just graph realizations in disguise). -You asked specifically about reducible/irreducible eigenspaces. In general, it is quite tricky to determine whether the eigenspaces of a graph are irreducible (without computing all irreducible subspaces). But there is one case for which it is easy: distance-transitive graphs. For these, the eigenspaces are exactly the irreducible subspaces of $\mathrm{Aut}(G)$. This basically follows from Proposition 4.1.11 (p. 137) in "Distance Regular Graphs" by Brouwer, Cohen and Neumaier. -Their proof is in a purely represenation theoretic language, but in the preprints I also discuss some more elementary approaches. -Finally, I can think about some connections to rigidity theory. -One might consider only the deformations of a graph realization that preserves the symmetry of the structure. -Whether such deformations exist depends on the decomposition of the permutation-representation of $\mathrm{Aut}(G)$ into irreducible representations (in particular, their multiplicities). -To connect this to spectral graph theory, one can observe that if a realization is rigid (i.e. it cannot be deformed without loosing symmetry), and irreducible, then one can show that it satisfies $(*)$ (it is not necessarily spectral, but almost). -Of course, for distance-transitive graphs, this implies that the realization is spectral. - -Here are some older posts of mine that might be related: - -[1] directly related: Representations of the automorphism group of graphs via spectral graphs theory -[2] how to get the irreducible subspaces when the eigenspaces are not irreducible: Determining the irreducible invariant subspaces of a permutation action by computing eigenspaces of a matrix -[3] a simple construction (the arrangement space) that I always found helpful for organizing my thoughts when working in spectral graph theory, representation theory and geometry at the same time (check in particular the two last bullet points): Where have you encountered "arrangement spaces"?<|endoftext|> -TITLE: improved Sobolev embedding -QUESTION [5 upvotes]: This is probably not a research level question but I am struggling with the geometry. My question is related to whether some monotonicity can increase the range of exponents in the Sobolev embedding. -For instance on the unit ball, nonnegative radially symmetric functions which are nondecreaing in the radial direction should satisfy a Sobolev embedding for an improved range of exponents: Indeed such functions must be large on the full boundary of the ball, yet the trace theorem prevents this from happening. -So I will ask the question on a finite cone; let $ S \subset \subset S^{N-1}$ be some nice spherical cap. For explicitness let us assume that the $x_1$ axis cuts through the center of $S$ (assume e.g. that $S$ is a ball in $S^{N-1}$ centered at some $y\in S^{N-1}$ lying on the $x_1$ axis). Let $ \Omega:=\{x=r \theta: 02^*=\frac{2N}{N-2}$) - -I suspect the answer is `no' and I am attempting to disprove it considering the first strategy that comes to mind: Take $ 0\le \phi \in C_c^\infty(B_1)$ a smooth radially nonincreasing function and then translate this function so that its support is centered at $y$, and scale it in order to concentrate its support at $y$. This sequence of functions will presumably violate any alleged improved Sobolev embedding, up to proving that these functions really have the correct monotonicity. Geometrically it looks like to me that this is so, but my geometric intuition almost always fails me now. Any comments would be great. - -REPLY [3 votes]: The answer is indeed "no." To see this, instead of considering translations of radial bump functions, one needs to use bump functions whose level sets are deformed e.g. to ellipsoids. More precisely, denote $x \in \mathbb{R}^n$ by $(x_1,\,x')$ and let $h$ be any decreasing function on $\mathbb{R}$. Then for -$$H(x) := h(\Lambda^2 x_1^2 + |x'|^2)$$ -we have that $x \cdot \nabla [H(x-y)] \geq 0$ in the ellipsoid -$$\left\{\frac{(x_1 - 1/2)^2}{(1/2)^2} + \frac{|x'|^2}{(\Lambda/2)^2} \leq 1\right\},$$ -which contains a neighborhood of $y$ in $B_1$ when $\Lambda$ is large (but not $1$). Now fix $\Lambda$ large and choose $h$ smooth with $h(s) = 1$ for $s \leq 0 $ and $h(s) = 0$ for $s \geq 1$. Then the family $$\lambda^{\frac{n-p}{p}}H(\lambda (x-y))$$ -shows as $\lambda \rightarrow \infty$ that the exponent cannot be improved.<|endoftext|> -TITLE: Lie monoids as monoids internal to the category of smooth manifolds? -QUESTION [5 upvotes]: This question can be thought as a complement to this one. -Lie groups can be defined as groups internal to the category of smooth manifolds. Lie monoids, however, as a particular case of Lie semigroups, seem to deserve a much more complicated definition (see, for instance, 'Lie semigroups and their applications', by Hilgert and Neeb, section 1.4). -Briefly, these are thought as closed subsemigroups of Lie groups, satisfying an extra property. This property, on its turn, is related to the infinitesimal counterpart of the notion of Lie semigroup (in the above reference, the notion of 'Lie wedge', whose definition, consequently, must precede that of a Lie semigroup). - -What kind of difficulties appear if one tries to define a Lie monoid simply as a monoid internal to the category of smooth manifolds (or some related category)? - -A LITTLE BIT OR FURTHER DISCUSSION -Lie groupoids, on their turn, can be defined as groupoids internal to the category of smooth manifolds. Is there an analogous notion of 'Lie category', in which morphisms are allowed not to be isomorphisms? Of course, the same question holds for its infinitesimal counterpart. -I tried to find some reference dealing with such a notion, but couldn't. Though, it seems to be a reasonable one to consider even within the realm of Lie groupoid theory. For example, if one wants to allow distinct objects to have distinct automorphism groups, but still be connected by morphisms, this notion seems to be a necessary step. -In particular, that's the case if one wants to allow morphisms between distinct objects to be not only isomorphisms between their automorphism groups, but also covering maps between them. I can't think right now of a concrete example coming, say, from Physics, but it sounds possible that the 'internal symmetries' of a system might 'collapse' in this particular way. -Besides that, exactly as Lie groupoids can be considered as natural generalizations of Lie groups (even if this shouldn't be considered the most appropriate point of view, for many reasons...), the 'Lie categories' would be natural generalizations of Lie monoids. Indeed, a 'Lie category' with one object would amount precisely to a Lie monoid. -Any references will be appreciated. - -REPLY [5 votes]: There is indeed a notion of "Lie category", introduced a 1959 paper of Charles Ehresmann: Catégories topologiques et categories différentiables. This is accessible in his OEuvres Complètes, part I, pages 237–250. I can't say it's gained massive traction. Incidentally, I think this is the paper that introduced Lie groupoids, but under the name 'differentiable groupoids'. Most of the 19 citations recorded by MathSciNet seem to be citing it for this purpose. The name of Lie was originally attached to a special case, which you can still see if you read Mackenzie's book on Lie groupoids from the 1980s. (Note that there is a more recent notion of 'differentiable category' in a 2006 paper by Blute–Cockett–Seely that is unrelated, coming from a computer science perspective.) -One nice result is that the core of the underlying category—the maximal subgroupoid—is proved to be a Lie groupoid, though the proof is in a rather old style. This is analogous to the maximal subgroup of the Lie monoid of $n\times n$ matrices with multiplication is a Lie group.<|endoftext|> -TITLE: Noncommutative torus as a von Neumann algebra -QUESTION [6 upvotes]: Le $\theta$ be irrational. One can define the noncommutative torus $A_{\theta}$ as a universal algebra generated by two unitaries $u,v$ satisfying the relation $vu=e^{2 \pi i \theta} uv$. This is an abstract defnition: however one can show that this algebra is simple and can be concretely represented as a $C^*$-subalgebra of $B(L^2(\mathbb{T}))$ generated by $U$ and $V$ where $Uf(x)=e^{2\pi i x}f(x)$ and $Vf(x)=f(x+\theta)$. Denote this concrete algebra as $\mathfrak{A}$ and consider $\mathfrak{A}''$ which is von Neumann algebra. - -How to prove that $\mathfrak{A}''$ is a type $II_1$ factor (correct me if it isn't true)? - -REPLY [6 votes]: To support Ruy's answer: in my opinion the most natural representation of the quantum torus C*-algebra is the GNS representation coming from its tracial state. This can be explicitly described as the action on $l^2(\mathbb{Z}^2)$ given by $$Ue_{m,n} = e^{-i\hbar n/2}e_{m+1,n}$$ and $$Ve_{m,n} = e^{i\hbar m/2}e_{m,n+1}.$$ The von Neumann algebra they generate is indeed a $II_1$ factor. -I would even say this is the "quantum torus von Neumann algebra". There's more in Section 6.6 of my book Mathematical Quantization.<|endoftext|> -TITLE: Reference for openness of subspace of PSL(2,R) representation variety corresponding to Teichmüller space -QUESTION [6 upvotes]: $\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\Hom{Hom}$Let $S$ be a compact oriented surface with nonempty boundary. There are two variants of Teichmuller space for $S$ you might consider: - -The one that parameterizes finite-volume complete hyperbolic metrics on the interior of $S$. These correspond to discrete and faithful representations of the fundamental group of $S$ into $\PSL(2,\mathbb{R})$ that take the loops surrounding the punctures to parabolic elements. - -The one that parameterizes finite-volume complete hyperbolic metrics on $S$ with geodesic boundary. These correspond to (certain, not all as in 1) discrete and faithful representations of the fundamental group of $S$ into $\PSL(2,\mathbb{R})$ that take the loops surrounding the punctures to hyperbolic elements. - - -Let $U \subset \Hom(\pi_1(S),\PSL(2,\mathbb{R}))$ be the set of representations in either 1 or 2, so you obtain Teichmüller space from $U$ by quotienting out by the conjugation action of $\PSL(2,\mathbb{R})$. -Question: What is a good reference for the fact that $U$ is open? I know many good sources for the corresponding fact when $S$ is a closed oriented surface, where in fact we can replace $\PSL(2,\mathbb{R})$ by an arbitrary Lie group (a theorem of Weil — here we require the representation to be discrete, faithful, and cocompact). But I don't know a source that does these variants. - -REPLY [3 votes]: You know, I also have been looking for a reference for openness in the type preserving setting and haven't been able to find it. (Actually, I'd like to have a theorem like this that works for orbifolds, even.) -At worst, I think you can prove it using the same arguments as in Weil's paper, though. Namely, Weil's proof (phrased in the $\mathbb H^2$ case) essentially involves taking a huge compact subset $K\subset \mathbb H^2$ that includes a fundamental domain of the image of a discrete, faithful representation $\rho$, letting $\Delta \subset \pi_1 S$ be a finite subset that includes all $\gamma\in \pi_1 S$ such that $\rho(\gamma)$ translates $K$ anywhere near itself, and then for $\rho'\approx \rho$, showing that $K / \rho'(\Delta)$ is a compact hyperbolic $2$-orbifold with $\rho'$-holonomy, implying that $\rho'$ is discrete and faithful. If $\rho$ has parabolics, one can instead take $K$ large enough so that it projects to a compact core of the quotient. Then $K / \rho'(\Delta)$ will be an (incomplete) hyperbolic $2$-orbifold with $\rho'$-holonomy, but since $\rho'$ is type preserving, one can ensure that the ends of this incomplete orbifold have parabolic holonomy, and hence the orbifold is contained in a complete finite volume orbifold. -Haven't really thought through the details, though. -Edit: Ah, it also follows from Theorem 1.1 in Bergeron and Gelander - A note on local rigidity, for example.<|endoftext|> -TITLE: Discrete entropy of the integer part of a random variable -QUESTION [10 upvotes]: Let $X$ be a real valued random variable. Of course, the integer part $\lfloor X \rfloor$ of $X$ is a discrete random variable taking values in $\mathbb{Z}$. We can therefore define its discrete entropy -\begin{equation} -H(\lfloor X \rfloor) = - \sum_{n\in\mathbb{Z}} \mathbb{P}( \lfloor X \rfloor = n ) \log( \mathbb{P}( \lfloor X \rfloor = n ) ), -\end{equation} -which is in $[0,\infty]$ as a sum of nonnegative terms, since $- x \log x \geq 0$ for any $0 \leq x \leq 1$ (with the convention $0\log 0 = 0$). -I am looking for sufficient conditions such that $H(\lfloor X \rfloor ) < \infty$. For instance, is it sufficient to know that $X$ has a finite absolute moment $\mathbb{E}[|X|^p] < \infty$ for some $p>0$? Any condition of this type, possibly weaker, is welcome. -Motivation: There are strong connection between the differential entropy of $X$ (assuming $X$ has a pdf whose differential entropy is well-defined) and the discrete entropy of $\lfloor nX \rfloor$ when $n\rightarrow0$. This was the main topic of the 1959 paper from Alfred Rényi intitled On the dimension and entropy of probability distributions: I am questioning the assumptions under which the discrete entropy is well-defined. - -REPLY [11 votes]: Using (say) decimal notation, ASCII encoding, and a delimiter symbol such as a space or comma, as well as the law of large numbers, one can almost surely encode $N$ independent copies of $\lfloor X \rfloor$ using $O( N {\bf E} \log( 2 + |X| ) ) + o(N)$ bits. Applying the Shannon source coding theorem, we conclude that -$$ {\bf H}( \lfloor X \rfloor ) \ll {\bf E} \log(2 + |X| )$$ -which by Jensen's inequality also gives -$$ {\bf H}( \lfloor X \rfloor ) \ll_p \log(2 + {\bf E} |X|^p)$$ -for any $0 < p < \infty$.<|endoftext|> -TITLE: Can this ultrafilter convergence condition be expressed as a compactness condition? -QUESTION [6 upvotes]: Suppose that $X$ is a topological space. Let us say that an ultrafilter $\mathcal U$ on the Boolean algebra $C_X$ of clopen subsets of $X$ is partition-prime if whenever $X = \amalg_{i \in I} X_i$ is a clopen partition of $X$ we have $X_i \in \mathcal U$ for some $i \in I$. -The set $X'$ of partition-prime ultrafilters on $X$ forms a topological space under the usual Stone topology with basis $[V] = \{ \mathcal U \in X' \mid V \in \mathcal U\}$ for $V \in C_X$, and we have a continuous map $\eta \colon X \to X'$ sending $x$ to $\{ V \in C_X \mid x \in V \}$. This is clearly the reflector onto a certain full subcategory of topological spaces; the game is to figure out which one. -It is easy to see that $\eta$ is injective just when $X$ is totally disconnected, and that $\eta$ is a subspace inclusion just when $X$ is zero-dimensional Hausdorff. We would like to know when, moreover, $\eta$ is surjective, and hence a homeomorphism. -This will happen precisely when each partition-prime ultrafilter on $C_X$ is of the form $\eta(x)$ for some $x \in X$. It is easy to re-express this condition (under the assumption that $X$ is zero-dimensional Hausdorff) as saying that $\eta$ is surjective precisely when each partition-prime ultrafilter on the full powerset $\mathcal P X$ converges. -My question is whether there is a natural way of re-expressing this convergence condition as a compactness condition. The obvious guess is that this is the same as ultraparacompactness: every open cover of $X$ can be refined to a disjoint clopen cover. -It is not so difficult to show that our convergence condition is implied by ultraparacompactness. Indeed, if $X$ is ultraparacompact, and $\mathcal U$ is a completely partition-prime ultrafilter on $\mathcal P X$ with no convergent point, then the open sets not in $\mathcal U$ cover $X$; we can refine this cover to a disjoint cover $X = \amalg_i X_i$; and so $X_i \in \mathcal U$ for some $i$, contradicting the fact that $X_i \subseteq U$ for some open $U \notin \mathcal U$. -The other direction is much less clear. Suppose $X$ is zero-dimensional Hausdorff and ultraparacompact, and suppose towards a contradiction that $X = \bigcup_i U_i$ were an open cover with no disjoint refinement. Clearly it does no harm to assume that this is a cover by clopens, and the assumption of no disjoint refinement implies in particular that there is no finite subcover (since a finite cover by clopens can always be refined to a disjoint one). So the sets $V_i = U_i^c$ generate a proper filter $\mathcal F$ with no adherent point. Moreover, any clopen partition $X = \amalg_j X_j$ must fail to refine the cover $(U_i)_{i \in I}$, and so there must exist some part $X_j$ of the partition which meets each set in $\mathcal F$. Clearly, this is a necessary condition for $\mathcal F$ to be extendable to a partition-prime ultrafilter. If it were also sufficient we would be done: for then we could choose a partition-prime $\mathcal U$ extending $\mathcal F$ which, since $\mathcal F$ has no adherent point, would not converge: a contradiction. -So, my claimed characterisation of the property that every partition-prime ultrafilter converges as ultraparacompactness would follow if we could prove the following ultrafilter lemma: - -If $\mathcal F$ is a proper filter such that for any clopen partition $X = \amalg_j X_j$ -there is some $X_j$ which meets each set in $\mathcal F$, then there is a partition-prime -ultrafilter extending $\mathcal F$. - -An obvious attempt at a transfinite argument works well at successor stages, but falls over at limit ordinals. So there remain three possibilities: this is true; this is false; this is independent of the axioms of set theory, and I don't know which of these it is. -Another way of looking at this ultrafilter lemma is via locale theory. Given any locale $L$, we can obtain a new locale $L'$ by taking sheaves on the Boolean algebra of complemented elements of $L$, for the topology whose covers are all clopen partitions. Now $L \to L'$ is a reflector into zero-dimensional ultraparacompact (aka "strongly zero dimensional") locales. Clearly the passage $X \mapsto X'$ described above sends $X$ to the space of points of the strongly zero-dimensional reflection $\mathcal O(X)'$: which would thus be ultraparacompact so long as $\mathcal O(X)'$ were spatial. -Now clearly not every strongly zero-dimensional locale is spatial (take any atomless complete Boolean algebra). But it's possible that the strongly zero-dimensional reflection of any spatial locale is spatial, and this is in fact exactly equivalent to the ultrafilter lemma I quote above. -Any ideas, either about the ultrafilter lemma, or about a different characterisation of the spaces of the form $X'$, would be gratefully received! - -REPLY [2 votes]: Your condition amounts to saying that the uniformity $\mathcal{C}$ generated by all clopen partitions is complete: 'partition-prime' is the same as '$\mathcal{C}$-Cauchy'. -It is also implied by the condition that the space is $\mathbb{N}$-compact, which means that every clopen ultrafilter with the countable intersection property is fixed: if $\mathcal{U}$ is $\mathcal{C}$-Cauchy then it has the countable intersectionproperty; if $\{U_n:n\in\omega\}$ is a decreasing sequence in $\mathcal{U}$ with empty intersection (and $U_0=X$) then $\{U_n\setminus U_{n+1}:n\in\omega\}$ is a clopen partition with no member in $\mathcal{U}$. -In New properties of Mrowka's space $\nu\mu_0$ Kulesza showed the space $\nu\mu_0$ from the title is $\mathbb{N}$-compact; it is not strongly zero-dimensional, hance not ultraparacompact.<|endoftext|> -TITLE: Negative surgeries on negative knots -QUESTION [5 upvotes]: This question is two-fold. -The first question is rather specific: what are some small examples of negative surgeries on negative knots that give rise to the same 3-manifold? I know one class of examples coming from Borromean rings. By performing $-1/m$ and $-1/n$ surgery on two components of the Borromean rings, we get the double twist knot $K_{m,n}$ which is negative. Now, $-1/l$ surgery on $K_{m,n}$ is just $-1/l$, $-1/m$ and $-1/n$ surgery on the Borromean rings, and by the symmetry of the Borromean rings, it is the same as $-1/m$ surgery on $K_{l,n}$ and $-1/n$ surgery on $K_{l,m}$. I would like to know some other simple examples (preferably with knots with small number of crossings). -The second question is a bit vague: what is known about the class of 3-manifolds obtained as negative surgeries on negative knots? I am curious to know if there are some theorems saying this class of 3-manifolds are "nice" in some way. Any kind of input would be greatly appreciated. - -REPLY [3 votes]: In general, to find explicit examples for the first part of your question is a hard problem, sometimes impossible. Actually, it is related to the notion of cosmetic surgeries, see Ni and Wu's paper, and further articles. -You may predict conjectures or obtain obstructions due to Thurston's theorem: all but finitely many surgeries on a hyperbolic knot result in hyperbolic manifolds. -On the other hand, as Kegel said, L. Moser completely classified surgeries along torus knots as follows: -Theorem: Let $K$ be an $(r,s)$ torus knot in $S^3$ and let $Y$ be the $3$-manifold obtained by performing a $(p,q)$-surgery along $K$. Set $\sigma =rsp−q$. -(a). If $|\sigma|>1$, then $Y$ is the Seifert manifold $\Sigma(\alpha_1, \alpha_2, \alpha_3)$ over $S^2$ with three exceptional fibers of multiplicities $\alpha_1=s, \alpha_2=r$ and $\alpha_3=|\sigma|$. -(b). If $\sigma =±1$, then $Y$ is the lens space $L(|q|,ps^2)$. -(c). If $\sigma =0$, then $Y$ is the connected sum of lens spaces $L(r,s) \#L(s,r)$. -EDIT: Considering mirror symmetry of knots and following the common convention on surgeries, we have for $n \geq 1$, - -$\Sigma(r,s,rsn-1)$ is obtained by $(-1,n)$-surgery along the left-handed $(r,s)$ torus knot. -$\Sigma(r,s,rsn+1)$ is obtained by $(-1,n)$-surgery along the right-handed $(r,s)$ torus knot. - -Note that these are only integral homology spheres obtained by surgery on a torus knot in $S^3$.<|endoftext|> -TITLE: Time of peak of an SIR epidemic -QUESTION [6 upvotes]: I've learned some classical results on the peak and the attack rate of an idealized epidemic which evolves according to a SIR model - -$\dot{s} = -\beta\cdot i \cdot s$ -$\dot{i} = +\beta\cdot i \cdot s - i/\delta$ -$\dot{r} = +i/\delta$ - -with infection rate $\beta$ and duration of infectiosity $\delta$, the basic reproduction number being $R_0 = \beta \cdot \delta$. -The classical results I have learned: -For $s(0) \approx 1$ the maximum of $i(t)$ is given by -$$i_{max} = \frac{R_0 - \ln{R_0} - 1}{R_0}$$ -(see e.g. Hethcote's The Mathematics of Infectious Diseases, Theorem 2.1, p. 607) -and the attack rate $r_\infty = \lim_{t\rightarrow \infty}r(t)$ is given by -$$r_\infty = 1 + \frac{W(-R_0 e ^{-R_0})}{R_0}$$ -with the Lambert W function (see e.g. here, Proposition 1.10, p. 3) -What I still haven't found is a reference for the time $t_{max}$ at which $i(t)$ reaches its maximum (i.e. $i(t_{max}) = i_{max}$) when $i(0) = i_0$ is given and $r(0) = 0$. -Find here some plots with a single patient 0 in a completely susceptible population of 10,000, i.e. $i_0 = 0.0001$. A period of 90 days is displayed. $[\delta] =$ 1 day, $[\beta] = $ 1/day. - -The peaks for $\beta_1\delta_1 = \beta_2\delta_2$ – for example $[4,1.5]$ and $[6,1]$ or -$[2,1.5]$ and $[4,0.75]$ or $[2,1]$ and $[4,0.5]$ – have the same $R_0 = \beta\delta$ and thus roughly(1) the same $i_{max}$, but they differ in $t_{max}$. So even though $t_{max}$ decreases with increasing $R_0$ for fixed $\beta$ or $\delta$ (which is quite natural), it cannot depend on the product $\beta\delta$ alone (as $i_{max}$ does) but must depend also on the quotient $\beta/\delta$ or maybe the difference $\beta - 1/\delta$. (1) Roughly because of errors due to finite time steps $\Delta t$ instead of infinitesimal $dt$. -To get a better picture of the heights and timings of the peaks, these are all the curves above overlayed: - - -A good algebraic approximation for $t_{max}$ as a function of $\beta$ and $\delta$ would be welcome (assuming that no closed formula exists) – or simply a reference. (Consider $i_0$ a fixed parameter and $r(0) = 0$.) - -Note that $\beta/\delta$ has a unit of 1/day2 and thus of an acceleration. - -REPLY [5 votes]: Seems like you fell in love with those equations and, especially, with the $I$ component of them:-). So let me try to show you how you can derive as many approximations as you want yourself, test them against simulations and (if you are lucky) find a few to your liking. Again, I'll put everything in the numerator: $\dot S=-\beta IS, \dot E=\beta IS-\lambda E, \dot I=\lambda E-\delta I, \dot R=\delta I$. I'll also normalize to $\lambda+\delta=1$ (time scaling) and denote $\rho=\lambda\delta\in(0,\frac 14]$ after such normalization. The total population will be normalized to $1$. -We shall assume that we are in the situation when the initially infected and exposed portions are very small and everyone is susceptible -Note that for a while (when $S\approx 1$), you run just a linear system on $I$ and $E$. Anything can happen here: for instance, if $E=0$, then $I$ initially goes down and there is no way its graph can match your idea of a peaking curve on that interval. We want to eliminate the decaying part of the solution from the initial data. -Fortunately, the linear theory is easy: you expect that all growth is determined by the eigenvector corresponding to the largest eigenvalue. The eigenvalues for the EI part are (under my normalization) $-0.5\pm\sqrt{0.25+(\beta-\delta)\lambda}$. So, you decompose the vector $(E,I)$ into the parts proportional to the eigenvectors and take the $I$-component of the part corresponding to the positive eigenvalie. That is your $I_{eff}$. You can now presume that starting with $I_{eff}$ and $E_{eff}$, you can run your approximate curve, whatever you decide it to be, infinite time in both directions. Thus, if you settled on some analytic curve $I(t)=F(t)$ approximating your solution that has the maximum at the origin and satisfies the equation on the whole line, then you just solve the equation $F(-t_{max})=I_{eff}$. -Now about how to find decent curves that describe the pandemic coming from $-\infty$ with $S=1,I=E=R=0$ there and going to $+\infty$ with $I=E=0$ there. First of all, determine the quantities that you know exactly. There is actually just one such quantity: the full integral $J=\int_{-\infty}^\infty I(t)dt$. It has two meanings. On one hand, $\delta J=R(+\infty)$. On the other hand, $e^{-\beta J}=S(+\infty)$. Then we get our first equation: -$$ -e^{-\beta J}+\delta J=1, -$$ -which can be solved in a unique way for $J>0$. So, from now on, I'll treat $J$ as a known quantity available for using in other formulae. -Next, generally speaking, we need to decide on some parametric family of curves $F_p(t)$, where $p$ is a set of parameters, that can be used for curve fitting. We need at least 2 free parameters since, after my normalization, we have $2$ degrees of freedom in the choice of $\beta,\lambda,\delta$. However, having just 2 formal parameters from the start seems too restrictive because we then need to guess everything just right. On the other hand, introducing too many parameters is also bad because we shall get too many complicated equations. So, perhaps, three or four would be a good choice. Note that we already have one nice to use relation for $p$: -$$ -\int_{-\infty}^{\infty}F_p(t)\,dt=J\,. -$$ -so ideally this integral should be at least approximately computable in terms of $p$. -Let's see what else we can discern before deciding on any particular kind of curve. -Write $I=\frac{I_0}{\psi}$ where $I_0$ is the maximal value attained at $0$. Then we can successively express $E$ and $S$ in terms of $\psi$ and its derivatives. This algebraic exercise yields -$$ -\lambda E=\delta I+\dot I=I_0\left[\frac \delta\psi-\frac{\dot\psi}{\psi^2}\right] -\\ -\beta\lambda SI=\lambda \dot E+\lambda(\lambda E)=I_0\left[\frac\rho\psi-\frac{\dot\psi}{\psi^2}-\frac{\ddot\psi}{\psi^2}+2\frac{\dot\psi^2}{\psi^3}\right] -\\ -\beta\lambda S=\rho-\frac{\dot\psi}{\psi}-\frac{\ddot\psi}{\psi}+2\frac{\dot\psi^2}{\psi^2} -$$ -(I used the normalization $\lambda+\delta=1$ in the process). -Now denote $B=\beta I_0$. The last equation that we haven't used (that for $S$) reads in this notation -$$ --\dddot\psi-\ddot\psi+5\frac{\dot\psi\ddot\psi}{\psi}-4\frac{\dot\psi^3}{\psi^2}= --B\left[\rho-\frac{\dot\psi}{\psi}-\frac{\ddot\psi}{\psi}+2\frac{\dot\psi^2}{\psi^2} -\right]\,, -$$ -which looks a bit scary until you realize that for every exponent $a\in \mathbb R$ (with one exception, which I leave to you to find), it has an asymptotic solution $ce^at+b+\dots$ at each infinity (the exponent is presumed to grow at the infinity here and $\dots$ stand for the decaying terms). To see it, just plug this form into the equation and find $b$ that eliminates all growth and constants (it depends on $a$, of course). -Now, it seems a good idea to presume that our actual solution of the IVP $\psi(0)=1,\dot\psi(0)=0$ also has this asymptotic form. Moreover, from the expression for $\beta\lambda S$, which can be rewritten as -$$ -\rho-\frac{\dot\psi}{\psi}-\frac{d}{dt}\frac{\dot\psi}{\psi}+\frac{\dot\psi^2}{\psi^2} -$$ -we can immediately see the equations for the exponents. Indeed, if $\psi\asymp e^{at}$, then $\frac{\dot\psi}{\psi}\to a$ and the derivative of it goes to $0$, so at $\pm\infty$, we have for the corresponding exponents -$$ -\delta-a_{\pm}+a_{\pm}^2=\beta\lambda S(\pm\infty) -$$ -and we know that $S(-\infty)=1$, $S(+\infty)=e^{-\beta J}$. So, these quadratic equations allow us to find $a_-$ as the unique negative root $0.5-\sqrt{0.25+(\beta-\delta)\lambda}$, which not surprisingly is just minus the "onset exponent" we found earlier. For $a_+$ we have two choices but it turns out that it is the smaller root that we really need (both are positive). Thus, our $F_p$ should be asymptotic to $e^{a_\pm t}$ at infinities at least approximately, which gives us two more equations for $p$. Finally, it would be nice to have our ODE satisfied at least at the point of the maximum to have the local behaviour near $0$ not too weird. This is the fourth equation. Thus, we should have at least 4 parameters not to sacrifice anything. We already have one: $I_0$, or, which is the same, $B$. It seems rather natural to take the two exponents $a_{\pm}$ as the other two, especially because we can force their values to be exactly right without much trouble. Thus, we need one more. -Here you have options. I played with a few and finally settled on the function of the kind -$$ -\psi(t)=c+c_+e^{a_+t}+c_-{ea_-t} -$$ -where $c\in[0,1)$ and $c_\pm$ can be immediately determined from $c$ and the conditions $\psi(0)=1,\dot\psi(0)=0$. The differential equation at the point $0$ of maximum becomes then a nice algebraic relation between $c$ and $B$ (quadratic, to be exact) once $a_{\pm}$ are known. -The nightmare equation then becomes the very first one: $\int F_p=J$. This requires to integrate $\frac 1{\psi}$ and there is no nice formula. However, there is a nice approximation: when $c=0$, we can find the full integral using residue technique, and when $a_-=-a_+$, we can find the dependence of $c$ exactly, so we just assume that it expands to the other values approximately. The quick numeric check shows that this assumption is not as idiotic as one can think, so we get the final relation, which, if you put everything together, reads -$$ -B\frac{2}{\sin(\pi t)t^t(1-t)^{1-t}} -\frac 1{a_+-a_-}\frac {\arctan(\frac{\sqrt{1-2c}}c)}{\sqrt{1-2c}}=\beta J -$$ -where $t=a_+/(a_+-a_-)$ (or something like that: when $c>\frac 12$, you need to replace the arctangent by the difference of logarithms: it is the same analytic function but the algebra is not programming-friendly here). Of course, you are more than welcome to experiment with other forms of the solutions. -Now, once you solve the resulting system, you can plug everythin in and see if the curve matches the simulation. Here are a few pictures: the black curve is the numeric solution of the ODE, the red curve is the fitting one (of the above kind), the green one is the best symmetric approximation, the blue line is the error in the equation (scaled in some reasonable way), the orange dot is the predicted time of maximum, the magenta horizontal line is the predicted maximum, etc. -Can one learn anything from this exercise? Honestly, I have no idea. I just wanted to show you how such things can be done, so you can try yourself. The approximate parametric form I suggested here is quite simple, but determining the parameters from $\beta,\lambda,\delta$ requires solving a few transcendental equations. The fit is pretty good though up to $\frac{\beta}{\delta}\approx 20$ uniformly in $\lambda$.<|endoftext|> -TITLE: $H^{p,q}(X)$ versus $H^{q}(X, \bigwedge^p TX)$ -QUESTION [7 upvotes]: Let $X$ be a Kahler manifold. To $X$ one can associate the cohomology groups $H^{p,q}(X)$, and $H^{(0,q)}(X, \bigwedge^p TX)$ with $TX$ being the holomorphic tangent bundle of $X$. -Is there a general relationship between these two cohomologies? -For instance, it is claimed in a footnote on page 30 in Hori, Iqbal, and Vafa - D-branes and mirror symmetry, that the Euler characters of the two agree up to a sign. It would be nice to know the origin of this. -Second we know that when $X$ is Calabi–Yau the two cohomologies match. For the more general case when $X$ is Kähler it would be good to know a precise way to measure the mismatch. - -REPLY [7 votes]: One reason that $H^p(X, \bigwedge^q TX)$ will not be as well behaved as $H^p(X, \bigwedge T^{\ast} X)$ is that it is not deformation invariant, and thus not topological. In other words, if we have a connected base $B$, and a flat projective family $\mathcal{X}$ over $B$ with smooth fibers, then will have the same Hodge numbers. This will not be true for $H^q(X, \bigwedge^p TX)$. Note that, in particular, this means that $H^q(X, \bigwedge^p TX)$ is not topological, since all fibers of such a family will be diffeomorphic. -Let $X_1$ be $\mathbb{P}^2$ blown up at $3$ general points, and let $X_2$ be $\mathbb{P}^2$ blownup at $3$ collinear points. I claim that $H^0(X_1, T) \cong \mathbb{C}^2$, but $H^0(X_2, T) \cong \mathbb{C}^3$. The surfaces $X_1$ and $X_2$ are diffeomorphic, and it is easy to make a smooth flat projective family over $\mathbb{A}^1$ with most fibers $X_1$ and some fibers $X_2$. This seems like bad news. -Recall that $H^0(\mathbb{P}^2, T)$ is $8$-dimensional. Writing $z_1$, $z_2$, $z_3$ for the homogenous coordinates on $\mathbb{P}^2$, it is spanned by $z_i \tfrac{\partial}{\partial z_j}$, modulo the relation $\sum z_k \tfrac{\partial}{\partial z_k}=0$. If we write a section of $T\mathbb{P}^2$ as $\sum A^i_j z_i \tfrac{\partial}{\partial z_j}$, then this section vanishes at the point $(x_1 : x_2 : x_3)$ if and only if $(x_1, x_2, x_3)$ is an eigenvector of the matrix $A^i_j$. -Let $U_j$ be the locus of $\mathbb{P}^2$ away from the points which are blown up in $X_j$. So $U_j$ is an open subset of both $\mathbb{P}^2$ and $U_j$, and we can restrict sections of the tangent bundle from the complete surfaces to $U_j$. Since $U_j$ is $\mathbb{P}^2$ remove a codimension $2$ locus, $H^0(\mathbb{P}^2, T) \cong H^0(U_j, T)$. On the other hand, I believe that a section $\sigma$ of $H^0(U_j, T)$ will extend to a section of $H^0(X_j, T)$ if and only if, considering $\sigma$ as a section on $\mathbb{P}^2$, that section vanishes at the blown up points (see computation below). -Thus, $H^0(X_j, T)$ will be sections $\sum A^i_j z_i \tfrac{\partial}{\partial z_j}$, subject to the condition that the matrix $A$ has eigenvectors at the blown up points, and modulo the relation $\sum z_k \tfrac{\partial}{\partial z_k}=0$. -If we blow up the points $(1:0:0)$, $(0:1:0)$, $(0:0:1)$, we are requiring that the matrix $A$ be diagonal. So $H^0(X_1, T)$ is matrices of the form $\left[ \begin{smallmatrix} \ast&0&0 \\ 0&\ast&0 \\ 0&0&\ast \\ \end{smallmatrix} \right]$ modulo scalar multiples of the identity. Three general points in $\mathbb{P}^2$ are equivalent, up to $PGL_3$, to these three points, so this is the general case. -On the other hand, if we blow up $(1:0:0)$, $(0:1:0)$ and $(1:1:0)$, then we are requiring that $(\ast:\ast:0)$ be an eigenspace of $A$. This means that we are looking at matrices of the form $\left[ \begin{smallmatrix} \lambda&0&\ast \\ 0&\lambda&\ast \\ 0&0&\ast \\ \end{smallmatrix} \right]$ , modulo scalar multiples of the identity. This vector space is one dimension larger. - -I had trouble finding a reference for the claim I made about vector fields extending to the blow up iff they vanish at they blown up point, so here is a proof: The statement is local, so I'll check it in the affine plane. A vector field on $\mathbb{C}^2$ corresponds to a derivation from $\mathbb{C}[x,y]$ to itself. Namely, the vector field $f(x,y) \tfrac{\partial}{\partial x} + g(x,y) \tfrac{\partial}{\partial y}$ gives the unique derivation with $D(x) = f$ and $G(y) = g$. -Such a vector field extends to the plane blown up at $(0,0)$ if and only if the derivation maps the rings $\mathbb{C}[x,y/x]$ and $\mathbb{C}[x/y,y]$ to themselves. I'll do the computation for the first case; it is enough to compute the derivation on the generators of the ring. -We have $D(x) = f \in \mathbb{C}[x,y] \subset \mathbb{C}[x,y/x]$, so there is no problem there. We then have -$$D(y/x) = \frac{x D(y) - y D(x)}{x^2} = \frac{x g(x,y) - y f(x,y)}{x^2}.$$ -Writing $f(x,y) = \sum f_{ij} x^i y^j$ and $g(x,y) = \sum g_{ij} x^i y^j$, we have -$$\frac{x g(x,y) - y f(x,y)}{x^2} = \frac{g_{00}}{x} - \frac{f_{00} y}{x^2} + \dots$$ -where the ellipses are terms that are in $\mathbb{C}[x,y] \langle 1, y/x, y^2/x^2 \rangle \subset \mathbb{C}[x,y/x]$. -So the derivation takes $\mathbb{C}[x,y/x]$ to itself if and only if $f_{00} = g_{00} = 0$, as desired.<|endoftext|> -TITLE: Gelfand pair, weakly symmetric pair, and spherical pair -QUESTION [6 upvotes]: I am a bit confused with the relations among Gelfand pairs, weakly symmetric pairs, and spherical pairs defined in the book "Harmonic analysis on commutative spaces" written by professor Joseph A. Wolf. -For convenience, let me recall the definitions in this book, and just consider connected groups $G$. -Definition 1 Let $G$ be a connected Lie group, and $K$ a compact subgroup. If the algebra $L^1(K\backslash G/K)$ is commutative under convolution, then $(G,K)$ is called a Gelfand pair. -Definition 2 Let $G$ be a connected Lie group, and $K$ a compact subgroup. If there exists an automorphism $\sigma$ of $G$ such that $\sigma(g)\in Kg^{-1}K$ for all $g\in G$, then $(G,K)$ is called a weakly symmetric pair. -Definition 3 Let $G$ be a complex reductive linear algebraic group, and $H$ a reductive subgroup. Denote by $\mathfrak{g}$ and $\mathfrak{h}$ the Lie algebras of $G$ and $H$ respectively. If there exists a borel subalgebra $\mathfrak{b}$ in $\mathfrak{g}$ such that $\mathfrak{b}+\mathfrak{h}=\mathfrak{g}$, then $(G,H)$ is called a spherical pair. -Now on page 281 of Wolf's book, there are two results: Theorem 12.6.10 and Theorem 12.6.11. -Let $G_\mathbb{C}$ be a connected complex reductive algebraic group, and $H_\mathbb{C}$ a reductive algebraic subgroup. Suppose that $G$ is a real form of $G_\mathbb{C}$ such that $H:=G\cap H_\mathbb{C}$ is a compact real form of $H_\mathbb{C}$. Then $(G_\mathbb{C},H_\mathbb{C})$ is a spherical pair if and only if $(G,H)$ is a weakly symmetric pair (by Theorem 12.6.10) if and only if $(G,H)$ is a Gelfand pair (by Theorem 12.6.11). -Thus, suppose that we have a real reductive group $G$ with its compact subgroup $K$, then $(G,K)$ is a Gelfand pair if and only if $(G,K)$ is a weakly symmetric pair. But I do not think that the two definitions are equivalent. As far as I know, weakly symmetric pairs are Gelfand pairs, but there exist Gelfand pairs which are not weakly symmetric pairs. -Hence, I think that I am probably misunderstanding the definitions, the theorems, or the relations among three pairs. I shall be grateful if experts may give any comments. - -REPLY [4 votes]: I like this question! I originally thought it was much less subtle, and so posted some ill informed guesses in the comments. -Although the notions of "weakly symmetric" and "Gelfand pair" differ in general (Lauret - Commutative spaces which are not weakly symmetric), they coincide for reductive groups. This is Theorem 7.3 of Akhiezer and Vinberg - Weakly symmetric spaces and spherical varieties.<|endoftext|> -TITLE: Status of a conjecture of C.T.C. Wall? -QUESTION [15 upvotes]: In Wall's paper Unknotting tori in codimension one and spheres in codimension two, he states the following conjecture: - -Any $h$-cobordism of $S^3 \times S^1$ to itself is diffeomorphic to $S^3 \times S^1 \times I$. - -What is the status of this conjecture? - -REPLY [9 votes]: The conjecture has been solved. This is Theorem 16.1 in C.T.C. Wall. (1999). Surgery on Compact Manifolds, Second edition. Mathematical Surveys and Monographs, -Vol. 69. Here, two proofs of this theorem are given: the first one is an application of a more general method presented in the book, while the second one relies on some geometrical properties of $S^3 \times S^1$.<|endoftext|> -TITLE: $\omega_1$-approximation property for Sacks iteration— contradiction in literature? -QUESTION [12 upvotes]: The following is a folklore result. - -Suppose $P$ is a countable support iteration of nontrivial forcings, $\langle P_\alpha, \dot{Q}_\alpha : \alpha < \omega_1 \rangle$. Then there is a complete embedding of $\mathrm{Add}(\omega_1)$ into $P$. - -The forcing to add a Cohen subset of $\omega_1$ fails the $\omega_1$-approximation property, since it produces a “fresh” sequence— a sequence such that all initial segments are in the ground model. -In the 1979 paper, “Iterated perfect-set forcing,” Baumgartner and Laver seem to make a contrary claim. Lemma 6.2 states that the countable support iteration of Sacks forcing produces no fresh sequences of length some ordinal of uncountable cofinality. This is key to their argument that iterating Sacks forcing up to a weakly compact forces the tree property at $\omega_2$. -I do not see a flaw in their argument. Is the folklore claim correct? How is this resolved? - -REPLY [11 votes]: I don't see why the "folklore result" holds. The analogous result for finite support iteration is as follows. -$\textbf{Fact}$: If $\langle P_i, Q_j: i \leq \alpha + \omega, j < \alpha + \omega \rangle$ is a finite support iteration of non-trivial forcings, then $P_{\alpha + \omega}$ adds a Cohen real over $V^{P_{\alpha}}$. -Its proof goes as follows. WLOG, $\alpha = 0$. Fix $q_k, r_k \in V^{P_{\alpha + k}}$ such that $\Vdash_{P_{\alpha + k}} q_k, r_k \in Q_k \wedge q_k \perp_{Q_k} r_k$. Define $c \in V^{P_{\omega}} \cap 2^{\omega}$ by $c(k) = 1$ iff $(\exists p \in G(P_{\omega}))(p(k) = q_k)$. To see that $c$ is Cohen over $V$, suppose $D \in V$ dense in $2^{< \omega}$ and $p \in P_{\omega}$. Fix $n$ such that $p \in P_n$. Extend $p$ to $p' \in P_n$ in $n$ steps such that for each $k < n$, $p \upharpoonright k \Vdash_{P_k} (p(k) \perp_{Q_k} q_k) \text{ or } (p_k \leq_{Q_k} q_k)$. Let $s \in 2^n$ be determined by this $p'$. Choose an extension $s'$ of $s$ in $D$. The rest should be clear. -In the case of countable support iteration, the passage from $p$ to $p'$ could be problematic since one might have to extend $p$ infinitely many times. This also shows that the folklore result would work if we iterate countably closed forcings. In any case, Baumgartner-Laver result implies that the folklore result is false. The closest related result about countable support iteraton is the following. Any countable support iteration of non-trivial forcings of length $\omega_1$ collapse the continuum to $\omega_1$. This is the Sublemma inside Lemma 6.3 in the Baumgartner-Laver paper. - -REPLY [10 votes]: Exactly. The rumour (not folklore ;) ) is even wrong if you iterate Cohen forcing (on $\omega$ !!) with countable support $\omega_1$ many times. Let $P$ denote the iteration of Cohen forcings. It follows that any complete embedding $F$ from $\text{Add}(\omega_1)$ into $P$ is an independent permutation of coordinates and $0$ and $1$'s. Indentify $2^\omega$ with the corresponding max antichain in $ \text{Add}(\omega_1)$ and wlog assume that for every $x \in 2^\omega$ it holds that $F(x)= \inf_{n \in \omega} F(x \restriction n)$. But then $F[2^\omega]$ has a uniform, countable support $A$. Wlog let $A=\omega$. But there is $p\in P$ such that $p$ codes the first Cohen-real horizontally into $p\restriction \omega\, (\cdot)\, (0)$. As the Cohen-real differs from any ground model real, the embedding cannot be complete.<|endoftext|> -TITLE: Spectral symmetry of a certain structured matrix -QUESTION [16 upvotes]: I have a matrix -$$ A= \begin{pmatrix} 0 & a & d & c\\ \bar a & 0 & b & d \\ \bar d & \bar b & 0 & a \\ \bar c & \bar d & \bar a & 0 \end{pmatrix} $$ -As you can see, the matrix is always self-adjoint for any $a, b, c, d \in \mathbb C$. -But it has a funny property (that I found by playing with some numbers): -If $a,b,c$ are arbitrary real numbers and also $d$ is real, then the spectrum of $A$ is in general not symmetric with respect to zero. To illustrate this, we take $d := 2$, $a := 5$, $b := 3$, $c := 4$ then the eigenvalues are -$$\sigma(A):=\{10.5178, -6.54138, -3.51783, -0.458619\}$$ -But once I take $d \in i \mathbb R$, the spectrum becomes immediately symmetric. In fact, $d := 2 i$, $a := 5$, $b := 3$, $c := 4$ leads to eigenvalues -$$\sigma(A)=\{-9.05607, 9.05607, -0.993809, 0.993809\}$$ -Is there any particular symmetry that only exists for $d \in i\mathbb R$ that implies this nice inflection symmetry? -I am less interested in a brute-force computation of the spectrum than of an explanation of what symmetry causes the inflection symmetry. - -REPLY [37 votes]: For real $a,b,c$ and imaginary $d$ the matrix $A$ has chiral symmetry, meaning it anticommutes with a matrix $X$ that squares to the identity: -$$X=\left( -\begin{array}{cccc} - 0 & 0 & 0 & -i \\ - 0 & 0 & i & 0 \\ - 0 & -i & 0 & 0 \\ - i & 0 & 0 & 0 \\ -\end{array} -\right),\;\;XA+AX=0,\;\;X^2=I.$$ -Hence the spectrum of $A$ has $\pm$ symmetry: -$$\det (\lambda-A)=\det(\lambda X^2-XAX)=\det(\lambda+X^2A)=\det(\lambda+A),$$ -so if $\lambda$ is an eigenvalue then also $-\lambda$. - -REPLY [16 votes]: An equivalent trick : Let $J:= \operatorname{diag}(1,i,-1,-i)$. Then $J^*AJ=iB$ where $B$ is real and skew-symmetric. Hence the spectrum of $iB$ (thus that of $A$) comes by pairs $\pm\lambda$.<|endoftext|> -TITLE: How badly can the GCH fail globally? -QUESTION [20 upvotes]: It's known that we have global failures of GCH---for example, where $\forall \lambda(2^\lambda = \lambda^{++})$---given suitable large cardinal axioms. -My question is whether we can have global failures of GCH where there is a weakly inaccessible cardinal between $\lambda$ and $2^\lambda$ for each $\lambda$. Similarly, whether we can have a cardinal fixed point between $\lambda$ and $2^\lambda$. I'd also be interested in whether $2^\lambda$ can be weakly inaccessible/a cardinal fixed point, for every $\lambda$. - -REPLY [12 votes]: In the Foreman-Woodin model The generalized continuum hypothesis can fail everywhere. - for each infinite cardinal $\kappa, 2^\kappa$ is weakly inaccessible. -This answers your last question. The answer to the first two questions can be yes as well. In the case of Foreman-Woodin model, they start with a supercompact $\kappa=\kappa_0$ and infinitely many inaccessibles $\kappa_n, n<\omega,$ about it. They first force to get $2^{\kappa_n}=\kappa_{n+1}$ preserving $\kappa$ supercompact, and this is reflected below for all cardinals. So if for example each $\kappa_n$ is measurable, then what you get in the final model is that for each infinite cardinal $\lambda, 2^\lambda$ has been measurable in $V$, in particular there are both weakly inaccessible and cardinal fixed points between $\lambda$ and $2^\lambda.$ -See also the paper A model in which every Boolean algebra has many subalgebras - by Cummings and Shelah, where they build a model in which for each infinite cardinal $\kappa, 2^\kappa$ is weakly inaccessible and $Pr(2^\kappa)$ holds. Here $Pr(\lambda)$ is in some sense a large cardinal property (for example it holds if $\lambda$ is a Ramsey cardinal). For its definition see the paper.<|endoftext|> -TITLE: Unit group of octonions over finite fields -QUESTION [12 upvotes]: One can define the algebra $A(K)$ of octonions over an arbitrary field $K$, see for example the command OctaveAlgebra in GAP: https://www.gap-system.org/Manuals/doc/ref/chap62.html . -When $K$ is a finite field, this is a finite dimensional $K$-algebra and thus has finitely many elements. Let $A_q$ denote the octonions over a field with $q$ elemetns. - -Question 1: What is the number of units in $A_q$? Can one even describe the (possibly non-associative) group of units up to isomorphism? - -For $q=2$ the order is 120 and for $q=3$ the order is 4320. In both cases it is indeed a group according to GAP. -Question 2 is motivated by Will Sawin's comment (I forgot the unit "group" might not be associative): - -Question 2: For which $q$ is the unit "group" of $A_q$ associative? - -It would be interesting to see what the smallest $q$ is such that the unit "group" is not associative. - -REPLY [14 votes]: This is all worked out in the article "A class of simple Moufang loops" by L.J. Paige. The short answer is that the loop of units has size $q^3(q^4-1)(q-1)$, and is not associative for any $q$. The example given by Paige (lemma 3.5) is given in terms of Zorn vectors as -$$\left[\begin{pmatrix} -1 & (0,0,1)\\ -(0,0,0) & 1 \\ -\end{pmatrix}\begin{pmatrix} -1 & (1,0,0)\\ -(0,0,0) & 1 \\ -\end{pmatrix}\right]\begin{pmatrix} -0 & (0,1,0)\\ -(0,-1,0) & 1 \\ -\end{pmatrix}=\begin{pmatrix} -0 & (1,1,1)\\ -(-1,-1,1) & 2 \\ -\end{pmatrix}$$ -and -$$\begin{pmatrix} -1 & (0,0,1)\\ -(0,0,0) & 1 \\ -\end{pmatrix}\left[\begin{pmatrix} -1 & (1,0,0)\\ -(0,0,0) & 1 \\ -\end{pmatrix}\begin{pmatrix} -0 & (0,1,0)\\ -(0,-1,0) & 1 \\ -\end{pmatrix}\right]=\begin{pmatrix} -1 & (1,1,1)\\ -(-1,0,1) & 1 \\ -\end{pmatrix}$$ -so these two products cannot be equal over any characteristic. For $q=2$ we obtain the smallest simple nonassociative Moufang loop, which has order 120. -The article actually shows that a certain subloop modulo its center is a simple Moufang loop. At the time before Paige's result, the only simple Moufang loops known where the simple groups. Liebeck later proved the converse: Every finite simple Moufang loop that is not a group corresponds to such a subloop of octonions over some $\mathbb F_q$. In particular we shouldn't expect a simple classification.<|endoftext|> -TITLE: Explanation for why an ideal fluid doesn't have increasing entropy? -QUESTION [5 upvotes]: The equations of motion for a very simple ideal fluid (specifically a calorically perfect, monatomic, ideal gas) are \begin{align*}\dot{\rho}+\nabla \cdot (\rho u)=0 \;&\text{(mass conservation)} \\ \dot{(\rho u)}+\nabla \cdot (\rho u u) + \nabla p=0 \;&\text{(momentum conservation)} \\ \dot{(\rho e)} +\nabla \cdot (\rho ue+\rho p)=0 \;&\text{(energy conservation)} \\ e=\frac{1}{2}u^2+\frac{3}{2}p \;&\text{(equation of state)}\end{align*} where $\rho$ is the density, $u$ the velocity, $p$ the pressure, and $e$ the total energy (including the internal energy). -I've noticed that these equations are time reversible, i.e. if we have a solution on the time interval $[0,T]$, then by simply sending $u \to -u$, $t \to -t$ we get a solution on $[-T,0]$. From the point of view of thermodynamics, specifically the fact that total entropy is (weakly) increasing, this only makes sense if entropy is constant. -EDIT: In response to some of the comments I've deleted the example of gas expansion, since as pointed out, this wasn't strange. However I'd like to mention that a very simple model of particle collisions in a gas gives rise to the above equations: -Assume particles are interacting through collisions only (i.e. not through 'long range' forces), and there are sufficient collisions occuring that to a good degree of accuracy, the distribution of velocities of particles at any point is isotropic (after subtracting the mean velocity). For example if the velocity distribution of all particles at a given point is always a spherical gaussian this would be the case. Under just this assumption, the above equations follow. -I'm not denying that they are time reversible, and so must have constant entropy. It's just that I have no intuitive explanation for this, other than simply computing the equations. The statistical model mentioned is not time reversible, in fact it relies heavily on frequent collisions, and so time reversibility is a rather suprising fact. -Is there some other explanation, perhaps more intuitive than simply computing the equations, that explains time reversibility? - -REPLY [8 votes]: This is a very important issue, to which an answer must be made in mathematical terms, rather than by waving hands. -Yes, the Euler system (conservation of mass, momentum and energy) is time-reversible. So where is the error when we say Entropy is non-decreasing, but the system is time-reversible, therefore the entropy must be constant along trajectories ? -The point is that the Cauchy problem (i.e. solving the PDEs together with imposing an initial data) is not uniquely solvable. It is so when the initial data is smooth enough, but only for some finite time interval. For rather general smooth initial data, the smooth solution exists only for a finite time interval $(0,T_{\max})$. As $t\rightarrow T_\max$, some first derivative becomes infinite somewhere (generic behaviour). Beyond $T_\max$, the solution is not any more smooth ; it is at best piecewise smooth, with $\rho,u,e,p$ being discontinuous accross hypersurfaces. These discontinuities are known as shock waves and contact discontinuities. -It turns out that once shock waves develop, the Euler system is not any more sufficient to select a unique solution. There are actually infinity many, among which only one has a physical sense. The way to recognize that one, and to select it from a mathematical perspective, is to add a so-called entropy criterion. This is nothing more than saying that when a particle crosses a shock, then its entropy increases. -This entropy criterion is expected to guaranty the uniqueness of the solution (this is still an open problem). But, being an inequality, is not compatible with the time reversal. This is why the Cauchy problem, at far as physicall meaningful solutions are concerned, is irreversible, despite the apparent time-reversibility of the Euler system. -Edit. To answer Michael's concerns, the constancy of entropy along smooth solutions is just the well accepted fact in thermodynamics that smooth flows are time-reversible. Of course, if you have in mind a finer description, the situation will be different. The mesoscopic level is described by a kinetic equation, say that of Boltzmann, which is irreversible: the entropy does increase whenever the local distribution of the gas deviates from the Gaussians (= Maxwell's equilibria). Thus the only reversible model is at the microscopic level, where particles obey to Newton's law and interact through short range forces (or hard spheres dynamics). -To come back to the Euler system, the boundary conditions at discontinuities are not ad hoc. They just express that the mass, momentum and energy are conserved.<|endoftext|> -TITLE: Consistency strength of lifting through a lot of collapsing -QUESTION [5 upvotes]: What is the consistency strength of the following situation? - -$j : V \to M$ is an elementary embedding definable from parameters in $V$, with critical point $\kappa$. -$\mathbb P$ is a forcing that collapses all ordinals between $\kappa$ and $j(\kappa)$. -$j$ can be lifted through $\mathbb P$. - -One can deduce that $j(\kappa)$ is regular in $V$. The only examples I know of such liftings involve almost-huge cardinals. - -REPLY [5 votes]: $\text{AD}^{L(\mathbb R)}$ suffices. The situation actually holds in the model $H = \text{HOD}^{L(\mathbb R)}$. We will have $\kappa = \omega_1$ and $j : H\to \text{Ult}(H,U)$ equal to the ultrapower of $H$ by the club measure $U$ over $\omega_1$ as computed in $L(\mathbb R)$ (using all functions in $L(\mathbb R)$). -For any number $n$, the $\Sigma_n$-satisfaction predicate of $L(\mathbb R)$ with ordinal parameters is definable over $H$ from its restriction to ordinals less than $\Theta$, so any subclass of $H$ that is ordinal definable over $L(\mathbb R)$ is definable from parameters over $H$. In particular, $j$ is definable from parameters over $H$. -Let $N$ be a $\mathbb P_\text{max}$-extension of $L(\mathbb R)$. -Note that $H = \text{HOD}^N$ by the homogeneity and definability of $\mathbb P_\text{max}$. Let $\mathbb P\in H$ be the Vopenka algebra of $N$ for adding a subset of $\omega_2$ to $H$. There is a set $A\subseteq \omega_2$ such that $N= L[A]$, and so $N = H[G_A]$ where $G_A\subseteq \mathbb P$ is the $H$-generic ultrafilter associated to $A$. -In $N$, $\text{NS}_{\omega_1}$ is saturated. Let $G\subseteq P(\omega_1)\setminus\text{NS}_{\omega_1}$ be $N$-generic, and in $N[G]$ let $i : N\to \text{Ult}(N,G)$ be the generic ultrapower embedding associated to $G$ (using functions in $N$). -Now as usual, we cite a theorem due to Woodin: $j = i\restriction H$. This follows from Theorem 4.53 in The Axiom of Determinacy, Forcing Axioms, and the Nonstationary Ideal. -Now in $H$, we have the situation you were looking for with $\kappa = \omega_1.$ Note that $i(\omega_1) = (\omega_2)^N$ by saturation, which means that all $H$-cardinals between $\kappa$ and $j(\kappa)$ are collapsed to $\kappa$ in $N$. Moreover $j$ lifts through the forcing $\mathbb P$ (to $i$) by construction.<|endoftext|> -TITLE: Extending Sobolev function on Riemannian manifold -QUESTION [9 upvotes]: Let $(M, \mu, d)$ be a geodesically complete non-compact Riemannian manifold such that measure $\mu$ is volume doubling, i.e. \begin{equation}\label{VD}\mu(B(x, 2r))\leq C\mu(B(x, r))\end{equation} for some constant $C>0$ and also $M$ satisfying the $L^{2}$-Poincare inequality $$\frac{1}{\mu(B(x, r))}\int_{B(x, r)}{|u-u_{B(x, r)}|^{2}d\mu}\leq cr^{2}\frac{1}{\mu(B(x, \delta r))}\int_{B(x, \delta r)}{|\nabla u|^{2}d\mu}$$ -for all $u\in W^{1, 2}(B(x, r))$ with $\delta>1$ and $\nabla u$ being the weak gradient of $u$. -For a fixed point $x_{0}\in M$ and $\alpha, \beta\in \mathbb{R}_{+}$ consider the annuli $$P_{\alpha, \beta}=\{x\in M:\alpha0$ such that for every $x\in S$ and $00}\frac{r^{-1}}{\mu(B(x, r))}\int_{B(x, r)\cap S}{|u-u_{B(x, r)\cap S}|d\mu}$$ can be extended to a function $\widetilde{u}\in CW^{1, 2}(M)$ such that $$\|\widetilde{u}\|_{CW^{1, 2}(M)}\leq C(\|u\|_{L^{2}(S)}+\|u_{1, S}^{\#}\|_{L^{2}(S)}),$$ where $CW^{1, 2}(M)$ is the Calderon-Sobolev space which coincides with the classical Sobolev space $W^{1, 2}(M)$ if one assumes volume doubling and $M$ satisfying the $L^{2}$-Poincare inequality. -Thanks in advance for your help! - -REPLY [8 votes]: If the annulus is small, then it is basically an Euclidean annulus and there is an extension operator for Sobolev spaces. However, if the annulus is large is may happen that it goes around a "neck" in a manifold and meets along the boundary as the picture shows. - -Then a smooth function on the annulus can approach to $0$ from one side of the common part of the boundary and to $1$ from the other side. Such a function has no Sobolev extension at all.<|endoftext|> -TITLE: Hadwiger number of a graph: Question about the original article from 1943 -QUESTION [5 upvotes]: I am analyzing Hadwiger's original article (Hadwiger, Hugo (1943), "Uber eine Klassifikation der Streckenkomplexe", Vierteljschr. Naturforsch. Ges. Zurich, 88: 133–143) for my work related to the Hadwiger Conjecture in graph theory. -This article is in German, and Hadwiger's terminology from 1943 is very different from current graph theory terminology. For example, he says "Komplex" for what we now call a graph, and he says "Simplex S(n)" for what we now call the complete graph $K_n$ over $n$ vertices. Also, when he says "K(k)", now we would call this a graph with Hadwiger number $k$ (for him, K(k) is not the complete graph). -In this article, he defines what we now call the Hadwiger number ${\rm had}(A)=k$, the size of the largest complete graph $K_k$ that can be obtained by contracting edges of $A$ (where $A$ is an undirected graph). -His original definition is - -Ein Komplex A heisst ein K(k), wenn er sich auf einen S(k), -aber nicht auf einen S(k+1) zusammenziehen lässt. Ein K(k) ist notwendig ein zusammenhängender Komplex. - -I am translating this into current terminology "An undirected graph $A$ has Hadwiger number $k$, if the complete graph $K_k$, but not $K_{k+1}$, can be obtained by contracting edges of $A$. It is necessarily a connected graph." -In the next sentences he says - -Die Eigenschaft, ein K(k) zu sein, kennzeichnet eine Art des höheren Zusammenhangs, die durch die natürliche Zahl k gegeben ist. - -What does "höheren Zusammenhangs" mean here? My current terminology translation would be "The property to have Hadwiger number $k$ characterizes a higher degree of connectedness, which is given by the natural number $k$". -But my second, more deliberate translation sounds more logical to me: "The property to have Hadwiger number $k$ characterizes a more abstract idea of connectedness, which can just be expressed by the natural number $k$". But it is more of an interpretation than the first version. -I have asked two German native speaker scientists, but they were not sure as they are no experts in graph theory. It would be great if someone could help me with this. - -REPLY [4 votes]: Let me give it a try. As a disclaimer, English is not my mother tongue, so my translation might have linguistic flaws. -First of all, I would say the sentence is hard to translate and it is a bit informal, i.e. it is not a rigorous mathematical statement. In my view, this sentence gives an informal motivation why it it is useful to look a "K(k)" graphs, i.e. graphs with Hadwiger number $k$. Secondly, I think your 2nd version is relatively close to what this sentence means, but I think the meaning goes beyond connectedness. -My suggestion is to translate it in the context of the beginning of the whole paragraph where you took the quotes from. The first sentence in this paragraph reads: -"Im folgenden sprechen wir von einer Möglichkeit der Klassifikation der -Streckenkomplexe, die besonders im Hinblick auf das Problem der chromatischen Zahl von besonderem Interesse zu sein scheint." -(My translation: In the following, we speak of a possible classification of graphs, which seems to be of particular interest regarding the problem of chromatic number.) -So this is setting the scene really broadly, and announcing that the following definition could be relevant for the chromatic number problem, and in particular for the famous 4-color-problem. -Next, he defines the Hadwiger number (his "K(k)" graphs), and then there is the sentence you are asking about. Here is my suggested translation: - -The property of having Hadwiger number $k$ ($k$ a natural number) characterizes a deeper connection between those graphs. (in the sense of: deeper than the ordinary notion of connectendness)<|endoftext|> -TITLE: Geometric interpretation of the Weyl tensor? -QUESTION [16 upvotes]: The Riemann curvature tensor ${R^a}_{bcd}$ has a direct geometric interpretation in terms of parallel transport around infinitesimal loops. -Question: Is there a similarly direct geometric interpretation of the Weyl conformal tensor ${C^a}_{bcd}$? -Background: My understanding is that the Weyl conformal tensor is supposed to play a role in conformal geometry analogous to the role of the Riemann curvature tensor in (pseudo)Riemannian geometry. For instance, it is conformally invariant, and (in dimension $\geq 4$) vanishes iff the manifold is conformally flat, just as the Riemann curvature tensor is a metric invariant and vanishes iff the manifold is flat. The two tensors also share many of the same symmetries. So it would be nice to have a more hands-on understanding of the Weyl tensor when studying conformal geometry. -Notes: - -I'd be especially happy with a geometric interpretation which is manifestly conformal in nature, referring not to the metric itself but only to conformally invariant quantities like angles. - -I'm also keen to understand any subtleties which depend on whether one is working in a Riemannian, Lorentzian, or more general pseudo-Riemannian context. - -REPLY [14 votes]: There is such an interpretation, with a few caveats. Essentially, there is a canonical connection on a certain vector bundle for which the "principal part" of the curvature is the Weyl tensor in dimensions $n\geq4$, and the Cotton tensor when $n=3$. I will describe this from the point of view of the tractor calculus, but avoid introducing unnecessary bundles where needed. This can also be described using the Fefferman–Graham ambient metric or using Cartan connections. This summary mostly follows Bailey–Eastwood–Gover, though Armstrong and articles written by Gover are also good references. I use abstract index notation throughout. -First, we define conformal densities. Given a conformal manifold $(M,c)$, a conformal density of weight $w\in\mathbb{R}$ is an equivalence class of pairs $(g,f)\in c\times C^\infty(M,c)$ with respect to the equivalence relation $(g,f)\sim(e^{2\Upsilon}g,e^{w\Upsilon}f)$. Let $\mathcal{E}[w]$ denote the space of conformal densities of weight $w$. We similarly define $\mathcal{E}^i[w]$ as the space of equivalence classes of pairs $(g,v^i)\in c\times\mathfrak{X}(M)$ with respect to the equivalence relation $(g,v^i)\sim(e^{2\Upsilon}g,e^{w\Upsilon}v^i)$. Here $\mathfrak{X}(M)$ is the space of vector fields on $M$. -Next, we define the space of sections of the standard tractor bundle. Fix a metric $g\in c$. Define $\mathcal{T}_g^A=\mathcal{E}[1]\oplus\mathcal{E}^i[-1]\oplus\mathcal{E}[-1]$. Given another metric $\hat g \mathrel{:=} e^{2\Upsilon}g\in c$, we identify $(\sigma,v^i,\rho)\in\mathcal{T}_g^A$ with $(\hat\sigma,\hat v^i,\hat\rho)\in\mathcal{T}_{\hat g}^A$ if -$$ \begin{pmatrix} \hat\sigma \\ \hat v^i \\ \hat\rho \end{pmatrix} = \begin{pmatrix} \sigma \\ v^i + \sigma\Upsilon^i \\ \rho - \Upsilon_j v^j - \frac{1}{2}\Upsilon^2\sigma \end{pmatrix} . $$ -(Recall these are densities, so exponential factors are suppressed.) The space of sections $\mathcal{T}^A$ is the result after making this identification. Note that the top-most nonvanishing component is actually conformally invariant modulo multiplication by an exponential factor. Because of this, we call the top-most nonvanishing component the projecting part. -There is a canonical connection on (the vector bundle whose space of sections is) $\mathcal{T}^A$, the standard tractor connection, which, given a choice of metric $g\in c$, is given by the formula -$$ \nabla_j \begin{pmatrix} \sigma \\ v^i \\ \rho \end{pmatrix} = \begin{pmatrix} \nabla_j\sigma - v_j \\ \nabla_j v^i + \sigma P_j^i + \delta_j^i\rho \\ \nabla_j\rho - P_{ji}v^i \end{pmatrix} . $$ -Here $P_{ij}=\frac{1}{n-2}\left( R_{ij} - \frac{R}{2(n-1)}g\right)$ is the Schouten tensor and $n=\dim M$. It is straightforward to check that this is well-defined, in the sense that it is independent of the choice of matrix $g\in c$. -Given a metric $g\in c$, it is straightforward to compute that -$$ (\nabla_i\nabla_j - \nabla_j\nabla_i)\begin{pmatrix} \sigma \\ v^k \\ \rho \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ C_{ij}{}^k & W_{ij}{}^k{}_l & 0 \\ 0 & -C_{ijl} & 0 \end{pmatrix} \begin{pmatrix} \sigma \\ v^l \\ \rho \end{pmatrix} . $$ -This is conformally invariant by construction. The "3-by-3" matrix is the tractor curvature, and its projecting part is $W_{ij}{}^k{}_l$ when $n\geq4$ and $C_{ij}{}^k$ when $n=3$. Standard interpretations of holonomy then give the interpretation of the Weyl tensor in terms of parallel transport around infinitesimal loops that I indicated in the first paragraph. -Finally, given your bullet points, let me emphasize that the signature of $c$ plays no role here, and everything is manifestly conformally invariant. -Added in response to a comment. There are many geometric motivations for introducing the standard tractor bundle. One is that the conformal group of the sphere is $\operatorname{SO}(n+1,1)$, so it makes sense that the right replacement of the tangent bundle of a conformal $n$-manifold should be a bundle of rank $n+2$, as is the standard tractor bundle. Note that the metric on $\mathcal{T}$ has signature $(n+1,1)$, assuming we start with a conformal manifold of Riemannian signature (if $c$ has signature $(p,q)$, the metric on the standard tractor bundle has signature $(p+1,q+1)$). -Another motivation comes from the ambient metric. First, note that the flat conformal sphere $(S^n,c)$ (i.e., the conformal class of the round $n$-sphere) can be identified with the positive null cone $\mathcal{N}$ centered at the origin in $\mathbb{R}^{n+1,1}$. This is done by noting that the projectivization of $\mathcal{N}$ is $S^n$ and identifying sections of $\pi\colon\mathcal{N}\to S^n$ with metrics in the conformal class $c$ by pullback of the Minkowski metric. (Incidentally, this leads to a proof that $\operatorname{SO}(n+1,1)$ is the conformal group of $S^n$.) In this case, a fiber $\mathcal{T}_x$ of the standard tractor bundle is identified with $T_p\mathbb{R}^{n+1,1}$ for some $p\in\pi^{-1}(x)$; this is made independent of the choice of $p\in\pi^{-1}(x)$ by identifying tangent spaces at points subject to a homogeneity condition matching that of $\mathcal{E}^i[-1]$ above. The standard tractor connection is then induced by the Levi–Civita connection in Minkowski space, after making some identifications. -For a general conformal manifold $(M^n,c)$ of Riemannian signature, Fefferman and Graham showed that there is a "unique" Lorentzian manifold $(\widetilde{\mathcal{G}},\widetilde{g})$ which is "formally Ricci flat" and in which $(M^n,c)$ isometrically embeds as a null cone. Here formally Ricci flat means that the Ricci tensor of $\widetilde{g}$ vanishes to some order, depending on the parity of $n$, along the null cone, and I write unique in quotes because the metric is only determined as a power series to some order along the cone, and this up to diffeomorphism. One recovers the standard tractor bundle and its canonical connection from that of $(\widetilde{\mathcal{G}},\widetilde{g})$ as in the previous paragraph. See Fefferman–Graham for details, or Čap–Gover for a detailed description of the relation between the tractor calculus and the ambient metric, including the identifications I didn't detail. A similar construction for other signatures works, consistent with what is described in the previous paragraph.<|endoftext|> -TITLE: Sign of the function $f(n)=\sum_{k=1}^n\frac{\mu(k)}{k}$ -QUESTION [10 upvotes]: It is well-known that the Mertens function $M(n)=\sum_{k=1}^n\mu(k)$ changes sign infinitely many times when $n\rightarrow +\infty$. Let $f(n)=\sum_{k=1}^n\frac{\mu(k)}{k}$, then $\lim\limits_{n\rightarrow +\infty}f(n)= 0$. -Question: Does the function $f(n)=\sum_{k=1}^n\frac{\mu(k)}{k}$ also change sign infinitely many times when $n\rightarrow +\infty$ ? - -REPLY [17 votes]: Yes it does. To see this, note that by partial summation, -$$\frac{1}{\zeta(s + 1)} = s\int_{1}^{\infty}\sum_{n \leq x} \frac{\mu(n)}{n} x^{-s} \, \frac{dx}{x}$$ -for all $\Re(s) > 0$. Now let $\Theta$ denote the supremum of the real part of the zeroes of $\zeta(s)$, and suppose in order to obtain a contradiction that there exists some $\varepsilon > 0$ and $x_{\varepsilon} > 1$ such that -$$\sum_{n \leq x} \frac{\mu(n)}{n} < x^{-1 + \Theta - \varepsilon}$$ -for all $x > x_{\varepsilon}$. Then Landau's lemma (Lemma 15.1 of Montgomery-Vaughan) states that if $\sigma_c$ is the infimum of $\sigma \in \mathbb{R}$ for which -$$\int_{1}^{\infty} \left(x^{-1 + \Theta - \varepsilon} - \sum_{n \leq x} \frac{\mu(n)}{n}\right) x^{-\sigma} \, \frac{dx}{x}$$ -is convergent, then -$$\int_{1}^{\infty} \left(x^{-1 + \Theta - \varepsilon} - \sum_{n \leq x} \frac{\mu(n)}{n}\right) x^{-s} \, \frac{dx}{x}$$ -is holomorphic in the right half-plane $\Re(s) > \sigma_c$ but not at the point $\sigma_c \in \mathbb{R}$. On the other hand, this integral is equal to -$$\frac{1}{s + 1 - \Theta + \varepsilon} - \frac{1}{s\zeta(s + 1)}$$ -for $\Re(s) > 0$ and hence for $\Re(s) > \sigma_c$ by analytic continuation. However, this expression has a pole at $s = -1 + \Theta - \varepsilon$ and no other poles on the real line segment $\sigma > -1 + \Theta - \varepsilon$, yet by the definition of $\Theta$, there are poles in the strip $-1 + \Theta - \varepsilon < \Re(s) \leq -1 + \Theta$. Thus a contradiction is obtained, and so it follows that -$$\sum_{n \leq x} \frac{\mu(n)}{n} = \Omega_{+}\left(x^{-1 + \Theta - \varepsilon}\right).$$ -The same method shows that -$$\sum_{n \leq x} \frac{\mu(n)}{n} = \Omega_{-}\left(x^{-1 + \Theta - \varepsilon}\right),$$ -which implies an infinitude of sign changes. Moreover, with more work, one can show that -$$\sum_{n \leq x} \frac{\mu(n)}{n} = \Omega_{\pm}\left(\frac{1}{\sqrt{x}}\right),$$ -and if one is sufficiently enthusiastic, then under the assumption of the Riemann hypothesis and the linear independence hypothesis, -$$\limsup_{x \to \infty} \sqrt{x} \sum_{n \leq x} \frac{\mu(n)}{n} = -\liminf_{x \to \infty} \sqrt{x} \sum_{n \leq x} \frac{\mu(n)}{n} = \infty.$$ -The "true" rate of growth is probably the following: -$$0 < \limsup_{x \to \infty} \frac{\sqrt{x}}{(\log \log \log x)^{5/4}} \sum_{n \leq x} \frac{\mu(n)}{n} < \infty, \quad -\infty < \liminf_{x \to \infty} \frac{\sqrt{x}}{(\log \log \log x)^{5/4}} \sum_{n \leq x} \frac{\mu(n)}{n} < 0.$$ -See also these two answers of mine.<|endoftext|> -TITLE: Minimal pizza cutting -QUESTION [20 upvotes]: Given a circle, we want to divide it into $n$ connected equally sized pieces. In such a way that the total length of the cutting is minimal. What can we say about the solution for each $n$. Are they unique (up to some symmetry). Do all cuttings arise from the intersection of three straight segments with angle 120º? -These are conjectured to be the best solutions for $n \in \{2, 3, 4, 7\}$ - -Case $n= 2$ goes as follows: For simplicity the radius is $1$. If there is only one or no points on the circumference, then we know the minimal curve is the circumference, which has length $\frac{2\pi}{\sqrt{2}}$ bigger than 2 and we are done. Hence, we have at least two points in the circumference. If they are in the same diameter, the shortest curve is the line, so the cutting has at least length 2. -Now, pick a diameter parallel to $\overline{AC}$, which leaves both points on the same side. Since the curve connecting both needs to have area $\pi/2$ in needs to contain at least one point on the other side of the diameter. - -Now, $\overline{AB}$ is bounded below by a straight line, same with $\overline{BC}$. And the sum of those two is smaller than $\overline{AO} + \overline{OC}$ where $O$ is the center of the circle. -For $n\in\{3, 4, 7\}$ I don't have a proof and they might not be minimal. -As for the conditions on the cuttings, probably something like picewise $C^2$ is enough. - -REPLY [15 votes]: Here is recent paper coauthored by Cox (whom Yoav Kallus cited), -with a different focus: - -Headley, Francis, and Simon Cox. "Least-perimeter partition of the disc into $N$ regions of two different areas." arXiv:1901.00319 (2019). -arXiv abstract. - -Their first figure illustrates Yoav's points that -the arcs "meet in threes at $120^\circ$, but not necessarily straight," -and that "the arcs meet the boundary perpendicularly":<|endoftext|> -TITLE: When do two ultrafilters yield isomorphic ultrapowers? -QUESTION [12 upvotes]: Fix a cardinal $\lambda$$\newcommand{\cU}{\mathcal U}\newcommand{\cV}{\mathcal V}$. Consider the equivalence relation on $\beta\lambda$ given by $\cU\sim \cV$ when for all first-order structures $M$ we have $M^{\cU}\cong M^{\cV}$. -By considering the structure $(\lambda,A)_{A\subseteq \lambda}$, one can show that the $\sim$-classes have at most $2^\lambda$ elements (arguing as here). -If we consider the action of the symmetric group $S(\lambda)$ on $\beta\lambda$, it is easy to see that for each $\cU\in \beta\lambda$ and $f\in S(\lambda)$, we have $\cU\sim f(\cU)$. -Is the converse also true, namely, if $\cU\sim \cV$, are they necessarily conjugate by a permutation of $\lambda$? If not, is it true for $\lambda=\aleph_0$ (or any $\lambda$)? - -Edit: Given that the answer is positive (for signature of size $2^\lambda$), it would also be interesting to know whether the equivalence still holds if we define $\cU\sim \cV$ when $M^{\cU}\cong M^{\cV}$ for all first-order structures $M$ whose signature is either - -strictly smaller than $\lambda$, or -of size at most $\lambda$, or -strictly smaller than $2^{\lambda}$. - -REPLY [9 votes]: If one allows an arbitrary signature, the answer to this question is fairly well-known. Consider the structure $(\lambda,A)_{A\subseteq \lambda}$. Let $(M,R_A)_{A\subseteq \lambda}$ be the common ultrapower by $\mathcal U$ and $\mathcal V$. Note that $A\in \mathcal U$ if and only if $[\text{id}]_U \in R_A$ by the definition of an ultrapower. Fix functions $f,g:\lambda\to \lambda$ such that $[f]_\mathcal V = [\text{id}]_\mathcal U$ and $[g]_\mathcal U = [\text{id}]_\mathcal V$. We have $f_*(\mathcal V) = \mathcal U$: for any $A\subseteq \lambda$, -$A\in \mathcal U$ if and only if $[\text{id}]_\mathcal U\in R_A$ or equivalently -$[f]_\mathcal V\in R_A$, which by definition means $\{\alpha < \lambda : f(\alpha) \in A\}\in \mathcal V$ or $f^{-1}[A]\in \mathcal V$; that is, $A\in f_*(\mathcal V)$. -Similarly $g_*(\mathcal U) = \mathcal V$. It -follows that $(g\circ f)_*(\mathcal V) = \mathcal U$. -A fundamental theorem (discovered independently by Rudin, Keisler, Blass Katetov, Frolík, and maybe others, proof below) states that for -any ultrafilter $\mathcal W$ over $X$ and any $h :X\to X$, if $h_*(\mathcal W) = \mathcal W$ -then $[h]_\mathcal W =[\text{id}]_\mathcal W$ -Therefore $[g\circ f]_\mathcal V = [\text{id}]_\mathcal V$. Hence there is a set $A\in \mathcal V$ such that $g\circ f\restriction A$ is the identity. In other words, $f$ is one-to-one on a set in $\mathcal V$. It is then not hard to modify $f$ on a null set to make it a permutation. -Proof that $h_*(\mathcal W) = \mathcal W$ implies $[h]_\mathcal W= [\text{id}]_\mathcal W$: Assume not, and so without loss of generality $h(x) \neq x$ for all $x\in X$. Consider the graph $G$ with vertex set $X$ and edge set $E = \{\{x,y\}\in [X]^2 : h(x) = y\}$. We claim $G$ is $3$-colorable. Any finite connected induced subgraph $H$ of $G$ with $n$ vertices contains at most $n$ edges (as $x\mapsto \{x,f(x)\}$ is a partial surjection), and hence contains at most one cycle. Therefore removing at most one edge of $H$ yields an acyclic and hence 2-colorable graph, and this easily implies $H$ is $3$-colorable. By compactness, $G$ is $3$-colorable. Therefore there is a partition $\{A_0, A_1, A_2\}$ of $X$ such that $G\restriction A_n$ is discrete for $n =0,1,2$. This means that $h^{-1}[A_n]\cap A_n\neq \emptyset$ for $n =0,1,2$. Therefore if $A_n\in \mathcal W$, then $A_n\notin h_*(\mathcal W) =\mathcal W$, contradiction. -The question for countably incomplete ultrafilters and finite signatures is far more interesting and seems to be sensitive to set theoretic hypotheses. The answer seems to be yes assuming Woodin's HOD Conjecture. Details on request. -Details: I can actually answer the finite signature question positively without the HOD Conjecture, although my proof looks like overkill. I need a lemma. -Lemma. Suppose $i,j : V_\alpha\to N$ are elementary embeddings that are continuous at regular cardinals $\delta_0 < \delta_1 < \alpha$, and suppose there is a partition $\vec S$ of $\{\alpha < \delta_1 : \text{cf}(\alpha) = \delta_0\}$ into $\delta_1$ stationary sets such that $i(\vec S) = j(\vec S)$. Then $i\restriction \delta_1 = j\restriction \delta_1$. -Given this, we proceed as follows. Consider the structure $M = (V_\alpha,\in,\delta_0,\delta_1,\vec S,f)$ where $\delta_0 > \lambda$ is regular, $\delta_1 \geq 2^\lambda$ is regular, $\vec S$ is a stationary partition as in the lemma, and $f$ is a surjection from the cardinal $2^\lambda$ onto $P(\lambda)$. The ultrapowers of this structure by $\mathcal U$ and $\mathcal V$ coincide, and so we can identify them. Let's say the ultrapower is $(N,E,d_0,d_1,\vec T,g)$. (The signature has one relation symbol, interpreted as $E$, and four constant symbols.) -Let $i,j:V_\alpha \to N$ be the ultrapower embeddings associated to $\mathcal U$ and $\mathcal V$. These embeddings are continuous at $\delta_0$ and $\delta_1$ since these cardinals are regular and above the underlying set $\lambda$ of $\mathcal U$ and $\mathcal V$. Since $i(\vec S) = \vec T = j(\vec S)$, the hypotheses of the lemma are true, so $i\restriction \delta_1 = j\restriction \delta_1$. In particular, these embeddings agree on the ordinal $2^\lambda$. -Now $i[P(\lambda)] = i(f)[i[2^\lambda]] = g[i[2^\lambda]] = g[j[2^\lambda]] = j[P(\lambda)]$. Inverting the transitive collapse shows $i\restriction P(\lambda) = j\restriction P(\lambda)$. This suffices to run the easy argument from the arbitrary signature case. -Proof of Lemma. Let $d_0 = i(\delta_0)$, $d_1 = i(\delta_1)$, $\vec T = i(\vec S) = \langle T_a : a < d_1\rangle$. We run an argument due to Solovay to show that $j[\delta_1]$ is equal to the set $\{a < d_1 : T_a\text{ meets every $\delta_0$-club in $d_1$}\}$. By symmetry (and since $j(\vec S) = \vec T$), we have the same characterization of $j[\delta_1]$, and this proves $i[\delta_1] = j[\delta_1]$, which easily implies the lemma. -(For the record, a set $C\subseteq d_1$ is $\delta_0$-club if it is cofinal in $d_1$ and any increasing $\delta_0$-sequence of elements of $C$ has a supremum in $N$ and this supremum belongs to $C$. Two(?) examples of $\delta_0$-clubs : $i[\delta_1]$ and $j[\delta_1]$.) -First assume $T_a$ meets every $\delta_0$-club in $d_1$, and we will show $a\in i[\delta_1]$. The point is that $T_a$ meets $i[\delta_1]$. Take $\xi < \lambda$ with $i(\xi)\in T_a$. Note that $\text{cf}(\xi) = \delta_0$, so there is some $\alpha < \lambda$ with $\xi \in S_\alpha$. Hence $\xi \in i(S_\alpha) = T_{i(\alpha)}$. Since $\xi \in T_a\cap T_{i(\alpha)}$ and the $T_b$ are pairwise disjoint, $a = i(\alpha)$. -Conversely let us show that $T_{i(\alpha)}$ (i.e., $i(S_\alpha)$) meets every $\delta_0$-club in $\delta_1$. Towards this fix such a $\delta_0$-club $C$. Then the usual argument shows that $C\cap i[\delta_1]$ is a $\delta_0$-club in $d_1$. This means $i^{-1}[C]$ is a $\delta_0$-club in $\delta_1$ in the usual sense. Therefore there is some $\xi \in S_\alpha\cap i^{-1}[C]$. Now $i(\xi)\in i(S_\alpha)\cap C$, so $i(S_\alpha)$ meets $C$, as desired. This proves the lemma.<|endoftext|> -TITLE: Is it the referee's responsibility to verify results from arXiv preprints used in the refereed paper? -QUESTION [31 upvotes]: I'm refereeing a Banach spaces paper and it looks pretty good. I'm about ready to recommend it for publication. -However, its main result depends crucially on some other results that are in preprints on the arxiv. Is it the referee's responsibility to verify those results too? Or should I just alert the editor that we should wait for publication of those preprints? Or something else? -It's not that I'm trying to be lazy exactly. I'm just super busy with other things, and so want to be efficient with my time. - -REPLY [54 votes]: I'm going to use the word "I" in this answer since there is no universally agreed-upon standard for what a referee should do. -I feel that the referee's only job is to make an informed recommendation to an editor as to whether or not a paper should be accepted. The extent to which that includes verifying that a paper is correct is a bit subtle -- you want to be able to vouch for the paper's correctness, but ultimately the correctness of the paper is the author's responsibility. My personal interpretation of that mandate is that I should understand and believe all the arguments of the paper, though there have been times when I have not verified things like enormous calculations, in which case I make that caveat in my report. I should also have no good reason to doubt the results of anything that is cited. If I have such a doubt (and I frequently have doubts about the correctness of both the published and the unpublished literature!), then my job is not to go and referee those papers as well. Instead, in my report I give an honest account to the editor of my concerns, and let them decide how they want to handle them.<|endoftext|> -TITLE: If two smooth manifolds are homeomorphic, then their stable tangent bundles are vector bundle isomorphic -QUESTION [9 upvotes]: I am currently reading Kervaire-Milnor's paper "Groups of Homotopy Spheres I", Annals of Mathematics, and I am trying to prove (or disprove) the following result. The more elementary the proof, the better. - -If two smooth manifolds are homeomorphic, then their stable tangent -bundles (i.e. the Whitney sum of the tangent bundle with the trivial -line bundle) are vector bundle isomorphic. - -I am trying to prove this as an intermediate step to give an alternative proof for KM's Theorem 3.1: Every homotopy sphere is $s$-parallelizable. - -REPLY [8 votes]: Let me add something to Michael Albanese's great answer to see this question in a broader context. -Novikov proved that the rational Pontryagin classes are homeomorphism invariants (in fact, this was one of the achievements for which he received the Fields medal in 1970). The integral Pontryagin classes, however, are not invariant under homeomorphism, see Chapter 4.4 of "The Novikov Conjecture" by Kreck and Lück. -Some polynomials in the Pontryagin classes are even homotopy invariant: for instance, $p_1$ of a closed oriented $4$-manifold $M$ agrees (by Hirzebruch's signature theorem) with $3\sigma(M)$ times the fundamental class in cohomology (where $\sigma(M)$ denotes the signature of $M$), which is invariant under homotopy equivalence. -The famous Novikov conjecture asks whether certain so-called "higher signatures" are also invariant under homotopy equivalence. It is one of the most important open questions in topology.<|endoftext|> -TITLE: Tips to organize a successful math workshop -QUESTION [15 upvotes]: I am a PhD student from India working on representations of quantum groups. I want to organize a workshop on Hopf Algebra and Quantum groups but there are only 2 or 3 specialists in India currently working on it. So I googled some professors outside India who are currently working on this field. Now the problem is how to approach them to participate as a speaker. Since I have no experience organizing such a workshop I want some help. I donot know what offers I have to make to them. I mean giving air fares both ways and lodging and broading will be enough or do I need to arrange some honorarium. Also there is another problem. To propose a workshop, I have to give details of the speakers, the topics they are covering and outcome of the proposed workshop. After recieving all the data the working committee is either going to accept or reject the proposal. If they accept it then they are going to give the funding. Now it may happen that the funds given may not be enough or in worse case they can reject the proposal. Now if some professor gets interested at the prospect and later find that the proposal is rejected, what they will think about me. -So I will be very grateful if someone comes up some piece of advise how to proceed. - -REPLY [17 votes]: The Lorentz Center has some advice that you might find useful, I have organized several workshops there and followed a route similar to the one you describe. -Tentative answers to your specific questions: - -Since you can only invite speakers conditional on the acceptance of the workshop proposal, it is crucial that you invite them well in advance, so that they can block their calendar without interfering with other commitments. I would invite at least a year in advance --- of course now with COVID a two-year advance seems more reliable. Don't worry about "what they will think of me" if the proposal gets rejected: if you explain in advance that the invitation depends on the acceptance by some agency they will understand. - -I would initially offer to reimburse "local expenses" (= hotel and meals) and explain that you have some funds for travel expenses which you can distribute according to need. Some speakers may be able and willing to use their own funds for travel. You could further motivate this by explaining that if speakers can help in this way with the expenses you will be able to support junior participants. - -Honorarium? No, for academic lectures that is unusual and not needed --- at least not for lectures delivered "in real life". The idea being that the hospitality which the host will offer (e.g. taking you out to a nice restaurant) is sufficient compensation. Since COVID I have seen a modest honorarium (a few hundred $) being offered for online lectures, mainly to secure commitment from the speaker.<|endoftext|> -TITLE: Fixed space of maximal torus and Weyl group -QUESTION [6 upvotes]: Let $G$ be a compact connected Lie group and $T\subset G$ a maximal torus. Let $V$ be a representation of $G$ and $U=\{v\in V: tv=v\textrm{ for all }t\in T\}$. For any $g\in N(T)$ we have for all $t\in T$ and $v\in U$ that $g^{-1}tgv=v \Rightarrow t(gv)=gv$. This shows that for all $v\in U$ we have $gv\in U$ as well. From this we can define a representation of the Weyl group $W$ on $U$. I have the following two questions: - -Does this $W$-module structure on $U$ depend on the choice of the maximal torus $T$? - -Assume that $V$ is irreducible and $U$ nontrivial. Is there a way to understand when $U$ is irreducible as $W$-module? Is it always the case? - -REPLY [7 votes]: This paper of Humphreys addresses your second question (the first is answered in the comments - the $W$-module structure is independent of the choice of torus): https://people.math.umass.edu/~jeh/pub/zero.pdf -Here is a quote from the paper (section 1.4): - -Indeed, it is usually unclear how to determine directly whether or not -the W-module $L_\lambda(0)$ [i.e. $U$] is simple, even if its dimension is compatible with -simplicity. - -The $W$-module $U$ is not always irreducible. Indeed in type $A_2$ (i.e. $G=SU(3)$), there is a formula for the dimension of $U$ in Section 2.2, which shows that the dimension may be unbounded.<|endoftext|> -TITLE: The smallest primitive root modulo powers of prime -QUESTION [5 upvotes]: I wrote a program to calculate the minimal primitive root modulo $p^a$ where $p > 2$ is a prime, by enumerating $g$ from $2$ and checking whether it's a primitive root, but I forgot to check $\gcd(g, p) = 1$. However, it still worked in all the test cases. -So is it true that the smallest primitive root modulo $p^a$ is smaller than $p$? -P.S. I think this should be right because the smallest primitive root modulo $p$ is $O(\log^6 p)$ (assuming the generalized Riemann hypothesis), which is much smaller than $p$ when $p$ is large enough. But I have no idea how to prove this. - -REPLY [5 votes]: This is known, see -https://arxiv.org/abs/1908.11497 -where it is show for squares of primes. Higher powers then follow from other elementary arguments<|endoftext|> -TITLE: Foundational results dependent on/equivalent to the continuum hypothesis or its negation? -QUESTION [5 upvotes]: I remember at a certain point early in my mathematical studies learning that the Axiom of Choice is equivalent to the following statement on Cartesian products: - -If $\{ X_i \}_{i \in I}$ is any collection of nonempty sets indexed by an index set $I$, then $\prod_{i \in I} X_i$ is nonempty. - -To me, this settled the question of whether to use Axiom of Choice in practical contexts (although it's still interesting to consider systems of math where it doesn't hold, and the interdependence of various other theorems/results/lemmas/axioms on $AC$). -My first question is: -Question 1--Is there any similarly fundamental lemma or theorem which depends on the continuum hypothesis or its negation? That is, are there any basic facts in set theory, topology, measure theory, etc. which are (a) "self-evident" and (b) equivalent to $CH$ or $\lnot CH$? -I would also be interested in hearing if such a statement existed for $GCH$ or its negation $\lnot GCH$, although to me $GCH$ seems "less likely" to be true than $CH$ just because it makes a much broader statement over the class of all cardinals, whereas $CH$ is a relatively narrow statement about the relationship of two cardinals $2^{\aleph_0}$ and $\aleph_1$. -Currently, the two "simplest" results (that I know of) in this vein that would directly depend on $CH$ or $\lnot CH$ are: - -Wetzel's problem - -Whether or not $\Bbb{R}^\omega$ is normal in the box topology - - -But neither of these seem intuitively true or false, much less so essential that we had better accept them one way or another if we want to get any serious math done in the related field. -I'm aware that attempts have been made to resolve $CH$ one way or another (e.g. Freiling's axiom of symmetry) that are basically trying to reduce $CH$ to such an obviously true/false statement of general set theory/topology/measure theory. So I have a follow-up: -Question 2--What seem to be the obstacles to finding such a resolution of $CH$ or $\lnot CH$? That is, why is it so difficult to make concrete and testable statements (i.e. not trivial things like "There exists an element of $2^{2^{\aleph_0}}$ which is neither countable nor of size $\mathfrak{c}$") dependent on $CH$'s truth or falsity? And, should this difficulty be taken as evidence one way or the other for $CH$? Slash, is it actually considered evidence one way or the other for $CH$? -For instance: every Borel set is either of size $\aleph_0$ (if countable) or of size $2^{\aleph_0}$ (if uncountable). Is our difficulty in constructing a set of intermediate cardinality (as opposed to the ease with which we can construct a non-measurable set) evidence that no such intermediate-cardinality set exists? -I'll also mention that I take the "Platonic view" of $CH$. That is, I believe that despite the existence of models of set theory where either $CH$ or $\lnot CH$ holds, the statement - -"If $S = 2^\Bbb{N}$ is the set of all subsets of $\Bbb{N}$, then for $A \subset S$ any subset of $S$, either $A$ is countable, or there exists a 1-1 correspondence between $A$ and $S$" - -has a canonical and demonstrable true/false answer. - -REPLY [6 votes]: As it is stated in the comments, one reference is Sierpinski's book, Hypothese Du Continu, though it is not in English. -Another reference is Propositions Equivalent to the Continuum Hypothesis. -See also Some propositions equivalent to the continuum hypothesis and The continuum hypothesis (CH) and its equivalent. -You may be also interested in Eliminating the Continuum Hypothesis. -Let me also state one equivalent of CH. I have taken it from Interactions between (set theory, model theory) and (algebraic geometry, algebraic number theory ,...): -Let $R$ be a ring and $D(R)$ its unbounded derived category. Let $D^c(R)$ be the full subcategory of compact objects (in the explicit example below it is spanned by bounded complexes of f.g. projective modules). We say that $D(R)$ satisfies Adams representability if any cohomological functor $D^c(R)^{op}\rightarrow Ab$, i.e. additive and taking exact triangles to exact sequences, is isomorphic to the restriction of a representable functor in $D(R)$ (in particular it extends to the whole $D(R)$), and any natural transformation between restrictions of representable functors $D^c(R)^{op}\rightarrow Ab$ is induced by a morphism in $D(R)$ between the representatives. -Let $\mathbb C\langle x,y\rangle$ be the ring of noncommutative polinomials on two variables. The statement '$D(\mathbb C\langle x,y\rangle)$ satisfies Adams representability' is equivalent to the continuum hypothesis. -For another interesting equivalent of CH see: Reductions between certain incidence problems and the continuum hypothesis.<|endoftext|> -TITLE: Is it possible to calculate the parallel transport on a loop from the Riemann curvature? -QUESTION [6 upvotes]: I admit I am not a differential geometer (a probabilist actually). However recently I get interested and I would like to have more intuitions and insight of what is the Riemann curvature. -This is the way I see it so far (please correct me if I am wrong): - -We start from a connection $\nabla$. -This defines a parallel -transport along paths $\gamma$ that is a linear application on the -vector bundle. -If $\gamma$ is a loop, this application may be different to the identity. -The Riemann curvature is different from the identity when $\gamma$ is a very small loop (at first order). - -This kind of definition seems very similar to the one of the rotational (as presented in physics classes). - -We start from a vector field (one form) $u$, - -If $\gamma$ is loop, $I=\oint_{\gamma}u\cdot d\gamma$ can be different to $0$. - -The rotation $\operatorname{rot}(u)$ is this value $I$ when $\gamma$ is a very small loop (at first order). - - -and we have the wonderful Stokes Theorem which for the $\operatorname{rot}$ follows very naturally from this definition (we glue small loops together to get a big loop) $$\oint_\gamma u \cdot d\gamma = \iint_\mathcal{S}\operatorname{rot}(u)\cdot d\sigma $$ with $\mathcal{S}$ a surface delimited by $\gamma$. -So here is my question: Does there exist an equivalent for the Riemann curvature? That is: can one calculate the parallel transport of $\nabla$ along a loop $\gamma$ from the Riemann curvature on $\mathcal{S}$? - -REPLY [11 votes]: In the case that the Riemannian manifold $M$ in question has dimension $2$ and is oriented and $\gamma([0,1])\subset M$ is the piecewise-$C^1$ oriented boundary of a compact domain $S\subset M$, we have the famous Gauss-Bonnet Theorem, which asserts that the holonomy around $\gamma$ is equal to rotation by the angle -$$ -\theta = \int_S K\,dA. -$$ -Thus, yes, in this case, the holonomy around $\gamma$ can be 'computed' from the Riemann curvature tensor. -This may seem a little unsatisfying because the parallel transport around $\gamma$ is defined using only information in an open neighborhood of $\gamma$ (and, of course, one can get away with less than that), but the above formula uses information that could, a priori, come from very far away from the image $\gamma([0,1])$. However, simple examples show that, even for surfaces, one can have a closed curve with arbitrary holonomy for which the metric is flat on a neighborhood of the curve. Thus, the holonomy cannot be computed purely locally from the Riemann curvature tensor. -(By the way, Ben's cautionary remark is based on a $\gamma$ that is not the boundary of any surface, so there couldn't be a formula of the kind you are seeking that would address his 'counterexample'.) -Once one goes to higher dimensions, even for Riemannian $3$-manifolds$(M,g)$, there is no formula known that would start with the Riemann curvature tensor and construct 'something' that could be integrated over every oriented compact surface $S$ with connected boundary $\partial S$ so as to yield the element of $\mathrm{SO}(T_{p}M)$ that is the holonomy of the oriented curve $\partial S$ at the boundary point $p\in \partial S$. -Actual nonexistence would be hard to prove without making some assumptions about what form the 'something' constructed out of the Riemann curvature tensor might take. However, if you make reasonable assumptions, you can rule things out. -For example, it is not hard to show that there is no universal $2$-form on $(M^3,g)$ constructed polynomially out of the Riemann curvature tensor (even when one is allowed to use $g$ and $\nabla$ as well) that has the property that its integral over any oriented compact surface $S\subset M$ with circular boundary $\partial S$ gives even the angle of the rotation of the Riemannian holonomy around $\partial S$. (Note that this angle in $[0,\pi]$ does not depend on the point $p\in\partial S$ that one choses as the 'initial point'.)<|endoftext|> -TITLE: Searching for an early, highly theoretical, even philosophical, math paper on models or small-world networks -QUESTION [6 upvotes]: All I can remember is that it was very high-level / abstact and kind of philosophical, explaining (the discovery or interdependence of) small world networks. I assume that it was +50 years old and 'might' be an iconic paper, but maybe not - surely it was by far not as popular as the, already mentioned, paper "small world networks and their dynamics" by Duncan J. Watts & Steven H. Strogatz, but also it was on a completely different level of math. -I think I found it once on Azimuth Blog from John Carlos Baez. But it is very large and I don't even find "small world network" in the search (it might have also been mentioned by someone in the comment section). - -REPLY [4 votes]: I am answering my own question, because I actually did find it again (after 5 years and many tries). -It was not even close to being iconic nor +50y (it just looked old) nor specific on small networks :) - -Rosen, R. (1993). On models and modeling. Applied mathematics and computation, 56(2-3), 359-372. PDF - -Thanks @Carlo Beenakker for bringing up the Milgram paper!<|endoftext|> -TITLE: Intuition for categorical fibrations? -QUESTION [9 upvotes]: I think I have a pretty good intuitive understanding of most types of fibrations of quasicategories: - -a (trivial) Kan fibration is a bundle of (contractible) spaces with equivalent fibers, -a left/right fibration is a bundle of spaces with covariant/contravariant functors between fibers, -a (co)Cartesian fibration is the same as left/right but now the fibers are $\infty$-categories, -an inner fibration is bundle of $\infty$-categories with correspondences between fibers. - -One major exception is the class of categorical fibrations. I know they are the fibrations in the Joyal model structure on sSet but that description isn't very illuminating to me. I feel this is problematic since categorical fibrations are central to the theory of $\infty$-operads, which I am trying to learn at the moment. -What would be the best way to describe categorical fibrations in a similarly intuitive way? - -REPLY [9 votes]: Categorical fibrations are not particularly meaningful in their own right. Luckily, there is a characterization in the most interesting case, of categorical fibrations $p:Q\to R$ between quasicategories. Namely such a map $p$ is nothing more than an inner fibration and an isofibration, that is, it is weakly orthogonal to the inclusion of either endpoint into $E[1]$, the nerve of the isomorphism category. Alternative characterizations are that the restriction of $p$ to the cores is a Kan fibration, which might be the most intuitive description, or that $p$ induces an isofibration on homotopy categories. This approach to categorical fibrations is used throughout Riehl and Verity's work and can also be found in Rezk's notes "Stuff about Quasicategories." -It may be worth remarking here that every functor of $\infty$-categories is equivalent to an inner fibration with the same codomain, so that the most invariant and conceptual way of thinking about a categorical fibration may be simply as an isofibration, full stop. This is in close analogy to the canonical model structure on the category of small categories.<|endoftext|> -TITLE: Largest eigenvalue of finite band random matrices -QUESTION [10 upvotes]: Let $\mathbf{M}_n$ be an $n \times n$ symmetric matrix -$$ -\mathbf{M}_n = \begin{cases} -X_{j-i,i}\ &\text{if }i\leq j\leq r+i\\ -0\ &\text{if }r+i< j\leq n\end{cases} -$$ -for some fixed $r>0$, and the random variables $\{X_{i,j}\}$ are assumed real, positive, i.i.d., and have finite mean and variance. -As an example, for $r=1$ and $n=4$ we have, -$$\mathbf{M}_4 = -\begin{pmatrix} -X_{0,1} & X_{1,1} & 0 & 0\\ -X_{1,1} & X_{0,2} & X_{1,2} & 0 \\ -0 & X_{1,2} & X_{0,3} & X_{1,3} \\ -0 & 0 & X_{1,3} & X_{0,4} -\end{pmatrix}$$ -I was wondering if something is known about the asymptotic of $\lambda_1(\mathbf{M}_n)$, i.e., the largest eigenvalue of $\mathbf{M}_n$, in the limit $n \to \infty$. In particular, is something is known about the deviation of $\lambda_1(\mathbf{M}_n)$ from its mean, i.e., -$$ -\Pr\left[|\lambda_1(\mathbf{M}_n)-\mathbb{E}\lambda_1(\mathbf{M}_n)|\geq t\right]\leq ? -$$ -I was wondering whether there is a general concentration bound, e.g., for non-identical matrices, which subsumes the above case. - -REPLY [2 votes]: I start with this simple remark: the tridiagonal matrix $$A_k=\begin{pmatrix}0 & 1 & & & \\ 1 & 0 & 1 & & \\ & 1 & 0 & \ddots & \\ & & \ddots & & 1 \\ & & & 1 & 0\end{pmatrix}$$, $A_k\in \mathbb{R}^{(k+1)\times (k+1)}$ have largest eigenvalue $\lambda_\max (A_k) =2\cos{\frac{\pi}{k+2}}$. -We will focus on the submatrices with large entries of $M_n$. When there are $k$ consecutive large entries :$ \forall i\leq k$ $X_{a+i}\geq C $ for some $a$, we will assume that $X_{a+i} = C$ for all $i$ and write $CA_k$. This is obiously not true but it is just to simplify the discussion. -We then have $$ M_n = \begin{pmatrix}\ddots & \\ & C_1A_{k_1} \\ & & \ddots \\ & & & C_2 A_{k_2} \\ & & & & \ddots \\ & & & & & . \end{pmatrix} $$ -where $\ddots$ have small entries (let say $\mathcal{O}(1)$) and $C_i\gg 1$. The largest eigenvalue will come from these submatrices -$$\lambda_\max (M_n) \approx \max_j \lambda_{\max}(C_j A_{k_j})=\max_j 2 C_j\cos(\frac{\pi}{k_j+2})$$ -For large $n$ the behaviour will depend on the tail of the random variable $X_1$. -We first consider the case of polynomial tail : $\mathbb{P}(X \geq K)\sim \frac{1}{K^\alpha}$. -For any $k$, $\lambda_{\max}(C A_{k})\geq K\Leftrightarrow C \geq \frac{K}{2\cos(\frac{\pi}{k+2})}$ and we estimate $$\mathbb{P}(\forall i\leq k, X_k \geq \frac{K}{2\cos(\frac{\pi}{k+2})}) = \Big(\frac{2\cos(\frac{\pi}{k+2})}{K} \Big)^k$$ For $K\rightarrow \infty$, one can see that the case $k=1$ have the much larger probability and we deduce that in this situation it is enougth to consider only $k=1$ submatrices. -Conclusion for polynomial tail we have -$$\lambda_\max (M_n) \approx \max_j X_j \sim n^{1/\alpha}$$ (Because there are $n$ iid $X_j$, we set $K=n^{1/\alpha}$ such that $\mathbb{P}(X_1 \geq K)=\frac{1}{n}$). -We now consider the case of exponential tail : $\mathbb{P}(X \geq K)\sim \exp(-\gamma K)$. -We estimate $$\mathbb{P}\Big(\forall i\leq k, X_k \geq \frac{K}{2\cos(\frac{\pi}{k+2})}\Big) = \exp\Big(-\frac{\gamma k K}{2 \cos(\frac{\pi}{k+2})} \Big)$$ Still here for $K\rightarrow \infty$, the case $k=1$ have the much larger probability. -Conclusion for exponential tail we have -$$\lambda_\max (M_n) \approx \max_j X_j \sim \frac{\log(n)}{\gamma}$$ (we set $K$ such that $\mathbb{P}(X_1 \geq K)=\frac{1}{n}$). -We continue with the case of sup-exponential tail : $\mathbb{P}(X \geq K)\sim \exp(-K^\gamma)$. -We have $$\mathbb{P}\Big(\forall i\leq k, X_k \geq \frac{K}{2\cos(\frac{\pi}{k+2})}\Big) = \exp\Big(-\frac{ k }{2^\gamma \cos(\frac{\pi}{k+2})^\gamma}K^\gamma \Big)$$ Here there is a $k^*$ that maximize $\frac{k}{\cos(\frac{\pi}{k+2})^\gamma}$ which have the much larger probability for $K\rightarrow \infty$. -We also set $K$ such that this event is of order $1/n$ and then for sup-exponential tail we have -$$\lambda_\max (M_n) \sim \frac{2\cos(\frac{\pi}{k^*+2})}{(k^*)^\frac{1}{\gamma}}\log(n)^{\frac{1}{\gamma}}$$ -Finally in case of bounded $X$, for any $\epsilon>0$, and $k$, we can find $a$ such that $\forall i\leq k, X_{a+i}\geq \|X\|_\infty-\epsilon$ with probability that goes to $1$ as $n\rightarrow \infty$. Then $$2 \|X\|_\infty \geq \lambda_\max (M_n) \geq 2 (\|X\|_\infty-\epsilon) \cos(\frac{\pi}{k+2}) $$ and we get $\lambda_\max (M_n) \rightarrow 2 \|X\|_\infty$.<|endoftext|> -TITLE: Counterexamples against all odds -QUESTION [38 upvotes]: What are some examples of conjectures proved to be true generically (i.e. there is a dense $G_{\delta}$ of objects that affirm the conjecture) but are nevertheless false? -Also, it would be cool to see examples where the conjecture was proved true with probability 1, but was nevertheless false. -Of course one can manufacture statements from the spaces themselves, but I'm mostly interested in actual conjectures from contemporary mathematics whose resolution exhibited this pattern. -I am curious about this situation because sometimes, although we cannot find them easily, the "counterexample space" for a given conjecture may be quite large, yet inaccessible due to the limitations of existing techniques. For example, Tsirelson's conjecture and the Connes Embedding Conjecture were recently proven false, and although we cannot yet concretely construct a counterexample I see no reason to believe that counterexamples will necessarily be terribly rare objects...once the techniques are available to construct them. (These may be fighting words.) -The present question inquires about a distinct situation where it has been proved that a randomly selected object will not provide a counterexample. As dire as this sounds, the situation may be advantageous in that the construction of a counterexample may have to be much more surgical, and so one may see more clearly a way to build one. I'm wondering if my intuition about this is valid, based on recent history. -The question is just a passing curiosity, really, but I think someone may have a good story or two that will educate. - -REPLY [3 votes]: Let $\mathsf{A} = (A_1,\dots,A_m)$ be a tuple of $d \times d$ matrices. The joint spectral radius (JSR) of $\mathsf{A}$ is $\mathrm{JSR}(\mathsf{A}) := \lim_{n\to\infty} \sup_{i_1,\dots,i_n} \|A_{i_1} \dots A_{i_n}\|^{1/n}$, where $\|.\|$ is any norm on $\mathrm{Mat}(d\times d) = \mathbb{R}^{d^2}$. -The JSR was introduced by Rota and Strang in 1960. In the case of a single matrix ($m=1$), the JSR is equal to the spectral radius, that is, the biggest modulus of an eigenvalue of the matrix. -For equivalent definitions of the JSR, see e.g. Jungers' monograph. -The finiteness conjecture of Lagarias and Wang (1995) asserted that for any tuple $\mathsf{A} = (A_1,\dots,A_m)$, there is a product $A_{i_1} \dots A_{i_n}$ of some finite length $n$ whose spectral radius is exactly equal to $[\mathrm{JSR}(\mathsf{A})]^n$. This conjecture was disproved in 2001 by Bousch and Mairesse. More counterexamples were constructed later, e.g. here, here, and here. -However, it is conjectured (see Conjecture 8 by Maesumi) that if $m \ge 2$ and $d\ge 2$, then the counterexamples to the finiteness conjecture form a subset of $\mathbb{R}^{d^2m}$ of zero Lebesgue measure, so the finiteness conjecture is almost always true. This conjecture is supported by numerical evidence (see e.g. here), but so far remains entirely open.<|endoftext|> -TITLE: The connections between Kolmogorov complexity and mathematical logic -QUESTION [5 upvotes]: We know that Kolmogorov Cmplexity (KC) has connections to mathematical logic since it can be used to prove the Gödel incompleteness results (Chaitin's Theorem and Kritchman-Raz). Are there any other striking application of Kolmogorov complexity to mathematical logic (outside KC itself of course)? -Relatively simple examples like the ones I mentioned are preferred, but more complicated illustrations are also very welcome! -(This is cross posted from MSE, because I wasn't getting any useful replies there) - -REPLY [3 votes]: Four pointers to the literature from the last 25 years on applications of Kolmogorov complexity to mathematical logic: -Applications of Kolmogorov complexity to computable model theory (2007). - -In this paper we answer the following well-known open question in -computable model theory. Does there exist a computable not -ℵ$_0$-categorical saturated structure with a unique computable -isomorphism type? Our answer is affirmative and uses a construction -based on Kolmogorov complexity. - -Logical operations and Kolmogorov complexity (2002). - -Conditional Kolmogorov complexity can be understood as the complexity -of the problem $Y\rightarrow X$, where $Y$ is the problem “construct -$y$” and $X$ is the problem “construct $x$”. Other logical operations -($\wedge,\lor,\leftrightarrow$) can be interpreted in a similar way, -extending Kolmogorov interpretation of intuitionistic logic and Kleene -realizability. - -The Kolmogorov expression complexity of logics (1997). - -We introduce the Kolmogorov variant of Vardi's expression complexity. -We define it by considering the value of the Kolmogorov complexity -$C(L[{\cal A}])$ of the infinite string $L[{\cal A}]$ of all truth -values of sentences of $L$ in ${\cal A}$. The higher is this value, -the more expressive is the logic $L$ in ${\cal A}$. - -Kolmogorov complexity and the second incompleteness theorem (1995). - -It is well known that Kolmogorov complexity has a close relation with -Gödel’s first incompleteness theorem. In this paper, we give a new -formulation of the first incompleteness theorem in terms of Kolmogorov -complexity, that is a generalization of Kolmogorov’s theorem, and -derive a semantic proof of the second incompleteness theorem from it.<|endoftext|> -TITLE: Modified energy method for transformed Fokker-Planck equation (tricky integration by parts…) -QUESTION [11 upvotes]: I came across Villani's paper titled "Hypocoercive diffusion operators" and couldn't figure out a computation that is skipped in that paper. Specifically, consider the following transformed Fokker-Planck equation, where $h(t,x,v)$ is the unknown, $(x,v) \in \mathbb{R}^n \times \mathbb{R}^n$, $V(x)$ is some potential force: $$\partial_t h + v\cdot \nabla_x h - \nabla V(x)\cdot \nabla_v h = \Delta_v h - v\cdot \nabla_v h.$$ Notice that the Laplacian $\Delta_v$ is only a partial Laplacian in the sense that it only acts on the velocity variables $v$, and for the usual $L^2$ energy $\int h^2 d\mu$, where $d\mu = f_\infty(x,v) dxdv$ and $f_\infty(x,v) = \frac{\mathrm{e}^{-\left(V(x)+\frac{|v|^2}{2}\right)}}{Z}$ with $Z$ a normalization constant making $f_\infty$ a probability density in $(x,v) \in \mathbb{R}^n \times \mathbb{R}^n$, and we easily have $\frac{1}{2} \frac{d}{dt} \int h^2 d\mu = -\int |\nabla_v h|^2 d\mu$. Then the author says under suitable assumptions on $V$, we can find suitable constants $a,c, K>0$ so that $$\frac{d}{dt}\left(\int h^2 d\mu + a\int |\nabla_x h|^2 d\mu + c\int |\nabla_v h|^2 d\mu \right) \leq -K\left(\int |\nabla_v h|^2 d\mu + \int |\nabla_v\nabla_x h|^2 d\mu + \int |\nabla_v\nabla_v h|^2 d\mu\right). $$ However, I have no clue why the above inequality holds (and justifying it in 1D should be enough for me, i.e., in the case $(x,v) \in \mathbb{R}\times\mathbb{R}$). What I did is the following (in 1D setting). Set -$$I(t):=\left(a\int |\nabla_x h|^2 d\mu + c\int |\nabla_v h|^2 d\mu \right).$$ -Then -\begin{align*} -\frac 12\frac{dI}{dt} &= -a\int |\partial_v\partial_x h|^2 d\mu - c\int |\partial_v\partial_v h|^2 d\mu - c\int |\partial_v h|^2 d\mu\\ -&\quad \color{red}{+ a\int \partial_x h \partial_x\left(V'(x)\partial_v h\right) - v\partial_xh\partial_{xx}h~d\mu} \\ -&\quad \color{red}{+c\int V'(x)\partial_vh\partial_{vv}h - \partial_vh\left(\partial_x h+v\partial_v\partial_xh\right)~d\mu} -\end{align*} -But I have no clue as to the treatment of the terms in red. Any help would be greatly appreciated! -Edit: I have also asked this question on Math Stack Exchange (the link is https://math.stackexchange.com/questions/3782421/modified-energy-method-for-transformed-fokker-planck-equation-tricky-integratio) and no satisfying answer is given as well. - -REPLY [4 votes]: There is a worked-out proof in page 10 and following of Hérau's lecture notes. The detailed steps are for $n=1$, $V=0$, but I assume once that is understood, the more general case would follow smoothly. -As a short-hand notation we write $ \|\partial_x h \|^2=\int (\partial h/\partial x)^2\,d\mu$ and $\langle\partial_x h,\partial_v h\rangle=\int (\partial h/\partial x)(\partial h/\partial v)\,d\mu$. We will make use of the identities -$$\langle f,\partial_v g\rangle=\langle(-\partial_v+v)f,g\rangle,$$ -$$\langle v\partial_x f,f\rangle=0$$ -$$[\partial_v ,v\partial_x]=\partial_x,$$ -and the Cauchy-Schwartz + Young inequality -$$2|c\langle\partial_v f,\partial_x f\rangle|\leq 2c \|\partial_v f \|\, \|\partial_x f \|\leq c^2 \|\partial_v f \|^2+ \|\partial_x f \|^2.$$ -The Fokker-Planck equation for $n=1$, $V=0$ reads -$$\partial_t h+Lh=0,\;\;L=v\partial_x + (-\partial_v+v)\partial_v .$$ -The adjoint of $L$ is -$$L^\ast=-v\partial_x + (-\partial_v+v)\partial_v.$$ -The resulting derivatives are -$$-\frac{1}{2}\frac{d}{dt} \|h \|^2= \|\partial_v h \|^2,$$ -$$-\frac{1}{2}\frac{d}{dt} \|\partial_x h \|^2= \|\partial_v\partial_x h \|^2,$$ -$$-\frac{1}{2}\frac{d}{dt} \|\partial_v h \|^2=\langle\partial_x h,\partial_v h\rangle+ \|(-\partial_v +v)\partial_v h \|^2=\langle\partial_x h,\partial_v h\rangle+ \|\partial_v^2 h \|^2+ \|\partial_v h \|^2.$$ -So the derivative $dI/dt$ in the OP reduces to -$$\frac{dI}{dt}=-2a \|\partial_v\partial_x h \|^2-2c\biggl(\langle\partial_x h,\partial_v h\rangle+ \|\partial_v^2 h \|^2+ \|\partial_v h \|^2\biggr)$$ -$$\qquad\leq -2a \|\partial_v\partial_x h \|^2-2c \|\partial_v^2 h \|^2-(2c-c^2) \|\partial_v h \|^2+ \|\partial_x h \|^2.$$ -It remains to bound $ \|\partial_x h \|^2$. -In the lecture notes they do this by adding the mixed term $b\langle\partial_x h,\partial_v h\rangle$ to the left-hand-side of the inequality, which then gives a term $-b \|\partial_x h \|^2$ on the right-hand-side to dominate. Let me work that out, using the derivative$^{\ast}$ -$$\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle=- \|\partial_x h \|^2-2\langle\partial_v\partial_x h,\partial_v\partial_v h\rangle-\langle\partial_x h,\partial_v h\rangle,$$ -hence -$$\frac{d}{dt}J\equiv\frac{d}{dt}\biggl( \|h \|^2+a \|\partial_x h \|^2+b\langle\partial_x h,\partial_v h\rangle+c \|\partial_v h \|^2\biggr)=$$ -$$\qquad=-2(c+1) \|\partial_v h \|^2-2a \|\partial_v\partial_x h \|^2-b \|\partial_x h \|^2-2c\|\partial_v\partial_v h\|^2-2b\langle\partial_v\partial_x h,\partial_v\partial_v h\rangle-(b+2c)\langle\partial_x h,\partial_v h\rangle.$$ -The first four terms on the right-hand-side have a fixed sign, the last two terms can be bounded by the Cauchy-Schwartz + Young inequality. - -$^\ast$ The result for $(d/dt)\langle\partial_x h,\partial_v h\rangle$ given on page 10 of the cited lecture notes is mistaken. Here is a derivation: - -\begin{align} -\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle&=-\langle\partial_x Lh,\partial_v h\rangle-\langle\partial_x h,\partial_v Lh\rangle\\ -&=-\langle\partial_x h,(L^\ast\partial_v +\partial_v L)h\rangle,\\ -L^\ast\partial_v +\partial_v L&=\bigl(-v\partial_x+(-\partial_v+v)\partial_v\bigr)\partial_v+\partial_v\bigl(v\partial_x+(-\partial_v+v)\partial_v\bigr)\\ -&=\partial_x+\partial_v+2(-\partial_v+v)\partial_v\partial_v,\\ -\Rightarrow\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle&=-\|\partial_x h\|^2-\langle\partial_x h,\partial_v h\rangle-2\langle\partial_x h,(-\partial_v+v)\partial_v\partial_v h\rangle\\ -&=-\|\partial_x h\|^2-\langle\partial_x h,\partial_v h\rangle-2\langle\partial_v\partial_x h,\partial_v\partial_v h\rangle. -\end{align}<|endoftext|> -TITLE: Topological objects associated to Steinerberger's 4-regular graphs -QUESTION [8 upvotes]: Very recently, in arXiv:2008.01153, Steinerberger has associated to any sequence $(x_n)_{n\in\mathbb{N}}$ of distinct real numbers a 4-regular graph. -In the case irrational multiples, like $x_n=n\sqrt{2} \pmod{1}$, the plots in $\mathbb{R}^2$ seem to show the projection of a certain genus-g surface (see page 2 of the preprint). [edit:06-sept-2020: I had written that these were plots in $\mathbb{R}^3$, which is actually not the case, apologies.] - -is that indeed the case, i.e. does a limit shape as $n$ goes to infinity exist ? What type of literature (e.g. keywords, theorems) one should be looking at to establish it ? - -REPLY [5 votes]: The Steinerberger article seems to be aimed at tests about randomness. The question you are raising, which is given as an initial observation in this article, is a topic in topological graph theory. - -The following link should give you helpful references for your question: Reference for topological graph theory (research / problem-oriented) . - -REPLY [4 votes]: Here is a short Mathematica script that computes the graph and plots it with some standard function -f[n_] := Mod[n * Sqrt[2]//N, 1]; - -n = 200; -seq = f /@ Range[1,n]; -map = PositionIndex[seq]; -sort = map[#][[1]] & /@ (Sort@seq); - -edge1 = Partition[Range[1,n], 2, 1] ~ Join ~ {{n,1}}; -edge2 = Partition[sort, 2, 1] ~ Join ~ {{sort[[-1]], sort[[1]]}}; -G = Graph[Join[edge1, edge2]] - - -GraphPlot3D[G, GraphLayout->"SpectralEmbedding"] - - -GraphPlot3D[G, GraphLayout->"SpringElectricalEmbedding"] - - -It seems to resemble some kind of genus 1 surface. - -But seems to have nothing to do with $\sqrt2$. If I replace $\sqrt 2$ with $\pi$, the result still looks like a torus: - -Apparently, all we need is that the number is irrational.<|endoftext|> -TITLE: Good reference for topological Hochschild homology -QUESTION [11 upvotes]: I want to start reading topological Hochschild homology(THH) as well as topological cyclic homology (TC). -I have read the Hochschild homology and cyclic homology from the book Cyclic homology by J. Loday. This is very fantastic written book. Can someone suggest me some good reference like this for THH/TC? - -REPLY [8 votes]: Ib Madsen's survey - -MR1474979 (98g:19004) Madsen, Ib Algebraic K-theory and traces. -Current developments in mathematics, 1995 (Cambridge, MA), 191–321, -Int. Press, Cambridge, MA, 1994. - -was written to be such an introduction, but is 25 years old. Its technical foundations are given in terms of FSPs (functors with smash product). A more recent source is the book - -MR3013261 Dundas, Bjørn Ian; Goodwillie, Thomas G; McCarthy, Randy -The local structure of algebraic K-theory. Algebra and Applications, -18. Springer-Verlag London, Ltd., London, 2013. xvi+435 pp. ISBN: 978-1-4471-4392-5; 978-1-4471-4393-2 - -which works with $\Gamma$-rings (= monoids in $\Gamma$-spaces). Both of these predate the connection to crystalline and syntomic cohomology.<|endoftext|> -TITLE: Module spectrum maps up to stable homotopy -QUESTION [7 upvotes]: Let $R$ be a commutative ring spectrum, $M$ and $N$ be a $R$-module spectra. -Let us consider $R$-module maps from $M$ to $N$ up to stable homotopy, that is maps $M \to N$ such that the composites $R \wedge M \to M \xrightarrow{f} N$ and $R \wedge M \xrightarrow{1 \wedge f} R \wedge N \to N$ are equal in the stable homotopy category. -Now suppose that $M = R \wedge X$ is a free $R$-module. -Is it true that the set of stable homotopy classes of $R$-module maps up to homotopy in the previous sense, $[M, N]_R = [R \wedge X, N]_R$, is in the natural bijection with stable homotopy classes of all maps $[X, N]$? -I believe that something like that (or even stronger) holds for genuine $R$-module maps, but what about homotopy $R$-module maps if we are interested in the stable homotopy classes of maps only? - -REPLY [4 votes]: In any symmetric monoidal category $C$, the free $R$-module functor $R\otimes -: C\to Mod_R(C)$ is left adjoint to the forgetful functor $U:Mod_R(C)\to C$. Hence, $Hom_{Mod_R(C)}(R\otimes X,N) \cong Hom_C(X,U(N))$. Since the stable homotopy category is symmetric monoidal, the result you asked for follows.<|endoftext|> -TITLE: Basic prerequisite (topics) to read current research in Diophantine equation for an independent researcher -QUESTION [10 upvotes]: I have completed studying Galois theory, Fermat's Last Theorem for Regular prime and some number theoretic complex analysis (prime number theorem), and basic linear forms in logarithm. -What else should I complete reading to be eligible to read contemporary literature in Diophantine Equation (Exponential) ? -My graduation was in engineering, so I am from non-math (I mean serious math!) background, and I am not going to a university in near future, but wish to conduct research as an independent researcher. -There might be a lack of specification in my question, so if possible, adjust your answer according to that, also feel free to edit. - -REPLY [5 votes]: If I understand correctly, the keywords for the type of number theory you're interested in are "Diophantine approximation" and "transcendental number theory." Are you aware of Florian Luca's article on Exponential Diophantine Equations? If not, that might be a good place to start. You might also like M. Ram Murty and Purusottam Rath's book Transcendental Numbers. -Getting to the research frontier is always a tricky business. Michel Waldschmidt's Diophantine Approximation on Linear Algebraic Groups will give you a sense of what some of the major open problems in the subject are, but that's a very demanding book. There undoubtedly exist lower-hanging fruit, but it may be hard to find such fruit without an experienced advisor to guide you.<|endoftext|> -TITLE: Arc connectedness of product spaces -QUESTION [5 upvotes]: Is arc connected-ness well-behaved with respect to products? -That is - - -$\prod X_\alpha$ is arc connected iff $X_\alpha$ is arc connected $\forall \alpha$ - -In this question on MathStackexchange, an answer is provided only for the reverse implication, that is - - -If $X_\alpha$ is arc connected $\forall \alpha$, then $\prod X_\alpha$ is arc connected - -However, I wasn't able to get an answer for the forward implication, nor have I been able to find it in any book or from Googling. So, is the forward implication true, or is there a counterexample for the same? - -REPLY [7 votes]: This is false when the spaces are not Hausdorff. Let $X$ be the line with two origins $\{O_1,O_2\}$ and $Y$ be the usual line. Then $X\times Y$ is arc connected because you can pick an arc that starts at $(O_1,y_1)$ travels outside $\{O_1,O_2\}\times Y$ and then comes back to $(O_2,y_2)$, but $X$ itself is not arc connected.<|endoftext|> -TITLE: Showing subgroups with equal Lie algebras are equal -QUESTION [9 upvotes]: Let $k$ be a field. It might as well be algebraically closed, but I do not want to assume that it has characteristic $0$. I will write "group" for "affine group scheme over $k$", not assuming smoothness. -Two groups can have the same Lie algebras without being equal. For example, if $k$ has characteristic $2$, then every maximal torus in $\operatorname{SL}_2$ has the same Lie algebra as the centre $\mu_2$. Even two smooth groups can have the same Lie algebras without being equal: for example, all maximal tori in $\operatorname{SL}_2$ have the same Lie algebra. At least it is true that, if a smooth group $H$ is contained in a connected group $G$, and their Lie algebras are equal, then $H$ equals $G$; and so, if two connected subgroups $H_1$ and $H_2$ of $G$ have equal Lie algebras and smooth intersection, then they are equal. -I'm looking more for a result in line with Borel - Linear algebraic groups, Theorem 13.18(4)(d): given a maximal torus $T$ in a smooth, reductive group $G$, and a root $\alpha$ of $T$ in $G$, there is a unique smooth, connected subgroup of $G$ that is normalised by $T$ and whose Lie algebra is the $\alpha$-weight space of $T$ on $\operatorname{Lie}(G)$. The key ingredients here are reductivity and the torus action. -So I'm interested in any more general results of this sort that allow one to deduce equality of groups from equality of their Lie algebras. If that's too broad, I'll focus a bit: suppose that $G$ is a smooth, reductive group; $H_1$ and $H_2$ are smooth, connected, reductive subgroups; and $T$ is a torus in $H_1 \cap H_2$ that is not necessarily maximal in $G$, but is maximal in both $H_1$ and $H_2$. In this setting, if the Lie algebras of $H_1$ and $H_2$ are equal, then can we conclude that the groups are equal? -EDIT: I forgot to add, in case it helps, that, in my situation, $\operatorname C_G(T)^\circ$ (the connectedness automatic if $G$ itself is connected) is a torus. - -REPLY [5 votes]: $\DeclareMathOperator\Ad{Ad}\DeclareMathOperator\Cent{C}\DeclareMathOperator\GL{GL}\DeclareMathOperator\Lie{Lie}$The key point is not, as I expected, whether $\Cent_G(T)^\circ$ is a torus, but whether it equals $\Cent_G(\Lie(T))^\circ$. Certainly it is contained in the latter group, so this is the same as asking whether $T$ centralises $\Cent_G(\Lie(T))^\circ$. -If we do not require this, then we may adapt a construction by @WillSawin, pointed out by @MikhailBorovoi, to give a counterexample that is quite close to the one I attempted in the comments. Specifically, we give connected, reductive subgroups $H_1$ and $H_2$ of $G = \GL_4$ that contain a common maximal torus $T$ (for which $\Cent_G(T)^\circ$ is itself a maximal torus in $G$), and satisfy $\Lie(H_1) = \Lie(H_2)$, but $H_1 \ne H_2$. Namely, let $t$ be any non-scalar diagonal matrix in $\GL_2$, and put $H_1 = \left\{\begin{pmatrix} g & 0 \\ 0 & g^{[p]} \end{pmatrix} \mathrel\colon g \in \GL_2\right\}$ and $H_2 = \left\{\begin{pmatrix} g & 0 \\ 0 & t g^{[p]}t^{-1} \end{pmatrix} \mathrel\colon g \in \GL_2\right\}$, where $g^{[p]}$ is the matrix obtained by raising every entry of $g$ to the $p$th power. -Next we prove that, if $H_1$ and $H_2$ are connected, reductive subgroups of a common group $G$ that contain a common maximal torus $T$, and satisfy $\Lie(H_1) = \Lie(H_2)$, and if in addition $T$ centralises $\Cent_G(\Lie(T))^\circ$, then $H_1$ must equal $H_2$. As suggested by @MikhailBorovoi, it suffices to show that, for every root $b$ of $T$ in $\Lie(H_1) = \Lie(H_2)$, the corresponding root subgroups of $b$ in $H_1$ and $H_2$ are equal. Let $\mathfrak u$ be the common $b$-root subspace of $\Lie(H_1) = \Lie(H_2)$. Then we have $T$-equivariant isomorphisms $e_{i\,b} \colon \mathfrak u \to H_i$ such that $\Ad(e_{i\,b}(X))Y$ equals $Y - \mathrm db(Y)X$ for all $X \in \mathfrak u$ and all $Y \in \Lie(T)$. That is, $e_{1\,b}(X)e_{2\,b}(X)^{-1}$ lies in $\Cent_G(\Lie(T))$ for all $X \in \mathfrak u$, and hence, since $\mathfrak u$ is connected, in $\Cent_G(\Lie(T))^\circ$. Since this group is centralised by $T$, we see upon conjugating $e_{1\,b}(X)e_{2\,b}(X)^{-1}$ by $t$ that it equals $e_{1\,b}(b(t)X)e_{2\,b}(b(t)X)^{-1}$, for all $X \in \mathfrak u$ and all $t \in T$. In particular, $e_{1\,b}(X)e_{2\,b}(X)^{-1}$, as a function of $X$, is constant on $\mathfrak u \setminus \{0\}$, and hence, since it is continuous, is constant on $\mathfrak u$; but its value at $X = 0$ is the identity.<|endoftext|> -TITLE: An internalized version of Tennenbaum's Theorem -QUESTION [9 upvotes]: Tennenbaum's celebrated 1959 theorem (see here for a reference) is certainly one of the key theorems in mathematical logic. Not so much for its proof, but because it helps "isolating" $N$ as something very special: it is the only countable model of Peano arithmetic which is recursive. -But, does it? -Suppose I live inside any countable model $M$ of Peano: I think I am actually living in the standard natural numbers, and addition/multiplication are the standard ones (and so are all the recursive function, etc). So, it would seem that from the point of view of $M$, other PA-models are not recursive. To be a bit more precise, let me state this: -Internalized Tennenbaum Theorem (ITT): Let $M$ be a countable model of Peano. Then, for any other countable model $N$ not isomorphic to $M$, $N$ is not recursive in $M$, ie it is not $\Delta_1$-definable in M. - -Question: Can ITT be proved in, say, ZFC? If not, what is the -obstruction? - -Post Scriptum. -Thanks to Emil Jerabek for his suggestion: rather than the original misleading name, Derived Tennenbaum, use Internal (or Internalized). -The original name generated some confusion, see the comments of Francois Dorais, -thus I decided to rename the question. The recursivity required is IN the model, not FROM the model. - -REPLY [2 votes]: To move this off the unanswered queue, let me summarize the situation as correctly explained by the comments above: -The standard proof of Tennenbaum's theorem goes through inside $\mathsf{PA}$: $\mathsf{PA}$ proves that there is no $\Delta_1$ description of a model of $I\Sigma_1$. (As usual, $\mathsf{PA}$ can be replaced with something vastly weaker here; at a glance, already $I\Sigma_1$ should be enough.) -One key point here is that $\mathsf{PA}$ can quantify over $\Delta_1$ descriptions since this only involves reference to a bounded truth predicate. (Something like "There is no definable structure such that [stuff]" would have to be expressed as a scheme, but that's not an issue here.) -On the other hand, $\mathsf{PA}$ cannot express "the structure defined by the formula tuple $\Phi$ satisfies $\mathsf{PA}$" (at least not as cleanly as one might hope - we'd need to talk about explicitly-Skolemized structures, and that's a whole annoying rabbit hole to go down). This is why I've used $I\Sigma_1$ as the "Tennenbaum target:" as a finitely axiomatizable theory, $\mathsf{PA}$ has no trouble talking about its satisfaction (or not) in a defined structure, again by virtue of the satisfactoriness of a bounded truth predicate.<|endoftext|> -TITLE: Recent uses of applied mathematics in pure mathematics -QUESTION [27 upvotes]: In this answer Yves de Cornulier mentioned a talk about the possible uses of persistent homology in geometric topology and group theory. Persistent homology is a tool from the area of topological data analysis, specifically designed to extract information from empirical data and used for various applications, ranging from changes in brain function under drugs to the study of fluid flows to the merging of maps of distinct scales, along with many others. So this definitely belongs to the realm of applied mathematics, and it being used in pure mathematics is very interesting. -It goes without saying that many applications inspired a lot of research in pure mathematics, both in order to establish the foundations for the tools used in applies mathematics and just for the sake of studying interesting objects that appear in such interactions. I am talking specifically about the applied tools themselves being used in research in pure mathematics. -As an example, interval arithmetics was used in the solution of Smale's 14th problem and in the proof of Kepler conjecture (the latter also used a lot of linear programming). -Going back in time, we find that a lot of methods that were initially developed mainly for some specific application, such as celestial mechanics, the stereometry of wine barrels or heat transfer, became the standard tools in pure mathematics. Now it seems that the flow of methods and techniques is mostly one-way, from pure mathematics to the applied. But not completely one-way, hence the question: - -What are the recent uses of the tools from applied mathematics to the problems in pure mathematics? - -If one requires a more specific indication what does "recent" mean, let's say last 30 years (but I would be delighted to hear about older examples as well). - -REPLY [17 votes]: If mathematical developments in physics count as "applied mathematics" there are many examples --- as requested by the OP here is a recent one (< 30 years old) and an older one: - -Gauge theory spawned fundamental resuls in -differential geometry, see for example A proof via the Seiberg-Witten -moduli space of Donaldson's theorem on smooth 4-manifolds with -definite intersection forms (1995) -Dirac delta functions were introduced to simplify the normalization of quantum mechanical wave functions, and were formalized by Laurent Schwartz in the theory of distributions.<|endoftext|> -TITLE: "Universal" triangulated category -QUESTION [6 upvotes]: Let $\mathcal{C}$ be some category. One way to map this category into a triangulated category is to take the category of simplicial objects $s\mathcal{C}$ (which is an $\infty$-category), take its stabilization $\text{Stab}(s\mathcal{C})$ and take the homotopy category $\text{Ho}(s\mathcal{C})$ of the simplicial category (which is triangulated since it is the homotopy category of a stable $\infty$-category). Then we get a natural functor -$$\mathcal{C}\rightarrow \text{Ho}(\text{Stab}(s\mathcal{C})).$$ -My question is: -Does $\text{Ho}(\text{Stab}(s\mathcal{C}))$ satisfy some universal property? -That is, is it the "universal triangulated category" associated to $\mathcal{C}$ in some sense, i.e. if $\mathcal{T}$ is a triangulated category and $\mathcal{C}\rightarrow \mathcal{T}$ satisfying some properties, does this factor through $\mathcal{C}\rightarrow \text{Ho}(\text{Stab}(s\mathcal{C}))?$ -If this isn't a "universal triangulated category," does there exist such a construction? - -REPLY [13 votes]: I will give a partial answer. I note that the OP has asked a LOT of questions recently (I count 12 so far in the first 9 days of August), and many of them are good questions on which much research has already been done. I would encourage the OP to slow the rate of question-asking, to spend more time reading the references that have been provided, and to think carefully in future questions to avoid easily avoided problems like those that have been raised in the comments. -Now to the answer. Morally, what the OP is suggesting is exactly the kind of thing we love to do as homotopy theorists, but the devil is in the details. Specifically, in this case, the devil is in "...and $\mathcal{C} \to \mathcal{T}$ satisfying some properties..." The issue is that there could be multiple "obvious" ways to stabilize $\mathcal{C}$, and the functor $F:\mathcal{C} \to \mathcal{T}$ has to know that you mean the one you suggested. For example, suppose $\mathcal{C}$ is the empty category. Then whatever conditions you have in mind will probably be satisfied vacuously, and you're asking for a triangulated category $Ho(Stab(C))$ that is supposed to admit a map from every triangulated category $\mathcal{T}$. That's probably not what you really meant. -That said, homotopy theorists have thought long and hard along the direction you have in mind. I recommend the following papers: - -Dugger's Universal homotopy theories: given any small category $C$, create a "universal" model category $UC$, which is essentially the free homotopy theory generated by $C$. -Hovey's Spectra and symmetric spectra in general model categories: given a model category $C$ and a Quillen endofunctor $G$, create the stabilization $Sp(C,G)$ where $G$ becomes a Quillen equivalence, just like if $C = Top$, $G$ is the suspension functor, and $Sp(C,G)$ is spectra. -Lurie's Higher Topos Theory, which expands the work above into the realm of infinity categories, e.g., so that $Sp(C,G)$ is a stable $\infty$-category, when you start with a presentable $\infty$-category $C$. Similarly, Dugger's construction can be made to work to produce a presentable $\infty$-category $UC$. -Hovey's book on model categories: chapter 6 shows how to start from a pointed model category and produce a pre-triangulated category in a universal way. But note that when Hovey says "pre-triangulated" that doesn't mean the same as when other authors say "pre-triangulated." Here the structure that would need to be preserved by $F$ has to do with fiber and cofiber sequences, and you need $C$ to be pointed to define these. -Beligiannis and Reiten's Homological and homotopical aspects of torsion theories: Section 5 shows how, given a left, right, or pretriangulated category, there is a universal stabilization that preserves the old partial triangulated structure. -There are many, many other papers in this vein. For example, building on Beligiannis's paper Relative homological algebra and purity in triangulated categories (from 2000), Balmer and Stevenson just this year showed how to take a triangulated category, take a quotient (making it no longer triangulated), and then stabilize in a nice way. - -The point is that you have to know what structure of $\mathcal{C}$ to you want to be preserved by $F: \mathcal{C} \to \mathcal{T}$. For example, if you want to assume that $\mathcal{T}$ can be realized by a stable $\infty$-category, then you can probably get a positive answer by making sure $F$ plays nicely with the eventual $\infty$-category structure on $sC$. -However, it's not true that every triangulated category comes from a stable $\infty$-category (Muro and others have constructed counterexamples), so the universal properties shown by Lurie do not provide an affirmative answer to your question in general. Broadly speaking, the collection of triangulated categories breaks down into two types: those that are "geometric/topological" (e.g., the homotopy category of a stable $\infty$-category) and those that are algebraic. Unless you have some way to connect the geometric-type triangulated categories to the algebraic-type (a problem that many have thought about and none have solved to my knowledge) via whatever conditions you place on $F$, then your question is unlikely to have a positive answer in the generality in which you asked it. Hope this helps!<|endoftext|> -TITLE: Injective integer polynomial is injective modulo some prime -QUESTION [15 upvotes]: Let $Q\in \mathbb{Z}[x]$ be a polynomial defining an injective function $\mathbb{Z}\to\mathbb{Z}$. Does it define an injective function $\mathbb{Z}/p\mathbb{Z}\to\mathbb{Z}/p\mathbb{Z}$ for some prime $p$? - -REPLY [32 votes]: Consider $Q(x)=x(2x-1)(3x-1)$. This gives an injective map $\mathbb Z\to \mathbb Z$, because $n -TITLE: Finding zero-one vectors in the row space of a matrix -QUESTION [5 upvotes]: Suppose that $M$ is a square matrix with all elements on its main diagonal equal to $1$, and every row containing exactly two off-diagonal elements equal to $-1$; all other elements are equal to $0$. The kernel of $M$ is nonzero and, indeed, contains a vector with all its coordinates nonzero. Does it follow that the row space of $M$ contains a (nonzero) zero-one vector? -In case it matters, the sum of all elements in every column of $M$ is nonpositive, and $M_{ij}M_{ji}=0$ whenever $i\ne j$. - -REPLY [3 votes]: This is true: - -For any square matrix $M$ with all elements on its main diagonal equal to 1, and every row containing exactly two off-diagonal elements equal to −1 (with all other elements are equal to 0), the row space of $M$ contains a nonzero zero-one vector. Moreover, there is a linear combination of the rows with the coefficients $0$ and $-1$ only which yields such a vector. - -This is in fact the main lemma of this preprint; see, on the other hand, this MO problem for the explanations.<|endoftext|> -TITLE: Gromov's articles suitable for master students -QUESTION [14 upvotes]: I'm a master student and I have read "Monotonicity of the volume of intersection of balls" by Gromov and it was a great experience. When trying to fill the gaps, I often end up finding some very beautiful ideas. I want to keep reading Gromov because he inspires me and was delighted by the ideas that he show on that article, however, I don't know what articles of him are also suitable for a master student. I know some functional analysis, have taken a graduate course on differential geometry, measure theory, complex analysis, commutative algebra, algebraic topology, algebraic geometry, dynamical systems, fourier analysis on number fields and embeddings of finite spaces. -What Gromov's articles do you recommend to read? - -REPLY [15 votes]: I agree with Alexandre Eremenko in that most can be hard to read. But I think you can get a lot out of trying to understand them, as long as you're willing to black-box certain parts which may be inaccessible. Here's a summary of a few major articles. At the least, they all have significant parts which may be understandable, depending on your specific background. -"Curvature, diameter and Betti numbers" is a foundational contribution to Riemannian manifolds of positive sectional curvature, saying essentially that they cannot be arbitrarily topologically complicated. The tool is a Morse theory for the distance function. If you take the Toponogov theorem as given, it should be accessible with some understanding of homology; there is a review of spectral sequences in an appendix. I think it doesn't require too much Riemannian geometry other than the statement of the Toponogov theorem. -"Groups of polynomial growth and expanding maps" has some technical content, but the core idea (in section 6) is purely to do with metric spaces. It is applied by viewing finitely generated groups as examples of metric spaces. I don't have a good understanding of all of the group theory needed to prove the "main theorem", such as the Tits alternative, but the essential ideas of the proof, and many of the technical parts, are interesting (in and of themselves) and accessible. The key idea is to rescale the metric space structure of a finitely generated group to get a new metric space whose group of isometries is a Lie group; this "discrete" to "continuous" passage is somewhat unexpected and fundamental, and is the main point of interest for me. -"A.D. Aleksandrov spaces with curvatures bounded below" (with Yuri Burago and Grigori Perelman) is also largely to do with metric spaces. The idea is to reconstruct some of the theory of sectional curvature in Riemannian geometry by only using a metric space structure. Many parts of it should be accessible, especially with the book "A course in metric geometry" by Burago-Burago-Ivanov as a companion. -"Filling Riemannian manifolds" is very long, and I've never tried to read the whole thing, as it can be read in different parts, and there are already interesting and accessible ideas in the first few sections, to do with "filling radius" and "filling volume" and its relation to systolic inequalities; the only main prerequisite to begin with is simplicial homology. Interestingly a similar filling radius was studied at the same time in a different context in Schoen & Yau's "The existence of a black hole due to condensation of matter" -"The classification of simply connected manifolds of positive scalar curvature" (with Blaine Lawson) proves that certain topological operations on a manifold preserve the existence of Riemannian metrics with positive scalar curvature. Gromov and Lawson prove it by hand by basically elementary means. I've heard that it may have an error in it, but I don't know where it's supposed to be (I've never read it carefully). Schoen & Yau earlier proved the same result in "On the structure of manifolds with positive scalar curvature" by PDE methods that are simpler for me. Taking that theorem as given, Gromov and Lawson give some topological corollaries in cobordism theory that don't appear in Schoen & Yau - this part is too advanced for me. -"Convex symplectic manifolds" (with Yakov Eliashberg) is quite accessible, I recall it not needing too much past standard differential geometry. I think it could definitely be read with any of the standard introductory symplectic geometry textbooks as a companion. The main theorem says that if there is some weak algebraic equivalence between symplectic manifolds, then provided that they are "convex" in Eliashberg and Gromov's sense, one can construct certain symplectomorphisms which realize the equivalence. -The survey article "Spaces and questions" is also interesting to look through. There are also a number of interesting articles on Gromov's website in the non-pure math sections, such as "Mendelian Dynamics and Sturtevant’s Paradigm" and "Crystals, proteins, stability and isoperimetry". Based on your interest in "inspiring and delightful," and distinct from the standard math genre, I would also recommend some of his "ergo" articles like https://www.ihes.fr/~gromov/wp-content/uploads/2018/08/ergo-cut-copyOct29.pdf.<|endoftext|> -TITLE: Different definitions for integral de Rham cohomology classes -QUESTION [12 upvotes]: Suppose that $S$ is a compact orientable surface. In this case, the top de Rham cohomology space $H^2(S)\cong \mathbb{R}$, with the isomorphism given by integration on $2$-forms along $S$. -Now, one can define integral cohomology classes as those cohomology classes $a$ so that $\int_S a \in \mathbb{Z}$. On the other hand, one can also define integral cohomology classes as those classes corresponding to integral Cech cohomology classes $\check{H}^2(S,\underline{\mathbb{Z}})$ in the following way: -If $\omega$ is a closed $2$-form, we can find an open cover $\mathfrak{U}$ of $S$ and $1$-forms $\alpha_U$ on each $U$ so that $\omega|_U = d\alpha_U$. Now, choose functions $f_{UV}$ so that $df_{UV} = \alpha_U - \alpha_V$. The cocycle -$$ -f_{UVW} = f_{UV} + f_{VW} - f_{UW} -$$ -satisfies that $df_{UVW}=0$, so $f_{UVW} \in \check{H}^2(S,\underline{\mathbb{R}})$. -With this in mind, I claim that $[\omega]$ is integral if and only if one can choose $(\alpha_U)$ and $(f_{UV})$ such that $(f_{UVW}) \in \check{H}^2(S,\underline{\mathbb{Z}})$. -I want to see why these two definitions coincide. More precisely, I would like to see an explicit proof on why, given that $\int_S \omega \in \mathbb{Z}$, I can choose the $f_{UV}$ so that $f_{UVW} \in \mathbb{Z}$. Please, I prefer an explicit proof of this fact, rather than invoking Poincaré Duality/de Rham theorem. - -REPLY [12 votes]: $\def\RR{\mathbb{R}}\def\ZZ{\mathbb{Z}}$I've considered assigning this when I've taught sheaf cohomology but it always seemed a little too hard. Let's see if I can do it. I'll be a little more general while I am at it and do the case of a smooth compact oriented $n$-fold. Choose a triangulation $S$ of the $n$-fold; let $F_j$ be the set of $j$-dimensional faces. -For each vertex $u \in F_0$, let $U(u)$ be the star shaped open neighborhood of $u$ as in the OP's answer. The $U(u)$ give an open cover of $X$. For any $u_0$, $u_1$, ..., $u_j$ in $F_0$, the intersection $U(u_0) \cap \cdots \cap U(u_j)$ is empty if $(u_0, \dots, u_j)$ are not the vertices of a face, and this intersection is a contractible open set which I'll call $U(\sigma)$ if $(u_0, \dots, u_j)$ are the vertices of a face $\sigma$ in $F_j$. Thus, the Cech complex of $\underline{\RR}$ is identified with the simplicial cohomology complex -$$\RR^{F_0} \to \RR^{F_1} \to \cdots \RR^{F_{n-1}} \overset{d_{n-1}}{\longrightarrow} \RR^{F_n}.$$ -For any $n-1$ dimensional face $\tau$, there are two $n$-faces $\sigma_1$ and $\sigma_2$ containing $\tau$. Letting $e_{\tau}$ be the basis function corresponding to $\tau$, we have $d_{n-1}(e_{\tau}) = e_{\sigma_1} - e_{\sigma_2}$. (I am being sloppy about signs, but the fact that we are on an oriented manifold will make it all work out in the end.) So (using that our manifold is connected) the cokernel of $d_{n-1}$ is clearly $\RR$, and an explicit map from $\RR^{F_n}$ to the cokernel sends a function $f \in \RR^{F_n}$ to $\sum_{\sigma \in F_n} f(\sigma)$. -Let $\Omega^p$ be the sheaf of smooth $p$-forms, and let $Z^p$ be the subsheaf of closed $p$-forms. Note that $Z^0 = \underline{\RR}$, so we have just computed that $H^n(X, Z^0) \cong \RR$. The Poincare lemma gives short exact sequences $Z^p \to \Omega^p \to Z^{p+1}$ for $0 \leq p \leq n$, so we get boundary maps -$$H^0(X, Z^n) \to H^1(X, Z^{n-1}) \to \cdots \to H^n(X, Z^0) \cong \RR.\quad (\ast)$$ -In the case of a surface, the OP has given explicit descriptions of these maps in his answer. -By the usual argument with partitions of unity, $H^q(X, \Omega^p)$ vanishes for $q>0$, so all these maps are isomorphisms except the first one. The first map, in turn, is surjective with kernel $d H^0(X, \Omega^{n-1})$. So the image of the first map is $H^n_{DR}(X)$, and all the other $H^q(X, Z^{n-q})$ are isomorphic to $H^n_{DR}(X)$. Our goal, given an $n$-form $\omega$, is to show that the composition of all these maps gives $\int_X \omega$. -Note that a class in $H^q(X, Z^{n-q})$ is given by a Cech representative $( \eta_{\sigma} )_{\sigma \in F_q}$, where $\eta_{\sigma}$ is a closed $(n-q)$-form on $U(\sigma)$. -Choose a regular CW subdivision $S^{\perp}$ of $X$ dual to the triangulation. That means the poset of faces of $S^{\perp}$ is dual to that of $S$ and each $j$-face $\sigma$ in $S$ crosses the dual $n-j$ face $\sigma^{\perp}$ transversely in one point. An explicit way to do this is to take the barycentric subdivision of $S$ and draw the "obvious" dual faces. If we choose an ordering of $F_0$, that gives an orientation to every face $\sigma$ of $F_q$, and then we can use the global orientation of $X$ to orient $\sigma^{\perp}$. -I claim that the composite isomorphism $(\ast)$ from $H^{q}(X, Z^{n-q})$ sends $(\eta_{\sigma})_{\sigma \in F_q}$ to -$$\sum_{\sigma \in F_q} \int_{\sigma^{\perp}} \eta_{\sigma}.$$ -Let's see what this means for $q=n$. Each $\eta_{\sigma}$ is a closed $0$-form on $U(\sigma)$. A closed $0$-form is locally constant function and $U(\sigma)$ is connected, so we just have a real number for each $\sigma$ in $F_n$ and we can thus think of $\eta$ as a vector in $\RR^{F_n}$. Each $\sigma^{\perp}$ is just a point in the interior of $\sigma$. So we are just summing up the values of $\eta$ on the $n$-faces, and this is the map $\RR^{F_n} \to \RR$ that we described before. -Let's next see what this means for $q=0$. Each $\eta_{\sigma}$ is an $n$-form on $\sigma$, and the condition that $(\eta_{\sigma})$ is a Cech co-cycle says that $\eta_{\sigma}$ is the restriction of a global $n$-form $\omega$ on $X$. The $n$-faces $\sigma^{\perp}$, for $\sigma \in F_0$, partition $X$. So -$$\sum_{\sigma \in F_q} \int_{\sigma^{\perp}} \eta_{\sigma} = \sum_{\sigma \in F_q} \int_{\sigma^{\perp}} \omega|_{\sigma^{\perp}} = \int_X \omega.$$ -Thus, we just need to show that, if $(\eta_{\sigma})$ represents a class in $H^q(X, Z^{n-q})$ and $\delta_q$ is the boundary map $H^q(X, Z^{n-q}) \to H^{q+1}(X, Z^{n-q-1})$, then -$$\sum_{\sigma \in F_q} \int_{\sigma^{\perp}} \eta_{\sigma} = \sum_{\tau \in F_{q+1}} \int_{\tau^{\perp}} \delta_q(\eta)_{\tau}. \quad (\dagger)$$ -Whew! Okay, let's remember how the boundary map in sheaf cohomology works. Let $(\eta_{\sigma})$ be a cocycle for $H^q(X, Z^p)$. Since each $\sigma$ is contractible, we can lift each $\eta_{\sigma}$ to a $p-1$ form $\theta_{\sigma}$ with $d(\theta_{\sigma}) = \eta_{\sigma}$. Let $\tau$ be a $q+1$ face of our triangulation. Then -$$\delta_q(\eta)_{\tau} = \sum_{\sigma \subset \tau} \pm \theta_{\sigma},$$ where the sign involves the relative orientation of $\sigma$ and $\tau$. -We want to show $(\dagger)$. Plugging in the above description of the Cech co-boundary, the right hand side is -$$\sum_{\tau \in F_{q+1}} \int_{\tau^{\perp}} \sum_{\sigma \subset \tau} \pm \theta_{\sigma}.$$ -Pulling the sum out of the integral and switching order of summation, we have -$$\sum_{\sigma \in F_q} \sum_{\tau \supset \sigma} \int_{\tau^{\perp}} \pm \theta_{\sigma}.\quad (\heartsuit)$$ -Now, the subdivisions $S$ and $S^{\perp}$ are dual, so $\tau \supset \sigma$ if and only if $\sigma^{\perp} \subset \tau^{\perp}$ or, in other words, $\tau^{\perp} \subset \partial(\sigma^{\perp})$. All the signs work out perfectly, so that $(\heartsuit)$ is -$$ \sum_{\sigma \in F_q} \int_{\partial(\sigma^{\perp})} \theta_{\sigma}.$$ -By Stokes' theorem, -$$\int_{\partial(\sigma^{\perp})} \theta_{\sigma} = \int_{\sigma^{\perp}} d(\theta_{\sigma}) = \int_{\sigma^{\perp}} \eta_\sigma.$$ -We have now recovered the left hand side of $(\dagger)$. - -The OP only asked for top cohomology, but I think other cohomological degrees are similar. Once again, we have maps -$$H^0(X, Z^k) \to H^1(X, Z^{k-1}) \to \cdots \to H^k(X, Z^0)$$ -giving isomorphisms -$$H^k_{DR}(X) \cong H^1(X, Z^{k-1}) \cong \cdots \cong H^k(X, Z^0) \cong H^k(X, \RR). \quad (\diamondsuit)$$ -We'd like to know that a class $\omega$ in $H^k_{DR}(X)$ is represented by a class in $H^k(X, \ZZ)$ if and only if $\omega$ pairs to an integer against every integer chain in $H_k(X, \ZZ)$; it is enough to test against chains coming from the triangulation $S$. Let $c = \sum_{\rho \in F_k} c_{\rho} \rho$ be a $k$-chain. We want to how to pair all the spaces in $(\diamondsuit)$ against $c$. Let $\eta$ be a $q$-cocycle for $Z^{k-q}$. I believe the same argument as before shows that $\langle c, \eta \rangle$ is -$$\sum_{\sigma \in F_q} c_{\rho} \sum_{\rho \in F_k} \int_{\sigma^{\perp} \cap \rho} \eta_{\sigma}. $$ -In particular, if $q=k$, then $\sigma^{\perp} \cap \rho$ is a single point when $\rho = \sigma$ and otherwise $0$. So, viewing the Cech cohomology $H^k(X,\mathbb{R})$ as the cohomology of -$$\RR^{F_0} \to \RR^{F_1} \to \cdots \RR^{F_{n-1}} \overset{d_{n-1}}{\longrightarrow} \RR^{F_n}$$ -and the simplicial cohomology $H_k(X, \ZZ)$ as the homology of -$$\ZZ^{F_0} \leftarrow \ZZ^{F_1} \leftarrow \cdots \leftarrow \ZZ^{F_n},$$ -the pairing between $H^k(X,\mathbb{R})$ and $H_k(X, \ZZ)$ is induced by the obvious pairing between $\RR^{F_k}$ and $\ZZ^{F_k}$. -We then want to show that, if a cocycle in $\RR^{F_k}$ pairs integrally against all cycles in $\ZZ^{F_k}$, then that cocyle is cohomologous to one in $\ZZ^{F_k}$. That sounds like some easy linear algebra, although I don't see a one line proof.<|endoftext|> -TITLE: Local cross-sections for free actions of finite groups -QUESTION [5 upvotes]: Let $G$ be a finite group, let $X$ be a locally compact Hausdorff space, and let $G$ act freely on $X$. It is well-known that the canonical quotient map $\pi\colon X\to X/G$ onto the orbit space $X/G$ admits local cross-sections. More precisely, for every $z\in X/G$ there are an open set $U$ in $X/G$ containing $z$, and a continuous function $s\colon U\to X$ such that $\pi\circ s$ is the identity on $U$. In particular, there is an open cover of $X/G$ consisting of sets where a local cross-section can be defined. -Question: is there a finite open cover of $X/G$ consisting of sets where a local cross-section can be defined? -(This is the same as asking whether the Schwarz genus of the fiber map $X\to X/G$ is finite.) -The answer is "yes" if $X$ (or at least $X/G$) is finitistic, so in particular whenever $X$ has finite covering dimension, and clearly also whenever $X$ is compact. I wonder if it is true in general. - -REPLY [2 votes]: There is a general cohomological lower bound for the Schwarz genus of a map $p:E\to B$. Namely, if there are cohomology classes $x_1,\ldots , x_k\in H^*(B)$ such that $0=p^*(x_i)\in H^*(E)$ for all $i=1,\ldots , k$ and $x_1\cup\cdots \cup x_k \neq 0$, then the genus of $p$ is greater than $k$. Here the coefficients are completely arbitrary, in particular can be twisted. (This is a generalisation of the cup-length lower bound for Lusternik-Schnirelmann category, since the LS-category of a space $X$ is equal to the genus of any fibration over $X$ with contractible total space.) -So you can get many counter-examples using this cohomological criterion. In fact, whenever $X$ is a contractible CW-complex then it is a model for $EG$, and $X/G$ is a model for $BG$. The cup-length of $BG$ is always infinite for a finite group $G$ (with appropriately chosen, possibly twisted coefficients). This generalises the example in Hannes Thiel's answer.<|endoftext|> -TITLE: Applications of the Infinitesimal Lifting Property -QUESTION [6 upvotes]: This question was posted to MSE but didn't get any answers, so I am posting it here. Original post -Hartshorne in his book gives the 'Infinitesimal Lifting Property' as an exercise in chapter 2, section 8 and mentions this to be very important in the deformation theory of nonsingular varieties. For completeness, I record the statement below: - -Let $ k $ be an algebraically closed field and $ A $ a finitely generated $ k $-algebra such that $ \operatorname{Spec} A$ is a nonsingular variety over $ k $. Let $ B $ be a $ k $-algebra and $ B' $ a square-zero extension of $ B $ by $ I $, i e., there is an exact sequence $$ 0 \rightarrow I \rightarrow B' \xrightarrow{\pi} B \rightarrow 0 $$ where $ B' $ is a $ k $-algebra and $ I $ is an ideal such that $ I^2 = 0 $. Let $ f : A \rightarrow B $ be a $ k $-algebra homomorphism. Then there is a lift $ g : A \rightarrow B' $, i.e. a $ k $-algebra homomorphism $ g $ such that $ \pi \circ g = f $. - -As someone starting with deformation theory, I would like to know how/why this result is very important as well as some applications of this result. Does this result have applications while studying moduli problems/moduli spaces? -For instance, Hartshorne gives one application: For $ X $ a nonsingular variety over $ k $ and $ \mathcal{F} $ a coherent sheaf on $ X $, the set of Infinitesimal extensions of $ X $ by $ \mathcal{F} $ upto isomorphism is in one to one correspondence with $ H^1(X, \mathcal{F} \otimes \mathcal{T}_X) $ where $ \mathcal{T}_X $ is the tangent sheaf. - -REPLY [7 votes]: This is known as the formal criterion for "formal smoothness." In this stacks project entry they prove that a morphism of schemes (in your case $\text{Spec }A \to \text{Spec }k$) is smooth if and only if it's formally smooth and locally finite presentation. -Aside from philosophical importance, it's often easier/more intuitive to check this formal criterion than to check the dimension of $\Omega_{\text{Spec }A/\text{Spec }k}$ or use a Jacobian. In the same stacks project tag, they say: - -Michael Artin's position on differential criteria of smoothness (e.g., Morphisms, Lemma 01V9) is that they are basically useless (in practice). - -Let's suppose $\text{Spec }A$ were instead a moduli space $\overline{M}$, e.g. of curves. Then to check $\overline{M}$ is smooth (if we know finite presentation), we need only consider an infinitesimal extension $S \subseteq S'$ coming from $B' \to B$ as in your question, and try to extend a curve over $S$ to a curve over $S'$. -If you want to build the cotangent complex of $X = \text{Spec } A$ (or equivalently the "normal sheaf") but $A$ is not smooth, the first step is to replace $A$ by a smooth $k$-algebra mapping to it, say $k[A]$. Even if some $A \to B$ doesn't factor through $B'$, $k[A] \to A \to B$ certainly will (by choosing a set-theoretic preimage in $B'$ of the image of $A$ in $B$) and so the "problem" obstructing a factorization of $A \to B$ can be traced to the kernel of $k[A] \to A$. I highly recommend this stacks project article that carries this out as concretely as possible. -You can get specific cohomological obstructions to such a factorization by continuing with a simplicial resolution: $\cdots k[k[A]] \rightrightarrows k[A] \to A$. One can even think of this as "covering $A$ by smooth algebras" in a topological sense using this Jonathan Wise article. -EDIT: I can be a bit more precise about the connection to ordinary differentials. Suppose $f : A \to B$ is fixed and we're trying to find a map $\widetilde{f} : A \to B'$ such that $A \overset{\widetilde{f}}{\to} B' \to B$ is $f$. First, pullback along $f$ to get another squarezero extension -$$0 \to I \to B' \times_B A \to A \to 0.$$ -Our old search for $\widetilde{f}$ translates to finding a section of the map $B' \times_B A \to A$. We've reduced to the case $f = id_B : A = B$. -Given one section $s$ of $B' \to B$, this splits the underlying sequence of modules and lets us write $B' =_{modules} B \oplus I$. But what's the ring structure? One computes $(b, i)*(b', i') = (bb', bi' + b'i)$, i.e. $B' = B + I\epsilon$ is the trivial squarezero extension! -If I have two sections $s, t$ of $B' \to B$, they'll give very different isomorphisms $B' \simeq B + I \epsilon$, inducing an automorphism $\varphi$ of $B + I \epsilon$ over $B$. Such automorphisms are precisely derivations! Indeed $\varphi$ must be the identity on $B$ and send $I$ to itself, but $\varphi - id_{B'}$ will be a map from $B \to I$ (using co/kernel univ props on the short exact sequence) which you can check is a derivation. -If sections exist, they all differ by a unique derivation $s-t$. A fancy way to say this is that "sections form a pseudo-torsor under $\text{Der}(B, I)$." If sections really do (locally) exist, you call it a plain torsor.<|endoftext|> -TITLE: Novikov-Wall non-additivity theorem with twisted coefficients -QUESTION [12 upvotes]: Let $Y$ be a compact manifold and let $\pi_1(Y) \to \mathbb{Z}^n= \langle t_1,\ldots,t_n\rangle$ be a homomorphism. Extend it to the group rings $\mathbb{Z}[\pi_1(Y)] \to \mathbb{Z}[ t_1,\ldots,t_n]$ and evaluate it in a certain $\omega \in (S^1)^n \subset \mathbb{C}^n$. We obtain a homomorphism -$$ \psi: \mathbb{Z}[\pi_1(Y)] \to \mathbb{C}$$ which endows $\mathbb{C}$ of a $(\mathbb{C},\mathbb{Z}[\pi_1(Y)])$-bimodule structure. I will indicate this bimodule as $\mathbb{C}^\omega$. -What I call $\omega$-twisted homology of $Y$ will be the homology of the complex $$ \mathbb{C}^\omega \otimes_{\mathbb{Z}[ \pi_1(Y)]} C^{ \text{CW}}_*(\widetilde{Y}) $$ -and I denote it with $H_*(Y;\mathbb{C}^\omega)$, where $\widetilde{Y}$ is the universal cover. -The cohomology of the cochain complex $$\text{Hom}_{\text{Mod}-\mathbb{Z}[\pi_1(Y)]}( \text{inv}(C_{*}(\widetilde{Y})), \mathbb{C}^\omega)$$ -is the $\omega$-twisted cohomology of $Y$. Here $ \text{inv}(C_{*}(\widetilde{Y}))$ indicates $C_{*}(\widetilde{Y})$ with the same additive structure, but the action of $\mathbb{Z}[\pi_1(Y)]$ is now on the right because we precompose it with $g \mapsto g^{-1}$ for $g \in \pi_1(Y)$. -If $Y$ is of even dimension $2k$ there is an intersection form on $H_k(Y;\mathbb{C}^\omega)$, defined more or less as usual: -$$\phi:H_k(Y;\mathbb{C}^\omega) \to H_k(Y, \partial Y;\mathbb{C}^\omega) \xrightarrow{\text{PD}} H^k(Y;\mathbb{C}^\omega) \xrightarrow{\text{ev}} \text{inv}(\text{Hom}_{\mathbb{C}}(H_k(W;\mathbb{C}^\omega),\mathbb{C})) $$ -The Poincaré Duality is an isomorphism in this context as well, and it is defined starting from the following isomorphism: -denote $Y'$ the space $Y$ endowed with the dual cell decomposition w.r.t. $Y$. Then there is a chain complex isomorphism: -$$ C_{n-*}(\widetilde{Y}) \to \text{Hom}_{\text{Mod}-\mathbb{Z}[\pi_1(Y)])}(\text{inv}(C_*(\widetilde{Y'}, \widetilde{\partial Y'})), \mathbb{Z}[\pi_1(Y)])$$ -$c' \mapsto [-,c']$ and -$$[c,c']= \sum_{\gamma \in \pi_1(Y)} (c \cdot \gamma c') \gamma $$ -where $(c \cdot \gamma c')$ is the integer intersection number of $c$ and $\gamma c'$. -I am trying to adapt the proof of Wall for the non-additivity theorem of signatures (from his article "Non Additivity of the Signature" of 1969) to the case of homology with twisted coefficients in the $(\mathbb{Z}[\pi_1(Y)],\mathbb{C})$-bimodule $\mathbb{C}$. -Most of the proof works exactly the same thanks to the properties of twisted homology, but I am really having difficulties in adapting the final geometric argument to this setting. The setting of the theorem is: -Let $Y$ be an oriented connected compact $4k$-manifold and let $X_0$ be an oriented compact $4k-1$-manifold, properly embedded into $Y$ so that $\partial X_0= X_0 \cap \partial M$. Suppose that $X_0$ splits $Y$ into two manifolds $Y_-$ and $Y_+$. For $\varepsilon= \pm$, denote by $X_\varepsilon$ the closure of $\partial Y_\varepsilon \setminus X_0$, which is a compact $4k-1$-manifold. Let $Z$ denote the compact $4k-2$-manifold $$Z= \partial X_0 = \partial X_+ = \partial X_-. $$ -The manifolds $Y_+$ and $Y_-$ inherit an orientation from $Y$. Orient $X_0$, $X_+$ and $X_-$ such that -$$ \partial Y_+ = X_+ \cup (-X_0)$$ and $$ \partial Y_- = X_0 \cup (-X_-)$$ -and orient $Z$ such that $$Z= \partial X_- = \partial X_+ = \partial X_0. $$ -I would like to prove -Novikov-Wall non additivity theorem: -In the situation above, -$$\text{sign}_\omega(Y)= \text{sign}_\omega(Y_+) + \text{sign}_\omega(Y_-) + \text{Maslov}(L_-,L_0, L_+) $$ -where $L_\varepsilon= \ker (H_{2k-1}(Z; \mathbb{C}^\omega) \to H_{2k-1}(X_\varepsilon ; \mathbb{C}^\omega)) $ for $\varepsilon=-, +, 0$. -Here $\text{sign}_\omega$ means the signature of the twisted intersection form. -In the final part of the proof we need to calculate the signature (which should give us the Maslov index term) of the twisted intersection form restricted to a subspace $L$ of $H_{2k}(Y, \partial Y; \mathbb{C}^\omega)$ which is isomorphic to $$\frac{L_0 \cap ( L_-+L_+)}{(L_0 \cap L_+)+ (L_0\cap L_-)} .$$ -The idea is to calculate the signature of the intersection form on $L$ by expressing it in terms of $Z$'s own skew-hermitian intersection pairing. -Here is where I get stuck. I understand how to represent an element $b \in L_0 \cap ( L_-+L_+)$ with a $2k$-cycle $ \xi + \eta + \zeta$ in $Y$, where $\xi, \eta, \zeta$ belong respectively to $Z_{2k}(X_+,Z; \mathbb{C}^\omega), Z_{2k}(X_0,Z; \mathbb{C}^\omega), Z_{2k}(X_-,Z; \mathbb{C}^\omega)$ and I know that given $b, b' \in L_0 \cap ( L_-+L_+)$ I want to calculate the twisted intersection form of $\xi + \eta + \zeta $ and $\xi'+\eta'+ \zeta'$, however, I have no idea how to compute it. - -REPLY [3 votes]: In a paper of W. Neumann, I noticed a reference to the thesis of W. Meyer, Die Signatur von lokalen Koeffizientensystemen und Faserbündeln. Neumann says that Meyer discusses some details Wall's non-additivity result with local coefficients. I don't have access to Meyer's work (it's in the series Bonner Math. Schriften). You could try Meyer's paper, Die Signatur von Flächenbündeln. Math. Ann. 201 (1973), 239–264 but there's not much on the subject in there. -Perhaps this will be of some use.<|endoftext|> -TITLE: Example of a mathematician who had problems with peer review system? -QUESTION [9 upvotes]: I am wondering if there is some example of a famous or well-known mathematician who often had trouble with peer review, or who often had to publish in obscure journals because referees didn't 'get' what they were saying (there could be any reasons for these troubles). -Now, you might say this is paradoxical (a well-known mathematician is probably well-known because they typically don't have serious problems with referees), but there is a definite example in physics, in the shape of Hannes Alfven. Alfven was renowned especially for his work on plasma physics and was awarded the Nobel Prize for Physics, but even after winning the Nobel, he typically had problems with the peer review process (especially in the US), complaining that referees would automatically reject his papers because they could not understand the formalism he was using, forcing him to publish many of his articles in obscure journals. - -REPLY [21 votes]: There does not seem to be a direct mathematical analogue of Alfven. Nobody who has won a Fields Medal or an Abel Prize has made well-publicized complaints about how they have had an unduly difficult time with the peer-review system. -Some partial analogues have been mentioned in the comments. Louis de Branges had trouble getting people to take his proof of the Bieberbach conjecture seriously, but there was a clear reason: de Branges had a well-deserved reputation for making repeated but incorrect claims to have solved famous open problems. Fourier's work met with resistance, but there were some legitimate objections that some crucial arguments were not fully clear; while the Mochizuki saga is still in progress, it seems that the clarity of certain crucial arguments is still in dispute. Galois also ran into difficulties, although the story has been exaggerated, and Galois was not a recognized figure at the time. -Why there is no direct analogue of Alfven is an interesting question. There could be a difference in culture. Mathematicians will often complain privately of getting an unfair rejection, but complaining too loudly or publicly about the lack of recognition of one's own work is generally regarded as whining, and is frowned upon. Mathematicians also take the general attitude that there are three primary reasons for rejection: (1) the work is wrong; (2) the writing is unclear; (3) the result is not regarded by community as sufficiently interesting. The first two are considered to be the fault of the author, and mathematicians tend to take a somewhat fatalistic attitude toward the third. Under these prevailing assumptions, it is hard to be viewed as taking the moral high ground when voicing a complaint about a rejection. -One example of someone making a public complaint was Friedrich Wehrung (search for the word "rejection"). His primary complaint, however, was that the entire subfield of lattice theory was being looked down upon unfairly. Similarly, you can find public discussion about how the peer-review system unfairly treats certain groups of people. But again, if you're looking for an individual analogue of Alfven, I don't think there exists a very good example.<|endoftext|> -TITLE: Weyl algebra as an Azumaya algebra over its centre -QUESTION [7 upvotes]: Assume that $k$ is an algebraically closed field of positive characteristic $p$. On page 3 (page 6 of the PDF file) of Bezrukavnikov, Mirković, and Rumynin - Localisation of Modules for a semisimple Lie Algebra in prime characteristic, we have the following sentence: - -The sheaf $D_X$ of crystalline differential operators on a smooth variety $X$ over $k$ has a non-trivial center, canonically identified with the sheaf of functions on the Frobenius twist $T^∗X^{(1)}$ of the cotangent bundle. Moreover $D_X$ is an Azumaya algebra over $T^∗X^{(1)}$. - -Instead of going through the general proof given, I only want to understand, in as simple a manner as possible, the situation when $X$ is the affine $n$-space over $k$. In this case, $D_X$ is simply the Weyl algebra and the Azumaya property, if I understand correctly, means that the quotient of the Weyl algebra by its centre is isomorphic to some matrix algebra over $k$. Is there a way to construct such a matrix algebra and a corresponding isomorphism to the quotient explicitly? Any help, even in the case of the affine line, would be highly appreciated. -P.S. If my understanding is incorrect, could you please point out the flaw(s) and how the question could be turned into something reasonable? -Major Edit It has been pointed out that my understanding of the Azumaya property is incorrect. But my question remains the same: is there a direct way to prove the claim of the paper in the case when $X$ is the affine $n$-space (or even the affine line) over $k$. - -REPLY [3 votes]: But my question remains the same: is there a direct way to prove the -claim of the paper in the case when is the affine -space (or even -the affine line) over . - -Yes, read the proof of Proposition 1 in "The Jacobian Conjecture is stably equivalent to the Dixmier Conjecture" by Alexei Belov-Kanel and Maxim Kontsevich. -If you are only interested in the $n=1$ case $A=k \langle X,Y \mathrel| XY-YX=1 \rangle$, then give $X$ degree $1$ and $Y$ degree $-1$, so that $A$ becomes a $\mathbb{Z}$-graded ring. It is even strongly graded, meaning $A_1\cdot A_{-1} = A_0$, so all graded info is determined by the zero-part $A_0=k[XY]$. In particular, if you divide out a graded maximal ideal you get a strongly graded ring with part of degree $0$ a field and finite over its center, and these must be central simple algebras (in this case, just $p \times p$ matrices) proving that $A$ is what is called a graded Azumaya algebra. Now, for the fun part, as there exist homogeneous central identities, a graded Azumaya algebra is a genuine Azumaya algebra. Done.<|endoftext|> -TITLE: Why is density and separability needed for uniqueness of weak (time) derivatives? -QUESTION [5 upvotes]: Let $X,Y$ be Banach spaces with $X \subset Y$. Recall that $u \in L^1(0,T;X)$ has weak derivative $g \in L^1(0,T;Y)$ if -$$\int_0^T u(t)\phi'(t) = -\int_0^T g(t)\phi(t) \qquad\forall \phi \in C_c^\infty(0,T).$$ -Suppose that $u$ also has a weak derivative $h \in L^1(0,T;Z)$ where $Y \subset Z$. -In Boyer and Fabrie's book on Navier-Stokes, page 95, he states that if $Y \subset Z$ is dense and $Z'$ is separable, then $g=h$. My question, why is the density and separability needed? Isn't the argument this simple: -Since $g$ and $h$ are weak derivatives of $u$, we have -$$\int_0^T (g(t)-h(t))\phi(t) = 0\qquad\forall \phi \in C_c^\infty(0,T)$$ -and by the fundamental lemma of the calculus of variations, it follows that $g(t) = h(t)$ in $Z$ for almost every $t$. -Isn't this enough? What do I miss? Does anyone know another source for this uniqueness claim where the derivatives lie in different spaces? - -REPLY [4 votes]: $\def\bbR{\mathbb R}\def\inc{\subseteq}$The requirements on density or separability are superfluous because of the following -Lemma. Let $J$ be a real open interval, and let $E$ be any real or complex Banach space. Let the function $f$ in $L^1(J,E)$ be such that $\int_J(\varphi\,f)=0_E$ holds for all compactly supported smooth $\varphi:J\to\bbR$. Then $f(t)=0_E$ holds for almost all $t\in J$. -Proof. Let $x\mapsto\|x\|$ be a norm for $E$. Having $f$ a.e. the limit of a sequence of simple functions, there are a Lebesgue null set -$N_1\inc J$ and a separable closed linear subspace $S$ in $E$ such that $f(t)\in S$ holds for all $t\in J\setminus N_1$. By Lemma 8.15.1 (p. 573) in R. E. Edward's Functional Analysis there is a countable set $D$ in the unit ball of the dual of $E$ such that -$\|x\|=\sup\{|u(x)|:u\in D\}$ holds for all $x\in S$. By classical results, it follows existence of a Lebesgue null set $N_0\inc J$ such that $u\circ f(t)=0$ holds for all $t\in J\setminus N_0$ and $u\in D$. It follows that $\|f(t)\|=0$ holds for all $t\in J\setminus N_0$. -I do not know whether there is a published reference where the above Lemma would be explicitly stated or proved.<|endoftext|> -TITLE: A question on moduli space of Hitchin's equations -QUESTION [5 upvotes]: I am reading Hitchin's Self-Duality paper. In section 5 (page 85), he is trying to prove that $Dim H^1=12(g-1)$. In doing so, he defines an operator $d^*_2+d_1$, where $d^*_2$ and $d_1$ are given by -$d_1\dot{\psi}=(d_{A}\dot{\psi},[\Phi, \dot{\psi}])$ -$d_2(\dot{A},\dot{\Phi})=(d_A\dot{A}+[\dot{\Phi},\Phi^*]+[\Phi,\dot{\Phi}^*], d^{\prime\prime}_{A}\dot{\Phi}+[\dot{A}^{0,1},\Phi])$ -Then, he claims that $(d^*_2+d_1)(\psi_1,\psi_2)=0$ if and only if -$d^{\prime\prime}_{A}\psi_1 -+[\Phi^*,\psi_2]=0$ -$d^{\prime}_{A}\psi_2 -+[\Phi,\psi_1]=0$ -I am not able to derive this fact, and I spent quiet a lot of time on this, but unfortunately was not able to prove it. He says that he obtains this by calculating the explicit form of adjoint of $d_2$. I was not able to perform this calculation. I am new to the subject, and I would really appreciate any help, or ideas on how to prove this. Thanks! -P.S. I know that a co-differential is defined by $ d^{*}=(−1)^{n(k-1)+1}*d*:\Omega^{k}\to \Omega^{k-1}$ where $*$ in the definition is Hodge star, and this is the adjoint of exterior derivative with respect to $L^2$ norm. But how would one apply hodge operator in this setting, or even use $L^2$ to get $d^*$. - -REPLY [2 votes]: The operator -$$ d_2^*+d_1\colon \Omega^0(M,ad P\otimes\mathbb C)\oplus\Omega^0(M,ad P\otimes \mathbb C)\to\Omega^{0,1}(M,ad P\otimes \mathbb C)\oplus\Omega^{1,0}(M,ad P\otimes \mathbb C)$$ -is just the sum of $d_1$ and $d_2^*,$ and it suffices to describe these two operators using the identification $$\Omega^{0,1}(M,ad P\otimes\mathbb C)=\Omega^1(M,ad P).$$ -We decompose $\psi_1\in\Omega^0(M,ad P\otimes\mathbb C)$ into real and imaginary parts $$\psi_1=\omega+i \eta$$ for $\omega,\eta\in \Omega^0(M,ad P),$ and let $d_1$ act on the real part ($\omega$) and $d_2^*$ on the imaginary part ($\eta$). -Then, we have $$d_1(\psi_1,\psi_2)=((d_A\omega)^{0,1},[\Phi,\omega])=(d_A''\omega,[\Phi,\omega])$$ -by definition. The dual of the operator $$d_A''\colon\Omega^{1,0}(M,ad P\otimes \mathbb C)\to \Omega^2(M,ad P\otimes \mathbb C); \; \psi_2\mapsto d_A''\psi_2$$ is by Serre duality the operator $$d_A''\colon \Omega^0(M,ad P\otimes \mathbb C)\to \Omega^{0,1}(M,ad P\otimes \mathbb C);\; \psi_2\mapsto d_A''\psi_2,$$ and using the hermitian metric and the fact that $d_A$ is unitary the adjoint operator gets identified with -$$d_A'\colon \Omega^0(M,ad P\otimes \mathbb C)\to \Omega^{1,0}(M,ad P\otimes \mathbb C);\; \psi_2\mapsto d_A'\psi_2.$$ -The adjoint of the operator -$$\phi\in\Omega^2(M,ad P)\mapsto ([\phi,\Phi^*]+[\Phi,\phi^*])^{0,1}\in\Omega^{0,1}(M,ad P)$$ becomes -$$\psi_2\mapsto -[\psi_2,\Phi^*]$$ ( taking the $i-$ part into account). -It remains to describe the operator $d_2^*$ acting on the imaginary part $i\eta$ of $\psi_1.$ As above, this becomes $$d_2^*((i\eta))=(id_A''\eta,i[\Phi,\eta]).$$ Putting together the pieces proves the claim.<|endoftext|> -TITLE: natural metrics for proof length -QUESTION [9 upvotes]: I am trying to make my way into Homotopy Type Theory(HoTT) where a mathematician may view proofs as paths. Intuitively, this leads me to the idea of a metric on the space of mathematical propositions. Has this been developed? -Specifically, is there a way to analyse short proofs as geodesics within the space of -mathematical propositions from the perspective of HoTT? If so, might -this metric be formulated using Kolmogorov Complexity? - -REPLY [4 votes]: Let me attempt a longer answer. (@Andrej Bauer 's answer is mostly about learning homotopy type theory, as opposed to delving deeper into your question.) -Unsurprisingly, the answer is still no. One could try to define the length of a path by the size of the smallest of its witnesses. But that definition relies on syntax, i.e. you need to have a 'language' in which you express your witnesses. The thing is, while HoTT does give rise to programming languages rather naturally (see my work with Amr Sabry), they still are not really 'canonical'. While $\left(\infty,1\right)$-topos certainly point in the right direction, it's not settled yet if we don't in fact want a bit more structure (see the work of Shulman, Riehl, etc if you want to dive in the really deep end of that). So we don't even know what outer structure to work in, never mind the internal language that we'll end up having inside that structure. -Even if all of that settled down, why would you expect the internal language to be Turing-complete? That is one of the cornerstones of the validity of Kolmogorov Complexity. Without it, length becomes a rather whimsical notion. Even with Kolmogorov Complexity, length remains a 'fuzzy' notion, because it's only defined up to a constant. So even if it existed, it would not tell you much about 'short' proofs, it only really tells you something interesting when you have one proof which is significantly shorter than the others. Certainly there is no hope that such a length notion would be a metric, never mind carve out geodesics. -Nevertheless, I do hold out some hope that there will be some notion of 'size' that will turn out to be informative and meaningful. It's just not going to be simple, or as informative as one would like.<|endoftext|> -TITLE: Gödel's ontological proof & Benzmüller's work -QUESTION [10 upvotes]: For a decade or so, Christoph Benzmüller from Berlin has explored Gödel's ontological proof (and variants) of existence of God. He uses the proof assistant Isabelle/HOL. He recently posted a preprint, which was highlighted by the cover of the French magazine Science et Vie. -Well, I am not familiar with AI, yet even less with applications to metaphysics. But many practitioners of MO must be. I should like to know how serious is Benzmüller's work considered in this community. Is it controversial or is this considered a respectable research activity ? - -REPLY [9 votes]: Gödel’s proof has the nice feature that one can cleanly separate the logical core of the argument (which is uncontroversial—but see the next paragraph) from its alleged application to theology (which of course is going to be controversial). My opinion is that the main significance of Gödel’s proof is that it shows that there is something to the ontological argument; many people (including my teenage self), when first encountering the ontological argument, perceive it to be total nonsense. Immanuel Kant’s “existence is not a predicate” objection was taken by many to be a decisive refutation. So showing that the ontological argument isn’t completely free of content is already a significant intellectual achievement. -The work of Benzmüller (and Paleo) is interesting because it showed that a (fixable) mistake in Gödel’s proof was more serious than people had previously realized. See The Inconsistency in Gödel’s Ontological Argument: A Success Story for AI in Metaphysics for more details.<|endoftext|> -TITLE: If $n = 18k+5$ is composite, there are at least 9 divisors of $\phi(n)$ which do not divide $n-1$ -QUESTION [6 upvotes]: If $n$ is a composite of the form $18k+5$, there at least 9 divisors of $\phi(n)$ which do not divide $n-1$. Is this true in general or if not, what is the smallest counter example? The conjecture has been verified for $n \le 1 \times 10^7$. -Related question. -Note: The question was posted in MSE a year ago but got no answer. Hence posting in MO - -REPLY [3 votes]: Positive answer to this question would imply that there are no solutions to Lehmer's totient problem of the form $18k+5$. I'm not aware of known restrictions of this kind.<|endoftext|> -TITLE: Manifolds with boundary admitting no closed embedded minimal hypersurface -QUESTION [5 upvotes]: The following Theorem is proved in the paper entitled "Compactness of the space of embedded minimal surfaces with free boundary in three-manifolds with nonnegative Ricci curvature and convex boundary", by A. Fraser and M. Li: - -Let $M^n$ be a compact $n$-dimensional Riemannian manifold with nonempty boundary $\partial M$. Suppose $M$ has nonnegative Ricci curvature and the boundary $\partial M$ is strictly mean convex with respect to the inward unit normal. Then, $M$ contains no smooth, closed, embedded minimal hypersurface. - -My question is: what are examples of compact Riemannian $3$-manifolds with nonnegative scalar curvature (but not nonnegative Ricci curvature) and mean convex boundary that don't admit closed embedded minimal surfaces? - -REPLY [6 votes]: A solid torus should work. Choose cylindrical coordinates $(r,\theta, \lambda)$, $0\leq r \leq r_0 < \pi/2, 0\leq \theta \leq 2\pi, 0\leq \lambda \leq l$, where we equate $(r,\theta,0)\sim (r,\theta, l)$ and $(0,\theta, \lambda)\sim (0,0,\lambda)$. Put a Riemannian metric on this solid torus of the form $dr^2+ f(r)^2 d\theta^2 + g(r)^2 d\lambda^2$, and $f(r)=\sin(r), g(r)=\cosh(\epsilon r)$, where $0 < \epsilon$ is small. -The sectional curvatures of such a metric are computed in Lemma 2.3 of this paper as -$$K_{\theta\lambda}=-\frac{f'g'}{fg},\ K_{r\theta}=-\frac{f''}{f},\ -K_{r\lambda}=-\frac{g''}{g}$$ -and the mean curvature of the level $r$ torus is -$$\frac12(\frac{f'}{f}+\frac{g'}{g}).$$ -From the sectional curvatures, we get the scalar -curvature as -$$R= -2 (\epsilon \cos(r) \sinh(r) -\sin(r) \cosh(\epsilon r) +\epsilon^2 \sin(r) \cosh(\epsilon r))/fg,$$ -and mean curvature of the level surface at height $r$ as -$$ \frac12(\cos(r)\cosh(\epsilon r) +\epsilon\sin(r) \sinh(\epsilon r))/fg.$$ -We see that the level surfaces $r=c$ are mean convex tori, and the scalar curvature is positive for $\epsilon$ and $r_0$ small. Hence this metric contains no closed minimal surface: the maximal $r$ value for such a surface would be tangent to a level surface which is mean convex, contradicting the maximum principle. -Here's how I found this metric: given your criteria, the double of the manifold admits a metric with positive scalar curvature (this is a trick of Hubert Bray; cf. also Pengzi Miao). Such manifolds are connect sums of space forms and $S^2\times S^1$. The reflection symmetry quotient then gives a handlebody (a space form with non-trivial fundamental group cannot admit a reflection symmetry with fixed set a surface). Hence the simplest non-trivial case is a solid torus. The above metrics are invariant under the $S^1\times S^1$ action on a solid torus (a ``double-warped product"). The $(r,\theta)$ slice is a spherical cap, and the $(r,\lambda)$ slices are scaled hyperbolic metrics on a cyinder. Then we adjust $\epsilon$ and $r_0$ to make the scalar curvature positive, the positive sectional curvature of the spherical cap dominating the negative curvature of the hyperbolic annulus. I suspect one can use the techniques in Codá-Marques' paper to realize any handlebody with such properties.<|endoftext|> -TITLE: Zero of the exponential p-adic -QUESTION [6 upvotes]: Consider the $p$-adic exponential defined over $\mathbb C_p$. One knows $\exp$ is analytic in the domain $\mathcal D=\{z\in\mathbb C_P\mid v_p(z)>\frac1{p-1}\}$. Does it exist an element $z_0\in\mathcal D$ such that $\exp(z_0)=0$? -Thanks in advance. - -REPLY [8 votes]: It’s hard to see in what sense it could be true that the exponential function is “defined over $\Bbb C_p$”, since the logarithm is defined on the whole open unit disk there, and has so very many zeros. -If you look closely, you can see that for all $z\in\mathcal D$, we have $v_p(e^z - 1)=v_p(z)$. This obtains quite independently of any multiplicativity of the exponential.<|endoftext|> -TITLE: Closed convex hull in infinite dimensions vs. continuous convex combinations -QUESTION [6 upvotes]: tl;dr: When is the closed convex hull of a set $K$ equal to the set of "continuous" convex combinations of $K$? -I am essentially asking for the most general, infinite-dimensional analogue of this related question. -Update: I forgot to specicy that $K$ is compact. As @GeraldEdgar points out below, for noncompact $K$, the answer is trivially "no". -Suppose $K\subset E$ where $E$ is a topological vector space (as far as I can tell, this is the most general kind of space for which this question makes sense). Obviously we can define the closed convex hull $\overline{\text{conv} K}$ of $K$ as usual. Now consider the set -$$ -K^* = -\{ \int_K x\,d\mu(x) : \mu \in\mathcal{P}(K)\}, -$$ -where $\mathcal{P}(K)$ is the set of (say, Borel) probability measures over $K$ and integral here is to be understood in the weak (Pettis) sense. -I would like to know when $\overline{\text{conv} K} = K^*$. If $E$ is finite-dimensional, there is equality. What are the most general assumptions on $E$ and $K$ for which this equality continues to hold? -(For the curious, the inspiration for this question came from trying to understand when $K^*$ is compact.) - -REPLY [6 votes]: No. Even in one dimension. Say $K$ is the open interval $(0,1)$. Show $0 \notin K^*$. Let $\mu$ be a probability measure with support contained in $(0,1)$. Indeed, -$$ -r(\mu) := \int_K x\,d\mu(x) -$$ -is the integral of a positive function. That is, $x > 0$ a.e. So $\int_K x\,d\mu(x) > 0$. Similarly $1 \notin K^*$. -In a locally convex topological vector space $E$, if there is any extreme point of $M = \overline{\text{conv} K}$ that does not already belong to $K$, then it also does not belong to $K^*$. So what if $K$ is the set $\text{ex}\; M$ of extreme points of a closed convex bounded set $M$? Can we recover $M$ as $K^*$? -A very nice little book that discusses this situation is -Phelps, Robert R., Lectures on Choquet’s theorem, Lecture Notes in Mathematics. 1757. Berlin: Springer. 124 p. (2001). ZBL0997.46005. -Choquet's theorem tells us roughly that every point of a compact convex set $M$ is of the form $r(\mu)$ for some probability measure concentrated on the set $\text{ex}\; M$ of extreme points of $M$. - -Plug -My first publication to attract any notice was this one, where there is a generalization of Choquet's theorem to certain closed bounded noncompact sets $M$. -Edgar, G. A., A noncompact Choquet theorem, Proc. Am. Math. Soc. 49, 354-358 (1975). ZBL0273.46012.<|endoftext|> -TITLE: Riemann hypothesis for exponential sum -QUESTION [7 upvotes]: Recently I've heard about the Riemann hypothesis for one-variable exponential sums, which states as - -For a polynomial $f\in\mathbb{F}_{p^k}[x]$ of degree $d$ and a character $\chi$ of $(\mathbb{F}_{p^k},+)$, provided $(d,p)=1$, we have$$\left|\sum_{x\in\mathbb{F}_{p^k}}\chi(f(x))\right|\le(d-1)\sqrt{p^k}$$ - -I also know the $L$-function associated to $f$ is$$L(f,T)=\exp\left(\sum_{n\ge1}S_n(f,\chi)\frac{T^n}{n}\right)$$ -where $S_n(f,\chi)=\sum_{x\in\mathbb{F}_{p^{kn}}}\chi(\operatorname{tr}_{\mathbb{F}_{p^{kn}}/\mathbb{F}_{p^{k}}}(f(x)))$. -My question is, what is the relation of the Riemann hypothesis for one-variable exponential sums, and Riemann hypothesis for the associated $L$-function? I guess they are equivalent forms, but I can't prove it. -My thought: This is similar to the relation in the elliptic curve version. The Hasse theorem - -For an elliptic curve $E$ over $\mathbb{F}_p$, $|\#E(\mathbb{F}_p)-p-1|\le 2\sqrt p$. - -is equivalent to the Riemann hypothesis for elliptic curve: - -The zeros of $\zeta(E,s)$ has real part $\frac1{2}$. - -This equivalence is due to computing the zeta function $\zeta(E,s)$, involving Riemann-Roch theorem. But as for the exponential sum case, I have no idea how to compute the $L$-function associated to $f\in\mathbb{F}_{p^k}[x]$. -Any help will be appreciated. - -REPLY [4 votes]: One can show by a nontrivial but elementary argument that $L(f,T)$ is a polynomial in $T$ of degree $d-1$. -The Riemann hypothesis in this case says all zeroes of $L(f,T)$ have absolute value $p^{-k/2}$. -It follows from the Riemann hypothesis that we can write $L(f,T) = \prod_{i=1}^{d-1} (1 - \alpha_i T)$ where $|\alpha_i|= p^{k/2}$. The bound $S_n(f,x) \leq (d-1) \sqrt{p^{kn}}$ follows from this by taking logs. -Conversely, if we have the bound $S_n(f,x) \leq (d-1) \sqrt{p^{kn}}$ for all $n$, we can check that the roots have absolute value $\geq p^{-k/2}$ using the radius of convergence for the power series, and then check that they have absolute value $p^{-k/2}$ by using the functional equation. Unlike in the elliptic curve case, we need all $n$ here instead of just one.<|endoftext|> -TITLE: Analytic approximations of smooth vector fields -QUESTION [5 upvotes]: Let $M$ be the set of smooth divergence-free vector fields $u$ on $\mathbb{R}^3$ with -$$|\partial_x^{\alpha} u(x)| \leq C_{\alpha K}(1+|x|)^{-K}$$ -on $\mathbb{R}^3$ for any $\alpha,K$. -Further, we consider the subset of analytic functions $M_0 \subset M$. The question is now, whether for each $u \in M$ and $\varepsilon >0$, there is an $v \in M_0$, such that -$$|u(x)-v(x)| < \varepsilon$$ for each $x \in \mathbb{R}^3$. -As a second question consider $M$ to be the set of smooth divergence-free vector fields $u$ on $\mathbb{R}^3$ with -$$u(x+e_j) = u(x)$$ -for all $x \in \mathbb{R}^3$ and $1 \leq j \leq 3$. -Again, consider the subset of analytic functions $M_0 \subset M$. The question here is now again, whether for each $u \in M$ and $\varepsilon >0$, there is an $v \in M_0$, such that -$$|u(x)-v(x)| < \varepsilon$$ for each $x \in \mathbb{R}^3$. -These questions are fairly similar to the Stone-Weierstrass theorem, but it seems it is not possible to derive it directly from it. - -REPLY [4 votes]: I believe the most natural approach to this particular question is via Fourier analysis. In the periodic case we have the series -$$u(x)=\sum_{k\in\mathbb{Z}^3}u_k e^{2\pi i (k,x)},$$ -and the condition $\nabla\cdot u=0$ simply means $(u_k,k)=0$. -Taking -$$v(x)=\sum_{|k|0$ we can approximate $u$ as close as we want because the Fourier series of a smooth function converges uniformly and very quickly. Obviously, a polynomial $v(x)$ is analytic. -The same trick works in the first case too except the Fourier series has to be replaced by the Fourier integral, and the fact that $v$ is analytic may be a little less obvious (but still true).<|endoftext|> -TITLE: A class of simplicial complexes defined by arithmetic properties -QUESTION [11 upvotes]: The purpose of my question is to ask about properties in a certain class of 3-dimensional (and other odd dimensional) simplicial complexes. I will first describe the construction in 3 dimension and then in general odd dimensions. -Let $n,g$ and $h$ be integers. Consider the pure simplicial complex $K(n;g,h)$ on the vertex set $[n]=\{1,2,\dots,n\}$ whose facets are described by quadruples $\{a,b,c,d\}$ such that $$b-a=g ({\rm mod}~ n)$$ and $$d-c=h ({\rm mod}~ n)$$. -Let us also consider a variant $L(n;g,h)$ where we further assume that the ordering of $\{a,b,c,d\}$ is cyclic. (Namely, $a -TITLE: Seifert matrices of cable knots -QUESTION [6 upvotes]: Let $K_{p,q}$ be a $(p,q)$-cable of the non-trivial knot $K$ in $S^3$. Let $V_{p,q}$ and $V$ denote the Seifert matrices of $K_{p,q}$ and $K$, respectively. -Is it possible to obtain a closed formula for the matrix $V_{p,q}$ in terms of $V$? - -REPLY [4 votes]: It also appeared in the article of H. Seifert as Theorem II: - -Seifert, H. (1950). On the homology invariants of knots. The Quarterly Journal of Mathematics, 1(1), 23-32.<|endoftext|> -TITLE: Is there a known asymptotic for $A(X):= \sum_{1 \leq i,j \leq X} \frac{1}{\mathrm{lcm}(i,j)}$? -QUESTION [11 upvotes]: My guess is that there exists a constant $C$ such that $A(X) \sim C (\log X)^2$. - -REPLY [10 votes]: If we are talking about "elementary", then just multiply the original sum by the sum of inverse squares and note that if we have three numbers $a=a'd, b=b'd, n^2$ where $(a',b')=1$, $d,n$ are arbitrary, then $A=a'n, B=b'n, d$ are arbitrary $3$ integers with the product $LCM(a,b)n^2$. The original triple sum has the bounds $a'd, b'd\le X$, $n$ formally unrestricted, but restricting it to $[1,N]$ for fixed $N$ changes the triple sum $1+O(1/N)$ times. But then we can squeeze the bounds on $A,B,d$ between $Ad\le X, Bd\le X$ (treating the implicit $n=(A,B)$ as unrestricted) to get the lower bound and $Ad\le NX, Bd\le NX$ (assuming $n\le N$ now) to get the upper bound, so we get an answer for the triple sum of $\frac 1{ABd}$ between roughly speaking $\frac13\log^3 X$ and $\frac13\log^3(XN)$ with additive error $O(\log^2(XN))$ (no number theory in this sum!), which have the same asymptotic up to $1+O(\frac{\log N}{\log X})$. Taking $N$ about $\sqrt{\log X}$, we get the total multiplicative error $1+O(\frac{1}{\sqrt{\log X}})$, which is, of course, suboptimal but who cares. :-)<|endoftext|> -TITLE: discontinuous functions on the Sobolev borderline -QUESTION [10 upvotes]: The Sobolev embedding theorem implies that every function of class $W^{k,p}$ on a reasonable $n$-dimensional domain is continuous if $kp > n$. Cases with $kp=n$ are known as "borderline" cases. In my question I'm going to focus on the case $p=2$ for functions on either ${\mathbb R}^n$ or ${\mathbb T}^n$, so that the Sobolev spaces $H^k = W^{k,2}$ have a nice description in terms of Fourier transforms or series, but answers concerning more general Sobolev spaces are also welcome. -It seems to be tricky to find concrete examples of discontinuous functions that are Sobolev borderline cases. Some searching turned up an example in $H^1({\mathbb R}^2)$, but I have been unable to find a "simple" example in what I intuitively expect to be the easiest case, namely $H^{1/2}(S^1)$, and I was surprised that none of the textbooks I could think to search in give one. My first impulse was to try classic discontinuous functions like the square wave and sawtooth whose Fourier series are easy to compute: these just miss the mark, as they turn out to be in $H^s(S^1)$ for all $s < 1/2$ but not for $s=1/2$. The one thing I have tried that worked was writing down an explicit Fourier series like -$$ -f(x) := \sum_{k=2}^\infty \frac{e^{2\pi i k x}}{k \ln k}, -\qquad \text{ (here $x \in S^1 := {\mathbb R} / {\mathbb Z}$)} -$$ -which one can easily check is in $H^{1/2}(S^1)$, and one can then use summation by parts to estimate $\sum_{k=N}^\infty \frac{e^{2\pi i kx}}{k \ln k}$ for large $N$ and small $|x|$ and thus prove $\lim_{x \to 0} f(x) = \infty$. One can do something similar with a Fourier transform and integration by parts to find a function in $H^{1/2}({\mathbb R})$ that is continuous everywhere except at $x=0$, where it blows up. But this type of construction is a lot trickier than what I was hoping for; expressing a function as a conditionally convergent series or improper integral does not give me the feeling that I can get my hands on it. -So, first question: does anyone know a simpler example of something that is discontinuous and belongs to $H^{1/2}(S^1)$ or $H^{1/2}({\mathbb R})$? Or other interesting examples of Sobolev borderline functions that can be understood without having to search the exercises in Baby Rudin for hints? -Followup question, admittedly a little vague: if you don't know more concrete examples, is there any deep reason why they don't exist, i.e. why every function I can think to write down in a reasonable way turns out to fall short of the borderline case? - -REPLY [7 votes]: There are plenty of examples of discontinuous Sobolev function in $W^{1,n}(\mathbb{R}^n)$. For example $f(x)=\log|\log|x||$ defined in a neighborhood of zero. -Now take $n=2$ and restrict the function to the $x$-axis. You will get a discontinuous function in the trace space which is $H^{1/2}(\mathbb{R})$. -You can use this function to construct quite strange examples. -Taking $x\mapsto f(x-a)$ you can place singularity at any point $a$. -Modifying this example you can assume that $\Vert f\Vert_{1,n}<\epsilon$ -and that the function has support in a small neighborhood of $a$. -If $\{a_i\}_i$ is a countable and dense subset of $\mathbb{R}^n$, -and $f_i$ is a function with the singularity as above at the point $a_i$ and $\Vert f_i\Vert_{1,n}<2^{-i}$, then the series -$$ -f=\sum_{i=1}^\infty f_i -$$ -converges to a function in $W^{1,n}$, because it is a Cauchy series in the norm and $W^{1,n}$ is a Banach space. The function $f$ will have singularities located on a dense subset of $\mathbb{R}^n$ and in particular the essential supremum of $f$ over any open set will be equal $+\infty$. -You can also take $\{a_i\}_i$ to be a dense subset in a subspace $\mathbb{R}^{n-1}$ of $\mathbb{R}^n$ and a similar construction will give you a function that is bad when restricted to that subspace. The trace belongs to $W^{1-1/n,n}(\mathbb{R}^{n-1})$. In particular if $n=1$ you get such a function in $W^{1/2,2}(\mathbb{R})=H^{1/2}(\mathbb{R})$.<|endoftext|> -TITLE: Intuitively, what does a graph Laplacian represent? -QUESTION [58 upvotes]: Recently I saw an MO post Algebraic graph invariant $\mu(G)$ which links Four-Color-Theorem with Schrödinger operators: further topological characterizations of graphs? that got me interested. It is about a graph parameter that is derived from the Laplacian of a graph. Its origins are in spectral operator theory, but it is quite strong in characterizing important properties of graphs. So I was quite fascinated by the link it creates between different branches of mathematics. -I went through other posts on MO that discuss this topic as well, and in the meantime I read a few linked articles that work with the graph Laplacian. I understand that they view an (undirected) graph as a metric graph embedded in a surface, and the metric on the graph is approximated by Riemannian metrics which give the edge distance along the edges, and which is close to zero everywhere else on the surface. The eigenvalues of the surface Laplacian approximate the eigenvalues of the graph Laplacian, and a lot of surprisingly useful conclusions follow, about connectivity and embeddability of the graph, and even about minor-monotonicity. -I have gained a technical understanding of what is happening and how these eigenvalues (and their multiplicity) are determined, using the graph Laplacian. I also have a basic understanding of the role of a Laplacian in differential geometry, like the Laplacian of a function $f$ at a point $x$ measures by how much the average value of $f$ over small spheres around $x$ deviates from $f(x)$, or I think of it to represent the flux density of the gradient flow of $f$. - -But I am failing to gain or develop such an intuition for the graph Laplacian. Conceptually or intuitively, what does a graph Laplacian represent? I am trying to understand, how can it be so powerful when applied to graphs? (I am aware that the graph Laplacian can be defined using the graph adjacency matrix, but I was unable to link this with my differential geometry intuition) - -REPLY [17 votes]: This is just a long comment, adding to the excellent answers above. -There is a great article from László Lovász "Discrete and Continuous: -Two sides of the same?", written around 2000 (https://web.cs.elte.hu/~lovasz/telaviv.pdf) which might be of interest to you. In chapter 5 of this article, Lovász covers the graph Laplacian. He explains the relation to random walks on graphs and also the link to the Colin de Vérdière graph invariant which sparked your interest (your link in the OP). -In your OP, you are asking how can the graph Laplacian be so powerful when applied to graphs? I think two quotes from this article could be of special interest to you, because quote (1) relates to the "power" and quote (2) relates to where the "limitations" were in applying the graph Laplacian. -About the "power": - -Quote (1) -"The Laplacian makes -sense in graph theory, and in fact it is a basic tool. Moreover, the study of the discrete and -continuous versions interact in a variety of ways, so that the use of one or the other is almost -a matter of convenience in some cases. (...) Colin de Verdière’s invariant created much interest among graph theorists, because of its -surprisingly nice graph-theoretic properties. (...) Moreover, planarity of graphs can be -characterized by this invariant: $\mu(G) \leq 3$ if and only if G is planar. -Colin de Verdière’s original proof of the “if” part of this fact was most unusual in graph -theory: basically, reversing the above procedure, he showed how to reconstruct a sphere and a -positive elliptic partial differential operator $P$ on it so that $\mu(G)$ is bounded by the dimension -of the null space of $P$, and then invoked a theorem of Cheng (...) asserting that this dimension is at most $3$. - -About the "limitations": - -Quote (2) -"Later Van der Holst (...) found a combinatorial proof of this fact [$\mu(G) \leq 3$ if and only if G is planar]. While this may seem as a step backward (after all, it eliminated the necessity of the only application of partial differential equations in graph theory I know of), it did open up the possibility of characterizing the next case. Verifying a conjecture of Robertson, Seymour, and Thomas, it was shown by Lovász and Schrijver (...) that $\mu(G) \leq 4$ if and only if G is linklessly embedable in $\mathbb R^3$."<|endoftext|> -TITLE: Graph embeddings in the projective plane: for the 35 forbidden minors, do we know their Colin de Verdière numbers? -QUESTION [17 upvotes]: The Graph Minor Theorem of Robertson and Seymour asserts -that any minor-closed graph property is determined by a finite set -of forbidden graph minors. It is a broad generalization e.g. of the Kuratowski-Wagner theorem, which characterizes planarity in terms of two forbidden minors: the complete graph $K_5$ and the complete bipartite graph $K_{3,3}$. -Embeddability of a graph in the projective plane is such a minor-closed property as well, and it is known that there are 35 forbidden minors that characterize projective planarity. All 35 minors are known, a recent reference from 2012 is, for example, https://smartech.gatech.edu/bitstream/handle/1853/45914/Asadi-Shahmirzadi_Arash_201212_PhD.pdf. -A classical reference is Graphs on Surfaces from Mohar and Thomassen, -Johns Hopkins University Press 2001. -I am interested in the Colin de Verdière numbers for these 35 forbidden minors and have searched for them for a while now, but could not find anything. - -Question: So I wondered whether the Colin de Verdière graph invariants for the whole set of these 35 forbidden minors are actually known? I would be grateful for any reference. - -UPDATE: -Updating this question based on a great comment from Martin Winter. As he points out, the Colin de Verdière number $\mu$ is known and $\mu=4$ for a handful of these 35 forbidden minors, e.g. the disjoint unions of $K_5$ and $K_{3,3}$. -Interestingly, as outlined in his answer to a related question (Algebraic graph invariant $\mu(G)$ which links Four-Color-Theorem with Schrödinger operators: further topological characterizations of graphs?), it follows that the Colin de Verdière invariant cannot provide a full characterization of graph embeddings e.g. in the projective plane. - -REPLY [8 votes]: Here's a table containing the Colin de Verdière numbers: -Name Graph6 μ Reason -K33 + K33 4 (components linklessly embeddable) -K5 + K33 4 (components linklessly embeddable) -K5 + K5 4 (components linklessly embeddable) -K33 . K33 4 (apex) -K5 . K33 4 (apex) - -K5 . K5 4 (apex) -B3 G~wWw{ 4 (apex) -C2 H~wWooF 4 (apex) -C7 G~_kY{ 4 (apex) -D1 Is[CKIC[w 4 (apex) - -D4 H~AyQOF 4 (apex) -D9 I]op_oFIG 4 (apex) -D12 H^oopSN 4 (apex) -D17 G~_iW{ 4 (apex) -E6 Is[BkIC?w 4 (apex) - -E11 I]op_oK?w 4 (apex) -E19 H~?guOF 4 (apex) -E20 H~_gqOF 4 (apex) -E27 I]op?_NAo 4 (apex) -F4 Is[?hICOw 4 (apex) - -F6 Is[@iHC?w 4 (apex) -G1 4 (apex) -K35 4 (apex) -K45-4K2 4 (apex) -K44-e 5 (Petersen family and -2 argument) - -K7-C4 4 (apex) -D3 G~sghS 4 (apex) -E5 H]oxpoF 5 (Petersen family and -2 argument) -F1 H]ooXCL 4 (apex) -K1222 4 (apex) - -B7 4 (apex) -C3 4 (apex) -C4 4 (apex) -D2 4 (apex) -E2 4 (apex) - -Let me give justification. Graphs with $\mu \leq 3$ are planar, hence embeddable on the projective plane. So all the $35$ graphs have $\mu \geq 4$. Since apex graphs are linklessly embeddable, and linklessly embeddable graphs have $\mu \leq 4$, the apex graphs in this table have exactly $\mu = 4$. Also, a graph is linklessly embeddable iff its components are linklessly embeddable, so the first three graphs have $\mu = 4$. -The graphs in the Petersen family are not linklessly embeddable, so they have $\mu \geq 5$. $K_{4,4}-e$ is already in the Petersen family, and $\mathcal E_5$ contains $K_{3,3,1}$ as a subgraph. They both have $\mu \geq 5$. -To see they have $\mu \leq 5$, use Theorem 2.7 in [1]: If $G=(V,E)$ is a graph, and $v$ a vertex of $G$, then $\mu(G) \leq \mu(G-v)+1$. Since we can remove $2$ vertices from $K_{4,4}-e$ to make it planar (by making it $K_{3,3}-e$), it follows that $\mu(K_{4,4}-e) \leq \mu(K_{3,3}-e)+2 = 5$. Hence $\mu(K_{4,4}-e)=5$. The same line of reasoning applies to the graph $\mathcal E_5$. -[1] Van Der Holst, Hein, László Lovász, and Alexander Schrijver. "The Colin de Verdiere graph parameter." Graph Theory and Computational Biology (Balatonlelle, 1996) (1999): 29-85.<|endoftext|> -TITLE: Information density of proofs? -QUESTION [5 upvotes]: I am a CS person so please excuse the hand-waving. -Given a set of machine-represented proofs, each different (but not necessarily proving a different thing), what sort of information-theoretic statements could we make about these? -I would define the information "density" of a proof by simply losslessly compressing it using some standard means (like lz4 or rle or whatever is suited to this domain) and comparing the compressed size to the input size. The compression ratio. -For example: if we compress them all, would they all have similar compression ratios (meaning non-redundant/repeating information content)? -Given a set of different proofs, all proving the same thing (say like the prime number theorem), what is the most dense? The least dense? -Are there useful proofs that are not "information dense", meaning they have long repeating sequences or repeating structure that is easy to compress. These would look tedious to a human but would be discover-able by a machine. -Are there any papers that have looked at this? What area is this? My googling has failed me. - -REPLY [3 votes]: First, beware: this can only be done meaningfully within a fixed "language of proof". If you try to compare across different systems, you can get wildly different results. There is a whole domain of Proof Complexity which has lots of results about "expressivity". Adding features to your language (like let x = in ) can make a huge difference. Other subtler changes can still result in exponential differences in proof length. -Similarly, which compressor you use matters. Kolmogorov Complexity relies crucially on Turing-completeness to obtain a 'stable' measure of length. Normal compressors can be used to approximate it, and give surprising good results in practice. Paul Vitanyi has written a variety of papers on that topic. -Regarding a specific example, like the prime number theorem, an obvious question becomes: what results do you consider to be 'background' and which are part of the proof? The stable answer, again relying on Kolmogorov Complexity, which I explain in Understanding expression simplification [ISSAC 2004], is to in fact consider the proof length to contain all the previous results needed to express the proof as part of its length. Then, to compare two proofs, you subtract their lengths, which takes care of 'quotienting out' all of the material that is common. However, if the two proofs take wildly different routes through mathematics, you might need to use finer concepts such as relative complexity, i.e. you compress the proofs on their own (but with the full library), and concatenated. See the works of Vitanyi for details and variants. -I would say that we currently don't really too much about this topic. We're still in the stage where we're finding uniform ways of writing down mathematical proofs. While the existing libraries are getting large, they are still not quite large enough to do convincing data mining, although that hasn't stopped many from doing so already. But they are aware that their results are very system-relative. -Some interesting early results are available on decently large databases: some use the whole of the arxiv as their data source. I quite like Discovering Mathematical Objects of Interest—A Study of Mathematical Notations for example.<|endoftext|> -TITLE: The cohomology of modular curves as a module over the Galois group -QUESTION [9 upvotes]: Consider the modular curve $\pi: X(N) \to X(1)$ where this map has Galois group $G = PSL_2(\mathbb Z/N\mathbb Z)$. In particular, $G$ acts on the singular cohomology $H^1(X(N),\mathbb Z)\otimes \mathbb C$ or in finite characteristic, on the etale cohomology group $H^1(X(N),\mathbb Z_\ell)\otimes_{\mathbb Z_\ell}\overline{\mathbb Q_\ell}$. -Do we know which irreducible representations of $G$ appear in the cohomology and with what multiplicities. Also, we can ask how the action of $G$ interacts with the Hecke operators, for instance. This seems to me to be very classical automorphic stuff but I have no knowledge about this area of math. Are there any friendly references? -Looking at the dimensions, I don't believe it is the regular representation. - -REPLY [6 votes]: Jared Weinstein's PhD thesis (http://math.bu.edu/people/jsweinst/jswthesis.pdf) is an excellent reference for this kind of thing. See section 3.4 in particular, where he computes the space $S_k(\Gamma(N), \mathbb{C})$ as a $\mathbb{C}[\mathrm{SL}_2(\mathbb{Z}/N)]$-module using an equivariant version of the Riemann--Roch formula.<|endoftext|> -TITLE: Probability of complex eigenvalues -QUESTION [6 upvotes]: I find this is the best site to post this question, even though I considered cs. -It is a Monte Carlo experiment over the set of 10.000 n×n matrices. -If a single matrix eigenvalue is complex then python numpy package will return all the eigenvalues as numpy.complex128 type, else it will return all eigenvalues as numpy.float64 type. -Here is the algorithm: -# Monte carlo experiment in numpy - -import numpy as np -from numpy import linalg as LA - -e=10000 # examples -# n - matrix rank -for n in range(1,10): - cc=0 # complex counter - for i in range (e): - m=np.random.randn(n,n)# m=np.random.random((n,n)) - w, _ = LA.eig(m) - if (type(w[0]).__name__ == "complex128"): - cc+=1 - print(f'matrix:{n}x{n}, total cases: {e}, complex cases: {cc}, ratio: {cc/e}') - - -The output of this algorithm is as follows: -matrix:1x1, total cases: 10000, complex cases: 0, ratio: 0.0 -matrix:2x2, total cases: 10000, complex cases: 2851, ratio: 0.2851 -matrix:3x3, total cases: 10000, complex cases: 6481, ratio: 0.6481 -matrix:4x4, total cases: 10000, complex cases: 8782, ratio: 0.8782 -matrix:5x5, total cases: 10000, complex cases: 9674, ratio: 0.9674 -matrix:6x6, total cases: 10000, complex cases: 9944, ratio: 0.9944 -matrix:7x7, total cases: 10000, complex cases: 9998, ratio: 0.9998 -matrix:8x8, total cases: 10000, complex cases: 9999, ratio: 0.9999 -matrix:9x9, total cases: 10000, complex cases: 10000, ratio: 1.0 - -It is very easy to understand the 1x1 case, since we take values from floating point numbers (set $\mathbb Q$) the eigenvalue will be the number itself (also in $\mathbb Q$). -For the case 2x2 and later, we may get complex eigenvalues ($\mathbb Z$). -I am counting the complex cases (where at least one eigenvalue is complex) and providing the ratio. -I would like to get some thoughts about this ratio possible the formula how to calculate one analytically. - -My thought is that the ratio is not dependent on floating point arithmetic. -I used the normal distribution $\sim \mathcal N(0,1)$, but one may test also the uniform distribution I commented. - -REPLY [10 votes]: The probability that a $n\times n$ real matrix (with elements that are independent random variables with standard normal distributions) has only real eigenvalues is given by -$$ 2^{-n(n-1)/4}$$ -Reference: A. Edelman, The Probability that a Random Real Gaussian Matrix has $k$ Real Eigenvalues, Related Distributions, and the Circular Law. Journal of Multivariate Analysis 60, 203-232 (1997).<|endoftext|> -TITLE: Is every countable discrete group a subgroup of a non discrete Lie group? -QUESTION [5 upvotes]: 1)Let $G$ be a countable discrete group. Can $G$ be embbeded in a locally connected Lie group? -2)let $G$ be a countable discrete group with a prescribed generating set and corresponding word metric. Can $G$ be embbeded isometrically in a locally connected Lie group with its left invariant metric? -Remark: We emphasis on the word "Locally connected" because of the example $G\subset \mathbb{R}\setminus\{0 \}$ with $G=\{\pm e^n\mid n\in \mathbb{Z} \}$ with multiplication. - -REPLY [6 votes]: Malcev proved that every finitely-generated matrix group $\Gamma$ (over any field) is residually finite, i.e. the intersection of all finite-index subgroups of $\Gamma$ is $\{1\}$. Baumslag-Solitar groups, such as $BS(2,3)= \langle a, b | ab^2 a^{-1} =b^3\rangle$, are among simplest examples of finitely generated groups which are not residually finite. A connected Lie group $G$ need not be linear (the universal covering group of $SL(2, {\mathbb R})$ is a standard example). However, the kernel of the adjoint representation $Ad_G$ of a connected Lie group $G$ is always contained in the center of $G$. Thus, if $\Gamma< G$ is a centerless subgroup, then the restriction of the adjoint representation $Ad_G$ to $\Gamma$ is faithful and, hence, $\Gamma$ is isomorphic to a matrix group. It is not hard to see that $BS(2,3)$ has trivial center. Thus, this group is not isomorphic to a subgroup of any connected Lie group. The same proof shows that $BS(2,3)$ is not isomorphic to a subgroup of a Lie group with finitely many components. -Remark. The standard definition of "locally connected" in topology is that every point should have a neighborhood basis consisting of connected subsets. Hence, each manifold (in particular, each Lie group) is, by definition, locally connected. Given your example, it seems that what you really had in mind is that a Lie group $G$ should have Alexandroff compactification $G\cup \{\infty\}$, such that $\infty$ admits a neighbourhood basis $U_i$ satisfying the condition that $U_i\cap G$ is connected. It is easy to see that this requirement is equivalent to the condition that $G$ is connected and 1-ended (equivalently, is neither compact nor a product of compact group with ${\mathbb R}$). I am not sure what to call this property, let's name it ($*$). Then every discrete countable group $\Gamma$ embeds in a Lie group with property ($*$), e.g., $G=\Gamma \times {\mathbb R}^2$.<|endoftext|> -TITLE: Emergence of the discrete from the continuum -QUESTION [11 upvotes]: An almost eternal theme in Mathematics is the approximation of the Continuum by the Discrete. This core idea goes back at least to Archimedes, and remains active to these very days (and quite likely for the next thousand years) . -You have a continuous structure, say a sphere, and you approximate it as the limit of a series of discrete objects, say polyhedra. -I do not need to provide more examples, I am sure you all have plenty. -But, there is also, though by no means so prominent, a reverse direction. -I am not able to date precisely when it made its debut in the history of mathematics, but it most certainly appears in the revolutionary work of Fourier: take a discrete function, say the Heaviside step function, and approximate it via series of trigonometric functions: - -So, here a discrete object is realized as a limit of continuous ones (in this case smooth functions). -I am especially intrigued by this possibility, which, pushed to the extreme, will depict a mathematical world where the Discrete is an emergent phenomenon, out of a Continuum -Thus I ask to everybody: - -can you list active research in the way of approximating discrete structures via smooth ones? - -For instance, a polyhedron via a series of smooth manifolds, or examples in analytic number theory, or patterns in finite combinatorics out of .....(fill the dots) . -Any thoughtful and possibly well documented answer will get my vote, regardless the domain chosen (in fact, the more examples I will harvest from heterogeneous disciplines, the happier I shall be). -On the other hand, to get the GREEN the stakes are higher: rather than single examples, a sketch of a general perspective on the Discrete as emerging from the Continuum -ADDENDUM: As pointed out by Andreas Blass, I have implicitly conflated two themes here: - -discrete as limit of the continuum -emergence of the discrete from some background - -Point 2) does not seem to necessarily imply point 1) and probably the same applies the other way around. Which one I am interested in? Easy: BOTH OF THEM. -But now that Andreas has already marked this point, the GREEN ANSWER would be, perhaps, a clarification on the relationship between 1 and 2 (inter alia) - -REPLY [5 votes]: Bill Lawvere has, for a long time now, promoted the idea that the discrete emerges as a limiting case by abstracting from the continuous. At the most general level, this manifests in the distinction between cohesive and constant (i.e. abstract) toposes of sets: -Quantifiers and Sheaves, Continuously Variable Sets, Variable Quantities and Variable Structures in Topoi, Toward the Description in a Smooth Topos of the Dynamically Possible Motions and Deformations of a Continuous Body, Cohesive Toposes and Cantor's 'lauter Einsen', etc... -This line of thought is worked out more precisely with the concept of Axiomatic Cohesion wherein, roughly speaking, a topos $\mathscr{E}$ is exhibited as cohesive relative to a base topos $\mathscr{S}$ via a string of adjoints between them describing how the discrete spaces $X \in \mathscr{S}$ sit inside the larger topos $\mathscr{E}$ of more general (cohesive, combinatorial, etc...) spaces. Lawvere gave some lectures on this topic in Como in 2008 and there are accompanying lecture notes. The nLab page for cohesive toposes is also quite helpful. -To bring things back down to earth a little bit, in Left and Right Adjoint Operations on Spaces and Data Types, he descirbes, in the last section, the following situation. -Suppose we have, in a cartesian closed category $\mathscr{C}$, a commutative ring object $R$ which we regard as the 'one-dimensional continuum'. We can form the correspinding 'complex numbers' ring $C = R[i]$ by defining complex multiplication on $R^2$ in the usual way. Inside of $C$ sits the multiplicative subgroup $S^1$ corresponding to the 'circle'. Finally, since we are in a cartesian closed category, we can extract from the map space $(S^1)^{(S^1)}$ the subspace $Z$ of those endomorphisms of $S^1$ which are group homomorphisms. This $Z$ can therefore be regarded as the 'integers' although it is important to note that it will not necessarily be the usual integers object $N[-1]$ derived from a natural numbers object $N$ in $\mathscr{C}$.<|endoftext|> -TITLE: Are invariant forms on homogeneous spaces necessarily closed? -QUESTION [8 upvotes]: Take a compact homogeneous space $G/K$, and a left $G$-invariant differential $k$-form $\omega \in \Omega^k(G/K)$. Will $\omega$ necessarily be closed? Might it even be harmonic when $G/K$ is endowed with a Riemannian metric? - -REPLY [8 votes]: Note that the answer depends on the pair $(G,K)$. -For example, if $K=\{e\}$, then one is asking whether the ring of left-invariant forms on $G$ consists only of closed forms. This only happens when $G$ is abelian. -On the other hand, if $M=G/K$ is a compact Riemannian symmetric space and $G$ is the identity component of the isometry group of $M$, then, indeed, every $G$-invariant form is closed and, in fact, the ring of $G$-invariant forms on $M$ is equal to the space of harmonic forms on $M$. This is a well-known result, but for a short proof, one can consult this note by Michael E. Taylor. -For example, when $M=\mathbb{CP}^n$ endowed with its Fubini-Study metric, one has $G = \mathrm{SU}(n{+}1)/\mathbb{Z}_{n+1}$, and the only $G$-invariant forms are (linear combinations of) powers of the Kähler form $\omega$. -As another example, if $K$ is compact and $M = (K\times K)/\Delta$, where $\Delta = \{ (k,k)\ |\ k\in K \}$, then -the $(K\times K)$-invariant forms on $M$ are simply the bi-invariant forms on $K$, which are all closed.<|endoftext|> -TITLE: Prove that a Boolean two-valued topos in which supports split is well-pointed -QUESTION [7 upvotes]: In Lawvere and Rosebrugh's Sets for Mathematics, they write - -It is a theorem [MM92] that a topos is well-pointed if and only if it is Boolean, two-valued, and supports split. - -[MM92] is a reference to Mac Lane and Moerdijk's Sheaves in Geometry and Logic. I have found the proof that a well-pointed topos is Boolean, two-valued and supports split (Propositions VI.7 and VI.8), but I cannot find the proof of the converse, there or anywhere else. Can anyone help me? - -REPLY [6 votes]: Given a Boolean, two-valued topos in which supports split, we want to prove it's well-pointed, meaning that, if $f,g:A\to B$ are distinct, then there is a point $p:1\to A$ such that $fp\neq gp$. Since $f$ and $g$ are distinct, their equalizer $e:E\to A$ is a proper subobject of $A$. (I'm ignoring the distinction between a subobject and a monomorphism representing it.) By "Boolean", $E$ has a non-zero complement $c:C\to A$. Since $C\neq0$, the support of $C$ is a non-zero subobject of $1$. By "two-valued", the support of $C$ is all of $1$. By "supports split" we have a morphism $p:1\to C$. I claim this point in $C$, considered as a point in $A$ (strictly speaking, the point $cp:1\to A$) does the job. Indeed, if $fcp$ were equal to $gcp$, then $cp$ would factor through the equalizer $E$ as well as its complement $C$, hence would factor through 0. But then we'd have a morphism $1\to0$, and the topos would be trivial, contrary to the assumption $f\neq g$<|endoftext|> -TITLE: Natural density of set of numbers not divisible by any prime in an infinite subset -QUESTION [10 upvotes]: Suppose $S$ is a subset of the primes with natural density $0 < \alpha < 1$ within the primes. If -$$D(X) := \{n \leq X \mid p \not \mid n \text{ for all } p \in S \}$$ -(so $D(X)$ is numbers at most $X$ not divisible by any $p \in S$), -then is there a nice form for the asymptotic value of $D(X)$ as a function of $X$? -I'm guessing this might be a standard result in analytic number theory but I unfortunately don't yet know much about the area. - -At least when $S$ is the inverse image of a union of Frobenius classes of $\text{Gal}(K/\mathbb{Q})$ for some number field $K$ under the Artin map, I have reason to believe that the answer is $\frac{X}{\log^{\alpha} X}$, but I don't know how to prove this either. I'm guessing the result holds in more generality though, hence the more general form of the question above - this specific case below the line reduces to the general question via Chebotarev density. - -REPLY [3 votes]: Under a very slight strengthening of the hypothesis, this can be done by elementary methods, namely the "Wirsing–Odoni method". The following version is Proposition 4 in this paper of mine with Finch and Sebah (I'm simplifying the hypotheses for clarity here): -Let $f$ be a multiplicative function satisfying $0\le f(n)\le 1$ for all $n$. -Suppose that there exist real numbers $\xi>0$ and $0<\beta<1$ such that -$$ -\sum_{p -TITLE: Is there a finitely generated group with the same structure as ZFC? -QUESTION [14 upvotes]: Is there a finitely generated computably presentable group $G$ on generator set $A$ and a computable function $f$ from first-order formulas to words on $A$ such that $\mathsf{ZFC}\vdash\sigma\leftrightarrow\tau$ iff $f(\sigma)$ and $f(\tau)$ represent the same element in $G$? - -REPLY [18 votes]: The relation $\text{ZFC}\vdash\varphi\leftrightarrow \psi$ is a $\Sigma_1^0$-definable equivalence relation on the set $\mathcal L$ of formulas in the language of set theory. It is a corollary of Theorem 3.2 of Neis-Sorbi's "Calibrating word problems of groups via the complexity of equivalence relations" that there is a finitely generated computably presentable group with generator set $A$ whose word problem, viewed as an equivalence relation $\sim$ on the set $W$ of words on $A$, is $\Sigma_1^0$-universal. As a consequence there is a reduction from the former equivalence relation to the latter, and this just means that there is a computable function $f : \mathcal{L}\to W$ such that $f(\varphi) \sim f(\psi)$ if and only if $\text{ZFC}\vdash\varphi\leftrightarrow\psi$, which is what you want.<|endoftext|> -TITLE: Rational characters of a number field are powers of norm -QUESTION [8 upvotes]: Consider a number field $K/\mathbb{Q}$ and the embedding of $K^* \hookrightarrow GL_n(\mathbb{Q})$. This is the set of rational points of a $\mathbb{Q}$-algebraic group $G \subseteq GL_n(\mathbb{C})$. Then is it true that any $\mathbb{Q}$-characters of $G$ will look like $g \mapsto \det(g)^k$ for some $k \in \mathbb{Z}$. That is, on $G_\mathbb{Q} = K^*$ the character will look like $x \mapsto N_\mathbb{Q}^K(x^k) $. -I heard someone make this remark in a discussion that all the rational characters on a number field are powers of norm. I have not been able to find a reference (or other non-norm characters!). - -REPLY [8 votes]: If $H$ denotes the multiplicative group defined over $K$, then $G=\mathrm{Res}_{K/\mathbb{Q}}H$. By Section 2.61 of Milne's "Algebraic Groups - The Theory of Group Schemes of Finite Type over a Field", the group $G_{\overline{\mathbb{Q}}}$ obtained from $G$ by extension of scalars is isomorphic to the product of $H_\sigma$, where $\sigma$ runs through the embeddings $K\hookrightarrow\overline{\mathbb{Q}}$. It follows that the characters of $G$ defined over $\overline{\mathbb{Q}}$ are the maps $\prod_\sigma f_\sigma^\sigma$, where each $f_\sigma$ is a character of $H$ defined over $\overline{\mathbb{Q}}$. The characters of $G$ defined over $\mathbb{Q}$ are those maps $\prod_\sigma f_\sigma^\sigma$, which are fixed by $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. The condition "fixed by $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$" means that $f_\sigma$ is independent of $\sigma$, and it is defined over $K$. In other words, $f_\sigma$ is the map $x\mapsto x^k$ on $K^\times$ with some $k\in\mathbb{Z}$ independent of $\sigma$, which means that -$$\prod_\sigma f_\sigma^\sigma(x)=\prod_\sigma(x^k)^\sigma=N_{K/\mathbb{Q}}(x^k).$$<|endoftext|> -TITLE: Fuchsian groups and Eichler's result -QUESTION [9 upvotes]: Let $G$ be a Fuchsian group of first kind contained in $\text{PSL}_2(\mathbb{R})$. A result of Eichler says, there exists a finite set $S\subset G$ such that any $\gamma$ in $G$ can be written as a product $\prod_{i=1}^{k} \gamma_i,$ where each $\gamma_i$ are either in $S,$ or power of some parabolic element coming from $S,$ for some $k\ll \log ||\gamma||.$ Howerver, the original Eichler's paper is in German, (http://matwbn.icm.edu.pl/ksiazki/aa/aa11/aa11111.pdf) and this is giving me trouble to understand the proof. Does anyone know if I can find the proof written in English somewhere else ? - -REPLY [9 votes]: This follows from Theorem 2(i) and Theorem 4 in The structure of words in discrete subgroups of $\mathrm{SL}(2,\mathbb{C})$, by Beardon. - -Since it isn't explicitly stated, I will roughly summarize/explain how you get the result. -Lets $D$ be a convex fundamental polygon for $G$, $S^*$ the associated generating set, and let $S$ be the set $S^*$ with the natural parabolics added (generators of maximal parabolic subgroup at each ideal vertex). For example, the natural generating set for $\mathrm{PSL}_2(\mathbb Z)$ will have $S^*=S$ since the parabolic is already there (guessing Eichler was inspired by Euclidean algorithm). If you have the "usual domain" for a Fuchsian group corresponding to a complete, once punctured, finite volume, hyperbolic torus, $S^*$ will be $\{A,B,A^{-1},B^{-1}\}$. The fundamental domain has four parabolic vertices, which get identified, so adding $P=ABA^{-1}B^{-1}$, its cyclic permutations, and inverses will give $S$. In this torus example consider the parabolic vertex corresponding to $P$, v, and note that -$$ -D, AD, (AB)D, (ABA^{-1})D, (ABA^{-1}B^{-1})D=PD -$$ -all contain $v$. More generally you have that $v \in (P^k W) D$ where $W$ is an initial segment of $P$. -Beardon defines decomposition of elements in Fuchsian groups into chunks $C_i$ which split into two types: type I are elements which are not longer than some constant $m$ and type II are elements longer than $m$. Theorem 3 tells you that type II $C_i$ are basically parabolics in the sense that there is a parabolic vertex $v \in D$ such that -$$ v \in D, A_1D,\dots,(A_1 \cdots A_n )D=C_i D$$ -where $A_i \in S^*$. That means that $C_i= (P^k W)D$ where $P$ is the parabolic at $v$ and $W$ is some initial segment of $P$. -Theorem 2 tells you that there are log many $C_i$ compared to norm and Theorem 4 gives the bounds if you "collapse" the type II pieces/split into parabolic part.<|endoftext|> -TITLE: Fibrations in complex geometry -QUESTION [5 upvotes]: Let $X^n$ be a compact Kähler manifold with $K_X$ semi-ample, i.e., a sufficiently high power of $K_X$ is basepoint free. The associated pluricanonical system $| K_X^{\ell} |$ furnishes a birational map $$f : X \dashrightarrow \mathbb{P}^{\dim H^0(X, K_X^{\ell})-1}$$ onto some normal projective variety $Y \subset \mathbb{P}^{\dim H^0(X, K_X^{\ell})-1}$ of dimension $\kappa(X)$. Here, $\kappa(X)$ denotes the Kodaira dimension of $X$. The semi-ampleness of $K_X$ further implies that $K_X^{\ell} \simeq f^{\ast} \mathcal{O}(1)$. In particular, for every $y \in Y$ that is not contained in the discriminant locus of $f$, $K_X^{\ell} \vert_{f^{-1}(y)} \simeq \mathcal{O}_{X_y}$. Since $f$ is a submersion near $f^{-1}(y)$, we have the adjunction-type relation $K_{f^{-1}(y)} \simeq K_X \vert_{f^{-1}(y)}$ and therefore the fibres of $f$ are Calabi--Yau manifolds of dimension $n- \kappa(X)$. -In the Kähler geometry literature, it is common to refer to this map $f$ as a Calabi--Yau fibration. My question may be extremely obvious, but nevertheless: -Question: Is this a fibration in the sense of homotopy theory, i.e., does this map satisfy the homotopy lifting property? - -REPLY [6 votes]: I would say that the answer is in general no. -Think of an elliptic surface $X$ with Kodaira dimension $1$ and whose elliptic fibration contains a cuspidal curve. Then the general fibre is not homotopically equivalent to the special one (the former is homeomorphic to $S^1 \times S^1$, the latter to $S^1$, in particular their fundamental groups are different), whereas all the fibres of a Hurewicz fibration have the same homotopy type. -Edit. Actually, any elliptic surface $X$ with Kodaira dimension $1$ and whose elliptic fibration contains a nodal curve also provides a counterexample. In fact, a nodal cubic is homeomorphic to a torus "with one cycle shrunk away”; in particular, it has the homotopy type of $S^1 \vee S^2$ and the previous argument applies.<|endoftext|> -TITLE: Eigenvalues of the complement of a graph -QUESTION [10 upvotes]: Let $A$ and $\widetilde A$ be the adjacency matrices of a graph $G$ and of its complement, respectively. - -Is there any relation between the eigenvalues of $A + \widetilde A$ and the eigenvalues of $A$ and $\widetilde A$? - -Also, do $A$ and $\widetilde A$ have the same set of eigenvectors? - - -Thank you. - -REPLY [15 votes]: Edit (bis). There are two answers, depending on whether loops about vertices are allowed or not. In addition, the case of regular graphs is completely described. - -If loops are allowed - -The relation between matrices is -$$A+{\widetilde A}=J$$ -where $J={\bf1}{\bf1}^T$ is the all-ones matrix. The first consequence is that the sum of the eigenvalues of $A$ and ${\widetilde A}$ equals $|V|$ where $V$ is the set of vertices. -A second consequence concerns multiple eigenvalues. If $\lambda$ is an eigenvalue of $A$, of multiplicity $m\ge2$, then $-\lambda$ is an eigenvalue of ${\widetilde A}$, of multiplicity larger than or equal to $m-1$. Just consider the intersection of the eigenspace with the hyperplane ${\bf1}^\bot$. In particular, this is a case where $A$ and ${\widetilde A}$ share common eigenvectors. - -If loops are not allowed - -Here -$$A+{\widetilde A}=K:=J-I_V$$ -The sum of the eigenvalues of $A$ is the opposite of that of ${\widetilde A}$. -If $\lambda$ is an eigenvalue of $A$, of multiplicity $m\ge2$, then $-1-\lambda$ is an eigenvalue of ${\widetilde A}$, of multiplicity larger than or equal to $m-1$. Again, this is a case where $A$ and ${\widetilde A}$ share common eigenvectors. - -Regular graphs - -If a graph is regular and connected (thanks to Emil for having pinned the point), then $\bf1$ is an eigenvector, with eigenvalue $d$, the degree of each vertex. It is a simple eigenvalue because $A$ is irreducible (connectedness). The other eigenspaces are contained in $\bf1^\bot$, because $A$ is symmetric. Thus eigenvectors of $A$ remain eigenvectors for $\widetilde A$, with the same multiplicity. The correspondance between eigenvalues is $\lambda\rightarrow-1-\lambda$. -Remark also that $d$ is the Perron eigenvalue of $A$, $n-1-d$ being that of $\widetilde A$. We thus deduce -$$\lambda\in D(0;d)\cap D(-1;n-1-d)$$ -for all the other eigenvalues of $A$.<|endoftext|> -TITLE: Primality test for $N=4p+1$ -QUESTION [6 upvotes]: Can you prove or disprove the following claim: - -Let $N=4p+1$ where $p$ is an odd prime number , let $T_n(x)$ be the nth Chebyshev polynomial of the first kind and let $F_n(x)$ denote an irreducible factor of degree $\varphi(n)$ of $T_n(x)$ . If there exists an integer $a$ such that $F_{p}(a) \equiv 0 \pmod{N} $ then $N$ is a prime. - -You can run this test here. An incomplete list of primes $p$ such that $4p+1$ is prime can be found here . I have verified this claim for $p \in [3,30000)$ with $a \in [1,100]$ . - -REPLY [8 votes]: The claim is true, and it holds more generally for every odd integer $p\geq 3$; the assumption that $p$ is prime is not needed. By the known factorization of Chebyshev polynomials, -$$F_p(x/2)=\prod_{\substack{1\leq m\leq 2p-1\\(m,4p)=1}}(x-\zeta^m-\zeta^{-m}),$$ -where $\zeta\in\mathbb{C}$ is a primitive $4p$-th root of unity. The splitting field of $F_p(x/2)$ is the subfield of the $4p$-th cyclotomic field fixed by complex conjugation; it is of degree $\varphi(p)$. -Assume that $q\nmid 2p$ is a prime number such that the reduction of $F_p(x/2)$ mod $q$ has a root in $\mathbb{F}_q$. The roots of $F_p(x/2)$ in $\overline{\mathbb{F}_q}$ are of the form $\xi^m+\xi^{-m}$, where $\xi\in\overline{\mathbb{F}_q}$ is a primitive $4p$-th root of unity. By assumption, the Frobenius automorphism $t\mapsto t^q$ fixes one of these roots, which is only possible when $q\equiv\pm 1\pmod{4p}$. It follows that, for any $a\in\mathbb{Z}$, the prime factors of $F_p(a)$ coprime to $2p$ are congruent to $\pm 1$ modulo $4p$. In particular, if $4p+1$ divides $F_p(a)$, then the only prime factor of $4p+1$ can be itself, i.e., $4p+1$ is prime.<|endoftext|> -TITLE: Is simultaneous similarity of matrices independent from the base field? -QUESTION [7 upvotes]: Suppose that $F$ is a subfield of a field $E$ and, for -$n\times n$ matrices $A_1,\dots,A_m, B_1,\dots,B_m$ -over $F$, there exists a matrix $T\in{\rm GL}_n(E)$ -such that $T^{-1}A_iT=B_i$ for all $i$. - -Does this imply that such a matrix $T$ can be chosen from ${\rm GL}_n(F)$? - -It is easy to see that the answer is - -yes if $m=1$; -and yes if the field $F$ is infinite. - -REPLY [5 votes]: This question is answered in comments: - -"As everyone is saying, this follows from Noether-Deuring. -See mathoverflow.net/questions/28469/hilbert-90-for-algebras for a quick proof. -I also asked this question a while back math.stackexchange.com/questions/305696." - -        – -David E Speyer -Thanks to all!<|endoftext|> -TITLE: The stationary reaping number $\mathfrak{r}_{cl}$ -QUESTION [10 upvotes]: Let $\kappa$ be at least inaccessible (but measurable is what I am primarily interested at the moment). Let $x,y \in [\kappa]^\kappa$ both be stationary. We say that $y$ stationary-splits $x$ iff $x \cap y$ and $x\setminus y$ are both stationary. -Define $\mathfrak{r}_{cl}:=\min \{\vert \mathcal{R}\vert \colon \,\, \forall x \in \mathcal{R} \,\, x \, \text{is stationary} \, \land \neg \exists y \in [\kappa]^\kappa \,\, \mathcal{R} \, \, \text{is stationary-split by} \,\, y\}$. -Of course, $\mathfrak{r}_{cl}$ has to be infinite and $\mathfrak{r}_{cl}= 2^\kappa$ in the $\kappa$-Sacks model. But can ZFC prove a nontrivial lower bound ($\omega_1,\, \kappa \,\, \text{or even} \,\, \kappa^+$) for $\mathfrak{r}_{cl}$ ?? -Note that this question is motivated by investigating cardinal characteristics on the 'higher' Cantor/ Baire space modulo the non-stationary (not the bounded) ideal. - -REPLY [8 votes]: I think that $\omega_1$ is certainly a lower bound, and that this is the case for any regular, uncountable $\kappa$. To see this, suppose that $\langle x_n \mid n < \omega \rangle$ is a sequence of stationary subsets of $\kappa$. We'll find a $y$ that stationary-splits every $x_n$. We'll do this by constructing a sequence $\langle x_n^* \mid n < \omega \rangle$ such that $x_n^*$ is a stationary subset of $x_n$ and $x_n^* \cap x_m^* = \emptyset$ for all $n < m < \omega$. Then we can let $y_n$ be a stationary subset of $x_n^*$ such that $x_n^* \setminus y_n$ is also stationary, and let $y = \bigcup_{n < \omega} y_n$. -The construction of $\langle x_n^* \mid n < \omega \rangle$ can be done by recursion on $n$, where we maintain the following additional recursion hypothesis: for all $n \leq m < \omega$, the set $x_{m,n} := x_m \setminus \bigcup_{k < n} x_k^*$ is stationary. Fix $n < \omega$ and suppose that $\langle x_k^* \mid k < n \rangle$ has been constructed. By hypothesis, $x_{n,n}$ is stationary; partition it into $\omega_1$-many disjoint stationary subsets, $\langle x^\alpha_{n,n} \mid \alpha < \omega_1 \rangle$. For each $m > n$, there is at most one $\alpha < \omega_1$ such that $x_{m,n} \setminus x^\alpha_{n,n}$ is nonstationary, so we can choose an $\alpha^* < \omega_1$ such that $x_{m,n} \setminus x^{\alpha^*}_{n,n}$ is stationary for all $m > n$, and then set $x_n^* = x^{\alpha^*}_{n,n}$, and continue to the next step. -It's not immediately clear to me how to achieve a better lower bound. This particular construction breaks down at limit steps, but it seems conceivable that a more clever version of the argument might yield a lower bound of $\kappa$ or even $\kappa^+$.<|endoftext|> -TITLE: Small ideas that became big -QUESTION [69 upvotes]: I am looking for ideas that began as small and maybe naïve or weak in some obscure and not very known paper, school or book but at some point in history turned into big powerful tools in research opening new paths or suggesting new ways of thinking maybe somewhere else. -I would like to find examples (with early references of first appearances if possible or available) of really big and powerful ideas nowadays that began in some obscure or small paper maybe in a really innocent way. What I am pursuing with this question is to fix here some examples showing how Mathematics behave like an enormous resonance chamber of ideas where one really small idea in a maybe very far topic can end being a powerful engine after some iterations maybe in a field completely different. I think that this happens much more in mathematics than in other disciplines due to the highly coherent connectedness of our field in comparison with others and it is great that Mathematics in this way give a chance to almost every reasonable idea after maybe some initial time required to mature it in the minds, hands and papers of the correct mathematicians (who do not necessarily have to be the same that first found that idea). -Summarizing, I am looking for ideas, concepts, objects, results (theorems), definitions, proofs or ways of thinking in general that appeared earlier in history (it does not have to be very early but just before the correct way of using the idea came to us) as something very obscure and not looking very useful and that then, after some undetermined amount of time, became a really powerful and deep tool opening new borders and frontiers in some (maybe other) part of the vast landscape of mathematics. -Edit: I really do not understand the aim in closing this question as it is actually at research level. I am clearly asking for tools that developed into modern research topics. I recognize that some answers are not research level answers, but then you should downvote the answer, not the question. I am really surprised by this decision as one of the persons that vote to close suggested it for publication in a place where it is clear that some of the most valuable answers that this question has received would have never occur precisely because the site that this person suggested is not research oriented. I do not imagine people on HSM answering about species or pointfree topology sincerely as these topics are really current research and not history (and I am interested mainly in current research topics). I do not agree with the fact that a limitation in reading understanding of some people can be enough to close a legitimate question, a question that it is worth for us as mathematicians to do and to show to other people that think that mathematics is useful and powerful the day after being published ignoring thus the true way mathematics is done, with its turnabouts and surprises; a discipline where a simple idea has the power to change the field as $0$ did, as the positional systems did, as sheaves did, or as species did. I am really sad for this decision. It is a pity that so many mathematicians regret the actual way in which their field develops, reject to explain and expose this behavior and hide themselves from this kind of questions about the internal development of ideas in mathematics. I challenge all those who voted to close this question as off-topic to look in HSM for any mention about "locale theory" there. - -REPLY [3 votes]: Another possibility could be the problem of the brachistochrone, a famous but one might think relatively innocent problem, which then led to the development of the calculus of variations.<|endoftext|> -TITLE: Australian Mathematical Society journal rankings -QUESTION [15 upvotes]: I apologize for a question that is not about mathematics, but I believe it is of interest to research mathematicians, and I believe there may be people on MathOverflow who can answer it objectively. If it is deemed unacceptable, I can survive. -For many years, I (and many others I know) have used a ranking of mathematics journals produced by the Australian Mathematical Society in 2009, formerly found here https://austms.org.au/Rankings/AustMS_final_ranked.html. -Without wishing to get into a debate on the usefulness or methodology of ranking journals, I found the list useful for judging what journals to submit papers to, and for justifying to deans or hiring committees the quality of journals that I or others (e.g. job, tenure applicants) have published in. I understand that it was getting out of date and I certainly did not agree with every grade there. But since it was made by mathematicians, rather than some trite formula, and since it used a simple A*/A/B/C grading scheme, it was solid and easy to reference and cite. For example, I planned to use it to help justify to deans an upcoming tenure decision. -It was also the top rated answer to this MO question about journal rankings. Unfortunately, the link above is now dead. - -Question 1: Does anyone know if this ranking is permanently gone from the internet? If the society has "disavowed" it as incorrect or out of date? If they are revising it? Or if the link has simply changed? - -I was able to find a "cached" version, so I still have access to the information. But I do not know how long this will be up, and it detracts from any semblance of authority if it is no longer hosted on a reputable website. -If this list is gone forever, - -Question 2: Do people have suggestions for a replacement with similar features? (Made by a reputable institution, with input from mathematicians rather than a trite formula, and easy to explain to non-mathematicians.) - -I understand that the second question has some overlap with the prior MO question linked above. - -REPLY [4 votes]: The system officially used in Finland is Jufo. It has four grades in the ranking, from 0 (lowest) to 3 (highest). The funding of Finnish universities partly depends on the amount of publications in those journals, with coefficients per publication being 0.1 - 1 - 3 - 4 (i.e., a publication in a level 3 journal is worth 4 publications in a level 1 journal, and 40 publications in a level 0 one.) -The ranks are re-allocated every 5 years by a committee based on suggestions from the academics; also there are some constraints (if one journal goes up, some other has to go down). This naturally creates some distortions, e. g., a new strong journal will lag in ranking, which means that nobody has incentive to publish there, which means nobody will be strongly pushing to move it up the ranking etc. For example, "Forum of mathematics, Pi" is only level 1. But other than that, the ranking is reasonable.<|endoftext|> -TITLE: Automorphism induced by an automorphism of the base -QUESTION [8 upvotes]: Let us consider a closed Riemann surface $\Sigma_b$ of genus $B$, and let $\Delta \subset \Sigma_b \times \Sigma_b$ be the diagonal. -If $G$ is a finite group, then any group epimorphism $$\varphi \colon \pi_1( \Sigma_b \times \Sigma_b - \Delta) \to G$$ induces, by Grauert-Remmert Extension Theorem, the existence of a compact complex manifold $X$ (actually, a complex projective surface), endowed with a Galois cover $$\pi \colon X \to \Sigma_b \times \Sigma_b$$ branched at most over $\Delta$. -Let us now denote by $a$ the involutory automorphism of $\Sigma_b \times \Sigma_b$ given by $a(x, \, y)=(y, \, x)$; it leaves $\Delta$ (pointwise) invariant, so we may ask the following - -Question. Under which conditions on $\varphi$ the automorphism $a \colon \Sigma_b \times \Sigma_b \to \Sigma_b \times \Sigma_b $ lifts to an automorphism $\bar{a} \colon X \to X$? - -REPLY [5 votes]: In this post I will work in the greatest generality I can think of. In particular, fundamental group means étale fundamental group, but for varieties over $\mathbf C$ the same argument carries through using the topological fundamental group instead. - -Lemma. Let $X$ and $Y$ be separated normal integral schemes, let $f \colon Y \to X$ be a finite and finitely presented separable Galois cover with group $G$, and let $a \colon X \to X$ be an automorphism. Let $U \subseteq X$ be the dense open locus where $f$ is étale, let $V = f^{-1}(U)$, let $\bar y \to V$ be a geometric point with image $\bar x \to U$, and let $\phi \colon \pi_1(U,\bar x) \twoheadrightarrow G$ be the surjection corresponding to the $G$-cover $V \to U$. Then the following are equivalent: - -There exists an automorphism $b \colon Y \to Y$ lifting $a$; -There exists a dominant rational map $b \colon Y \to Y$ lifting $a$; -The isomorphism $a$ takes $U$ to itself, and the pullback $V' \to U$ of $V \to U$ along $a$ is isomorphic to $V \to U$ (as étale $G$-covers of $U$); -For any choice of path $[\gamma] \in \pi_1(U,\bar x, a^*\bar x)$, the subgroups $\ker \phi$ and $\ker(\phi a_* \gamma_*)$ of $\pi_1(U,\bar x)$ are conjugate (see proof for precise statement). - -Moreover, the set of such lifts is a $G$-bitorsor via pre- and post-composition of deck transformations. - -Proof. For (1) $\Leftrightarrow$ (2), note that a dominant rational lift $b$ is automatically an automorphism. Indeed, given a commutative diagram -$$\begin{array}{ccc}Y & \stackrel{b}\dashrightarrow & Y \\ \downarrow & & \downarrow \\ X & \underset a\to & X,\!\end{array}\tag{1}\label{1}$$ -multiplicativity of function field degrees shows that $b$ is birational. Since $Y$ is the integral closure of $X$ in $K(Y)$, we conclude that $b$ is an isomorphism since normalisation is a functor. -Thus for (2) $\Leftrightarrow$ (3), we know that $b$ gives an isomorphism $V \to V$ lifting $a|_U \colon U \to U$. This is exactly the same thing as an isomorphism $V \to V'$ over $U$, where $V'$ is the pullback -$$\begin{array}{ccc}V' & \to & V \\ \downarrow & & \downarrow \\ U & \stackrel a\to & U.\!\end{array}$$ -Finally, for (3) $\Leftrightarrow$ (4), we note that the cover $V' \to U$ corresponds to the surjection -$$\pi_1(U,a^*\bar x) \stackrel{a_*}\to \pi_1(U,\bar x) \stackrel \phi\twoheadrightarrow G.$$ -Any choice of path $[\gamma] \in \pi_1(U,\bar x, a^*\bar x)$ gives an identification -\begin{align*} -\gamma_* \colon \pi_1(U,\bar x) &\stackrel\sim\longrightarrow \pi_1(U,a^* \bar x)\\ -[\alpha] &\longmapsto [\gamma^{-1}] \cdot [\alpha] \cdot [\gamma], -\end{align*} -well-defined up to conjugation. Under this identification, the surjection $\pi_1(U,a^*\bar x) \twoheadrightarrow G$ above corresponds to the surjection $\pi_1(U,\bar x) \twoheadrightarrow G$ given by $\phi a_* \gamma_*$. The induced cover is isomorphic to the cover $V \to U$ given by $\phi$ if and only if the kernels are conjugate (see e.g. [Munkres, Thm. 79.4] in the topological setting), proving (3) $\Leftrightarrow$ (4). -The final statement follows for example because $\operatorname{Isom}_X(Y,Y')$ is naturally a $G$-bitorsor, as $G$ agrees with both $\operatorname{Aut}_X(Y)$ and $\operatorname{Aut}_X(Y')$. (See also this post for a general discussion of Galois covers of normal schemes.) $\square$ - -References. -[Munkres] J. R. Munkres, Topology (second edition). Pearson, 2018.<|endoftext|> -TITLE: On the functional equation $f(xf(y))=\frac{f(f(x))}y$ on arbitrary groups -QUESTION [7 upvotes]: In this answer, it was shown that there is no function $f\colon\mathbb Q_{+}^{*}\to\mathbb Q_{+}^{*}$ such that -\begin{equation} - f(xf(y))=\frac{f(f(x))}y \label{1}\tag{1} -\end{equation} -for all $x$ and $y$ (in $\mathbb Q_{+}^{*}$), where $\mathbb Q_{+}^{*}$ is the set of all (strictly) positive rational numbers. -The only properties of $\mathbb Q_{+}^{*}$ used in the proof were that $\mathbb Q_{+}^{*}$ is a abelian group with respect to the multiplication and $x^{-1}\ne x$ for some $x\in\mathbb Q_{+}^{*}$. -The question now is this: - -Can the stated result be extended to non-abelian groups? - -A somewhat similarly looking functional equation was discussed here. - -REPLY [7 votes]: For not necessarily abelian groups, we can interpret the division in (1) as the right or left division. -Let $G$ be any group. The answer to the question is given by - -Theorem: The following three conditions are equivalent to one another: -(right): there is a function $f\colon G\to G$ such that -\begin{equation*} - f(xf(y))=f(f(x))y^{-1} \label{r}\tag{$r$} -\end{equation*} -for all $x$ and $y$ (in $G$); -(left): there is a function $f\colon G\to G$ such that -\begin{equation*} - f(xf(y))=y^{-1}f(f(x)) \label{l}\tag{$l$} -\end{equation*} -for all $x$ and $y$; -(involutive) $x^{-1}=x$ for all $x$ (and hence $G$ is abelian). - -Proof of implication ($x^{-1}=x$ for all $x$)$\implies$ $G$ is abelian: For all $x$ and $y$ we have $xy=(xy)^{-1}=y^{-1}x^{-1}=yx$, so that $xy=yx$, as claimed. $\Box$ -Proof of implications (involutive)$\implies$(right) and (involutive)$\implies$(left). Suppose the (involutive) property holds, so that $G$ is abelian. Let $f(x)=x$ for all $x$. Then $f(xf(y))=xy=f(f(x))y^{-1}$ for all $x,y$, so that condition \eqref{r} holds. Since $G$ is abelian, condition \eqref{l} holds as well. $\Box$ -Proof of implication (right)$\implies$ (involutive): Substituting $x=1$ in \eqref{r}, we get $f(f(y))=f(b)y^{-1}$ (for all $y$), where -\begin{equation*} - b:=f(1). -\end{equation*} -So, $f(f(x))=f(b)x^{-1}$, and now \eqref{r} yields $f(xf(y))=f(b)x^{-1}y^{-1}$. Substituting here $y=1$, we get $f(xb)=f(b)x^{-1}$ or, equivalently, -\begin{equation*} - f(z)=cz^{-1} \label{2}\tag{2} -\end{equation*} -for all $z$, where $c:=f(b)b$. Now \eqref{r} can be rewritten as -\begin{equation*} -yc^{-1}x^{-1}=xc^{-1}y^{-1}. \label{r'}\tag{$r'$} -\end{equation*} -Substituting here $y=1$ and $x=c$, we get $c^{-2}=1$, that is, $c^{-1}=c$. Taking now any $z\in G$ and letting $y=zxc$, we rewrite \eqref{r'} as $z=z^{-1}$, which means $G$ has the (involutive) property. -$\Box$ -Proof of implication (left)$\implies$ (involutive): Substituting $x=1$ in \eqref{l}, we get $f(f(y))=y^{-1}f(b)$, where $b=f(1)$, as before. -So, $f(f(x))=x^{-1}f(b)$, and now \eqref{l} yields $f(xf(y))=y^{-1}x^{-1}f(b)$. Substituting here $y=1$, we get $f(xb)=x^{-1}f(b)$ or, equivalently, -\begin{equation*} - f(z)=bz^{-1}d -\end{equation*} -for all $z$, where $d:=f(b)$. Now \eqref{l} can be rewritten as -\begin{equation*} -bd^{-1}yb^{-1}x^{-1}=y^{-1}bd^{-1}xb^{-1}. \label{l'}\tag{$l'$} -\end{equation*} -Substituting here $y=1$ and $x=b$, we get $b^{-2}=1$, that is, $b^{-1}=b$. -Substituting in \eqref{l'} $y=b$ and $x=d$, we get $bd^{-2}=b^{-1}=b$ and hence $d^{-2}=1$, that is, $d^{-1}=d$. -Now \eqref{l'} becomes -\begin{equation*} -bdybx^{-1}=y^{-1}bdxb. \label{l2}\tag{$l''$} -\end{equation*} -Substituting here $y=b$ and $x=1$, we get $bd=db$. -Taking now any $z\in G$ and letting $y=bdz$, so that $z=bdy$, we rewrite \eqref{l2} as -\begin{equation*} -zbx^{-1}=z^{-1}xb. -\end{equation*} -Substituting here $x=b$, we get $z=z^{-1}$, which means $G$ has the (involutive) property. $\Box$ -The theorem is now completely proved. -Remark: As seen from \eqref{2}, if any one of the three equivalent conditions -- (right), (left), or (involutive) -- holds, then the solutions of equation \eqref{r} and/or, equivalently, \eqref{l} are precisely those of the form $f(x)=cx$ for some $c$ and all $x$. -Corollary: Neither the (right) nor the (left) property can hold for any non-abelian group $G$.<|endoftext|> -TITLE: Best texts on Lie groups for number theorists -QUESTION [18 upvotes]: What are the most comprehensive textbooks on the structure of Lie groups and their infinite-dimensional representations if one is interested in their applications to number theory (so covering discrete subgroups and automorphic representations)? - -REPLY [8 votes]: I think the most comprehensive reference would be the following conference proceedings (Proceedings in Symposia in Pure Mathematics) : - -Automorphic Forms, Representations, and L-functions, Parts 1&2, vol. 33 -Motives, Parts 1&2, vol.51 -Representation Theory and Automorphic Forms, vol. 61 - -However you might be also interested in the following books. An introduction to the Archimedean representation theory is given in "Representation Theory of Semisimple Groups" by Knapp. A slightly more advanced book is "Continuous Cohomology, Discrete Subgroups, and Representations of Reductive Groups" by Borel and Wallach. -There is also a book "An Introduction to Automorphic Representations with a view toward Trace Formulae" by Getz and Hahn. Another recent introductory book is "Eisenstein Series and Automorphic Representations with Applications in String Theory" by Fleig, Gustafsson, Kleinschmidt, Persson.<|endoftext|> -TITLE: Deciding if $\mathbb{Z}\ltimes_A \mathbb{Z}^5$ and $\mathbb{Z}\ltimes_B \mathbb{Z}^5$ are isomorphic or not -QUESTION [14 upvotes]: I asked this in this MSE question but I didn't get answers. I think maybe here someone can help me. -I have the two following groups -$G_A=\mathbb{Z}\ltimes_A \mathbb{Z}^5$, where $A=\begin{pmatrix} 1&0&0&1&0\\0&-1&0&0&0\\0&0&-1&0&0\\0&0&0&0&-1\\0&0&0&1&0\end{pmatrix}$ and -$G_B=\mathbb{Z}\ltimes_B \mathbb{Z}^5$, where $B=\begin{pmatrix} 1&0&0&1&0\\0&-1&0&1&0\\0&0&-1&0&0\\0&0&0&0&-1\\0&0&0&1&0\end{pmatrix}$. -The product is given (for example in $G_A$) by $(k,m)\cdot(\ell,n)=(k+\ell, m+A^k n)$. - -Problem: Decide if $G_A$ is isomorphic to $G_B$ or not. - -My thoughts: I think strongly that they are not isomorphic but I couldn't prove it. The matrices $A$ and $B$ are both of order 4, they're not conjugate in $\mathsf{GL}(n,\mathbb{Z})$ (neither $B$ and $A^{-1}$) but they are conjugate in $\mathsf{GL}(n,\mathbb{Q})$. In some other cases, I've seen that they're not isomorphic by computing the abelianization, but in this case both have the same abelianization, namely $\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$. Even worse, both have 1 as an eigenvalue. -In my previous MO question there is a counterexample for the implication "$G_A\cong G_B\Rightarrow A\sim B^{\pm 1}$" so I cannot use that. -Thanks! - -REPLY [6 votes]: Here is Derek Holt's computation done in GAP: -gap> LoadPackage("anupq"); -gap> F := FreeGroup("a","b","c","d","e","t");; -gap> AssignGeneratorVariables(F); -gap> comms := List(Combinations(GeneratorsOfGroup(F){[1..5]},2),Comm);; -gap> G1 := F/Concatenation(comms, -> [Comm(a,t),b^t*b,c^t*c,d^t*a^-1*e^-1, e^t*d]);; -gap> G2 := F/Concatenation(comms, -> [Comm(a,t),b^t*b,c^t*c,d^t*b^-1*a^-1*e^-1,e^t*d]);; -gap> Pq(G1:Prime:=2,ClassBound:=2); - -gap> StructureDescription(last); -"(C4 x C4 x C4 x C2) : C4" -gap> Pq(G2:Prime:=2,ClassBound:=2); - -gap> StructureDescription(last); -"C2 x ((C4 x C4 x C2) : C4)"<|endoftext|> -TITLE: Alternate proofs that hyperbolic plane can’t be isometrically immersed in $\mathbb{R}^3$ -QUESTION [23 upvotes]: A famous theorem of Hilbert says that there is no smooth immersion of the hyperbolic plane in 3-dimensional Euclidean space. The expositions of this that I know of (in eg do Carmo’s book on curves/surfaces, and in Spivak vol 3) are very analytic and non-geometric, with lots of delicate formulas. However, in Thurston’s book on 3d geometry and topology, he suggests that this is really a geometric fact that should be proved by looking at the lines of curvature and how they twist around as you go off to infinity. -Question: does anyone know any alternate proof of this (or at least novel expositions of the usual ones) that emphasize the geometry and minimize the pages and pages of formulas? I don’t care about minimizing smoothness assumptions (I would rather have analytic simplicity!). - -REPLY [8 votes]: There's a nice discussion of local isometric immersions of hyperbolic planes in Robert McLachlan's expository article: -McLachlan, Robert, A gallery of constant-negative-curvature surfaces, Math. Intell. 16, No. 4, 31-37 (1994). ZBL0812.53003. -See also his accompanying website gallery and Richard Palais' gallery of pseudospherical surfaces. Ghys has a nice survey too in french, but I didn't see how to fill in the sketch of proof that he gives of Hilbert's theorem. - -I'm not sure what Thurston had in mind referring to the twisting of the lines of curvature. -Following Hilbert's proof, the asymptotic curves give natural coordinates on an isometrically immersed hyperbolic surface which form a Chebyshev net. In these coordinates, the metric (first fundamental form) has the form -$$\mathrm{I} = \left(\begin{array}{cc}1 & \cos(\omega) \\\cos(\omega) & 1\end{array}\right), $$ -where $\omega$ is the angle between the asymptotic curves. -Note that the area form is $dA=\sqrt{det(\mathrm{I})}dxdy=\sin(\omega)dxdy$. -A nice proof of this is given in the Proposition on p. 192 of Ghys' paper (he says, translated: "Let us try to demonstrate it by avoiding too many calculations. . ."). He shows that the shape operator acts on vectors in the two asymptotic directions by rotation by $\pi/2$ and $-\pi/2$. Then giving a local parameterization by coordinates parallel to the asymptotic directions, he uses this to show that the mixed second partial derivative of this parameterization is orthogonal to the surface, which is equivalent to the Chebyshev net condition. I'm not sure if there's an intuitive way to understand this. One thing to note is that the asymptotic curves have constant torsion ($\pm 1$), which I think is related to the result about the action of the shape operator. -McLachlan and Ghys describe a Chebyshev net as a piece of clothing with fibers going in two directions, where the fiber lengths and their crossings remain rigid, but the angles between them can vary. Then any such piece of clothing can be draped over a piece of a surface by varying the fiber angles, also like a fish net or fishnet stocking (indeed, Chebyshev considered the fitting of clothes in the presentation he made in 1878). See also Terng and Uhlenbeck "The Geometry of Solitons". - -Then the second fundamental form has the formula -$$\mathrm{I\!I} = \left(\begin{array}{cc}0 & \sin(\omega) \\\sin(\omega) & 0\end{array}\right). $$ -The $0$ on the diagonal are equivalent to taking asymptotic curves for the coordinates. One checks that -$\frac{det(\mathrm{I\!I})}{det(\mathrm{I})}+-1 = 0,$ -agreeing with Gauss' Theorema Egregrium. -Now McLachlan notes that the Gauss-Codazzi-Mainardi equations imply that the angle $\omega$ satisfies the sine-Gordon equation and has $0 < \omega <\pi $, since the angle between the coordinates must be non-zero. -$$\frac{\partial^2\omega}{\partial x\partial y}= \sin(\omega).$$ -On any rectangle in the coordinate chart, the sine-Gordon equation implies that the area is bounded by $2\pi$ ("Hazzidakis' Formula"). Let the rectangle $X=[x_1,x_2]\times[y_1,y_2]$, then -$$\omega(x_1,y_1)-\omega(x_1,y_2)-\omega(x_2,y_1)+\omega(x_2,y_2) = \iint_X \frac{\partial^2 \omega}{dxdy} dx dy \\ = \iint_X \sin(\omega) dx dy = \iint_{\psi(X)} dA = Area(X).$$ -Since $0<\omega < \pi$, the area is bounded by $2\pi$. -Now suppose that a global isometric immersion exists $\varphi:\mathbb{H}^2\to \mathbb{E}^3$. -Take a point $p$ in the hyperbolic plane, then we have canonical -Chebyshev coordinates about $p$, since the asymptotic -curves are complete (first follow one asymptotic line through $p$ a distance $x$, then follow the transverse asymptotic curve a distance $y$ to get to the coordinates $(x,y)$ to get $\psi:\mathbb{R}^2\to \mathbb{H}^2$). If we pull back the metric to this chart, -then we cannot get a complete metric because the area is bounded -by $2\pi$, but the area of the hyperbolic plane is infinite. -Hence by Hopf-Rinow there must be some geodesic line $\gamma:\mathbb{R}\to \mathbb{R}^2$ in this chart that is finite length, so that $\underset{t\to ∞}{\lim} \psi(\gamma(t))=q\in \mathbb{H}^2$. But along this curve then we must have $\omega$ getting arbitrarily close to $0$ or $\pi$, since if $\omega$ remains bounded -away from these, then the length of the tangent vector is bounded away from $0$, so the curve would have infinite length. The principal curvatures are $\cot(\omega)\pm \csc(\omega)=\frac{\cos(\omega)\pm 1}{\sin(\omega)}$, and thus as $\omega\to 0$ or $\omega \to \pi$, the principal curvatures get arbitrarily close to $0$ and $\infty$. But by completeness of the immersion, the limit point $\underset{t\to ∞}{\lim} \varphi(\psi(\gamma(t)))=\varphi(q)$ would have to exist with finite principal curvatures, a contradiction.<|endoftext|> -TITLE: Weibel's H-book, Milnor's exact sequence for spectral sequence of filtered complex, Theorem 5.5.5 -QUESTION [6 upvotes]: This is a question which I asked on StackExchange first, but might be more suited here. -I got stuck on the proof of Theorem 5.5.5 in Weibel's book. Not only that, but I also could not even find the statement of said theorem in any other source, so I am completely at a loss how to proceed. -The result is called the Eilenberg-Moore Filtration Sequence for complete complexes. Interestingly, I was unable to find this result under this name anywhere else. - -Suppose the filtered chain complex $C$ is complete with respect to it's filtration $F_*C$, i.e $C=\lim\limits_{\leftarrow} C/F_pC.$ In this case the exact sequence $$0\longrightarrow\mathbin{\lim\limits_{\longleftarrow}}^1H_{n+1}(C/F_pC)\longrightarrow H_n(C)\longrightarrow\mathbin{\lim\limits_{\longleftarrow}} H_n(C/F_pC)\longrightarrow0$$ -takes the form -$$0\longrightarrow\bigcap F_pH_n(C)\longrightarrow H_n(C)\longrightarrow H_n(C)/\bigcap F_pH_n(C)\longrightarrow0$$ -and $${\lim\limits_{\longleftarrow}} H_n(C/F_pC)\cong {\lim\limits_{\longleftarrow}} H_n(C)/F_pH_n(C). $$ - -The first exact sequence is Milnor's $\mathbin{\lim\limits_{\longleftarrow}}^1$ exact sequence (Theorem 3.5.8): - -Let $C_i$ be a tower of chain complexes satisfying the Mittag-Leffler condition, and set $C=\mathbin{\lim\limits_{\longleftarrow}} C_i$. Then there is an exact sequence for each n: -$$0\longrightarrow\mathbin{\lim\limits_{\longleftarrow}}^1H_{n+1}(C_i)\longrightarrow H_n(C)\longrightarrow\mathbin{\lim\limits_{\longleftarrow}} H_n(C_i)\longrightarrow0$$ - -The proof then proceeds as follows: - -Taking the limit of the exact sequence of towers -$$0\longrightarrow F_pH_*(C)\longrightarrow H_*(C)\longrightarrow H_*(C)/F_pH_*(C)\longrightarrow0$$ -we find that there is a monomorphism $H_*(C)/\bigcap F_pH_*(C)\to {\lim\limits_{\longleftarrow}} H_*(C)/F_pH_*(C)$. Also, there is an exact sequnce -$$0\longrightarrow H_*(C)/F_pH_*(C)\to H_*(C/F_pC),$$ -which also gives a monomorphism ${\lim\limits_{\longleftarrow}} H_*(C)/F_pH_*(C)\to \lim\limits_{\longleftarrow} H_*(C/F_pC)$. This combines into a monomorphism $H_*(C)/\bigcap F_pH_*(C)\to\lim\limits_{\longleftarrow} H_*(C/F_pC)$. - -Then Weibel says to "combine this with the $\mathbin{\lim\limits_{\longleftarrow}}^1$ sequence of 3.5.8". And I don't get this last step at all. It seems we have not yet connected $\mathbin{\lim\limits_{\longleftarrow}}^1$ and $\bigcap F_pH_n(C)$. -This result is used quite a bit in later sections, so I would really like to understand it. I would be thankful for any help here. - -REPLY [5 votes]: You have constructed, up to this point, a monomorphism $H_n C / \bigcap F_p H_n C \to \lim H_n(C/F_p C)$, and this is compatible with the map from $H_n C$. This allows you to construct the right-hand square, and then assemble all of, the following map of exact sequences: -$$ -\require{AMScd} -\begin{CD} -0 -@>>> -\bigcap F_pH_n(C) -@>>> -H_n(C) -@>>> H_n(C)/\bigcap F_pH_n(C) -@>>> -0\\ -@. -@VVV -@| -@VVV -@.\\ -0 -@>>> -\mathbin{\lim\limits_{\longleftarrow}}^1H_{n+1}(C/F_pC) -@>>> -H_n(C) -@>>> -\mathbin{\lim\limits_{\longleftarrow}} H_n(C/F_pC) -@>>>0. -\end{CD}$$ -The middle map is an isomorphism and you have shown that right-hand vertical map is injective. Applying the snake lemma, we find that all the vertical maps are isomorphisms.<|endoftext|> -TITLE: Infinitely many counterexamples to Nash-Williams's conjecture about hamiltonicity? -QUESTION [5 upvotes]: Question from 2013 -gives one counterexample to Nash-Williams's conjecture about hamiltonicity -of dense digraphs. -Later, we found tens of counterexamples on more than 30 vertices -and believe there are infinitely many counterexamples. -Define $K_{x_1,x_2,...x_n}$ to the complete multipartite digraph -with partitions $x_i$ and every edge is oriented in both directions. -Let $L=\max x_i$. -Conjecture 1: as $n,L$ vary, there are infinitely many counterexamples - -Q1 Does this give infinitely many counterexamples? - -sagemath code for $K_{1,1,2,5}$: -G1=graphs.CompleteMultipartiteGraph((1,1,2,5)).to_directed() -sage: print G1.edges(False) -[(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (1, 0), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (2, 0), (2, 1), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (3, 0), (3, 1), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (4, 0), (4, 1), (4, 2), (4, 3), (5, 0), (5, 1), (5, 2), (5, 3), (6, 0), (6, 1), (6, 2), (6, 3), (7, 0), (7, 1), (7, 2), (7, 3), (8, 0), (8, 1), (8, 2), (8, 3)] - -For counterexample on 15 vertices take $x_i=(1, 1, 1, 2, 2, 8)$. -Added The suggested counterexamples are wrong and were the result of a program bug. - -REPLY [4 votes]: These examples are symmetric digraphs, i.e. graphs. For graphs, the Nash-Williams conjecture just becomes Chvatal's theorem (If $G$ is a graph on $n\geq 3$ vertices with degree sequence $d_1\leq d_2\leq \dots\leq d_n$ and for all $1\leq i -TITLE: Creation and annihilation operators in QFT -QUESTION [8 upvotes]: As I said before, I'm not a QFT expert but I'm trying to understand the basics of its rigorous formulation. -Let's take Dimock's book, where the foundation of QM and QFT is discussed. If we consider, say, two particles, one living in a Hilbert space $\mathcal{H}_{1}$ and the other in another Hilbert space $\mathcal{H}_{2}$, the description of the state of the two-particle system is given in terms of the tensor product $\mathcal{H}^{(2)}=\mathcal{H}_{1}\otimes \mathcal{H}_{2}$. Of course, we could go furhter and study a system $\mathcal{H}^{(N)}=\mathcal{H}_{1}\otimes \cdots \otimes \mathcal{H}_{n}$. If all the particles are identical, then $\mathcal{H}_{1}=\cdots = \mathcal{H}_{n} \equiv \mathcal{H}$ and we must take into account symmetric and anti-symmetric subspaces of $\mathcal{H}^{(N)}$, which correspond to the fact that particles may be either bosons or fermions, respectivelly. At this point, one defines symmetrization and anti-symetrization operators. The next step is to consider a system of an arbitrary number of particles. At this point, one defines Fock spaces $\mathcal{F}^{\pm}(\mathcal{H}) = \bigoplus_{n=0}^{\infty}\mathcal{H}_{n}^{\pm}$ for bosons and fermions. Also, one defines creation and annihilation operators $a(h)$ and $a^{\dagger}(h)$ on $\mathcal{F}^{\pm}(\mathcal{H})$. -Now, as far as I understand, this is all quantum mechanics, not QFT. However, these ideas seem to find analogues in QFT, and this is the point where I get confused. -On section I.5 of Feldman, Trubowitz and Knörrer's book there is a quick discussion on (fermionic) QFT and it is stated that, in this context, creation and annihilation operators are special families $\{\varphi^{\dagger}(x,\sigma):\hspace{0.1cm} x \in \mathbb{R}^{d}, \hspace{0.1cm} \sigma \in \mathcal{S}\}$ and $\{\varphi(x,\sigma):\hspace{0.1cm} x \in \mathbb{R}^{d}, \hspace{0.1cm} \sigma \in \mathcal{S}\}$ on a Hilbert space $\mathcal{H}$. This is very different than the creation and annihilation operators mentioned above. For instance, these are now families of operators indexed by $x$ and $\sigma$. I believe this is a reflection of the fact that we passed from QM to QFT. But I'm really lost here and I don't know what's the difference between these two constructions and definitions. Can anyone help me, please? I'm mainly interested in understanding the second approach, since the first one I believe I understand (at least sufficiently well). If, in addition, you could suggest some reference where these ideas of Feldman, Trubowitz and Knörrer are discussed in more details and with rigor, I'd appreciate! -ADD: Based on Feldman, Trubowitz and Knörrer's book, it seems to me that the understanding of these objects (to be more precise, the objects they briefly describe in the first 2 pages of section I.5) is fundamental to understand the formulation of a bunch of QFT models (at least for fermions). Thus, I'd appreciate if someone could elaborate a little more on the structure behind these creation and annihilation operators and its connections to the quantum case that is needed to understand the rest of the discussion on FTK's book. In other words, I think I just need to better understand these first definitions (and how are they connected with the usual quantum case I (seem) to know) to be able to understand the rest of the text. - -REPLY [7 votes]: The connection can be seen by taking $H = L^2(\mathbb{R}^3)$ in the first explanation. This is the Hilbert space of a nonrelativistic, spinless, three-dimensional particle. By direct summing the symmetric (antisymmetric) tensor powers of $H$ we get the Hilbert space of an ensemble of noninteracting Bosonic (Fermionic) nonrelativistic, spinless, three-dimensional particles, known as Fock space. The $n$th tensor power represents the states in which $n$ particles are present. -Now we have "creation" and "annihilation" operators which take states in the $n$th tensor power into the $(n \pm 1)$st tensor power. For each state $h$ in the original Hilbert space $H$ there is a creation operator which tensors with $h$ and symmetrizes (antisymmetrizes), taking the $n$th tensor power into the $(n+1)$st, and its adjoint which goes in the opposite direction and removes a tensor factor of $h$. -In the physics literature one usually works with idealized creation/annihilation operators for which the state $h$ is a fictional Dirac delta function concentrated at some point of $\mathbb{R}^3$. This is what is described in your second explanation. As is usual in physics, the Hilbert space is unspecified, but in the case of free fields it corresponds to the Fock space in the first explanation. -Fock space is inadequate to model interacting fields (indeed, here the mathematical issues become deep and fundamentally unresolved). However, it is not trivial; for instance, one can study free quantum fields against a curved spacetime background and derive Hawking radiation, the Unruh effect, etc. Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics by Wald is an excellent, mathematically rigorous explanation of this setting. -In QFT the intuition is that one has a separate Hilbert space at each point of space, and one takes their tensor product to get the Hilbert space of the entire field. I indicated how, intuitively, the Fock space models a "measurable tensor product" of a family of harmonic oscillators (Bosonic case) or two-state systems (Fermionic case) indexed by all the points of space in my answer here. See Section 2.5 of my book Mathematical Quantization for a full explanation.<|endoftext|> -TITLE: Iterated Gentzen: or, a Sith objection to the proof of consistency of PA -QUESTION [7 upvotes]: $\DeclareMathOperator\PRA{PRA}\DeclareMathOperator\WF{WF}\DeclareMathOperator\Con{Con}\DeclareMathOperator\PA{PA}$Preamble: In the year … in a galaxy far far away, a nasty Sith named Darth Dubious (DD) asks a Jedi, Obi Wan Mathobi (OWM), about the consistency of PA: -DD: How do you know that PA is consistent? -OWM: Don't you know that many centuries ago a great Mathematician from Terra by the name of Gentzen proved its consistency? -DD: Of course I do. He proved that $\PRA + \WF(\epsilon_0) \vdash \Con(\PA)$, but …. -OWM: But what? Are you going to tell me the usual story that the proof is not finitistic enough? -DD: I dare not, Master Mathobi. I happily concede that the proof is valid. Yet, I have still a doubt lingering in my brain: how do you know that -$G(0) =\PRA + \WF(\epsilon_0)$ itself is not contradictory? -As I said, the proof seems quite acceptable to me, but not any argument that $G(0)$ is consistent because it has a model in some very infinitistic theory such as ZF. -Rather, let us say that we apply Gentzen's argument to $G(0)$, and that we can prove the following: -$\PRA + \WF(\alpha_0) \vdash \Con( G(0))$. -Now, I would surmise that $\alpha_0 > \epsilon_0$, right? -OWM: It would seem so. I suspect (but I am not sure) that else $\PA$ would be able to prove the consistency of $G(0)$. Need to ask some other Jedis more skillful in Ordinal Analysis …. -DD: Ok, waiting for them I tell you where I am going, although methinks you know it already: I am going to -repeat my argument again, and create a chain of theories $G(i)$ such that each one Gentzen-proves the consistency of the previous one by an ever greater countable ordinal. If necessary, we can iterate beyond $\omega$. Now, is this series of countable recursive ordinals cofinal in the set of all countable recursive ordinals? If yes, I am afraid you ask for too much, because then I would have to accept induction all the way to $\omega^{CK}_1$. -If, on the other hand, it does not, I would like to know which is the upper bound. -THAT ordinal is the actual price to pay to secure $\PA$'s consistency. - -Question: What is wrong with DD's argument? Or, if it is sound, any clues on the upper bound ordinal which would secure the consistency of the entire chain of iterated Gentzen theories - -ADDENDUM I started with $\PA$ but, mutatis mutandis, you can begin from $\operatorname Q$, in which case instead of $\epsilon_0$ you use $\omega$. - -REPLY [4 votes]: Mathobi is making a different argument in the comments on this post than in the question itself: in the post Mathobi is considering how far we need to justify transfinite induction to prove $G(i)$ consistent over PRA, while in the comments you mentioned Mathobi's hope would be supported if PA+Con(PA) had proof-theoretic ordinal $\varepsilon_0$. In favor of the position in the comments, the theories PA, PA+Con(PA), PA+Con(PA+Con(PA))), etc. (for finite iterations) all have the same strength, $\varepsilon_0$. However, we can show $G(1)$ is quite a bit stronger than PA+Con(PA), and each $G(i+1)$ is stronger than $G(i)$. -But first, we need to distinguish a few concepts: there's more than one formulation of proof-theoretic ordinal, the most commonly used one is $\vert\, .\,\vert_{sup}$, and $\vert\, .\,\vert_{Con}$ is the rendering of Gentzen's result in your question, $\textrm{PRA+TI}(\alpha)\vdash\textrm{Con(T)}$. (These are described in more detail in Rathjen's "The Realm of Ordinal Analysis".) However $\vert\, .\,\vert_{Con}$ is only useful with respect to a "natural" ordinal representation system fixed in advance, so for the rest of this answer we fix a representation system such as one based on Veblen's functions. Let's also define theories $H(0)=\textrm{PA}$, and $H(i+1)=\textrm{PA+Con}(H(i))$. -Here we get our first result comparing proof-theoretic ordinals of these theories, which is that for all $i<\omega$ we have $\vert H(i)\vert_{sup}=\varepsilon_0$. Since each $H(i)$ is a recursive theory, $\textrm{Con}(H(i))$ is a $\Pi_1^0$-sentence, and by proposition 2.24 from Rathjen's "The Realm of Ordinal Analysis", we have $\vert\textrm{PA+}\phi\vert_{sup}=\varepsilon_0$ where $\phi$ is any true arithmetical sentence. So we have $\vert\textrm{PA}\vert_{sup}=\vert \textrm{PA+Con}(H(i))\vert_{sup}$ for all $i<\omega$, and all those proof-theoretic ordinals are $\varepsilon_0$. -This doesn't fully answer the original question though, which was framed in terms of $\vert G(i)\vert_{Con}$ instead of $\vert H(i)\vert_{sup}$. The next step to answering the main question is utilizing this result to compare $\vert H(i)\vert_{Con}$, which also gives us that $\vert H(i)\vert_{Con}=\varepsilon_0$ for all $i<\omega$. (I can provide proof of this fact.) -However, failure to replicate this result for the hierarchy $G(i)$ shows the discrepancy between Mathobi's points: while for all $i<\omega$ we have $\vert H(i)\vert_{Con}=\vert H(i+1)\vert_{Con}$, in contrast we have $\vert G(i)\vert_{Con}<\vert G(i+1)\vert_{Con}$! (This is because otherwise $G(i+1)=\textrm{PRA+TI}(\vert G(i)\vert_{Con})=\textrm{PRA+TI}(\vert G(i+1)\vert_{Con})$ would prove its own consistency.) As a result $\textrm{PRA+TI}(\varepsilon_0)$ is quite a bit stronger than PA+Con(PA) at least in terms of $\vert\, .\,\vert_{Con}$.<|endoftext|> -TITLE: Ideal of the boundary of $G/U \subset \overline{G/U}$ -QUESTION [5 upvotes]: Let $G$ be a semi simple algebraic group, $B \subset G$ is a Borel subgroup and $U \subset B$ is the unipotent radical of $B$. We can consider the variety $G/U$. Let us also denote $\overline{G/U}:=\operatorname{Spec}(\mathbb{C}[G/U])$. It is known that the natural morphism $G/U \rightarrow \overline{G/U}$ is an open embedding. Let $\partial{G/U}$ be the boundary of $G/U$ inside $\overline{G/U}$. Note now that $\mathbb{C}[G/U]=\bigoplus_{\mu} V(\mu)$, where the sum runs through dominant characters $\mu$ of $G$ (we fix some maximal torus $T \subset B$, here $V(\mu)$ is the irreducible representation of $G$ with highest weight $\mu$). -Claim: the ideal of $\partial{G/U} \subset \overline{G/U}$ is generated by $V(\mu)$ with $\mu$ being regular (strictly dominant). How to prove this claim? Maybe there are any references? - -REPLY [8 votes]: Here is one way to see it, via classifying $G$-invariant radical ideals. (This has the bonus that it implicitly describes the boundary.) -Lemma: $G$-invariant ideals $I$ of $\mathbb{C}[G/U]$ are in bijection with sets of weights $S$ so that for $\lambda\in S$ and $\mu > \lambda$, $\mu\in S$. Such an ideal is radical iff for all $\lambda\notin S,$ we have $n\lambda\notin S$ for all positive integers $n$. -To see this, note that $G$-invariance tells you that $I$ must split as a sum $$\displaystyle\bigoplus_{\lambda\in S}V(\lambda)$$ for some set $S$. Now if $\lambda\in S,$ the multiplication map $V(\mu-\lambda)\otimes V(\lambda)\rightarrow V(\mu)$ is surjective and hence $\mu > \lambda$ must also be in $S$. -The statement about radical ideals follows similarly. -From this statement, you can see that the minimal nonzero $G$-invariant radical ideal (which necessarily cuts out the boundary) corresponds to taking $S$ the set of all regular weights.<|endoftext|> -TITLE: Example of a function with a curious property -QUESTION [8 upvotes]: Denote by $L^1(0,1)$ the space of Lebesgue integrable functions on the interval $(0,1)$. -$\textbf{Question:}$ Does there exist a function $F:(0,1)\rightarrow\mathbb{R}$ such that: - -$\frac{F(x)}{x}\in L^1(0,1)$, -$\frac{F'(x)}{x}\in L^1(0,1)$, -$\frac{F(x)}{x^2}\notin L^1(0,1)$? - -I'm guessing that the answer is positive and the point is to construct $F$ such that $F$ and $F'$ behave suitably near zero. It seems quite delicate. I checked that $F$ cannot be a polynomial or a power function (since then $F'\simeq \frac{F}x$, thus conditions 2 and 3 cannot hold simultaneously). -I would appreciate any hints! - -REPLY [11 votes]: There is no such function. First of all, $|F(a)-F(b)|\leqslant \int_a^b |F'(x)|dx\to 0$ when $a,b\to 0$. So $F$ has a limit $c$ at point 0. If $c\ne 0$, then 1) fails. So $\lim_{x\to 0} F(x)=0$. -Next, $$|F(a)|\leqslant \int_{0}^a|F'(x)|dx\leqslant a\int_{0}^a\frac{|F'(x)|}x dx=o(a),\quad\text{when}\quad a\to 0.\quad (1)$$ -Now -$$ -\int_a^b \frac{F(x)}{x^2}dx=\frac{F(a)}a-\frac{F(b)}b+\int_a^b \frac{F'(x)}xdx. \quad(2) -$$ -Consider two cases: - -$F$ has fixed sign near 0. Then choosing $a,b$ close to 0 we conclude from (1) and (2) that $\int \frac{F(x)}{x^2}dx$ converges at 0, but this is equivalent to the convergence of $\int \frac{|F(x)|}{x^2}dx$ which we need. - -$F$ has infinitely many zeroes in any neighborhood of 0. Then choosing $(a_k,b_k)$ being inclusion-maximal intervals of the open set $\{x:F(x)\ne 0\}$ and applying (2) for $a=a_k,b=b_k$ we bound $\int_0^c \frac{|F(x)|}{x^2}dx$ via -$\int_0^c \frac{|F'(x)|}{x}dx$. Here $c=b_1$, for example.<|endoftext|> -TITLE: On a limit involving a transform of the chromatic polynomial -QUESTION [6 upvotes]: I was playing around with the chromatic polynomial (denoted here by $\chi_G(x)$) and I have made the following conjecture. -Let $(G_n)_{n \ge 1}$ be a sequence of graphs with $v(G_n) \to \infty$ ($v(G_n)$ denotes the number of vertices of $G_n$) and $e(G_n) \to \infty$ ($e(G_n)$ denotes the number of edges of $G_n$). -For each $x \neq 0$, let us define the following transform of the chromatic polynomial of $G_n$ -$$ - \psi_{G_n}(x) = \frac{x^{v(G_n)}}{e(G_n)^{v(G_n)}} \chi_{G_n}\left( \frac{e(G_n)}{x} \right). -$$ -The conjecture is that for each fixed real number $x \neq 0$, we have $\psi_{G_n}(x) \to \exp(-x)$ as $n$ goes to infinity. -I have checked the conjecture for a few sequences of graphs: for example, $G_n$ being the complete graph $K_n$, for $G_n$ being a tree on $n$ vertices and for $G_n$ being a collection of $n$ independent edges (a matching on $2n$ vertices). -Does anyone know if this is well-known? -PS: I am not sure if the conditions on $v(G_n)$ and $e(G_n)$ are the right one. Any comments on this are welcome as well. - -REPLY [6 votes]: Here is a heuristic argument which perhaps someone can make -rigorous. I write $v_n=v(G_n)$ and $e_n=e(G_n)$. Let -$$ \chi_{G_n}(x) = - x^{v_n}-c_{n,v_n-1} x^{v_n-1}+c_{n,v_n-2}x^{v_n-2}-\cdots. $$ -I claim that for fixed $k\geq 0$, -$$ \lim_{n\to\infty} \frac{c_{n,v_n-k}}{e_n^k} = \frac{1}{k!}. $$ -One can prove this by noting that (by the Broken Circuit Theorem, for -instance, which shows that $c_{n,v_n-k}$ increases as we add more -edges to $G_n$) $c_{n,v_n-k}$ is bounded below by its value when $G_n$ -is a tree, and is bounded above by its value when $G_n$ is a complete -graph. The claimed result is easily verified for trees and complete -graphs (in the latter case, using known asymptotics for the Stirling -numbers of the first kind). Perhaps there is a more direct proof, but -in any case, if we don't worry about justifying interchanging limits -and sums, we get -$$ \lim_{n\to\infty} \frac{x^{v_n}}{e_n^{v_n}}\chi_{G_n}\left( - \frac{e_n}{x}\right) = \sum_{k\geq 0} \lim_{n\to\infty} - \frac{(-1)^k c_{n,v_n-k}x^k}{e_n^k} $$ -$$ \qquad = \sum_{k\geq 0} \frac{(-1)^k x^k}{k!} = \exp(-x). $$<|endoftext|> -TITLE: Ideal norm in orders -QUESTION [9 upvotes]: Let $\overline{T}$ be a Dedekind ring such that $\overline{T}/\overline{I}$ is finite for every nonzero ideal $\overline{I}$ of $\overline{T}$. Let $T$ be a subring of $\overline{T}$ with the same total ring of fractions (i.e. an order). -Let $I$ be an ideal of $T$ and let $\overline{I} = I\overline{T}$. The norm $N_T(I)$ of $I$ is defined to be the cardinality of $T/I$. -Question: Is there a formula relating $N_T(I)$ and $N_{\overline{T}}(\overline{I})$? -For example, it seems plausible that the discrepancy is measured by some "tor" group. -Remarks: - -If $I$ is projective then $N_T(I)$ and -$N_{\overline{T}}(\overline{I})$ are equal. -Localization reduces the problem to the case when -$T$ is local (and $\overline{T}$ is semi-local), and both $I$ and the conductor of $T$ are proper ideals. -(Thanks to Luc Guyot) If $T$ is a Bass ring ($\leftrightarrow$ every intermediate ring $T \subset R \subset \overline{T}$ is Gorenstein $\leftrightarrow$ every ideal is generated by two elements), and $T = \{a \in \overline{T} : a I \subset I \}$, then by [2, Proposition 5.8] $I$ is projective. It follows that $N_T(I)$ and $N_{\overline{T}}(\overline{I})$ are equal (by the first remark). -(generalization of third remark) if $T$ is a Gorenstein integral domain and $T = \{a \in \overline{T} : a I \subset I \}$, then $I$ is projective. This follows from combining Theorem 6.2(4) with Proposition 7.2 of [1]. It follows that $N_T(I)$ and $N_{\overline{T}}(\overline{I})$ are equal (by the first remark). - -[1] H. Bass, "On the ubiquity of Gorenstein rings", 1963. -[2] L. Levy and R. Wiegand, "Dedekind-like behavior of rings with 2-generated ideals", 1985. - -REPLY [2 votes]: I am recording for the benefit of others what is to my knowledge the full extent of what is known about the general problem. Luc Guyot has provided a nice and explicit answer for the case of quadratic orders. -I do not mark this post as "the answer" as the original question has not yet been answered. -Let the discrepancy of a $T$-ideal $I$ be defined as $ds(I) = N_{\overline{T}}(\overline{I})/N_T(I)$ (non-standard definition). -When does $ds(I) = 1$? -The following theorem is the main tool of the paper [1]. The statement uses the module index notation of [2]. -Theorem [1; Theorem 1]: - -$[\overline{T}:\overline{I}] \subset [T:I]$. -$[\overline{T}:\overline{I^{-1}}] \subset [I:T]$. -$[{T}:{I^{-1}}] \subset [\overline{I}:\overline{T}]$. - -Moreover, the following are equivalent: - -Any subset relation among (1), (2), (3) is an equality. -All subset relations among (1), (2), (3) is an equality. -$I$ is invertible. - -This theorem has the following corollaries for the "discrepancy". Recall that the different of $T$ is defined to be $\mathfrak D_{T} = (T^\vee)^{-1}$ where $T^\vee$ is the dual of $T$ for the trace form. -Corollary: $ds(I) \geq 1$ with equality if and only if $I$ is invertible. -Corollary: The following are equivalent: - -The discrepancy of $\mathfrak D_{T}$ is $1$. -For every ideal $I$ of $T$, $ds(I) = 1$ if and only if $T = (I:I)$. -$T$ is Gorenstein. - -Everything in these corollaries follows immediately from the theorem except the second point of the second corollary which follows from the well-known equivalence $T=(I:I) \iff I \text{ invertible}$ when $T$ is Gorenstein (cf. e.g. [3; Proposition 5.8] or [4; Proposition 2.7]). -Quadratic case -[Following the notation in Luc Guyot's answer] -Using the above corollaries we revisit the quadratic case. The discrepancy is invariant under homotheties and so we may assume the ideal $I$ is primitive ($e = 1$). By [5; Lemma 6.5], the ideal $I$ satisfies $R = (I:I)$ if and only if $\gcd(a,b,c) = 1$. Indeed, the formula for the discrepancy in Luc Guyot's answer is precisely $\gcd(a,b,c)$. (By the remark in Luc Guyot's answer, we even have $ds(I) = f/f'$ where $f$ is the conductor of $T$ and $f'$ is the conductor of $(I:I)$.) Thus the formula $ds(I) = c(q_I)$ is consistent with the second corollary. -Upper bound -We will derive an upper bound for $ds(I)$ which is independent of $I$. I assume that $T$ is a domain for simplicity. We may suppose that $T \neq \overline{T}$ and set $S = \overline{T}$. Let $\mathfrak f$ denote the conductor of $T$. -Upper bound: For any T-fractional ideal $I$, $ds(I) \leq |S/T||S/\mathfrak f|.$ -Two $T$-fractional ideals are in the same genus if they are locally isomorphic; equivalently, there exists an invertible T-ideal which multiplies one ideal into the other. -Claim: Any $T$-fractional ideal $I$ is in the same genus as a $T$-fractional ideal $J$ such that $\mathfrak f \subset J \subset S.$ -Proof: Let $P$ be a prime ideal of $T$ and let $S_P$ denote the integral closure of $T$ (integral closure commutes with localization). It suffices to construct a $T_P$-fractional ideal which is isomorphic to $I_P$ such that $\mathfrak f_P \subset J_P \subset T_P$ where subscript denotes tensoring with $T_P$. $S_P$ is a finite product of local Dedekind rings so it is a PID. Hence $I_PS_P = \alpha S_P$ for some $\alpha$ in $Quot(T)$. Let $J_P = \alpha^{-1}I_P$. Then $J_P \subset S_P$, but also $$J_P \supset J_P \mathfrak f_P = J_P S_P \mathfrak f_P = \mathfrak f_P.$$ -Claim: The discrepancy $ds(I)$ is constant on genera. -Proof: This is proven by localizing and using that an invertible ideal of $T$ is locally principal (this latter fact follows from [5; Proposition 2.3]). -Putting these claims together, we have that for $I$ any $T$-fractional ideal, $ds(I) = ds(J)$ for some $T$-fractional ideal $J$ such that $\mathfrak f \subset J \subset S$. From [1; Theorem 1], $|T/J| \leq |S/SJ|$. We also have $S\mathfrak f = \mathfrak f \subset SJ \subset S$, and so $|S/SJ| \leq |S/\mathfrak f|$. Write $M' = M/\mathfrak f$ for any module containing $\mathfrak f$. Putting the inequalities together we have -$$ds(I) = |S/SJ|/|T/J| \leq |S/\mathfrak f|/ |T/J| = |S'|/(|T'|/|J'|) = |S/T| |J/\mathfrak f| .$$ -The last term is bounded from above by $|S/T| |S/\mathfrak f|$. -Conclusion -The discrepancy function satisfies the inequality, $1 \leq ds(I) \leq |\overline{T}/T||\overline{T}/\mathfrak f|$, for any $T$-fractional ideal $I$, and admits an explicit and natural formula in terms of conductors in the quadratic case. -However it appears to be unknown whether the discrepancy function can be given a "closed form" in general (e.g., an expression in terms of the conductor of $T$, the differents or discriminants of $T$ and $\overline{T}$, Ext or Tor groups over $T$ or $\overline{T}$). -References: -[1] I. Del Corso, R. Dvornicich, Relations among Discriminant, Different, and Conductor of an Order, 2000. -[2] A. Fröhlich, Local fields, from J. W. S. Cassels and A. Fröhlich, Algebraic number theory, 1967. -[3] L. Levy and R. Wiegand, Dedekind-like behavior of rings with 2-generated ideals, 1985. -[4] J. Buchmann and H. W. Lenstra, Jr., Approximating rings of integers in number fields, 1994. -[5] V. M. Galkin, $\zeta$-functions of some one-dimensional rings, 1973.<|endoftext|> -TITLE: Is it possible to find a (nonsquare) integer which is a quadratic residues modulo a given infinite list of primes? -QUESTION [8 upvotes]: I'm wondering if it's possible, given a prime p and an infinite list of primes $q_1$, $q_2$, ... to find an integer d which (1) is not a square mod p, but (2) is a square mod $q_i$ for all i. Always, sometimes, never? Probably sometimes --- what are some conditions? In the application I have in mind, the $q_i$ are all the prime divisors of the numbers $p^{2^n}-1$ as n ranges from 1 to infinity, but that's somewhat flexible. -(The application, by the way, involves taking a p-adic interpolation of exponentiation of rational integers, and extending it to rings of integers in towers of number fields.) -[ETA: I forgot to mention that d should also be a square mod 8 for the application, which rules out the answer of -1 given below.] - -For a finite list, d can be constructed using the Chinese Remainder Theorem, but that doesn't seem to help here. - -Given d, quadratic reciprocity gives an infinite set of primes for which d is a square, but I need the primes specified first. - -Grunwald-Wang says, if I understand it correctly, that condition (1) implies that d is not a square modulo $q$ for infinitely many primes $q$, but doesn't say anything about primes which d is a square for. - -The Chebotarov Density Theorem seems to imply that the set of possible d has density zero, but doesn't rule out (or imply) that one such d exists. - - -Thanks for any help, sources, or advice! -----Josh - -REPLY [5 votes]: It depends on the given list of primes. A simpler but necessary condition is that there be a $d$ so that all the primes of the list (greater than $d$) are concentrated in a few congruence classes $\bmod 4d.$ We can stick to odd prime divisors since everything is a quadratic residue $\bmod 2.$ -If the list is all primes congruent to $1 \bmod 4$ then $-1$ is a common quadratic residue. That probably doesn't seem very exciting. -If the list is all odd prime divisors of $3^{2^n}-1$ as $n$ ranges over the positive integers then $-1$ is again a common quadratic residue. That is the kind of thing you were mentioning. But the reason is that all those primes are $1 \bmod 4$ -If I am not mistaken, and for the same reason, $-1$ is a common quadratic residue of of the prime divisors of $p^{2^n}-1$ as $n$ ranges over the integers starting at $2.$ -For certain primes , such as $5,7,17,19,31,53,59$ we can expand the list to all prime divisors of $p^{2^n}-1$ with the exception of $3.$ In general it is sufficient to discard any divisors of $p^2-1$ which are $3 \bmod 4.$ -The facts behind this are - -$p^{2^n}-1=(p-1)(p+1)(p^2+1)(p^4+1)\cdots(p^{2^{n-1}}+1)$ -every odd factor of $p^{2^m}+1$ is of the form $2^{m+1}q+1$ -$-1$ is a quadratic residue for primes which are $1 \bmod 4.$ - - -Think first about this (easy) question. For fixed $d$ what are the odd primes $q$ such that $d$ is a quadratic residue $\bmod q?$ Call this set $G_d.$ We may assume that $d$ is squarefree. -Then the members of $G_d$ are the prime divisors of $d$ along with those primes in a union of certain congruence classes $\bmod 4d.$ Half of the classes $(r \bmod 4d)$ with $\gcd(r,4d)=1$ -In some cases ($d$ even or $d$ odd with all divisors $1 \bmod 4$) it suffices to consider congruence classes $\bmod 2d$. However what is written is still correct. I will ignore your $p$ on the assumption that the goal was to rule out $d$ being a square. -Then the specific $d$ works for a particular instance of your problem, precisely if the chosen list is one of the uncountably many infinite subsets of $G_d.$ -On the other hand, suppose it is given that the members of the list (other than the divisors of $d$ in the list, if any) are chosen from some $k \ll \phi(d)$ of the congruence classes $\bmod 4d$. Then, if the $k$ are chosen at random, the chance that $d$ will work is less than $2^{-k}$. -So starting from a list $\mathbf{q}=q_1,q_2,\cdots$ the first question is "Is there some reason to suspect that there is an $M$ so that all the members of $\mathbf{q}$ (prime to $M$) are concentrated in a few of the congruence classes $\bmod M?$" If that does not happen, then there is no hope. If it does happen for a certain $M,$ then chances still may be low. -So it very much depends on where $\mathbf{q}$ comes from. -By the way, the problem of finding a $d$ which is a quadratic non-residue relative to all $q \in \mathbf{q},$ is equally difficult.<|endoftext|> -TITLE: How to define jet bundles algebraically? -QUESTION [11 upvotes]: Let $B\to X$ be a surjective submersion over the smooth integral scheme $X$ over $\mathbb{C}$. Associated to this we have in the $C^\infty$ world the notion of the $k$ jet-bundles $J_k(B)$, which are affine bundles over $X$. I wonder what the best way to define this notion in the setting of algebraic geometry is. If $B=TX$ there is an algebraic description on Wikipedia which makes sense, but I cannot figure out how to give a satisfactory definition in general. -I am fine with restricting to the case where $B$ is the spectrum of a sheaf of $\mathcal{O}_X$ algebras. -Edit: I am talking here about jet bundles in the sense of e.g. this wikipedia article. - -REPLY [6 votes]: According to Mustață, Jet schemes of locally complete intersection canonical singularities, the $m$-th jet scheme of (the $\mathbb{C}$-scheme) $X$ is the scheme $J_m X$ over $X$ representing the functor $\mathrm{Sch} \to \mathrm{Set}$ given by -$$ S \mapsto \mathrm{Hom}(\Delta^m\times S,X)$$ -where $\Delta^m:=\mathrm{Spec}\frac{\mathbb{C}[t]}{\langle t^{m+1} \rangle}\;.$ -So the closed points of $J_m X$ are the morhisms of schemes $\Delta^m\to X$. And the fiber of $J_m X\to X$ over $x\in X$ has as set of closed points the set $\mathrm{Hom}(\mathcal{O}_{X,x}\;,\mathbb{C}[t]/\langle t^{m+1} \rangle)$. -Another way of seeing it is: let $\mathbf{F}:\mathrm{Sch}\to\mathrm{Sch}$ be the functor $\mathbf{F}(S)=\Delta^m\times S$, than it has a right adjoint given by $X\mapsto J_m X$. -Remark. According to the above construction we have $J_1 X=TX$, the tangent bundle (or total space of tangent sheaf if $X$ isn't smooth).<|endoftext|> -TITLE: How to study to learn differential geometry for applying it to statistics -QUESTION [5 upvotes]: Basically I want to learn information geometry or specifically the application of differential geometry in statistics to do a project. I am from a statistical background and have a knowledge about real analysis, several variable calculus, linear algebra. -One of my professors told me that the first three chapters from Do Carmo's Differential geometry would be sufficient. -Can someone assure me if that's enough or do I need to learn Riemannian geometry. -And If I need to learn Riemannian geometry then what should be my path for learning. -I don't want to learn rigorous mathematics. I just want to apply it to statistics. - -REPLY [2 votes]: I think Do Carmo is a good option. Personally, I'm a fan of John Lee's Introduction to Smooth Manifolds and its sequel Riemannian Manifolds. While these are written at a higher level, they really emphasize the geometric picture at work. -I think the survey by Nielsen is a good article and I found it very helpful to get a broad overview of IG. However, I would not recommend using it to learn differential geometry. Most books about information geometry take a very idiosyncratic approach to geometry, which can give rise to various misunderstandings. These are not a big deal if you are already familiar with differential geometry but are more of a problem if you are trying to learn it. -Both of these works are well worth reading if you are interested in IG, but I'll give an example of what I mean. Both Amari's book and the survey article by Nielsen state that the holonomy of a flat connection is trivial (although they don't use this language). In information geometry, the flat connections of interest are generally on exponential families (where this ends up being true). However, in general, the holonomy of a flat connection is not zero (it's induced by fundamental group). Furthermore, for this result, the connection must be both curvature- and torsion-free (not just curvature-free). Statistical manifolds are generally taken to have torsion-free connections, so this is not an issue in applications. These are relatively minor points if you are familiar with differential geometry, but would be misleading for someone learning it.<|endoftext|> -TITLE: Do distance functionals separate probability measures? -QUESTION [13 upvotes]: Let $(\Omega,d)$ be a compact metric space and $\mathcal P(\Omega)$ its space of Borel probability measures. Let $D=\{ d_p\mid p\in\Omega\}$ where $d_p(x)=d(p,x)$ be the set of all "distance functionals". As usual, we can think of $D$ acting on $\mathcal P(\Omega)$ (or vice versa) via integration i.e. $\langle d_p,\mu\rangle = \int_\Omega d_p(x)\,\mathrm d\mu(x)$. -Title Question - -Does $D$ acting on $\mathcal P(\Omega)$ via integration separate points? - -Or equivalently, - -If $\mu,\nu \in \mathcal P(\Omega)$ and $\langle d_p,\mu\rangle = \langle d_p,\nu\rangle$ for all $p\in \Omega$, then must $\mu=\nu$? - -Alternative Formulations -There are a few other ways to frame the question as well. -Probabilistic Formulation -Rewriting all integrals as expectations the question becomes, - -If $\mathbb E_{X\sim\mu}[d_p(X)] = \mathbb E_{Y\sim\nu}[d_p(Y)]$ for all $p\in \Omega$, then must $\mu=\nu$? - -In other words, does knowing the expected distance to a point for all points determine the measure? -Geometric Formulation -Recall that the 1-Wasserstein distance is given by $W_1(\mu,\nu) = \inf_{\gamma\in\Gamma(\mu,\nu)} \int_{\Omega\times\Omega} d(x,y) \,\mathrm d\gamma(x,y)$ where $\Gamma(\mu,\nu)$ is the set of couplings between $\mu$ and $\nu$ i.e. Borel probability measures on $\Omega\times\Omega$ with marginals $\mu$ and $\nu$ respectively. Since the product measure $\delta_p\otimes\mu$ is the unique coupling between a Dirac delta measure $\delta_p$ and $\mu$, we have that -$$W_1(\delta_p,\mu)=\int_{\Omega\times\Omega} d(x,y)\,\mathrm d(\delta_p\otimes\mu)(x,y)=\int_\Omega d(p,y)\,\mathrm d\mu(y)=\langle d_p,\mu\rangle$$ -Now the question can be stated geometrically as - -If $W_1(\delta_p,\mu)=W_1(\delta_p,\nu)$ for all $p\in \Omega$, then must $\mu=\nu$? - -In other words, does knowing the $W_1$ distance to the extreme points of $\mathcal P(\Omega)$ completely determine the probability measure? -Integral Transform Forumlation -Define the distance transform of $\mu\in\mathcal P(\Omega)$ as the function $\phi_\mu:\Omega\to\mathbb R$ given by $\phi_\mu(p) = \int_\Omega d(p,x)\,\mathrm d\mu(x)$. The question can now be restated as, - -Is the distance transform injective on $\mathcal P(\Omega)$? - -Moreover, by the geometric formulation we have $\phi_\mu(p) = W_1(\delta_p,\mu)$. We will use the weak-$*$ topology for $\mathcal P(\Omega)$ (which coincides with the $W_1$ topology). Since the map $p\mapsto \delta_p$ is an embedding of $\Omega$ into $\mathcal P(\Omega)$, it follows that $\phi_\mu:\Omega\to\mathbb R$ is continuous. Denote the distance transform by $\Phi(\mu)=\phi_\mu$. Since $\mathcal P(\Omega)$ is compact Hausdorff and $C(\Omega)$ is Hausdorff we can restate the question as - -If $\Phi:\mathcal P(\Omega)\to C(\Omega)$ is continuous, is it an embedding? - -Final Thoughts -Are any of these equivalent statements true? I have unfortunately only been able to reformulate the question and have not identified any clear proof, though I wouldn't be surprised if there is an easy one I'm overlooking. The geometric formulation of the problem leads me to believe that $D$ does indeed separate points in $\mathcal P(\Omega)$. However, if the answer is affirmative then I feel the resulting nice properties of $\Phi$ would make it something that would be easy to look up. Any insight would be appreciated. -Update: In light of George Lowther's elegant 4-point counter-example and Pietro Majer's affirmative answer for $\Omega=[0,1]$, it would be interesting to better understand what factors determine whether the underlying metric space yields an affirmative answer. -George's counter-example can be extended to counter-examples where $\Omega$ is a sphere (with intrinsic metric). Thus, requiring $\Omega$ to be positive-dimensional, a manifold, connected, path-connected, simply-connected, etc, will not make the issue go away. On the other hand, Pietro suspects that the answer is again affirmative in the case when $\Omega$ is a compact convex subset of Euclidean space. - -REPLY [7 votes]: On the positive side, the answer is affirmative if $\Omega$ is the unit interval $[0,1]$ with its standard distance. In this case $\phi_\mu$ is a convex $1$-Lipschitz function (in fact, it is also defined for all $p\in\mathbb{R}$, with $\phi'(p)=\mathrm{sgn} p$ for $p\notin[0,1]$), with left and right derivatives -$$\phi_-'(p)=\mu[0,p)-\mu[p,1]= 2\mu[0,p)-1$$ -$$\phi_+'(p)=\mu[0,p] -\mu (p,1]= 2\mu[0,p] -1=1-2\mu(p,1]$$ -so that $\mu$ is determined on all intervals, hence on all Borel subsets. -Conversely, note that any convex function $\phi$ as above -may be written in the form $\phi(p)=\int_{[0,1]}|t-p|dm(t)$ for some Borel probability measure $m$ on $[0,1]$. This because $g:= \frac{1}{2}\big(1-\phi_+'\big) $ is a nonnegative bounded cadlag function, so there is a Borel probability function $m$ such that $g(p)=m(p,1]$, whence $\phi(p)=\int_{[0,1]}|t-p|dm(t)$ follows easily from the above relations. -I'd guess the answer is also affirmative for $\Omega$ a convex compact set of $\mathbb{R}^n$ with the Euclidean distance.<|endoftext|> -TITLE: Any continuous map is homotopic to one assuming fixed values at finitely many points -QUESTION [5 upvotes]: Let $X$ and $Y$ be topological spaces. Assume $X$ is locally contractible and has no dense finite subset. Assume $Y$ is path-connected. -Given $n$ pairs of points $(x_i, y_i)$ where $x_i\in X$ and $y_i\in Y$ for $1\leq i\leq n$ and a continuous map $f:X\to Y$ can we find a continuous map $g:X\to Y$ homotopic to $f$ such that $g(x_i)=y_i$? - -REPLY [13 votes]: Let $X$ be the real line with a doubled origin and $Y$ be $\Bbb R$, and let $f$ be the projection map that collapses the two origins $0^+$ and $0^-$ to $0$. Then any map $g: X \to Y$ satisfies $g(0^+) = g(0^-)$ because $\Bbb R$ is Hausdorff. Therefore, $f$ is not homotopic to any map that sends these two points to distinct ones. -Your question is closely related to the inclusion $\{x_1,\dots,x_n\} \subset X$ having the homotopy extension property. In particular, if it is the inclusion of a neighborhood deformation retract, then such homotopies exist. In the example above, each point individually has a contractible neighborhood but the two origins together do not have a neighborhood that retracts back onto them.<|endoftext|> -TITLE: Coefficients of the characteristic polynomial of the map on algebraic de Rham cohomology -QUESTION [9 upvotes]: Let $k$ be an algebraically closed field. Let $V$ be a smooth projective variety over $k$. For a map $\phi:V\to V$, do the coefficients of the characteristic polynomial of $\phi^*:H^ i_{dR}(V/k)\to H^ i_{dR}(V/k)$ lie in the prime subfield of $k$? - -REPLY [5 votes]: Suppose that $\mathrm{char}\, k=p>0$. It is easy to give an example of a stack $V$ with an endomorphism that violates this property. Take $V=B\alpha_p$, the scaling action of $\mathbb{G}_m$ on $\alpha_p\subset \mathbb{G}_a$ induces an action on $B\alpha_p$ so, in particular, the group $k^{\times}$ acts on $V$. By Proposition 4.12 of https://arxiv.org/pdf/1909.11437.pdf an element $\lambda\in k^{\times}$ acts on the $1$-dimensional space $H^1_{dR}(V/k)$ by $\lambda^p$ so any $\lambda\in k^{\times}\setminus\mathbb{F}_p$ gives an example of an endomorphism with characteristic polynomial with coefficients outside of $\mathbb{F}_p$. -Let's now approximate this example by an actual smooth projective variety. Choose a finite subgroup $\Gamma\subset k^{\times}$ not contained in $\mathbb{F}_p^{\times}$ and consider the group scheme $G=\Gamma\ltimes \alpha_p$ where the semi-direct product is formed using the scaling action of $\Gamma\subset \mathbb{G}_m(k)$ used above. There exists a smooth complete intersection $X$ of dimension $3$ with a free action of $G$. Put $V=X/\alpha_p$ which is equipped with the remaining action. The $\Gamma$-equivariant morphism $V\to B\alpha_p$ induces a $\Gamma$-equivariant isomorphism $H^1_{dR}(B\alpha_p)\simeq H^1_{dR}(V)$ by weak Lefschetz so any $\varphi\in\Gamma\setminus \mathbb{F}_p^{\times}$ gives a counterexample. For a reference see e.g. proof of Theorem 1.2 in https://arxiv.org/pdf/1909.11437.pdf , the original reference for the construction of $X$ is Proposition 4.2.3 in http://www.numdam.org/article/AST_1979__64__87_0.pdf -These cohomology classes come from torsion in crystalline cohomology and this is in fact necessary to get a counterexample: the characteristic polynomial of $\phi$ on the subquotient $$(H^i_{cris}(V/W(k))/\text{torsion})/p$$ of the space $H^i_{dR}(V/k)$ coincides with that on $H^i_{cris}(V/W(k))/\text{torsion}$ which has integer coefficients by Theorem 2(ii) of Katz-Messing's "Consequences of the Riemann Hypothesis".<|endoftext|> -TITLE: Extensions of (semi-)abelian schemes -QUESTION [6 upvotes]: Let $S$ be a regular Noetherian scheme, and let $U\subset S$ be the complement of a divisor. If $A\to B$ is an isogeny of abelian schemes over $U$, and $A$ extends to a semi-abelian scheme over $S$, does $B$ also extend to a semi-abelian scheme over $S$? -I'm happy to assume that the degree of the isogeny is invertible in $\mathcal{O}_S$. - -REPLY [2 votes]: Under the hypothesis that the degree of the isogeny is invertible in $S$, the answer is yes; I attempted to sketch a proof below under slightly more general assumptions on the base. [On the other hand, if the degree of the isogeny is not invertible in $S$, the answer is no. There are counterexamples over regular rings of dimension $2$. You can look at page 192 of the book by Faltings-Chai for a counterexample in equal characteristic $p$; or to chapter 6 of De Jong-Oort's "On extending families of curves" for an example in mixed characteristic.] -Statement: let $S$ be a normal noetherian integral scheme, $U\subset S$ a non-empty open, $f\colon A\to B$ an isogeny of $U$-abelian schemes, of degree $m$ invertible on $S$. Suppose $A$ extends to a semiabelian scheme $\mathcal A/S$; then $B$ extends to a semiabelian scheme $\mathcal B/S$, and $f$ extends to an isogeny $\mathcal A\to \mathcal B$ (i.e. a fibrewise finite surjective homomorphism). -Proof: the $m$-torsion $\mathcal A[m]$ is a quasi-finite etale group scheme over $S$, and the inclusion $\mathcal A[m]\to \mathcal A$ is a closed immersion, as semiabelian schemes are separated. -Let $K/U$ be the kernel of $f$, a finite etale group scheme. The closed immersion $K\to A[m]$ is etale hence an open immersion. We can therefore write $A[m]$ as a disjoint union of schemes $K\sqcup Z$. -Claim: the schematic closure of $K$ inside $\mathcal A[m]$ (or $\mathcal A$), is etale over $S$. Proof of claim: Write $\mathcal A[m]$ as the disjoint union $Y_1\sqcup Y_2\sqcup \ldots Y_n$ of its connected components. As $S$ is geometrically unibranch, so is $\mathcal A[m]$, hence each $Y_i$ is also an irreducible component. In particular, the restrictions $Y_{1 ,U},\ldots,Y_{n,U}$ are the irreducible (and connected) components of $A[m]$, and by reordering the indexes we can write $K=Y_{1,U}\sqcup \ldots \sqcup Y_{k,U}$, $Z=Y_{k+1,U}\sqcup\ldots\sqcup Y_{n,U}$. It follows that $\mathcal A[m]$ is the disjoint union of $\overline K$ and $\overline Z$, the schematic closures of $K$ and $Z$. This proves the claim. -It is clear that the multiplication map of $\mathcal A$ maps $\overline K\times \overline K$ inside $\overline K$, hence $\overline K$ is a subgroup of $\mathcal A$. The sheaf quotient $\mathcal B=\mathcal A/\overline K$ is a smooth group algebraic space extending $B$; it is separated because quotient of a separated group scheme by a closed subgroup; one argues easily that the fibres are semiabelian varieties. It remains to show that it is a scheme; here you can apply theorem XI 1.13 from Raynaud's "Faisceaux amples sur les schemas en groupes et les espaces homogenes" which shows that $\mathcal B/S$ is quasi-projective. It is here that the hypothesis that $S$ is normal, noetherian, integral is used.<|endoftext|> -TITLE: Characterisation of finite dimensional C*-algebras? -QUESTION [7 upvotes]: $\DeclareMathOperator\Spec{Spec}$Let $A$ be a finite dimensional $*$-algebra over $\mathbb C$. -(Namely, an associate algebra equipped with an involution $*:A\to A$ satisfying $(ab)^*=b^*a^*$ and $(\lambda a)^*=\bar\lambda a^*$.) - -Assume that for $\forall a\in A$ we have $\Spec(a^*a)\subset\mathbb R_+$. Does it follow that $A$ is a C*-algebra? - -Here, the spectrum $\Spec(x)$ of an element $x$ is the set of scalars $\lambda\in \mathbb C$ such that $x-\lambda$ is not invertible. - -REPLY [7 votes]: Let $V$ be a complex vector space equipped with an involutive anti-linear star operation (e.g. a C*-algebra whose multiplication has been forgotten). Equip $V$ with the identically zero multiplication, namely -$xy=0$ -for all $x$ and $y$ in $V$. Then the unitization of $V$ is a counter-example. In fact, every element $a$ of $V$ is nilpotent so $\text{spec}(a) = \{0\}$. Consequently the spectrum of any element of the form $a-\lambda$ is $\lambda$ from where one easily checks the required condition. -However $a^*a=0$ for every $a$ in $V$, so $\tilde V$ cannot possibly be a C*-algebra.<|endoftext|> -TITLE: Why should the number of $\mathbb{F}_q$ points on degree $d$ curves $C\subset \mathbb{P}_{\mathbb{F}_q}^n$ decrease as $n$ increases? -QUESTION [17 upvotes]: This question concerns some counterintuitive results (to me at least) regarding the number of points on a projective curve over a finite field. Namely, if one fixes the degree of the curve, but increases the dimension of the ambient projective space, one can get tighter bounds on the number of $\mathbb{F}_q$ points on the curve, despite there being a larger number of $\mathbb{F}_q$ points in the ambient space. Let me make this more precise with two examples. -Let $C\subset \mathbb{P}^n_{\mathbb{F}_q}$ be a projective curve of degree $d$. Suppose $C$ is nondegenerate in the sense that it is not contained in any smaller projective space $\mathbb{P}^k_{\mathbb{F}_q}$, $k2$; recently Beelen and Montanucci show that if $C\subset \mathbb{P}^3_{\mathbb{F}_q}$ is nondegenerate then in fact -$$ -\#C(\mathbb{F}_q)\leq (d-2)q+1. -$$ -They further conjecture than if $C\subset \mathbb{P}^n_{\mathbb{F}_q}$, the general bound should be -$$ -\#C(\mathbb{F}_q)\leq (d-n+1)q+1. -$$ -This is reminiscent of a phenomenon from work of Bucur and Kedlaya. For example: a random smooth curve in $\mathbb{P}^2_{\mathbb{F}_q}$ is expected to have -$$q+1$$ -points over $\mathbb{F}_q$ as its degree grows to infinity. A random complete intersection of two smooth degree $d$ surfaces in $\mathbb{P}^3_{\mathbb{F}_q}$ is expected to have -$$ -q+1 - \frac{q^{-2}(1+q^{-1})}{1+q^{-2}-q^{-5}} < q+1 -$$ -points over $\mathbb{F}_q$, again as $d\to\infty$. -These results are counterintuitive to me, as the number of points in the ambient projective space grows (exponentially) as $n$ does, so in particular it seems to me that it should be easier for curves to have $\mathbb{F}_q$ points when they are embedded in larger projective spaces. Does anyone have any intuition as to why the opposite should be true? -References: -Beelen and Montanucci: A bound for the number of points of space curves over finite fields -Bucur and Kedlaya: The probability that a complete intersection is smooth -Homma: A bound on the number of points of a curve in projective space over a finite field - -REPLY [8 votes]: One way to get some intuition comes from looking at the (weaker) combinatorial bound. Suppose you had a nondegenerate curve $C$ in some projective space $\mathbb P^n$. Suppose that that $L$ is a subspace of codimension $2$ in $\mathbb P$ and that $|C\cap L|=m$. The higher the dimension $n$ gets, the higher value we are allowed to pick for $m$. Indeed we can always find at least $n-1$ points in $C$ that span a $\mathbb P^{n-2}$. -Bezout tells you that for any hyperplane $H$ that contains $L$, the number of points of $C$ that lie in $H$ and don't lie in $L$ is at most $d-m$. Since the number of such hyperplanes is $q+1$, independent of the dimension, we get -$|C|-m\le (q+1)(d-m)$ -or equivalently from rearranging terms -$$|C|\le (d-m)q+d.$$ -For $m=n-1$ this gives the bound $|C|\le (d-n+1)q+d$ for all nondegenerate curves $C$. Of course this is weaker than the conjecture and theorems that you mention in the post, but (1) it holds true for all curves including the one that violates the Sziklai bound (2) it already exhibits the phenomenon "bound gets tighter as $n$ goes up".<|endoftext|> -TITLE: Reference request: Diophantine equations -QUESTION [9 upvotes]: I am looking for a textbook, or preferably lectures, on the subject of Diophantine equations. I am familiar with the basic principles of modular arithmetic, conics and the Hasse Principle, and the basics of elliptic curves, Mordell's Theorem etc (though I'm not up to the point where I can understand the proof). -What I need is something that takes me beyond the basics. Something which will teach me the advanced theory, and also teach me about diophantine surfaces (not just curves). - -REPLY [3 votes]: To the books mentioned above I would add one more: - -Rational and Nearly Rational Varieties (Cambridge Studies in Advanced Mathematics) by J. Kollár, K. E. Smith, and A. Corti. - -The authors present a more or less elementary approach to the rationality questions using a mix of classical and modern methods.<|endoftext|> -TITLE: Balls in Hilbert space -QUESTION [16 upvotes]: I recently noticed an interesting fact which leads to a perhaps difficult question. If $n$ is a natural number, let $k_n$ be the smallest number $k$ such that an open ball of radius $k$ in a real Hilbert space of sufficiently large dimension or infinite dimension contains $n$ pairwise disjoint open balls of radius 1. (The dimension of the -Hilbert space is irrelevant as long as it is at least $n-1$ since it can be replaced by the affine subspace spanned by the centers of the balls.) We obviously have $k_1=1$ and $k_2=2$, and it is easy to see that $k_3=1+\frac{2}{\sqrt{3}}\approx 2.1547$. The interesting fact is that $k_n\leq 1+\sqrt{2}\approx 2.414$ for all $n$, since in an infinite-dimensional Hilbert space an open ball of this radius contains infinitely many pairwise disjoint open balls of radius 1 [consider balls centered at points of an orthonormal basis]. The obvious questions are: (1) What is $k_n$? This may be known, but looks difficult since it is related to sphere packing. (2) Is $k_n$ even strictly increasing in $n$? (3) Is $k_n<1+\sqrt{2}$ for all $n$, or are they equal for sufficiently large $n$? (4) Is it even true that $\sup_n k_n=1+\sqrt{2}$? It is not even completely obvious that $k_n$ exists for all $n$, i.e. that there is a smallest $k$ for each $n$, but there should be some compactness argument which shows this. I find it interesting that the numbers $1+\frac{2}{\sqrt{3}}$ and $1+\sqrt{2}$ are so close but the behavior of balls is so dramatically different. -I suppose the question is also interesting in smaller-dimensional Hilbert spaces: let $k_{n,d}$ be the smallest $k$ such that an open ball of radius $k$ in a Hilbert space of dimension $d$ contains $n$ pairwise disjoint open balls of radius 1. Then $k_{n,d}$ stabilizes at $k_n$ for $d\geq n-1$. What is $k_{n,d}$? (This my be much harder since it is virtually the sphere-packing question if $n>>d$.) - -REPLY [8 votes]: For convenience of notation, let me write the expectation $\mathop{\mathbb{E}}_i t_i$ to denote the average $(\sum_{i=1}^n t_i)/n$. -If I understand your construction correctly, you have disjoint balls of radius $1$ centered at $x_i = \sqrt{2} e_i$ contained in a ball of radius $1+\sqrt{2}$ centered at $y = 0$. This construction, which places $n$ balls tightly packed at the vertices of a regular simplex, is optimal in terms of the positions $x_i$. For the exact optimal bound for your problem, you should pick $y=\mathop{\mathbb{E}}_i x_i$ to get the radius $$\boxed{k_n = 1+\sqrt{2 (1-1/n)}}.$$ -The claim that placing the $x_i$ at the vertices of a regular $(n-1)$-simplex and $y$ at the centroid of this simplex is optimal has been proven many times before in many different contexts. For example, it is implied by a bound known by various substrings of "the Welch-Rankin simplex bound" in frame theory. Here's a simple direct proof: -By the triangle inequality, a ball of radius $1+r$ centered at $y$ contains a ball of radius $1$ centered at $x_i$ iff $\lVert x-y\rVert \le r$. Two balls of radius $1$ centered at $x_i$ and $x_j$ are disjoint iff $\lVert x_i - x_j \rVert \ge 2$. Therefore, your problem asks to minimize $1 + \max_i \lVert y-x_i\rVert$ subject to $\min_{i\ne j} \lVert x_i - x_j\rVert \ge 2$. -Working with squared distances is easier. The maximum squared distance $\max_i \lVert y-x_i\rVert^2$ is surely at least the average $\mathop{\mathbb{E}}_i \lVert y-x_i\rVert^2$. This average is minimized when $y$ is itself the average $\mathop{\mathbb{E}}_i x_i$, in which case it equals $\mathop{\mathbb{E}}_i \mathop{\mathbb{E}}_j \lVert x_i-x_j\rVert^2/2$. Each term where $i=j$ contributes $0$ to this expectation, while each term where $i\ne j$ contributes at least $2$, so overall this expectation is at least $2(n-1)/n$. Thus the maximum squared distance $\max_i\lVert y-x_i\rVert^2$ is at least $2(n-1)/n$ and thus $1+r \ge 1+\sqrt{2(n-1)/n}.$ We can check that the optimal configuration mentioned before achieves this bound either by direct calculation or by noting that it achieves equality in every step of our argument.<|endoftext|> -TITLE: Write $(4x+1)^n$ as the linear combination of certain polynomials -QUESTION [6 upvotes]: Let -$$ -P_m(x):=\begin{cases}4x+1\quad&\ \text{if}\ m=1,\\ -0\quad&\ \text{if}\ m=2,\\ -8x^m+(x+1)^{m-3}(2x+1)^3\quad&\ \text{if}\ m\geq3.\end{cases} -$$ -How to prove that for any positive odd integer $n$, there exist integers $a_1^{(n)},a_2^{(n)},\ldots,a_n^{(n)}$ such that -$$ -(4x+1)^n=\sum_{k=1}^n a_k^{(n)}P_k(x). -$$ - -REPLY [14 votes]: I enjoyed very much this question! My solution contains two ideas, each of which addresses one of two distinct subproblems: - -show that the coefficients $a_m^{(n)}$ are integers; -show that the coefficients $a_m^{(n)}$ exist. - -The subproblem (1) is not completely obvious because the polynomials $P_m$ are not monic. -However we are lucky here! The simple idea here is to renormalize the variable as $x=y/4$ and clear the denominators appropriately, noticing that there is a fortunate 2-adic coincidence that makes this approach work. -The subproblem (2) is not obvious and quite annoying because $P_0=P_2=0$. In other words, there is no nonzero $P_m$ of degree $0$ and $2$, so a priori the linear elimination may leave us with a remainder of degree at most 2. The idea to go here is more hidden (and nicer!). It starts by essentially rewriting the problem as -$$ -(4x+1)^n = 8x^3 A(x) + (2x+1)^3 A(x+1) -$$ -and finding some hidden symmetry. It may even be the case that this observation may be useful to find the coefficients $a_m^{(n)}$ explicitly. - -So, let's start the proof. After the change of variable $x=y/4$ the polynomial $(4x+1)^n$ becomes $(y+1)^n$, which has integer coefficients in the variable $y$. -Instead the polynomials $P_m(x)$ for $m\geq 3$ become: -$$ -P_m(y/4) = \frac 8 {4^m} y ^m + \left(\frac{y+4} 4 \right)^{m-3} \left(\frac{y+2} 2 \right)^3. -$$ -Now it is tempting to multiply over by $4^m/8$ to clear all the denominators, but instead we will multiply by $4^{m-2}$, that is, one less factor of 2 than the greedy one. -So, for $m\geq 3$ define $Q_m(y):=4^{m-2} P_m(y/4)$ and we compute that -$$ -Q_m(y) = \frac 1 2 y^m + (y+4)^{m-3}\frac{(y+2)^3} 2 -$$ -Now, you should note, expanding the binomials and the products, that all the terms $y^k$ with $k -TITLE: Is $C^{*}$-algebra the most modern way to study QFT? -QUESTION [7 upvotes]: I am not an expert on either QFT or $C^{*}$-algebras, but I'm trying to learn the basics of QFT. In all books/papers and other materials that I know, QFT is studied mainly using a lot of functional analysis and distribution theory, but I know that some algebraic constructions are also being used, and in this context $C^{*}$-algebras seem to be the most modern tool. So, what should an inexperienced student like me know about these approaches to QFT and statistical mechanics? What's the role of $C^{*}$-algebras and other algebraic methods in those theories? What are the problems they fit better? If I'd like to study QFT, do I have to learn $C^{*}$-algebra? Are there problems in which algebraic methods don't fit well? Are there problems in which either approach is fruitful? What does one lose by not knowing these algebraic constructions? -ADD: I work with rigorous statistical mechanics but I'm trying to learn some QFT because...well, these are two related areas at some level. However, I don't know yet what or how much I need to learn about QFT. I have a background in functional analysis and distribution theory, but not in $C^{*}$-algebra. As an unexperienced student, it will be very useful to get a general picture, i.e. what are the problems one is trying to solve in QFT and where do each of these approaches come into play. I think each of these tools are applicable for different kinds of problems or even different subareas of the theory, but I don't know for sure. - -REPLY [7 votes]: My PhD work used C*-algebras quite heavily, so I guess I can claim some expertise there, but I'm not an expert in QFT. That will be the main perspective of my answer. -A good starting point for this discussion is the Stone-von Neumann theorem, a foundational result in both operator algebras and quantum mechanics. The setup is basically the Heisenberg uncertainty principle, which asserts that the operations of measuring the position $x$ and the momentum $p$ of a quantum system don't commute: -$$[x,p] = 2\pi i h$$ -An important mathematical question about quantum mechanics in its early history was: what kind of objects are $x$ and $p$? Physicists want them to be self-adjoint operators on some Hilbert space, but you can prove rigorously that no pair of bounded operators have this property. This result belongs to the representation theory of Lie algebras - essentially, the Lie algebra with two generators and the relation above has no representation by bounded self-adjoint operators on Hilbert space. -Stone and von Neumann's idea was to focus on the Lie group rather than the Lie algebra; the relation above is the derivative at 0 of the following relation between time evolution operators $U(t)$ and $V(s)$: -$$U(t) V(s) = e^{-ist} V(s) U(t)$$ -The Lie group generated by such $U$ and $V$ is called the Heisenberg group, and the Stone-von-Neumann theorem asserts that that this group has a unique unitary representation on Hilbert space, up to unitary equivalence (and some adjectives that I won't go into here). This provides a nice foundation for basic quantum mechanics which unifies the Heisenberg and Schrodinger pictures of the theory into one set of axioms. -To handle more complicated quantum systems, we need to generalize to more operators satisfying possibly more complicated relations. Here's how this generalization works: - -Start with a locally compact group $G$; for the original Stone-von-Neumann theorem, $G = \mathbb{R}$. -The Fourier transform determines and isomorphism $C^*(G) \to C_0(\hat{G})$, where $C^*(G)$ is the group C*-algebra and $\hat{G}$ is the Pontryagin dual. -Such an isomorphism is equivalent to a unitary representation of the crossed-product algebra $C_0(G) \rtimes G$. -All irreps of this C*-algebra are unitarily equivalent. - -So now we have quantum mechanics for systems with many particles. But what about QFT? The basic reason why QFT is hard, as I understand it, is that the Stone-von-Neumann theorem is no longer true. -For ordinary quantum mechanics, the classical phase spaces are finite dimensional manifolds - for instance, the classical phase space of a single particle flying around in $\mathbb{R}^3$ is $\mathbb{R}^6$. The classical analog of the phase space in quantum field theory, however, is the space of paths in $\mathbb{R}^3$, which is some sort of infinite dimensional manifold. This means infinitely many operators with infinitely many commutation relations, and the corresponding infinite dimensional Lie groups, to the extent that they even exist, have a much more complicated representation theory. -So now I can try to answer your question. Operator algebras were more or less invented in order to provide a nice model for quantum mechanics. The nice property that this model has - namely, that there is only one realization of it up to unitary equivalence - is no longer true in QFT. So one (implicit) goal of a lot of work in QFT is to cope with this situation and search for better foundations. I have no idea if C*-algebras are the best or most modern way to think about QFT - probably not - but a good place to start for a student is to learn the Stone-von-Neumann theorem in some reasonable generality since we can blame a lot of the difficulty of QFT on its absence.<|endoftext|> -TITLE: Non-alternating knot diagrams -QUESTION [5 upvotes]: (I asked this on -MSE -a few days ago without a clear resolution.) -Start with a closed, self-intersecting curve, where every crossing is -transverse. Now form something like the opposite of an alternating knot -diagram as follows. Starting anywhere, traverse the curve, and -at each previously unvisited crossing, go over/above. If the crossing has been previously visited, leave the assigned crossing designation. -Two examples are shown below. (a) is clearly the unknot. -(b) is also the unknot, perhaps not as obviously. - - -Red circle indicates starting point, arrow the traversal direction. - -I expected these diagrams to obviously represent the unknot, but -I am not seeing a clear proof. So: - -Q. Prove (or disprove) that such a knot diagram always represents -the unknot. - -REPLY [6 votes]: Let us parametrize the plane curve by $\gamma:[0,1]\to\mathbb R^2$ and assume $\gamma(0)=\gamma(1)=(0,0)$. Then your curve is the knot diagram of the knot which is parametrized by $K:[0,2]\to\mathbb R^3$ given by -$$K(t)=\begin{cases}(\gamma(t),1-t)&\text{if }t\leq 1,\\(0,0,t-1)&\text{if }t>1.\end{cases}$$ -(essentially, imagine suspending your knot on a stick, such that the rope goes down at a uniform speed.) Then we can "unwind" this knot. Namely, since $\gamma$ only goes through $(0,0)$ at the endpoints, we can write $\gamma(t)$ in polar coordinates by $(r(t),\phi(t))$ with $r,\phi$ continuous on $(0,1)$. We can then unknot $K$ by the following sequence of knots $K_s$, which starts with an unknot and ends with $K$, written in cylindrical coordinates: -$$K_s(t)=\begin{cases}(r(t),s\phi(t),1-t)&\text{if }t\leq 1,\\(0,0,t-1)&\text{if }t>1.\end{cases}$$<|endoftext|> -TITLE: Definability of ordinals in various signatures -QUESTION [6 upvotes]: Recently, I've been studying what the definable subsets of the countable ordinals "look like" from the perspective of bare-bones first order logic (not set theory) equipped with various ways to "access" the structure of the ordinals. -For example, we may have a signature consisting only of a 2-arity relational symbol $S$ which we interpret in a structure $\mathcal{A}$ with underlying set $\omega_1$ as the set of $(\alpha,\beta)$ such that $\beta$ is the successor of $\alpha$. We can then ask questions about which subsets of $\mathcal{A}$ are definable by first-order logic sentences with this signature, where a subset $S\subset\mathcal{A}$ is considered definable if there is a first order logic sentence $\phi(x)$ for which the set of satisfying assignments of $x$ is $S$. In our example, we can define the set of all countable successor ordinals via the formula $\exists y:S(y,x)$. -We can also ask questions like "what is the smallest ordinal $\alpha$ such that $\alpha$ is undefinable in the sense that $\{\alpha\}$ is undefinable" and such. In the example above, it's clear to see that in fact no ordinal is definable, so the smallest undefinable ordinal is zero. I am particularly interested in how the smallest undefinable ordinal grows as we have stronger and stronger signatures. For example, I have been able to convince myself that with the signature $\{<\}$ with the obvious interpretation in $\omega_1$ as the "less than relation", the smallest undefinable ordinal is $\omega^\omega$ (though I haven't written my argument out formally yet). -My question is: has anyone studied questions like these? Is it known what the smallest definable ordinal is for various other signatures, like $\{ADD(x,y,z)\}$ which is true on all $x,y,z$ so that $x+y=z$, or even other signatures with multiplication, exponentiation, veblen functions, or more? Are there any known generalizations of these ideas? Any help or related literature would be appreciated. - -REPLY [6 votes]: I do not have enough reputation to add a comment. The following paper may be useful for you. It contains results extending the work of Tarski, Mostowski, and Doner, as well as some very nice historical overview and references. -Buchi, Siefkes - The Complete Extensions of the Monadic Second Order Theory of Countable Ordinals. -Weak monadic second order logic appears already in Ehrenfeucht's original work. Even if you are exclusively interested in first order results, (weak) monadic second order logic can play a role. -For example the first order theory of ordinal addition coincides with the first order theory of ordinal addition inside $\omega^{\omega^{\omega}}$ (by Ehrenfeuct), while $(\omega^{\omega^{\omega}},+)$ is a reduct of a generalised power of $(\omega,+)$ with 'exponent' being the weak monadic second order version of $(\omega^{\omega},<)$ (the Feferman-Vaught theorem is the correct tool to understand this). For more details there is Thomas - Ehrenfeucht, Vaught, and the decidability of the weak monadic theory of successor, the details here are all correct but I think the conclusions have some issues. -There is also more recent work on the automata side such as Cachat - Tree Automata Make Ordinal Theory Easy. I know nothing about the content of this but if you want a comprehensive overview of the area, this is maybe a starting point.<|endoftext|> -TITLE: Stability of displacement interpolation in optimal transport -QUESTION [7 upvotes]: Let $(X,d)$ be a complete separable metric space, and let $(\mathcal{P}_2 (X), W_2)$ be the space of probability measures on $X$ with finite second moments, equipped with the 2-Wasserstein distance. It is known that discrete measures are dense inside $(\mathcal{P}_2 (X), W_2)$ - namely, given any $\mu \in \mathcal{P}_2 (X)$, and $\delta>0$, one can find a discrete measure $\mu_\delta$ with $W_2 (\mu, \mu_\delta)<\delta$. -Now, let $\mu_0, \mu_1 \in \mathcal{P}_2 (X)$, and let $\mu_t$ be a $W_2$ geodesic connecting $\mu_0$ and $\mu_1$ (a.k.a. $\mu_t$ is a [not necessarily unique] displacement interpolation between $\mu_0$ and $\mu_1$). Is the displacement interpolation stable under discrete approximation? That is, can one pick discrete $\mu_{0,n}, \mu_{1,n}$ such that $\mu_{t,n}$ is close to $\mu_t$ for all $t\in[0,1]$? - -REPLY [3 votes]: The displacement interpolation $\mu_t$ should not be fixed a priori, due to nonuniqueness of Wasserstein Geodesics. Thus, the correct question should be: fix the approximating sequences $(\mu_{0,n}),(\mu_{1,n})$ and $W_2$ geodesics $\mu_{t,n}$, and ask if there exists one $\mu_t$ close to $\mu_{t,n}$ for $t \in [0,1]$.<|endoftext|> -TITLE: Conceptual explanation for the sign in front of some binary operations -QUESTION [8 upvotes]: In several situations, I've seen that given a binary operation on a graded module $m:A\otimes A\to A$, a new operation $M(x,y)=(-1)^{|x|}m(x,y)$ is defined so that it satisfies some properties. -One example of this happens in Homotopy G-algebras and moduli spaces, where for a binary operation $m\in\mathcal{O}(2)$ such that $m\circ m=0$ for some operad $\mathcal{O}$, an associative product is defined by $xy=(-1)^{|x|+1}m\{x,y\}$, where the brace notation stands for the brace algebra structure on $\mathcal{O}$. In this case, the explanation I've been able to deduce is that this is necessary for the brace relation (equation (2) in the paper) to imply associativity of the product $xy$. In this case the sign $(-1)^{|x|}$ works for this purpose too. -Another more direct instance of this situations occurs in Cartan homotopy formulas and the Gauss-manian connection in cyclic homology, where given an $A_\infty$-algebra with $m_i=0$ for $i>2$, one gets a dg-algebra by defining again $xy=(-1)^{|x|}m_2(x,y)$. In this case this is because the author uses a convention for $A_\infty$-algebras in which the equations only have plus signs, so some extra sign is needed to produce the associativity relation and the Leibniz rule. So the reasons are very similar to the previous case even though the construction is simpler because there is no brace algebra here. -And another extra example for which I don't have any reference is in the case of Lie algebras. When one defines a generator of the operad of graded Lie algebras, often one takes $l(x,y)=(-1)^{|x|}[x,y]$ instead of directly defining $l$ as the bracket. If I remember correctly this was needed to obtain the Jacobi identity in purely operadic terms. - -So it looks like it's very common to add that sign in order to make some relations hold. What I would like to know if there is a more conceptual explanation of why this holds systematically. Maybe it's just that it works when writing down the equations, but I'm looking for a more general intuition. -My motivation is generalizing this idea to maps of higher arity. More precisely, given an $A_\infty$-multiplication $m\in\mathcal{O}$ such that $m\circ m=0$, I want to define an $A_\infty$-structure $M$ on $\mathcal{O}$ that satisfies the sign convention -$$\sum_{n=r+s+t}(-1)^{rs+t}M_{r+1+t}(1^{\otimes r}\otimes M_s\otimes 1^{\otimes t})=0.$$ -(There is also another possible convention where $rs+t$ is replaced by $r+st$) -So this is very similar to Getzler's paper where he defines $M_j(x_1,\dots, x_j)=m\{x_1,\dots x_j\}$, and this structure maps satisfy the relation $M\circ M=0$ but with all plus signs. So I need to modify these maps by some signs in a similar way as the associative case. Of course I can try to sit down and write the equations and find some necessary conditions for the signs and maybe find a pattern. But if there is a conceptual explanation for the associative case and the lie algebras, then maybe there is an easier way to find out what the signs I need are. - -REPLY [4 votes]: As Gabriel C. Drummond-Co commented, it has to do with suspensions that are implicit. I'll do it with the example of Gerstenhaber and Voronov and the others should follow similarly. Let us denote $M_2(x,y)=x\cdot y$ the product that we want to define based on the brace $m\{x,y\}$. If we define it as a map $(s\mathcal{O})^{\otimes 2}\to s\mathcal{O}$ (suspension as graded vector spaces), then the natural thing to do is using the brace $m\{-,-\}:\mathcal{O}^{\otimes 2}\to \mathcal{O}$, but to do so one has to compose with suspensions and desuspensions. Namely, $M_2(x,y)=s(m\{(s^{-1}x,s^{-1}y)\})$. And it's applying $(s^{-1})^{\otimes 2}(x,y)$ what makes the sign $(-1)^{|x|}$ appear. If we use $(s^{\otimes 2})^{-1}$ instead then we get the original sign $(-1)^{|x|+1}$.<|endoftext|> -TITLE: On a certain expansion in term of Schur functions -QUESTION [6 upvotes]: This question is related to this other one -A Schur positivity conjecture related to row and column permutations -by Richard Stanley (thanks to Sam Hopkins for letting me know about it). -Consider a Young subgroup $S_{\lambda}$ of the symmetric group $S_n$, corresponding to some integer partition $\lambda$ of $n$. Let $\tau$ be some permutation and define the symmetric function -$$ -F(\tau)=\sum_{\sigma\in S_{\lambda}}p_{c(\tau\sigma)} -$$ -where $p_{\mu}$ is the usual power sum symmetric function and $c(\rho)$ denotes the integer partition given by the cycle-type of the permutation $\rho$. -Q: What is known about the Schur function expansion of $F(\tau)$, given the double coset class of $\tau$ for the Young subgroup? - -REPLY [4 votes]: One fact is that $F(\tau)$ is Schur positive if and only if $\tau\in S_\lambda$. More generally, if $K$ is any coset (left or right) of any subgroup $G$ of $S_n$, then $\sum_{\sigma\in K}p_{c(\sigma)}$ is Schur positive if and only if $K=G$. The only known proof for the "if" part requires representation theory; see Enumerative Combinatorics, vol. 2, page 396. For the "only if" part, it is easy to see that if a nonzero linear combination $\sum_{\lambda\vdash n} a_\lambda p_\lambda$ of power sums is Schur positive, then $a_{1^n}>0$.<|endoftext|> -TITLE: When does a triangulated category have a heart? -QUESTION [7 upvotes]: Suppose $\mathcal{T}$ is a triangulated category. What are the conditions $\mathcal{T}$ must satisfy in order to have a t-structure? If a t-structure exists, which further conditions would ensure that $\mathcal{T}$ is the derived category of its heart? -My question is motivated by the ongoing search for an abelian category of mixed motives for which several constructions of triangulated categories exist. In this context, is it the case that -(1) the aforementioned conditions are met by one or all of the existing triangulated categories so the existence of the abelian category is assured and the remaining issue is one of construction of a t-structure, or -(2) the conditions are not known to be satisfied by any of the existing triangulated categories so even the existence of a t-structure is unknown, or -(3) no such conditions are known, i.e., the answer to my questions in the first paragraph is "don't know!", at least in that generality. -I believe from my reading that option (1) is not true, but I've included it just to make sure. Thanks! - -REPLY [5 votes]: A silly remark is that "trivial" $t$-structures always exist. You should probably say that you want a bounded or a non-degenerate $t$-structure. As far as I remember, non-zero negative $K$-groups of $T$ should give an obstruction for the former condition if you believe that the heart is noetherian or something like this. Note also that -these groups for $DM_{gm}$ are isomorphic to that of Chow motives; see Sosnilo, Vladimir, Theorem of the heart in negative K-theory for weight structures. Doc. Math. 24 (2019), 2137–2158. -As for comparing $DM_{gm}$ with $D^b(MM)$: try to read (the introduction to?) Positselski, Leonid, Mixed Artin-Tate motives with finite coefficients. Mosc. Math. J. 11 (2011), no. 2, 317–402.<|endoftext|> -TITLE: How did the Baker-Gill-Solovay paper come to be? -QUESTION [12 upvotes]: How did the Baker-Gill-Solovay paper come to be? Why were those three people talking together about "Relativizations of the $P=?NP$" question, and what was their collaboration like for the paper submitted July 16, 1973? -The paper itself, as published in the 1975 SIAM Journal of Computation, does not cite any prior work of any of Ted Baker, John Gill or Robert Solovay. -Furthermore, it says that half of the famous result (theorem 1, an oracle $A$ such that $P^A = NP^A$) "was also discovered, independently, by Albert Meyer with Michael Fischer and by H. B. Hunt III", and the other half (theorem 3, an oracle $B$ such that $P^B \neq NP^B$) "was obtained independently by Richard Ladner". Apparently we would have gotten the BGS result in some form without any of the three named authors. -For what it's worth, here are webpages about Baker (from Florida State), Gill (from Stanford), and Solovay (from Wikipedia). Here is a book about the JSEP, an organization listed as funding Gill, with detail on Stanford in 1973 in the area of acoustic microscopy but not in logic. -All in all I see few historical hints, but the BGS result is well-enough known to seem worth a couple paragraphs of history here. Does anyone have good information? Or want to contact the people involved? Has this been written about elsewhere already? - -REPLY [11 votes]: Apparently, we would have gotten at least half of the BGS result without any of the three named authors and also without any of the 4 people they credit, all we needed was Dekhtiar. -The Annals of the History of Computing (1984) has a historical account by Trakhtenbrot of the proof by Dekhtiar (1969) that we can have $P^A\ne NP^A$. -Trakhtenbrot also explains that the $P^A\ne NP^A (\exists A)$ question was for him the main question they had been investigating, and was not viewed as a relativization of something else. - -$P\ne NP$ says that there is no way to short-circuit an exhaustive search through a mathematical space defined by the input string; -$P^A\ne NP^A (\exists A)$ says that there is no way to short-circuit an exhaustive search through a mathematical space defined by a combination of (i) the input string and (ii) an external database.<|endoftext|> -TITLE: Finding the asymptotic of the function $\Lambda(x):=\sum_{1 \leq m,n \leq x \,\land \,\gcd(m,n)=1} \frac{1}{mn}$ -QUESTION [6 upvotes]: Inspired by this question Is there a known asymptotic for $A(x):=\sum_{1\leq i,j \leq X} \frac{1}{\text{lcm}(i,j)}$? I tried to find the asymptotic of the following function. -$$ -\Lambda(x)=\sum_{\substack{ 1 \leq m,n \leq x \\ \text{gcd}(m,n)=1}} \frac{1}{mn}. -$$ -My approach: -$$ -\left(\sum_{1\leq k \leq x} \frac{1}{k}\right)^2=\sum_{1\leq l \leq x} \frac{\Lambda\big(\frac{x}{l}\big)}{l^2}\label{1}\tag{1} -$$ -Now, -$$ -f(x)=\left(\sum_{1\leq k \leq x} \frac{1}{k}\right)^2≈(\ln(x)+\gamma)^2 -$$ -From, \eqref{1} can establish the approximate identity -$$ -2f(x)-\Lambda(x)≈ 2\int_{1}^{x} \frac{\Lambda(\frac{x}{t})}{t^2} dt -\label{2}\tag{2}$$ -or, -$$ -2f(x)-\Lambda(x)≈ \frac{2}{x}\int_{1}^{x} {\Lambda(\varphi)} d\varphi -$$ -Using the Newton-Leibniz rule we get -$$x\Lambda'(x)+3\Lambda(x)≈4(\ln(x)+\gamma)+2(\ln(x)+\gamma)^2$$ -Solving this differential equation we get, -$$ -\Lambda(x)≈\frac{2}{3}\ln^2(x)+\left(\frac{8}{9}+\frac{4}{3}\gamma\right)\ln(x)+\left(\frac{2}{3}\gamma^2+\frac{8}{9}\gamma-\frac{8}{27}\right)+\frac{c_1}{x^3} -$$ ($c_1$ is the integral constant, for large $x$ this term can be neglected). -My question: Is the asymptotic formula correct? If not, then how to find the asymptotic of the function $\Lambda(x)$? -Is the method correct? -Edit: Though the answer comes wrong with the relation \eqref{2} , but if we use the identity involving the equation $A(x)$ instead of ${\zeta_x}^2(1)=\tau(x)$, then we get the correct answer (the leading term). The approximation \eqref{2} works well here. See my answer below. - -REPLY [15 votes]: We have, for $x\geq 2$, -\begin{align*} -\sum_{\substack{ 1 \leq m,n \leq x \\ \mathrm{gcd}(m,n)=1}} \frac{1}{mn} -&=\sum_{1 \leq m,n \leq x}\frac{1}{mn}\sum_{k\mid\mathrm{gcd}(m,n)}\mu(k)\\ -&=\sum_{1\leq k\leq x}\mu(k)\sum_{\substack{ 1 \leq m,n \leq x \\ k\mid\mathrm{gcd}(m,n)}} \frac{1}{mn}\\ -&=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}\left(\sum_{1\leq m\leq x/k}\frac{1}{m}\right)^2\\ -&=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}\left(\log\frac{x}{k}+\gamma+O\left(\frac{k}{x}\right)\right)^2\\ -&=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}\left(\log^2\frac{x}{k}+2\gamma\log\frac{x}{k}+O(1)\right)\\ -&=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}\left(\log^2 x-2\log x\log k+2\gamma\log x+O(\log^2 k)\right)\\ -&=S_1(x)(\log^2 x+2\gamma\log x)-S_2(x)(2\log x)+O(1), -\end{align*} -where -\begin{align*} -S_1(x)&:=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}=\frac{6}{\pi^2}+O\left(\frac{1}{x}\right),\\ -S_2(x)&:=\sum_{1\leq k\leq x}\frac{\mu(k)\log k}{k^2}=\frac{36\zeta'(2)}{\pi^4}+O\left(\frac{\log x}{x}\right). -\end{align*} -We conclude that, for $x\geq 2$, -$$\sum_{\substack{ 1 \leq m,n \leq x \\ \mathrm{gcd}(m,n)=1}} \frac{1}{mn}= -\frac{6}{\pi^2}\log^2 x+C\log x+O(1),$$ -where -$$C:=\frac{12\gamma}{\pi^2}-\frac{72\zeta'(2)}{\pi^4}=1.3947995\dots$$<|endoftext|> -TITLE: Complete target and complete fibers imply complete source? -QUESTION [5 upvotes]: Let $f:X\to Y$ be a surjective morphism of smooth irreducible varieties over $\mathbb{C}$. Assume further that $Y$ is complete and that every fiber $f^{-1}(y)$ for $y\in Y$ is complete and irreducible. Does it necessarily follow that $X$ is complete as well? If no, what additional assumptions can we put so that this follows? -Edit: According to the remark of vrz, I added the assumption that each fiber is irreducible (which implies connectedness). If I don't require equidimensionality, is there also a counterexample? - -REPLY [8 votes]: Let $k$ be a field. Let $Y$ be a separated, finite type $k$-scheme that is geometrically connected and normal. Let $f:X\to Y$ be a separated, finite type morphism from a geometrically connected and reduced $k$-scheme to $Y$ such that the fiber over every geometric point of $Y$ is connected and proper. -Proposition. The morphism is proper. -Proof. By Nagata compactification, there exists a dense open immersion $i:X\hookrightarrow \overline{X}'$ into a proper $k$-scheme. The product morphism, $$(i,f):X\to \overline{X}'\times_{\text{Spec}\ k}Y,$$ is also an open immersion between separated, finite type $k$-schemes. Thus, the closure of the image is also a separated, finite type $k$-scheme. Denote this closure by $\overline{X}$. Also denote the restriction to $\overline{X}$ of the second projection by -$$\overline{f}:\overline{X}\to Y.$$ -By construction $\overline{f}$ is a proper, surjective morphism. Thus, there exists a Stein factorization, $$\overline{X}\xrightarrow{h} Z\xrightarrow{g} Y,$$ where $h$ has geometrically connected fibers and where $g$ is a finite surjective morphism. Since $X$ is a dense open subscheme of $\overline{X}$ that is connected, also $\overline{X}$ is connected. Therefore $Z$ is also connected. Similarly, since $X$ is reduced, also $Z$ is reduced. Since the geometric generic fiber of $f$ is connected and dense in the geometric generic fiber of $\overline{f}$, also the geometric generic fiber of $\overline{f}$ is connected. Thus, $g$ is birational. Since $Y$ is normal, by Zariski's Main Theorem, the morphism $g$ is an isomorphism. In other words, the fiber of $\overline{f}$ over every geometric point is connected and proper. -The fiber of $f$ over that same geometric point is an open subscheme of the fiber of $\overline{f}$. By hypothesis, it is also proper, and thus it is a closed subscheme of the fiber of $\overline{f}$. Since the fiber of $\overline{f}$ is connected, the fiber of $\overline{f}$ equals the fiber of $f$ for every geometric point of $Y$. For every geometric point of $\overline{X}$, for the image geometric point of $Y$, the geometric point of $\overline{X}$ is a point of the fiber of $\overline{f}$ over that geometric point of $Y$. Thus, it is also a point of $X$. In other words, the open subscheme $X$ of $\overline{X}$ equals all of $\overline{X}$. Therefore, also $f$ equals $\overline{f}$, so that the morphism $f$ is proper. QED<|endoftext|> -TITLE: Number of unramified quadratic extensions of a number field -QUESTION [8 upvotes]: Is there a general formula for the number of unramified quadratic extensions of a number field $K$? -When $K$ is quadratic, this is known (by genus theory) to be $2^{\omega(\Delta_K)-1}$, where $\omega(n)$ denotes the number of distinct prime factors of $n$ and $\Delta_K$ is the discriminant of $K$. I'm interested in results for when $K$ is of higher degree. -It seems like this problem might be much harder and is maybe adjacent to understanding the two-torsion of the class group $\text{Cl}_K$ (which seems hard when $K$ is not quadratic), but I'm pretty new to the area and could be totally off-base. Is there any hope of a more direct approach? - -REPLY [7 votes]: The answer seems to be no. - -The number of unramified quadratic extensions of $K$ is equal to the number of index-two subgroups of the ideal class group $\text{Cl}_K$ by class field theory. -The index-two subgroups of $\text{Cl}_K$ correspond to the non-zero elements of $\text{Hom}(\text{Cl}_K, \mathbb{Z}/2\mathbb{Z})$. -$\#\text{Hom}(\text{Cl}_K, \mathbb{Z}/2\mathbb{Z}) = \#\text{Cl}_K[2]$ by Pontryagin duality, as pointed out to me by @RP_ and @abx in the comments. -The problem of computing (or even bounding) the size of $\#\text{Cl}_K[2]$ when $K$ is not a quadratic extension appears to be under active study and seems to be a challenging problem in general.<|endoftext|> -TITLE: Existence of finite limits of quasi-coherent modules on a scheme -QUESTION [5 upvotes]: Defining a quasi-coherent module $\mathcal{M}$ on a scheme $X$ to be a compatible family of modules $(\mathcal{M}(x))_{x \in X(A), A \in \textbf{Rings}}$ (as in here), is there a straightforward way to show the existence of (finite) limits (and that it forms an abelian category)? -One possible way, of course, should be to show that this definition gives rise to a category equivalent to the category of quasi-coherent sheaves of modules on the small Zariski-site associated to $X$, but that feels like a rather dirty solution. -The problem, I guess, is that taking pullbacks of sheaves of modules (generally) doesn't commute with taking limits so that the limit isn't defined "fibrewise"; colimits work fine for exactly that reason. -Another argument that a friend of mine explained to me seems to be that, denoting the in the above way defined category of modules as $\textbf{Mod}(X)$, one has -$$\textbf{Mod}(X) = \varprojlim_{A \in \textbf{Aff}/X} \textbf{Mod}(A)$$ -where the ($2$-)limit is taken in the $(2,1)$-category of categories, functors and natural isomorphisms. -Now the argument would be that $\textbf{Mod}(A)$ is a locally presentable category, (certain?) limits of locally presentable categories are locally presentable, and locally presentable categories admit arbitrary limits. -I was still wondering whether there wouldn't be a more elementary way to for example directly construct kernels and finite products of modules when defined this way. -I'd appreciate any thoughts! -//Edit: Ok another way seems to be to first show that one can glue quasi-coherent modules along Zariski-coverings and then do everything locally. I guess that's fine for me, but I'd still be interested in seeing other elementary arguments if anyone has one! - -REPLY [3 votes]: So I was lush's friend who he had originally asked this question, and I had some concerns, specifically because I gave the same answer as Riza, then realized that it gave incorrect answers if you follow the direct nLab construction. The point is that the limit of a diagram in the limit has to be computed first pointwise in the lax limit as above, then you have to apply a coreflector into the actual limit. -For example, if I have a cartesian square of locally presentable categories -$$\begin{matrix} -P&\xrightarrow{f^\prime_!}&C_1\\ -g^\prime_!\downarrow &\ulcorner&\downarrow g_!\\ -C_2&\xrightarrow{f_!}&C_0 -\end{matrix}$$ -and a diagram $d:D\to P$, I can compute $P$ as a colocalization of the lax limit of this diagram (the category of not-necessarily-cartesian sections of the associated cartesian fibration over the span category $\operatorname{Span}$). Let's denote this lax limit by $L$. Then we have an adjunction $P\leftrightarrows L$, where the left adjoint $P\to L$ is fully faithful. This tells us that the limit in $P$ is computed as the image under the coreflector $L\to P$ of the limit in $L$, which is actually indeed the pointwise limit together with the connecting maps -$$g_! \lim (f^\prime_! \circ d)\to \lim (g_! \circ f^\prime_! \circ d)=\lim (f_! \circ g^\prime_! \circ d) \leftarrow f_!\lim(g'_!\circ d).$$ -So to form the true limit, I have to apply the coreflector to this formal diagram (viewed as an object of the lax limit). -This gives you a formula to compute the limit now of such a diagram, but actual existence of limits is following from the fact that this fibre product is presentable (plus the thing about arbitrary products still being presentable). -To finish working out the example, the coreflector then gives you the fibre product in $P$ -$$ \lim(f^{\prime \ast}\lim (f^\prime_! \circ d)\to f^{\prime\ast}g^\ast\lim (g_! \circ f^\prime_! \circ d)=g^{\prime\ast}f^\ast\lim (f_! \circ g^\prime_! \circ d) \leftarrow g^{\prime\ast}\lim(g'_!\circ d)).$$ -but in order for this formula to make sense, you first needed to know that limits in $P$ existed, and that's because $\operatorname{Pr}^L$ admits limits that agree with limits in $\mathbf{Cat}$. -Note: I've used the categorical convention for left and right adjoints (lower shriek and upper star, rather than upper star and lower star) in $\operatorname{Pr}^L$ rather than the algebro-geometric convention, because it is clearer in this case. -Edit: It looks like lush's question here was slightly different from the one we discussed in private. My mistake. Riza's answer is correct for flat covers (this is a theorem, but it is completely obvious for open immersions, as desired).<|endoftext|> -TITLE: Periods of the harmonic conjugate and a Dirichlet problem on a multiply connected domain -QUESTION [5 upvotes]: Any harmonic function $u$ on a simply connected domain in $\mathbb{R}^2$ is the real part of a holomorphic function. If the domain is multiply connected, then this is no longer true: the harmonic conjugate of $u$ may have periods. -I wonder if the following is true: let $C_1, \ldots, C_n$ be the components of the boundary of a domain, and let functions $f_i \colon C_i \to \mathbb{R}$ be given. Do there exist constants $a_1, \ldots, a_n$ such that the solution to the Dirichlet problem $\Delta u = 0$, $u \mid_{C_i} = f_i + a_i$ is the real part of a holomorphic function? -This is equivalent to the non-degeneracy of a matrix (periods of the conjugates of the solutions to several Dirichlet problems, the $i$-th problem being $f_j = \delta_{ij}$), and maybe follows from some classical facts about Riemann surfaces, but I am a stranger to the field... - -REPLY [2 votes]: The answer is yes and you can find it in the book [1], chapter 1, §4, theorem 4.3, pp. 20-22. Precisely Wen, by constructing a suitable harmonic function and its harmonic conjugate, proves that on a $(N+1) $-connected domain in $\Bbb C$ whose connected components of the boundary $\Gamma$ are $C_0,\ldots,C_N$, there exists a unique analytic function $F(z)$ such that -$$ -\begin{cases} -\Re F(t) =\Phi(t) + h(t) & t\in\Gamma=\bigcup_{i=0}^NC_i\\ -\Im F(a) = b & a\in C_0 -\end{cases} -$$ -where - -$\Phi(t)\in C^{1,\alpha}(\Gamma)$ is a real valued function on the boundary of $D$ (the requirement can be relaxed in $\Phi(t)\in C^{0,\alpha}(\Gamma)$ by a suitable approximation as explained in [1], p. 20 footnote 2) -The function $h(t)$ is a simple function defined as -$$ -h(t)= -\begin{cases} -0, & t\in C_0 \\ -h_j & t\in C_j, \; j= 1,\ldots, N. -\end{cases} -$$ -where the $\{h_j\}_{j=1,\ldots,N}$ are indeterminate real constants -$b$ is a real constant. - -Reference -[1] Guo-Chun Wen, Conformal mappings and boundary value problems, translated from the Chinese by Kuniko Weltin (English), Translations of Mathematical Monographs 106, Providence, RI: American Mathematical Society (AMS). viii, 303 p. (1992), ISBN 0-8218-4562-4, MR1187758, Zbl 0778.30011<|endoftext|> -TITLE: Completion w.r.t. ideal generated by a part of regular system of parameter -QUESTION [5 upvotes]: Let $R$ be a $d$-dimensional Noetherian regular local $k$-algbera ($k$ any field of char($k$) = 0, $d \geq$ 2). Let $x, y$ be a part of regular system of parameter for $R$. Let $I = (x, y)$ be a ideal generated by $x$ and $y$ and $\hat{R}$ denote the completion of $R$ with respect to $I$. -Is it true that $\hat{R} = \frac{R}{I}[[x, y]]$? -I know the very special case of it is true when $d = 2$ and $I$ is the maximal ideal. The above statement seems correct but I am not entirely sure. Any help would be great. - -REPLY [2 votes]: Suppose additionally that $R$ is essentially of finite type over $k$. -Then $\hat{R}$ contains a subring that maps isomorphically on to $R/I$. -Proof: Note that $R/I$ is regular and essentially of finite type over $k$. Hence it is $0$-smooth over $k$ (Matsumura, Commutative Ring Theory, Theorem 30.3). We can therefore lift the identity map of $R/I$ to a $k$-algebra map $f_2 : R/I \to R/I^2$. Repeating the argument, we get a $k$-algebra map $f_j : R/I \to R/I^j$ lifting $f_{j-1}$, for every $j \geq 3$. Thus we get a map $R/I \to \hat{R}$ such that the composite $R/I \to \hat{R} \to R/I = \hat{R}/I\hat{R}$ is the identity map of $R/I$. - -Older answer, left here so that the comments below this make sense. -$\hat{R}$ is a flat $R$-module (Matsumura, Commutative Ring Theory, Theorem 8.8) but $\frac{R}{I}[[x,y]]$ is not, since every non-zero element of $I$ is a zero-divisor on it.<|endoftext|> -TITLE: Confusion around the reflection equation algebra -QUESTION [6 upvotes]: I have encountered several occurrences of the so called reflection equation algebra (REA) but depending on where I find them, I feel like I get slightly different objects. In all cases there is a quasi-triangular Hopf algebra lurking in the background. In what follows $V$ will always be a vector space of dimension $n$. Here is a list of the different occurrences I encountered: - -Let $H$ be a quasi-triangular Hopf algebra with $R \in H \otimes H$ its universal $R$-matrix (here we possibly have completions but it does not really matters). The reflection algebra is then as vector space the restricted dual $H^\circ$. That is the subalgebra of the full dual spanned by the so called matrix coefficients. The algebra structure comes from the algebra structure of the full dual but twisted by the universal $R$-matrix. I think this is sometimes called the braided dual of $H$. -See for example definition 4.12 of https://arxiv.org/pdf/math/0204295.pdf - -Let $R: V \otimes V \rightarrow V \otimes V$ be an endomorphism of $V \otimes V$ satisfying the Yang-Baxter equation. Then the reflection equation algebra if the algebra generated by elements $(a_{ij})_{1 \leq i,j \leq n}$ with relation -$$RA_1RA_1 = A_1RA_1R$$ -where $A$ is the matrix $n \times n$ having the generating elements as coefficients and $A_1 = A \otimes Id$. -I think here the generating elements are somewhat thought to be elements of $V^{\ast} \otimes V$. This was found at the very beginning of the introduction of https://arxiv.org/pdf/1806.10219.pdf - -This one is a special example. Here the Hopf algebra lurking in the background is $U_q(\frak{sl_2})$ and the $R$-matrix is given by -$$ -\begin{pmatrix} -q & 0 & 0 & 0\\ -0 & 1 & q-q^{-1} & 0 \\ -0 & 0 & 1 & 0\\ -0 & 0 & 0& q -\end{pmatrix}.$$ In this case it is the algebra generated by the elements $(a_{ij})_{1 \leq i,j \leq 2}$ with relation : -$$R_{21}A_1RA_2 = A_2R_{21}A_1R$$ and also -$$ a_{11}a_{22}- q^2a_{12}a_{21} = 1.$$ -This algebra is often denoted by $\mathcal{O}_q(SL_2)$ or sometimes $\mathcal{O}(Rep_q(SL_2))$. This appeared as Example 1.23 in https://arxiv.org/pdf/1908.05233.pdf and also as Definition 2.1. in https://arxiv.org/pdf/1811.09293.pdf (be aware of the foot note to get back what I wrote). - - -I can see how some of those are related, for example the third one is almost a specific case of the second one but there is one more relations. -In the first one matrix elements can be thought as being in $W^{\ast} \otimes W$ for any representation $W$ of $H$. In the case where any finite dimensional representation of $H$ can be seen as subrepresentation of a tensor product of the standard representation $V$, then it is actually generated only by the matrix coefficients coming from $V$.It then looks a lot like what we have in 2). However, there is still the a relation missing if one specializes to the case $H = U_q(\frak{sl2})$ to get the same as in 3). And what if there is a representation of $H$ that is not a subrepresentation of a tensor product of the standard one? -QUESTION: -Are all of those actually the same thing or am I missing something? I am a bit confused on what people actually call the reflection equation algebra. Is there some kind of nice definition for any quasi-triangular Hopf algebra $H$ that englobes all the above "examples"? - -REPLY [3 votes]: Let me first note that the reflection matrix, which you denote by $A$, is often called K-matrix, cf its graphical representation with | a 'wall' and < the 'worldline' of particle bouncing off the wall. The graphical form of the equation can already be found in Cherednik, Factorizing particles on a half-line and root systems (1984) https://link.springer.com/article/10.1007/BF01038545. The notation $K$ might be due to Sklyanin, Boundary conditions for integrable quantum systems (1988), https://iopscience.iop.org/article/10.1088/0305-4470/21/10/015. -The reflection (equation) algebra is the reflection-equation analogue of the Yang--Baxter algebra: to any choice of finite-dimensional vector space and R-matrix obeying the Yang--Baxter equation (and suitable other properties, such as braiding unitarity and an 'initial condition') one can associate a unital associative algebra generated by the operator-valued (noncommutative) entries of the K-matrix obeying the reflection equation. -If one would replace the reflection (`$RKRK$') equation by the $RLL$-equation one instead arrives at the Yang--Baxter algebra, which is the operator algebra closely related to the FRT (or R-matrix) presentation of quantum affine algebras. -Re 3: The FRT presentation says nothing about the quantum determinant, so to get $SL_n$ you need to impose $qdet = 1$ separately, which is your last equation in 3. The version of the reflection equation that you give there can sometimes be simplified: Suppose that the R-matrix is symmetric in the sense that $P R P = R$ with $P$ the permutation. Then $R_{21} = R_{12}$ in the usual tensor-leg notation. In such cases all R-matrices in the reflection equation can be written using just $R$. (Graphically the need for $R_{21}$ is clear, though.) -Re 2: These authors work with the braid-like version of the R-matrix, often denoted by $\check{R}$. Namely, suppose that $R$ obeys the YBE -$$ R_{12}(u,v) \ R_{13}(u,w) \ R_{23}(v,w) = R_{23}(v,w) \ R_{13}(u,w) \ R_{12}(u,v) \ , $$ -where I have assumed that the R-matrix might depend on a spectral parameter associated to each copy of the auxiliary space in general. (This is for the affine case, but helps highlighting the structure of the equation.) -Then both of $P \ R$ and $R \ P$ obey the braid-like version of the YBE -$$ \check{R}_{12}(u,v) \ \check{R}_{23}(u,w) \ \check{R}_{12}(v,w) = \check{R}_{23}(v,w) \ \check{R}_{12}(u,w) \ \check{R}_{23}(u,v) \ . $$ -You always have to check which version is used. In the paper you cite in 2 it's the latter, which is why both $A$s have the same subscript. -Re 1: I believe that the proper algebraic interpretation of Sklyanin's construction of representations of the K-matrix as the double-row monodromy matrix, constructed from a K-matrix with scalar entries and an L-operator, is as a coideal subalgebra, see Kolb and Stokman, Reflection equation algebras, coideal subalgebras, and their centres, https://arxiv.org/abs/0812.4459. -You might also be interested in the recent paper by Appel and Vlaar, Universal k-matrices for quantum Kac-Moody algebras, https://arxiv.org/abs/2007.09218<|endoftext|> -TITLE: What is the roadblock in the discovery of new taxicab numbers? -QUESTION [11 upvotes]: The $n$-th taxicab number, denoted $\text{Ta}(n)$, is the smallest integer that can be expressed as a sum of two positive integer cubes in $n$ different distinct ways. -$\text{Ta}(1) = 2 = 1^3 + 1^3$ is trivial, and the infamous $\text{Ta}(2) = 1729$ was known as early as the 17th century, much before the well-known Hardy-Ramanujan story. -$\text{Ta}(3)$ was found by John Leech in 1957. After no further discoveries for three decades, the quest for more taxicab numbers seems to have gained traction around the same time computer-assisted proofs became more widespread. Rosenstiel, Dardis and Rosenstiel found $\text{Ta}(4)$ in 1989; Dardis found $\text{Ta}(5)$ in 1994 and this was later confirmed by Wilson in 1999; and finally Calude et al. announced $\text{Ta}(6)$ in 2003 which was later verified by Hollerbach in 2008. -The best information we have regarding other taxicab numbers are the upper bounds for $\text{Ta}(7)$ through $\text{Ta}(12)$ provided by Boyer in 2006-2008. There seems to have been a relatively rapid succession in the discovery of taxicab numbers from early 1990s until mid-2000s. One would imagine, the quality of the computational tools we have access to nowadays would only have accelerated the search -- but the quest seems to be silent since Boyer's upper bounds. Why is this? - -REPLY [20 votes]: There are a few issues here. -(1) It is relatively easy to show that Ta($n$) exists, for example by using a point of infinite order on an elliptic curve $x^3+y^3=mz^3$ to show that there is at least one number with $n$ distinct representations. However, the number tends to be divisible by a large cube, or alternatively, the $(x,y)$ pairs tend to have large $\gcd(x,y)$. -(2) So let's define $\operatorname{Ta}^*(n)$ to be the smallest that can be expressed as a sum of two relatively prime positive integer cubes in $n$ different distinct ways. Then $\operatorname{Ta}^*(2)=1729$,$\operatorname{Ta}^*(3)=15170835645$ (Vojta), $\operatorname{Ta}^*(4)=1801049058342701083$ (Gascoigne, Moore, independently), and there is some reason to believe that $\operatorname{Ta}^*(5)$ doesn't exist. (Or in any case, at some point $\operatorname{Ta}^*(n)$ doesn't exist.) -(3) To get back to your question, the size of $\operatorname{Ta}(n)$ probably (maybe?) grows exponentially with $n$. And increased computer power, even with Moore's law, has a hard time keeping up with a problem whose computational complexity grows exponentially. So for example, if increasing from $n$ to $n+1$ makes the taxicab search space grow by a factor of 100, and if it took 2 years of computer time to find $T(n)$, it's going to require a much faster computer to compute $T(n+1)$. - -It looks as if $\operatorname{Ta}(n)$ may be growing superexponentially in $n$, although of course there isn't a lot of data, and the values for $7\le n\le 12$ are upper bounds. But the last line of this table is suggestive. -\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline -n & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline -\dfrac{\log\operatorname{Ta}(n)}{n} &3.73 & 6.10 & 7.39 & 7.69 & 8.59 & 9.34 & 9.99 & 10.52 & 11.25 & 12.03 & 13.02 \\ \hline -\dfrac{\log\operatorname{Ta}(n)}{n\ln(n)} & -5.38 & 5.55 & 5.33 & 4.78 & 4.79 & 4.80 & 4.80 & 4.79 & 4.89 & 5.02 & 5.24 -\\ \hline -\end{array}<|endoftext|> -TITLE: A better way to explain forcing? -QUESTION [67 upvotes]: Let me begin by formulating a concrete (if not 100% precise) question, and then I'll explain what my real agenda is. -Two key facts about forcing are (1) the definability of forcing; i.e., the existence of a notion $\Vdash^\star$ (to use Kunen's notation) such that $p\Vdash \phi$ if and only if $(p \Vdash^\star \phi)^M$, and (2) the truth lemma; i.e., anything true in $M[G]$ is forced by some $p\in G$. -I am wondering if there is a way to "axiomatize" these facts by saying what properties forcing must have, without actually introducing a poset or saying that $G$ is a generic filter or that forcing is a statement about all generic filters, etc. And when I say that forcing "must have" these properties, I mean that by using these axioms, we can go ahead and prove that $M[G]$ satisfies ZFC, and only worry later about how to construct something that satisfies the axioms. - -Now for my hidden agenda. As some readers know, I have written A beginner's guide to forcing where I try to give a motivated exposition of forcing. But I am not entirely satisfied with it, and I have recently been having some interesting email conversations with Scott Aaronson that have prompted me to revisit this topic. -I am (and I think Scott is) fairly comfortable with the exposition up to the point where one recognizes that it would be nice if one could add some function $F : \aleph_2^M \times \aleph_0 \to \lbrace 0,1\rbrace$ to a countable transitive model $M$ to get a bigger countable transitive model $M[F]$. It's also easy to grasp, by analogy from algebra, that one also needs to add further sets "generated by $F$." And with some more thought, one can see that adding arbitrary sets to $M$ can create contradictions, and that even if you pick an $F$ that is "safe," it's not immediately clear how to add a set that (for example) plays the role of the power set of $F$, since the "true" powerset of $F$ (in $\mathbf{V}$) is clearly the wrong thing to add. It's even vaguely plausible that one might want to introduce "names" of some sort to label the things you want to add, and to keep track of the relations between them, before you commit to saying exactly what these names are names of. But then there seems to be a big conceptual leap to saying, "Okay, so now instead of $F$ itself, let's focus on the poset $P$ of finite partial functions, and a generic filter $G$. And here's a funny recursive definition of $P$-names." Who ordered all that? -In Cohen's own account of the discovery of forcing, he wrote: - -There are certainly moments in any mathematical discovery when the resolution of a problem takes place at such a subconscious level that, in retrospect, it seems impossible to dissect it and explain its origin. Rather, the entire idea presents itself at once, often perhaps in a vague form, but gradually becomes more precise. - -So a 100% motivated exposition may be a tad ambitious. However, it occurs to me that the following strategy might be fruitful. Take one of the subtler axioms, such as Comprehension or Powerset. We can "cheat" by looking at the textbook proof that $M[G]$ satisfies the axiom. This proof is actually fairly short and intuitive if you are willing to take for granted certain things, such as the meaningfulness of this funny $\Vdash$ symbol and its two key properties (definability and the truth lemma). The question I have is whether we can actually produce a rigorous proof that proceeds "backwards": We don't give the usual definitions of a generic filter or of $\Vdash$ or even of $M[G]$, but just give the bare minimum that is needed to make sense of the proof that $M[G]$ satisfies ZFC. Then we "backsolve" to figure out that we need to introduce a poset and a generic filter in order to construct something that satisfies the axioms. -If this can be made to work, then I think it would greatly help "ordinary mathematicians" grasp the proof. In ordinary mathematics, expanding a structure $M$ to a larger structure $M[G]$ never requires anything as elaborate as the forcing machinery, so it feels like you're getting blindsided by some deus ex machina. Of course the reason is that the axioms of ZFC are so darn complicated. So it would be nice if one could explain what's going on by first looking at what is needed to prove that $M[G]$ satisfies ZFC, and use that to motivate the introduction of a poset, etc. -By the way, I suspect that in practice, many people learn this stuff somewhat "backwards" already. Certainly, on my first pass through Kunen's book, I skipped the ugly technical proof of the definability of forcing and went directly to the proof that $M[G]$ satisfies ZFC. So the question is whether one can push this backwards approach even further, and postpone even the introduction of the poset until after one sees why a poset is needed. - -REPLY [30 votes]: This is an expansion of David Roberts's comment. It may not be the sort of answer you thought you were looking for, but I think it is appropriate, among other reasons because it directly addresses your question - -if there is a way to "axiomatize" these facts by saying what properties forcing must have. - -In fact, modern mathematics has developed a powerful and general language for "axiomatizing properties that objects must have": the use of universal properties in category theory. In particular, universal properties give a precise and flexible way to say what it means to "freely" or "generically" add something to a structure. -For example, suppose we have a ring $R$ and we want to "generically" add a new element. The language of universal properties says that this should be a ring $R[x]$ equipped with a homomorphism $c:R\to R[x]$ and an element $x\in R[x]$ with the following universal property: for any ring $S$ equipped with a homomorphism $f:R\to S$ and an element $s\in S$, there exists a unique homomorphism $h:R[x]\to S$ such that $h\circ c = f$ and $h(x) = s$. -Note that this says nothing about how $R[x]$ might be constructed, or even whether it exists: it's only about how it behaves. But this behavior is sufficient to characterize $R[x]$ up to unique isomorphism, if it exists. And indeed it does exist, but to show this we have to give a construction: in this case we can of course use the ring of formal polynomials $a_n x^n + \cdots + a_1 x + a_0$. -From this perspective, if we want to add a function $F : \aleph_2\times \aleph_0 \to 2$ to a model $M$ of ZFC to obtain a new model $M[F]$, the correct thing to do would be to find a notion of "homomorphism of models" such that $M[F]$ can be characterized by a similar universal property: there would be a homomorphism $c:M\to M[F]$ and an $F : \aleph_2\times \aleph_0 \to 2$ in $M[F]$, such that for any model $N$ equipped with a homomorphism $f:M\to N$ and a $G : \aleph_2\times \aleph_0 \to 2$ in $N$, there is a unique homomorphism $h:M[F]\to N$ such that $h\circ c = f$ and $h(F) = G$. -The problem is that the usual phrasing of ZFC, in terms of a collection of things called "sets" with a membership relation $\in$ satisfying a list of axioms in the language of one-sorted first-order logic, is not conducive to defining such a notion of homomorphism. However, there is an equivalent formulation of ZFC, first given by Lawvere in 1964, that works much better for this purpose. (Amusingly, 1964 is exactly halfway between 1908, when Zermelo first proposed his list of axioms for set theory, and the current year 2020.) In Lawvere's formulation, there is a collection of things called "sets" (although they behave differently than the "sets" in the usual presentation of ZFC) and also a separate collection of things called "functions", which together form a category (i.e. functions have sets as domain and codomain, and can be composed), and satisfy a list of axioms written in the language of category theory. (A recent short introduction to Lawvere's theory is this article by Tom Leinster.) -Lawvere's theory is usually called "ETCS+R" (the "Elementary Theory of the Category of Sets with Replacement"), but I want to emphasize that it is really an entirely equivalent formulation of ZFC. That is, there is a bijection between models of ZFC, up to isomorphism, and models of ETCS+R, up to equivalence of categories. In one direction this is exceedingly simple: given a model of ZFC, the sets and functions therein as usually defined form a model of ETCS+R. Constructing an inverse bijection is more complicated, but the basic idea is the Mostowski collapse lemma: well-founded extensional relations can be defined in ETCS+R, and the relations of this sort in any model of ETCS+R form a model of ZFC. -Since a model of ETCS+R is a structured category, there is a straightforward notion of morphism between models: a functor that preserves all the specified structure. However, this notion of morphism has two defects. -The first is that the resulting category of models of ETCS+R is ill-behaved. In particular, the sort of "free constructions" we are interested in do not exist in it! However, this is a problem of a sort that is familiar in modern structural mathematics: when a category is ill-behaved, often it is because we have imposed too many "niceness" restrictions on its objects, and we can recover a better-behaved category by including more "ill-behaved" objects. For instance, the category of manifolds does not have all limits and colimits, but it sits inside various categories of more general "smooth spaces" that do. The same thing happens here: by dropping two of the axioms of ETCS+R we obtain the notion of an elementary topos, and the category of elementary toposes, with functors that preserve all their structure (called "logical functors"), is much better-behaved. In particular, we can "freely adjoin a new object/morphism" to an elementary topos. -(I am eliding here the issue of the replacement/collection axiom, which is trickier to treat correctly for general elementary toposes. But since my main point is that this direction is a blind alley for the purposes of forcing anyway, it doesn't matter.) -The second problem, however, is that these free constructions of elementary toposes do not have very explicit descriptions. This is important because our goal is not merely to freely adjoin an $F:\aleph_2\times \aleph_0 \to 2$, but to show that the existence of such an $F$ is consistent, and for this purpose we need to know that when we freely adjoin such an $F$ the result is nontrivial. Thus, in addition to characterizing $M[F]$ by a universal property, we need some concrete construction of it that we can inspect to deduce its nontriviality. -This problem is solved by imposing a different niceness condition on the objects of our category and changing the notion of morphism. A Grothendieck topos is an elementary topos that, as a category, is complete and cocomplete and has a small generating set. But, as shown by Giraud's famous theorem, it can equivalently be defined as a cocomplete category with finite limits and a small generating set where the finite limits and small colimits colimits interact nicely. This suggests a different notion of morphism between Grothendieck toposes: a functor preserving finite limits and small colimits. Let's call such a functor a Giraud homomorphism (it's the same as a "geometric morphism", but pointing in the opposite direction). -The category of Grothendieck toposes and Giraud homomorphisms is well-behaved, and in particular we can freely adjoin all sorts of structures to a Grothendieck topos -- specifically, any structure definable in terms of finite limits and arbitrary colimits (called "a model of a geometric theory"). (To be precise, this is a 2-category rather than a category, and the universal properties are up to isomorphism, but this is a detail, and unsurprising given the modern understanding of abstract mathematics.) Moreover, the topos $M[G]$ obtained by freely adjoining a model $G$ of some geometric theory to a Grothendieck topos $M$ -- called the classifying topos of the theory of $G$ -- has an explicit description in terms of $M$-valued "sheaves" on the syntax of the theory of $G$. This description allows us to check, in any particular case, that it is nontrivial. But for other purposes, it suffices to know the universal property of $M[G]$. In this sense, the universal property of a classifying topos is an answer to your question: - -when I say that forcing "must have" these properties, I mean that by using these axioms, we can go ahead and prove that $M[G]$ satisfies ZFC, and only worry later about how to construct something that satisfies the axioms. - -Only one thing is missing: not every Grothendieck topos is a model of ETCS+R, hence $M[G]$ may not itself directly yield a model of ZFC. We solve this in three steps. First, since ZFC satisfies classical logic rather than intuitionistic logic (the natural logic of categories), we force $M[G]$ to become Boolean. Second, by restricting to "propositional" geometric theories we ensure that the result also satisfies the axiom of choice. Finally, we pass to the "internal logic" of the topos, which is to say that we allow "truth values" lying in its subobject classifier rather than in the global poset of truth values $2$. We thereby get an "internal" model of ETCS+R, and hence also an "internal" model of ZFC. -So where does the complicated machinery in the usual presentation of forcing come from? Mostly, it comes from "beta-reducing" this abstract picture, writing out explicitly the meaning of "well-founded extensional relation internal to Boolean sheaves on the syntax of a propositional geometric theory". The syntax of a propositional geometric theory yields, as its Lindenbaum algebra, a poset. The Boolean sheaves on that poset are, roughly, those that satisfy the usual "denseness" condition in forcing. The "internal logic" valued in the subobject classifier corresponds to the forcing relation over the poset. And the construction of well-founded extensional relations translates to the recursive construction of "names". -(Side note: this yields the "Boolean-valued models" presentation of forcing. The other version, where we take $M$ to be countable inside some larger model of ZFC and $G$ to be an actual generic filter living in that larger model, is, at least to first approximation, an unnecessary complication. By comparison (and in jesting reference to Asaf's answer), if we want to adjoin a new transcendental to the field $\mathbb{Q}$, we can simply construct the field of rational functions $\mathbb{Q}(x)$. From the perspective of modern structural mathematics, all we care about are the intrinsic properties of $\mathbb{Q}(x)$; it's irrelevant whether it happens to be embeddable in some given larger field like $\mathbb{R}$ by setting $x=\pi$.) -The final point is that it's not necessary to do this beta-reduction. As usual in mathematics, we get a clearer conceptual picture, and have less work to do, when working at an appropriate level of abstraction. We prove the equivalence of ZFC and ETCS+R once, abstractly. Similarly, we show that we have an "internal" model of ETCS+R in any Grothendieck topos. These proofs are easier to write and understand in category-theoretic language, using the intrinsic characterization of Grothendieck toposes rather than anything to do with sites or sheaves. With that done, the work of forcing for a specific geometric theory is reduced to understanding the relevant properties of its category of Boolean sheaves, which are simple algebraic structures.<|endoftext|> -TITLE: Restriction of a branched cover to its branch locus -QUESTION [13 upvotes]: Assume that we have a smooth, compact, complex surface $X$, and a smooth and irreducible divisor $B \subset X$. Let $G$ be a finite group. For every group epimorphism $$\varphi \colon \pi_1(X-B) \to G,$$ by Grauert-Remmert extension theorem there is a smooth complex surface $Y$ and a Galois cover $$f \colon Y \to X,$$ -with Galois group $G$ and branched at most over $B$. -Since $B$ is smooth, setting $R =f^{-1}(B) \subset Y$ we see that the restriction $$f|_R \colon R \to B$$ is an unramified Galois cover, with Galois group $H=G/G_R$, where $G_R$ is the stabilizer of the curve $R$. Such a Galois cover must correspond in turn to a group homomorphism $$\psi \colon \pi_1(B) \to H,$$ that is surjective if and only if $R$ is irreducible. - -Question. How we can recover, in a purely algebraic way, the map $\psi$ from $\varphi$ and from the homomorphisms (induced by the inclusion maps) $$i_* \colon \pi_1(X-B) \to \pi_1(X), \quad j_* \colon \pi_1(B) \to \pi_1(X)?$$ - -Here "in a purely algebraic way" means (for instance) that, if I have implemented the three homomorphisms $\varphi \colon \pi_1(X-B) \to G$, $i_*$ and $j_*$ in a software like GAP4, there should be, at least in principle, a finite sequence of commands providing $\psi \colon \pi_1(B) \to H$. -I expect this to be possible, since $\varphi$ completely determines $f \colon Y \to X$, and so completely determines the restriction $f|_R \colon R \to B$. - -REPLY [11 votes]: It is useful to reformulate the question in its natural differential topology setting, leaving unneeded geometric considerations aside. It is also natural to consider the analog of the problem in all dimensions. -So assume that we are given a closed, orientable, connected, smooth $n$-manifold $X$, and a closed, orientable, connected, smooth, codimension-$2$ submanifold $B \subset X$. We adopt the basic the notation used in the question. Let $G$ be a finite group. For every group epimorphism $$\varphi \colon \pi_1(X-B) \to G$$ there is a closed, orientable, connected, smooth $n$-manifold $Y$ and a Galois (or ``regular'') ramified covering map $$f \colon Y \to X,$$ -with deck transformation group $G$ that is branched at most over $B$. -Since $B$ is smooth, setting $R =f^{-1}(B) \subset Y$ we see that the restriction $$f|_R \colon R \to B$$ is an unramified cover. The question seeks an explicit description of this covering map. -Among the issues that arise when trying to give such an explicit description are that $R$ need not be connected, that $f|_R:R \to B$ need not be a Galois covering, and that $B$ and $X-B$ cannot have the same base point. -The additional piece of data needed to clarify things is the normal bundle of the branch set and its boundary, a circle bundle over $B$. With this extra piece of information one can effectively answer the question. -We will from this point of view - -Characterize when $R$ is connected; -Characterize when $f$ is -actually ramified; -Characterize when $R \to B$ is Galois; -Show that on each component of $R$ the restriction of the branched covering is in fact always a Galois covering, with an explicit Galois group. - -Let $N$ denote a small tubular neighborhood of $B$ in $X$, which has the structure of a $2$-disk bundle over $B$. Let $D$ denote a 2-disk fiber, with boundary $C = D \cap \partial N$, a linking circle to $B$. Then $\partial N$ is a circle bundle over $B$, with typical fiber $C$. -This circle bundle is determined by its Euler class in $H^2(B;\mathbb{Z})$ and determines an exact sequence of homotopy groups (in which we suppress mention of the required base points) -$$ -1 \to \pi_2(\partial N) \to \pi_2(B) \to \pi_1(C) \to \pi_1(\partial N) \to \pi_1(B)\to 1. -$$ -The image of $\pi_1(C)$ in $\pi_1(\partial N)$ lies in the center because of our orientability assumption. The only case in the dimension range $n\leq 4$ that $\pi_2(B)\neq 1$ is when $n=4$ and $B=S^2$. In all other low-dimensional cases it reduces to a central extension of $\pi_1(B)$ by $\mathbb{Z}$. -In general the assertion that $R$ is connected is the same as requiring that $f^{-1}(\partial N)$ be connected. And that translates into the homomorphism -$$ -\varphi j_*:\pi_1(\partial N) \to G -$$ -being surjective, where $j:\partial N \to X-B$ is the inclusion. -The condition that actual ramification occurs, translates into the condition that the homomorphism -$$ -\varphi i_*:\pi_1(C) \to G -$$ -is nontrivial, where $i:C \to X-B$ is the inclusion. -In general the image of $\varphi j_*:\pi_1(\partial N)\to G$ gives the group of deck transformations on any one of the path components of the pre-image of the circle bundle $\partial N$ in $Y$. -It follows that for each component $R_k$ of the pre-image of the branch set, the projection $R_k\to B$ is a Galois covering with group of deck transformations isomorphic to -$$ -\varphi j_*(\pi_1(\partial N))/ \varphi i_*(\pi_1(C)). -$$ -The components of $R$ are permuted transitively by the action of $G$ on $Y$. The full ramification covering $R\to B$ is the quotient map for the action of $G$ restricted to $R$. The covering $R\to B$ will be Galois if and only if the image $\varphi i_*(\pi_1(C))$ is a normal subgroup of $G$, in which case the group of the covering is $G/ \varphi i_*(\pi_1(C))$. -Note, by the way, that since the image of $\pi_1(C)$ is central in $\pi_1(\partial N)$, it follows that if there is nontrivial ramification and $G$ has trivial center, then the pre-image of the branch set cannot be connected.<|endoftext|> -TITLE: Are (completely) positive maps approximated by normal (completely) positive maps? -QUESTION [8 upvotes]: Let $\mathcal{H}$ denote a Hilbert space and $B(\mathcal{H})$ denote the algebra of all bounded operators on $\mathcal{H}$. By recognizing the (Banach) dual of $B(\mathcal{H})$ with the double dual of trace-class operators, one can show using standard result of Banach space theory that, any bounded linear functional $\phi$ on $B(\mathcal{H})$ can be approximated in weak$^*$ topology by (bounded) trace-class operators. In other words, $\phi$ is approximated by normal linear functionals on $B(\mathcal{H})$. My question is the following: -If the linear functional $\phi$ is positive, can $\phi$ be approximated by positive normal linear functionals in weak$^*$ topology? -Moreover, can this be generalized to completely positive maps? The topology here in consideration is bounded-weak topology. More specifically, if $M$ is a von Neumann algebra, then can every completely positive map $\Phi:M\to B(\mathcal{H})$ be approximated by normal completely positive maps in bounded-weak topology? -Some reference would be appreciated on these topics as I am new to them. Thank you. - -REPLY [8 votes]: Also the answer to the second question is yes, and the approximation may be chosen to converge in the point-ultrastrong$^*$ topology. -First, by choosing a net of finite rank orthogonal projections $p_i \in B(\mathcal{H})$ such that $p_i \rightarrow 1$ strongly, the completely positive maps $\Phi_i : M \rightarrow B(p_i H) : \Phi_i(a) = p_i \Phi(a) p_i$ converge to $\Phi$ in the point-ultrastrong$^*$ topology. So it suffices to deal with completely positive maps $\Phi : M \rightarrow M_n(\mathbb{C})$. This can be found in [BO, Corollary 1.6.3]. By [BO, Proposition 1.5.14], -$$\omega : M_n(\mathbb{C}) \otimes M \rightarrow \mathbb{C} : \omega(A) = \sum_{i,j} \Phi(A_{ij})_{ij}$$ -is a positive functional. Choose a net $\omega_k$ of normal positive functionals on $M_n(\mathbb{C}) \otimes M$ that converge pointwise to $\omega$. Again by [BO, Proposition 1.5.14], there is a corresponding net of completely positive maps -$$\Phi_k : M \rightarrow M_n(\mathbb{C}) : (\Phi_k(a))_{ij} = \omega_k(e_{ij} \otimes a) \; .$$ -By construction, the maps $\Phi_k$ are normal and they converge to $\Phi$ in the point-norm topology. -[BO] N.P. Brown and N. Ozawa, C$^*$-algebras and finite-dimensional approximations. -Graduate Studies in Mathematics 88. American Mathematical Society, Providence, 2008.<|endoftext|> -TITLE: Is the weight in Serre's conjecture "minimal"? -QUESTION [9 upvotes]: Serre's conjecture says that given any odd, irreducible, continuous representation $\rho:G_{\mathbb{Q}}\rightarrow GL_2(\overline{\mathbb{F}_p})$ there is some eigenform $f$ of weight $k(\rho)$, level $N(\rho)$, and nebentype $\epsilon(\rho)$, such that $\rho$ is isomorphic to the mod $p$ representation $\bar \rho_f$ associated to $f$. Up until now, I have intuitively thought of the Serre weight $k(\rho)$ as being the minimal weight among all weights of eigenforms whose representations are isomorphic to $\rho$. I am no longer sure whether this is correct: I am currently reading Edixhoven's survey of Serre's conjecture, and it seems like one might be able to cook up an example where the Serre weight $k(\bar\rho_f)$ of some weight $k$ eigenform $f$ is actually greater than the weight of $f$, which seems strange to me... Can this happen? I.e., are there examples where the Serre weight of a representation coming from a weight $k$ eigenform is actually greater that $k$? -To give this a bit more context, fix $p\nmid N$ and let $f \in S_{k}(\Gamma_0(N))$ be an eigenform with mod $p$ representation $\bar\rho_f$. Fix an inertia subgroup $I_p\subseteq G_{\mathbb{Q}}$ at $p$ and write $I_{p,w}\subseteq I_p$ for its wild subgroup. If $\bar\rho_f\mid_{I_{p,w}}$ is nontrivial, we can uniquely write -$$ -\bar\rho_f\mid_{I_{p}}=\begin{pmatrix}\chi^\beta &* \\ 0& \chi^{\alpha} \end{pmatrix} -$$ -where $\chi$ is the mod $p$ cyclotomic character and $\alpha$, $ -\beta$ are integers such that $0\leq \alpha\leq p-2$, $1\leq\beta\leq p-1$. As stated in the Edixhoven article, setting $a=\min(\alpha,\beta)$ and $b=\max(\alpha,\beta)$, we define $k(\bar\rho_f)$ to be $1+pa+b+p-1$ if $\chi^{\beta-\alpha}=\chi$ and $\bar \rho_f\mid_{G_{\mathbb{Q}_p}} -\otimes \chi^{-\alpha}$ is not finite at $p$, otherwise we define it to be $1+pa+b$. -Now consider the case where $p=3$, and suppose that we have some weight 6 eigenform $f$ on $\Gamma_0(N)$ such that -$$ -\bar\rho_f\mid_{I_{3}}=\begin{pmatrix}1 &* \\ 0& \chi \end{pmatrix} -$$ -and $\rho_f\mid_{G_{\mathbb{Q}_3}} -\otimes \chi^{-1}$ is not finite at $3$. Then the formula above would say that $f$ has Serre weight $k(\bar \rho_f)=8$, which is larger than the weight of $f$. I am of course assuming that we can find such a form satisfying these conditions, but even so, I see no obvious reason why this shouldn't occur... - -REPLY [4 votes]: I believe a modular form as you describe indeed cannot exist. I think it's easier to think about these issues if they're translated into the representation-theoretic language of Serre weights. Associated to $\overline{\rho}$ is a set of Serre weights, i.e., of irreducible mod $3$ representations of $\mathrm{GL}(2,\mathbf{F}_3)$. If $\overline{\rho}|_{D_3}$ is a tres ramifiee extension of $\chi$ by $\chi^2 = 1$, then this set is the singleton $\{ \mathrm{det} \otimes \mathrm{Sym}^2\}$, where $\mathrm{Sym}^a$ denotes the $a$th symmetric power of the standard representation. (Since we're in two dimensions you can think about this in terms of crystalline lifts: your $\overline{\rho}|_{D_3}$ will lift to a crystalline extension of cyclotomic by cyclotomic^4 (but not of cyclotomic by cyclotomic^2, because of the tres ramifiee condition). So it has a crystalline lift with Hodge-Tate weights $(s,t) = (4,1)$ [but not $(2,1)$] and the numerology is that the Serre weight associated to this lift is $\mathrm{det}^t \otimes \mathrm{Sym}^{s-t-1}$.) -Now the Breuil-Mézard conjecture for $\mathrm{GL}(2,\mathbf{Q}_3)$, which is known (and due to Shen-Ning Tung in this case), has the consequence that if $\overline{\rho}|_{D_3}$ has a crystalline lift of weights $(0,k-1)$ (in particular if $\overline{\rho}$ comes from a modular form of weight $k$ and level prime to $p$) then $\{ \mathrm{det} \otimes \mathrm{Sym}^2\}$ must be a Jordan-Hölder factor of $\mathrm{Sym}^{k-2}$. -If I haven't miscalculated, the Jordan-Hölder factors of $\mathrm{Sym}^4$ are $\{\mathrm{Sym}^2, \mathrm{det}, \mathrm{det}^2\}$, whereas the the Jordan-Hölder factors of $\mathrm{Sym}^6$ are $\{\mathrm{Sym}^2, \mathrm{det} \otimes \mathrm{Sym}^2, \mathrm{det}^2\}$, so indeed $k=8$ is the minimal weight where such $\overline{\rho}$ can be found.<|endoftext|> -TITLE: A "subtle" isomorphism testing problem: $\mathbb{Z}\ltimes_{A} \mathbb{Z}^5\cong \mathbb{Z}\ltimes_{B}\mathbb{Z}^5$ or not? -QUESTION [9 upvotes]: EDIT: I've made a mistake with the matrices. Now it is corrected. -A couple of days ago I asked this question. There, answerers gave me excellent hints to solve that case and others too. But I've found two matrices for which I had to distinguish the corresponding groups and I couldn't solve the problem with any of those techniques (see below). -I'm almost done with my task of analyzing these matrices and groups and I think the following are the last examples which I've to distinguish. -Let $A=\begin{pmatrix} 1&0&0&0&0\\0&0&-1&0&0 \\ 0&1&-1&0&0\\ 0&0&0&0&-1\\0&0&0&1&1\end{pmatrix}=1\oplus A'$ and $B=\begin{pmatrix} 1&0&0&0&0\\ 0&0&-1&1&0\\0&1&-1&0&0\\0&0&0&0&-1\\0&0&0&1&1\end{pmatrix}=1\oplus B'$. - -Question: Are isomorphic $G_A=\mathbb{Z}\ltimes_A \mathbb{Z}^5$ and $G_B=\mathbb{Z}\ltimes_B\mathbb{Z}^5$? Well again I think they're not. - -Thoughts and advances: -$\bullet$ $B$ is not conjugate to $A$ or $A^{-1}$ in $\mathsf{GL}_5(\mathbb{Z})$ but they are in $\mathsf{GL}_5(\mathbb{Q})$. They are both of order 6 and have 1 as eigenvalue. -$\bullet$ I computed the 2 and 3 exponencial central classes up to 11 (as the answerers taught me in the previous question) and result in isomorphic pQuotients. The presentations are: -> GA := Group a^t=a, b^t=b^-1*c^-1, c^t=b, d^t=d*e^-1, e^t=d>; -> -> GB := Group a^t=a, b^t=b^-1*c^-1, c^t=b, d^t=b*c*d*e^-1, e^t=b*c*d>; - - -$\bullet$ I've found in this paper Corollary 8.9 (cf Prop 4.2 and Def 4.3) that if I had $\mathbb{Z}\ltimes_{A'} \mathbb{Z}^4$ and $\mathbb{Z}\ltimes_{B'}\mathbb{Z}^4$ then those semidirect products wouldn't be isomorphic because $B'\not\sim A',(A')^{-1}$ in $\mathsf{GL}_5(\mathbb{Z})$ (and because neither have 1 as eigenvalue) but I don't know how to relate these semidirect products with the original ones I have. -$\bullet$ $G_A^{ab}\cong G_B^{ab}\cong \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}_3$. Also I tried to compute the quotients $G/\gamma_i(G)$ (for $i\geq 2$) where $\gamma_i=[\gamma_{i-1}(G),G]$ and $\gamma_1=[G,G]$ and all of them are isomorphic. -$\bullet$ Thinking of $\Gamma_A=(G_A/Z(G_A))$ and $\Gamma_B=(G_B/Z(G_B))$ I get $\Gamma_A\cong \mathbb{Z}_6\ltimes_{A'}\mathbb{Z}^4$ and $\Gamma_B\cong \mathbb{Z}_6\ltimes_{B'}\mathbb{Z}^4$ and I computed the abelianization ($\mathbb{Z}_6\oplus\mathbb{Z}_3$) and pQuotients here too but I couldn't distinguish them either. -> Gamma_A := Group (c,d), t^6, a^t=a^-1*b^-1, b^t=a, c^t=c*d^-1, d^t=c>; -> - -> Gamma_B := Group (c,d), t^6, a^t=a^-1*b^-1, b^t=a, c^t=a*b*c*d^-1, d^t=a*b*c>; - -I hope someone can help me again with this. - -REPLY [7 votes]: Claim. The groups $G_A$ and $G_B$ are not isomorphic. -We will use the following lemma. -Lemma. Let $\Gamma_A = G_A/Z(G_A) = C_6 \ltimes_{A'} \mathbb{Z}^4$ and $\Gamma_B = G_B/Z(G_B) = C_6 \ltimes_{B'} \mathbb{Z}^4$, where $C_6 = \langle \alpha \rangle$ is the cyclic group of order $6$ and $A'$ and $B'$ are obtained from $A$ and $B$ respectively by removing the first row and the first column. -Let $M_A \Doteq [\Gamma_A, \Gamma_A]$ and $M_B \Doteq [\Gamma_B, \Gamma_B]$ be the corresponding derived subgroups considered as $\mathbb{Z}[C_6]$-modules where $\alpha$ acts as $A'$ on $M_A$ and as $B'$ on $M_B$. Then we have the following $\mathbb{Z}[C_6]$-module presentations: -$$M_A = \langle x, y \vert \, (\alpha^2 + \alpha + 1)x = (\alpha^2 - \alpha + 1)y = 0\rangle -$$ and -$$ -M_B = \langle x \,\vert \, (\alpha^4 + \alpha^2 + 1)x = 0\rangle -$$ - -Proof. Use the description of $M_A$ as $(A' - 1_4) \mathbb{Z}^4$ and observe how $A'$ transforms the column vectors of $A' - 1_4$. Do the same for $M_B$. - -We are now in position to prove the claim. - -Proof of the claim. It suffices to show that $\Gamma_A$ and $\Gamma_B$ are not isomorphic. An isomorphism $\phi: \Gamma_A \rightarrow \Gamma_B$ would induce an isomorphism $M_A \rightarrow M_B$ of Abelian groups. As we necessarily have $\phi((\alpha, (0, 0, 0, 0))) = (\alpha^{\pm 1}, z)$ for some $z \in \mathbb{Z}^4$ and since the presentations of the above lemma remain unchanged if we replace $\alpha$ by $\alpha^{-1}$, the isomorphism $\phi$ would induce an isomorphism of $\mathbb{Z}[C_6]$-modules. This is impossible because $M_A$ cannot be generated by less than two elements whereas $M_B$ is cyclic over $\mathbb{Z}[C_6]$. Observe indeed that $M_A$ surjects onto $\mathbb{F}_4 \times \mathbb{F}_4$ where $\mathbb{F}_4 \simeq \mathbb{Z}[C_6]/(2, \alpha^2 + \alpha + 1)$ is the field with four elements. - - -Addendum 1. -Let $G$ be finitely generated group $G$. -We denote by $d(G)$ the rank of $G$, i.e., the minimum number of generators of $G$. -For these two special instances, we actually have $d(G_A) = 4$ and $d(G_B) = 3$. -In general, it can be difficult to compute the rank of a group, but some knowledge is available for $G_A$ and some other split extensions by cyclic groups, see [1, Corollary 2.4] and [2, Theorem A and Section 3.1]. -Let us set $G_A = \mathbb{Z} \ltimes_A N_A$ with $N_A \Doteq \mathbb{Z}^n$. Then $N_A$ can be endowed with the structure of a $\mathbb{Z}[C]$ module where $C \subset G_A$ is the infinite cyclic group generated by $a \Doteq (1, (0, \dots, 0)) \in G_A$ which acts on $N_A$ via conjugation, or equally, via multiplication by $A$. -Let $R$ be a unital ring and let $M$ be a finitely generated $R$-module. -We denote by $d_R(M)$ the minimum number of generators of $M$ over $R$. Then the aforementioned results implies that $$d(G_A) = d_{\mathbb{Z}[C]}(N_A) + 1.$$ -Let us denote by $(e_1, \dots, e_n)$ the canonical basis of $\mathbb{Z}^n$. -For $A$ and $B$ as in OP's question, it is easy to derive the following $\mathbb{Z}[C]$-module presentations: $$N_A = \langle e_1, e_2, e_4 \, \vert (a - 1)e_1 = (a^2 + a + 1)e_2 = (a^2 - a + 1)e_4 = 0 \rangle$$ and -$$N_B = \langle e_1, e_5 \, \vert (a - 1)e_1 = (a^4 + a^2 + 1)e_5 = 0 \rangle.$$ -From these presentations and the above rank formula, we can easily infer the claimed identities, that is, $d(G_A) = 4$ and $d(G_B) = 3$. -Addendum 2. Let $C_A$ be the cyclic subgroup of $G_A$ generated by $a \Doteq (1, (0, \dots, 0))$ and $K_A$ the $\mathbb{Z}[C_A]$-module defined as in Johannes Hahn's answer (and subsequently mine) to this MO question. -Let $\omega(A)$ be the order of $A$ in $\text{GL}_n(\mathbb{Z})$, that we assume to be finite, and set $e_0 \Doteq (\omega(A), (0, \dots, 0)) \in G_A$. Let us denote by $(e_1, \dots, e_n)$ the canonical basis of $\mathbb{Z}^n \triangleleft G_A$. -It has been established that the pair $\{K_A, K_{A^{-1}}\}$ of $\mathbb{Z}[C]$-modules is an isomorphism invariant of $G_A$, where $C = C_A \simeq C_{A^{-1}}$ with the identification $a \mapsto (1, (0, \dots,0)) \in G_{A^{-1}}$. It can be used to address the previous example and this one as well. -For the instances of this MO question, straightforward computations show that -$$K_A = K_{A^{-1}}= \langle e_0, e_1, e_2, e_4 \, \vert \, (a - 1)e_0 = (a - 1)e_1 = (a^2 + a + 1)e_2 = (a^2 - a + 1)e_4 = 0\rangle$$ and -$$K_B = \langle e_0, e_1, e_5 \, \vert \, -(a - 1)e_0 = (a -1)e_1 = (a^4 + a^2 + 1)e_5 = 0\rangle.$$ -Since $d_{\mathbb{Z}[C_A]}(K_A) = 4$ and $d_{\mathbb{Z}[C_B]}(K_B) = 3$ the groups $G_A$ and $G_B$ are not isomorphic. - -[1] G. Levitt and V. Metaftsis, "Rank of mapping tori and companion matrices", 2010. -[2] L. Guyot, "Generators of split extensions of Abelians groups by cyclic groups", 2018.<|endoftext|> -TITLE: Linkage and Cohen-Macaulay-ness -QUESTION [5 upvotes]: Suppose I have a reduced l.c.i. scheme with two irreducible components: $X = Y \cup Z$. I want to say that if $Y$ is Cohen-Macaulay then $Z$ is as well. -I think this follows from Eisenbund Theorem 21.23 (which has a typo: the first $J = (0:_A I)$ should be deleted). Or from Peskine and Szpiro, "Liaison des variétés algébriques," Proposition 1.3, which is essentially the same. -Am I understanding correctly? - -REPLY [4 votes]: The question is local. So, let $R$ be a local ring which is Gorenstein. $I,J\subset R$ define $Y,Z$ as in your question. Then you have an exact sequence $0\to I\to R\to R/I\to 0$ and we are assuming that $R/I$ is Cohen-Macaulay. Notice that all $R,R/I,R/J$ have the same dimension $d$. Dualizing, one gets $0\to\omega_{R/I}\to R\to R/J\to 0$. This implies that the depth of $R/J\geq d-1$. By going modulo a general set of $d-1$ elements in the maximal ideal, one can reduce to the case where $d=1$. Now dualize again to get, $0\to \operatorname{Hom}_R(R/J,R)\to R\to R/I\to\operatorname{Ext}^1_R(R/J,R)\to 0$. It is clear by naturality, that the map $R\to R/I$ is onto and thus the ext is zero. This says that depth of $R/J>0$ which is what we wanted.<|endoftext|> -TITLE: Finite simple groups with three conjugacy classes of maximal local subgroups -QUESTION [5 upvotes]: $\DeclareMathOperator\PSL{PSL}$In [1] it was proved that - -A finite nonsolvable group $G$ has three conjugacy classes of maximal subgroups if and only if $G/\Phi(G)$ is isomorphic to $\PSL(2,7)$ or $\PSL(2,2^q)$ for some prime $q$. -This implies that, among finite simple groups, only only $\PSL(2,7)$ and $\PSL(2,2^q)$ have three conjugacy classes of maximal subgroups. - -My question: I wonder if we can also find all finite simple groups with three conjugacy classes of maximal local subgroups. -A subgroup is a local subgroup if it is the normalizer of some nontrivial subgroup of prime power order. A proper local subgroup is a maximal local subgroup if it is maximal among proper local subgroups. -Maximal subgroups are not necessarily local, and maximal local subgroups are not necessarily maximal subgroups. I know that the three non-conjugate maximal subgroups of $\PSL(2,4)=A_5$ and $\PSL(2,7)$ are local respectively, but is it true that $\PSL(2,2^q)$ has three conjugacy classes of maximal local subgroups for each prime $q$? And how can I find all simple groups with such property? -Any help is appreciated! -Reference: -[1] Belonogov, V. A.: Finite groups with three classes of maximal subgroups. Math. Sb., 131, 225–239 (1986) - -REPLY [2 votes]: If you want to use CFSG, I think this is doable (and may even be doable without CFSG if you use H. Bender's classification of finite groups with a strongly embedded subgroup, with some additional work). -For sporadic groups, is a matter of checking. In an alternating group $G$, there are three non-conjugate maximal local subgroups, -$N_{G}(\langle (123) \rangle)$, $N_{G}( \langle (12)(34), (13)(24) \rangle )$ and $N_{G}(\langle (12345) \rangle)$, and for $n \geq 7$, it is easy to construct maximal local subgroups not conjugate to any of these. -For simple groups of Lie type of defining characteristic $p$, then for rank at last three, there are at least three conjugacy classes of maximal $p$-locals ( which are parabolics here) which are also non-conjugate maximal local subgroups. -Also, (with a few exceptions), the normalizer of the maximal torus ,$T$, of the Borel is contained in a maximal local subgroup which is not conjugate to any parabolic. -Hence the real work is in dealing with simple groups of Lie type of defining characteristic $p$ and of rank at most $2$, and this should be manageable.<|endoftext|> -TITLE: $\mathrm{mod}\:p$ Galois representation with respect to Zariski topology -QUESTION [5 upvotes]: Let $G$ be the absolute Galois group of some number field. Can there be a semisimple continuous representation $G\to GL_n(\overline{\mathbb{F}_p})$ (the latter has Zariski topology) with infinite image? - -REPLY [7 votes]: No. A quick proof uses the existence of Haar measure on compact topological groups like the Galois group. -The kernel would be a closed subgroup of the Galois group with infinite index, and thus would have Haa measure $0$. However, because $GL_n (\overline{\mathbb F_p})$ is countable, countably many translates cover the Galois group, so the Galois group would have measure $0$, contradicting the fact that it has measure $1$. -I'm sure a more direct proof that avoids Haar measure can be made to work as well.<|endoftext|> -TITLE: Ferrand pushouts for algebraic stacks -QUESTION [6 upvotes]: Given algebraic spaces $X$, $Y$, $Z$ with a finite morphism $Y \rightarrow X$ and a closed immersion $Y \hookrightarrow Z$, the pushout $P \cong X \amalg_Y Z$ exists as an algebraic space (cf. Temkin and Tyomkin - Ferrand pushouts for algebraic spaces, Theorem 6.2.1 (ii),(b)). -Does this still hold if everywhere above we replace "algebraic space" by "algebraic stack" (or "DM stack")? - -REPLY [10 votes]: Yes, this is exactly Theorem A.4 in my old preprint Compactification of tame Deligne–Mumford stacks which is long overdue to appear on the arXiv. The proof is rather terse but fairly standard (compare with Appendix A of Jack Hall's Openness of versality via coherent functors). A more detailed proof will also appear in the upcoming paper: -Artin algebraization for pairs with applications to the local structure of stacks and Ferrand pushouts (now on arXiv:2205.08623) -which is joint with Jarod Alper, Jack Hall and Daniel Halpern-Leistner. There we prove the existence of pushouts of affine morphisms and closed immersions in the category of (quasi-separated) algebraic stacks. -PS. At least some authors call pushouts of finite morphisms and closed immersions for "pinchings". Ferrand considered the more general case of affine morphisms, hence the name Ferrand pushouts.<|endoftext|> -TITLE: Equivariant (co)homology of flag manifolds, convolution algebra and nil hecke algebra? -QUESTION [8 upvotes]: For a complex reductive group $G$ and its Borel subgroup $B$, it seems to be well-known that the equivariant homology group $H^G_*(G/B\times G/B)$ forms a nil-Heck algebra -$$NH=\Bbbk[y_i,\partial_{j}]_{{1\leq i\leq n}\atop{1\leq j\leq n-1}}\big/\left<\begin{array}{c} -\partial_i\partial_{i+1}\partial_i=\partial_{i+1}\partial_{i}\partial_{i+1}\\ -\partial_{i}\partial_j=\partial_j\partial_i, |i-j|\geq 2\\ -\partial_i^2=0\end{array},\quad -\begin{array}{c}y_j\partial_j=\partial_j y_{j+1}\\ -y_{j+1}\partial_j=\partial_j y_{j}\\ -y_j\partial_i=\partial_iy_j, |i-j|\geq 2 -\end{array}\right>$$ -under convolution with the Schubert cells $X_w$ corresponding to the symbol $\partial_w$. Besides, its action over the equivariant cohomology group $H_G^*(G/B)=H_T(pt)=\Bbbk[x_1,\ldots,x_n]$ is the Demazure operator. -But I did not find any reference for this fact even for the definition of convolution. I only saw the usual homology (Borel--Moore homology) version and the K-theory version in Representation Theory and Complex Geometry by Neil ChrissVictor Ginzburg. Besides, they refer without proofs. Maybe it can be defined by sheaf theory, but then how to compute with the Schubert cells? Since $H_G(G/B\times G/B)=H_T(G/T)$, it has Schubert cells. -In the cohomology case, we can define the convolution in a proper way to be -$$H^*_G(B\times A)\times H^*_G(C\times B)\stackrel{p_1^*\otimes p_3^*}\longrightarrow H_G^*(C\times B\times A)\otimes H_G^*(C\times B\times A)\stackrel{\smile}\longrightarrow H_G^*(C\times B\times A)\stackrel{(p_2)_*}\longrightarrow H_G^*(C\times A)$$ -The last map is the Gysin push forward when $B$ is smooth compact. The problem of homology is that there is no intersection product for $EG\times_G C\times B\times A$ since it is infinite dimensional. Moreover when I compute the convolution over equivariant cohomology, it does not give the a proper isomorphism $H_G^*(G/B\times G/B)\to NH$. -My question is, are there any references for the fact that $H^G_*(G/B\times G/B)\cong NH$ under convolution and references for the definition of convolution algebra in equivariant homology? Further I also wonder if there is an isomorphism from cohomology to $NH$? - -REPLY [3 votes]: I did more computation recently, and I got what I desired. -Firstly, to be exact, it should be the cohomology group rather than homology group, and presentation in the question is wrong, it should be -$$\Bbbk\left_{{1\leq i\leq n}\atop{1\leq j\leq n-1}}\bigg/ -\left<\begin{array}{c} -\partial_i\partial_{i-1}\partial_i=\partial_{i-1}\partial_i\partial_{i-1},\\ -|i-j|\geq 2, \quad \partial_i\partial_j=\partial_j\partial_i,\\ -\partial_i^2=0. \end{array}\begin{array}{c} -X_iX_j=X_jX_i,\\ -\partial_iX_j-X_{s_i(j)}\partial_i\\ -=\delta_{i,j}-\delta_{i+1,j}. -\end{array}\right>$$ -I was mislead by Kumar's definition (Kac-Moody Groups, their Flag Variety and Representation Theory) of Nil-Hecke ring and the definition of convolution in homology. - -To prove this, one can first do it in nonequivairant case, the $G$-orbits of $G/B\times G/B$ are one-to-one correspondent to $B$-orbit of $G/B$, i.e. Schubert cells. -The Poincar'e duality of each, say $\partial_w$, with respect to the Schuber cell $BwB/B$, acts on $H^*(G/B)$ by the Demazure operator $\partial_w$. To check this, it suffices to do the intersection product, where they all intersects transversally. -The $X_i=X_i\partial_e$, where $H^*(G/B)$ acts on $H^*(G/B\times G/B)$ by the first projection acts on $H^*(G/B)$ by left multiplication $X_i$. -Now relation is easy to check, by a standard topological argument, it is an isomorphism (for example, Harish–Leray). Actually, $H^*(G/B\times G/B)$ is actually the subalgebra generated by left mutiplications and Demazure operators in $\operatorname{End}_{\Bbbk}(H^*(G/B))$. -To deal with equivariant case, we first do it in $T$-equivariant case, it is harmless since $H_G^*(X)\to H_T^*(X)$ is always injective ($\operatorname{char} \Bbbk=0$). -There no longer exists Poincar'e duality, but the pairing of cells also gives a well-defined cohomology class. The computation of the result of paring of cells in nonequivariant case can be directly move to equivariant case. As a result, so it also acts as Demazure operator. -The rest is completely the same to nonequivariant case. Actually, $H_G^*(G/B\times G/B)$ is actually the subalgebra generated by left mutiplications and Demazure operators in $\operatorname{End}_{H_G^*(pt)}(H^*(G/B))$. The actions are all $H_G^*(pt)$ map by the associativity of convolution -$$H_G^*(G/B\times G/B)\stackrel{\displaystyle\curvearrowright}{\phantom{\square}} - H_G^*(G/B\times pt)\stackrel{\displaystyle\curvearrowleft}{\phantom{\square}} H_G^*(pt\times pt). $$ - - -The last two points are wrong. The real reason is, over characteristic zero case, $H_G(G/B)$ is known to be free of rank $\dim H(G/B)$ over $H_G(pt)$. So bthe convolution algebra is eaxctly of rank $\dim H(G/B)^2$ over $H_G(pt)$. This is the main point.<|endoftext|> -TITLE: Multiplication maps for big line bundles -QUESTION [7 upvotes]: In Birational Geometry of Algebraic Varieties, Kollar and Mori write that for a line bundle "being big is essentially the birational version of being ample" (page 67). Recall that a line bundle $L$ on a projective variety $X$ of dimension $d$ is big if -$$ \limsup_{n \to \infty } \dfrac{H^0(X,L^n)}{n^d} \neq 0.$$ -In other words, the rate of growth of the spaces of global sections is as big as possible. Big line bundles tend to exhibit behavior analogous to ample line bundles. I will give a couple of examples. In what follows, let $X$ be a variety over the complex numbers and let $L$ be a line bundle on $X$. - -Suppose $X$ is normal. If $L$ is ample, some power of $L$ defines an embedding in a projective space. Analogously, if $L$ is big, some power of $L$ defines a map - -$$ \varphi_m: X \dashrightarrow H^0(X,L^m)$$ -that is birational onto its image (Positivity in Algebraic Geometry I, page 139). - -If $L$ is ample, some power of $L$ is globally generated. On the other hand, if $L$ is big, some positive power of $L$ is generically globally generated; that is, the natural map - -$$ H^0(X,L^m) \otimes \mathcal{O}_{X} \rightarrow L^m$$ -is generically surjective (Positivity in Algebraic Geometry I, page 141). -Now, to get to my question, recall that if $L$ is ample, there exists a natural number $m$ such that the multiplication maps -$$ H^0(X,L^a) \otimes H^0(X,L^b) \rightarrow H^0(X,L^{a+b}) $$ -are surjective for $a,b \geq m$ (Positivity in Algebraic Geometry I, page 32). -Question: Do big line bundles have a property analogous to the surjectivity of multiplications maps? -It is not clear to me what this property should be, but I would hope that these multiplication maps have eventually high rank in some suitable sense. - -REPLY [2 votes]: If $R(L)=\oplus H^0(mL)$ is not finitely generated, the above surjectivity will fail, however it will hold "asymptotically" for any big line bundle $L$. In fact, by Fujita's approximation of big classes (see eg. Lazarsfeld's Positivity book Theorem 11.4.4), for any $\epsilon >0$ there is a birational modification $f:X'\to X$ such that $f^*L=A+E$ where $A$ is an ample $\mathbb Q$-divisor and $E$ is an effective $\mathbb Q$-divisor such that ${\rm vol}(A)>{\rm vol}(L)-\epsilon$. Thus, by the ample case, there is an $m>0$ such that $H^0(aA)\otimes H^0(bA)\to H^0((a+b)A)$ is surjective for all $a,b\geq m$ sufficiently divisible (so that $aA$ and $bA$ are Cartier). Since $f_*\mathcal O _{X'}(aA)\subset \mathcal O _X(aL)$, we see that if $$V_{a,b}={\rm Im} \left( H^0(aL)\otimes H^0(bL)\to H^0((a+b)L))\right),$$ -then $\dim V_{a,b}/h^0((a+b)L)>(1-\epsilon)$ for $m\gg 0$.<|endoftext|> -TITLE: A basic question about Young symmetrizers -QUESTION [5 upvotes]: This is probably elementary for experts on the representation theory of the symmetric group, but I did not find the answers I need by a cursory look at the usual textbooks (they could be there, but I gave up trying to decipher conflicting notations and conventions). -Let $\lambda$ be an integer partition of $n$. A Young tableau $T$ is a bijective filling of the corresponding Young diagram with the numbers $1,2,\ldots,n$. -For a permutation $\sigma$, let $\sigma T$ denote the tableau obtained by replacing each entry $i$ by $\sigma(i)$. Standard tableaux are the ones where entries increase in each row and column. -For a Young tableau $T$, let $C(T)$ denote the group of permutations which preserve the columns of $T$, and let $R(T)$ the group of permutations which preserve the rows of $T$. In the group algebra $\mathbb{C}\mathfrak{S}_n$ of the symmetric group -define, as usual, the elements -$$ -P(T)=\sum_{\sigma\in R(T)} \sigma -$$ -and -$$ -N(T)=\sum_{\sigma\in C(T)} {\rm sgn(\sigma)}\ \sigma\ . -$$ -Finally, the convention for Young symmetrizer that I will use is -$$ -Y(T)=P(T)N(T)\ . -$$ -Q1: Is it always true that for two different standard Young tableaux $T,T'$, of the same shape $\lambda$, we have $Y(T)Y(T')=0$? -Q2: Let $T$ be a standard Young tableau and let $\alpha\in C(T)$, $\beta\in R(T)$ be such that $\alpha\beta T$ is also standard. Does this necessarily require $\alpha=\beta=Id$? - -REPLY [8 votes]: For Q1 the answer in general is no. Young symmetrizers can be used to give a decomposition of $\mathbb C[S_n]$ into a direct sum of minimal left ideals but in general they are not pairwise orthogonal. One can actually characterize precisely when $Y(T)Y(T')\neq 0$ holds: (i) the underlying shape of $T$ and $T'$ needs to be the same (ii) every row of $T$ must intersect every column of $T'$ in at most one element. -So an explicit example where they fail to be orthogonal is -$$T=\begin{matrix} - 1 & 3 & 5 \\ - 2 & 4 & \\ - \end{matrix} \qquad, \qquad T'=\begin{matrix} - 1 & 2 & 3 \\ - 4 & 5 & \\ - \end{matrix}$$ -See Orthogonal sets of Young symmetrizers by Stembridge for more details. -For Q2 the answer is no. It is possible for $\alpha\beta T$ to be a different standard Young tableaux. For example you can take -$$T=\begin{matrix} - 1 & 2 & \\ - 3 & 4 & \\ - 5 & & \\ - \end{matrix}$$ -and also $\alpha=(24)(35)$, $\beta=(34)$.<|endoftext|> -TITLE: "Tietze-like transformations" for defining interesting bijections between algebraic structures -QUESTION [5 upvotes]: Consider the following two definitions of the natural numbers: - -The natural numbers are the algebraic structure $\mathbb{N}_1$ generated by one constant, $0$ and one unary function, $S$ (and no relations). -The natural numbers are the monoid $(\mathbb{N}_2, 0, +)$ with presentation $\langle 1 \mid \rangle$. - -These two definitions are equivalent, in the sense that there exists a certain "nice" bijection between the structures they define: namely, the unique function $f : \mathbb{N}_1 \to \mathbb{N}_2$ with $f(0) = 0$ and $f(S(x)) = f(x) + 1$, which is a bijection. -How could we prove that the bijection $f$ satisfying those two equations really does exist? One option, of course, is to take your favorite set theory, define all of these objects formally, and use first-order logic to construct a proof. -However, it's also possible to show that this bijection exists without using set theory or logic at all. The method is essentially the same as using Tietze transformations to define an isomorphism between the groups generated by two group presentations. -Groups and Tietze transformations -Consider the following two group presentations (which I'm writing using deliberately bulky notation). First: - -$a$ -$b$ -$ab$ = $ba$ -$a^3 = b^2$ - -And second: - -$c$ - -Both of these presentations present the infinite cyclic group. If we want to construct an isomorphism, then using set theory and first-order logic would be overkill. Instead, we can simply use Tietze transformations, as shown: - -Add a generator $c$ with definition $c = b a^{-1}$ (5 and 6 below). -Add a relation $c^3 = b$ (7 below). Proof: $c^3 = (b a^{-1})^3 = b^3 a^{-3} = b^3 b^{-2} = b$. -Add a relation $c^2 = a$ (8 below). Proof: $c^2 = (b a^{-1})^2 = b^2 a^{-2} = a^3 a^{-2} = a$. -Remove the relation $c = b a^{-1}$ (6 below). Proof: $c = c^3 c^{-2} = b a^{-1}$. -Remove the relation $ab = ba$ (3 below). Proof: $ab = c^2 c^3 = c^3 c^2 = ba$. -Remove the relation $a^3 = b^2$ (4 below). Proof: $a^3 = (c^2)^3 = (c^3)^2 = b^2$. -Remove the generator $a$ with definition $a = c^2$ (1 and 8 below). -Remove the generator $b$ with definition $b = c^3$ (2 and 7 below). - - -$a$ -$b$ -$ab = ba$ -$a^3 = b^2$ -$c$ -$c = b a^{-1}$ -$c^3 = b$ -$c^2 = a$ - -After all of these transformations have been completed, the only item remaining is item 5, which is the generator $c$. -So, using the Tietze transformations, we have constructed an isomorphism $f$ from the first group to the second group, with $f(a) = c^2$ and $f(b) = c^3$. -Generalizing -Define a generic presentation as an algebraic theory. We refer to the free algebra of the theory as "the algebra generated by the presentation." -The first definition of the natural numbers above ($\mathbb{N}_1$) is formalized as this generic presentation: - -$0$ (a generator which is a nullary operation) -$S(-)$ (a generator which is a unary operation) - -And the second definition of the natural numbers ($\mathbb{N}_2$) is formalized like so: - -$0$ -$P(-,-)$ -$P(0,x) = x$ -$P(x,0) = x$ -$P(x,P(y,z)) = P(P(x,y),z)$ -$1$ - -As mentioned at the beginning of this question, there is a bijection $f : \mathbb{N}_1 \to \mathbb{N}_2$ with $f(0) = 0$ and $f(S(x)) = P(f(x), 1)$. How can we construct this bijection? -Much as we did with the infinite cyclic group above, we can construct this bijection using a sequence of transformations which are similar to the Tietze transformations. -However, the Tietze transformations themselves are not quite sufficient for this purpose. In addition to the four Tietze transformations, we need to add two additional "Tietze-like transformations" to our toolbox. Specifically, in addition to adding (or removing) a constant along with a single equation defining it, I think we need to be able to add (or remove) a function symbol along with a set of equations defining it. (I think we can require the set of equations to be a primitive recursive function definition; I haven't worked out the details.) -Furthermore, two of the Tietze transformations need to be altered to make them more powerful. Specifically, the Tietze transformations allow us to add or remove a relation if we can prove that relation from the other relations using a simple proof by substitution. We need to alter these so that we are also permitted to use inductive proofs of equality. (Again, I haven't worked out the details.) -The resulting "toolset" consists of six Tietze-like transformations: adding or removing a (constant) generator; adding or removing a function; and adding or removing a relation (potentially using an inductive proof). These six transformations are sufficient to construct the desired bijection between $\mathbb{N}_1$ and $\mathbb{N}_2$. -Below is the construction. Once again, it consists of a sequence of Tietze-like transformations, starting with the first presentation and ending with the second one. - -Add a generator $1$ with definition $1 = S(0)$ (3 and 4 below). -Add a generator $P(-,-)$ with definition $P(x,S(y)) = S(P(x,y))$ and $P(x,0) = x$ (5, 6, and 7 below). -Add a relation $P(x,1) = S(x)$ (8 below). Proof: $P(x,1) = P(x,S(0)) = S(P(x,0)) = S(x)$. -Add a relation $P(0,x) = x$ (9 below). The proof is by induction. The $0$ case: $P(0,0) = 0$. The $S$ case: $P(0,S(x)) = S(P(0,x)) = S(x)$. -Add a relation $P(x,P(y,z)) = P(P(x,y),z)$ (10 below). The proof is by induction. The $0$ case: $P(x,P(y,0)) = P(x,y) = P(P(x,y),0)$. The $S$ case: $P(x,P(y,S(z))) = P(P(x,y),S(z))$ (details omitted). -Remove the relation $1 = S(0)$ (4 below). Proof: $1 = P(0,1) = S(0)$. -Remove the relation $P(x,S(y)) = S(P(x,y))$ (6 below). Proof: $P(x,S(y)) = P(x,P(y,1)) = P(P(x,y),1) = S(P(x,y))$. -Remove the generator $S(-)$ with definition $S(x) = P(x,1)$ (2 and 8 below). - - -$0$ -$S(-)$ -$1$ -$1 = S(0)$ -$P(-,-)$ -$P(x,S(y)) = S(P(x,y))$ -$P(x,0) = x$ -$P(x,1) = S(x)$ -$P(0,x) = x$ -$P(x,P(y,z)) = P(P(x,y),z)$ - -When we work through the above list of transformations, we start with items 1 and 2, and we add items 3 through 10, and then we remove items 2, 4, 6 and 8, leaving items 1, 3, 5, 7, 9, and 10. This list of items is identical to the second presentation above, so we have successfully constructed the bijection. -Summary and question -There are 6 "Tietze-like transformations" that we've used to construct the desired bijection between the two definitions of the natural numbers above: - -Adding a relation which can be proved from the other relations. -Removing a relation which can be proved from the other relations. -Adding a (nullary) generator along with a relation defining it. -Removing a (nullary) generator along with a relation defining it. -Adding a generator with any arity along with a set of equations constituting a primitive recursive definition of that generator. -Removing a generator with any arity along with a set of equations constituting a primitive recursive definition of that generator. - -Transformations 1 through 4 are the Tietze transformations; 5 and 6 are new. (Of course, 3 and 4 are special cases of 5 and 6.) -I'm sure that I'm not the first person to come up with this idea. Have these "Tietze-like transformations" been studied before? - -REPLY [3 votes]: Tietze transformations for arbitrary algebraic theories (with respect to their presentations) have been considered in Malbos–Mimram's Homological Computations for Term Rewriting Systems, in the context of rewriting systems (that is, equations are considered directed). They consider (Definition 7) two operations (and their converses): - -Adding a superfluous operation. Add a new operation $f : n$ and a rewrite $R : t \Rightarrow f(x_1, \ldots, x_n)$ for some term $x_1, \ldots, x_n \vdash t$. -Adding a derivable relation. For terms $u, v$ that are interderivable (via rewriting), add a new relation $R : u \Rightarrow v$. - -They state (Proposition 8) that two algebraic theories $P$ and $Q$ are isomorphic (and hence have the same models) iff they are Tietze equivalent in that $Q$ may be derived from $P$ through a series of Tietze transformations. (Though they do not give a proof in the paper.) -Their second Tietze operation (and its converse) correspond to your operations 1 and 2. However, their first operation (and its converse) are simpler than your operations 3 through to 6.<|endoftext|> -TITLE: Do the isomorphism classes of indecomposable objects in $R{\text{-mod}}$ form a set? -QUESTION [10 upvotes]: Let $R$ be a unital (associative) ring. Consider the category $R\text{-mod}$ of unitary left $R$-modules. Set $\text{Indec}(R)$ to be the class of all isomorphism classes of indecomposable objects in $R\text{-mod}$. - -Is $\text{Indec}(R)$ a set? If not, for which ring $R$ is $\text{Indec}(R)$ a set, and for which ring $R$ is $\text{Indec}(R)$ a proper class? - -Clearly, if $R$ is semisimple, then $\text{Indec}(R)$ coincides with the set $\text{Irred}(R)$ of isomorphism classes of simple $R$-modules. I am not sure what would happen if $R$ is nonsemisimple. -If possible, I would like to request a reference that discusses a more general result. That is, if $\mathscr{C}$ is an arbitrary abelian category, and $\text{Indec}(\mathscr{C})$ is the class of all isomorphism classes of indecomposable objects in $\mathscr{C}$. How do we tell when $\text{Indec}(\mathscr{C})$ is a set? -Just as in the case of $R\text{-mod}$, the same observation holds: if $\mathscr{C}$ is semisimple, then $\text{Indec}(\mathscr{C})$ is identical to the class $\text{Irred}(\mathscr{C})$ of isomorphism classes of simple objects in $\mathscr{C}$. However, I am very certain that there are nonsemisimple abelian category $\mathscr{C}$ such that $\text{Indec}(\mathscr{C})$ is a proper class. Such an example is very welcome. $\phantom{aaa}$ Edit: With help from YCor, at least when $R=\mathbb{Z}$, $\text{Indec}(\mathbb{Z})$ is a proper class. -Remark. I am even more interested in the case where $R$ is a (not necessarily finite-dimensional) $\mathbb{K}$-algebra for some field $\mathbb{K}$. We may assume that $R$ is countable-dimesional. Even more specifically, I would like to know what happens if $R$ is the universal enveloping algebra of some (not necessarily finite-dimensional) Lie algebra $\mathfrak{g}$ over some field $\mathbb{K}$. Again, we may assume that $\mathfrak{g}$ is countable-dimensional. However, anything that can elaborate me on how to find out when $\text{Indec}(\mathscr{C})$ is a set will be greatly appreciated. - -REPLY [12 votes]: In Conjecture $1_{\infty}$ of -Simson, Daniel, On large indecomposable modules, endo-wild representation type and right pure semisimple rings., Algebra Discrete Math. 2003, No. 2, 93-118 (2003). ZBL1067.16029, -Simson conjectures that a right noetherian ring either - -is right pure semisimple (which would imply it is right artinian, and conjecturally would imply that it is right artinian of finite representation type), or -has indecomposable modules of arbitrarily large cardinality. - -So if this conjecture is true, then the answer to the question "for which noetherian rings is $\operatorname{Indec}(R)$ a proper class" is "almost all of them". -Simson proved this conjecture for several classes of finite dimensional algebras (e.g., finite dimensional local $k$-algebras with residue field $k$, and group algebras of finite groups) in -Simson, Daniel, On Corner type endo-wild algebras., J. Pure Appl. Algebra 202, No. 1-3, 118-132 (2005). ZBL1151.16014. -For non-noetherian rings the answer will be more complicated, since, for example, there are non-noetherian rings for which every indecomposable module is simple.<|endoftext|> -TITLE: On the equation $a^6+b^6+c^6=d^2$ -QUESTION [12 upvotes]: I have been studying the equation $a^6+b^6+c^6=d^2$, trying to find rational solutions. I know it is a K3 surface, with high Picard rank, so there should be rational or elliptic curves on it. -When Elkies found solutions to the equation $a^4+b^4+c^4=d^4$, he started by using the simpler equation $r^4+s^4+t^2=1$. Taking inspiration from this, I looked at the equation $(2):y^2=x^3+z^6+1$. Noting the two trivial solutions $(x,y)=(-1,z^3)$ and $(x,y)=(-z^2,1)$, and taking the sum of the two points (in the elliptic curve addition sense), and multiplying to remove fractions yields the parametric equation: -$(3z^6+9z^5+15z^4+17z^3+15z^2+9z+3)^2=(2z^4+4z^3+5z^2+4z+2)^3+(z^2+z)^6+(z+1)^6$ -This would result in a solution to my original problem, if $2z^4+4z^3+5z^2+4z+2$ were a square. This yields an equation $u^2=2z^4+4z^3+5z^2+4z+2$. By inspection I found the solution $(z,u)=(-1,\pm1)$. Unfortunately, if $z=-1$, then the equation above collapses into $1^2=1^6+0^6+0^6$, which is trivial. However, this equation can be converted into Weierstrass form, resulting in: $y^2=x^3-x^2-8x-4$, with the point $(-1,1)$ taken to be the point at infinity, and the other point $(-1,-1)$ taken to the point $(-2,0)$. -However, this elliptic curve has only those two rational points on it, therefore no solutions to the original equation can be obtained. Other points I have found on equation (2) by using chord and tangent methods starting from those two initial points result in parametric equations with a polynomial of too high order to result in an elliptic curve. -What are some other more fruitful approaches to this problem? Note that I am aware of other questions on mathoverflow providing some solutions. However, I am looking for a way to generate infinitely many solutions. This would preferably be with a parametric equation, however I'll also be happy with an elliptic curve and a rational point of infinite order. -If at all possible I would appreciate hints in the right direction over full solutions. I'm wanting to use this to grow my expertise and problem solving ability in this area. I'll update this question with any future attempts that are worth mentioning. - -REPLY [5 votes]: The mentioned equation has infnitely many solutions. See the paper by A. Bremner and myself: A. Bremner, M. Ulas, On $x^a\pm y^b \pm z^c \pm w^d = 0, 1/a + 1/b + 1/c + 1/d = 1$, Int. J. Number Theory, 7(8) (2011), 2081-2090.<|endoftext|> -TITLE: Norms in quadratic fields -QUESTION [14 upvotes]: This should be well-known, but I can't find a reference (or a proof, or a counter-example...). Let $d$ be a positive square-free integer. Suppose that there is no element in the ring of integers of $\mathbb{Q}(\sqrt{d})$ with norm $-1$. Then I believe that no element of $\mathbb{Q}(\sqrt{d})$ has norm $-1\ $ -(in fancy terms, the homomorphism $H^2(G,\mathscr{O}^*)\rightarrow H^2(G,\mathbb{Q}(\sqrt{d})^*)$, with $G:=\operatorname{Gal}(\mathbb{Q}(\sqrt{d})/\mathbb{Q})=$ $\mathbb{Z}/2 $, is injective). Is that correct? If yes, I'd appreciate a proof or a reference. - -REPLY [5 votes]: Dirichlet's version of Gauss composition is in the book by Cox, (page 49 in first) with a small typo corrected in the second edition. -For our purpose, duplication, it has a better look to equate $a=a'$ from the start, with $\gcd(a,b) = 1$ sufficing, -$$ \left( ax^2 +bxy+ acy^2 \right) \left( aw^2 +bwz+ acz^2 \right) = c X^2 + b XY + a^2 Y^2 $$ -where -$$ X = axz + ayw+byz \; \; , \; \; \; Y = xw - c yz $$ -so that the square of $\langle a,b,ac \rangle$ is $\langle c,b,a^2 \rangle.$ -Today's question concerns $c=-1$ -$$ \left( ax^2 +bxy -ay^2 \right) \left( aw^2 +bwz -az^2 \right) = - X^2 + b XY + a^2 Y^2 $$ -where -$$ X = axz + ayw+byz \; \; , \; \; \; Y = xw + yz $$ -so that $$\langle a,b,-a \rangle^2 = \langle -1,b,a^2 \rangle.$$ -We also see Stanley's fact that the discriminant is the sum of two squares, $b^2 + 4 a^2$ the way I wrote things. -By the Gauss theorem on duplication, $ \langle -1,b,a^2 \rangle$ is in the principal genus -Furthermore, we now know that the principal form is $SL_z \mathbb Z$ equivalent to -$$ \langle 1,b,-a^2 \rangle $$ -The principal form may not integrally represent $-1$ but does so rationally. -As to being in the same genus, we can use Siegel's definition of rational equivalence without essential denominator. -$$ -\left( -\begin{array}{rr} - 0 & 1 \\ - -a^2 & -b \\ -\end{array} -\right) -\left( -\begin{array}{rr} - 1 & \frac{b}{2} \\ - \frac{b}{2} & -a^2 \\ -\end{array} -\right) -\left( -\begin{array}{rr} - 0 & -a^2 \\ - 1 & -b \\ -\end{array} -\right) = \; a^2 \; -\left( -\begin{array}{rr} - -1 & \frac{b}{2} \\ - \frac{b}{2} & a^2 \\ -\end{array} -\right) -$$ -$$ -\left( -\begin{array}{rr} - b & 1 \\ - -a^2 & 0 \\ -\end{array} -\right) -\left( -\begin{array}{rr} - -1 & \frac{b}{2} \\ - \frac{b}{2} & a^2 \\ -\end{array} -\right) -\left( -\begin{array}{rr} - b & -a^2 \\ - 1 & 0 \\ -\end{array} -\right) = \; a^2 \; -\left( -\begin{array}{rr} - 1 & \frac{b}{2} \\ - \frac{b}{2} & -a^2 \\ -\end{array} -\right) -$$<|endoftext|> -TITLE: Online, evolving, collaborative foundational text projects -QUESTION [34 upvotes]: There are two online, evolving, collaborative "foundational text" projects for research mathematicians that I am aware of: -(1) The Stacks Project for algebraic geometry -(2) Kerodon for categorical homotopy theory -They are sort of Bourbaki for the internet age. I'd like to know if there are others of the same nature that are ongoing or in the offing. -Please note that I am not looking for texts or monographs available online which is why I have highlighted the adjectives in the first line. - -REPLY [2 votes]: For number theory, there is this very interesting project similar to the Stacks project that seems to cover in great detail all of the main subfields of modern number theory: -https://github.com/holdenlee/number-theory<|endoftext|> -TITLE: A functional equation in two complex variables -QUESTION [5 upvotes]: Let $X$ be a compact metric space, or just $X=\mathbb T$, the unit circle, if it helps. We consider only continuous, complex-valued functions on $X$. - -Let $\varepsilon >0$. Is there $\delta > 0$ such that for any given functions $f, g$ on $X$ that are nowhere zero and any function $d$ with $\|d\|_\infty\leqslant \varepsilon$ (the supremum norm) there are two functions $a,b$ such that $fa + gb +ab = d$ with $\|a\|_\infty, \|b\|_\infty \leqslant \delta$? - -I have an abstract argument for the circle but I feel there should be an elementary solution, maybe modulo Urysohn's lemma. - -REPLY [8 votes]: $Hello$, Tomasz! (for some reason the MO prohibits saying "Hi" or "Hello" in the normal text mode). Nice to see you back. Apparently you are still asking the same question whether a function $H$ close to the product $fg$ can be represented as a product $FG$ where $F$ is close to $f$ and $G$ is close to $g$ but now just in the continuous category. -The answer is "Not necessarily" even for the closed unit disk $\mathbb D=\{|z|\le 1\}$. The obvious counterexample would be $f(z)=Mz,g(z)=M\bar z$, $H=M^2|z|^2+\varepsilon$ with huge $M$, but you tried to exclude it by demanding that $f$ and $g$ vanish nowhere. However, it doesn't save the day. Indeed, consider any non-negative continuous functions $\varphi, \psi:\mathbb D\to[0,1]$ such that $\varphi=0$ on $[0,1]$ and $\varphi=1$ outside a small neighborhood of $[0,1]$ while $\psi$ has the same properties with respect to the interval $[-1,0]$. Put $f(z)=Mz\varphi(z)-\frac{\varepsilon}{6M}, g(z)=M\bar z\psi(z)+\frac{\varepsilon}{6M}$. Then the product $fg$ is $\varepsilon$-close to a strictly positive function $H(z)=M^2|z|^2\varphi(z)\psi(z)+\frac\varepsilon 2$. However, if $H=FG$ and $F,G$ are $M/10$ close to $f$ and $g$ respectively, then the argument of $F$ should essentially follow that of $f$ on the left semicircle where $|f|$ is large and then that of $1/g$ on the right semicircle where $|g|$ is large (because $F=H/G$ and we control the arguments of both $H$ and $G$), i.e., it cannot deviate much from that of $z$ anywhere, so the winding number will be $1$ and $F$ will be forced to have a zero inside $\mathbb D$, which is impossible since $H>0$ everywhere. -Of course, if $X$ is the circle, this effect is excluded and the answer becomes "Yes", but you said that you knew it yourself, so I'll stop here for now.<|endoftext|> -TITLE: Lines on an anticanonical K3 on a Fano 3-fold -QUESTION [13 upvotes]: Let $Y_d$ be a Fano threefold of Picard rank $1$ and index $2$ (eg cubic 3fold). There is a natural anticanonical map $Y_d\to \mathbb{P}^{d+1}$. Smooth sections of the anticanonical bundle are $K3$ surfaces, so we can ask the following - -Question: does such a general $K3$ surface contain lines in $\mathbb{P}^{d+1}$? - -REPLY [15 votes]: If $Y$ is a Fano threefold with Picard rank $1$, then its general anticanonical element $X \in |-K_Y|$ is a $K3$ surface with Picard rank $1$. In particular, $X$ contains no lines. -This is explained (for instance) at p. 797 of -C. F. Doran, A. Harder, A. Y. Novoseltsev, A. Thompson: Calabi–Yau threefolds fibred by high rank lattice polarized K3 surfaces, Mathematische Zeitschrift 294 (2020).<|endoftext|> -TITLE: Non-isomorphic complex Lie groups with the same exceptional Lie algebra for $\mathfrak{g_2,f_4,e_6,e_7,e_8}$? -QUESTION [8 upvotes]: An exceptional complex Lie algebra is a simple Lie algebra whose Dynkin diagram is of exceptional (nonclassical) type. There are exactly five such Lie algebras: $\mathfrak{g}_{2}$, ${\mathfrak {f}}_{4}$,${\mathfrak {e}}_{6}$, -${\mathfrak {e}}_{7}$, ${\mathfrak {e}}_{8}$; their respective dimensions are 14, 52, 78, 133, 248. -See https://en.wikipedia.org/wiki/Exceptional_Lie_algebra -Usually, given a complex Lie algebra, - -there could be non-isomorphic connected complex Lie groups with the same given Lie algebra. - -For example, the ${\rm SO}(N)$ and ${\rm Spin}(N)$ can have the same Lie algebra ${\mathfrak {so}}_{n}$, but they are non-isomorphic Lie groups -because ${\rm SO}(N)={\rm Spin}(N)/(\mathbb{Z}/2\mathbb{Z})$ has smaller center than ${\rm Spin}(N)$. -In particular, ${\rm SO}(5)$ and ${\rm Sp}(2)\simeq {\rm Spin}(5)$ are non-isomorphic Lie groups with isomorphic Lie algebras ${\mathfrak {so}}_{5}\simeq{\mathfrak{sp}}_2$. -Questions: - -It is commonly said that the Lie groups with given Lie algebra $\mathfrak{g}_{2}$, ${\mathfrak {f}}_{4}$,${\mathfrak {e}}_{6}$, -${\mathfrak {e}}_{7}$, ${\mathfrak {e}}_{8}$ are $G_2$, $F_4$, $E_6$, $E_7$, $E_8$. However, do we have non-isomorphic Lie groups with the same exceptional Lie algebra $\mathfrak{g}_{2}$, ${\mathfrak {f}}_{4}$,${\mathfrak {e}}_{6}$, -${\mathfrak {e}}_{7}$, ${\mathfrak {e}}_{8}$? - -What are the centers $Z(G)$ of these Lie groups $G$? For $G_2$, $F_4$, $E_6$, $E_7$, $E_8$ and possibly others with the same given Lie algebra? - -What are the homotopy groups $$\pi_d(G)$$ of these Lie groups $G$ for lower dimensions? say $d=0,1,3,4,5,...$? - - -We already know that $\pi_2(G)=0$ for any Lie group. -I appreciate your patience, comments and answers - -REPLY [15 votes]: I prefer to use the language of algebraic groups. -All algebraic groups and Lie algebras are defined over $\Bbb C$. -1. Let ${\mathfrak g}$ be a semisimple Lie algebra. -Consider the automorphism group ${\rm Aut\,}{\mathfrak g}$, -its identity component $G^{\rm ad}:=({\rm Aut\,}{\mathfrak g})^0$, -and the group of outer automorphisms ${\rm Out\,} {\mathfrak g}:=({\rm Aut\,} {\mathfrak g})/({\rm Aut\,} {\mathfrak g})^0$. -We say that $G^{\rm ad}$ is the adjoint group (or the group of adjoint type) with Lie algebra ${\mathfrak g}$. Note that $Z(G^{\rm ad})=\{1\}$. -2. Starting with a semisimple Lie algebra ${\mathfrak g}$, -one can construct the simply connected group $G^{\rm sc}$ with Lie algebra ${\mathfrak g}$; -see Steinberg, Lectures on Chevalley groups, AMS, 2016. -Note that $\pi_1(G^{\rm sc})=\{1\}$. -This algebraic group $G^{\rm sc}$ has the following universal property: -for any algebraic group $H$ with Lie algebra ${\mathfrak h}$ -and for any homomorphism of Lie algebras $\varphi_{\rm Lie}\colon {\mathfrak g}\to{\mathfrak h}$, -there exists a unique homomorphism of algebraic group $\varphi\colon G^{\rm sc}\to H$ inducing $\varphi_{\rm Lie}$. -3. For any connected algebraic group $G$ with Lie algebra ${\mathfrak g}$, there exists a canonical surjective homomorphism -$$\rho\colon G^{\rm sc}\to G $$ -inducing the identity isomorphism on ${\mathfrak g}$; see above. -We have -$$\pi_1(G^{\rm sc})=\{1\},\quad \pi_1(G)={\rm ker}\,\rho.$$ -On the other hand, we have a canonical surjective homomorphism -$${\rm Ad}\colon G\to G^{\rm ad}\subseteq {\rm Aut\,} {\mathfrak g}$$ -with kernel $Z(G)$. -Write -$$C=Z(G^{\rm sc})=\pi_1(G^{\rm ad}).$$ -The homomorphism -$$ {\rm Ad}\colon G\to G^{\rm ad}$$ -induces a homomorphism -$$i\colon \pi_1(G)\to\pi_1(G^{\rm ad})=C.$$ -Moreover, the homomorphism -$$\rho\colon G^{\rm sc}\to G$$ -induces a homomorphism -$$j\colon C=Z(G^{\rm sc})\to Z(G).$$ -In this way we obtain a short exact sequence -$$1\to\pi_1(G)\overset{i}{\longrightarrow} C\overset{j}{\longrightarrow} Z(G)\to 1.$$ -Conversely, for each subgroup $F\subseteq C$ one can associate a connected semisimple group -$ G_F:=G^{\rm sc}/F$ with Lie algebra ${\mathfrak g}$, with fundamental group $\pi_1(G_F)=F$, and with center $Z(G_F)=C/F$. -In this way we obtain a canonical bijection between the set of subgroups of $C$ up to conjugation by ${\rm Out\,} {\mathfrak g}$ -and the set of isomorphism classes of connected semisimple algebraic groups with Lie algebra ${\mathfrak g}$. -It is known that ${\rm Out\,} {\mathfrak g}$ is canonically isomorphic to ${\rm Aut\,} {\rm Dyn}({\mathfrak g})$, -where ${\rm Dyn}({\mathfrak g})$ is the canonical Dynkin diagram of ${\mathfrak g}$. -4. Let us return to our exceptional simple Lie algebras. -The group $C=C({\mathfrak g})$ can be found, for instance, in tables in the book by Bourbaki "Lie Groups and Lie Algebras, Chapters 4-6", -or in the book by Onishchik and Vinberg "Lie Groups and Algebraic Groups", Springer-Verlag, 1990. -For ${\mathfrak g}_2$, ${\mathfrak f}_4$, and ${\mathfrak e}_8$ we have $C({\mathfrak g})=\{1\}$. -Thus there is only one (up to isomorphism) algebraic group $G^{\rm sc}({\mathfrak g})=G^{\rm ad}({\mathfrak g})$ with Lie algebra ${\mathfrak g}$. -For ${\mathfrak g}={\mathfrak e}_6$ we have $C({\mathfrak g})\simeq {\Bbb Z}/3{\Bbb Z}$. This group has no nontrivial subgroups. -Thus there are exactly two connected algebraic groups (up to isomorphism) $E_6^{\rm sc}$ and $E_6^{\rm ad}$ with Lie algebra ${\mathfrak e}_6$. -We have -$$Z(E_6^{\rm sc})=\pi_1(E_6^{\rm ad})\simeq{\Bbb Z}/3{\Bbb Z}.$$ -For ${\mathfrak g}={\mathfrak e}_7$ we have $C({\mathfrak g})\simeq {\Bbb Z}/2{\Bbb Z}$. This group has no nontrivial subgroups. -Thus there are exactly two connected algebraic groups (up to isomorphism) $E_7^{\rm sc}$ and $E_7^{\rm ad}$ with Lie algebra ${\mathfrak e}_7$. -We have -$$Z(E_7^{\rm sc})=\pi_1(E_7^{\rm ad})\simeq{\Bbb Z}/2{\Bbb Z}.$$ -5. The real forms of a connected algebraic group of an exceptional type correspond bijectively to the real forms of -(or real structures on) its Lie algebra. -My favorite way to classify those is via Kac diagrams. -See Table 7 in the book by Onishchik and Vinberg. -The number of real forms is 2 for ${\mathfrak g}_2$, 3 for ${\mathfrak f}_4$, 3 for ${\mathfrak e}_8$, 4 for ${\mathfrak e}_7$, 5 for ${\mathfrak e}_6$. -These real forms are listed also in Table V in Chapter X of Helgason's book "Differential Geometry, Lie Groups, and Symmetric Spaces" -(Helgason lists all non-compact forms). -Helgason classifies real forms using the original method of Kac with infinite dimensional Lie algebras. -Onishchik and Vinbeg use another method, which gives exactly the same answer (the same Kac diagrams).<|endoftext|> -TITLE: Groups that act transitively on $\mathrm{Gr}(k,\Bbb R^n)$ but not transitively on $\mathrm{Gr}(k+1,\Bbb R^n)$ -QUESTION [8 upvotes]: Is it known for which $n, k\in\Bbb N$ there exists a matrix group $\Gamma\subseteq\mathrm{GL}(\Bbb R^n)$ that - -acts transitively on $\mathrm{Gr}(k,n)$, i.e., on the $k$-dimensional subspaces of $\Bbb R^n$, but -acts not transitively on $\mathrm{Gr}(k+1,n)$, i.e., on the $(k+1)$-dimensional subspaces of $\Bbb R^n$? - -For example, $\mathrm U(n)$ (acting on $\Bbb C^n\cong\Bbb R^{2n}$) acts transitively on $\mathrm{Gr}(1,\Bbb R^{2n})$ but not on $\mathrm{Gr}(2,\Bbb R^{2n})$. - -REPLY [4 votes]: This is only a partial answer, and it's based on YCor's comment about the groups that act transitively on spheres. What is missing, as YCor commented, is knowing that if $G\subset\mathrm{GL}(n,\mathbb{R})$ acts transitively on $\mathrm{Gr}(k,\mathbb{R}^n)$ for some $k$ with $12$, the oriented $2$-plane bundle $E_2$ has nonzero Euler class in $H^2(\mathrm{Gr}^+(k,\mathbb{R}^n),\mathbb{Z})\simeq \mathbb{Z}$. In particular, it is not the sum of two line bundles (which would be trivial). -Now suppose that $G$ acts transitively on $\mathrm{Gr}^+(2,\mathbb{R}^n)$ and let $H\subset G$ be the subgroup that fixes $e_0\in \mathrm{Gr}^+(2,\mathbb{R}^n)$. Then $H$ acts on the $2$-plane $e_0\subset\mathbb{R}^n$ and hence on $\mathrm{Gr}^+(1,e_0)\simeq S^1$. -If the orbits of $H$ on $\mathrm{Gr}^+(1,e_0)$ are open, then there is only one orbit, so that $H$ acts transitively on $\mathrm{Gr}^+(1,e_0)$. Since $e_0$ was arbitrary, it follows that $G$ acts transitively on $\mathrm{Gr}^+(1,\mathbb{R}^n)$, as desired. Meanwhile, if there were a non-open orbit, then fixing one such $H$-orbit $X_0\subset \mathrm{Gr}^+(1,e_0)\subset \mathrm{Gr}^+(1,\mathbb{R}^n)$ and looking at its $G$-orbit, i.e., -$$ -X = \{ ([g v], g(e_0)\ |\ [v]\in X_0, g\in G\ \} -$$ -gives a bundle $X\to\mathrm{Gr}^+(2,\mathbb{R}^n)$ with discrete fibers over the simply-connected base $\mathrm{Gr}^+(2,\mathbb{R}^n)$. Hence it is topologically trivial, so that there exists a section $S\subset X$ that is the projectivization of a line bundle $L\subset E_2$, contradicting the fact that $E_2$ has no rank $1$-subbundles. Thus, this cannot occur, and the proof for $k=2$ is complete. -In the case $k=3$, one has to deal with the cases $n=4$ and $5$ separately. When $n=4$, the group $G = \mathrm{SU}(2)$ acts transitively on $\mathrm{Gr}^+(3,\mathbb{R}^4)\simeq S^3$, but does not act transitively on $\mathrm{Gr}^+(2,\mathbb{R}^4)$. When $n=5$, applying duality and the above argument, one sees that if $G$ acts transitively on $\mathrm{Gr}^+(3,\mathbb{R}^5)$, then it acts transitively on $\mathrm{Gr}^+(4,\mathbb{R}^5)$, but then, by the paper YCor cites, this says $G$ must contain a copy of $\mathrm{SO}(5)$ and hence acts transitively on $\mathrm{Gr}^+(k,\mathbb{R}^5)$ for all $k$. -Thus, we can assume that $n\ge 6$. In this case, we have that $H^p(\mathrm{Gr}^+(3,\mathbb{R}^n),\mathbb{Z})=0$ for $p = 1, 2, 3$, but that, not only is $H^4(\mathrm{Gr}^+(3,\mathbb{R}^n),\mathbb{R})\not=0$, we have that $p_1(E_3)\not=0$. By a characteristic class argument, it follows that the bundle $E_3\to\mathrm{Gr}^+(3,\mathbb{R}^n)$ does not split as the sum of a line bundle and a $2$-plane bundle. -Now, I claim that if $G\subset\mathrm{GL}(n,\mathbb{R})$ acts transitively on $\mathrm{Gr}^+(3,\mathbb{R}^n)$ then it acts transitively on $\mathrm{Gr}^+(2,\mathbb{R}^n)$. To see this, suppose that $G$ acts transitively on $\mathrm{Gr}^+(3,\mathbb{R}^n)$, and let $H\subset G$ be the stabilizer of $e_0\in \mathrm{Gr}^+(3,\mathbb{R}^n)$, and consider the action of $H$ on $e_0\simeq\mathbb{R}^3$. -First, suppose that $H$ has only open orbits in $\mathrm{Gr}^+(1,e_0)\simeq S^2$. Then $H$ acts transitively on $\mathrm{Gr}^+(1,e_0)$ and hence $G$ acts transitively on $\mathrm{Gr}^+(1,\mathbb{R}^n)$. -Second, suppose that $H$ has a 0-dimensional orbit $X_0\subset\mathrm{Gr}^+(1,e_0)$. Then, constructing the bundle $X\to\mathrm{Gr}^+(3,\mathbb{R}^n)$ with discrete fibers in the projectivization of $E_3$ as was done above for $E_2$, we see that $X$ has a section and that this can be used to construct a splitting of $E_3$ as the sum of a line bundle and a $2$-plane bundle, which is known to be impossible. -Finally, suppose that all of the orbits of $H$ on $\mathrm{Gr}^+(1,e_0)$ are either $1$- or $2$-dimensional. Not all of the orbits can be $1$-dimensional because $S^2$ has no foliation, and there cannot be more than a finite number of components of the union of the $1$-dimensional orbits (and there must be at least 1 component, otherwise we would be in the first case already dealt with). Thus, there are only a finite number of components of the union of the $2$-dimensional orbits. Each such component, which is homogeneous under a connected, finite dimensional Lie group, must have Euler characteristic either $0$ or $1$. Thus, there must exist exactly two components that have Euler characteristic 1, i.e., are contractible disks, and these must map to a single component $D_0$ in $\mathrm{Gr}(1,e_0)$. Using the contractibility of this component, we can now construct a rank 1 subbundle $L$ of $E_3\to \mathrm{Gr}^+(3,\mathbb{R}^n)$ with the property that the projectivization of $L_e$ lands in the $e$-fiber of the $G$-orbit of the contractible orbit $D_0$. Again, this is impossible because $E_3$ does not split nontrivially as a sum. -Remark: As far as I know, the only time that $G$ acts transitively on $\mathrm{Gr}(k,\mathbb{R}^n)$ with $4\le k\le n/2$ is when $G$ contains a copy of $\mathrm{SO}(n)$. When $k=3$, there is $\mathrm{Spin}(7)\subset\mathrm{SO}(8)$ -and, when $k=2$, we have $\mathrm{G}_2\subset\mathrm{SO}(7)$ and $\mathrm{Spin}(7)\subset\mathrm{SO}(8)$. Neither $\mathrm{Spin}(9)$ nor $\mathrm{Spin}(9,1)$ -in their $16$-dimensional representations act transitively on any $\mathrm{Gr}^+(k,\mathbb{R}^{16})$ except when $k=0,1,15,16$. -I believe that one could continue the above line of argument, showing that, if $G$ acts transitively on $\mathrm{Gr}^+(4,\mathbb{R}^n)$ for $n\ge 8$, then it must act transitively on $\mathrm{Gr}^+(3,\mathbb{R}^n)$, and so on. However, I think that this argument will get more complicated as $k\le n/2$ increases, and it's probably not the right approach for general $k$.<|endoftext|> -TITLE: Why is uncomputability of the spectral decomposition not a problem? -QUESTION [33 upvotes]: Below, we compute with exact real numbers using a realistic / conservative model of computability like TTE. -Assume that there is an algorithm that, given a symmetric real matrix $M$, finds an eigenvector $v$ of $M$ of unit length. -Let -$$M(\epsilon) = \begin{cases} -\left[\begin{matrix}1 & \epsilon\\ \epsilon & 1\end{matrix}\right] -,& \epsilon \geq 0 \\ -\left[\begin{matrix}1 - \epsilon & 0\\0 & 1 + \epsilon\end{matrix}\right] -,& \epsilon \leq 0 -\end{cases}$$ -and assume that it's possible to find an eigenvector $v$ of $M(\epsilon)$. - -If $\epsilon > 0$ then $v$ must necessarily be $\pm \frac 1 {\sqrt 2}\left[\begin{matrix}1\\1\end{matrix}\right]$ or $\pm \frac 1 {\sqrt 2}\left[\begin{matrix}-1\\1\end{matrix}\right]$. Observe that in all four cases, the $L^1$ norm of $v$ is $\sqrt 2$. - -If $\epsilon < 0$, then $v$ must necessarily be $\pm\left[\begin{matrix}1\\0\end{matrix}\right]$ or $\pm\left[\begin{matrix}0\\1\end{matrix}\right]$. Observe that in all four cases, the $L^1$ norm of $v$ is $1$. - - -It's easily determinable whether the $L^1$ norm of $v$ is less than $\sqrt 2$ or greater than $1$. Therefore we can decide whether $\epsilon \leq 0$ or $\epsilon \geq 0$, which is impossible! -In a way, this is strange, because many sources say that the Singular Value Decomposition (SVD) and Schur Decomposition (which are generalisations of the Spectral Decomposition) are numerically stable. They're also widely used in numerical applications. But I've just tested the examples above for small $\epsilon$ using SciPy and got incorrect results. -So my question is, how do numerical analysts get around this problem? Or why is this apparently not a problem? -I could venture some guesses: While finding eigenvectors of general matrices may be impossible, it is possible to find their eigenvalues. Also, it's possible to "shift" a problematic matrix by some small $\epsilon$ so that its eigendecomposition is computable. - -REPLY [21 votes]: This is primarily an issue of backwards vs. forwards stability. Good SVD algorithms are backwards stable in the sense that the computed singular values and singular vectors are the true singular values and singular vectors of a slightly perturbed problem. You may see this by noting that while $P$ may change drastically as you change $\epsilon$, the product $PDP^T$ changes negligibly. -The SVD is not forwards stable when the singular values have small spectral gap, as your example demonstrates and other answers here discuss in more detail. -For more on backwards and forwards stability, see, e.g., this post and the links therein: -https://math.stackexchange.com/a/78907/3060 -SCIPY uses LAPACK; some details on the stability of the algorithm are provided here: -https://www.netlib.org/lapack/lug/node97.html<|endoftext|> -TITLE: Does every set have a rigid self-map? -QUESTION [10 upvotes]: The question was asked on Mathematics Stackexchange -but has remained unanswered so far. -A self-map is a map $f:X\to X$ from a set $X$ to itself. There is an obvious notion of morphism, and thus of isomorphism and automorphism, of self-maps. [A morphism from $f:X\to X$ to $g:Y\to Y$ is a map $\phi:X\to Y$ such that $g\circ\phi=\phi\circ f$.] -A self-map is rigid if it has no non-trivial automorphism. -The question is in the title: - -Does every set have a rigid self-map? - -Clearly, the existence of a rigid self-map of a given set $X$ depends only on the cardinality $|X|$ of $X$. -There is an obvious notion of the coproduct $f:X\to X$ of a family $f_i:X_i\to X_i$ of self-maps. [The set $X$ is the disjoint union of the $X_i$ and $f$ coincides with $f_i$ on $X_i$.] Any self-map is the coproduct of its indecomposable components, and a self-map is rigid if and only if its indecomposable components are rigid and pairwise non-isomorphic. (In the sequel I use the expression "component" instead of "indecomposable component". Moreover the identity of the empty set does not count as a component.) -We claim: - -(1) If $|X|\le2^{2^{\aleph_0}}$, then $X$ has a rigid self-map. - -Define $f:\mathbb N\to\mathbb N$ by $f(i)=\max(i-1,0)$. Then $f$ is rigid. Moreover, for each $n\in\mathbb N$ the map $f$ induces a rigid self-map of the set $\{0,1,\ldots,n\}$. This proves that (1) holds for $|X|\le\aleph_0$. -It remains to prove that $X$ has a rigid self-map when $\aleph_0<|X|\le2^{2^{\aleph_0}}$. -This will follow from Lemmas 1, 2 and 3 below. -Lemma 1. Let $X$ be an infinite set and $\Sigma$ a set of non-isomorphic rigid surjective indecomposable self-maps of $X$. Assume $|\Sigma|>|X|$. Then the coproduct of the elements of $\Sigma$ is a rigid surjective self-map of a set of cardinality $|\Sigma|$. -This is obvious. -Lemma 2. Let $f$ be a rigid surjective self-map of an infinite set $X$, and $Y$ a set satisfying $|X|\le|Y|\le2^{|X|}$. Then $Y$ has a rigid self-map. -Proof. Let $X'$ be a set disjoint from $X$ and $\phi:X'\to X$ a bijection. For each subset $S$ of $X'$ put $X_S=X\sqcup S$ (disjoint union) and define $f_S:X_S\to X_S$ by setting $f_S(x)=f(x)$ for $x\in X$ and $f_S(s)=\phi(s)$ for $s\in S$. -It suffices to show that the coproduct $g:Y\to Y$ of the $f_S:X_S\to X_S$ (where $S$ runs over all the subsets of $X'$) is rigid. -Let $h:Z\to Z$ be a component of $g$. Then $h$ is a component of $f_S$ for some $S$. It is easy to see that there is a unique component $f_0:X_0\to X_0$ of $f$ such that, if we set $S_0:=S\cap\phi^{-1}(X_0)$, then $h$ is equal to -$$ -f_{0,S_0}:X_{0,S_0}\to X_{0,S_0}, -$$ -where $f_{0,S_0}$ is defined as $f_S$ was defined above (replacing the bijection $\phi:X'\to X$ with the bijection $\phi^{-1}(X_0)\to X_0$ induced by $\phi$). -Let -$$ -f_{1,T_1}:X_{1,T_1}\to X_{1,T_1} -$$ -be another component of $g$, corresponding to a subset $T$ of $X'$, and let -$$ -\psi:X_{0,S_0}\to X_{1,T_1} -$$ -be an isomorphism from $f_{0,S_0}$ to $f_{1,T_1}$. Since $X_0$ and $X_1$ are the respective images of $f_{0,S_0}$ and $f_{1,T_1}$ by surjectivity of $f$, the isomorphism $\psi$ maps $X_0$ onto $X_1$ and $S_0$ onto $T_1$. By rigidity of $f$ we have $X_0=X_1$ and $\psi(x)=x$ for all $x\in X_0$. Let $s$ be in $S_0$. It suffices to show $\psi(s)=s$. Set $x=\phi(s)\in X_0$. Then $\psi$ maps the fiber of $\phi$ above $x$ to itself, but $s$ is the only point in this fiber. This completes the proof of Lemma 2. -Lemma 3. Let $A$ be the set of all increasing self-maps of $\mathbb N$ such that $a(0)\ge1$. Then there is a family of pairwise non-isomorphic rigid surjective indecomposable self-maps -$$ -(f_a:X_a\to X_a)_{a\in A}, -$$ -where each $X_a$ is an infinite subset of $\mathbb N^2$. -Proof. Define the subset $X_a$ of $\mathbb N^2$ by the condition that $(i,j)\in X_a$ if $i\in a(\mathbb N)$ or if $j=0$, and define $f_a:X_a\to X_a$ by setting -$\bullet\ f_a(i,j)=(i,j-1)$ if $j\ge1$, -$\bullet\ f_a(i,0)=(i-1,0)$ if $i\ge1$, -$\bullet\ f_a(0,0)=(0,0)$. -Let us fix $a\in A$ and sketch the proof that $f_a$ is a rigid self-map of $X_a$. -The point $(0,0)$ is the only fixed point. The points of the form $(i,0)$ with $i\ge1$ are characterized by the fact that they have ancestors which have two parents, and any two distinct such points are at different distances to $(0,0)$. Therefore the points $(i,0)$ are fixed by any automorphism of $f_a$. The point $(a(n),j)$ with $j\ge1$ has no ancestor with two parents, its first descendent with two parents is $(a(n),0)$, which is fixed by the automorphisms of $f_a$, the point $(a(n),j)$ is at distance $j$ from $(a(n),0)$, and these properties characterize $(a(n),j)$. Thus $(a(n),j)$ is fixed by the automorphisms of $f_a$. -This argument shows also that the $f_a$ are pairwise non-isomorphic. The other statements are clear. -[If $y=f(x)$ we say that $x$ is a parent of $y$. If $y=f^n(x)$ for $n\in\mathbb N$ we say that $x$ is an ancestor of $y$ and $y$ a descendent of $x$.] -set-theory - -REPLY [11 votes]: Yes. Actually, this was part of my first answer to this question, but this was a digression there (and I also posted there another answer to the same question which addressed it and was accepted). So I'm copying this digression here and will delete the initial answer to that question to avoid a duplicate. -Fact. For every set $X$ there exists $f\in X^X$ whose centralizer in $\mathrm{Sym}(X)$ is reduced to $\{\mathrm{id}_X\}$ -It relies on the following second fact: there exists (for $X\neq\emptyset$) a rooted tree structure on $X$ whose automorphism group is trivial. Indeed, granting this, and denoting $v_0$ the root, for a vertex $v$ define $f(v)$ as $v_0$ if $v_0=v$, and as the unique vertex in $[v_0,v]$ at distance 1 to $v$ otherwise. Then $f\in X^X$ and its centralizer in $\mathrm{Sym}(X)$ is the automorphism group of the corresponding rooted tree, which is reduced to $\{\mathrm{id}_X\}$. -To prove the second fact, if $X$ is finite just take a linear tree rooted at an extremal vertex. If $X$ is infinite, by an elementary but very tricky argument (see this answer by user "bof"), there actually exist for every infinite cardinal $\kappa$, $2^{\kappa}$ pairwise non-isomorphic trees of cardinal $\kappa$ each with trivial automorphism group. [Interestingly the induction really requires proving that there are $>\kappa$ such trees, and not only a single one.]<|endoftext|> -TITLE: Looking for citable reference for a well-known fact about tensor product of finite dimensional algebras over an algebraically closed field -QUESTION [7 upvotes]: Let $K$ be an algebraically closed field and let $A$ and $B$ be finite dimensional algebras over $K$. Let $e_1,\ldots, e_n$ be orthogonal primitive idempotents of $A$ summing to $1$ and $f_1,\ldots, f_m$ be orthogonal primitive idempotents of $B$ summing to $1$. Then the $e_i\otimes f_j$ form a set of orthogonal primitive idempotents of $A\otimes_K B$ summing to $1$. Moreover, $(A\otimes_K B)(e_i\otimes f_j)\cong (A\otimes_K B)(e_{i'}\otimes f_{j'})$ if and only if $Ae_i\cong Ae_{i'}$ and $Bf_j\cong Bf_{j'}$. Note that this need not hold if $K$ is not algebraically closed. -I know how to prove these things but I want to cite this in a paper I'm writing and couldn't find this spelled out in any of my usual references on finite dimensional algebras. I don't really want to put a proof in my paper as it is a bit far a field. Actually, what I want to use this for is to conclude that the basic algebra of $A\otimes_K B$ is the tensor product of the basic algebras of $A$ and $B$ (again with $K$ algebraically closed). -I would greatly appreciate any reference, particularly, to a book. - -REPLY [5 votes]: A treatment of this can be found in chapter IV.11. in the book "Frobenius algebras I" by Skowronski and Yamagata with proofs in the case $A=B^{op}$. But the proofs work exactly the same in the general case with the exact same proofs so it can be used as a reference for the general case I think.<|endoftext|> -TITLE: Chernoff-style concentration inequality for k-tuples -QUESTION [5 upvotes]: I'm looking for a seemingly natural generalization of a Chernoff bound. -In many scenarios, we have a distribution $D$ with support $\mathsf{Supp}(D)$, and some event $E \subset \mathsf{Supp}(D)$ telling us whether a property of a sample from $D$ holds (i.e. $a \in E$ iff $a\sim D$ has the property we want). Denoting $p =\Pr_{a\sim D}[a \in E]$, we use Chernoff to say something of the kind: if I draw $n$ independent samples from $D$, then with probability at least $1-\exp(-\delta^2pn/2)$, my multiset $A = \{a_1, \cdots, a_n\}$ of samples will be "$\delta$-good", where "$\delta$-good" means that if I fix this multiset $A$ once for all and sample (uniformly over the multiset) $a$ from $A$, then $a \in E$ will hold with probability $(1-\delta)p$. This is the standard Chernoff bound for Bernouilli random variables. -In my scenario, I need a generalization of the above, where the event is over $k$-tuples of samples from $D$ (i.e. $E \subset \mathsf{Supp}(D^k)$). Let $p = \Pr_{(a_1, \cdots, a_k)\sim D^k}[(a_1, \cdots, a_k) \in E]$. Suppose that for $i=1$ to $k$, I draw $n$ independent samples $(a_{i,1}, \cdots, a_{i,n})$ from $D$, which form a multiset $A_i$. I want to be able to make statements of the form: with probability at least 'something', the multisets $(A_1, \cdots, A_k)$ will be "$\delta$-good", where "$\delta$-good" means that if I fix $(A_1, \cdots, A_k)$ once for all and uniformly sample a $k$-tuple $(a_1, \cdots, a_k)$ from $A_1 \times \cdots \times A_k$, then $(a_1, \cdots, a_k)\in E$ will hold with probability at least $(1-\delta)p$. -Of course, the standard Chernoff bound does not apply anymore (it would apply if, instead, I had fixed a single multiset $A$ of $n$ random $k$-tuples sampled from $D^k$). Other concentration bounds I'm familiar with, such as Azuma's inequality or McDiarmid's bounded difference inequality, do not seem to apply either. -Question: is any such bound known in the literature, or does it follow from any standard concentration bound? Any pointer would be welcome. To be clear, I crucially need Chernoff-level strength: Markov or anything of the kind wont do. I've tried to derive a bound of this kind, first from standard concentration bounds with limited dependence (e.g. McDiarmid), and I've searched a bit the literature, both without success. Before trying to establish it from first principles, I figured it would be better to ask first, since it looks like something people should have considered before. --- -EDIT - answering the comments of kodlu - -Do you have any other constraints on your function $f$? Lipschitz type? Subgaussian type? - -Are you referring to the function $f$ that I initially used to define the event $E$? If so, why would this function being Lipschitz or Subgaussian matter in any way? Note that $f$ has nothing to do with the function we want to be Lipschitz when applying e.g. McDiarmid's inequality. For example, if you consider the case $k=1$ (which is the base case I'm trying to generalize), then whatever $f$ is, the resulting bound is exactly a bound on a sum of independent Bernouilli random variables - that is, the function is just a direct sum, and $f$ is just what defines whether the event happened or not. I understand that my choice of notations might have been confusing, I hope switching to $E$ as suggested by dohmatob makes things better. - -What makes you think you'll get concentration in such an arbitrary setting in a product space? Do you have any experimental evidence? - -My intuition is that there should be such a bound - now, that is barely more than an intuition. I have some kind of experimental evidence, but only for the very specific context I'm actually working on, though I believe that such a bound should hold in a more general setting (which is why I refrained from describing my precise and confusing setting). -In case it helps, anyway (and simplifying a bit): in the concrete setting I'm working on, a sample from $D$ is a length-$t$ vector of bits (for some parameter $t$) where each entry is sampled independently and is biased towards $0$, and the event over a $k$-tuple of samples $(a_1, \cdots, a_k)$ is defined as follows: the fraction of positions $i \in [1, t]$ such that at least one $a_j$ contains a $1$ at position $i$ belongs to $[1/10, 9/10]$. I'm trying to show that this event happens often enough it I fix $k$ multisets of samples as I described above, and sample one entry of the $k$-tuple from each multiset. -In this setting, yes, I have some weak kind of experimental evidence, coming from the fact that this bound captures the hardness of attacking a cryptographic primitive with a restricted family of attacks (well, at least a part of the analysis requires this bound). Since it's a primitive some people tried to break with these attacks and failed, it appears likely that there is such a bound. - -REPLY [4 votes]: Theorem 2 in [1] gives a bound of $1-\frac{4e^{-\delta^2n/8}}{\delta}$. I think you can incorporate $p$ in the bound since the proof of that theorem uses the standard Chernoff bound. -[1] Yakov Babichenko, Siddharth Barman, Ron Peretz (2017) Empirical Distribution of Equilibrium Play and Its Testing Application. Mathematics of Operations Research 42(1):15-29. http://dx.doi.org/10.1287/moor.2016.0794<|endoftext|> -TITLE: Estimate for an Airy integral -QUESTION [7 upvotes]: Let me define for $x\in\mathbb R$, -$ -F(x)=\int_{\mathbb R} e^{-π t^2}\cos(x t^3) dt. -$ -I claim that $F(x)>0$ for all $x\in\mathbb R$. -Well, it is obvious for $x=0$ since $F(0)=1$ and also for $x$ near $0$ by continuity of $F$. -I guess that for $x$ large, a van der Corput method or a version of the stationary phase method should give the result. How can I prove this at "finite distance", in a situation where no asymptotic method could help? - -REPLY [9 votes]: The most elementary way is to exploit the periodicity and compare positive and negative contributions. Changing variable in the integral, with $\tau=xt^3$, and integrating by parts, you will find an analogous statement for -$$\int_0^\infty f(\tau)\sin(\tau)d\tau>0,$$ -for some positive and decreasing function (also depending on $x$) $f$, which makes the claim evident since any positive contribute on $[2k\pi,(2k+1)\pi]$ is larger in absolute value than the successive negative contribution on $[(2k+1)\pi,(2k+2)\pi]$. - -REPLY [9 votes]: $$ -\frac{F(x)}2=\int_0^\infty e^{-\pi t^2}\cos(x t^3) dt=\int_0^\infty \frac{e^{-\pi t^2}}{3xt^2}d\sin(x t^3)=-\int_0^\infty \left(\frac{e^{-\pi t^2}}{3xt^2}\right)'\sin(x t^3)dt\\=\int_0^\infty \frac1{3xt^2}\left(\frac{e^{-\pi t^2}}{3xt^2}\right)'d\left(1-\cos(x t^3)\right)=-\int_0^\infty \left(\frac1{3xt^2}\left(\frac{e^{-\pi t^2}}{3xt^2}\right)'\right)'\left(1-\cos(x t^3)\right)dt\\= -\int_{0}^\infty\frac{2e^{-\pi t^2}(2\pi^2t^4+5\pi t^2+5)}{9x^2t^6}\left(1-\cos(x t^3)\right)dt\geqslant 0. -$$<|endoftext|> -TITLE: A vector field whose flow has constant singular values -QUESTION [5 upvotes]: $\newcommand{\tr}{\operatorname{tr}}$ -$\renewcommand{\div}{\operatorname{div}}$ -Let $D \subseteq \mathbb{R}^2$ be the closed unit disk. Given a vector field $X$ on $D$, let $\psi_t$ be its flow. -Does there exist a (non-homothetic) divergence-free vector field $X \in \Gamma(D)$ whose associated flow $\psi_t \in \operatorname{diff}(D)$ has singular values which depend only on $t$ and not on the point in $D$? Equivalently, I want $|(d\psi_t)_p|^2$ to be independent of $p$. -By requiring that $X$ is not homothetic, I am excluding the possibility of $L_x g=\lambda g$ for some constant $\lambda$. - -Write $f(t,p)=|(d\psi_t)_p|^2=\tr_g\big((\psi_t^*g)_p\big)$. -A necessary and sufficient condition on $X$ is that $$\frac{\partial }{\partial t}f(t,p)=\tr_g(\psi_t^*L_Xg)$$ would be independent of $p$, for all $t$. Can we write this as an explicit PDE on $X$? -Some partial information is obtained from differentiating twice at $t=0$: -$$ -\frac{\partial^2 }{\partial t^2}\left|_{t=0}\right.\tr_g\big((\psi_t^*g)_p\big)=\operatorname{tr}_g(\mathcal{L}_X(\mathcal{L}_Xg))=2X(\operatorname{div}X)+2\|\nabla X\|^2+2\operatorname{tr}(\nabla X\circ\nabla X)= -$$ -$$ -2\|\nabla X\|^2+2\operatorname{tr}(\nabla X\circ\nabla X), -$$ -so $\|\nabla X\|^2+\operatorname{tr}(\nabla X\circ\nabla X)$ must be constant. - -$X(r,\theta)=\log r \frac{\partial}{\partial \theta}$ is an example on $D\setminus\{0\}$: -Its flow is $\psi_t: (r,\theta)\mapsto (r,\theta+t\log r)$; -$[d\psi_t]_{\{ \frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta}\}}=\begin{pmatrix} 1 & 0 \\\ t & 1\end{pmatrix}$ -is independent of $p$. - -REPLY [8 votes]: The vector fields $X$ that satisfy this condition are highly constrained, as there exist only a finite-dimensional family of $C^5$ solutions in a neighborhood of any given point in the plane. Here is how one can see this: -Using the standard coordinates, we have $g = dx^2 + dy^2$. Suppose that $X$ is defined on a simply-connected domain $D$ in the plane. Then, since $X$ is divergence-free, it can be written in the form -$$ -X = u_y\,\frac{\partial}{\partial x} - u_x\,\frac{\partial}{\partial y} -$$ -for some function $u$ on $D$. Define $\mathcal{L}_X^2g = \mathcal{L}_X(\mathcal{L}_X g)$ and, inductively $\mathcal{L}_X^{k+1}g = \mathcal{L}_X(\mathcal{L}^k_X g)$. As has been observed, we have $\mathrm{tr}_g(\mathcal{L}_X g) = 0$, since $X$ is divergence free. A direct computation gives -$$ -\mathrm{tr}_g(\mathcal{L}^2_X g) = 2\left((u_{xx}-u_{yy})^2 + (2u_{xy})^2\right) = 8c^2 -$$ -for some constant $c\ge0$. If $c=0$, then either $X = a_0\,(y-y_0)\partial_x - a_0\,(x-x_0)\partial_y$ for some constants $a_0,a_1,a_2$, and the flow of $X$ is rotation around $(x_0,y_0)$ or $X = a_1\partial_x + a_2\partial_y$ and hence the flow of $X$ is simply translation. -Thus, from now on assume that $c>0$. Then the Hessian matrix of $u$ is of the form -$$ -\begin{pmatrix} u_{xx} & u_{xy}\\u_{xy} & u_{yy}\end{pmatrix} -= \begin{pmatrix} h + c\,\cos v & c\,\sin v\\c\,\sin v & h-c\,\cos v\end{pmatrix} -$$ -for some functions $h$ (unique) and $v$ (unique up to an additive constant that is an integral multiple of $2\pi$). The identities $(u_{xx})_y = (u_{xy})_x$ and $(u_{xy})_y = (u_{yy})_x$ imply that -$$ -\mathrm{d}h = c\bigl(v_y\,\cos v-v_x\,\sin v\bigr)\,dx - + c\bigl(v_x\,\cos v+v_y\,\sin v\bigr)\,dy, -$$ -and, since $c\not=0$, the identity $\mathrm{d}(\mathrm{d}h) = 0$ gives -$$ -(v_{xx}-v_{yy}+2v_xv_y)\,\cos v + (2v_{xy}-{v_x}^2+{v_y}^2)\,\sin v = 0. -$$ -Thus, the Hessian of $v$ can be parametrized as -$$ -\begin{pmatrix} v_{xx} & v_{xy}\\v_{xy} & v_{yy}\end{pmatrix} -= \begin{pmatrix} -q+p\,\sin v - v_xv_y & \tfrac12({v_x}^2-{v_y}^2)-p\,\cos v\\ -\tfrac12({v_x}^2-{v_y}^2)-p\,\cos v & q-p\,\sin v + v_xv_y\end{pmatrix} -$$ -for some functions $p$ and $q$. Now, a straightforward calculation yields that -$$ -\mathrm{tr}_g(\mathcal{L}^3_X g) = 0 -\quad\text{while}\quad -\mathrm{tr}_g(\mathcal{L}^4_X g) = 32\,c^4 - 8c^2\,(2h-u_xv_y+u_yv_x)^2. -$$ -It follows that we must have $h = b + \tfrac12(u_xv_y-u_yv_x)$ for some constant $b$. Differentiating this equation and using the various formulae above allows one to solve for $p$ and $q$ in the form. -$$ -p = \frac{P(\cos v, \sin v, v_x,v_y,u_x,u_y)}{\bigl(({u_x}^2-{u_y}^2)\cos v +2u_xu_y\sin v\bigr)} -$$ -and -$$ -q = \frac{Q(\cos v, \sin v, v_x,v_y,u_x,u_y)}{\bigl(({u_x}^2-{u_y}^2)\cos v +2u_xu_y\sin v\bigr)}, -$$ -where $P$ and $Q$ are certain polynomials in their arguments, which I won't write out here. Moreover, using these formulae, we find that -$$ -\begin{align} -\mathcal{L}_X g -&= 2c\bigl(\sin v\,dx^2-2\,\cos v\,dx\,dy-\sin v\,dy^2\bigr)\\ -\mathcal{L}^2_X g -&= 4c\bigl((c{+}b\,\cos v)\,dx^2+2b\sin v\,dx\,dy + (c{-}b\,\cos v)\,dx^2\bigr) -\end{align} -$$ -and -$$ -\mathcal{L}^3_X g = 4(c^2{-}b^2)\,\mathcal{L}_X g, -$$ -from which it follows by induction that $\mathcal{L}^{k+2}_X g = 4(c^2{-}b^2)\,\mathcal{L}^k_X g$ for all $k\ge 1$, so that -$$ -\mathrm{tr}_g(\mathcal{L}^{2k+1}_X g) = 0 -\quad\text{and}\quad -\mathrm{tr}_g(\mathcal{L}^{2k+2}_X g) = 8c^2 \bigl(4(c^2{-}b^2)\bigr)^k -$$ -for all $k\ge0$. -Moreover, because the equations express all of the second derivatives of $u$ and $v$ algebraically in terms of $u$, $v$ and their first derivatives, it follows that any solution $(u,v)$ that is $C^2$ is real-analytic. Thus, any $C^2$ solution $(u,v)$ of the equations above defines a vector field $X$ whose flow has singular values that depend only on $t$. -Now, it is easy to show that the equation $({u_x}^2-{u_y}^2)\cos v +2u_xu_y\sin v = 0$ cannot hold on any open domain. Moreover, given constants $(r_1,r_2,s,s_1,s_2)$ that satisfy $({r_1}^2-{r_2}^2)\cos s +2r_1r_2\sin s \not= 0$, and fixing constants $b$ and $c\not=0$, there is at most one solution on a neighborhood of $(x,y) = (0,0)$ of the above equations such that -$$ -\bigl(u_x(0,0),u_y(0,0),v(0,0),v_x(0,0),v_y(0,0)\bigr) = (r_1,r_2,s,s_1,s_2). -$$ -In fact, one finds that, in the case $b=0$, there exists such a local solution, as the PDE system is involutive (i.e., Frobenius). Unfortunately, there is good evidence that the 'generic' solution with $b=0$ cannot be written in terms of elementary functions. Sometimes, though, one can express a solution implicitly: As an example of a particular solution with $b=0$, one can show that when $x+i\,y$ is parametrized by a holomorphic parameter $w$ such that -$$ -dx + i\,dy = \mathrm{e}^{w^2}\,dw, -$$ -then, setting -$$ -u_x + i\,u_y = \overline{w}\,\mathrm{e}^{w^2}, -$$ -one finds that, for all $t$, the time $t$ flow of the unimodular vector field $X = u_y\,\partial_y - u_x\,\partial_x$ has constant singular values. (This is most easily checked in the $w$ coordinate, where $X = 2i\left(w\,\partial_{\bar w} - \bar w\,\partial_w\right)$ and $g = \mathrm{e}^{w^2+\bar w^2}\,dw\circ d\bar w$.) -When $b\not=0$, it turns out that there are more restrictions on the initial conditions. In fact, an analysis shows that, when $b\not=0$, either $v$ is constant, or else the equation $(u_xv_x+u_yv_y)=4(c^2{-}b^2)$ must hold. -When $v$ is constant, the vector field $X$ is an affine vector field -whose flow is a $1$-parameter family of (unimodular) affine transformations of the plane, which is clearly a solution. -When $v$ is not constant and $b\not=0$, it turns out that one can integrate the equations explicitly. One finds that, up to a rigid motion in the plane, these solutions can be written in polar coordinates $(r,\theta)$ (where $g = dr^2 + r^2\,d\theta^2$) in the form -$$ -X = a_1\left(r\,\frac{\partial}{\partial r}-2\theta\,\frac{\partial}{\partial\theta}\right) + (a_2 - 2b\log r)\,\frac{\partial}{\partial\theta} -$$ -where $a_1, a_2, b$ are constants and $c^2=a_1^2+b^2$. Note that the case $a_1=a_2=0$ is, up to a constant multiple, the example provide by the OP.<|endoftext|> -TITLE: The double dual of the unitization of a $C^*$-algebra -QUESTION [7 upvotes]: I am studying the proof that if $A$ is a $C^*$-algebra such that $A^{**}$ is a semidiscrete vN algebra, then $A$ has the completely positive approximation property (CPAP). I was able to handle the unital case, but I am stuck in the non-unital setting: The authors (N. Brown and N. Ozawa) suggest that one should prove that if $A^{**}$ is semidiscrete then so is $(\tilde{A})^{**}$ and then conclude by proving that if $\tilde{A}$ has the CPAP then so does $A$. -My problem is this: I can't prove that the double dual of the unitization will be semidiscrete. I cannot understand the double dual of the unitization in relevance to the double dual of $A$ at all. The authors state that $(\tilde{A})^{**}\cong A^{**}\oplus\mathbb{C}$ and mention that it is furthermore true that if $B$ is any $C^*$-algebra with a (closed, two-sided) ideal $J$, then $B^{**}\cong J^{**}\oplus(B/J)^{**}$. First of all, does $\cong$ mean as vector spaces or as $C^*$-algebras? How can one prove this isomorphism? Extra bonus question: If all double duals involved are endowed with their ultraweak topologies, is $\cong$ a homeomorphism? - -REPLY [9 votes]: Believe it or not, these are $*$-isomorphisms as C${}^*$-algebras. If $J$ is a closed two-sided ideal of $B$ then $J^{**}$ is a weak* closed two-sided ideal of $B^{**}$, and every weak*-closed two-sided ideal of a von Neumann algebra is a direct summand. I suppose these are good exercises. The supremum in $B^{**}$ of an approximate unit for $J$ will be a central projection $p$ such that $pB^{**} = J^{**}$. -Bonus, any $*$-isomorphism of von Neumann algebras is automatically a weak* homeomorphism. That is because it preserves order and hence must be normal.<|endoftext|> -TITLE: Does every map $K(\mathbb{Z}, n) \to K(\mathbb{Z}/m, n + k)$ factor through $K(\mathbb{Z}/m, n)$? -QUESTION [22 upvotes]: Recall that a cohomology operation is a natural transformation $H^n(-; \pi) \to H^{n+k}(-; G)$ defined on CW complexes. - -Does every cohomology operation $H^n(-; \mathbb{Z}) \to H^{n+k}(-; \mathbb{Z}/m)$ factor through $H^n(-; \mathbb{Z}/m)$? - -The cohomology operations $H^n(-; \mathbb{Z}) \to H^n(-; \mathbb{Z}/m)$ are all multiples of the map induced on cohomology by the quotient map $\mathbb{Z} \to \mathbb{Z}/m$. In particular, if the above question has a positive answer, then for any such cohomology operation $\theta$, we have $\theta(mx) = 0$. -One can rephrase the above question in terms of Eilenberg-MacLane spaces: - - -Does every map $K(\mathbb{Z}, n) \to K(\mathbb{Z}/m, n + k)$ factor through $K(\mathbb{Z}/m, n)$? - - -I would actually be content with a positive answer to the following broader question: - - -Does every map $K(\mathbb{Z}, n) \to K(\mathbb{Z}/m, n + k)$ factor through $K(\mathbb{Z}/r, n)$ for some $r$? - - -I initially asked question 1 in the Homotopy Theory chatroom. Piotr Pstrągowski's comments, see here, explain why there is a positive answer for $m = 2$. - -REPLY [2 votes]: I Steenrod's 1957 Colloquium Lectures, published as -Steenrod, Norman E. -Cohomology operations, and obstructions to extending continuous functions. -Advances in Math. 8 (1972), 371–416. - -he ends Section 17 with: - -There are certain elementary cohomology operations which are taken for -granted but must be mentioned in order to state the main result. These -are: addition, cup products, homomorphisms induced by homomorphisms -of coefficient groups, and Bockstein coboundary operators associated -with exact coefficient sequences $0 \to G' \to G \to G'' \to 0$. Then -the main result becomes: -The elementary operations and the operations $Sq^i$, $\beta_2$, $P_p^i$, $\beta_p$ generate all reduced power operations by forming -compositions. - -Thereafter, at the end of Section 21, he writes: - -Using the full strength of Cartan's result, Moore [18] has shown that -all cohomology operations, whose initial coefficient groups are finitely generated, are generated by the cohomology operations listed -at the end of Section 17. - -Here, reference [18] is: "J. Moore, Seminar notes 1955/1957, Princeton University." -Moore's result implies that the answer to Question 1 is "yes", because any Bockstein operation will vanish on any integral cohomology class. Unfortunately, I do not know if Moore's seminar notes are available somewhere.<|endoftext|> -TITLE: Embedding any graph into a vertex-transitive graph of the same chromatic number -QUESTION [6 upvotes]: If $G=(V,E)$ is a simple, undirected graph, is there a vertex-transitive graph $G_v$ such that $\chi(G) = \chi(G_v)$ and $G$ is isomorphic to an induced subgraph of $G_v$? - -REPLY [5 votes]: For $k\in\mathbb N$ the random $k$-chromatic countably infinite graph is vertex transitive and contains an isomorphic copy of every $k$-colorable countable graph as an induced subgraph. I suppose this can be generalized somehow to uncountable graphs and infinite chromatic numbers, but I don't think anyone is interested in that. Instead, I'm guessing you are interested in the case where $G$ is a finite graph, and you want $G_v$ to be finite as well. I believe that can be done. -For $k,n\in\mathbb N$ let $V_{k,n}=\{0,1,\dots,nk-1\}$ and let -$$S_{k,n}=\{t\in V_{k,n}:t\lt\frac{nk}2\text{ and }t\text{ is not a multiple of }k\}.$$ -For any set $T\subseteq S_{k,n}$ let $G_{k,n,T}$ be the graph with vertex set $V_{k,n}$ and edges $\{x,x+t\}$ (addition modulo $nk$) where $t\in T$. -Plainly $G_{k,n,T}$ is vertex transitive and $k$-colorable. Moreover, given any $k$-colorable finite graph $G$, for sufficiently large $n$ we can construct a set $T\subseteq S_{k,n}$ so that $G_{k,n,T}$ contains an isomorphic copy of $G$ as an induced subgraph. -Suppose $G$ is a $k$-colorable graph of order $p$; let $V(G)=\{v_1,v_2,\dots,v_p\}$, and let $c:V(G)\to\{0,1,\dots,k-1\}$ be a proper coloring of $G$. Let $n=2^{p+1}$. -For $i=1,2,\dots,p$, let $x_i=(2^i-2)k+c(v_i)\in V_{k,n}$. -Let $T=\{x_i-x_j:i\gt j,\ v_iv_j\in E(G)\}$. -Then $T\subseteq S_{k,n}$, and the mapping $v_i\mapsto x_i$ is an isomorphism between $G$ and an induced subgraph of $G_{k,n,T}$. (Note that the $\binom p2$ differences $x_i-x_j$, $1\le j\lt i\le p$, are pairwise distinct.)<|endoftext|> -TITLE: A result on Lie group actions on 15-dimensional spheres? -QUESTION [19 upvotes]: In this interview by Eric Weinstein to Roger Penrose, Timestamp 1:24:05., what result is the host talking about? -Transcription of the relevant part: - -"If you have two sets of symmetries, known as Lie groups, that act transitively on the same sphere in usual position, then either their intersection acts transitively on that sphere, or the dimension of that sphere is $15$. And I believe the intersection of the groups looks like the electro-strong group. So it's very close to the... particle spectrum of theoretical physics... pulled out of nowhere just by talking about sphere transitive group actions" - -Edit: it seems like the host is trying to recall a particular result. Given how bizarre and peculiar the result seems to be (in line with dimension $4$ being special for differentiable structures on Euclidean spaces, or dimension $7$ in the case of exotic spheres), I would like to know if it's a real thing and, in case it's real, what's the exact statement. -In particular, - -I don't care if the exact quoted statement is true or false; -I only want to know if there's a result that sounds very similar to that one and is actually true and, if you're aware of such a result, what's its exact statement. - -REPLY [35 votes]: My guess is that Weinstein was thinking of this fact, but didn't get it out correctly: -For every $n\not=15$, there is a compact Lie group $H_n\subseteq\mathrm{SO}(n{+}1)$ that acts transitively on the $n$-sphere such that any Lie group $G$ that acts transitively and effectively on the $n$-sphere contains a subgroup $G'$ that acts transitively on the $n$-sphere and is conjugate to $H_n$ in $\mathrm{Diff}(S^n)$. -There are two non-isomorphic subgroups, $\mathrm{Spin}(9)$ and $\mathrm{Sp}(4)$ of $\mathrm{SO}(16)$, both of dimension $36$, that act transitively on $S^{15}$ such that any Lie group $G$ that acts transitively on $S^{15}$ contains a subgroup $G'$ that is conjugate to (exactly) one of these two subgroups in $\mathrm{Diff}(S^{15})$. -Note: -$\bullet$ For $m\not=0,3$, $H_{2m}\simeq \mathrm{SO}(2m{+}1)$, -while $H_0\simeq\mathrm{O}(1)$ and $H_6 \simeq \mathrm{G}_2$, -$\bullet$ for $m\not=0$, $H_{4m+1}\simeq \mathrm{SU}(2m{+}1)$ while $H_1\simeq\mathrm{SO}(2)$, and -$\bullet$ for $m\not=4$, $H_{4m-1}\simeq\mathrm{Sp}(m)$. -This follows from Borel's classification of the Lie groups acting transitively on spheres. -N.B.: The phrase 'and effectively' in the above statement is needed to rule out the following kinds of (ineffective) actions: First, $\mathbb{Z}$ has a transitive action on $S^0 = \{-1,1\}\subset\mathbb{R}$ but has no subgroup isomorphic to $\mathrm{O}(1)\simeq \mathbb{Z}_2$. Second, the simply-connected cover of $H_1=\mathrm{SO}(2)$ is isomorphic to $\mathbb{R}$, and it acts transitively on $S^1$ without containing a subgroup isomorphic to $H_1 = \mathrm{SO}(2)$. Third, for $m\not=0,3$, $H_{2m}\simeq\mathrm{SO}(2m{+}1)$ has a nontrivial double cover $\mathrm{Spin}(2m{+}1)$ that acts transitively on $S^{2m}$ but does not contain a subgroup isomorphic to $H_{2m}$.<|endoftext|> -TITLE: The period of Fibonacci numbers over finite fields -QUESTION [18 upvotes]: I stumbled upon these very nice looking notes by Brian Lawrence on the period of the Fibonacci numbers over finite fields. In them, he shows that the period of the Fibonacci sequence over $\mathbb{F}_p$ divides $p$ or $p-1$ or $p+1$. -I am wondering if there are explicit lower bounds on this period. Is it true, for instance, that as $p \to \infty$, so does the order? -A quick calculation on my computer shows that there are some "large" primes with period under 100. -9901 66 -19489 58 -28657 92 - -REPLY [4 votes]: The question above is about lower bounds, but I allow myself to comment about upper bounds: -$\pi(n)$, the period function of the Fibonacci sequence mod $n$, satisfies $\pi(n)\leq 6n$ and equality holds iff $n=2\cdot 5^k$ for some $k\geq 1$. -This fact is well known. In the 90's it was considered here as a puzzle to the monthly readers. $\pi(n)$ was also discussed in an elementary fashion in the 60's in this monthly paper. -But really, I want to share a little observation which forms a generalization of the above mentioned fact: -denoting, for an element $g\in \mathrm{GL}_2(\mathbb{Z})$, by $\rho_g(n)$ the order of the image of $g$ in $\mathrm{GL}_2(\mathbb{Z}/n)$, $\rho_g(n)\leq 6n$. This is a generalization because $\rho_g(n)=\pi(n)$ for -$g= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$. -Note that if $\det(g)=-1$ then $\rho_g(n)=2\rho_{g^2}(n)$, thus it is enough to prove -that for $g\in \mathrm{SL}_2(\mathbb{Z})$, $\rho_g(n)\leq 3n$. -Let me now fix $g\in \mathrm{SL}_2(\mathbb{Z})$, denote $\rho(n)=\rho_g(n)$ -and prove that indeed $\rho(n)\leq 3n$. -First note that, for natural $p$ and $n$, if $p$ divides $n$ then $\rho(pn)$ divides $p\rho(n)$. -This follows by expanding the right hand side of -$ g^{p\rho(n)}=(g^{\rho(n)})^p=(1+nh)^p$ -and note that it is 1 mod $pn$. -By induction we conclude that for every $k>1$, $\rho(p^k)$ divides $p^{k-1}\rho(p)$. -Assume now $p$ is a prime and note that $\rho(p)$ divides either $p-1,p+1$ or $2p$. -Indeed, if $\bar{g}\in \mathrm{SL}_2(\mathbb{F}_p)$ is diagonalizable over $\mathbb{F}_p$ -then its eigenvalues are in $\mathbb{F}_p^\times$ and their orders divides $p-1$, -else, if $\bar{g}$ is diagonalizable over $\mathbb{F}_{p^2}$ then -its eighenvalues $\alpha,\beta$ are conjugated by the Frobenius automorphism, thus -their order divides $p+1$ because -$\alpha^{p+1}=\alpha\alpha^p=\alpha\beta=\det(\bar{g})=1$, -else $\bar{g}$ has a unique eigenvalue, which must be a $\pm 1$ by $\det(\bar{g})=1$, thus -$\bar{g}^2$ is unipotent and its order divides $p$. -For $p=2$, in the last case, there was no reason to pass to $g^2$, -as $-1=1$ in $\mathbb{F}_2$, thus $\rho(2)$ is either 1,2 or 3. -From the above two points, we conclude that for every odd prime $p$ and natural $k$, $\rho(p^k)$ divides $p^{k-1}(p-1)$, $p^{k-1}(p+1)$ or $2p^k$. -All these numbers are even and bounded by $2p^k$, -thus $\mathrm{lcm}\{\rho(p^k),2\} \leq 2p^k$. -For $p=2$ we get that $\rho(2^k) \leq 2^{k-1}\cdot 3$. -Fix now an arbitrary natural $n$. -Write $n=2^km$ for an odd $m$ and decompose $m=\prod_{i=0}^r p_i^{k_i}$. Then -\begin{align*} -\rho(m)= \mathrm{lcm}\{\rho(p_i^{k_i}) \mid i=0,\dots r\} -\leq \mathrm{lcm}\{\mathrm{lcm}\{\rho(p_i^{k_i}),2\} \mid i=0,\dots r\} =\\ -2\mathrm{lcm}\{\frac{\mathrm{lcm}\{\rho(p_i^{k_i}),2\}}{2} \mid i=0,\dots r\} \leq 2\prod_{i=0}^r \frac{\mathrm{lcm}\{\rho(p_i^{k_i}),2\}}{2}\leq 2\prod_{i=0}^r p_i^{k_i} =2m -\end{align*} -and we get -$$ \rho(n) = \rho(2^km) \leq \rho(2^k) \cdot \rho(m) \leq 2^{k-1}\cdot 3 \cdot 2m = 3\cdot 2^km=3n. $$ -This finishes the proof that $\rho(n)\leq 3n$. - -As always, it is interesting to analyze the case of equality. -For $g\in \mathrm{SL}_2(\mathbb{Z})$ we have $\rho_g(n)=3n$ for some $n$ -iff $\mathrm{tr}(g)$ is odd and not equal $-1$ or $-3$. -If $g$ satisfies this condition, then $n$ satisfices $\rho_g(n)=3n$ iff $n=2st$, for some odd $s\geq 3$, $t\geq 1$ where -every prime factor of $s$ divides $\mathrm{tr}(g)+2$, -every prime factor of $t$ divides $\mathrm{tr}(g)-2$ and $g$ is not $\pm 1$ -modulo any of these prime factors. -For $g$ satisfying $\det(g)=-1$, using the identity $\mathrm{tr}(g^2)=\mathrm{tr}(g)^2-2\det(g)$, we get that $\rho_g(n)=6n$ for some $n$ -iff $\mathrm{tr}(g)$ is odd and in this case, -$n$ satisfices $\rho_g(n)=6n$ iff $n=2st$, for some odd $s\geq 3$, $t\geq 1$ where -every prime factor of $s$ divides $\mathrm{tr}(g)+4$, -every prime factor of $t$ divides $\mathrm{tr}(g)$ and $g$ is not $\pm 1$ -modulo any of these prime factors. -Specifically for $g=\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$, -$\det(g)=-1$, $\mathrm{tr}(g)=1$ is odd, 5 is the only prime factor of $\mathrm{tr}(g)+4$ and there is no prime factor for $\mathrm{tr}(g)$. -Since $g$ is not $\pm 1$ -modulo 5, we get that $\pi(n)=\rho_g(n)=6n$ -iff $n=2\cdot 5^k$ for some $k\geq 1$, as claimed above.<|endoftext|> -TITLE: Public personal ideas: using contributions from other people -QUESTION [7 upvotes]: Let's say I'm thinking about an unsolved mathematical problem for a hobby and I draw some conclusions of my own. I'd like to make these ideas public, allow anyone to use them absolutely freely (even without mentioning me) and maintain these ideas on my blog/notes website. -Now let's say a reader leaves a comment and contributes with something extra, advancing the quest to solving the unsolved mathematical problem. Maybe he/she improved an idea I already posted. I want to then continue with my hobby free to use ideas that my readers are posting, meaning, make it clear that I don't owe them anything as I use their ideas, it is therefore their responsibility if they choose to post something under this condition. I don't want to get to be in a situation where someone demands something from me because 'they came up with it'. Once some mathematical result R is communicated from person A to person B, B cannot simply "undo" this exchange even if he wants to. He cannot simply pretend he "doesn't know" R. -How could I approach this? Is it enough to mention my conditions on my website? Would I be legally bound to anything in such a situation or is this actually a non-problem and I'm overthinking it? -(I'm not sure what the proper place to ask this is, I've also posted it on the 'law' stackexchange.) - -REPLY [6 votes]: When it comes to legal rights, I think a disclaimer might suffice, though IANAL and you maybe should consult one. -When it comes to giving proper attribution and credit for ideas, I think this is not (or not only) something that an author owes to a specific originator of the ideas, but rather (or also) something an author owes to the community at large. In particular, if I read your blog and make significant use of your original ideas to write a paper, and don't cite (or otherwise acknowledge) your blog, I have committed plagiarism whether you care or not. Similarly, if I leave an important comment and you use it to write a paper, you are obliged to cite or acknowledge the comment whether I care or not. (Note this obligation persists even if I remain completely anonymous, even untraceably so!) This is because you have a duty to your readers to make as clear as possible the history of the ideas in your paper, in addition to anything you might "owe" me as originator of those ideas. -(There has been one situation where knowing that a particular theorem in a paper was due to an anonymous referee and not the author arguably led me to pursuing the right ideas in my own research.)<|endoftext|> -TITLE: Local fusion categories -QUESTION [5 upvotes]: A local fusion category ${\cal R}$ is a unitary fusion category equipped with a top-faithful surjective monoidal functor to the fusion category of vector spaces: $\beta: {\cal R} \to {\cal V}ec$. Here, top-faithful means that the functor $\beta$ is injective when acting on the morphisms. -What are those local fusion categories? -${\cal R}ep(G)$ and ${\cal V}ec_G$ are local fusion categories, for a finite group $G$. Are there other examples? Is there a classification? - -REPLY [11 votes]: Let $\mathcal{R}$ be a fusion category and $\beta : \mathcal{R} \to \mathrm{Vec}$ an additive monoidal functor. -I first claim that $\beta$ is automatically faithful. (I know why you use "top faithful" — in higher categories, you want faithfulness just on top-morphisms — but here in 1-category land "top faithful" is just faithful.) First, note that, since $\mathcal{R}$ is semisimple, every additive functor out of $\mathcal{R}$ is exact. Second, suppose $f : X \to Y$ is a nonzero map in $\mathcal{R}$. Then by composing with the pairing $Y \otimes Y^* \to 1$, you get a nonzero map $f^\# : X \otimes Y^* \to 1$. But $1$ is simple, so this map is a surjection. So $\beta(f^\#)$ is a surjection onto $\beta(1) = 1$, and so $\beta(f^\#) \neq 0$, so $\beta(f) \neq 0$. For further details, see Deligne's Catégories tannakiennes. -Thus your "local fusion category" is also called "fusion category with a fibre functor". These are fully understood: such a fusion category is canonically $\mathrm{Mod}(H)$ for a finite-dimensional semisimple Hopf algebra $H$. There are many places to see the details, so I will be telegraphic. As an algebra, $H$ is defined as the endomorphisms of $\beta$-as-a-functor. Then the Hopf structure on $H$ comes from the monoidality of $\beta$. -Your two examples correspond to $H = \mathbb{C}[G]$, the group ring, and $H = \mathcal{O}(G)$, the functions on $G$.<|endoftext|> -TITLE: Every subgroup is isomorphic to a normal subgroup -QUESTION [13 upvotes]: Let $G$ be a group such that, for every subgroup $H$ of $G$, there exists a normal subgroup $K$ of $G$, such that $H$ is isomorphic to $K$. Under such conditions, can we determine the structure of $G$ ? -This question comes from the discussion of Dedekind group in group theory, which is one such that all of its subgroups are normal. We know that a Dedekind group is an Abelian -group or a direct product of the quaternion group $Q_8$ and an Abelian group $A$, where $A$ has no elements with order $4$. -So my problem can be regarded as a promotion of Dedekind group that only requires isomorphism. I don't know how to deal with this case. - -REPLY [12 votes]: Let us call $G$ a generalised Dedekind group if every subgroup of $G$ is isomorphic to a normal subgroup of $G$. -As expected, Dedekind groups are generalised Dedekind groups. In addition, YCor has established that a finite generalised Dedekind group $G$ is nilpotent, since its $p$-Sylow subgroups are necessarily normal. -Note that the fundamental group of the Klein bottle, that is, $\mathbb{Z} \rtimes_{-1} \mathbb{Z} = \left\langle a, b \, \vert \, aba^{-1} = b^{-1} \right\rangle$ is an infinite generalised Dedekind group which is not nilpotent. -Here is a family of finite nilpotent groups which are generalised Dedekind groups but not Dedekind groups. - -Claim. Let $p$ be a prime number and let $u$ be an integer such that $u \equiv 1 \text{ mod } p \mathbb{Z}$ and $u^p \equiv 1 \text{ mod } p^2 \mathbb{Z}$. Let $$G(p, u) \Doteq \mathbb{Z}/p^2 \mathbb{Z} \rtimes_u \mathbb{Z}/p \mathbb{Z}$$ where the conjugation by $a \Doteq (0, 1 + p \mathbb{Z})$ induces the multiplication by $u$ on $\mathbb{Z}/p^2 \mathbb{Z}$, i.e., $aba^{-1} = b^u$ where $b \Doteq (1 + p^2 \mathbb{Z}, 0)$. Then $G(p, u)$ is a generalised Dedekind group. -If in addition $u \not\equiv 1 \text{ mod } p^2 \mathbb{Z}$, then $G(p, u)$ is not Dedekind. - - -Proof. -Let $H$ be a subgroup of $G(p, u)$. It is easy to check that $H$ can be generated by two elements $(x, y) = (b^k, b^la^m)$ with $k \in \{0, 1, p\}$ and $l, m \in \mathbb{Z}$. -Let us show first that $G(p, u)$ is a generalised Dedekind group. -If $k = 0$ or $m \equiv 0 \text{ mod } p\mathbb{Z}$ , then $H$ is a cyclic subgroup of order at most $p^2$ and hence isomorphic to either $\{1\}$, $\langle b \rangle$ or $\langle b^p \rangle$ which are normal in $G(p, u)$. Indeed, we have $(b^la^m)^p = -b^{l(1 + u^m + \dotsb + u^{m(p-1)})}$ with -$1 + u^m + \dotsb + u^{m(p-1)} \equiv p \bmod{p\mathbb{Z}}$, hence the order of $b^la^m$ is at most $p^2$. -If $k = 1$, then $H = \langle b, a^m \rangle$ which is easily seen to be normal in $G(p, u)$. -Let us assume eventually that $k = p$ and $m \not\equiv 0 \text{ mod } p\mathbb{Z}$. Then we have $H = \langle b^p, b^l a^m \rangle$. -If $(b^l a^m)^p \neq 1$, then $b^l a^m$ generates $H$ so that $H$ is isomorphic to a cyclic normal subgroup of $G(p, u)$. Otherwise $\langle b^p \rangle \cap \langle b^l a^m \rangle = \{1\}$. Since $\langle b^p \rangle$ is central in $G(p, u)$, the subgroup $H$ is isomorphic to the Abelian normal subgroup $\langle b^p, a \rangle \triangleleft G(p, u)$. -Thus we have established that $G(p, u)$ is a generalised Dedekind group. -We complete the proof by observing that $G(p, u)$ is Abelian if and only if $u \equiv 1 \text{ mod } p^2 \mathbb{Z}$. - -Taking $p = 2$ and $u = 3$ yields the dihedral group of order $8$ which is the smallest finite generalised Dedekind ring which is not Dedekind. -Taking $p = 3$ and $u = 4$ provides us with an example of a $3$-group which is a generalised Dedekind but not Dedekind. - -Addendum. Geoff Robinson and YCor have observed in the comments attached to the question that, more generally, the groups of order $p^3$, for $p$ a prime number, are generalised Dedekind groups. If $p \neq 2$, there are only $2$ isomorphism classes of non-Abelian groups of order $p^3$: the class of $G(p, u)$ (which does not depend on $u$) and the class of the Heisenberg group modulo $p$, that is, $H(\mathbb{Z}/p\mathbb{Z}) \Doteq (\mathbb{Z}/p\mathbb{Z})^2 \rtimes_A \mathbb{Z}/p\mathbb{Z}$ with $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. The proof becomes obvious once we have observed that the centers of these two groups are cyclic of order $p$ and that groups of order $p^2$ are Abelian. If $p = 2$, the latter two classes collapse in one, the isomorphism class of the dihedral group of order $8$, and a new class must be considered, namely the isomorphism class of the quaternion group.<|endoftext|> -TITLE: Proofs of circle packing theorem -QUESTION [8 upvotes]: Circle packing theorem is a famous result stating that for every connected simple planar graph $G$ there is a circle packing in the plane whose intersection graph is $G$ https://en.wikipedia.org/wiki/Circle_packing_theorem. -I know that this result has many proofs and I want to read one of them, but don't understand how to start (for quite a while). The article in wiki gives a reference to Thurston notes, but the proof comes only in the last section and I am not sure if this is the simplest approach. I like these notes very much, but was never able to read them till the end. So I wonder if there are some simple proofs of this result nowadays. Can you advise something? - -REPLY [3 votes]: Books are written on the subject, so, finding a proof (which are many by now) shouldn't be a problem. I also enjoyed greatly Rohde's tribute to Schramm that explains in very nice way some ideas that Schramm introduced into the area; following references from there one should be able to find more detailed accounts.<|endoftext|> -TITLE: References about Nomizu Conjecture -QUESTION [5 upvotes]: tI want to look for some references about Nomizu Conjecture(if $R(X,Y)R = 0$ Then $\nabla R = 0$),is there anyone know some papers/references about Nomizu Conjecture or the progress about Nomizu Conjecture. - -REPLY [8 votes]: Luis A. Florit, Wolfgang Ziller, Manifolds with conullity at most two as graph manifolds -Abstract states: "... [W]e show that Nomizu's conjecture, well known to be false in general, is true for manifolds with finite volume."<|endoftext|> -TITLE: Interesting "epimorphisms" of $E_\infty$-ring spectra -QUESTION [7 upvotes]: $\newcommand{\Mod}{\mathbf{Mod}} \newcommand{\map}{\mathrm{map}_{E_\infty-A}}$ Suppose $i:A\to B$ is a map of $E_\infty$-ring spectra. It induces a functor of $\infty$-categories $\Mod_B\to\Mod_A$ by restriction of scalars. -A reasonable question is to ask when this is fully faithful; studying the counit of the restriction-extension of scalars adjunction, it's pretty easy to check that this is the case if and only if $B\otimes_A B\to B$ (the "multiplication" map) is an equivalence. -By studying its sections, if I'm not mistaken, one checks that this is the case if and only if the two inclusions $i_0,i_1: B\to B\otimes_A B$ are equivalent as maps of $E_\infty$-$A$-algebras. -For this it suffices that $A\to B$ be an "epimorphism" of $E_\infty$-$A$-algebras (and I think it's actually equivalent), that is, that $\map(B,-)\to \map(A,-)$ be an inclusion of components; since $\map(A,-) \simeq *$, this amounts to saying that $\map(B,C)$ is empty or contractible for all $C$. -For instance, this happens if $B$ is a localization of $A$ at a certain set of classes $S\subset \pi_*(A)$ (for instance $\mathbb{S\to Q, Z\to Q}, ku\to KU,$ etc.) -My question is: - -Are there interesting cases where this happens but it's not a localization in the above sense ? - -In the $1$-categorical case, this question was asked about epimorphisms of commutative rings (for which $\Mod_B\to \Mod_A$ is fully faithful if and only if $A\to B$ is an epimorphism), and there are examples that are neither quotients nor localizations. -Here, quotients usually do not satisfy this property, as "$x=0$" becomes additional structure (e.g. $\mathbb F_p\otimes_\mathbb Z\mathbb F_p \simeq \mathbb F_p[\epsilon], |\epsilon|=1$ as $E_1$-algebras), so it seems reasonable to ask what "epimorphisms" can look like in this setting. - -REPLY [2 votes]: If $A$ is an $E_\infty$ ring spectrum and $i : A \to B$ is any map of $A_\infty = E_1$ ring spectra such that the multiplication $\mu : B \wedge_A B^{op} \to B$ is an equivalence, then $B \simeq LA$ where $L$ is some smashing Bousfield localization on the category of $A$-modules. In particular, $B$ will be $E_\infty$ and $i$ is an $E_\infty$ map. Taking $A = S$ and $L = L_n$ to be the Bousfield localization with respect to the Johnson-Wilson theory $E(n)$, for $0 < n < \infty$, gives examples that are not given by algebraic localization at any set $S$ of classes in $\pi_*(A)$. The case $n=1$ corresponds to localization at ($p$-local) topological $K$-theory, with $B = L_1 S$ closely related to the image-of-$J$ spectrum. See Definition 1.18 of Ravenel's 1984 Amer. J. Math. paper for the notion of a smashing localization, and Proposition 9.3.3 in my AMS Memoir for the stated relation to "smashing maps".<|endoftext|> -TITLE: Why is the Dyck language/Dyck paths named after von Dyck? -QUESTION [10 upvotes]: The Dyck language is defined as the language of balanced parenthesis expressions on the alphabet consisting of the symbols $($ and $)$. For example, $()$ and $()(()())$ are both elements of the Dyck language, but $())($ is not. There is an obvious generalisation of the Dyck language to include several different types of parentheses. -It seems to me that the first time the term "Dyck language" is used to describe this language (and its generalisation) is in [Chomsky, N.; Schützenberger, M. P. The algebraic theory of context-free languages. 1963 Computer programming and formal systems, pp. 118–161]. Furthermore, all sources online agree that the "Dyck" in question is Walther von Dyck, who introduced the notion of a group presentation in 1882. -However, in the above paper, I can only see a weak reason as to why this language is named after von Dyck. A paragraph directly following the definition reads: The Dyck Language $D_{2n}$ on the $2n$ letters $x_{\pm i} \: (1 \leq i \leq n)$ [...] is a very familiar mathematical object: if $\varphi$ is the homomorphism of the free monoid generated by $\{ x_{\pm i}\}$ onto the free group generated by the subset $\{ x_i \mid i > 0\}$ that satisfies identically $(\varphi x_i)^{-1} = \varphi x_{-i}$, then $D_{2n}$ is the kernel of $\varphi$. -This alternate characterisation is obviously related to presentations, and thus has some connection with von Dyck. However, I am uncertain whether this is the full reason as to why it is named after him. Perhaps there is an intermediate study of the Dyck language inbetween the work of von Dyck and Chomsky-Schützenberger which makes this connection stronger? Thus, my question: -Why is the "Dyck language" named after von Dyck? -Of course, the same question might as well be asked about "Dyck paths" in combinatorics, closely related to the Catalan numbers, but it seems to me quite clear that Dyck paths were named after the Dyck language. -Any thoughts would be appreciated! - -REPLY [7 votes]: Diekert and Lange, in Variationen über Walther von Dyck und Dyck-Sprachen, quote a personal communication from Chomsky that attributes the name Dyck language to Schützenberger's 1962 paper on Certain elementary families of automata, although there only the letter "D" is used. -They note the link between the parenthetical structure of Von Dyck's free groups and the push-down automata that were studied in the 1950's and formalized by Schützenberger in On context-free languages and push-down automata (1963, where the "D" is now written in full as Dyck).<|endoftext|> -TITLE: Intuition behind choosing a specific test function -QUESTION [11 upvotes]: I am learning about elliptic PDEs using the book by Chen & Wu, especially on the maximum principle. The author uses the De Giorgi iteration technique to establish the weak maximum principle for elliptic operators under some conditions. I attached the statement here, and you can also see the proof in this link. - - -Before asking my question, I will describe the scheme of the proof briefly. As the main lemma of the De Giorgi iteration, the following is presented. - -Lemma. Suppose $\varphi(t)$ is a nonnegative decreasing function on $[k_0, \infty)$ with -$$ \varphi(h) \leq \frac{C}{(h-k)^\alpha} \varphi(k)^\beta$$ -for $h>k\geq k_0$, where $\alpha>0, \beta>1$. Then, for -$$d = C^{1 /\alpha} \left[ \varphi(k_0)\right]^{(\beta -1)/{\alpha}} 2^{\beta / (\beta -1)},$$ -we have -$$ \varphi(k_0 + d) = 0.$$ - -Then, we want to apply this lemma to the measure of the sets -$$ A(k) = \left\lbrace x \in \Omega \ \vert \ u(x) >k \right\rbrace, \quad k \in \mathbb R$$ -in order to obtain an upper bound of the essential supremum of $u$ on $\Omega$. After some estimation and using the lemma, we can get the following result. - -Result. Let $\tilde C $ be the embedding constant of the Sobolev embedding of $W^{1,2}_0 (\Omega)$. Suppose $k_0 \geq l := \sup_{\partial \Omega} u^+$ satisfies -$${\tilde C}^2 \left\vert A(k_0) \right\vert^{2/{n}} \leq \frac{1}{2}.$$ -Then, -$$\DeclareMathOperator{\esssup}{\mathrm ess \, sup} \esssup_{\Omega} \leq k_0 + CF_0 \lvert \Omega\rvert^{(1/n) - (1/p)} =: k_0 + C \tilde{F}_0,$$ -where $F_0 = \frac{1}{\lambda} \left( \sum_i \lVert f^i \rVert_{L^p} + \lVert f \rVert_{L^{p_*}} \right)$ and $p_* = np/(n+p)$. - -To get $k_0$, we can first use the Chebyshev inequality on $u$. Then, we obtain some $k_0$ such that $k_0 \leq \sup_{\partial \Omega} u^+ + C \lVert u \rVert_{L^2}$, but this only guarantees the essential boundedness of $u$ on $\Omega$. Thus, we need to estimate further. -In order to obtain a better choice of $k_0$, the following test function is chosen: for $v = (u-l)^+$, -$$ \varphi = \frac{v}{M+ \epsilon + \tilde{F}_0 - v} \in W^{1,2}_0(\Omega),$$ -where $M = \esssup_{\Omega} u - l$. This gives a better estimate on $\left\vert A(k) \right\vert$ than the estimate by the Chebyshev inequality: for $l 0$ we have that $v := \log(M+A)-\log(w)$ is compactly supported and satisfies $|\nabla v|^2 \leq 1 + \Delta v$. Thus the integral of $|\nabla v|^2$ (hence $v^{2^*}$) is bounded in terms of the volume of the domain. -(3) The Bombieri-De Giorgi-Miranda interior gradient estimate for a solution $u$ to the minimal surface equation is based on the fact that the vertical component $\nu^{n+1}$ of the unit normal to the graph of $u$ is positive and superharmonic (on the graph). The proof uses the equation for $v := -\log(\nu^{n+1})$, which just as above contains a useful term quadratic in $|\nabla v|$. -(4) The Li-Yau proof of the Harnack inequality for a harmonic function $u$ is obtained by looking at the quantity $w := |\nabla (-\log u)|^2$. The key is that $w$ solves a differential inequality with the powerful term $\frac{2}{n}w^2$, which allows one to bound $w$ from above locally by a universal constant independent of $w$. -I am sure there are many other interesting examples, and I am not sure where the first instances of the "log trick" appeared. One final remark is that the estimate (4.9) can also quickly be inferred using the properties of the Green's function $G$ for uniformly elliptic operators (namely, $G \in L^p$ for $p < \frac{n}{n-2}$ and $\nabla G \in L^p$ for $p < \frac{n}{n-1}$, just like the Laplace case).<|endoftext|> -TITLE: Dehn surgery on $S^3$ along a Hopf link with rational surgery coefficients -QUESTION [5 upvotes]: Is there an exhaustive list of conditions satisfied by rational surgery coefficients assigned to the components of the Hopf link in $S^3$ such that the resulting 3-manifold by Dehn surgery acting on $S^3$ along this Hopf link is again $S^3$? - -REPLY [5 votes]: If you're doing surgery with coefficients $p/q$ and $p'/q'$, the condition is that ${pp}'-{qq}' = \pm 1$. -In order to see this, let's think of what we're doing: we're drilling out of $S^3$ two solid torus neighbourhoods of the two components, obtaining $M$, and then we're gluing in solid tori, $T$ and $T'$. Since $M$ is a thickened torus (because this is the Hopf link complement), if we glue $T'$ first, we get $M'$ which is again a solid torus, and then when we glue $T$ we get a lens space $L$. Therefore, we only need to figure out when $H_1(L)$ is trivial. -Since I'm a bit lazy, I'll do it with Mayer–Vietoris: we need to look at subspace generated by the two boundary slopes $s$ and $s'$ in $H_1(M) = H_1(T^2) = \mathbb{Z}^2$. We pick coordinates $\lambda, \mu$ on $H_1(M)$ given by Seifert longitude and meridian of one of the components. Since it's the Hopf link, these are also the meridian and Seifert longitude of the other component (note that I've swapped the order of the two, but that I haven't changed any orientation!). In this basis, the surgery slopes are $(p,q)$ and $(q',p')$, so $H_1(L)$ (which is the quotient $H_1(M)/\langle s, s'\rangle$, by Mayer–Vietoris) has order $\det \left(\begin{array}{cc} p & q'\\ q & p'\end{array}\right) = pp'-qq'$, so $H_1(L)$ is trivial if and only if $pp'-qq' = \pm 1$, as claimed. -With a little bit more care, one can pin down which lens space one gets by doing any rational surgery on the Hopf link, but right now I'm a bit lazy...<|endoftext|> -TITLE: Inverse of a Cauchy-like matrix -QUESTION [5 upvotes]: Consider $n\times n$ symmetric Cauchy-like matrix $M$ with elements $(M_{ij})_{i,j=1}^{n}$ given by -$$M_{ij} = \frac{1}{(n-i)!(n-j)!(2n-i-j+1)} = \displaystyle\int_{0}^{1}\frac{x^{n-i}}{(n-i)!} \frac{x^{n-j}}{(n-j)!}\:{\rm{d}}x.$$ -Is there a way to compute the elements of the inverse $(M^{-1})_{ij}$ analytically? - -REPLY [2 votes]: I was able to figure this out by viewing $M$ as a scaled Cauchy matrix. - -Theorem. $\left(M^{-1}\right)_{ij} = \dfrac{(n-i)!(n-j)!}{2n-i-j+1}\dfrac{\displaystyle\prod_{r=1}^{n}(2n-i-r+1)(2n-j-r+1)}{\left(\displaystyle\prod_{\stackrel{r=1}{r\neq i}}^{n}(r-i)\right) \left(\displaystyle\prod_{\stackrel{r=1}{r\neq j}}^{n}(r-j)\right)}$. - -Proof. Define $n\times 1$ vector $\alpha$ with elements $(\alpha)_{i} = 1/(n-i)!$. Then $M = \text{diag}(\alpha) N \text{diag}(\alpha)$, where $N_{ij} := 1/(2n-i-j+1)$. Hence $\left(M^{-1}\right)_{ij} = (n-i)!(n-j)!\left(N^{-1}\right)_{ij}$. -Now, write $N$ as a Cauchy matrix: $N_{ij} = 1/(a_i + b_j)$ where $a_i := n-i$, $b_j := n-j+1$. Then using the known result [1, Sec. 1.2.3, Exercise 41] for Cauchy matrix inverse: -$$\left(N^{-1}\right)_{ij} = \dfrac{1}{2n-i-j+1}\dfrac{\displaystyle\prod_{r=1}^{n}(2n-i-r+1)(2n-j-r+1)}{\left(\displaystyle\prod_{\stackrel{r=1}{r\neq i}}^{n}(r-i)\right) \left(\displaystyle\prod_{\stackrel{r=1}{r\neq j}}^{n}(r-j)\right)},$$ -the result follows. -[1] D. E. Knuth, The Art of Computer Programming. Volume 1: Fundamental Algorithms, 3rd ed. Addison- Wesley, 1997.<|endoftext|> -TITLE: Nonhomeomophic spaces with homeomorphic mapping cones -QUESTION [6 upvotes]: It is natural to ask if it is possible for the mapping cone $X\cup_\alpha CA$ -to be homeomorphic to the mapping cone $X\cup_\beta CB$ with $A$ and $B$ -nonhomeomorphic. Is there a standard go-to example for this? -I have vague memories that there are manifolds $M$ and $N$ that are not homeomorphic, but -$M\times \mathbb{R} \cong N \times \mathbb{R}$, and it seems like it might be a mere hop, skip, and a jump from there to an example. - -REPLY [6 votes]: The double suspension theorem says that if $Y$ is a homology $3$-sphere, then its double suspension $\Sigma^2 Y$ is homeomorphic to $S^5$. If we take $Y$ to be the Poincaré sphere, then $\Sigma Y$ is not a topological manifold, since the suspension points are not manifold points, and in particular $\Sigma Y$ is not homeomorphic to $S^4$. Taking these two spaces as $A$ and $B$ and maps to a point as $\alpha$ and $\beta$ gives a fairly well-known example.<|endoftext|> -TITLE: Original proof of Hilbert's syzygy theorem -QUESTION [8 upvotes]: Does anyone know an English reference for the original proof of Hilbert's syzygy theorem? The three proofs that I know use either: - -the theory of projective dimension and change of rings (plus a step to go from projective to free resolutions) -the symmetry of the Tor functors -Groebner bases - -None of these tools would have been available to Hilbert, and I guess his original proof was much more direct. But, unfortunately, the original reference is in German. Is there an English proof somewhere? - -REPLY [9 votes]: Actually, there is an English translation of Hilbert's -"Über die Theorie der algebraischen Formen" (Mathematische Annalen 36, 473--530, 1890), where the theorem is in Part III of that five-part paper. The translation is in pages 143--224 of "Hilbert's Invariant Theory Papers", Volume VIII of R. Hermann's "Lie Groups: History, Frontiers and Applications", Math. Sci. Press, Brookline, MA, 1978. I don't know if the book is available on-line.<|endoftext|> -TITLE: Equivalence of topological Hochschild homology and Mac Lane homology via an equivalence $QA\simeq HA \wedge_{\mathbb{S}} H\mathbb{Z}$ -QUESTION [33 upvotes]: Mac Lane homology is a homology theory for (not necessarily commutative) rings. Given a ring $A$, Eilenberg and Mac Lane define its cubical construction $QA$ to be a certain connective chain complex, whose homology is isomorphic to the stable homology of Eilenberg-Mac Lane spaces: $$H_i(QA)\cong H_{i+j}(K(A,j)) \text{ for } j \gg i.$$In fact, $QA$ is a dg-ring, and it comes with a ring map $QA \to A$ inducing an isomorphism on $H_0$. You can find the construction and basic properties of the cubical construction in [1] or [2, Chapter 13]. One then defines the Mac Lane homology of $A$ to be the Hochschild homology $$HML_i(A):= HH_i(QA,A)$$and similarly the Mac Lane cohomology is $HML^i(A):=HH^i(QA,A)$. -In 1992, Pirashvili and Waldhausen [3] proved that $HML_i(A)\cong THH_i(A)$, where the right-hand side is topological Hochschild homology. The proof goes by identifying both of them with a functor homology group. Subsequently in 1995, Fiedorowicz-Pirashvili-Schwaenzl-Vogt-Waldhausen [1] outlined a `brave new algebra' proof of this fact. In modern terms, they noticed that the Pirashvili-Waldhausen result would follow if one has a stable equivalence of $H\mathbb{Z}$-algebra spectra $$HQA \simeq HA \wedge_{\mathbb{S}} H\mathbb{Z}.$$Here, $H$ denotes the Eilenberg-Mac Lane functor taking dgas to $H\mathbb{Z}$-algebra spectra, and $\mathbb{S}$ is the sphere spectrum. (For concreteness, my preferred model of spectra is symmetric spectra in simplicial sets.) Indeed, if one has such a stable equivalence, then it follows by base change results that $HML(A)\simeq THH(A)$, for both homology and cohomology. -My question: Is it known that there is a stable equivalence $HQA \simeq HA \wedge_{\mathbb{S}} H\mathbb{Z}$ of $H\mathbb{Z}$-algebra spectra? -This may be a fact known to experts in $THH$, but I was unable to find anything in the literature. One can almost get there: because the homology of $QA$ is the stable cohomology of Eilenberg-Mac Lane spaces, it follows that one has $$\pi_i(HQA)\cong \pi_i(HA \wedge_{\mathbb{S}} H\mathbb{Z}).$$Moreover, both are $H\mathbb{Z}$-module spectra, hence wedges of Eilenberg-Mac Lane spectra, and so an isomorphism on $\pi_*$ actually lifts to a stable equivalence $HQA \simeq HA \wedge_{\mathbb{S}} H\mathbb{Z}$ of $H\mathbb{Z}$-module spectra. This argument already appears in [1], but it wasn't clear to the authors then, and it's certainly not clear to me now, how to lift this to a stable equivalence of $H\mathbb{Z}$-algebra spectra. I guess because $QA$ is connective, in theory one could write down both sides of the equation and match up the algebra structure somehow, but I am not sure if this is very tractable. -Thanks for your time. -References: -[1] Fiedorowicz, Z.; Pirashvili, T.; Schwänzl, R.; Vogt, R.; Waldhausen, F. Mac Lane homology and topological Hochschild homology. Math. Ann. 303 (1995) -[2] Jean-Louis Loday, Cyclic homology, Springer 1998 -[3] Pirashvili, Teimuraz; Waldhausen, Friedhelm. Mac Lane homology and topological Hochschild homology. J. Pure Appl. Algebra 82 (1992) - -REPLY [15 votes]: The answer to your question is yes, these two $H\mathbb{Z}$-algebras are equivalent (and if $A$ is commutative, they are equivalent as commutative $H\mathbb{Z}$-algebras). I am in fact in the process of writing this up with Maxime Ramzi. Hopefully the paper will be finished in a week or 2. -Update : The paper has now appeared on arXiv<|endoftext|> -TITLE: How to characterize the metrical relations in the uniform (4 4 4) tiling of the hyperbolic unit disk from a purely analytic point of view? -QUESTION [5 upvotes]: I wonder how can one describe the generators of the triangle group for the tesselation of Poincare unit disk by triangles with angles $\pi/4, \pi/4 , \pi/4 $ in terms of Mobius transformations. I also wonder how one can describe the metrical relations of such a network without using the conception of the unit disk as model for hyperbolic geometry; that is, by purely analytic methods, as Gauss probably did. -I'm not very familar with the theory of hyperbolic tesselations, so there may be inaccuracies in my understanding and even with the specific terminology i use. -Side remark: -My purpose is to verify an historical hypothesis i have on Gauss's tesselation of the unit disk as described in John Stilwell "Mathematics and its history". Looking at the relevant pages in Gauss's Nachlass (volume 8, p.102-105), I read that the commentor (Robert Fricke) on this fragment of Gauss says that Gauss's drawing (the (4 4 4) tesselation) is intended to be a geometrical illustration for composition of substitutions other then the fundamental generators of the modular group. The following sentences are a citation of Fricke about the substitutions Gauss used: - -Gauss has repeatedly dealt with composition of other substitutions of the group defined from these generators. In addition to the information in fragment [i], the following formula should also be mentioned $$\frac{[\alpha, \beta,\dots,v]\theta + [\beta,\gamma,\dots,v]i}{-i[\alpha,\beta,\dots,\mu]\theta+[\beta,\gamma,\dots,\mu]}$$ -which can be found in a booklet entitled “Cereri Palladi Junoni sacrum, Febr. 1805”. The continued fraction expansions of the two substitutions are given as examples:$$\frac{128\theta + 37i}{-45i\theta + 13}$$ $$ \frac{121\theta+36i}{-84i\theta+25}$$ - -Just to explain the words of Fricke, "the group defined from these generators" is the modular group (since the generators mentioned before this sentence are addition of $i$ and inversion with respect to the unit circle), and the $[,,\dots,]$ is the "Gaussian Brackets" notation (which Gauss defined in article 27 of Disquisitions Arithmeticae, and is connected directly to continued fractions). I think $\theta$ is just a notation for the complex variable that is transformed under the Mobius transformation. -Checking the determinants of these substitutions gave $-1$ for the first one and $+1$ for the second one, so this made me suspect that these are isometric Mobius transformations (the only thing that doesn't settle is that $a,b,c,d$ in the Mobius transformation should be real integers, not imaginary integers). One should also notice that the second Mobius transformation can be rewritten as: -$$\frac{121\theta+36i}{-84i\theta+25} = \frac{[3,2,1,3,3]\theta+[2,1,3,3]i}{-[3,2,1,3,2]i\theta+[2,1,3,2]}$$ -, in exact accordance with Gauss's formula which uses the Gaussian Brackets notation (here $\alpha=3, \beta=2,\gamma=1, v=3, \mu=2$). -In addition, the diagonal elements of the second Mobius transformation ($121$ and $25$) are both reduced to $1$ modulo $12$, while the off-diagonal elements reduce to $0$ modulo $12$ (if one allows imaginary sizes for $b,c$). In other words, this shows that the second Mobius transform belongs to a congruence subgroup of level 12 in $SL_{2}(Z[i])$. -Although it seems at first that these Mobius transformations are just examples of a general principle of certain continued fractions developement, immediately after that Fricke says: - -Both to explain the continued fraction development of the substitutions and to draw conclusions from the theory of functions, Gauss made use of the geometrical representation which has become the basis of the more recent theory of module functions. In the booklet just mentioned, Gauss drew the figure shown here. Since the above-mentioned continued fraction expansions of substitutions are also to be found, Gauss must have used the figure as a means to illustrate these continued fraction expansions. In fact, one has here the beginning of the well-known network of circular arc triangles, which is the basis of the theory of modulus functions. It's evident that Gauss generally understood the "principle of the symmetrical multiplication of curved triangles", which comes into consideration here, and even the character of the "natural limit" of a triangular network to be obtained in this way did not remain hidden... These are circular arc triangles of the angles $\frac{\pi}{4},\frac{\pi}{4},\frac{\pi}{4}$, and the orthogonal circle highlighted in the drawing represents their natural limit. In addition to the drawing, the following information was written by Gauss: -"Center of the first circle: $2^{\frac{1}{4}}$, radius of the first circle: $\sqrt{\sqrt{2}-1}$, center of the second circle: $\frac{1}{2}(\sqrt{\sqrt{2}+1}+\sqrt{\sqrt{2}-1})$, radius of the second circle:$\frac{1}{2}(\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1})$. - -Therefore, it seems that these Mobius substitutions are actually the generators for the (4 4 4) tiling of the hyperbolic disk. But this conclusion is a result of a very shallow reading of Fricke's comments and i lack the proffesional knowledge needed to verify my reading. In addition, there are two drawings in these pages (one on p.103 and the Gauss's tesselation on p.104), and i'm not sure to which drawing Fricke refers. -Historical significance of Gauss's results: -Being the first drawing of it's kind, the tessellation drawed by Gauss and his related results have planted some the seeds of Felix Klein's "Erlangen program" (with the other influences being Galois's theory of equations and Riemann's geometric ideas). Klein read Gauss's fragments very closely and seems to have been influenced by them, so i think it's not an exaggeration to say that Gauss's drawing was one of his sources of inspiration. Therefore i believe that for a correct historic appreciation of the roots of Erlangen program, it's important to know wethever the substitutions written by Gauss are related to his drawing or not. -Concerning the results stated by Gauss on the location and radiuses of the centres of the first and secondary circles in his tessellation, my posted answer already confirms them (although it might not be the original method of Gauss). The main issue that remains to be resolved is therefore the meaning of the Mobius transformations. -Questions -Since my question didn't get any response for a few months, I'll try to focus it: - -Are there any online programs that get as an input an element of the modular group and returns as output a representation of this element in terms of the generators $S,T$ of the modular group (as a word in S, T)? I need this in order to check several things about the Mobius transformations from Gauss's notebook. - -Is there a purely analytic way of arriving at Gauss's formulas for the radiuses of curvature of the first and second generation circles in his uniform tiling of the unit disk? I ask that because the derivation presented in my answer assumes that Gauss essentialy knew the Poincare disk model of the hyperbolic plane, an assumption that i find unprobable (the disk model is essentially a conformal model of the hyperbolic plane, and Gauss didn't think about conformal mappings until at least 1815). It's more likely that he arrived at this result in the context of his analytic investigations on fundamental domains and the action of modular group, and not in the context of his "meditations" on non-Euclidean geometry. - - -Update (September 25, 2021) -Since all my efforts up to now were fruitless, I think I need to change direction of thought. I think that since the fundamental triangle in Gauss's drawing (with angles $\pi/4,\pi/4, \pi/4$) is not related in any way to the fundamental domain of the modular group, the only way to explain the correctness of Gauss's results on the metrical relations in this network (from analytic point of view) is to assume it was derived from some form of automorphic function. Automorphic functions are a generalization of modular functions in which the acting group can be any general triangle group, and not just the modular group (which has structure of the $(2,3,\infty)$ triangle group). - -REPLY [3 votes]: This is a very partial answer that is intended to verify only some of the many statements in Gauss's fragment - the four statements about the network of curved triangles. In order to facilitate the whole deduction of Gauss's formulas as presented in this answer, i added the original drawing of Gauss (to help visualize the geometric relations). - -Radius and distance of the center of the first circles: -Theorem: Each of the eight curved triangles with one vertex at the origin $(0,0)$ of the Poincare unit disk has two straight sides which are portions of diameters of this disk and one curved side which is a portion of a circle orthogonal to the unit circle. The center of curvature of this orthogonal circle is at distance $2^{\frac{1}{4}}$ from the origin and it's radius of curature is $\sqrt{\sqrt{2}-1}$ (as in Gauss's fragment). -Proof: -The length $a$ of a side of an equilateral hyperbolic triangle with angles $\alpha=\pi/4,\beta=\pi/4,\gamma=\pi/4$ in the "real" hyperbolic plane (not the euclidean distance in the Poincare disk model of it) is according to the laws of hyperbolic trigonometry: -$$cosh(a) = \frac{cos\alpha}{1-cos\alpha} = \frac {1}{\sqrt{2}-1} = \sqrt{2}+1\implies tanh(a) = \sqrt{2}\sqrt{\sqrt{2}-1}\implies a = \frac{1}{2}ln(\frac{1+\sqrt{2}\sqrt{\sqrt{2}-1}}{1-\sqrt{2}\sqrt{\sqrt{2}-1}})$$ -The relation between the "euclidean" distance $r$ between each of the other two vertexes to the origin and the corresponding real hyperbolic distance $a$ is, according to the formulas of Cayley-Klein metrics: -$$ln(\frac{1+r}{1-r}) = a \implies \frac{1+r}{1-r} = \sqrt{\frac{1+\sqrt{2}\sqrt{\sqrt{2}-1}}{1-\sqrt{2}\sqrt{\sqrt{2}-1}}}\implies r = \sqrt{\sqrt{2}-1}$$. -Therefore the coordinates of the two other vertexes, after aligning the x-axis with one of the straight sides of the curved triangle, are: -$$(\sqrt{\sqrt{2}-1},0),(\frac{\sqrt{\sqrt{2}-1}}{\sqrt{2}},\frac{\sqrt{\sqrt{2}-1})}{\sqrt{2}})$$. -Now the equation of any circle orthogonal to the unit circle is of the form: -$$x^2+y^2+ax+by+1 = 0$$ -Substituting the x,y coordinates of the two points, one gets two linear equation with variables $a,b$, whose results are: -$a = -\frac{\sqrt{2}}{\sqrt{\sqrt{2}-1}}, b = a(\sqrt{2}-1)$ -Since the canonic form of the equation of the orthogonal circle is: -$$(x+a/2)^2+(y+b/2)^2= \frac{a^2+b^2}{4}-1$$ -, one gets the desired theorem by an easy calculation. Q.E.D -Radius and distance of the center of the secondary circles: -The principle of generation of the curved triangles network is, as with all tesselations of the plane (whetever it's geometry is euclidean, hyperbolic or spherical), the successive reflection of the triangles with respect to their sides; in this way we fill the whole plane with triangles. In euclidean geometry, one can generate the tiling by simply reflecting the triangles with respect to the sides; however, in hyperbolic geometry one needs to generalize the notion of reflection from reflection with respect to line to reflection with respect to a circle. -The notion required is therefore that of inversion with respect to a circle; the secondary circular arcs (of the "second generation triangles") are inversions of the straight sides of the "first generation triangles" with respect to the first circular arcs. The formula for the radius $r'$ of the inversion of a circle with radius $r$ with respect to a smaller circle with radius $k$ whose center is at distance $d$ from the center of the inverted circle is: -$$r' = \frac{k^2r}{d^2-r^2} = \frac{k^2}{d^2/r-r}$$ -In our problem: $k = \sqrt{\sqrt{2}-1}, d^2 = r^2+x^2 -2xrcos(5\pi/8), x = \sqrt{\sqrt{2}}, -cos(5\pi/8) = sin(\pi/8) = \sqrt{\frac{1-\frac{1}{\sqrt{2}}}{2}} $. -In this notation, $x$ is the distance of the center of the first circles from the origin, and $d$ is derived using the euclidean cosine theorem. One needs to calculate $r'$ in the limit where $r$ tends to infinity since the straight sides of the first triangle can be thought of as circles with infinite radius. -Therefore: -$$r' = lim_{r\to \infty}\frac{(\sqrt{\sqrt{2}-1})^2}{(r^2+x^2 -2xrcos(5\pi/8))/r - r} = \frac {\sqrt{2}-1}{-2xcos(5\pi/8)} = \frac{\sqrt{2}-1}{(\sqrt{2-\sqrt{2}})\cdot\sqrt{\sqrt{2}}} = \frac{\sqrt{2}-1}{\sqrt{2(\sqrt{2}-1)}} = \frac{\sqrt{\sqrt{2}-1}}{\sqrt{2}}$$ -one can easily verify that the result for $r'$ is equal to the formula given by Gauss (simply by squaring both sides). The result for the distances $x'$ of the centers of the secondary circles from the origin is then easily obtained if we keep in mind that, for any $n$th generation circles: $x^2_n-r^2_n = 1$. -Final remarks: -Besides one step of the calculation (the use of Cayley-Klein metric), the whole procedure of generation of this curved triangles network can be viewed through euclidean eyes; one needs the concept of inversion of circle in a smaller circle to get a recursive definition of the network. However, Gauss deliberately gives sizes of the first triangles that correspond to tessellation of the unit disk (and not, for example, to a disk of radius 2), and i can't see any way to calculate the required size of the first triangles (required in order to make the unit circle the natural border of this network) without using the conception of the Poincare disk model of the hyperbolic plane. -Therefore, this (very partial!) reconstruction of Gauss's results uses the much later terminology of models of hyperbolic plane, and makes far-reaching assumptions on Gauss's insights into non-euclidean geometry. I suspect there might be other ways by which Gauss stated this result, perhaps by his analytic insights into the j-invariant.<|endoftext|> -TITLE: Does this prove Collatz is a $\Sigma_1$ problem? -QUESTION [5 upvotes]: So I got an email from one of my colleagues on the Collatz Conjecture with a link to the article Computer Scientists Attempt to Corner the Collatz Conjecture by Kevin Hartnett in Quanta Magazine. -On digging through the conjecture, it seems it is not even known it is a $\Pi_1$ or $\Sigma_1$ statement. See the Math.StackExchange question Is the Collatz conjecture in $Σ_1/Π_1$?. For more information we can find details in the following thesis: - -Matthew Alexander Denend, Challenging variants of the Collatz Conjecture, Masters Thesis, The University of Texas at Austin 2018, doi:10.26153/tsw/1559. - -Does it mean they have already shown Collatz is a $\Sigma_1$ statement? - -REPLY [12 votes]: It seems I was wrong - see Emre Yolcu's comment below. - -My understanding is that that has not been accomplished (although the Quanta article is pretty vague so I could be misunderstanding the situation). -The Quanta article describes the following process: - -Whip up a rewriting system which always terminates iff Collatz is true. This has been successfully done - but note that the termination problem is a priori $\Pi^0_2$, just like Collatz. - -Try to find a collection of matrices satisfying some complicated constraints relating to that rewriting system. This is the task which SAT solvers are relevant for. However, they have not yet found an appropriate collection of matrices. - -(This is where I was wrong:) Even after finding such a collection, we're not done. All that this will accomplish is reduce Collatz to a particular problem about matrix multiplication (which the Quanta article doesn't state - moreover, it doesn't explain why that problem should be more tractable than the rewriting one or the original Collatz conjecture). - - -Re: that third bulletpoint, I think that there's a particular part of the article which is potentially confusing: - -“You try to find matrices that satisfy these constraints,” said Emre Yolcu, a graduate student at Carnegie Mellon who is working with Heule on the problem. “If you can find them, you prove [they’re] terminating,” and by implication, you prove Collatz. - -It would have been clearer to write "If you can find them, you then try to prove [they're] terminating, and if you can you prove Collatz." That is, finding a system of matrices satisfying the given constraints - which is indeed $\Sigma^0_1$ - is only the first step, and the remaining fact we need to prove is presumably still $\Pi^0_2$. -Actually it seems I got that exactly wrong! -That said, pending further elaboration from Emre we may only have a $\Sigma_1$ sentence which implies Collatz - I don't know if the nonexistence of an appropriate matrix family would imply that Collatz fails.<|endoftext|> -TITLE: If it quacks like an abelian variety over a finite field -QUESTION [14 upvotes]: Consider smooth projective varieties over a finite field. If a curve "looks like" an elliptic curve (i.e. has genus $1$) then it can be made into an elliptic curve. -Is there something similar in higher dimensions, i.e. if we fix the values of some invariants can we guarantee that the variety admits an algebraic group structure? What is sufficient for threefolds? - -REPLY [8 votes]: One possible answer to this could be Lang's theorem: it says that if $G/\mathbb{F}_q$ is a smooth connected algebraic group, then $H^1(\mathbb{F}_q,G)$ is trivial, or otherwise put every $G$-torsor has an $\mathbb{F}_q$-rational point. -This generalizes your example: if $X/\mathbb{F}_q$ is a smooth projective variety such that $X_{\bar{\mathbb{F}}_q}$ is isomorphic to an abelian variety, then $X$ is a torsor under its Albanese variety $A = Alb(X)$. -If $X$ has dimension $1$ then requiring the genus of $X$ to be $1$ is enough. -If $X$ has dimension $2$ then by the classification of surfaces it is enough to assume that e.g. the canonical bundle of $X$ is trivial and $X$ is not simply connected. -Edit: as pointed out below, the previous sentence is correct only if the characteristic of $\mathbb{F}_q$ is $\neq 2,3$. If we additionally assume that the second $l$-adic Betti number equals $6$, then we get a criterion in all characteristics.<|endoftext|> -TITLE: Splendid groups -QUESTION [13 upvotes]: The following definition has arisen naturally in two papers of mine. The papers are on rather unrelated topics; of course they are within my narrow interests, so there's some symbolic dynamics connection, but really in both cases I just needed to find good "generic elements / finite subsets" of the group, and I can't help but wonder if this has appeared elsewhere, or is even something I know but am just not recognizing. - -Definition. A topological group $G$ is splendid if for all compact $C \subset G$, there exists $g \in G$ such that $gCg \cap C = \emptyset$. - -I'll mention some sufficient conditions for being splendid and give some examples. First, abelian groups are usually splendid unless obviously not splendid, and this passes to preimages, as follows: - -Lemma. Let $G$ be a topological group that has a continuous abelian (surjective) quotient $(H,+)$ such that $\{2h \;|\; h \in H\} \leq H$ is non-compact. Then $G$ is splendid. - -For simplicity let's now concentrate on discrete infinite groups, so compact means finite. The previous lemma shows that e.g. free abelian groups, free groups and Thompson's $F$ are splendid, since they have positive rank abelianizations. Every finitely-generated infinite discrete left-orderable amenable group is indicable by a result of Morris, thus splendid by the previous lemma. (You can generalize easily to locally indicable, thus all discrete left-orderable amenable groups.) -Not every group is splendid: the infinite dihedral group $\langle a, b \;|\; a^2 = b^2 = e \rangle$ with the discrete topology is not splendid, namely pick $C = \{e, a\}$. Besides the above lemma, a way to show splendedness is that every non-splendid group that has an infinite abelian subgroup is at least a bit little dihedral (I worked this out just for this post, so take with a grain of salt): - -Lemma. Suppose $G$ is a non-splendid discrete infinite group. If $H \leq G$ is abelian, then there is a finite-index subgroup $H' \leq H$ and some $a \in G$ such that $h^a = h^{-1}$ for all $h \in H'$. - -Proof: Let $C$ be the finite set proving non-splendidness. Then $hCh = C \neq \emptyset$ for all $h \in H$. -Since $H$ is abelian, $H$ acts on $G$ from the left by $h * g = hgh$. -Since $h*C \cap C \neq \emptyset$, writing $S_a \leq H$ for the stabilizer of $a \in C$ and $T_{a,b}$ for the transporter $\{h \in H \;|\; ha = b\}$, we have -$$ H = \bigcup_{a, b \in C} T_{a, b} = \bigcup_{a, b \in C \\ T_{a,b} \neq \emptyset} h_{a, b} S_a $$ -for some choices $h_{a,b} \in H$. By a result of Neumann, some $H' = S_a$ has finite index in $H$, and for $h \in H'$ we have $hah = a \implies h^{-1} = aha^{-1}$. Square. -In particular a group is splendid if there is an element of infinite order such that no power of it is conjugate to its inverse. -For example this lemma applies to Thompson's $T$ (which has no nontrivial abelian quotients so the first lemma does not apply): Pick an element $f$ that fixes a unique interval $I \subset S^1$ and has derivative $>1$ after the right endpoint of $I$. Then $\langle f \rangle$ is of infinite order and every positive power $f^i$ has the same property as $f$, while no negative power $f^{-1}$ does. This property is preserved under conjugacy, so no $f^i$ is conjugate to its inverse. - -Question. Does this have a more standard, even if possibly less splendid, existing name? Alternative characterizations that are easier to verify? - -You can refine the definition as follows, for discrete groups: say $G$ is $k$-splendid if the splendidness condition holds for all $|C| \leq k$. Every infinite group is $1$-splendid, while the dihedral group is not $2$-splendid. A characterization of $k$-splendidness for small $k$ would also be of interest. -A list of my favorite groups and whether or not they are splendid is also of interest to me, I tried to compile one, but after working out those Thompson's's I got stuck on the Grigorchuk group. -Neumann, B. H., Groups covered by finitely many cosets, Publ. Math., Debrecen 3, 227-242 (1954). ZBL0057.25603. -Morris, Dave Witte, Amenable groups that act on the line., Algebr. Geom. Topol. 6, 2509-2518 (2006). ZBL1185.20042. - -REPLY [5 votes]: edit January 18th, 2021 -Final nail in the f.g. splendid coffin. Apparently, the dihedral group and its variants are indeed the only non-splendid groups. So maybe I should've called non-splendid groups splendid and vice versa, because this would roll off the tongue better then. I of course originally thought non-splendid groups would usually be some crazy monsters, and that the dihedral group was kind of an outlier. -Let's say an automorphism $\phi : H \to H$ of a group $H$ flips it if $\phi(h) = h^{-1}$ for all $h \in H$. Obviously a flipping automorphism can only exist on an abelian group, since $a^{-1}b^{-1} = (ba)^{-1} = \phi(ba) = \phi(b)\phi(a) = b^{-1}a^{-1}$. - -Theorem. A finitely-generated group $G$ is non-splendid if and only if it admits an internal automorphism that flips some non-trivial finite-index normal torsion-free abelian subgroup. - -Proof. We know from the previous edit (from yesterday) that any f.g. non-splendid group $G$ is virtually abelian, and a non-trivial finite-index normal torsion-free abelian subgroup implies virtual abelianity, so we may restrict to virtually abelian $G$. -Let $H \leq G$ be any non-trivial finite-index normal torsion-free abelian subgroup (so $H \cong \mathbb{Z}^d$ for some $d \geq 1$), which exists by intersecting the conjugates of any non-trivial finite-index torsion-free abelian subgroup. We show that $H$ is flipped by an inner automorphism of $G$ if and only if $G$ is non-splendid. Because the choice of $H$ is arbitrary, this shows the a priori stronger fact $\exists H: P \implies \mbox{non-splendid} \implies \forall H: P$. Let $F \subset G$ be the finitely many coset representatives for $H$. -Suppose first that conjugation by $t \in G$ flips $H$. Write $h^t = t^{-1} h t$. Pick $C = \{ f^{-1} t, tf \;|\; f \in F \}$. Now, for any $g \in G$ we can write $g = h f$ for $h \in H, f \in F$, and then -$$ g C g \ni hf \cdot f^{-1}t \cdot hf = h t h f = t \cdot h^t h \cdot f = tf \in C. $$ -So $G$ is not splendid. -Suppose then that for all $g \in G$, conjugation by $G$ does not flip $H$. Then there exists $h \in H$ which is not flipped by any $g \in G$, and in fact we can find $h$ such that conjugation by any $g \in G$ takes $h$ very far from $h^{-1}$. More precisely, for each $R$ there exists $h \in H$ such that $d(h^g, h) \geq R$ in the right-invariant word metric $d$ of $G$. -There are many ways to see the above, I guess you can geometrize and it's a basic lemma about finite reflection groups. A direct proof is also easy: Observe that it's enough to bound the word metric of $H$ since the distortion is bounded since $H$ is finite index. We only need to consider conjugation by each $f \in F$. Now take for each $f \in F$ an element $h_f$ not flipped by $f$, and consider products of the form $h_{f_1}^{n_1} \circ h_{f_2}^{n_2} \circ \cdots \circ h_{f_{|F|}}^{n_{|F|}}$ where the $n_i$ have different scales. Basically, take $h_{f_1}$ not flipped by $f_1$, so $h_{f_1}^{n_1}$ is conjugated very far from $(h_{f_1}^{n_1})^{-1}$, and take a massive $n_1$. If also $f_2$ does not flip $h_{f_1}$, we continue to $f_3$, if $f_2$ does flip $h_{f_1}$ then it does not flip $h_{f_1}^{n_1} \circ h_{f_2}^{n_2}$ where $n_2$ has a smaller scale, but $f_1$ does not flip this either since $n_1$ is much bigger than $n_2$. And so on. (In the geometric situation I suppose you can replace this by Baire or the fact $\mathbb{R^d}$ is not a union of finitely many $(d-1)$-dimensional subspaces.) -We can now use the elements $h$ given by the two paragraphs above to prove splendidness. Let $C$ be arbitrary, and suppose $C \subset B_R$, the ball of radius $R$ in the right-invariant word metric of $G$ around identity. Let $h$ be such that $d(h^a, h^{-1}) > 2R$ for all $a \in G$. If $hCh \cap C \neq \emptyset$ then $h^a h = a^{-1} h a h = a^{-1} b$ for some $a, b \in C$, so $d(h^a, h^{-1}) \leq 2R$, a contradiction. So $G$ is splendid. Square. -edit January 17th, 2021 - -Theorem. Suppose $G$ is a f.g. discrete group which is not splendid. Then $G$ is virtually abelian. - -Proof. We know from the first version of my answer (see below after the text original) that $G$ is virtually nilpotent. Then it's well-known that $G$ has a finite-index torsion-free nilpotent subgroup $H$. Let $C$ be the finite set contradicting splendidness. -First, let's think about what splendidness says when $H$ is a torsion-free subgroup. For any $g \in H$ and $i \in \mathbb{N}$, we have $g^i a g^i = b$ for some $a, b \in C$. Now, pigeon says $g^i a g^i = b$, $g^{i+k} a g^{i+k} = b$ for some $k > 0$, and we get $g^i a g^i = g^{i+k} a g^{i+k}$, in other words $g^{-k} = a g^k a^{-1}$. Clearly this $k$ is bounded, and by taking the least common multiple of the $k$ that occur, we find $a \in C$ such that for all $g \in H$, $(g^k)^a = g^{-k}$ (true for either convention for the $g^a$ notation, because inversion is an involution). -Note that in particular $H$ is now a torsion-free nilpotent group which admits an automorphism such that for all $g \in H$, $(g^k)^a = g^{-k}$. I claim that if $H$ is not abelian, this is impossible, which will conclude the proof. -To see this, suppose $H$ is not abelian. I claim that in this case there is a three-dimensional Heisenberg group inside: We use the convention $[x,y] = xyx^{-1}y^{-1}$ if it matters. Let $(H_i)_i$ be the lower central series of $H$, $H_n \neq 1, [H_n, H] = 1$. Take $1 \neq c = [a,b] \in H_n$ a non-trivial commutator, where $a \in H_{n-1}, b \in H$. Now, $c$ is superlinearly distorted in $\langle a,b,c \rangle$, so certainly $a$ and $b$ cannot satisfy any identity. If $\langle a,b,c \rangle$ satisfies an identity that is not true in the Heisenberg group, then commuting elements to normal form $a^k b^\ell c^m$ we see that $c$ satisfies an identity on its own, contradicting the torsion-freeness assuption. -Now it suffices to show that the three-dimensional Heisenberg group $H = \langle a,b,c \;|\; [a,b] = c, [a,c], [b,c] \rangle$ does not admit an automorphism $\phi : H \to H$ satisfying $\phi(g^k) = g^{-k}$ for all $g \in H$. This is geometrically obvious, go around a big cycle on the $\langle a,b \rangle$-quotient, and you jump up by the area in the central direction. If you apply the inverse elementwise, you get a cycle of the same area in the same direction (just reordered). So you don't flip the direction. -In algebra, take $[a^{km}, b^{km}] = c^{k^2 m^2}$. Now apply $\phi$ to get -$$ [a^{-km}, b^{-km}] = [\phi(a^{km}), \phi(b^{km})] = \phi([a^{km}, b^{km}]) = \phi(c^{k^2 m^2}) = c^{-k^2 m^2} $$ -But -$$ [a^{-km}, b^{-km}] = [a^{km}, b^{km}]^{b^{km} a^{km}} = [a^{km}, b^{km}] = c^{k^2 m^2} $$ -since $[a^{km}, b^{km}]$ is central. So $c^{-k^2 m^2} = c^{k^2 m^2}$ which is the desired contradiction. Square. -Presumably you could spell out in more detail what the situations are where you get non-splendidness. I suppose it's just that some element flips the torsion-free part of the finite-index abelian subgroup. Also, the proof essentially uses the observation that non-splendidness implies essentially that some element conjugates things to their inverses, giving that inversion is almost an automorphism. This is strange since it is of course a flip automorphism, which sounds like it should ultimately be the reason that non-splendid groups must be virtualy abelian. Maybe you could cut out the middle man (ICC quotients) by working more on the algebra. -original -Joshua Frisch pointed out to me in private communication that ICC $\implies$ splendid, which seems to give a good partial answer to the question. Let me restrict to f.g. infinite discrete groups $G$. -Let $\mathcal{D}$ be the class of quotients of the fundamental group of the Klein bottle $\langle h, a \;|\; h^a = h^{-1} \rangle$, where $h$ is mapped to an element of infinite order and $a$ to a nontrivial element. For example the quotient where $a^2 = e$ is added gives the dihedral group. - -Theorem. If $G$ f.g. infinite non-splendid, then $G$ is virtually nilpotent, and contains a group from $\mathcal{D}$. - -This is presumably not a characterization, but at least it reduces the problem to virtually nilpotent groups, which are much better understood than general ones. First, there's the following result of [1] (which is indeed a consequence of Neumann's theorem, but there's a nice algebraic trick that I missed). - -Theorem. If $G$ is an f.g. ICC group, then for every finite $Z \subset G$ there exists $g$ such that all solutions to $gxg^{\pm} = y$ with $x, y \in Z$ (if they exist) satisfy $x = y = e$. - -This gives ICC $\implies$ splendid. There's a minor complication with involutions, so I give a proof: - -Theorem. If $G$ is an f.g. ICC group, then it is splendid. - -Proof. Suppose ICC, and fix some symmetric generating set. Let $n$ be arbitrary and take $Z = B_N$ for $N = 2n$. By the previous theorem, for some $g \in G$, among $x, y \in Z$ the equation $gxg^{\pm} = y$ can only have solutions with $x = y = e$. If $g$ is not an involution, there aren't any such solutions either, and we are done. -If $g$ is an involution, then take any $h$ of word norm exactly $n$, and consider $gh$ instead. Now if $gh \cdot k \cdot gh = k'$ for $k, k' \in B_n$, then $g \cdot hk \cdot g = k' h^{-1}$ and here $hk, k'h^{-1} \in Z$, and this cannot happen by the assumption on $g$. So we have $gB_ng \cap B_n = \emptyset$. Square. -Next, there's a theorem of [1,2] (see also [3]). - -Theorem. Every f.g. infinite group is either virtually nilpotent, or has a non-trivial ICC quotient. - -And also let's prove a lemma. - -Lemma. If $G$ has a splendid quotient, it is splendid. - -Proof. Let $\phi : G \to H$ be the quotient map, and $C \subset G$ be finite. Then $\phi(C) = C'$ is also finite, and thus $hC'h \cap C' = \emptyset$ for some $h \in H$. Letting $\phi(g) = h$, we have $\phi(gCg) \cap \phi(C) = hC'h \cap C' = \emptyset$, so $gCg \cap C = \emptyset$ as well, and this means $G$ is splendid. Square. -We obtain the following: - -Theorem. Every non-splendid group is virtually nilpotent. - -Proof. If $G$ is not virtually nilpotent, then it has a non-trivial ICC quotient. That quotient is then splendid, and thus $G$ is splendid. Square. -Now we can prove the theorem I stated at the beginning of the post. -Proof. An infinite f.g. nilpotent group contains $H \cong \mathbb{Z}$, which is abelian. By the lemma in my post, there exists $a \in G$ and $h \in H \setminus \{e\}$ such that $h^a = h^{-1}$ for all $h \in H$. Then $h$ and $a$ give the quotient, taking $n$ the order of $a$. Square. -[1] Erschler, A.; Kaimanovitch, V. A., Arboreal structures on groups and the associated boundaries, Arxiv https://arxiv.org/abs/1903.02095] -[2] Duguid, A. M.; McLain, D. H., (FC)-nilpotent and (FC)-soluble groups, Proc. Camb. Philos. Soc. 52, 391-398 (1956). ZBL0071.02204. -[3] McLain, D. H., Remarks on the upper central series of a group, Proc. Glasg. Math. Assoc. 3, 38-44 (1956). ZBL0072.25702. -[4] Frisch, J.; Ferdowsi, P. V., Non-virtually nilpotent groups have infinite conjugacy class quotients, Arxiv https://arxiv.org/abs/1803.05064]<|endoftext|> -TITLE: Using linear logic in algebraic geometry and commutative algebra -QUESTION [16 upvotes]: In algebraic geometry and commutative algebra we often deal with many categories that are not topoi nor cartesian closed (or even locally CC), but are nevertheless closed monoidal. These include the category $\mathbf{Mod}_R$ of modules over a commutative ring $R$, the category $\mathbf{PShMod}_X$ of presheaves of modules over a scheme $X$ (perhaps also $\mathbf{ShMod}_X$?), and et cetera. -Of course, there are a lot of topoi occuring in algebraic geometry, but others are not. If a category is not a topos, one can't use "normal" intuitionistic reasoning inside the category (because its internal logic does not entail higher-order intuitionistic logic, or the Calculus of Constructions from a TCS point-of-view). Then, as a computer scientist, the natural idea would be: since the internal logic of a monoidal category is some form of linear logic, can we use linear logic fruitfully to study those categories, perhaps in the style of Blechschmidt? -I found this recent manuscript by Paul-André Melliès, which was apparently submitted to LICS 2020 but not accepted. I can't find anything else in this direction, so is this a problem that has been considered in the past? Does anyone have any pointers in this direction and/or perhaps related work? It seems that this has some connections with Tannaka duality, but this is not a topic I'm familiar with at all... - -REPLY [4 votes]: If Simon Henry has not found anything beyond the references you have already mentioned, I doubt very much there is anything out there (though one never knows, math is full of works kept in drawers, or published in some remote journal..). -So, the only option seems to be: let us reason together and see how far we can go. Before I start, let me tell you that your question rocks. I have also to declare that it is a bit loose, and that perhaps by making it a tad more focused we could find the magic thread to answering it. -Your line of thought is: -1.Monoidal closed categories come equipped with (a fragment of) linear logic, so it makes sense to leverage such a logic to describe "things" and "constructions" within those categories. -2.Moreover, you suggest that some cats such as R-MOD (the archetypal monoidal closed cat) and sheaves/presheaves of modules over a scheme are ubiquitous in Algebraic Geometry. Thus, assuming that linear logic tells us something about such cats, perhaps it is also useful to express meaningful procedures in Algebraic Geometry. -Breaking it down in these two steps has an advantage: if we can validate 1, we have some chance to also address 2. -Let us start with 1 then. -What you stated about linear logic basically says: linear logic is kind of the vanilla internal logic for monoidal closed cats (CMC in what follows). Now, as one would expect, this does not tell much as far as specific monoidal cats, only about general constructions and entities living in any generic CMC. -Take for instance the ring R of polynomials over a field (which is exactly the archetypal algebraic structure for classical AG). Now, consider the CMC of its modules. -Can I describe it by a list of linear logic axioms? -In other words, is there a linear theory which is valid in this cat or cats equivalent to it? I do not know the answer, but this is definitely the very first step I would take. -Assume one gets somewhere with 1, and now tries to tackle 2. -Here is what I would do: there is an entire field called Computational Algebraic Geometry, with gadgets like Grobner Basis and the like. - -Can one axiomatize these constructions in the internal linear theory (see previous point) of polynomial rings?* - -If that were possible, it would be quite interesting to computer scientists, for instance in that one could develop a sort of "Linear Prolog" to describe computational AG and computational commutative algebra. -There is something else, much bigger than this, but I shall stop here (but see my last question on quantale sub-object classifiers ) -One last reference: almost nobody has tried to take seriously Linear Logic as THE logic of constructive math. There are a few refs worth while though, one being Mike Shulman's article here<|endoftext|> -TITLE: Milnor excision for algebraic stacks -QUESTION [8 upvotes]: Recall that a commutative square of commutative rings -$$\begin{matrix} -A&\to&B\\ -\downarrow &&\downarrow\\ -A^\prime&\to&B^\prime\end{matrix}$$ -is called a Milnor square if the vertical maps are surjective and the square is both a pullback and pushout of rings. -It has been shown that Milnor squares give rise to diagrams that are still both pullbacks and pushouts in the category of all schemes after applying $\operatorname{Spec}$. Since it is standard that the $\operatorname{Spec}$ functor sends pushouts to pullbacks, perhaps a more interesting way to state this fact is that a functor of points representable by a scheme $S$ satisfies Milnor excision: -$$S(A)\simeq S(A^\prime)\times_{S(B^\prime)}S(B).$$ -Question: Is it true that Deligne-Mumford stacks or Artin stacks also satisfy Milnor excision? Is there a reference? -Note: Lurie shows that spectral Deligne-Mumford stacks satisfy a weaker condition in SAG chapter 16 called cohesion, which gives excision as above when all of the maps of rings in the square are surjective. I'm primarily interested in the spectral DM case, but a proof in the non-derived case should be enough to suss out what's going on. -Edit: It turns out that this is quite a hard problem in complete generality, but the case I care about is, specifically, quasicompact quasiseparated Deligne Mumford stacks (but no affine/quasiaffine diagonal!). Not sure if that makes things any easier. - -REPLY [9 votes]: In upcoming (now on arXiv:2205.08623) joint work with Jarod Alper, Jack Hall and Daniel Halpern-Leistner: -Artin algebraization for pairs with applications to the local structure of stacks and Ferrand pushouts -we prove more generally the existence of pushouts of affine morphisms along closed immersions in the category of (quasi-separated) algebraic stacks (Thm. 1.8). This in particular implies that Milnor squares are pushouts in the category of (quasi-separated) algebraic stacks. Let me sketch how this is proved: -Let $X=\operatorname{Spec} B$ and $Y=\operatorname{Spec} A$ and similarly for the primes so we have a cartesian square: -$\require{AMScd}$ -\begin{CD} -X' @>f'>> Y'\\ -@V g' V V @VV g V\\ -X @>>f> Y -\end{CD} -with $g$, $g'$ closed immersions. -By assumption, this is co-cartesian in the category of affine schemes. -To show that this is co-cartesian in the category of algebraic stacks, let $Z$ be an algebraic stack together with maps $u\colon X\to Z$ and $v\colon Y'\to Z$ and a $2$-isomorphism $ug'\cong vf'$. We can replace $Z$ with an open quasi-compact neighborhood of the images of $u$ and $v$ and assume that $Z$ is quasi-compact. -Let $p\colon Z_1\to Z$ be an affine smooth presentation. Consider the pull-backs along $u$, $ug'\cong vf'$ and $v$ and call these $X_1\to X$, $X'_1\to X'$ and $Y'_1\to Y'$. The easiest case is if $Z$ has affine diagonal. Then $p$ is affine and $X_1$, $X'_1$, $Y'_1$ are also affine. Then we can take the pushout of the three affine schemes resulting in $Y_1\to Y$. This gives us a map $Y_1\to Z_1\to Z$. One then observes that $Y_1\to Y$ is smooth (flatness is [Fer, Thm 2.2 (iv)] and finite presentation can be proven similarly and smoothness then follows by considering fibers). Then take $X_2=X_1\times_X X_1$ etc. We obtain two maps $Y_2\rightrightarrows Y_1\to Z_1\to Z$. Since $Y_2$ also is a pushout in the category of affine schemes (they are stable under flat base change by [Fer, Thm 2.2 (iv)]) these two maps coincide (*). By descent, we obtain a map $Y\to Z$. -(*) It remains to show that any two maps $Y\to Z$ fitting in the diagram are isomorphic up to unique 2-isomorphism. For this, one takes two maps and pull-back the diagonal of $Z$. This is then turned into an existence question. Again, if the diagonal is affine, it is immediate. -When the diagonal is not affine, then the $X_1$, $X'_1$ and $Y'_1$ above are merely algebraic spaces. One can take an étale affine presentation of $X_1$ and pull this back to $X'_1$. The crucial step is then to extend this to an étale presentation of $Y'_1$. This is where the Artin algebraization alluded to in the title comes in. It is also needed when you want to construct the pushout $Y$ of a diagram $X\leftarrow X'\rightarrow Y'$ of algebraic stacks (affine / closed immersion). -Edit: In [TT], the case where $\Delta_Z$ is (ind-)quasi-affine is handled. The crucial result is [TT, Thm 5.7/5.8] which in the setup above proves that $Y_1$ exists when $X_1$ is (ind-)quasi-affine. This settles the case when $Z$ is an algebraic space or a Deligne–Mumford stack with separated diagonal. The case where $f$ is finite/integral is easier and treated in [Fer] and [R, Thm. A.4]. Also see MO question Ferrand pushouts for algebraic stacks. -[Fer] Daniel Ferrand, Conducteur, descente et pincement, Bull. Soc. Math. France 131 (2003), no. 4, 553–585. -[R] David Rydh, Compactification of tame Deligne–Mumford stacks, preprint, https://people.kth.se/~dary/tamecompactification20110517.pdf -[TT] Michael Temkin and Ilya Tyomkin, Ferrand pushouts for algebraic spaces, Eur. J. Math. 2 (2016), no. 4, 960–983.<|endoftext|> -TITLE: Is there a formulation of Huygens' principle using the language of algebraic geometry? -QUESTION [8 upvotes]: Huygens' principle is a way to describe the propagation of light rays in a manifold $\mathcal{M}$. Suppose we start with an infinitesimal ball of light at position $X$ in a manifold. Let $B(X,t)$ be the set of points to which light rays can arrive at times less than or equal to $t$. Then $\partial B(X,t)=\Sigma(X,t)$ is a closed surface that constitutes the set of points to which light arrive precisely at time $t$. Consider both $\Sigma(X,t_-)$ and $\Sigma(X,t_+)$ -Consider also the class of surfaces $\mathcal{A}(t_--t_+)=\Sigma(Y,t_+-t_-)$ for each $Y\in \Sigma(X,t_-)$ -Then $\Sigma(X,t_+)$ is the unique closed surface in $\mathcal{M}$ that is tangent to each element in $\mathcal{A}(t_--t_+)$ -This is Huygens' theorem. -My question is whether this can be reframed in the language of algebraic geometry. My question is vague because I know very little in algebraic geometry. I guess I am trying to ask if there is a generalization of Huygens' principle to more exotic spaces like supermanifolds, for instance. - -REPLY [2 votes]: From your question, I guess that you start out with a "smooth Riemannian manifold $\mathcal{M}$". Algebraic geometry is, very roughly, the realm of zero loci of polynomial (or perhaps rational) equations, and polynomial (rational) maps between them. A general smooth Riemannian manifold need not belong to that realm, hence algebraic geometry would have very little to say about such an $\mathcal{M}$, or the Huygens' principle on it. -That said, I have some brief observations. - -The way you describe Huygens' principle corresponds to the notion of taking an envelope of curves or surfaces (the $\mathcal{A}$ family in your case). This is an operation that makes sense in algebraic geometry because the envelope of a polynomial family of algebraic varieties is also a variety, I believe. - -There is a couple of famous articles bringing together algebraic geometry and the Huygens' principle (or something very closely related to it) for constant coefficient hyperbolic equations (at order 2, such equations basically generalize the wave equation on Euclidean space, while at higher orders they are no longer describable by simple Riemannian geometry): - - - - -Atiyah, Michael F.; Bott, Raoul; Gårding, Lars, Lacunas for hyperbolic differential operators with constant coefficients. I, Acta Math. 124, 109-189 (1970). ZBL0191.11203. -Atiyah, Michael F.; Bott, Raoul; Gårding, Lars, Lacunas for hyperbolic differential operators with constant coefficients. II, Acta Math. 131, 145-206 (1973). ZBL0266.35045. - - - -As was already implied in the previous point, the notion of light ray (or more generally characteristic ray) only sometimes coincides with that of a geodesic in a Riemannian manifold. The notion itself is more closely tied with the finite speed of propagation for a hyperbolic equation. Such equations can be formulated on supermanifolds and for superfields. Roughly speaking, any supermanifold can be seen as a bundle of odd or super directions over a regular manifold (the body, cf. Serre-Swan theorem). Getting one's hands dirty, and thinking as an analyst, a hyperbolic equation in this setting can be represented as a hierarchical system of regular hyperbolic equations on the body, where the super directions appear in a filtered manner at higher orders of the hierarchy. At the base of the hierarchy, you have a regular hyperbolic PDE on the body without any super information entering into it, and it is only this equation that controls the speed of propagation and hence the Huygens' principle. This approach was pioneered by Choquet-Bruhat in the seminal paper - - - -Choquet-Bruhat, Yvonne, The Cauchy problem in classical supergravity, Lett. Math. Phys. 7, 459-467 (1983). ZBL0529.58039.<|endoftext|> -TITLE: Is Carlitz's paper correct about the number of similarity classes of commuting matrices? -QUESTION [18 upvotes]: L. Carlitz has a paper, Classes of pairs of commuting matrices over a finite field, that computes the number of simultaneous similarity classes of of pairs of commuting matrices in $\operatorname{Mat}_n(\mathbb F_q)$. Two pairs $(A,B)$ and $(A',B')$ are called simultaneously similar if there is $U\in \operatorname{GL}_n(\mathbb F_q)$ such that $A'=UAU^{-1}$, $B'=UBU^{-1}$. -However, from the proof (specifically, to go from equation (4) to (5) on p. 193), it seems that he implicitly uses the statement that $(A,B)$ and $(A,B')$ belong to different similarity classes whenever $B\neq B'$. But it is possible that there is $U\in \operatorname{GL}_n(\mathbb F_q)$ that commutes with $A$ and such that $UBU^{-1}=B'$. In this case, they belong to the same class. -Is that it? If this paper turns out to be wrong, is there any result about the same question? - -REPLY [17 votes]: It is wrong. See the correction published in AMM 71 (1964), issue 8, page 900. (You have to scroll down to the bottom of the last page to find this correction.) -Unfortunately this was published in the "Mathematical Notes" section, making it hard to find (as these notes are not individually indexed much of the time, and don't get separate DOIs). I've only managed to find it by looking at the back references to Carlitz's original paper (always a good first step if you suspect something is wrong in a paper; whoever cited it might too have noticed), and realizing that one of these back references also cites a correction.<|endoftext|> -TITLE: Theory of weak enrichment in higher categories -QUESTION [9 upvotes]: Has there been work towards a general theory of weak enrichment in higher categories? To be more pointed, has there been any work towards trying to make sense of statements such as - -There is a (weak) $(n+1,r+1)$-category $\mathcal V\mathbf{Cat}$ of categories enriched in a monoidal $(n,r)$-category $\mathcal V$. - -generalising the fact that there is a $2$-category of categories enriched in an ordinary monoidal category, and how I imagine there is a (weak?) $3$-category of categories enriched in a monoidal bicategory. -I know there's work by Gepner and Haugseng (and more in this direction) which makes sense of an $(\infty,1)$-category of categories enriched in a monoidal $(\infty,1)$-category, though from my (limited) understanding there is little care here for the non-invertible "enriched natural transformations" which would exist in the corresponding $(\infty,2)$-category of enriched categories, and a lot of the work is put into ensuring that the equivalences of enriched categories correspond to the correct notion of fully faithful and essentially surjective functors. -Besides these, and other texts that address special cases (e.g., using a notion of weak enrichment to create a theory of weak $n$-categories), I haven't found anything that addresses a more general notion of enrichment in higher monoidal categories. Perhaps I just don't know how to look for them, but it also leads me to wonder: is there reason to study weak enrichment in this way? Or conversely, is there no good reason to be concerned with this? (e.g., problems where this would come in handy don't come up often / don't exist, or the $(\infty,1)$-category theory of enrichment is sufficient for most practical purposes, etc.) -Edit: Harry Gindi pointed out that Gepner and Haugseng do provide some work in the direction of higher categorical structure. In particular (this is their example 7.4.11), if $\mathcal V$ is a $\Bbb E_2$-monoidal (presentable) $(\infty,1)$-category, then the $(\infty,1)$-category $\mathcal V\mathbf{Cat}$ of $\mathcal V$-enriched categories will be monoidal also; moreover, if $\mathcal V$ is closed, then so is $\mathcal V\mathbf{Cat}$. Especially, the latter is self-enriched to provide a "$\mathcal V$-$(\infty,2)$-category" of $\mathcal V$-enriched categories (in that between $\mathcal V$-enriched categories is a $\mathcal V$-enriched category of functors). -While this provides $\mathcal V\mathbf{Cat}$ with a nice self-enrichment in instances where the enriching category is nice, this is somehow giving more than I asked for given less than I have. For instance, if $\mathcal V$ is just an ordinary monoidal $1$-category, then $\mathcal V\mathbf{Cat}$ will always be a $2$-category, though with additional assumptions on $\mathcal V$ (take it to be a Bénabou cosmos, for instance) it can also be given an internal hom (making it a "$\mathcal V$-enriched $2$-category"). -If $\mathcal V$ is a general monoidal $(\infty,1)$-category, is it possible that $\mathcal V\mathbf{Cat}$ will to naturally an $(\infty,2)$-category? -Edit 2 (for clarification): While Harry's answer does a great job of elaborating on the work of Gepner and Haugseng, it doesn't really answer my question (as far as I know). I'm not trying to inductively define $(n+1,r+1)$-categories through iterated weak enrichment; rather, I'm just trying to see if there are constructions which, given for example a (nice?) monoidal $n$-category $\mathcal V$, produces the $(n+1)$-category $\mathcal V\mathbf{Cat}$ of categories enriched in $\mathcal V$. I mentioned Gepner and Haugseng's work because it's a step in this direction, but I don't think they were trying to address this generality. - -Since my post is getting a bit long, I'll keep a summary of the questions I'm really trying to ask: - -Is there existing work towards a general theory which describes the $(n+1,r+1)$-category of categories enriched in a monoidal $(n,r)$-category? -If not, is this because there is no merit to this endeavour in the first place, or is it because this is just a difficult thing to accomplish in general? - -REPLY [5 votes]: If you take a look at Gepner-Haugseng Corollary 5.7.12, the first thing you'll notice is that the functor $\mathbf{Cat}^{(-)}_\infty:\mathbf{MON}^{\operatorname{lax}}_∞\to \mathbf{CAT}_\infty$ is lax monoidal and sends $\mathcal{O}$-algebras to $\mathcal{O}$-monoidal categories for any symmetric operad $\mathcal{O}$. Another way to say this is that enrichment takes an $\mathcal{O}\otimes E_1$-monoidal category to an $\mathcal{O}$-monoidal category. -Now here's a neat thing to note: If we restrict this to the subcategory of presentably monoidal $\infty$-categories, we get the stronger statement that the functor $\mathbf{Cat}^{(-)}_\infty$ carries $\mathbf{Mon}^{\operatorname{pr},\operatorname{lax}}_\infty$ to $\mathbf{Pr}^L_\infty$ and is lax monoidal, sending the tensor product of presentably monoidal $\infty$-categories to the tensor product of presentable $\infty$-categories. -In particular, we now get the same statement as before, namely that for any symmetric operad $\mathcal{O}$, the functor $\mathbf{Cat}^{(-)}_\infty$ carries presentably $\mathcal{O}\otimes E_1$ monoidal categories to presentably $\mathcal{O}$-monoidal categories. Ok, so how does this answer your question? -Well, presentably $\mathcal{O}$-monoidal categories are closed, namely for each arity, we have an n-fold tensor product $\otimes^n:C^{\otimes n}\to C,$ and if we choose a family of $n-1$ objects $(c_1,\dots,c_{\hat{i}},\dots,c_n)$ (omitting the ith index), we obtain a colimit-preserving functor $C\to C$, which now admits an adjoint, the internal hom (if you aren't symmetric monoidal, these can all vary in complicated ways). -Now applying this to $\mathbf{Cat}^{\mathcal{V}}_\infty$, for $\mathcal{V}$ presentably $E_n$-monoidal, we see that it is canonically enriched over itself. Moreover, the assignment $\ast \mapsto \mathbb{1}_{\mathcal{V}}$ extends to a lax monoidal functor $\mathcal{S}\to \mathcal{V}$, which we can use to understand the 'underlying $(\infty,2)$-category $\widetilde{\mathbf{Cat}}^{\mathcal{V}}_\infty$'. -You can then iterate this procedure to produce an $E_{n-k}$ presentably-monoidal $(\infty,k)$-category of $k$-fold iterated enriched categories. -So I'm a bit confused what you're talking about. You do indeed get all of the correct kinds of natural transformations. There is a problem in the case where you choose $\mathcal{V}$ to be $\mathbf{Cat}_{\infty,n}$, but this is not an enriched story. -It is expected (or already proven, depending on who you ask) that there is another special biclosed $E_1$ presentably-monoidal structure on $\mathbf{Cat}_{\infty,n}$ for each $n\leq \omega$. This is called the Gray tensor product, or the lax Gray tensor product. This is not an enriched tensor product at all. It can't commute with or distribute over the Cartesian product. It is sui generis, and its right adjoints classify functors with lax or oplax natural transformations between them. This is an extremely important construction, but it has very little to do with enrichment. It looks like the easiest way to construct it is actually by first inducing it on $\mathbf{Cat}_{\infty,\omega}$, then inducing it on each finite $n$ by localization.<|endoftext|> -TITLE: What sets can be unraveled? -QUESTION [16 upvotes]: A set $X\subseteq\omega^\omega$ is unravelable iff there is a possibly larger set $A$ and a clopen set $Y\subseteq A^\omega$ (with respect to the product topology coming from the discrete topology on $A$) such that winning strategies for the game on $A$ with payoff set $Y$ can be converted to winning strategies for the game on $\omega$ with payoff set $X$ in a particularly simple way (see Martin's paper A purely inductive proof of Borel determinacy for the precise definition). By Gale–Stewart, unravelability implies determinacy; however, it is not the only route to proving the determinacy of a pointclass. -My question is whether there is a known upper bound on the complexity of an unravelable pointclass. To make this somewhat precise: - -Is it consistent with $\mathsf{ZFC}$ that every set in $\mathcal{P}(\mathbb{R})^{L(\mathbb{R})}$ is unravelable? - -The strongest results I can find are vastly weaker than this, namely that under large cardinal assumptions $\Pi^1_1$ sets are unravelable (Neeman 1,2; I remember seeing the same result for Borel-on-$\Pi^1_1$ sets, but I can't find a reference for it at the moment). On the other hand, I don't see an easy proof that this is anywhere close to the most unravelability we can expect, nor can I find this stated in the literature. - -REPLY [6 votes]: I emailed Itay Neeman, and he told me the following: - -As far as I know it's open. I don't think anything is known about -unraveling beyond what you can get from my methods. These give the -Suslin operation on $\Pi^1_1$ sets, and various iterations of that. In -terms of the large cardinal hierarchy it's still all just using -measures. Anything above that is open I think. -In particular I don't think it's known if $\Sigma^1_2$ sets can be -unraveled. - -For now I think that settles this question. (Itay later said that it's unclear whether we "should" expect unravelability of $\Sigma^1_2$ sets from large cardinals or not in the first place. So I think my main takeaway at this point is that unravelability is not best thought of as yet another tameness property, where the slogan "large cardinals make definable things tame" is expected to prevail.)<|endoftext|> -TITLE: A claim on partitioning a convex planar region into congruent pieces -QUESTION [5 upvotes]: Let us define a perfect congruent partition of a planar region $R$ as a partition of it with no portion left over into some finite number n of pieces that are all mutually congruent (ie any piece can be transformed into another piece by an isometry. We consider only cases where each piece is connected and is bounded by a simple curve). -Note: It is known there are convex planar regions - indeed quadrilaterals - which do not allow perfect congruent partition for any n ([1] proves a stronger result). -Claim: If a convex polygonal $R$ allows a perfect congruent partition of itself into $N$ non-convex pieces each with finitely many sides, then $R$ also allows a perfect congruent partition into $N$ convex pieces with finitely many sides. In other words, allowing the pieces to be non-convex polygons does not improve the chances of a convex planar region achieving a perfect congruent partition into $N$ pieces. -I know no proof, no counter example. One can consider replacing 'congruent' with 'similar' in the above question. Some more related thoughts are in [2]. -References: -1.https://www.research.ibm.com/haifa/ponderthis/challenges/December2003.html -2.https://arxiv.org/abs/1002.0122 - -REPLY [7 votes]: This picture looks like a counterexample with $N=2$ and $R$ a convex pentagon: - -This should work more generally starting from an $n \times (n+1)$ rectangles -for any integer $n>1$, removing two congruent right triangles that are -not isosceles (the picture shows $n=3$ with $2:5$ triangles). -Replacing the heavy lines with more complicated polygonal convex paths -yields convex polygons with more sides.<|endoftext|> -TITLE: Variety of negative Kodaira dimension contains a projective line -QUESTION [5 upvotes]: Does a smooth projective variety over an algebraically closed field of negative Kodaira dimension contain a projective line? -I do not want to assume any conjectures. - -REPLY [2 votes]: I remark that it is true in dimension at most $3$, over $\mathbb{C}$. (This probably isn't the easiest way to see it and is not a complete solution, but it was too short for a comment. ) -In dimension $2$ it follows from the fact that all of the minimal surfaces with negative Kodaira dimension are covered by smooth rational curves (they are either $\mathbb{P}^1$-bundles or $\mathbb{P}^2$). -In dimension $3$ it follows from the existence and classification of extremal contractions for smooth $3$-folds due to Mori. Let $X$ be a smooth $3$-fold with Kodaira dimension $-\infty$ then $X$ is uniruled (Theorem 6.1.8 (FV)). Hence by a result of Miyoaka and Mori there is a non-empty Zariski open $U \subset X$ such that each point in $U$ is contained in an irreducible curve $C \subset X$ with $K_{X}.C <0$ (Theorem 6.1.4 (FV)), in particular $K_{X}$ is not nef. -(by (FV) I mean the book "Fano varieties" by Iskovskikh and Prokhorov). -Hence by the contraction theorem, there is some extremal contraction. You may consult the list of extremal contractions which can occur. In all cases; other than when the image is a point, some fibre of the extremal contraction clearly contains smooth rational curves (For conic bundles and del Pezzo bundles one needs that a general fibre is smooth, which is true). If the image is a point, $X$ is Fano $3$-fold with $b_{2}=1$. Then since Fano $3$-folds are classified we may go through each of the $17$ prime Fano $3$-folds and check there is a smooth rational curve in each.<|endoftext|> -TITLE: Short research articles -QUESTION [12 upvotes]: I am a masters student. I am interested in short articles which have counter examples and very few references. I want to write a short and interesting article. -For example; One of the best known shortest and best academic paper articles I read is Counterexample to Euler's Conjecture on Sums of Like Powers by L. J. Lander and T. R. Parkin (Bull. Amer. Math. Soc. 72 (1966) p 1079, doi:10.1090/S0002-9904-1966-11654-3). -It has only one reference. It's really fascinating. -Is there any short articles in Mathematics and especially in Analysis/Complex Analysis? I am also looking at Counterexamples books for learning something and I searched open problems in Wikipedia and looked at undiscovered, newly valued and current topics. -So can you share these type of articles you read? I am curious about a good and interesting short article topic. What do you recommend to me about it? You can also give some references.** Thanks for your ideas and answers.** - -REPLY [8 votes]: For counterexamples in analysis, a good start would be "Counterexamples in Analysis" by Gelbaum and Olmsted, there's even a Dover edition.<|endoftext|> -TITLE: Number of reduced decompositions of the longest element of the Weyl group -QUESTION [9 upvotes]: Let $R$ be a reduced root system, $W$ the associated Weyl group, and $w_0 \in W$ the longest element of $W$. In general $w_0$ admits more than one reduced decomposition into a product of reflections, a number which we denote by $d_R$. Where can one find a list of values of $d_R$ for low-dimensional root systems? -For example are the explicit values of $d_R$ known for the exceptional root systems? - -REPLY [5 votes]: Around the time Fomin and I wrote this -paper, Tao Kai Lam applied the technique to type $D_n$. It -emerged that it was "natural" to weight a reduced -decomposition $\rho$ by $2^{d(\rho)}$, where $d(\rho)$ is -the number of simple reflections in $\rho$ that correspond -to the $n-2$ "nonbranch nodes" in the Coxeter diagram for -$D_n$. Using this weighting, there is a -nice product formula for the number of weighted reduced -decompositions of the longest element, which I unfortunately -have forgotten. I hope someone can redo this work. - -REPLY [5 votes]: This is easy to do in SageMath. E.g. the following code -G = WeylGroup("F4") -w = G.long_element_hardcoded() -print(w) -rw = w.reduced_words() -len(rw) - -outputs 2144892. If you want to look at some of these reduced words just examine the list rw. To create a list for classical types of different rank do -res = {} -for n in range(2,5): - G = WeylGroup(["A", n]) - w = G.long_element_hardcoded() - print("Calculating rank ", n) - res[n] = len(w.reduced_words())<|endoftext|> -TITLE: Frobenius algebras from symmetric polynomials -QUESTION [5 upvotes]: Let $K$ be a field of characteristic 0 (maybe it works for more general fields) and $K[x_1,...,x_n]$ the polynomial ring in $n$ variables. Let $e_1,e_2,...,e_n$ denote the elementary symmetric polynomials in $n$ variables (see for example https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial ). (One might also choose other symmetric polynomials such as https://en.wikipedia.org/wiki/Complete_homogeneous_symmetric_polynomial or https://en.wikipedia.org/wiki/Power_sum_symmetric_polynomial for this problem ). -For $n,t \geq 2$ let $A_{n,t}:=K[x_1,...,x_n]/(e_1^t,...,e_n^t)$. -$A_{n,t}$ is by definition a Frobenius algebra if there is a unique polynomial up to scalars $p \in A_{n,t}$ with $p x_i=0$ for $i=1,...,n$. This is a purely combinatorical condition. -For example for $n=t=2$, we have $A_{2,2}$ has vector space basis $1,x_1,x_2,x_1^2,x_1 x_2,x_1^3,x_1^2 x_2,x_1^4$ and indeed the unique longest polynomial in $A_{2,2}$ is $p=x_1^4$. Note that we have in $A_{2,2}$ $x_1^4=-x_2^4$ and unique means here really in the algebra $A_{2,2}$ and not as a polynomial in the polynomial ring. -I was able to prove by force that $A_{2,t}$ is a Frobenius algebra for $t \geq 2$. - -Question 1: Is $A_{n,t}$ for general $n,t$ a Frobenius algebra? - -In case this is true in general, there is probably a nice proof that avoid heavy computation (and might work for the other classes of symmetric polynomials such as the power sum symmetric polynomials). - -Question 2: Did those algebras appear in the literature already? Do they have other nice properties such as being Hopf algebras in special cases (maybe allowing fields of certain characteristics). - - -Question 3: Is there a generalisation of question 1 in case question 1 is true? - -For example small experiments suggest that even the algebras $K[x_1,...,x_n]/(e_1^{t_1},...,e_n^{t_n})$ are Frobenius as long as $t_1 \geq 2$ and $t_i \geq 1$. But maybe one can also do more, for example taking certain polynomials of the $e_i$. -The problem gets complicated quickly for larger $n$. Here is the case $n=3, t=2$ with the computer over the rationals($x_1=x,x_2=y,x_3=z$): -$A_{3,2}$ has vector space dimension 48 and longest polynomial for example $x^8y$. -A vector space basis via the computer is given as follows: - -[ [(1)*v1], [(1)*x], [(1)y], [(1)z], [(1)x^2], [(1)xy], [(1)xz], [(1)y^2], [(1)yz], -[(1)x^3], [(1)x^2y], [(1)x^2z], [(1)xy^2], [(1)xyz], [(1)y^3], [(1)y^2z], -[(1)x^3z], [(1)x^2y^2], [(1)x^2yz], [(1)x^4+(2)x^3z+(-2)xy^3], [(1)xy^3], -[(1)x^3y+(1)xy^3], [(1)xy^2z], [(1)y^4], [(1)x^4z], [(1)x^3y^2], [(1)x^3yz], -[(1)x^5+(2)x^4z+(-4)x^3y^2+(-4)x^3yz+(-2)x^2y^3], -[(1)x^4y+(2)x^3y^2+(2)x^3yz+(1)x^2y^3], [(1)x^2y^2z], [(1)x^2y^3+(1/2)xy^4], -[(1)xy^4], [(1)x^4yz], [(1)x^3y^3], [(1)x^5y+(2)x^4y^2+(2)x^4yz+(1)x^3y^3], -[(1)x^6+(4)x^5y+(2)x^5z+(4)x^4y^2+(4)x^4yz+(2)x^3y^3], [(1)x^2y^4], -[(1)x^4y^2+(2)x^3y^3+(1)x^2y^4], [(1)x^5z+(2)x^2y^4], [(1)x^6z+(-2)x^5y^2], -[(1)x^5y^2], [(1)*x^7+(3/2)x^6y+(3)x^5y^2], [(1)x^4y^3], -[(1)x^6y+(2)x^5y^2+(4)x^4y^3], [(1)x^7y], [(1)x^6y^2], -[(1)*x^8+(3/2)x^7y+(3)x^6y^2], [(1)x^8y] ] - -REPLY [5 votes]: In this commutative situation, a Frobenius algebra is the same as an -artinian Gorenstein ring. In general, if $\theta_1,\dots,\theta_n$ are -homogeneous elements of positive degree of $A=K[x_1,\dots,x_n]$ and if -$K[x_1,\dots,x_n]/(\theta_1,\dots,\theta_n)$ is artinian (i.e., a -finite-dimensional vector space in this situation), then -$A/(\theta_1,\dots,\theta_n)$ is Gorenstein (in fact, a complete -intersection, which is stronger). Since $A/(e_1,\dots,e_n)$ is artinian -the same is true for $A/(e_1^t,\dots,e_1^t)$, so the Gorenstein property -follows. In fact, the socle polynomial $p$ is given by $\prod_{1\leq -i -TITLE: GIT and singularities -QUESTION [7 upvotes]: Let $G$ be a complex reductive group acting on a complex affine variety $X$ and let $X // G = \operatorname{Spec}\mathbb{C}[X]^G$ be the GIT quotient. -Is there a relationship between the singular locus of $X$ and that of $X // G$? -Of course, $X//G$ can be highly singular while $X$ is smooth. But, for example, I was wondering if (or under what conditions) singular points of $X$ are mapped to singular points of $X // G$. -Edit. Spenser's nice comment below shows that the answer to the latter is no. But perhaps a better and more precise question is: If $X // G$ is non-singular at $y$, is there a non-singular $x \in X$ mapping to $y$? In other words, do all fibres of $X \to X // G$ at non-singular points contain a non-singular point of $X$? I'm willing to assume irreducibility or other nice properties. - -REPLY [5 votes]: This is an answer to the revised question. It is the simplest counterexample that I can think of where the reductive group is smooth and connected, where $X$ is normal and affine, and where $Y=X//G$ is smooth, even though there are fibers of the quotient map that are contained in the singular locus of $X$. -Let $Y$ be $\text{Spec}\ k[x,y,z]$, i.e., affine $3$-space. Let $G$ be the multiplicative group of units, $G=\text{Spec}\ k[u,u^{-1}]$. Let $X$ be $\text{Spec}\ k[x,y,z,s,t,t^{-1}]/\langle f \rangle$ where $f$ is the polynomial, $$f=s^2+t(xz-y^2).$$ Let the action of $G$ on $X$ be defined by $$\mu:G\times_{\text{Spec}\ k} X \to X, \ \ \mu(u,(x,y,z,s,t)) = (x,y,z,us,u^2t). $$ The ring of $G$-invariant polynomials is the subring, $$k[X]^G = k[x,y,z].$$ The quotient map is just the usual projection, $$q:X\to Y, \ \ q(x,y,z,s,t) = (x,y,z).$$ For the dense Zariski open $U = D(xz-y^2)\subset Y$, the inverse image $q^{-1}(U)$ is a $G$-torsor over $U$. -The singular locus of $X$ is the single $q$-fiber, $q^{-1}(0,0,0)$. Even though the origin is a smooth point of $Y$, every point of this $q$-fiber is a singular point of $X$.<|endoftext|> -TITLE: On sums of quadratic residues -QUESTION [12 upvotes]: Let $p>3$ be a prime. -We set $R=\{x\in\mathbb{Z}: (x/p)=1\}$, where $(\cdot/p)$ is the Legendre symbol. When $p\equiv3\pmod4$, by class formulae of imaginary quadratic fields $\mathbb{Q}(\sqrt{-p})$, we can easily obtain that -$$A_p:=\sum_{0 -TITLE: Where does the Laplace transform come from? -QUESTION [7 upvotes]: The Gelfand transform on the commutative Banach *-algebra $L^1(\mathbb{R})$ is just the Fourier transform. -Q. What can we say concerning the Laplace transform? - -REPLY [21 votes]: Set $A = L^1([0, \infty))$, equipped with the structure of a Banach $*$-algebra via convolution. The spectrum of this algebra is the half plane $\text{Re}(z) \geq 0$, and the Gelfand transform is the Laplace transform.<|endoftext|> -TITLE: Artin reciprocity via Shimura varieties -QUESTION [8 upvotes]: The point of Shimura varieties, as far as I've understood it, is that for a given Shimura datum $(G,D)$, there exist models, by which I mean that for congruence subgroups $\Gamma$ there exists a Shimura variety $X(\Gamma)$ defined over some number field. Hence we get a action of the absolute galois group $G_{\mathbb{Q}}$ on -$$V:=\lim_{\Gamma} H^*_{ét}(X(\Gamma),\mathbb{Q}_{\ell}).$$ -However, from an adelic point of view, we also get a (continuous) action of $G(\mathbb{A}_f)$ on the shimura variety, and so in fact $V$ is a representation of $G_{\mathbb{Q}}\times G(\mathbb{A}_f)$. The point now (from a Galois representation/ Langlandian point of view) is that for every representation $\rho:G_{\mathbb{Q}}\rightarrow \mathbb{\mathbb{Q}_\ell}^\times$, we can associate a representation of $G(\mathbb{A}_f)$ as -$$\text{Hom}(\rho,V).$$ -The main difficulty (as far as I understand it) is now to show that we can generate sufficently many representations that way to prove Langlands Program. My question is how this argument looks like in abelian case, i.e. what happens when $G=\text{GL}_1$? The Shimura varieties of tori are relatively simple to understand, namely we know that the Shimura variety associated to $\text{GL}_1$ is of the form -$$\mathbb{Q}^\times \backslash \mathbb{A}_f^\times/K$$ -for an open compact subgroup $K\subset \mathbb{A}_f^\times$ and is a finite étale scheme over some number field. How do we finish the proof from there on to get Artin's reciprocity? Or do we need to restrict ourself to the local case to even be able to complete the proof? - -REPLY [10 votes]: If $K$ is 'everything 1 mod N' for some N, then the canonical model of $\mathbf{Q}^\times_+ \backslash \mathbf{A}^\times_{\mathrm{f}} / K$ is exactly $\mu_N / \mathbf{Q}$, the scheme of $N$-th roots of unity. Any open compact $K$ will contain one of these, so $GL_1 / \mathbf{Q}$ Shimura varieties all look like quotients of $\mu_N$ for some $N$. -Hence the answer to "How do we finish the proof from there on to get Artin's reciprocity?" is "we prove the Kronecker--Weber theorem" [i.e. every abelian extension of $\mathbf{Q}$ is contained in a cyclotomic field]. -Perhaps it's disappointing that Shimura varieties don't tell you how to prove Kronecker--Weber. But they do something much more important: they tell you how to generalize Kronecker--Weber, pointing you towards a much more general (mostly conjectural) picture of which Kronecker--Weber is just one small part.<|endoftext|> -TITLE: Set-theoretic geology: controlled erosion? -QUESTION [11 upvotes]: I have to say that after the two last posts by Timothy Chow on Forcing I got so intrigued that I am trying to rethink the little I know about this formidable chapter of mathematics. -I have also to add that, although aware of the new field of set-theoretic geology, I am far from having a full grasp of it, so pre-emptive apologies to experts if I ask something that is either trivial or false. -Onto the point. Suppose I start from a transitive model of set theory $M$, and, rather than trying to expand it, I would rather do something opposite, namely the following: given an element $G$ of the model, try to "yank it out", to remove it surgically so that what remains is still a transitive model $M_0$. -In other words, try to establish $M= M_0[G]$. -Of course things are not so easy: I want to eliminate $G$ from $M$, but obviously I have to get rid of a lot of other sets in $M$ which are associated to $G$, for instance other sets which would imply its existence. Moreover, I have to choose judiciously whether or not $G$ is removable in such a way that after its removal (and of its "peers" ) the remaining set is still a model of $ZF$ of the same ordinal height. -I would call this operation selective erosion (if there is a canonical name for this operation please supply it) . -I understand that this may not be possible in some scenarios: for instance if $M$ is the minimal model, it is too "skinny" to allow for removals. But, unless intuition fails me, there should be plenty of "fat" models which should be liable to erosion. -MOTIVE -The way I look at this scenario is kind of the reciprocal of forcing: I would like to yank out some specific $G$ which codes some specific truths in $M$, for instance get rid of some map which collapses some cardinals. -QUESTION: - -Are there methods that can be employed to do the surgery I sketched -? Notice that I do not ask whether a model is liable to erosion, rather whether some specific sets can be removed, and if so how. - -NOTE: if I already know that $M$ is a forcing extension by $G$, then the problem is already trivially solved. Rather, suppose I only know that $M$ is a transitive model and someone comes along and gives me a $G$ in the model, and asks: is $G$ removable? I want to answer yes or no. Again, in some particular case the negative answer is obvious (example if G is an ordinal in $M$). But what about less trivial cases? -ADDENDUM: After the comments of Asaf, and especially after the great first answer by Jonas, time to take stock: The first thing that comes to my mind is that there are at least TWO candidate strategies to tackle this problem (and perhaps neither of them is the good one). You can call them BOTTOM-UP, which is the one I have sketched very loosely in my "debate" with Asaf, and the one which I would call TOP-DOWN which is the one advocated by Jonas. -Let us briefly recap them: - -BOTTOM UP. Start from a minimal model $W_0$ such that $A\notin W$ (for instance the constructibles in $M$) , and look at the set of extensions $W$ of the bottom $W_0$ such that $W[A] \neq M$, ordered by inclusion, then try to take the colimit of this ordered set (in other words, you hope that the sup of all of them is a model and does not contain A, but adding A you get M) . Is such a beast exists you found your A-eroded M -TOP DOWN See Jonas's answer (I would call it the "take the limit " method). - -Notice that both could be considered a form of selective geology: -1 is like growing the "earth", from some core, till a layer where A is present is reached. -2 is actually more in line with erosion, getting rid of as much as you can, as so aptly Jonas said. -So, either going from non A-grounds and looking for their union, or from A-grounds and looking for their intersection -PROBLEM: Both methods rely on looking at a certain ordered set of models of ZF in the "universe' M, and on certain lattice operations which can be performed. I have absolutely no clue whether such operations (taking sups or infs) are admitted in all cases (my gut feeling is no). -The story continues.... - -REPLY [9 votes]: What a fantastic question, and thanks to Asaf and Mirco for the great discussion in comments! I love the idea of “removing” a given set from a model of ZFC, to obtain a smaller model of ZFC - some kind of inner model method analogous to the outer model method of forcing. This may not be a complete answer, but I think that geology does offer a useful framework for attacking this question, at least when the “erosion” is strictly due to forcing (the more general question, when is a set removable at all while leaving behind a model of the same height, can be answered I think by looking to see whether the set in question is in $L$). -(Recall from set-theoretic geology: an inner model $W$ is a ground of our universe V if it is a transitive proper class satisfying ZFC, such there exists $G \in V$ which is generic over $W$ and $W[G]=V$. The foundational theorem of geology says the grounds of $V$ form a uniformly first-order definable collection of inner models in $V$). -Given a candidate set $A\in V$, we can ask whether $A$ is forcing-erodable by asking “Is there a ground $W$ that omits $A$”? Any such ground $W$ is a candidate for the model obtained by removing $A$ from $V$. -How do we identify a single, canonical inner model by removing $A$? In contrast to forcing, in which we want add as little as possible to $V$ in order to obtain $V[G]$, here we are doing the inverse - I argue that we want to remove the absolute maximum possible from $V$, while still retaining the property that everything we remove can be added back by adding $A$ itself. -For example, given a Cohen extension $V[c]$, we can eliminate $c$ by going to an inner model $V[c^\prime]$ that contains only the real $c^\prime$ that lies on the even digits of $c$... but this is unsatisfying, because although we removed $c$ it feels as though we only removed half of the information contained in $c$. To “erode $c$”, we want to go all the way down to the inner model $V$. -Geology gives us an approach. For a set $A \in V$, call a ground $W$ of $V$ an $A$-ground if: - -$A\notin W$ (we are eroding $A$) -$W[A] =V$ (we are not going ‘too far’ - everything we remove can be added back by adding $A$) - -Is there a minimal such $A$-ground? I am not certain of the answer, but the natural candidate is the intersection of all $A$-grounds (let’s call this the $A$-mantle). -Questions: If $M_A$ is the $A$-mantle, then - -is $M_A$ an $A$-ground ? If so, this is the right candidate for “eroding $A$ from $V$”. -If $M_A$ is not an $A$-ground, then is $M_A$ a model of ZFC? If that is the case, then does $M_A[A]=V$? - -These are analogous to the questions in geology “Is the Mantle a model of ZFC” and “Is the mantle necessarily a ground”.<|endoftext|> -TITLE: Public games for solving open problems -QUESTION [28 upvotes]: Many people in the world like playing games. Also, maybe we can design special interesting games for some open problems in mathematics (or physics). -In quantum mechanics, there are some games (for example, Quantum Moves; which it is about quantum control and hybrid intelligence About the Game) that if you play a game and can finish it, you solved an open problem. For example (simplified explanation), there is a bucket of water and a lot of obstacles (hills and pits). If you can move the bucket to the end of the road and there are specific amount of water on the bucket, you solved an open problem in quantum mechanics (you solved an equation in quantum mechanics). The game is so funny and everyone can play it without any knowledge about the quantum mechanics Quantum Moves. -My question is: -Are there similar games for solving open problems in mathematics (or in physics)? - -REPLY [26 votes]: I believe the start of citizen-science games for scientific research -was FoldIt, -asking users to find a low potential energy configurations of protein molecules. -Some publishable results were achieved.1 - -      - - -Wikipedia has a list of citizen-science projects here. -Only one is marked as "Mathematics": -SZTAKI Desktop Grid, -closed two years ago. -There are several labeled Physics, including -Quantum Moves -as mentioned by the OP. -Astronomy projects are prevalent, e.g., Planet Hunters. - -1Cooper S, Khatib F, Treuille A, Barbero J, Lee J, Beenen M, et al. (August 2010). "Predicting protein structures with a multiplayer online game." Nature. 466 (7307): 756–60.<|endoftext|> -TITLE: If $\mathcal C$ is a $\kappa$-accessible $\infty$-category, then is $Mor \mathcal C$ $\kappa$-accessible? -QUESTION [6 upvotes]: If $\mathcal C$ is a $\kappa$-accessible 1-category, then the category of morphisms $Mor \mathcal C$ is a $\kappa$-accessible 1-category, with the $\kappa$-presentable objects being those morphisms whose domains and codomains are each $\kappa$-presentable. -In the context of $\infty$-categories, the best result I know of is HTT Proposition 5.4.4.3, which shows that if $\mathcal C$ is a $\kappa$-accessible $\infty$-category and $\kappa \ll \tau$ (meaning that $\lambda < \tau \Rightarrow \kappa^\lambda < \tau$ and $\kappa < \tau$), then $Mor \mathcal C$ is $\tau$-accessible. -Lurie's proof, via HTT Lemma 5.4.4.2 (note that this lemma's proof has been revised since the printed edition), seems to really use the full strength of the assumption $\kappa \ll \tau$. Can this be improved to $\kappa = \tau$? Or at least to the "sharply below" relation $\kappa \triangleleft \tau$ familiar from the theory of accessible 1-categories? -This boils down to asking: if $\mathcal C$ is $\kappa$-accessible, then is every morphism of $\mathcal C$ a levelwise $\kappa$-filtered colimit of morphisms between $\kappa$-presentable objects? -In the case of 1-categories, a follow-your-nose argument works: you just take colimiting diagrams for the domains and codomains and factor the original map through stages of the colimit. I suspect that the same must be true in $\infty$-categories, with the same argument in principle working. But the question seems to be much more subtle $\infty$-categorically. - -REPLY [2 votes]: Marc Hoyois answered in the comments: the answer is affirmative, by HTT 5.3.5.15.<|endoftext|> -TITLE: Is $\limsup_{x\to\infty}\big(\sum\limits_{d|3^x-1}{1/d}\big)/\big(\sum\limits_{p -TITLE: Trace ideal of a projective module -QUESTION [5 upvotes]: In his 1969 paper "On projective modules of finite rank", Wolmer Vasconcelos writes - -Let $M$ be a projective $R$-module... The trace of $M$ is defined to be the image of the map $M \otimes_R \operatorname{Hom}_R(M, R) \to R$, $m \otimes f \to f(m)$; it is denoted by $\tau_R(M)$. If $M \oplus N = F$ (free), it is clear that $\tau_R(M)$ is the ideal of $R$ generated by the coordinates of all elements in $M$, for any basis chosen in $F$. It follows that for any homomorphism $R \to S$, $\tau_S(M \otimes_R S) = \tau_R(M) S$. - -A similar claim appears in his 1973 paper "Finiteness in projective ideals": - -We recall the notion of trace of a projective module $E$ over the commutative ring $A$. It is simply the ideal $J(E) = J = \Sigma f(E)$ where $f$ runs over $\operatorname{Hom}_A(E, A)$. Equivalently, $J$ is the ideal generated by the “coordinates” of all the elements of $E$ whenever a decomposition $E \oplus G = F$ (free) is given. Under the second interpretation, it follows that if $h \colon A \to B$ is a ring homomorphism, then $J(E \otimes_A B) = h(J(E)) B$. - -The first claim is easy to verify (albeit with a change of free module) as follows. If $F = M \oplus N$ has a basis $\{ v_i \}$, we can consider $F' = F \oplus R$, where $u$ is a generator for $R$. For any $f \colon F \to R$, $F'$ has a basis composed of $u$ and all $w_i := v_i - f(v_i)u$. With respect to this basis, $v_i = w_i + f(v_i) u$, hence the $u$-coordinate of $v_i$ is $f(v_i)$. So every homomorphism $M \to R$ is the restriction of a coordinate function on $F'$. -What is not clear to me is the reason for the second claim that $\tau_S(M \otimes_R S) = \tau_R(M) S$. The inclusion $\tau_R(M) S \subset \tau_S(M \otimes_R S)$ is obvious, so let me focus on the other one. -One can choose a decomposition $M \oplus N = F$ (free), so that $M \otimes_R S \oplus N \otimes_R S = F \otimes_R S =: F_S$, which is free over $S$. Up to adding a $S$ summand, one can also assume that every $f \colon M \otimes_R S \to S$ is the restriction of some coordinate function on $F_S$. But the coordinates on $F_S$ depend on the choice of a basis. If the basis is obtained from an $R$-basis of $F$, the claim is clear. But $F_S$ could have many choices of $S$-bases which are not derived from $R$. -I think I am missing something quite trivial, but I cannot see it right now, so I though I'd rather ask here. - -REPLY [7 votes]: The confusion is linguistic, as identified in the comments. -Lemma. Let $M$ be a projective $R$-module, and suppose $M \oplus N \cong F$ is free on a basis $\mathcal B$. For $b \in \mathcal B$, write $\varepsilon_b \colon F \to R$ for the 'dual' element taking $b$ to $1$ and all other basis elements to $0$. Then $\tau(M)$ is the ideal generated by $\varepsilon_b(m)$ for $b \in \mathcal B$ and $m \in M$. -(By abuse of notation, we write $\varepsilon_b(m)$ for what should properly be denoted $\varepsilon_b(m,0)$.) -Proof. Since $\varepsilon_b|_M$ is a homomorphism $M \to R$, we clearly have $\varepsilon_b(m) \in \tau(M)$ for all $b \in \mathcal B$ and all $m \in M$. We have to show that they generate. In the definition of $\tau(M)$, we may replace $\operatorname{Hom}(M,R)$ by $R^{\mathcal B} = \operatorname{Hom}(F,R) \twoheadrightarrow \operatorname{Hom}(M,R)$. Elements can be written as $f = (f_b)_{b \in \mathcal B}$, where $f_b = f(b)$ are constants. Now the idea is that $f(m)$ only depends on the coordinates of $f$ where $m$ is supported: -Let $f = (f_b)_{b \in \mathcal B} \in R^{\mathcal B}$ and $m = \sum_{b \in \mathcal B'} a_b b \in M$ for some finite subset $\mathcal B' \subseteq \mathcal B$. Write $f_{\mathcal B'}$ for the function whose $\mathcal B'$-coordinates agree with $f$ and whose other coordinates vanish. Then -$$f(m) = \sum_{b \in \mathcal B'} f(a_b b) = \sum_{b \in \mathcal B'} f_b \cdot a_b = \sum_{b \in \mathcal B'} f_b \cdot \varepsilon_b(m),$$ -so $f(m)$ is expressed as a combination of the $\varepsilon_b(m)$. $\square$ -Corollary. Let $M$ be a projective $R$-module, and let $R \to S$ be a ring homomorphism. Then -$$\tau\left(M \underset R\otimes S\right) = \tau(M)S.$$ -Proof. Write $M \oplus N \cong F$ for some $R$-module $N$ and a free $R$-module $F$. Then -$$\left(M \underset R\otimes S\right) \oplus \left(N \underset R\otimes S\right) \cong F \underset R\otimes S.$$ -If $F$ has basis $\mathcal B$, then the elements $b \otimes 1$ form a basis of $F \otimes_R S$. Moreover, $M \otimes_R S$ is generated by elements of the form $m \otimes 1$. Therefore, $\tau(M \otimes_R S)$ is exactly the ideal generated by $\varepsilon_{b \otimes 1}(m \otimes 1)$, which is $\tau(M)S$. $\square$<|endoftext|> -TITLE: In a locally presentable category, is every object (a retract of) the colimit of a chain of smaller objects? -QUESTION [13 upvotes]: Let $\mathcal C$ be an accessible category. For $C \in \mathcal C$, define the presentability rank $rk(C)$ of $C$ to be the minimal regular $\kappa$ such that $C$ is $\kappa$-presentable. Following Lieberman, Rosicky, and Vasey, say that $C$ is filtrable if it is the colimit of a chain $C = \varinjlim_{\alpha < \lambda} C_\alpha$ of objects $C_\alpha$ of lower presentability rank $rk(C_\alpha) < rk(C)$, and almost filtrable if it is a retract of the colimit $D$ of such a chain such that $rk(D) = rk(C)$. -Question: -Let $\mathcal C$ be an accessible category. Under what conditions can we say that every object $C \in \mathcal C$ of sufficiently large presentability rank is almost filtrable? Does it suffice to assume that $\mathcal C$ is locally presentable? -(Of course, if "chain" is replaced with "highly-filtered colimit", then no conditions are necessary.) -In the above-linked preprint are given various conditions for filtrability dependent on $rk(C)$, but they are not really focused on the locally presentable case. In this case, - -$rk(C)$ is always a successor (unless it's $\aleph_0$ or perhaps if it's smaller than the accessibility rank of $\mathcal C$); - -there's a basic argument which shows that if $rk(C)$ is the successor of a regular cardinal, then $C$ is almost filtrable (and the last Remark in the above-linked paper asserts that the retract can be removed with the fat small object argument). - - -But I'm not sure how to say anything when $rk(C)$ is the successor of a singular cardinal. -Motivation: -It's important to me to be able to handle all $C \in \mathcal C$ of sufficiently large presentability rank, because this opens up the possibility of a new sort of inductive argument in the theory of locally presentable categories: induction on presentability rank using decomposition by chains. This sort of induction should be particularly well-suited to applications related to the small object argument, which interacts well with chains but not with general highy-filtered colimits. - -REPLY [4 votes]: Since this result is derived by Lieberman, Rosicky, and Vasey as a corollary of some more sophisticated constructions with more sophisticated goals, I think it might be worth "compiling out" the proof here. It turns out to be not so bad. Let $\mathcal C$ be a locally $\lambda$-presentable category, and recall the following fact: - -For any $C \in \mathcal C$, if $\mathrm{rk}(C) > \lambda$, then $\mathrm{rk}(C) = \kappa^+$ is a successor. (Proof hint: $\kappa$ is the smallest cardinal such that $C$ is a retract of a $\kappa$-sized colimit of $\lambda$-presentable objects.) - -Theorem [Lieberman, Rosicky, and Vasey] -Let $\mathcal C$ be a locally $\lambda$-presentable category and let $C \in \mathcal C$ with $\mathrm{rk}(C) = \kappa^+ > \lambda$. Then $C$ is almost filtrable. -Proof: Write $C = \varinjlim_{i \in I} C_i$ as a colimit of $\lambda$-presentable objects. Then $C$ is a retract of the colimit of a $\kappa^+$-small subdiagram, so we may assume without loss of generality that the diagram $I$ is of cardinality $\kappa$. We may write $I = \cup_{\alpha < \mathrm{cf}(\kappa)} I_\alpha$ as the union of a $\mathrm{cf}(\kappa)$-sized increasing chain of subdiagrams of cardinality $|I_\alpha| <\kappa$. Setting $C_\alpha = \varinjlim_{i \in I_\alpha} C_i$, we have $C = \varinjlim_{\alpha < \mathrm{cf}(\kappa)} C_\alpha$, yielding the desired filtration.<|endoftext|> -TITLE: A meet-semilattice with top element that is not a lattice? -QUESTION [5 upvotes]: I am reading Francis Borceux’s “Handbook of Categorical Algebra I” and on page 135 it says - -In particular a finite version of 4.2.5 does not hold: a finitely complete and well-powered category certainly admits finite intersections of subobjects (see 4.2.3), but not in general finite unions of subobjects. Finite unions have been constructed in 4.2.5 using possibly infinite intersections. For a counterexample, just consider a ∧-semi-lattice with a top element which is not a lattice. - -I get the argument for not being able to construct a union through finite intersection. But I have trouble proving it with the mentioned counterexample. Specifically, I can’t picture a “ ∧-semi-lattice with a top element which is not a lattice” in my mind. Why can’t the top element be the union for a finite set of elements if there is no other “better” element? Does the construction of this counterexample have anything to do with (in)finiteness? - -REPLY [11 votes]: It is well-known that a finite meet-semi-lattice with a maximum element is a lattice. The reason is that we can define $a \vee b := \wedge \{c\colon \textrm{$c$ is an upper bound for $a,b$}\}$, where this set is non-empty (since we have a maximum) and finite (since the poset is finite), and finite meets exist by supposition that we have a meet-semi-lattice. -But this is not true for infinite posets. Let $P := (\{(a,b)\colon 0\leq a,b \leq 1\}\setminus \{(1,1)\}) \cup \{(a,a)\colon 1 < a \leq 2\}$, with the usual partial order $(a_1,b_1)\leq (a_2,b_2)$ iff $a_1 \leq a_2$ and $b_1 \leq b_2$. Then $P$ is a meet-semi-lattice (with $(a_1,b_1)\wedge (a_2,b_2)=(\mathrm{min}(a_1,a_2),\mathrm{min}(b_1,b_2)$) and it has a maximum element $(2,2)$. But $(1,0)$ and $(0,1)$ lack a join.<|endoftext|> -TITLE: Are there any identities for alternating binomial sums of the form $\sum_{k=0}^{n} (-1)^{k}k^{p}{n \choose k} $? -QUESTION [10 upvotes]: In equations (20) - (25) of Mathworld's article on binomial sums, identities are given for sums of the form $$\sum_{k=0}^{n} k^{p}{n \choose k}, $$ with $p \in \mathbb{Z}_{\geq 0}$. I wonder whether identities also exist for the alternating counterparts: $$\sum_{k=0}^{n} (-1)^{k}k^{p}{n \choose k} .$$ Furthermore, I'm interested in results for the same sum that is “cut off”, i.e. when the summands go from $k=0$ to some $D0$, then -$$s_{p,n}\sim b_{p,n}.$$<|endoftext|> -TITLE: Why are root data a natural candidate for classifying connected reductive groups? -QUESTION [10 upvotes]: For the purpose of this question, you may assume that we are working over the complex numbers. -Given a connected reductive group $G$, one can choose a maximal torus $T$, and then let $T$ act on the Lie algebra $\mathfrak{g}$ of $G$. One can use this action to define the root datum, which in turn is invariant of the choice of $T$, and use it to classify connected reductive groups. -The action of $T$ on $\mathfrak{g}$ is nice in that it has more information than just the action of $T$ on the Lie algebra $\mathfrak{t}$ of $T$, and yet is simple enough so as to decompose into one dimensional weight spaces. But that is a far cry from saying that this is a natural action to consider when trying to classify connected reductive groups! -Is there a deep reason that root data, or more generally the action of a maximal torus on the Lie algebra of $G$, is a natural thing to consider? Does it correspond to some cohomological invariant? Does it arise naturally? Or is this entire theory a fluke? -The proof does not seem to bring much insight into this story, in that it boils down to a series of reductions, which reinforces for me the suspicion that root data are not in and of themselves natural, but rather that this was a guess for a way to classify connected reductive groups that just happened to work out... - -REPLY [9 votes]: I can't give you a very deep reason for why root data appear in this context (because, let's face it, root systems spring out everywhere), but there are some very elementary reasons to why the action in question is very natural with regard to the classification. -Let me start with the following two considerations: - -When one tries to distinguish between the two objects, one usually looks for some simple properties which differ between them. For example, to show that two abstract groups are not isomorphic, one starts by comparing their orders, and procede by comparing the number of elements of a given order in each of them or which subgroups are there and how they fit together. -In a semisimple Lie algebra $L$ there is a Jordan decomposition, which tells that every elements $x$ is the sum of an $\operatorname{ad}$-semisimple element $x_s$ and an $\operatorname{ad}$-nilpotent part $x_n$. And there is a subalgebra consisting of semisimple elements (otherwise $L$ itself is nilpotent by Engel theorem). Such subalgebras are called toric, and it turns out they are always abelian. Thus when considered in their adjoint repesentation, the elements of a toric subalgebra form a commuting family of semisimple endomorphisms of $L$, hence are simultaneously diagonalizable, which is equivalent to $L$ decomposing into the direct sum of its weight subspaces, which gives rise to the root system. - -So combining these two considerations, to distinguish (and ultimately classify) semisimple Lie algebras we essentially take the simplest type of elements of $L$ (the semisimple ones) and look at how we can fit them together in $L$ (so that they from a subalgebra, and a maximal such). -This looks somewhat abstract, but really just mimics what can be easily seen in the examples, namely, in the classical semisimple Lie algebras. The standard constructions in their minimal representations are equipped with some very simple bases (for example, what comes first to mind for $\mathfrak{sl}_n$?), and there is a very natural maximal toric subalgebra $H$, namely, the diagonal matrices. The non-zero-weight subspaces are the spans of individual off-diagonal basis elements, and the root system captures their configuration. -Now getting back to algebraic groups, they can be roughly classified by the isomorphism classes of the corresponding Lie algebras, but additional information is needed to account for the center. The center sits inside the torus, so to incorporate this missing data into the rough classification one translates the adjoint action of the corresponding toric subalgebra to the adjoint action of the torus.<|endoftext|> -TITLE: Hadamard factorization of L-functions -QUESTION [20 upvotes]: I have already asked this question here in a different form, but really need an answer. -Let $L(s)$ be a "standard" $L$-function, say with Euler product, functional equation, etc... -(Selberg class if you like), of order 1, and let $\Lambda(s)$ be the completed $L$-function -with gamma factors. We thus have $\Lambda(k-s)=\omega\Lambda^*(s)$, where $\Lambda^*$ is -the "dual" Lambda function (example: if $L(s)$ corresponds to a Dirichlet character $\chi$, -$\Lambda^*$ corresponds to its conjugate), and $\omega$ root number of modulus 1. -Assume for instance that there are no poles. Since $\Lambda$ has order $1$ it has a Hadamard -product $$\Lambda(s)=ae^{bs}\prod_{\rho}(1-s/\rho)\;,$$ -where the product is over the zeros of $\Lambda$ and understood as the limit as $T\to\infty$ -of the product for $|\rho|R$, obtaining that the error term is -$$ -\ll \sum_{|\gamma|\le R} \frac{1}{R^2} + \sum_{R<|\gamma|} \frac{R}{|\gamma|^3} \ll \frac{\log qR}{R}, -$$ -upon recalling that there are $\ll \log q(|t|+1)$ zeros in an interval of length $1$ (we will recall this more precisely next). -Now the main term in (1) can be handled by partial summation. For $t>0$, put $N^+(t)$ to be the number of zeros of $\Lambda$ with imaginary part between $0$ and $t$, and $N^{-}(t)$ to be the number of zeros with imaginary part between $-t$ and $0$. Then both $N^+$ and $N^-$ satisfy by the argument principle the well known asymptotic formula (for $t\ge 1$) -$$ -N^+(t), N^{-} (t) = \frac{t}{2\pi} \log \frac{qt}{2\pi e} +O(\log (q(t+1))). -$$ -Thus for all $t>0$ -$$ -|N^+(t) - N^-(t)| = O(\log (q(2+t))). -$$ -Now by partial summation -\begin{align*} -\sum_{|\gamma|\le T} \frac{\gamma}{R^2+\gamma^2} &= \int_0^{T} \frac{t}{R^2+t^2} dN^+(t) - \int_0^T \frac{t}{R^2+t^2} dN^-(t) \\ -&= \frac{T}{R^2+T^2} (N^+(T)-N^-(T)) - \int_0^T (N^+(t)-N^-(t)) \Big( \frac{t}{R^2+t^2}\Big)^{\prime} dt \\ -&= O\Big(\frac{T\log qT}{R^2+T^2} \Big) + O\Big(\int_0^T (\log q(t+2)) \Big(\frac{1} {R^2+t^2} + \frac{2t^2}{(R^2+t^2)^2} \Big)dt \Big)\\ -&= O\Big( \frac{\log qR}{R}\Big), -\end{align*} -upon letting $T\to \infty$. -We conclude that the quantity in (1) is $O((\log qR)/R$, and so tends to $0$ as $R\to \infty$.<|endoftext|> -TITLE: About the function $\prod_{k \in \mathbb{N}}(1-\frac{x^3}{k^3})$ -QUESTION [21 upvotes]: I'm wondering if the function $$f(x)=\prod_{k \in \mathbb{N}}\left(1-\frac{x^3}{k^3}\right)$$ has a name, or if there are any properties (especially about derivatives of $f$) have studied so far. -I got inspired by proof of the Basel problem ($\frac{\pi^2}{6}=\sum \frac{1}{k^2}$) using the product form of $$\frac{\sin(x)}{x}=\pi \prod_{k \in \mathbb{N}} \left(1-\frac{x^2}{k^2 \pi^2}\right).$$ The function $f(x)$ above would be considered as analogue for the Apery constant and $\zeta(3k)$. - -REPLY [39 votes]: If one starts with the Weierstrass factorisation -$$ \Gamma(z) = \frac{e^{-\gamma z}}{z} \prod_{k=1}^\infty (1 + \frac{z}{k})^{-1} e^{z/k}$$ -of the Gamma function, applied to $z = -x, -\omega x, -\omega^2 x$ (where $\omega = e^{2\pi i/3}$ is the cube root of unity), and multiplies the three together, one obtains -$$ \Gamma(-x) \Gamma(-\omega x) \Gamma(-\omega^2 x) = \frac{1}{-x^3} \prod_{k=1}^\infty (1-\frac{x^3}{k^3})^{-1}$$ -and hence -$$\prod_{k=1}^\infty (1-\frac{x^3}{k^3}) = -\frac{1}{x^3 \Gamma(-x) \Gamma(-\omega x) \Gamma(-\omega^2 x)}.$$ -One can manipulate the right-hand side a little using the various functional equations of the Gamma function, but it does not have a dramatically simpler form (as one can already see from the zero set ${\bf N} \cup \omega {\bf N} \cup \omega^2 {\bf N}$, which is too weird to come from anything more elementary than a Gamma function). This is in contrast to the analogous identity -$$ \prod_{k=1}^\infty (1 - \frac{x^2}{k^2}) = -\frac{1}{x^2 \Gamma(-x) \Gamma(x)}$$ -where the Euler reflection formula (and $\Gamma(1-x) = -x \Gamma(-x)$) applies to simplify the right-hand side to $\frac{\sin(\pi x)}{\pi x}$ as you mention, and the zero set ${\bf N} \cup -{\bf N} = {\bf Z} \backslash \{0\}$ is now simple enough to be explainable via trigonometric functions. (The different structure of the zero sets also helps explain why the zeta function evaluated at even natural numbers is much more tractable than the zeta function evaluated at odd numbers; the former relates to the Fourier analytic structure of the integers, but the latter relates to the Fourier analytic structure of the natural numbers, which is far worse.)<|endoftext|> -TITLE: Information about Milnor conjecture -QUESTION [6 upvotes]: I'm a student of mathematics and I need know about the status of the Milnor conjecture (if there are partial results or if someone solved that). The statement is: - -A complete Riemannian manifold with non-negative Ricci curvature has a finitely generated fundamental group. - -If someone can help me with references to papers or anything I would be grateful. - -REPLY [10 votes]: According to David Roberts comment and the following paper it is open for dimensions $n\geq 4$. -Pan, Jiayin, A proof of Milnor conjecture in dimension 3, J. Reine Angew. Math. 758, 253-260 (2020). ZBL1432.53053. -There is a nice survey by Shen and Sormani that can be found in author homepage: -Shen, Zhongmin; Sormani, Christina, The topology of open manifolds with nonnegative Ricci curvature, Commun. Math. Anal., Conference 1, 20-34 (2008). ZBL1167.53309. -And a few related and partial attacks to the conjecture: -Paeng, Seong-Hun, On the fundamental group of manifolds with almost nonnegative Ricci curvature, Proc. Am. Math. Soc. 131, No. 8, 2577-2583 (2003). ZBL1040.53042. -Xu, Senlin; Deng, Qintao, The fundamental group of open manifolds with nonnegative Ricci curvature, Acta Math. Sin., Chin. Ser. 49, No. 2, 353-356 (2006). ZBL1120.53021. -Which is stronger than -Sormani, Christina, Nonnegative Ricci curvature, small linear diameter growth and finite generation of fundamental groups., J. Differ. Geom. 54, No. 3, 547-559 (2000). ZBL1035.53045.<|endoftext|> -TITLE: Interaction of plethysm with other operations -QUESTION [8 upvotes]: The plethysm $s_{\nu}[s_{\mu}]$ of two symmetric functions is the character of the composition of Schur functors $S^{\nu}(S^{\mu}(V))$. We know that this operation is linear and multiplicative in its first argument. But is there a way to develop - -$s_{\nu}[s_{\mu} + s_{\lambda}]$; - -$s_{\nu}[s_{\mu}s_{\lambda}]$; - - -in terms of plethysms $s_{\nu}[s_{\mu}]$ and $s_{\nu}[s_{\lambda}]$ ? I think I have heard of a formula for the first one, but I don't find it anymore! - -REPLY [6 votes]: In principle one can develop (1) using the coproduct in the ring of symmetric functions. By the Littlewood–Richardson rule, $\Delta(s_\nu) = \sum_{\alpha}\sum_\beta c^\nu_{\alpha\beta} s_\alpha \otimes s_\beta$ where $c^\nu_{\alpha\beta}$ is a Littlewood–Richardson coefficient, and correspondingly -$$s_\nu[s_\lambda + s_\mu] = \sum_{\alpha}\sum_\beta c^\nu_{\alpha\beta} s_\alpha[s_\lambda] s_\beta[s_\mu].$$ -Here the sum is over all partitions such that $|\alpha|+|\beta| = |\nu|$. -Somewhat similarly, -$$s_\nu[s_\lambda s_\mu] = \sum_{\alpha}\sum_{\beta} k^\nu_{\alpha\beta} s_\alpha[s_\lambda] s_\beta[s_\mu]$$ -where the sum is over all partitions $\alpha$ and $\beta$ of $|\nu|$ and $k^\nu_{\alpha\beta}$ is the Kronecker coefficient, most easily defined as the inner product $\langle \chi^\nu, \chi^\alpha \chi^\beta \rangle$ in the character ring of the symmetric group. Equivalently the$k^\nu_{\alpha\beta}$ are the structure constants for the internal product, usually denoted $\star$, on the ring of symmetric functions. These formulae can be found in MacDonald's textbook: see (8.8) and (8.9) on page 136, and hold replacing $s_\lambda$ and $s_\mu$ with arbitrary symmetric functions. -In practice, at least in my experience, this usually leads to a mess. One special case that's worth noting is when $\nu = (n)$, in which case the Littlewood—Richardson coefficient is non-zero only if $\alpha = (m)$ and $\beta = (n-m)$ for some $m \in \{0,1,\ldots, n\}$ and we get -$$s_{(n)}[s_\lambda + s_\mu] = \sum_m s_{(m)}[s_\lambda] s_{(n-m)}[s_\mu].$$ -This is the symmetric function version of $\mathrm{Sym}^n (V \oplus W) = \sum_{m=0}^n \mathrm{Sym}^m V \otimes \mathrm{Sym}^{n-m} W$ for polynomial representations of $\mathrm{GL}_d(\mathbb{C})$. There is a corresponding rule for exterior powers and so for $s_{(1^n)}$. -This also gives one indication that (2) is even harder: one related question was asked on MathOverflow. Example 3 on page 137 of MacDonald gives the special case for $\nu = (n)$, when $\chi^{(n)}$ is the trivial character, and so $\langle \chi^{(n)}, \chi^{\alpha}\chi^{\beta}\rangle = \langle \chi^{\alpha}, \chi^\beta\rangle = [\alpha=\beta]$. Hence -$$s_{(n)}[s_\lambda s_\mu] = \sum_{\alpha} s_\alpha[s_\lambda] s_\alpha[s_\mu]. $$ -Great care is needed when extending these rules to arbitrary symmetric functions. For instance, $s_\nu[-f] = (-1)^{|\nu|} s_{\nu'}[f]$ for any symmetric function $f$ and, as Richard Stanley points out in a comment below, the expression $s_\nu[f-f]$ should be interpreted as a plethystic substitution using the alphabets for $f$ and $-f$, not as $s_\nu[0]$; correctly interpreted, it can be expanded using the coproduct rule and the rule for $s_\nu[-f]$ just given.<|endoftext|> -TITLE: Who introduced the abstract definition of a DGA? -QUESTION [14 upvotes]: Differential graded algebras, or DGAs, are a basic object of study in many areas of modern mathematics. While they were present (implicitly at least) since the start of modern differential geometry, I would like to know where the abstract definition of a DGA was first written down, and by whom? - -REPLY [7 votes]: Search for the earliest appearance of "differential graded algebra" and "DGA" on MathSciNet. The earliest hit is DGA in a review of a 1954 paper of Cartan where DGA (or more precisely, the redundant term "DGA-algebra") is defined: "Sur les groupes d'Eilenberg-Mac Lane $H(\Pi,n)$ I. Méthode des constructions," Proc. Nat. Acad. Sci. U.S.A. 40 (1954), 467–471. In the same year there is an entry from the Cartan seminar: Séminaire Henri Cartan de l'Ecole Normale Supérieure, 1954/1955. Algèbres d'Eilenberg-MacLane et homotopie.<|endoftext|> -TITLE: On the width of the Catalan monoid and the rank of K-groups of the Furstenberg transformation group -QUESTION [5 upvotes]: The semigroup algebra of the Catalan monoid is isomorphic to the incidence algebra of $P_n$, where $P_n$ is the poset consisting of subsets of { 1,...,n } where for two subsets $X \leq Y$ if and only if $X$ and $Y$ have the same cardinality and if X= {x_1 < ... < x_k } and Y= {y_1 < ... < y_k } we have $x_i \leq y_i$ for $i=1,...,k$. This is for example proved in https://arxiv.org/pdf/1806.06531.pdf. Recall that the width of a poset is the maximum size of an antichain. -I noted that the width of the poset $P_n$ starts with 2, 3, 4, 6, 8, 13, 20, 32, 52, 90, 152 for $n=1,...,11$ and this leads to the sequence https://oeis.org/A084239 of the rank of K-groups of the Furstenberg transformation group of the $n$-torus. See table 1 in https://arxiv.org/pdf/1109.4473.pdf . - -Question 1: Is this true for all $n$? Is there a deeper explanation? Does the width have a homoogical interpretation for the Catalan monoid? - -The poset $P_n$ has $n+1$ connected components, for each of the $k$-subsets and one can restrict to find antichains in those subsets and put them together. But I am more curious whether there is a deeper connection to the $K$-group sequence or is this just random? -One might also ask about other nice properties of $P_n$. I noted that appending a minimum and maximum to $P_n$, one obtains a lattice. -The width of $P_n$ is equal to the maximal number of covers an element can have in the distributive lattice of order ideals $L(P_n)$ of $P_n$. - -Question 2: Does the incidence algebra of $L(P_n)$ have an algebraic meaning in relation to the Catalan monoid? - - -Question 3: Is the Coxeter matrix of $L(P_n)$ periodic? - -Question 3 has a positive answer for $n=1,2,3,4$ and the periods are given by 6,12,30,42 in that case. -(Small values suggest that also the Coxeter matrix of $P_n$ might be periodic, but that might be not so good evidence since the connected compoenent have not many points for small $n$.) - -REPLY [4 votes]: The distributive lattice $L(k,n-k) := [\varnothing,(n-k)^k]$, the interval between the empty partition and the rectangular shape $(n-k)^k$ in Young's lattice, is the same as the poset of subsets of $[n]:=\{1,2,\ldots,n\}$ of size $k$ ordered by $\{x_1<\cdots -TITLE: What can be said about a projective morphisms that admit decomposition theorem like smooth morphisms? -QUESTION [12 upvotes]: Let $f\colon X\to Y$ be a surjective morphism of smooth projective varieties. If the decomposition theorem for $f$ is given by $$Rf_*\mathbb{C} \simeq \bigoplus_i R^if_*\mathbb{C}[-i],$$ what are the necessary conditions the morphism $f$ must satisfy? Is there an example where such a morphism is not smooth but the decomposition theorem nonetheless looks like the above? -Edit: If I additionally assume that $R^if_*\mathbb{C}$ are local systems for all $i$, can one conclude that $f$ is smooth? I understand that the limit mixed Hodge structure is pure as there is no monodromy around the singular fibers. - -REPLY [2 votes]: Here is an example where $f$ is not smooth but $Rf_* \mathbb{C}$ behaves as if it were: -Let $X$ be a hyperelliptic surface and $f$ the natural morphism to $Y \cong\mathbb{P}^1$. All reduced fibres of $f$ are elliptic curves, but there is a nonzero number of nonreduced fibres, the number dependending on $X$. -The singular cohomology of $X$ is given by $H^0(X, \mathbb{C}) \cong H^4(X,\mathbb{C}) \cong \mathbb{C}$ and $H^1(X, \mathbb{C}) \cong H^3(X, \mathbb{C}) \cong \mathbb{C}^2$. Furthermore, the restriction map $H^1(X, \mathbb{C}) \to H^1(F, \mathbb{C})$ is an isomorphism for any fibre $F$ of $f$. -It is clear that $R^0 f_* \mathbb{C}_X \cong R^2f_* \mathbb{C}_X \cong \mathbb{C}_Y$, so let us consider $R^1f_* \mathbb{C}_X$. Since $H^1(X, \mathbb{C}) \cong \mathbb{C}^2$, we get a natural map $\mathbb{C}^2_Y \to R^1f_* \mathbb{C}$. By evaluating this on stalks and using the proper base change theorem, we see that this is an isomorphism. -Finally, since we know exactly what each sheaf $R^i f_* \mathbb{C}_X$ is, the same proof as in the case $f$ smooth can be used to show that $Rf_*\mathbb{C}_X$ decomposes as a direct sum of its (shifted) cohomology sheaves. -One may ask if a similar statement holds whenever all reduced fibres are smooth (and say $f$ is flat); I did not think about this. It would also be interesting to know if there are examples with non-smooth reduced fibres. Also, note that in the example $R^1 f_* \mathbb{Z}_X$ is not a local system.<|endoftext|> -TITLE: Is unirationality decidable? -QUESTION [10 upvotes]: When the answer to the Lüroth problem is affirmative, the genus (for curves) or Castelnuovo's criteria (for separably unirational surfaces) give computable invariants which decide if a given variety is unirational. -I am interested in the simplest cases where non-rational examples are known: namely surfaces over $\overline{\mathbb{F}_p}$ and 3-folds over $\mathbb{C}$. -Is it known whether there exists an algorithm to determine if a given variety of the above type is unirational? -A naive hope would be to produce a bound on the degree of a rational map $\mathbb{P}^n \to X$ however this is clearly impossible since rational maps $\mathbb{P}^n \to \mathbb{P}^n$ have unbounded degree. -The next best thing would be to put a computable upper bound on the minimal degree of such a rational map. Are such bounds known for the two cases I outlined above? Are such bounds sufficient, in principle, to provide an algorithm? - -REPLY [6 votes]: This is a widely studied problem, and I think with the current technology it looks unlikely this will be answered. Let me stick to characteristic $0$ for simplicity; the story gets much richer in positive characteristic (even if you add the adjective 'separably' everywhere). -Unirational varieties are rationally connected, which in turn implies $H^0(X,(\Omega_X^1)^{\otimes m}) = 0$ for all $m > 0$. Mumford conjectured that this vanishing conversely implies that $X$ is rationally connected, but I think little is known about this conjecture. At least if you believe Mumford (and maybe with a bound on which $m$ you have to try, analogous to Castelnuovo's criterion), then rational connectedness should be somewhat decidable. -However, a well-known open problem is whether every rationally connected variety is unirational (I think people expect this to be false ― I certainly do). The reason we can't answer this is that we don't have any obstructions that can distinguish between the two. Therefore, I think it's highly unlikely that we can give a characterisation that decides whether a variety is unirational. -I'm not sure what the status is on unirationality in families, but (stable) rationality is not a deformation invariant of smooth projective varieties by Hassett–Pirutka–Tschinkel. So discrete [i.e. locally constant in families] invariants like cohomology vanishing are not enough to detect (stable) rationality, and very likely the same is true for unirationality. - -As for surfaces over $\bar{\mathbf F}_p$, some K3 surfaces are unirational, but most are not, so once again it is not a deformation invariant. Shioda conjectured that if $X$ is a surface over an algebraically closed field $k$ of characteristic $p > 0$ with $\pi_1^{\operatorname{\acute et}}(X) = 0$, then $X$ is unirational if and only if $H^2_{\operatorname{cris}}(X/K)$ is supersingular, i.e. all slopes of Frobenius are $1$. (This is clearly a necessary condition, since $H^2_{\operatorname{cris}}(X/K) \hookrightarrow H^2_{\operatorname{cris}}(Y/K)$ is injective and preserves Frobenius actions if $Y \twoheadrightarrow X$ is a dominant map of smooth projective varieties; see e.g. Prop. 1.2.4 in Kleiman's paper Algebraic cycles and the Weil conjecutres in Dix Exposés.) -But this doesn't say anything about what happens if $\pi_1^{\operatorname{\acute et}}(X) \neq 0$, although it has to be finite and its order indivisible by $p$ (see e.g. this note by Chambert-Loir). So you're first supposed to compute the universal cover and then compute slopes ― I'm not sure if there is a more direct way to do this. Assuming Shioda's conjecture this gives a criterion.<|endoftext|> -TITLE: Equidistribution of $\{\alpha p\}$ for $p$ in an arithmetic progression -QUESTION [12 upvotes]: Let $\alpha$ be irrational. A famous theorem of Vinogradov says that $\{ \alpha p\}$ is equidistributed in $[0,1]$ as $p$ runs over all primes. -Let $a,q$ be natural numbers with $\gcd(a,q) = 1$. Then is the sequence $\{ \alpha p\}$ equidistributed in $[0,1]$, as $p$ runs over primes with $p \equiv a \bmod q$? -Almost certainly this must be known. So I'm looking for a precise reference in the literature as I need it in a paper. Ideally, it would be nice to have an effective version which makes explicit the speed of convergence (via the Erdős-Turán inequality, say). - -REPLY [14 votes]: I think the sought result follows from Vinogradov's theorem. By Weyl's criterion and the orthogonality of Dirichlet characters, the sought result can be reformulated as follows. For every nonzero integer $k$, and for every Dirichlet character $\chi$ modulo $q$, we have -$$\sum_{p -TITLE: Factorization in formal power series versus in convergent power series over the complexes -QUESTION [6 upvotes]: Let $R=\mathbb C\{x_1,...,x_n\}\subset S=\mathbb C [[x_1,...,x_n]]$ denote the ring of convergent, respectively formal, power series over $\mathbb C$. -Suppose $f\in R$ is irreducible in $R$. Does it remain irreducible in $S$? - -REPLY [4 votes]: As Arno Fehm points out, this follows from results in Nagata's Some Remarks on Local Rings II. Both $R$ and $S$ are UFD's, so $f$ is irreducible, in $R$ or $S$ respectively, if and only if the ideal it generates, in $R$ or $S$ respectively, is prime. At the bottom of page 1 of Nagata's paper, he states that, if $\mathfrak{p}$ is a prime ideal of $R$, then $\mathfrak{p}S$ is prime as well. -I found it hard to absorb all of Nagata's vocabulary; here is a route to get the desired claim from his results while skipping some of the sophisticated language. -Let $R'$ and $S'$ denote the versions of $R$ and $S$ with $n-1$ variables. Suppose that $f=gh$ for $f \in R$ and $g$, $h \in S$ nonunits. Use the Weierstrass preparation theorem to factor $f = pu$, $g=qv$ and $h = rw$ where $p \in R'[x_n]$, $q \in S'[x_n]$ and $r \in S'[x_n]$ are Weierstrass polynomials and $u \in R^{\times}$, $v \in S^{\times}$ and $w \in S^{\times}$ are units. Then $qr$ is a Weierstrass polynomial of $S$, and $vw \in S^{\times}$, so $f = (qr) (vw)$ and $f = pu$ are both Weierstrass factorizations in $S$. Since such factorizations are unique, we have $p = qr$ and $u = vw$. -Write $p(x_n) = x_n^a + \sum_{i=0}^{a-1} p_i x_n^i$, $q(x_n) = x_n^b + \sum_{i=0}^{a-1} q_i x_n^i$, $r(x_n) = x_n^c + \sum_{i=0}^{c-1} r_i x_n^i$. Then the $q_i$ and $r_i$ are polynomial combinations of the roots of $p$, so the $q_i$ and $r_i$ are integral over $R'$. But Nagata, in his proof of Theorem 5, shows that $R'$ is integrally closed in $S'$, and this proof is extremely concrete. So that shows that the $q_i$ and $r_i$ land in $R'$. Thus $q$ and $r \in R$, and we deduce that $f$ factors in $R$ as well.<|endoftext|> -TITLE: What are immediate applications of the classification of connected reductive groups? -QUESTION [18 upvotes]: After years of putting it off, I finally sat down, read, and understood the classification of connected reductive groups via root data. -That's a non-trivial theory! I'm hoping that now that I am done I can reap some benefits. What are some immediate applications/corollaries of this theory? Now that I understand this classification, what can I do with it? At the moment, I don't really feel like I can do anything I couldn't do before... -Number theoretic and algebro-geometric applications are great! But I'm open to anything. - -REPLY [2 votes]: Given a prime $p$ and a connected reductive algebraic group $G$ over $\mathbb{F}_p^{\mathrm{alg}}$ with a Frobenius map $F$, the fixed points $G^F$ are a finite group 'of Lie type'. The finite groups of Lie type are the main case in the Classification Theorem of Finite Simple Groups. They can all be obtained in a uniform way by this construction, except for the Suzuki and Ree groups. (Roughly speaking, these are also obtained using the data from root systems and Dynkin diagrams, but they require further automorphisms that do not descend from the algebraic group.) Various structural properties of finite groups of Lie type follow easily from their analogues in the algebraic group, for example that they have $BN$ pairs. -I'm surprised this wasn't already an answer. It seems close enough to an 'application' to me to be worth mentioning. I admit it is far from an 'immediate application'.<|endoftext|> -TITLE: How to write down a generic genus $g$ curve in $\mathbb{P}^n$ as an intersection of hypersurfaces? -QUESTION [7 upvotes]: Let $g, n \geq 1$ be positive integers. How can we describe a generic genus $g$ curve inside $\mathbb{P}^n$ as an intersection of hypersurfaces? -For example when $g = 1$ and $n = 2,3,4$, we have the following descriptions: - -$n = 2$: a generic genus 1 curve is given by a plane cubic; -$n = 3$: a generic genus 1 curve is given by the intersection of two quadrics; -$n = 4$: a generic genus 1 curve is given by the intersection of the five quadrics defined by $4 \times 4$ sub-Pfaffians of a $5 \times 5$ skew-symmetric of linear forms. - -As one can see, the $(g,n) = (1,4)$ case becomes complicated, as a generic genus one curve can no longer be expressed as a complete intersection. -For $g = (d-1)(d-2)/2$, we always have a nice description of genus $g$ curves in $\mathbb{P}^2$: in particular, a smooth plane curve of degree $d$ will have genus $g$. -Are there any other pairs of $(g,n)$ which admit an explicit description of generic genus $g$ curves as the intersection (possibly not complete) of hypersurfaces? - -REPLY [9 votes]: For $(g,n)=(g,g-1)$, i.e. the case of curves embedded by their complete canonical system for $g\geq 3$, one has a good understanding of the equations of a general $C$ for $g$ up to about $9$. The cases $g=3,4,5$ are classical and already described in the comments. The breakthrough was I believe Mukai's Curves and symmetric spaces I,II ($g=7,8,9$); there is a lot of followup work such as Ide-Mukai Canonical curves of genus 8, von Bothmer's Geometric Syzygies of Mukai Varieties and General Canonical Curves with Genus at most 8, etc. The main result (quoted from the last paper) is as follows. -Theorem (Mukai). Every general canonical curve of genus $7\leq g \leq 9$ is a general linear section of an embedded rational homogeneous variety $M_g$. General canonical curves of genus $6$ are cut out by a general quadric on a general linear section of a homogeneous variety $M_6$. -The equations of these homogeneous varieties can be written down reasonably explicitly using representation theory. -In a different direction, for $(g,n)=(1,m-1)$, one is talking about elliptic normal curves of degree $m$. Their equations are studied for example in Fisher's Pfaffian representations of elliptic normal curves, which generalises your examples for $g=1$. -For more general $(g,n)$ lots of things can happen. I recommend Eisenbud's A Mystery Variety in $P^3$ in the collection Computations in algebraic geometry with Macaulay 2.<|endoftext|> -TITLE: What is the meaning of the monodromy theorem in Hodge theory? -QUESTION [7 upvotes]: Let $f : X^m \to Y^n$ be an algebraic fiber space (between projective manifolds) whose discriminant locus is denoted by $E$. Let $U$ be a polydisk in $\mathbb{C}^n$ (with coordinates $(y_1, ..., y_n)$) such that $U \backslash E \simeq (\Delta^{\ast})^{\ell} \times \Delta^{n-\ell}$, where $1 \leq \ell \leq n$. Let $x_0$ be a point in $U\backslash E$. Associated to $x_0$ is the monodromy operator $$T_k : H^{m-n}(f^{-1}(x_0), \mathbb{C}) \longrightarrow H^{m-n}(f^{-1}(x_0), \mathbb{C})$$ for a loop based at $x_0$ around the $k$th copy of $\Delta^{\ast}$, where $1 \leq k \leq \ell$. -The monodromy theorem states that $T_k$ is quasi-unipotent, i.e., there are positive integers $m_k$ and $d_k$ such that $$(T_k^{m_k} - I)^{d_k} =0.$$ Here, $m_k$ is the least common multiple of the multiplicities of the irreducible components over the generic point of $\{ y_k =0 \}$. -Question: What does the quasi-unipotence of the monodromy transformation tell us qualitatively? - -REPLY [6 votes]: I'm not completely sure what you're after, but perhaps it's simply some insight. So let make a few remarks. For simplicity, assume $Y$ is a curve. (In general, you need to assume that $E$ has normal crossings, which you did implicitly.) Replace it by a small disk. So $f:X\to D$ becomes a $C^\infty$ fibre bundle when restricted to $D^*= D-\{0\}$, which can be replaced by the circle, since we can work up to homotopy. To construct a such a fibre bundle, you just take a manifold $F$ and a diffeomorphism $\phi:F\to F$, and the glue the ends of $F\times [0,1]$ using $\phi$. Now ask, given a fibre $F$ diffeomorphic to a projective manifold and a self diffeomorphism, when can it arise from a projective family? The monodromy theorem tells you that the answer is almost always no, even when $F$ is $2$-torus, because a necessary condition is that the eigenvalues of $\phi^*H^i(F)\to H^i(F)$ must be roots of unity for all $i$. I still find this statement pretty striking. -Does this help?<|endoftext|> -TITLE: Interpretation of "27" lines for cubic surface with rational double points -QUESTION [9 upvotes]: It is well known that a smooth cubic surface has $27$ distinct lines. Explicitly, if we choose a planar representation, i.e., blowup $\mathbb P^2$ at $6$ general points $p_1,...,p_6$, the $27$ lines are (1) $E_i$, $1\le i\le 6$, the exceptional divisors, (2) $F_{ij}$, $1\le i -TITLE: Spherical average of $\frac{1}{x}$ -QUESTION [5 upvotes]: Let $X_1,...,X_n$ be points on $\mathbb S^1.$ -We then define the expectation value $E(X)=\frac{1}{n}\sum_{i=1}^n X_i.$ -Let $\frac{dS(X_1)}{2\pi}$ be the normalized surface measure of $\mathbb S^1,$ i.e. $X_i$ are uniformly distributed random variables on the circle. -I am curious to know: -How does -$$\int_{(\mathbb S^1)^n } \frac{1}{\vert E(X) \vert}\frac{dS(X_1)}{2\pi}...\frac{dS(X_n)}{2\pi}$$ -scale with $n$? - -REPLY [4 votes]: The probability distribution $P(R)$ of $R=n|E(X)|$ was calculated by Kluyver (1906), it is given by -$$P(R)=\frac{1}{2\pi}\int_0^\infty [J_0(x)]^n J_0(rx)x\,dx.$$ -For $n\gg 1$ one has a Rayleigh distribution (here is derivation including higher order corrections): -$$P(R)=\frac{2R}{n}e^{-R^2/n}.$$ -The desired integral then becomes -$$I=\int_0^{\infty}\frac{n}{R}P(R)\,dR\rightarrow \sqrt{\pi n}$$ -in the limit $n\rightarrow\infty$.<|endoftext|> -TITLE: Fixed point scheme of finite group Cohen-Macaulay? -QUESTION [11 upvotes]: Let $X$ be a quasi-projective scheme over a field $k$. -Let $G$ be a finite group acting on $X$ whose order is invertible in $k$. -If $X$ is Cohen-Macaulay, can we conclude that the subscheme of fixed points $X^G$ is Cohen-Macaulay? - -REPLY [13 votes]: Here is a simpler example than the one I left before, using the same strategy. Let -$$X = \{ x_1 x_3 = x_1 x_4 = x_1 x_5 = x_2 x_4 = x_2 x_5 = x_3 x_5 = 0 \} \subset \mathbb{C}^5.$$ -This is the reduced union of four $2$-planes. Here is a projective picture, where $j$ represents the point where $x_j$ is the sole nonzero coordinate: -$$1 - 2 - 3 - 4 - 5.$$ -The graph above is shellable, so this is Cohen-Macaulay. -Now, let $C_2$ act on $X$ by $(x_1, x_2, x_3, x_4, x_5) \mapsto (x_1, x_2, - x_3, x_4, x_5)$. Then the fixed locus of $C_2$ (even scheme-theoretically) is -$$Y = \{ x_1 x_4 = x_1 x_5 = x_2 x_4 = x_2 x_5 = x_3 = 0 \}.$$ -This is the reduced union of two $2$-planes; we can visualize it as -$$1 - 2 \phantom{- 3 -} 4 - 5.$$ -That is a standard example of a non-Cohen-Macaulay ring.<|endoftext|> -TITLE: Looking for a paper on transfinite diameter by David Cantor -QUESTION [14 upvotes]: I have been reading about transfinite diameter and its applications to number theory and have been hunting for the following paper for quite a while: -Cantor D.: On an extension of the definition of transfinite diameter and some -applications, J. reine Angew. Math., vol. 316 (1980), pp. 160-207. -Unfortunately, I am an undergraduate and have no access to MathSciNet or other resources which provide papers for free. The only links I got results from are ones accepting a payment which is completely outside my financial capacity. From some advice I got in a previous similar reference-request post, I tried an interlibrary loan but due to the pandemic and no nearby libraries, that hasn't been successful so far. I would really appreciate a link to the full text of the above reference. Thank you. - -REPLY [19 votes]: Hopefully this works: -Cantor D.: On an extension of the definition of transfinite diameter and some applications - -Since you said that you had "been hunting for the following paper for quite a while," here are the specific steps that I took to find this paper. Probably this will make it seem simple in retrospect, but I don't want to give off the impression that this required a complex knowledge of databases. - -I googled the title of the paper - -I clicked the link to EUDML (with which I wasn't previously familiar) - -The sidebar says In Other Databases and I didn't find ZBMath helpful, so I used GDZ - -I've never heard of GDZ before, but I saw the paper listed on p. 160 of that document and, indeed, the paper was there - - -For future reference, there is also an often used website called sci-hub [dot] tw that can pull papers for which you have the DOI. You can read more about Sci-Hub on wikipedia and decide whether that is a route that you are willing to take when searching out future materials.<|endoftext|> -TITLE: Comparing sizes of sets of integers -QUESTION [14 upvotes]: Is there a total preorder $\lesssim$ on the power set of $\mathbb Z$ such that: - -$A [b,\infty)\cap\mathbb Z$ for all $a,b$ (it is biased to the left). - -REPLY [6 votes]: Yes, there is such a preorder. I will argue that there is a preorder on the space of bounded functions $\mathbb Z\to\mathbb R$ so that comparing indicator functions in this space does the job. A vector space preorder can be constructed from a suitable "positive cone", the set of non-negative elements, so the main task is to construct this cone. -Let $M$ be the set of non-negative, not identically zero, finitely-supported functions $\mathbb Z\to\mathbb R.$ Let $B$ be the real vector space of bounded functions $\mathbb Z\to\mathbb R.$ Let $a*\phi$ denote convolution of a function $a\in B$ by a function $\phi\in M.$ Define $a\sim b$ for $a,b\in B$ to mean that $a*\phi=b*\psi$ for some $\phi,\psi\in M.$ This is an equivalence relation because $M$ is a (commutative) monoid under convolution. The $\sim$-equivalence class $[0]$ of zero is a linear subspace of $B.$ -Define a good cone to be a set $C\subset B$ such that - -C1. $y\in C\iff z\in C$ whenever $y\sim z,$ and -C2. $C$ is a convex cone ($x,y\in C\implies \lambda x+\mu y\in C$ for $\lambda,\mu\geq 0$), and -C3. $C\cap (-C)=[0].$ - -Define $C_0$ to be the set of $x\in B$ such that $x\sim y$ for some non-negative function $y\in B.$ Because non-negative functions are closed under convolution by any $\phi\in M,$ the definition of $C_0$ simplifies slightly to $x*\phi$ being non-negative for some $\phi\in M.$ The set $C_0$ satisfies the good cone conditions: (C1) is obvious, for (C2) if $x*\phi$ and $y*\psi$ are non-negative and $\lambda,\mu\geq 0$ then $(\lambda x+\mu y)*\psi*\psi$ is non-negative, and for (C3) if $x*\phi$ is non-negative and $x*\psi$ is non-positive, then $x*\phi*\psi$ is identically zero so $x\sim 0.$ By Zorn's lemma there is a maximal good cone $C$ containing $C_0.$ -Consider $x\in B\setminus C.$ Define $C_x$ to be the set of $y\in B$ such that $y*\phi= x*\psi+c$ for some $\phi\in M$ and $\psi\in M\cup\{0\}$ and $c\in C.$ By maximality of $C,$ the set $C_x$ is not good. -(C1) holds: whenever $y*\eta=z*\zeta$ and $y*\phi= x*\psi+c$ we have $z*\zeta*\phi=x*\psi*\eta+c*\eta,$ which implies $z\in C_x.$ -(C2) holds: if $y*\phi= x*\psi+c$ and $y'*\phi'= x'*\psi'+c'$ and $\lambda,\mu\geq 0$ then $(\lambda y+\mu y')*\phi*\phi'=x*(\psi*\phi'+\psi'*\phi)+(c*\phi'+c'*\phi).$ -So (C3) fails: some $y\not\sim 0$ satisfies $y*\phi= x*\psi+c$ and $y*\phi'=-x*\psi'-c'.$ But then $$-x*\psi'*\phi-c'*\phi=y*\phi*\phi'=x*\psi*\phi'+c*\phi'$$ -which implies $x*(\psi*\phi'+\psi'*\phi)+(c*\phi'+c'*\phi)=0.$ If $\psi$ and $\psi'$ are both zero, then $c*\phi=-c'*\phi$ is in $C\cap (-C)$ contradicting $y\not\sim 0.$ So $\psi$ and $\psi'$ are not both zero, which means $-x\sim c*\phi'+c'*\phi\in C.$ -In other words $x\in -C.$ -We have shown that $C\cup (-C)=B.$ The cone $C$ defines a total vector order on $B/[0],$ but to answer the question we just need to define $S\lesssim T\iff 1_T-1_S\in C.$ Your condition 1 comes from $C_0\subset C$ and (C3). Your condition 2 comes from (C1) - shifting is convolution by a delta function. Your condition 3 comes from $1_{T\cup U}-1_{S\cup U}=1_T-1_S$ whenever $S\cap U=T\cap U=\emptyset.$<|endoftext|> -TITLE: Work on triply periodic minimal surfaces -QUESTION [8 upvotes]: I have seen in some engineering departments that they manufacture models of periodic minimal forms (characterised by equal and opposite curvature at every points on the surface). In pure mathematics, they are known as triply periodic minimal surfaces. -If I understand rightly, these have been observed experimentally in crystallography and polymer chemistry but I assume they must have been studied in differential geometry as well. The Wikipedia page mentions the classification of these surfaces as an open problem: has there been any recent progress on this? The -physics literature also mentions the possibility of constructing minimal surfaces with the properties of a quasicrystal (ie. minimal surfaces with a quasicrystalline order). Again, has there been any further geometric work on this construction? - -REPLY [6 votes]: This is an active research topic. I'm currently working on the front line towards a classification of TPMSs of genus 3 (TPMSg3s). My collaborators include Weber and Traizet. I also know a Japanese team working on the moduli space. -Recent progress include surprising discoveries of new examples: - -https://arxiv.org/abs/1804.01442 (j/w Weber, to appear in Trans. AMS) -https://arxiv.org/abs/1807.10631 (j/w Weber, to appear in Trans. AMS) - -and rigorous proof of some deformations of the gyroid. - -https://arxiv.org/abs/1901.04006 - -These examples are important because they reveal concrete singularities in the moduli space of TPMSg3s. So their existence basically proves that the classification is very complicated. -However, an ultimate classification of TPMSg3s is not completely hopeless. We are making big progress on the boundary of the moduli space, and we expect to see more new examples in the near future. I estimate about 5 years of hard works before we can finally evaluate the feasibility of the classification. I'm optimistic. -We also constructed uncountably many examples of non-periodic minimal surfaces that of great interest for material scientists and crystallographers. - -https://arxiv.org/abs/1908.06276 (j/w Traizet, to appear in SIAM J. Math. Anal.) - -Some of them can be regarded as "quasi-periodic", but not "quasi-crystallographic". I have been thinking about quasi-crystallographic minimal surfaces, but currently have no progress at all. -My involvement in the topic is mostly motivated by physics, and I'm actively collaborating with many experimental teams.<|endoftext|> -TITLE: A group where the Weil topology induced by the Haar measure does not coincide with the original topology -QUESTION [7 upvotes]: Let $(G,\tau)$ be a locally compact Hausdorff topological group that is $\sigma$-finite with respect to the Haar measure $\mu:\mathcal{B}(G)\to[0,\infty]$ ($\mathcal{B}(G)$ is the Borel $\sigma$-algebra for $G$). Define $\mathcal{B}\boldsymbol{a}(G)\subseteq \mathcal{B}(G)$ to be the Baire $\sigma$-ring in $G$ (the $\sigma$-ring generated by the compact $G_\delta$'s), and furthermore assume that $G\in\mathcal{B}\boldsymbol{a}(G)$ (i.e. $\mathcal{B}\boldsymbol{a}(G)$ is a $\sigma$-algebra). Let -$$\mathcal{A}=\{EE^{-1} \mid E\in \mathcal{B}\boldsymbol{a}(G), 0<\mu(E)<\infty\}.$$ Now forget about the topology $\tau$. It is well known that $\mathcal{A}$ forms a system of neighborhoods for $e$, which induces a topology $\tau_\mu$ in $G$ which makes it a Hausdorff topological group. This topology is called Weil's topology (see [1]). Under this topology $G$ is densely embeddable in a Hausdorff locally compact group $\overline{G}$, and the Haar integral in $\overline{G}$ coincides with the integral with respect to $\mu$ for all continuous functions of compact support contained in $G$. -It can be easily shown that $\tau \subseteq \tau_\mu$, and it was shown in [2] that adding the assumption that $\mathcal{B}\boldsymbol{a}({G})$ is analytic, $\tau_\mu\subseteq \tau$. -I am trying to come up with a simple example where $\tau_\mu\not\subseteq \tau$ (evidently in the case where $\mathcal{B}\boldsymbol{a}({G})$ is not analytic), but I have not been successful. Any ideas? -Refs: -[1] Halmos, Paul R., Measure theory. 2nd printing, Graduate Texts in Mathematics. 18. New York - Heidelberg- - Berlin: Springer-Verlag. XI, 304 p. DM 26.90 (1974). ZBL0283.28001. -[2] Mackey, George W., Borel structure in groups and their duals, Trans. Am. Math. Soc. 85, 134-165 (1957). ZBL0082.11201. - -REPLY [5 votes]: There are no such locally compact groups, because if $G$ is a locally compact group under the topology $\tau$, then the Weil topology $\tau_\mu$ defined by the Haar measure $\mu$ is the same as the original topology $\tau$. -To show $\tau_\mu$ is finer than $\tau$, let $N$ be a $\tau$-neighbourhood of $e$. Since the mapping $g \mapsto gg^{-1}$ is continuous $G \rightarrow G$, there is a neighbourhood $M$ of $e$ such that $MM^{-1} \subseteq N$. Since $G$ is locally compact, we can find a compact $G_\delta$ neighbourhood $K$ of $e$ such that $K \subseteq M$ and therefore $KK^{-1} \subseteq N$. Since $K$ is compact, $\mu(K) < \infty$, and since it contains an open set, $\mu(K) > 0$ and therefore $KK^{-1} \in \mathcal{A}$ and so $N$ is a $\tau_\mu$-neighbourhood of $e$. -The other direction holds by Weil's extension of Steinhaus's theorem, which states that if $\mu(E) > 0$ then $EE^{-1}$ is a $\tau$-neighbourhood of $e$. Weil proved this by what is now the standard argument that convolving $\chi_E$ by $\chi_{E^{-1}}$ produces a continuous function vanishing outside $EE^{-1}$ but taking the nonzero value $\mu(E)\mu(E^{-1})$ at $e$. - -For the more general question of topological groups with Haar measures, I do not know an example of a topological group $G$ with a left-invariant Radon measure $\mu$ such that the original topology $\tau$ differs from $\tau_\mu$. However, if we drop the requirement that $\mu$ be Radon there is a simple example. Take $G = \mathbb{Q}$, and let $\tau$ be its subspace topology in $\mathbb{R}$. The counting measure $\mu$ is an invariant measure on this group. However, the Weil topology $\tau_\mu$ defined by the counting measure on $\mathbb{Q}$ is easily seen to be the discrete topology, which is strictly finer than $\tau$.<|endoftext|> -TITLE: A possible error in Elliot's book "Probabilistic Number Theory" -QUESTION [7 upvotes]: In Elliot's book "Probabilistic Number Theory", there seems to be an inaccuracy. The author defines, for any sequence $a_n$, the quantity -$$V(p)=\sum_{r=0}^{p-1}\left|\sum_{\substack{n=1 \\n \equiv r\mathrm{mod}(p)}}^N a_n-p^{-1}\sum_{n=1}^N a_n -\right|^2$$ -He then asserts that, if $a_n$ assumes only the values 0,1, then -$$\sum_{p\leq Q}pV(p)\leq c_1 Q^2 \log(Q)\sum_{n=1}^N|a_n|^2$$ -where $c_1$ is some absolute constant. The issue is, this would imply that -$$\limsup_{N\to\infty}\sum_{p\leq Q}p\frac{V(p)}{N^2}\leq c_1 Q^2 \log(Q)\limsup_{N\to\infty}\frac{1}{N^2}\sum_{n=1}^N|a_n|^2=0$$ -which isn't always true. An easy counter-example is $a_n$ defined as $0$ when $n$ is even and $1$ when $n$ is odd, and $Q=2$. Namely, we have that -\begin{align*} -\sum_{p\leq Q}p\frac{V(p)}{N^2} &= 2\frac{V(2)}{N^2}\\ -&=\frac{2}{N^2}\sum_{r=0}^{1}\left|\sum_{\substack{n=1 \\n \equiv r\mathrm{mod}(2)}}^N a_n-\frac{1}{2}\sum_{n=1}^N a_n -\right|^2\\ -&=2\sum_{r=0}^{1}\left|\frac{1}{N}\sum_{\substack{n=1 \\n \equiv r\mathrm{mod}(2)}}^N a_n-\frac{1}{2N}\sum_{n=1}^N a_n\right|^2\\ -&=2\left|\frac{1}{N}\sum_{\substack{n=1 \\n \equiv 0\mathrm{mod}(2)}}^N a_n-\frac{1}{2N}\sum_{n=1}^N a_n\right|^2+2\left|\frac{1}{N}\sum_{\substack{n=1 \\n \equiv 1\mathrm{mod}(2)}}^N a_n-\frac{1}{2N}\sum_{n=1}^N a_n -\right|^2\\ -\end{align*} -Since -$$\lim_{N\to\infty}\frac{1}{2N}\sum_{n=1}^Na_n=\frac{1}{4}$$ -$$\lim_{N\to\infty}\frac{1}{N}\sum_{\substack{n=1 \\n \equiv 1\mathrm{mod}(2)}}^N a_n=\frac{1}{2}$$ -$$\lim_{N\to\infty}\frac{1}{N}\sum_{\substack{n=1 \\n \equiv 0\mathrm{mod}(2)}}^N a_n=0$$ -We see that -$$\lim_{N\to\infty}\sum_{p\leq Q}\frac{V(p)}{N^2}=2\left(\frac{1}{4}\right)^2+2\left(\frac{1}{4}\right)^2=\frac{1}{8}$$ -Numerical computations show that $\frac{1}{8} \neq 0$ and thus this is a contradiction. The paper cited for this result is locked behind a paywall so I cannot access it and see what the true theorem is. Does anyone know what the actual result should have been? Where is the typo? -The paper cited with the result is -Roth, Klaus F., On the large sieves of Linnik and Renyi, Mathematika, Lond. 12, 1-9 (1965). ZBL0137.25904. -SIDE QUESTION: -In the book, there are many inequalities given for the sum $\sum_{p -TITLE: Which cases of Beilinson-Bloch-Kato for elliptic motives are known? -QUESTION [6 upvotes]: Let $V$ be a semisimple geometric Galois representation of a number field. Then the Bloch-Kato conjectures state that -$$ -\operatorname{ord}_{s=0}{L(V^*(1),s)} = \operatorname{dim}{H^1_f(G_k,V)}-\operatorname{dim}{H^0(G_k,V)}. -$$ -Beilinson has similar conjectures relating the LHS to algebraic K-theory rather than Selmer groups. -If $E$ is an elliptic curve, and we set $V=h_1(E)=h^1(E)(1)$, then the conjecture above is equivalent to the statement that the analytic rank is the same as the rank of the $p$-adic Selmer group. Beilinson's conjecture in this case is equivalent to the assertion that the analytic rank equals the Mordell-Weil rank. -For $V$ of non-negative weight, the conjecture simply asserts that the Selmer group vanishes. Assuming the conjectured properties of $L$-functions, the conjectures for $V$ and $V^*(1)$ are equivalent (in particular, the case of weight $\le -2$ follows from the case of weight $\ge 0$). -My question is: which cases of this conjecture are known for $V=\operatorname{Sym}^k{h^1(E)}(n)$ for $E$ an elliptic curve? I know that many cases are known when $k=n=1$ due to the theory of Heegner points, Gross-Zagier, etc, but I'd like to know what's known outside that range. (I'm especially interested in cases where $w=k-2n=-2, -3, -4$). -Feel free to give reference, or even better, specific elliptic curves in LMFDB. - -REPLY [7 votes]: There are three approaches I know of to studying $H^1_{\mathrm{f}}(K, V)$, where $V = Sym^k(h^1(E))(n))$. All rely on $E$ being modular, so let me assume this henceforth (of course, this is no assumption if $K = \mathbf{Q}$, or for some other small-degree fields). - -Via "anticyclotomic" Euler systems, such as Heegner points (and the closely-related method of "arithmetic level-raising"). This works extremely well when $k = n = 1$, and $K$ is totally real (or $K$ is CM and $E$ is base-extended from $K^+$); under these hypotheses we know the BK conjecture holds, for any $p$, whenever the analytic rank is 0 or 1 (Zhang, Nekovar). More generally, this might potentially be accessible for any $n$ and $k = 2n-1$, although huge amounts of work would be needed to carry that out. However, it's entirely impossible to generalise this approach beyond the case of motivic weight $w = -1$. - -Via modularity-lifting theorems. This gives a way of studying Selmer groups of representations that have the shape $W \otimes W^*$, where $W$ is irreducible. More generally, if $W$ has some extra structure (e.g. self-duality) which forces $W \otimes W^*$ to be reducible, then you can get some information about the cohomology of the pieces. This gives you very nice control over $Sym^k(h^1(E))(n)$ for $k = 2$ and $n = 1$ (or by duality $n = 2$) (Diamond--Flach--Guo). More generally, one should be able to get some information about general $n$ and $k = 2n$ or $2n-2$ using the recent work of Newton--Thorne; Theorem 5.6 of this paper tells you something about $W\otimes W^*$ where $W = Sym^k(h^1(E))$, and this representation breaks up as a sum of $Sym^{2m}(h^1(E))(m)$ for $0 \le m \le k$. [Caveat: I'm not sure exactly what their method gives; it's possible that you need to twist by an odd quadratic character at some point.] However, this is again restricted to specific values of $w$; it won't tell you anything unless $w = 0$ or $w = -2$. - -Via "cyclotomic" Euler systems, such as Kato's Euler system. This method has the advantage that it can tell you something about general motivic weights (i.e. a fixed $k$ and any $n \in \mathbf{Z}$). However, it gives you a criterion for vanishing of the $H^1$ in terms of p-adic L-functions, and these are only indirectly related to complex $L$-functions outside the critical range (i.e. away from $w = -1$ in your case). Nonetheless, these p-adic L-functions are computable, so you can check explicitly whether they vanish in examples. For elliptic curves over $\mathbf{Q}$, an Euler system for $Sym^k E$ exists for $k = 1$ due to Kato, and for $k = 2$ [*] and $k = 3$ due to Zerbes and myself (building on work of lots of other people). So, for example, if $K = \mathbf{Q}$, and $k = 1, 2, 3$, this would give an approach to proving the vanishing of $H^1_{\mathrm{f}}(\mathbf{Q}, Sym^k(h^1(E))(n))$ for your favourite elliptic curve and a specific but arbitrary value of $n$ (and $p$), using only a finite amount of computation. - - -[*] Actually there is a caveat here -- embarrassingly, I forgot the statement of my own theorem! -- so the result as published only applies to $Sym^2(E)$ twisted by a non-trivial Dirichlet character. But the un-twisted case might also be accessible with some extra work.<|endoftext|> -TITLE: Complex projective manifolds are homeomorphic if homotopy equivalent -QUESTION [32 upvotes]: If two complex projective manifolds are homotopy equivalent are they homeomorphic? - -REPLY [31 votes]: For curves this follows from the classification of (2-dimensional topological) surfaces, and for simply-connected surfaces this follows from Freedman's theorem. -My former colleagues Anatoly Libgober and John Wood found examples of pairs of 3-folds which are complete intersections and are homotopy equivalent but not diffeomorphic, in fact have distinct Pontryagin classes. See Example 9.2. Since in this case $H^4(M;\mathbb{Z})\cong \mathbb{Z}$, this implies that the manifolds are not homeomorphic by the topological invariance of rational Pontryagin classes (see Ben Wieland's comment). -For the higher dimensional case see: -Fang, Fuquan, Topology of complete intersections, Comment. Math. Helv. 72, No. 3, 466-480 (1997). ZBL0896.14028.<|endoftext|> -TITLE: Non-isomorphic compact Kähler manifolds that are biholomorphic, symplectomorphic and isometric -QUESTION [11 upvotes]: Let $(M, \omega_M, J_M)$ and $(N, \omega_N, J_N)$ be compact Kähler manifolds. Denote $g_M=\omega_M(\cdot, J_M\cdot)$ and $g_N=\omega_N(\cdot, J_N\cdot)$. -Assume there is a diffeomorphism $\nu:M\to N$ such that $\nu^*(\omega_N)=\omega_M$, there is a diffeomorphism $\phi:M\to N$ such that $\phi^*(J_N)=J_M$ and there is an orientation-preserving diffeomorphism $\chi:M\to N$ such that $\chi^*g_N=g_M$. -Is there a diffeomorphism $\psi:M\to N$ such that $\psi^*(\omega_N)=\omega_M$ and $\psi^*(J_N)=J_M$ (and hence also $\psi^*(g_N)=g_M$)? - -REPLY [14 votes]: The answer is 'no, not necessarily'. -Consider the following example: Let $M=N=\mathbb{CP}^2$, let $(\omega_0,J_0)$ be the standard Fubini-Study Kähler structure on $M$. Now let $f$ be an arbitrary, but '$C^2$-small' smooth function on $M$, so that $\omega_0 + t\,\mathrm{i}\,\partial\bar\partial f$ is nondegenerate (and hence symplectic) for all $0\le t\le 1$. -Let $\omega_M = \omega_0 + \mathrm{i}\,\partial\bar\partial f$ and let $J_M= J_0$. Let $\omega_N=-\omega_M$ and let $J_N=-J_0$. Note that $g_M=g_N$, so $(M,g_M)$ and $(N,g_N)$ are isometric via the identity map. -When $f$ is chosen sufficiently generically, the isometry group of $g_M$ will consist of only the identity, so suppose this. -Note that $(M,J_M)$ and $(N, J_N)$ are biholomorphic, since $\mathbb{CP}^2$ is biholomorphic to its conjugate complex manifold. -By Weinstein's theorem, since $\omega_M$ and $\omega_0$ are $C^0$-close and cohomologous, there is a symplectomorphism between $(M,\omega_M)$ and $(M,\omega_0)$. Similarly, there is a symplectomorphism between $(N,\omega_N)$ and $(N,-\omega_0)$. Since, as has already been noted, $(M,\omega_0)$ and $(N,-\omega_0)$ are symmplectomorphic, it follows that $(M,\omega_M)$ and $(N,\omega_N)$ are symplectomorphic. -However, when $f$ is chosen sufficiently generically, the only map $\psi:M\to N$ that aligns the metrics $g_M$ and $g_N$ is the identity, which is neither a biholomorphism nor a symplectomorphism. -Remark: Of course, I woke up this morning and was struck by the fact that there is an even simpler example: Let $g$ be any metric on $S^2=\mathbb{CP}^1$ whose isometry group is trivial, let $J$ be one of the two $g$-orthogonal complex structures on $S^2$, and let $\omega_J$ be the associated area $2$-form on $S^2$. Let $(M,\omega_M,J_M)=(S^2,\omega_J,J)$ and let $(N,\omega_N,J_N)=(S^2,-\omega_J,-J)$. -Then $(M,\omega_M,J_M)$ and $(N,\omega_N,J_N)$ are biholomorphic, symplectomorphic, and isometric, but not isomorphic as Kähler manifolds. -Moreover, it is not difficult to choose an unramified rational curve in $\mathbb{CP}^2$ (it may have crossings, but that doesn't matter, for example, the generic irreducible cubic with one node but no cusp would certainly do) so that the metric on the normalized curve (which is an $S^2$ topologically) induced by the Fubini-Study metric on $\mathbb{CP}^2$ has no nontrivial isometries. Thus, one can even construct a pair of examples with the additional property requested by the OP that the metric be induced from the Fubini-Study metric by an immersion into $\mathbb{CP}^2$. (Using $\mathbb{CP}^3$ as a target, one could even arrange the map to be an embedding.)<|endoftext|> -TITLE: On Integrals of the Airy function -QUESTION [7 upvotes]: Let $Ai$ be the classical Airy function and let $(a_j)_{j\ge 1}$ be the strictly decreasing sequence of its zeroes: we have $a_{j+1}g$ on $(0,+\infty)$, then the graph of $v$ lies above the first hump of the graph of $u$ (i.e., $v>u$ between $0$ and the first positive zero of $u$). -The standard proof is by noticing that $u>v$ slightly to the right of $0$ (Taylor with the usual trick of considering $(1+\varepsilon)v$ instead of $v$ to break the tie at $0$) and considering the Wronskian $W(u,v)=\det\begin{bmatrix}u&v\\u'&v'\end{bmatrix}$ of $u$, $v$. We have $W'=(f-g)uv>0$ and, thereby, $W>W(0)=0$ as long as both $u,v>0$ and if $v$ tries to go below $u$ as long as $u>0$, it has to hit from above, so at the hitting point $v=u>0,v'\le u'$, i.e., $W(u,v)\le 0$, which is impossible. -Now just look at two consecutive humps of the Airy function and think of them as shot from the zero separating them. The lemma shows that a couple of reflections put one hump below the other one. The rest should be clear.<|endoftext|> -TITLE: Sum of the coefficients of the characteristic polynomial of periodic matrices -QUESTION [10 upvotes]: Let $M$ be an integer matrix with determinant equal to one (or maybe also minus one, but I did not do any tests for this case) and assume that $M$ is periodic, that is $M^n$ is the identity matrix for some $n$. Let $p_M$ denote the characteristic polynomial of $M$. - -Question 1: Is it true that then $p_M(1) \geq 0$ for periodic matrices $M$? - - -Question 2: Let $a_n$ be the largest period of a matrix in $Sl_n(\mathbb{Z})$. What is $a_n$? (or is there a good bound?) - -For $n=2$ it should be $a_n=6$. Is $a_n$ attained at a matrix with entries only in $\{-1,0,1\}$? - -REPLY [10 votes]: Q1: This was already given in the comments, but: a matrix $M \in GL_k(\mathbb{Z})$ of finite order $n$ must have rational normal form a block-diagonal matrix with blocks the companion matrices of cyclotomic polynomials $\Phi_d$ for $d | n$, so the problem reduces to the case of a single such matrix, which is to say the problem reduces to asking whether we always have $\Phi_d(1) \ge 0$. This is true, and in fact: - -Proposition: $\Phi_n(1)$ is equal to $p$ if $n = p^k$ is a prime power and equal to $1$ otherwise. - -Proof. $\Phi_p(x) = \frac{x^p - 1}{x - 1}$ and $\Phi_{p^k}(x) = \Phi_p(x^{p^{k-1}})$ so the computation in the prime power case is clear. For a general $n$ we have that if $p \nmid m$ then -$$\Phi_{pm}(x) = \frac{\Phi_m(x^p)}{\Phi_m(x)}$$ -and hence that $\Phi_n(1) = 1$ as soon as $n$ has more than one prime factor. $\Box$ -Q2: As before it suffices to consider block sums of companion matrices of cyclotomic polynomials. A block sum of companion matrices of cyclotomic polynomials $\Phi_{d_i}(x)$ is an element of $GL_n(\mathbb{Z})$ where $n = \sum \varphi(d_i)$ of order $\text{lcm}(\{ d_i \})$ so the problem is to optimize this (the cyclotomic polynomials satisfy $\Phi_n(0) = 1$ for $n \ge 2$ so all these block matrices lie in $SL_n(\mathbb{Z})$ also). This looks hard in general. -The corresponding problem of finding the largest order of an element of $S_n$ is a similar optimization problem but where $n = \sum d_i$. That sequence is Landau's function (A000793) but I don't know if this one has a name or is in the OEIS. -Edit #1: If $L(n)$ denotes this largest order then we have - -$L(1) = 2$ ($1 = \varphi(2)$) -$L(2) = 6$ ($2 = \varphi(6)$) -$L(3) = 6$ ($3 = \varphi(6) + \varphi(2)$) -$L(4) = 12$ ($4 = \varphi(6) + \varphi(4)$) -$L(5) = 12$ ($5 = \varphi(6) + \varphi(4) + \varphi(2)$) -$L(6) = 30$ ($6 = \varphi(10) + \varphi(6)$) - -which, if I haven't messed up, already shows that this sequence is not in the OEIS. On the other hand, it's not hard to see that $L(2k+1) = L(2k)$ for $k \ge 1$ since $\varphi(d)$ is even for $d \ge 2$ and $\varphi(2d) = \varphi(d)$ if $d$ is odd so it never helps us to add a $\varphi(2) = 1$ term to the sum. (We need to rule out the possibility that $L(n)$ is a power of $2$ but this shouldn't be hard.) So perhaps the OEIS has only the even terms $L(2n)$ somewhere; I haven't ruled that out yet. -An easy upper bound is that we can compute the exponent $E(n)$, namely the lcm of all orders of elements of finite order in $GL_n(\mathbb{Z})$, so that $L(n) | E(n)$. By considering each prime separately we have that -$$\nu_p(E(n)) = \text{max} \left\{ k : \varphi(p^k) = (p - 1) p^{k-1} \le n \right\} = \left\lfloor \log_p \frac{n}{p-1} \right\rfloor + 1$$ -and hence -$$E(n) = \prod_p p^{ \left\lfloor \log_p \frac{n}{p-1} \right\rfloor + 1}.$$ -This sequence is much easier to compute although the bound becomes increasingly bad. It does have the virtue of also being a bound on the exponent of any finite subgroup of $GL_n(\mathbb{Z})$. We again have $E(2k+1) = E(2k)$ for $k \ge 1$, and - -$E(1) = 1$ -$E(2) = E(3) = 6$ -$E(4) = E(5) = 2^3 \cdot 3 \cdot 5 = 120$ -$E(6) = E(7) = 2^3 \cdot 3^2 \cdot 5 \cdot 7 = 2520$ - -which also does not appear to be in the OEIS, with or without the terms doubled. The corresponding sequence for $S_n$ is $\text{lcm}(1, 2, \dots n)$ which is A003418 and the formula is similar except the exponent is more simply just $\lfloor \log_p n \rfloor$. -Edit #2: Okay, I computed that $L(8) = 60$ which was finally enough terms for me to find it: $L(2n)$ appears to be (up to some indexing issues) A005417 on the OEIS. A comment there suggests the following argument which makes $L$ a bit easier to compute than I had been thinking: if $\gcd(n, m) = 1$ and $\varphi(n), \varphi(m) \ge 2$ (so neither $m$ nor $n$ is equal to $2$) then we can always replace a $\Phi_{mn}(x)$ block with a $\Phi_n(x)$ block and a $\Phi_m(x)$ block, since $\varphi(mn) = \varphi(m) \varphi(n) \ge \varphi(m) + \varphi(n)$. So we only ever need to consider $\Phi_d(x)$ blocks where $d$ is a prime power or twice an odd prime power. A similar argument works in $S_n$. It follows (this is the OEIS comment) that -$$L(n) = \text{max} \left\{ \prod p_i^{e_i} : \sum (p_i - 1) p_i^{e_i - 1} \le n \right\}.$$ -Edit #3: The observation in the previous paragraph answers Q3: yes, the maximum is attained for a matrix with entries in $\{ -1, 0, 1 \}$, since the same is known about the cyclotomic polynomials $\Phi_d(x)$ where $d$ is a prime power or twice an odd prime power. Famously the cyclotomic polynomials are known to not always have coefficients in $\{ -1, 0, 1 \}$ and $\Phi_{105}(x)$ is the smallest counterexample, but that doesn't matter here. -Edit #4: Okay, here are some bounds. For a lower bound we clearly have $g(n) \le L(n)$. For an upper bound let $r_i = (p_i - 1) p_i^{e_i - 1}$, so that we can write the optimization problem defining $L(n)$ as -$$L(n) = \text{max}\left\{\prod \frac{p_i}{p_i - 1} r_i : \sum r_i \le n \right\}.$$ -We can bound this factor $\prod \frac{p_i}{p_i - 1}$ as follows. The primes occurring in this product are at worst the primes up to $n+1$, and I believe the asymptotic behavior of $\prod_{p_i \le n+1} \frac{p_i}{p_i - 1}$ should be $\log n$ but I don't see an extremely clean proof so I'll settle for the worse bound -$$\prod_{p_i \le n+1} \frac{p_i}{p_i - 1} \le \prod_{k=2}^{n+1} \frac{k}{k-1} = n+1$$ -which gives -$$L(n) \le \text{max} \left\{ (n+1) \prod r_i : \sum r_i \le n \right\}.$$ -We can now relax this optimization problem so that the $r_i$ can take real values, and then a standard Lagrange multiplier argument shows that, for any number $k$ of terms (which we've left unspecified), we want to take $r_i = r$ for some fixed $r$. This gives -$$L(n) \le \text{max} \left\{ (n+1) r^k : kr \le n, k \in \mathbb{N}, r \in \mathbb{R} \right\}$$ -and if we further relax $k$ to a real number then a standard calculus argument gives $r = e, k = \frac{n}{e}$, so -$$\boxed{ L(n) \le (n+1) \exp \left( \frac{n}{e} \right) }$$ -exactly paralleling the analogous but slightly simpler argument for Landau's function which gives $g(n) \le \exp \left( \frac{n}{e} \right)$. I would guess that in fact like $g(n)$ we should also have $\log L(n) \sim \sqrt{n \log n}$. I think the starting point is that the relaxation we used above is very inaccurate for large primes and for $p$ such that $(p-1)p > n$ the corresponding exponent is at most $1$.<|endoftext|> -TITLE: Decomposing a (co)limit by decomposing the indexing diagram -QUESTION [9 upvotes]: Let $D: I \to \mathcal C$ be a diagram, and suppose we have a colimit decomposition $I = \varinjlim_{j \in J} I_j$ in $Cat$. Then under certain conditions, we can decompose the colimit of $D$ as $\varinjlim_{i \in I} D_i = \varinjlim_{j \in J} \varinjlim_{i \in I_j} D_i$. But I've never seen general conditions along these lines spelled out for 1-categories. -Question 1: Is there some place where conditions making the above true are given in the 1-categorical setting? -For $\infty$-categories, there is Corollary 4.2.3.10 of Higher Topos Theory. Unfortunately, the formulation of the result is somewhat abstruse, being expressed in terms of the bespoke simplicial set denoted $K_F$ there (defined using 4 conditions in Notation 4.2.3.1). -As a result, I'm having the following problem: it seems to me that for any cocone of $\infty$-categories $(I_j \to I)_{j \in J}$, one should be able to construct a natural map $\varinjlim_{j \in J} \varinjlim_{i \in I_j} D_i \to \varinjlim_{i \in I} D_i$, and one would expect HTT 4.2.3.10 to imply that under the appropriate conditions, this map is an equivalence. But the formulation doesn't seem to easily lend itself to confirming this. -Question 2: Is the natural map $\varinjlim_{j \in J} \varinjlim_{i \in I_j} D_i \to \varinjlim_{i \in I} D_i$ constructed somewhere in reasonable generality? (Or else is it easy to construct from general machinery given somewhere?) -Question 3: Is there written somewhere an account of conditions (perhaps analogous to those of HTT 4.2.3.10) which ensure that this map is an equivalence? - -REPLY [9 votes]: Let $p \colon E \to J$ be the cocartesian fibration for the diagram $j \mapsto I_j$. Then the colimit over $E$ of $F \colon E \to C$ can always (assuming the appropriate colimits exist in $C$) be written as an iterated colimit: -$$ \mathrm{colim}_E \, F \simeq \mathrm{colim}_J \, p_! F \simeq \mathrm{colim}_{j \in J} \, \mathrm{colim}_{I_j} \, F|_{I_j} $$ -by first doing the colimit in two steps using the left Kan extension along $p$ and then that the inclusion $E_j \to E \times_J J_{/j}$ is cofinal since $p$ is cocartesian. -Now the colimit $I$ can be described as the localization of $E$ at the cocartesian morphisms. Since any localization is cofinal, this means there is a cofinal functor $q \colon E \to I$. For a functor $D \colon I \to C$, this means we have equivalences -$$ \mathrm{colim}_I \, D \simeq \mathrm{colim}_E \, Dq \simeq \mathrm{colim}_{j \in J} \, \mathrm{colim}_{I_j} \,D|_{I_j}. $$ - -REPLY [5 votes]: The way I always remember this stuff is as follows: - -Given a map $J \to \mathsf{Cat}$ form the associated cocartesian fibration $E \to J$. -By assumption, $I$ is the actual colimit (as opposed to the left lax one) so we have a (weak) localization $E \to I$. Weak localizations are final (and initial, in fact) so, to compute the colimit over $I$ is the same as computing it over $E$. -To compute the colimit over $E$ we may first left Kan extend to $J$. -Since $E \to J$ is cocartesian, the map $E_x \to E_{/x}$ is final, and we may replace $E_{/x}$ with $E_x=I_x$ in the formula for left Kan extensions. - -That gives the result. - -REPLY [4 votes]: I assume $\varinjlim_{j : \mathcal{J}} \mathcal{I}_j = \mathcal{I}$ is meant in the strict sense of 1-categories. Since $\textbf{Cat}$ is cartesian closed, -$$\textstyle [\mathcal{I}, \mathcal{C}] \cong \varprojlim_{j : \mathcal{J}} [\mathcal{I}_j, \mathcal{C}]$$ -where the limit on the RHS is also meant in the strict sense of 1-categories. Let $\lambda_j : \mathcal{I} j \to \mathcal{I}$ be the component of the colimit cocone in $\textbf{Cat}$. Then, we also get a limit formula for the hom-sets of $[\mathcal{I}, \mathcal{C}]$, namely, -$$\textstyle [\mathcal{I}, \mathcal{C}](D, \Delta T) \cong \varprojlim_{j : \mathcal{J}} [\mathcal{I}_j, \mathcal{C}](D \lambda_j, \Delta T)$$ -so if the relevant colimits exist in $\mathcal{C}$, -$$\textstyle \mathcal{C} \left( \varinjlim_\mathcal{I} D, T \right) \cong \varprojlim_{j : \mathcal{J}} \mathcal{C} \left( \varinjlim_{\mathcal{I}_j} D \lambda_j, T \right) \cong \mathcal{C} \left( \varinjlim_\mathcal{J} \varinjlim_{\mathcal{I}_j} D \lambda_j, T \right)$$ -as desired.<|endoftext|> -TITLE: Non-sequential spaces in the wild -QUESTION [11 upvotes]: TLDR: What are examples of (function-)spaces that are not sequential? When does this matter? -As a simple analyst, I am most happy if I can just work with sequences all the time. In most situations this is totally fine, as many many spaces one encounters in one's daily life are actually sequential (or even 1st countable, or even better metrisable). Now recently I was a bit shocked to find out that the seemingly familiar space of test-functions $\mathscr{D}(\mathbb{R}^d)$ (with its usual LF-topology) actually fails to be sequential. But hadn't I learned that one can verify whether a linear functional on $\mathscr{D}(\mathbb{R}^d)$ is a distribution by checking continuity with sequences? Well in that case it is true (Proposition 21.1 in Trèves TVS book), but only because we looked at linear functionals. -This got me thinking that there might actually be a bunch of spaces around, not pathological counterexamples, but real spaces one encounters in the wild, that fail to be sequential. In some cases, like above, this might not be problematic, but potentially for non-trivial reasons. In order to become more aware of these subtleties I would like to collect some more examples of this. -An answer should ideally contain the following: - -A concrete example or a class of examples of non-sequential spaces, which are widely used or naturally show up in analysis. My main interested lies in topological vector spaces that appear as function spaces in some context. The example should not be some 'patholocial counterexample' (this is of course a bit vague). -An instance of where it matters that the space is non-sequential. Or a warning, of when one needs to be more careful with it and use filters or nets. -Loop holes or special situations where it suffices to focus on sequences nevertheless. - -I'll make a start: - -Test-functions: $\mathscr{D}(\mathbb{R}^d)$ (with standard LF-topology) is not sequential. In particular a function $f:\mathscr{D}(\mathbb{R}^d)\rightarrow \mathbb{C}$ might be sequentually continuous, but not continuous (example by PhoemueX). However, if $f$ is linear then sequential continuity implies continuity (Corollary after Proposition 13.1 in Trèves' TVS book). The same is true for other LF-spaces. -Distributions: $\mathscr{D}'(\mathbb{R}^d)$ (with the strong topology) is not sequential. A sequence of distributions converges strongly if and only if it converges weakly, but this is not true when sequences are replaced by nets/filters. The same result (for sequences) holds in strong duals of Montel spaces (Corollary 1 to Proposition 34.6 in Trèves) -An infinite dimensional Banach space, equipped with the weak topology is not sequential. However, despite this we have that compactness = sequential compactness (Eberlein-Smulian theorem). - -Finally, here are some spaces that are sequential: $\mathscr{S}(\mathbb{R}^d)$ (Schwartz-space), $\mathscr{D}(M)$ (distributions on compact manifold $M$), the dual of a separable locally convex space with the weak$^*$-topology, ... - -REPLY [3 votes]: The sequentiality does not match well with an algebraic structure. For example, the following result of Banakh and Zdomskyy characterizes sequential topological groups with countable $cs^*$-character: - -Theorem. A topological group $G$ with countable $cs^*$-character is sequential if and only if $G$ is either metrizable or contains an open $\mathcal M\mathcal K_\omega$-subgroup. - -Let us recall that a topological space $X$ has countable $cs^*$-character if for every point $x\in X$ there exists a countable family $\mathcal F_x$ of subsets of $X$ such that for every neighborhood $O_x\subseteq X$ of $x$ and every sequence $\{x_n\}_{n\in\omega}\subseteq X$ that converges to $x$, there exist a set $F\in\mathcal F_x$ such that $F\subseteq O_x$ and $F$ contains infinitely many points of the sequence $(x_n)$. -A topological space $X$ is $\mathcal{MK}_\omega$ if there exists a countable cover $\mathcal C$ of $X$ by compact metrizable subspaces such that a subset $F\subseteq X$ is closed if and only if for every compact set $C\in\mathcal C$ the intersection $C\cap F$ is closed in $C$. -In this paper of Banakh and Repovs the above result of Banakh--Zdomskyy was extended to rectifiable spaces and topological left-loops. -In fact, the above theorem, is a corollary of the following result of Banakh: - -Theorem. If a perfectly normal topological group $G$ contains a topological copy of the Frechet-Urysohn fan $S_\omega$ and a closed topological copy of the metric fan $M$, then $G$ is not sequential. - -The metric fan is the subspace $$M=\{0\}\cup\{\tfrac1{n}+\tfrac{i}{nm}:n,m\in\mathbb N\}$$ of the complex plane. The Fr'echet-Urysohn fan is the set $M$ endowed with the strongest topology that coincides with the Euclidean topology on each subspace $K_m=\{0\}\cup\{\frac1{n}+\tfrac{i}{nm}:n\in\mathbb N\}$, $m\in\mathbb N$. It is easy to see that the Fr'echet-Urysohn fan is an $\mathcal{MK}_\omega$-space.<|endoftext|> -TITLE: loop space of a finite CW-complex -QUESTION [5 upvotes]: Let $X$ be a finite connected pointed CW-complex and $H_{\ast}(\Omega X)$ the integral homology of the loop space on $X$. Are the homology groups $H_{n}(\Omega X)$ finitely generated abelian groups for any $n$ ? -If the answer is negative, what are the sufficient conditions to impose on $\pi_{1}(X)$ such that the homology groups $H_{n}(\Omega X)$ turns out to be finitely generated ? -My goal is to collect different sufficient conditions on the fundamental group for which a positive answer holds. - -REPLY [14 votes]: This is true for finite $\pi_1$ and false for infinite $\pi_1$: Let $\widetilde{X}$ denote the universal cover of $X$, then $\Omega\widetilde{X}$ is the unit connected component of $\Omega X$, and $\Omega X = \coprod_{\pi_1(X)} \Omega\widetilde{X}$. So if $\pi_1$ is infinite, then certainly $H_0(\Omega X)$ is not finitely generated as others have noted in the comments, and indeed if $\Omega\widetilde{X}$ has any nontrivial homology group (which is true unless $\widetilde{X}$ is contractible), some higher homology group of $\Omega X$ will be an infinite direct sum of nontrivial abelian groups, so also not finitely generated. -If $\pi_1$ is finite, on the other hand, $\widetilde{X}$ is again a finite CW complex, so in that case it suffices to look at the simply-connected case. For a simply-connected finite CW-complex $X$, $H_*(\Omega X)$ indeed consists of finitely generated abelian groups, which goes back to Serre (and is proved easily using the spectral sequence named after him).<|endoftext|> -TITLE: Erroneous Wolfram result for $\sum_{k=1}^\infty (k^3 + a^3)^{-1}$, looking for correct formula -QUESTION [9 upvotes]: I was trying to get some interesting result for $\zeta(3)$, exploring the following function: -$$W(a) = \sum_{k=1}^\infty \frac{1}{k^3 + a^3}, \mbox{ with } \lim_{a\rightarrow 0} W(a) = \zeta(3).$$ -Let $w_1, w_2, w_3$ be the three roots (one real, two complex) of $(w+1)^3+a^3=0$, with $w_1=-(a+1)$. Also, $a$ is a real number. Using Wolfram Alpha (see here), I get -$$W(a)=\frac{-1}{3}\cdot\sum_{j=1}^3 W_j(a), \mbox{ with } W_j(a) = \frac{\psi^{(0)}(-w_j)}{(w_j+1)^2}.$$ -Here $\psi^{(0)}$ is the digamma function. The result is wrong because $W_1(a) \rightarrow \infty$ as $a\rightarrow 0^+$ while $W_2(a)$ and $W_3(a)$ remain bounded. Indeed using $a=0.0001$, Wolfram yields $W(a)\approx -2334.16$, see here. Surprising, with $a=0.01$ it yields $W(a)\approx 1.20206$ which is very close to the true result. -Surprisingly, Wolfram knows (see here) that -$$\lim_{a\rightarrow 0} W(a) = -\frac{\psi^{(2)}(1)}{2}.$$ -Of course (this is a well known fact), $\zeta(3)=-\psi^{(2)}(1)/2$ and thus Wolfram is correct this time. -My question: -What is going on with this computation (or is it me?), and what is the correct formula for $W(a)$? -Update -See the two answers below proving that I was wrong, and that the Mathematica formula I though was incorrect, is indeed right. Kudos Mathematica! You were successful at solving a nice problem involving a few challenging steps, and coming with a somewhat unexpected but neat formula involving derivatives of the digamma function instead of the classic $\zeta(3)$. -Final note -It is possible to use a different, simpler approach that does not involve complex numbers. Consider -$$V(a) =\sum_{k=1}^\infty (-1)^{k+1}\frac{1}{k(k^2-a^2)}.$$ -Wolfram is able to compute the limit of $V(a)$ as $a\rightarrow 0$, and returns the correct value $3\zeta(3)/4$, see here. It is easy to establish that -$$V(a)=\frac{1}{a^2} \Big[\int_0^\infty \frac{\cosh(ax)}{1+e^x} dx -\log 2\Big].$$ -To compute $\lim_{a\rightarrow 0} V(a)$, we apply L'Hospital Rule twice to the above expression, the denominator in this case being $a^2$. This yields -$$\lim_{a\rightarrow 0}V(a) = \frac{1}{2}\lim_{a\rightarrow 0}\int_0^\infty \frac{x^2\cosh(ax)}{1+e^x}dx =\frac{1}{2}\int_0^\infty \frac{x^2}{1+e^x}dx=\frac{3\zeta(3)}{4}.$$ -Here, we assume $a<1$. - -REPLY [10 votes]: I think the statement in the OP that $W_2(a)$ and $W_3(a)$ remain bounded when $a\rightarrow 0$ is mistaken, so that there is no inconsistency with the Mathematica result. -The three roots of $(w+1)^3+a^3=0$ are -$$w_1= -a-1,\;\; w_2= \tfrac{1}{2} \left(-i \sqrt{3} a+a-2\right),\;\;w_3= \tfrac{1}{2} \left(i \sqrt{3} a+a-2\right).$$ -Then the denominator $(w+1)^2$ vanishes for all three roots when $a\rightarrow 0$, while the numerator remains finite (equal to $-\gamma_{\rm Euler}$). -And indeed, a numerical check suggets that the Mathematica output is actually correct, and the erroneous numerical result for small $a$ is a numerical instability in the computation of the digamma function. See these two plots that compare the digamma expression (blue) with a numerical evaluation of the sum (gold), as a function of $a$. For $a\gtrsim 0.01$ the two answers are nearly indistinguishable. - -REPLY [8 votes]: We have the partial fraction decomposition -$$\frac{ca^2}{k^3+a^3}=\frac{-\omega}{k-a/\omega }+\frac{\omega -1}{k+a}+\frac{1}{k-a \omega},$$ -where $c:=3(\omega-1)$ and $\omega:=e^{i\pi/3}$. -Also, -$$\sum_{k=1}^n\frac1{k+b}=\ln n-\psi(1+b)+o(1)$$ -(as $n\to\infty$), where $\psi$ is the digamma function. -Collecting the pieces, for $a\in(-1,\infty)\setminus\{0\}$ we get -$$s(a):=\sum_{k=1}^\infty\frac1{k^3+a^3} -=\frac1{ca^2}\, -\left((1-\omega) \psi(1+a)+\omega\psi\left(1-a/\omega\right) --\psi(1-a \omega)\right).$$ -For $a\to0$, -$$s(a)=-\frac{\psi ^{(2)}(1)}{2}-\frac{\pi ^6 a^3}{945}+O\left(a^4\right) -=\zeta(3)-\frac{\pi ^6 a^3}{945}+O\left(a^4\right).$$ -Here is the graph $\{(a,s(a))\colon0 -TITLE: What are the eigenvalues of the curvature operator on $\mathbb{C}P(2)$? -QUESTION [6 upvotes]: I tried asking this on stackexchange but was unsuccessful. -On page 150 of section 4.5.3 of Peter Petersen's Riemannian Geometry it is noted that, given an orthonormal basis $X,iX,Y,iY$ for $T_p\mathbb{C}P^2$, the following basis diagonalizes the curvature operator $\mathfrak{R}:\Lambda^2T_p\mathbb{C}P^2 \to \Lambda^2T_p\mathbb{C}P^2 $: -\begin{align*} -&X \wedge iX \pm Y \wedge iY\\ -& X \wedge Y \pm iX \wedge iY \\ -& X \wedge iY \pm Y \wedge iX -\end{align*} -with eigenvalues lying in $[0,6]$. I have attempted the calculations using the O'Neill formula but get stuck. For example assume $\mathfrak{R}(X \wedge iX + Y \wedge iY)=c \left(X \wedge iX + Y \wedge iY\right)$. Then: -\begin{align*} -g(\mathfrak{R}(X \wedge iX + Y \wedge iY),X \wedge iX + Y \wedge iY)&=cg(X \wedge iX + Y \wedge iY,X \wedge iX + Y \wedge iY)\\ &=2c -\end{align*} -and by the definition of $\mathfrak{R}$: -\begin{align*} -&g(\mathfrak{R}(X \wedge iX + Y \wedge iY),X \wedge iX + Y \wedge iY) \\ &=R(X,iX,iX,X)+R(Y,iY,iY,Y)+2R(X,iX,iY,Y) \\ &=\sec(X,iX)+\sec(Y,iY)+2R(X,iX,iY,Y) \\ &=8+2R(X,iX,iY,Y) -\end{align*} -so: -\begin{align*}2c &= 8+2R(X,iX,iY,Y) \\ c&=4+R(X,iX,iY,Y) \end{align*} -At this point I do not know how to proceed. I know of a formula for expanding $R(X,iX,iY,Y)$ in terms of sectional curvatures but it is quite complicated. Alternatively, going back to the O'Neill formula we have: -\begin{align*}R(X,iX,iY,Y)=&\overline{R}(\overline{X},\overline{iX},\overline{iY},\overline{Y})+\frac14 \overline{g} ([\overline{iX},\overline{Y} ],[\overline{X},\overline{iY}])-\frac14 \overline{g}([\overline{X},\overline{Y}],[\overline{iX},\overline{iY}]) \\ &+ \frac12 \overline{g} ([ \overline{X},\overline{iX}],[\overline{iY},\overline{Y}] )\end{align*} -where $\overline{R}$ denotes the curvature tensor on $S^5$, $\overline{g}$ denotes the metric on $S^5$, and $\overline{V}$ denotes a horizontal lift. Since $X,iX,Y,iY$ are orthonormal and their lifts are also orthonormal I suppose $\overline{R}(\overline{X},\overline{iX},\overline{iY},\overline{Y})=0$ so we are left with -\begin{align*}R(X,iX,iY,Y)=&\frac14 \overline{g} ([\overline{iX},\overline{Y} ],[\overline{X},\overline{iY}])-\frac14 \overline{g}([\overline{X},\overline{Y}],[\overline{iX},\overline{iY}]) + \frac12 \overline{g} ([ \overline{X},\overline{iX}],[\overline{iY},\overline{Y}] )\end{align*} -Have I missed something that should make this easier to work out? -edit: I followed through with the first idea (expanding out $R(X,iX,iY,Y)$ in terms of sectional curvatures) and got that the eigenvalues are $0,0,2,2,2,6$. Given that $\mathbb{C}P(2)$ is Einstein with Einstein constant $6$ and Scalar curvature $12$, I am tempted to believe that the eigenvalues are correct (as their sum does equal $12$). - -REPLY [6 votes]: Perhaps, it would be easier to just compute it directly from the structure equations. For example, suppose we wanted to compute the eigenvalues of the curvature operator for $\mathbb{CP}^n=\mathrm{SU}(n{+}1)/\mathrm{U}(n)$. I claim that they are $0$ with multiplicity $n(n{-}1)$, $2$ with multiplicity $n^2{-}1$, and $2(n{+}1)$ with multiplicity $1$. Here's how to see this: Write the left-invariant form on $\mathrm{SU}(n+1)$ as -$$ -\lambda = \begin{pmatrix} -i\,n\,\rho & -^t\omega+i\,^t\eta\\ -\omega+i\,\eta &\alpha + i\rho\,I_n + i\,\beta\end{pmatrix} -$$ -where $\omega$ and $\eta$ are columns of height $n$, $\alpha = -\ ^t\alpha$ and $\beta = {}^t\beta$ while $\mathrm{tr}(\beta)=0$. The pullback of the metric on $\mathbb{CP}^n$ to $\mathrm{SU}(n{+}1)$ is the quadratic form $g = \ ^t\omega\circ\omega + ^t\eta\circ\eta$. The Maurer-Cartan equation $\mathrm{d}\lambda = -\lambda\wedge\lambda$ unpacks to -$$ -\mathrm{d}\begin{pmatrix}\omega\\ \eta\end{pmatrix} -=-\begin{pmatrix}\alpha & -(n{+}1)\rho I_n-\beta \\ - (n{+}1)\rho I_n+\beta & \alpha -\end{pmatrix}\wedge \begin{pmatrix}\omega \\ \eta\end{pmatrix} -$$ -together with the equations -$$ -\begin{align} -\mathrm{d}\rho &= -2/n\,{}^t\omega\wedge\eta\\ -\mathrm{d}\alpha + \alpha\wedge\alpha - \beta\wedge\beta -&= \omega\wedge {}^t\omega + \eta \wedge {}^t\eta\\ -\mathrm{d}\beta +\beta\wedge\alpha-\alpha\wedge\beta -&= \eta\wedge {}^t\omega- \omega\wedge {}^t\eta + (2/n) {}^t\omega\wedge\eta\, I_n\,. -\end{align} -$$ -Consequently, the matrix -$$ -\theta = \begin{pmatrix}\alpha & -(n{+}1)\rho I_n-\beta \\ - (n{+}1)\rho I_n+\beta & \alpha -\end{pmatrix} -$$ -is the Levi-Civita connection matrix for the metric $g$, and its curvature is -$$ -\begin{align} -\Theta &= \mathrm{d}\theta+\theta\wedge\theta\\ -&= \begin{pmatrix}\omega\wedge {}^t\omega + \eta \wedge {}^t\eta -& 2\,{}^t\omega\wedge\eta\,I_n -\eta\wedge {}^t\omega+\omega\wedge {}^t\eta \\ - -2\,{}^t\omega\wedge\eta\,I_n +\eta\wedge {}^t\omega- \omega\wedge {}^t\eta -& \omega\wedge {}^t\omega + \eta \wedge {}^t\eta -\end{pmatrix}. -\end{align} -$$ -Now, $\theta$ (and hence $\Theta$) takes values in the subalgebra -${\frak{u}}(n)\subset{\frak{so}}(2n)\simeq\Lambda^2(\mathbb{R}^{2n})$, -where ${\frak{u}}(n) = {\mathbb{R}}\cdot J + {\frak{su}}(n)$ with $J = \begin{pmatrix} 0_n & I_n\\ -I_n & 0_n\end{pmatrix}$ and ${\frak{su}}(n)$ being the skew-symmetric matrices of the form $\begin{pmatrix} a & b\\ -b & a\end{pmatrix}$ where $a$ is $n$-by-$n$ skew-symmetric and $b$ is $n$-by-$n$ symmetric and tracefree. It follows that the curvature operator $R$ (which is symmetric) annihilates everything in ${\frak{u}}(n)^\perp\subset {\frak{so}}(2n)$, a vector space of dimension $n(n{-}1)$, so these kernel eigenvectors all have eigenvalue $0$. Taking the $J$-trace of $\Theta$, we see that $R$ has $J$ as an eigenvector of eigenvalue $2(n{+}1)$. The subspace ${\frak{su}}(n)$ is irreducible and must be mapped by $R$ to a scalar multiple of itself, and pairing with a typical element, we see that the multiplier is $2$. Thus, $2$ is an eigenvalue of multiplicity $n^2{-}1$, the dimension of ${\frak{su}}(n)$.<|endoftext|> -TITLE: What are some foundational authors/papers in dynamical systems? -QUESTION [15 upvotes]: I have just begun my first dynamical systems class, and I would like to try out the advice in the top answer here. To summarize, the answer suggests that when studying a new field, one should look at the original papers, and supplement it with modern material. -My class is roughly following Introduction to Dynamical Systems by Brin and Stuck. We are starting in the second chapter, which is about topological dynamics (it is only 20 pages, but very dense). I was wondering if you could please recommend me some authors/papers which started the field? -Thank you very much. - -REPLY [2 votes]: Katok / Hasselblatt: Introduction to the Modern Theory of Dynamical Systems gives a good introduction.<|endoftext|> -TITLE: Replication crisis in mathematics -QUESTION [63 upvotes]: Lately, I have been learning about the replication crisis, see How Fraud, Bias, Negligence, and Hype Undermine the Search for Truth (good YouTube video) — by Michael Shermer and Stuart Ritchie. According to Wikipedia, the replication crisis (also known as the replicability crisis or reproducibility crisis) is - -an ongoing methodological crisis in which it has been found that many -scientific studies are difficult or impossible to replicate or -reproduce. The replication crisis affects the social sciences and -medicine most severely. - -Has the replication crisis impacted (pure) mathematics, or is mathematics unaffected? How should results in mathematics be reproduced? How can complicated proofs be replicated, given that so few people are able to understand them to begin with? - -REPLY [14 votes]: Has this crisis impacted (pure) mathematics, or do you believe that -maths is mostly immune to it? - -Immune to the replication problem, yes. But not immune to the attitudes which cause scientists to do unreplicable research in the first place. Some mathematicians will announce that a particular theorem has been proven, harvest the glory based on the fact that they have proved things in the past, and then never publish their results. Rota's Conjecture is one notorious example. Now we are in a situation where (a) nobody knows whether it is true and (b) nobody has worked on it for seven years, and probably (if it turns out that no proof actually exists) will not work on it for at least another decade. - -How should results in mathematics be reproduced? - -In science, it would be ideal if people dedicated research time to replicating published experimental results. This doesn't happen much because there is no glory to be gained by doing it. -The analogue in mathematics would be for people to publish new proofs of existing results, or expositions of existing proofs, which is happily much more common. I don't mean copying out well-known results in new language (Tom Leinster, The bijection between projective indecomposable and simple modules), I mean expository papers like this (Cao and Zhu, A complete proof of the Poincaré and geometrization conjectures, Asian J. Math. 10 (2006) pp. 165–492). -Even more noble are the people using proof assistant software to verify existing mathematics. - -How can we expect that increasingly complicated proofs are replicated -when so few people can understand them in the first place? - -I think our best hope is proof assistant software. Perhaps by the end of this century, we will he living in a world where no mathematician can replicate any reasonably cutting-edge proof, yet research is still happily chugging along.<|endoftext|> -TITLE: $π(x+y) - π(x) ≤ c·y/\ln(y)$ for some constant $c$? -QUESTION [9 upvotes]: (I posted this question on Math SE but it has had no answer for a year now so I would like to ask if anyone here can provide one.) -Thinking about the prime number theorem, I wondered whether it is known that there is some constant $c$ such that $π(x+y) ≤ π(x) + c·y/\ln(y)$ for every integers $x,y > 1$. I read that experts believe $π(x+y) ≤ π(x) + π(y)$ fails for some $y$, since it fails for $y = 3159$ if the k-tuple conjecture holds, but it is just barely false, so I am curious if it is known to be true if the inequality is relaxed by a constant factor. If so, is it also known that $π(x+y) ≤ π(x) + π(y) + c·\!\sqrt{y}·\ln(y)$ for some constant $c$? I simply do not know how to search for such conjectures, and neither Wikipedia nor Wolfram seem to state any results that would affirm or refute these two conjectures easily, so any references would be appreciated! - -REPLY [15 votes]: As mentioned in my comment, Montgomery and Vaughan (The Large Sieve, Mathematika 20 (1973) 119–132, doi:10.1112/S0025579300004708) showed an explicit version of the Brun--Titchmarsh inequality: -$$ -\pi(x+y) - \pi(x) \le \frac{2y}{\log y}. -$$ -The other question asked is likely false. Hensley and Richards (Primes in intervals, Acta Arithmetica 25 (1974) 375-391, EuDML) showed that the Hardy-Littlewood $k$-tuples conjecture contradicts the hypothesis that $\pi(x+y) -\pi(x) \le \pi(y)$. In fact their paper establishes that (on the $k$-tuples conjecture) one has (for large fixed $y$, and infinitely many $x$) -$$ -\pi(x+y) -\pi(x) \ge \pi(y) + (\log 2 -\epsilon) \frac{y}{(\log y)^2}. -$$ -Roughly the lower bound that they are obtaining is $2 \pi(y/2)$ (and some conjectural improvements over this are also discussed).<|endoftext|> -TITLE: Can prolates overlap more easily than oblates? -QUESTION [7 upvotes]: Context: -When modeling anisotropic particles, the two common types of shapes of interest are cylindrical and disk-like particles. For simplicity let us say we model these as prolates and oblates respectively (see below). - -With increasing aspect ratio, the oblate becomes more and more disk-like, while the prolate becomes needle-like (either way, as anisotropic spheroids, both have a positional and orientational degree of freedom). -When one is interested in the connectivity properties of configurations of such particles, where the connectivity is justified on physical grounds, given the core of particles is hard (they cannot overlap or go through one another), connectivity criteria are defined as follows: - -For any given particle geometry, the particle is coated by a contact shell of the same shape that covers the hard core. This shell may be very thin, but it allows us to establish a notion of connectivity, as in, when two adjacent particles have overlapping contact shells, we assume them to be connected. - - -Questions: - -I am trying to find out, whether purely on geometric grounds, that is by solely considering the difference in shape of an oblate to a prolate, it is possible to tell which type provides a more efficient connectivity. In other words, is it easier (more probable) for two prolates to become connected or for two oblates? Instead of "probability", one can imagine it in terms of "the number of ways two such given particles can become connected". - -If the question is vague, let us take an extreme scenario: suppose we have very high aspect ratios, that is, near slender needles (for prolates), and near disks (for oblates), in a box we throw two prolates, and once two oblates, can we tell which case is more likely to result in a configuration having overlaps? - - -I guess the question is partly also related to the packing problem of anisotropic particles, but I don't know if knowledge of how particles pack, directly tells us something about how they can connect. I understand if there's no clear-cut statement or answer that can be made here, but if you have an intuitive take on the matter, based purely on the geometry of the two particles at hand, it would be most definitely welcome as answer here. Any reference to the literature where such/similar comparison has been made would also be helpful. - -REPLY [3 votes]: This is not an answer; more an excuse to post the intriguing image below. -This Phys Rev article is quite recent, and perhaps its 46 references will help -focus your question. - -Jin, Weiwei, Ho-Kei Chan, and Zheng Zhong. "Shape-Anisotropy-Induced Ordered Packings in Cylindrical Confinement." Physical Review Letters 124, no. 24 (2020): 248002. -Journal link. - - -"Packing prolate and oblate spheroids into cylinders creates new types of packings not seen for perfect spheres."<|endoftext|> -TITLE: Representing relative homology classes orientable surfaces with boundary -QUESTION [5 upvotes]: Let $S$ be compact oriented surface without boundary. Then it is a classical result that a primitive class $\gamma \in H_1(S; \mathbb{Z})$ is always represented by a simple closed curve. It implies that any class $\beta \in H_1(S; \mathbb{Z})$ is represented by a disjoint union of simple closed curves (take $\beta = k \gamma$ with $\gamma$ primitive and consider $k$ parallel simple closed curves representing $\gamma$). -Let now $\Sigma$ be a compact oriented surface with non-empty boundary: -Is it true that I can always represent any element $\gamma \in H_1(\Sigma, \partial \Sigma; \mathbb{Z})$ by a disjoint union of simple closed curves and properly embedded arcs? - -REPLY [6 votes]: Yes, this can be done. You can do this directly for surfaces but it's as much a "codimension one" as a "dimension one" phenomenon and so useful to see the general argument. -For any compact oriented n-manifold with boundary, duality says that $H_{n-1}(M,\partial M) \cong H^1(M)$. From simple obstruction theory, $H^1(M)$ is identified with homotopy classes of maps to a circle. Assuming your manifold was smooth, take a smooth map $f$ representing your cohomology class and a regular value $p\in S^1$. (I guess I mean regular value for both $f$ and the restriction of $f$ to the boundary.) Then $f^{-1}(p)$ is a codimension one submanifold in the given homology class. -For surfaces, the submanifold is a union of properly embedded arcs and simple closed curves. If you want to do so, you can assume that every component is an arc.<|endoftext|> -TITLE: Any group is a quotient of an acyclic group? -QUESTION [12 upvotes]: As far as I know, for any group $G$ there exists an acyclic group $H$ such that $G$ is a subgroup of $H$. -I am wondering about the dual situation. Is any group $A$ a quotient of an acyclic group $B$ or more simply, given a group $A$ does it exist an acyclic group $B$ and a surjective homomorphism $B\rightarrow A$ ? - -REPLY [25 votes]: Acyclic groups must in particular have trivial abelianization, so all of their quotients must be perfect. -This is the only obstruction; A.J. Berrick shows in The acyclic group dichotomy (which I just found by googling!) that every perfect group is a quotient of an acyclic group of cohomological dimension $2$ (Proposition 2.3).<|endoftext|> -TITLE: When is a twisted form coming from a torsor trivial? -QUESTION [9 upvotes]: Consider a sheaf of groups $G$, equipped with a left torsor $P$ and another left action $G$ on some $X$. Form the contracted product $P \times^G X := (P \times X)/\sim$ where $\sim$ is the antidiagonal quotient: $(g.p, x)\sim (p, g.x)$. -Q1: When is $P\times^G X$ trivial? I.e., when do we have an isomorphism $P \times^G X \simeq X$? -Partial answer: $P \times^G X \simeq X$ over $[X/G]$ iff $P \times [X/G]$ is a trivial torsor over the stack quotient $[X/G]$. -Proof: We can rewrite $P \times^G X$ as a contracted product of two torsors $(P \times [X/G])\times^G_{[X/G]} X$. Then we contract with ``$X^{-1}$'' -- the inverse to contracting with $X$ as a torsor over $[X/G]$ and we win. (as in B. Poonen's Rational Points on Varieties, section 5.12.5.3) -Am I allowed to do this? This argument probably shouldn't have to appeal to algebraic stacks and may be somewhat dubious. -Q2: If I have one isomorphism $P \times^G X \simeq X$, can I choose another one that lies over $[X/G]$? Or at least is $G$-equivariant? -Q3: Is there a natural way to write the triviality of such a twisted form? -I first thought $P \times^G X \simeq X$ iff $P$ was trivial, which is clearly false for trivial actions on $X$. Then I was excited to have the pullback $* \to BG$ represent triviality of the twisted form $P \times^G X$ as well as the torsor $P$. Is there a natural representative of the sheaf of isomorphisms between $P \times^G X$ and $X$? -These can all be sheaves, although I'm primarily interested in $G = GL_n, PGL_n, SL_n$, etc. acting on $X = \mathbb{A}^n, \mathbb{P}^n$ as appropriate. More ambitious is $G = \text{Aut}(X)$ for even simple $X$. I'd be happy with answers in any level of generality. -Due Diligence Statement: I'm a novice in the area of "twisted forms" of varieties, so I apologize if the above is evident or obtuse. I checked all the "similar questions" listed here and couldn't find an answer. - -REPLY [2 votes]: This lemma might hold in a more general topos-theoretic context, but for "safety" I'm going to formulate it in a more restricted setting. -Lemma:Let $G$ be a group scheme with an action $\rho: G \times X \to X$ corresponding to a morphism $\varphi: G \to \mathrm{Aut}(X)$. Assume $G, \mathrm{Aut}(X)$ are smooth over a field $k$, let $S$ be a $k$-scheme and let $P$ be a $G$-torsor on $S$. Then $P \times^G X \simeq X \times S$ if and only if there is a morphism $\sigma: S \to \mathrm{Aut}(X)/G$ such that $P$ is the torsor of maps $\tilde{\sigma}: S \to \mathrm{Aut}(X)$ with $\tilde{\sigma} = \sigma \,\mathrm{mod} \, G$. -I'm requiring the groups to be smooth only so that the classifying stacks are algebraic. Rather than trying to precisely formulate and prove such a lemma here, I'll list 2 ways to think about it: -Morphism of classifying stacks version: The map $\varphi: G\to \mathrm{Aut}(X)$ induces a morphism $B\varphi: BG \to B\mathrm{Aut}(X)$, and one can check that the fiber of $B\varphi$ over the map $\gamma: S \to B\mathrm{Aut}(X)$ classifying $X$ is $S \times \mathrm{Aut}(X)/G$. -Exact sequence of cohomology groups version: At least when $G$ and $\mathrm{Aut}(X)$ are abelian, there will be an exact sequence -$$\cdots \to H^0(S,\mathrm{Aut}(X)/G) \to H^1(S, G) \to H^1(S, \mathrm{Aut}(X)) \to \cdots $$ -In general there might be a similar exact sequence ("of sets"), but I'm not confident enought with non-abelian cohomology groups to assert that.<|endoftext|> -TITLE: Kodaira–Spencer tensor of an isoperiodic deformation -QUESTION [6 upvotes]: Let $\alpha$ be a holomorphic 1-form on a curve $X$ of genus $g$, which we view as a map of sheaves $\alpha \colon T \to O$. The cokernel of this map is the structure sheaf $O_Z$ of the zero locus $Z \subset X$ of $\alpha$, which is a sum of $2g-2$ skyscraper sheaves (let zeroes $z_i$ of $\alpha$ be simple). It gives rise to the exact sequence $0 \to H^0(O) \to H^0(O_Z) \to H^1(T) \to H^1(O) \to 0$, where the range of the connecting homomorphism is the tangent space of the universal isoperiodic deformation (i. e. the deformations of complex structures on $X$ s. t. the cohomology class $[\alpha] \in H^1(X,\mathbb{C})$ still lies in the corresponding $H^{1,0}$ subspace). On the other hand, the space $H^0(O_Z)/\mathrm{const}$ has natural choice of coordinates, which integrate to the coordinate system on the base of the universal isoperiodic deformation, given by $\left\{\int_{z_0}^{z_i}\alpha\right\}_{i=1}^{2g-3}$ (called in the physics literature 'the relative periods'). -My question is the following: how to describe the Kodaira–Spencer tensor associated to a vector from $H^0(O_Z)$? I tried to write down the connecting homomorphism for the Čech complexes, but did not succeed, and did not bother myself again since it must be well-known anyway. However, I did not find a reference. - -REPLY [2 votes]: I suppose you are asking for how to write down pairing of basis elements $v_i\in H^0(O_Z)$ with $\omega\in H^0(\Lambda^1\otimes \Lambda^1)$. Here the simplest answer: locally around $z_i$ let our $\alpha=\alpha(z)dz$ and $\omega=\omega(z)dz^2$ then -$\langle \alpha, \omega\rangle=\oint\limits_{z_i} \frac{\omega dz^2}{\alpha dz}=\oint\limits_{z_i} \frac{\omega}{\alpha}dz$ -To see this let's recall that Serre duality is in fact local statement -- it is comming from Yoneda product in $\mathcal{Ext}$'s. -We want to write local pairing in Cech terms for covering of the curve with small disks $U_i$ around $z_i$ and complement $U$ to all $z_i$: -$$\mathcal{Ext}^0(O,\Lambda^1\otimes \Lambda^1)\otimes \mathcal{Ext}^0(O,O_Z)\xrightarrow{1\otimes\delta}$$ -$$\mathcal{Ext}^0(O,\Lambda^1\otimes \Lambda^1)\otimes \mathcal{Ext}^1(\Lambda^1,O)\xrightarrow{\cup} \mathcal{Ext}^1(O,\Lambda^1)$$ -The first is given by connecting homomorphism. By inspecting Cech double-complex for $T\to O\to O_Z$ and down-to-earth definition of $\delta$ we obtain that $\delta v_i$ is given by $\frac{1}{\alpha}\partial_z\in T(U_i\cap U)$ as Cech cocyle. -The second map is given by cup-product of Cech cocycles, so it is $\frac{\omega(z)}{\alpha}dz$ on $U_i\cap U$. -Finally we have explicit Cech cocycle as element of $\Lambda^1(U_i\cap U)$ and we need to evaluate $\check{H}^1(\Lambda)\xrightarrow{\int\limits_{C}}\mathbb{C}$. Well, it's given by residue and could be found in general treatment somewhere in Griffits&Harris book. For our curve $C$ it's simple -- take double-complex Cech-Dolbeau resolution of the constant sheaf $\mathbb{C}$, introduce a partition of unity $\rho,\rho_i$ subordinate to $U,U_i$. Then, starting from $a=\Lambda^1(U_i\cap U),\bar{\partial}a=0$ construct by gluing cohomologous element in $\Lambda^{1,1}(C)$ as $W=\bar{\partial}(\rho a)-\bar{\partial}(\rho_i a)\in \Lambda^{1,1}(C)$, by simple complex analysys we obtain that $\int\limits_{C}W= -\lim\limits_{\varepsilon\to 0} \int\limits_{C-D_{i,\varepsilon}}W= -\lim\limits_{\varepsilon\to 0} -\int\limits_{C-D_{i,\varepsilon}}-\bar{\partial}(\rho_i a)= -\lim\limits_{\varepsilon\to 0} -\oint\limits_{\partial D_{i,\varepsilon}}\rho_i a= -2\pi i\ res_{z_i}(a)$ -Here I denote by $D_{i,\varepsilon}$ small disk around $z_i$<|endoftext|> -TITLE: How to calculate inverse of sum of two Kronecker products with specific form efficiently? -QUESTION [5 upvotes]: I have a matrix with specific form of $A\otimes I + B\otimes J$ where $A$ and $B$ are general dense matrices, $n\times n$. $I$ is an $m\times m$ identity matrix. $J$ is a $m \times m$ dense matrix with 1 everywhere. -Is there any efficient way to calculate $(A\otimes I + B \otimes J)^{-1}$ efficiently? -If it is impossible, what if we assume $B$ is diagonal? - -REPLY [2 votes]: In more generality, we can write down an "efficient" inverse for both -$$A_1 \otimes A_2 + B_1 \otimes B_2 \quad \text{and} \quad A_1 \otimes B_2 + B_1 \otimes A_2$$ -whenever $B_1$ and $B_2$ are nonsingular. The matrix in the question is of the fist form with $A_1 = B$, $A_2=J$, $B_1 = A$, $B_2 = I$ (since, as noted by Denis Serre in his answer, the matrix can only be nonsingular when $A$ is). The idea is to simultaneously diagonalize each pair $A_i, B_i$ independently, using the generalized eigendecompositions, -$$A_i V_i = B_i V_i \Lambda_i$$ -where $V_i$ is the nonsingular matrix of eigenvectors and $\Lambda_i$ the diagonal matrix of eigenvalues. It's then immediate that the "generalized" Kronecker sums have generalized eigendecompositions -$$(A_1 \otimes A_2 + B_1 \otimes B_2)(V_1 \otimes V_2) = (B_1 \otimes B_2)(V_1 \otimes V_2)(\Lambda_1 \otimes \Lambda_2 + I \otimes I),$$ -$$(A_1 \otimes B_2 + B_1 \otimes A_2)(V_1 \otimes V_2) = (B_1 \otimes B_2)(V_1 \otimes V_2)(\Lambda_1 \otimes I + I \otimes \Lambda_2),$$ -whence -$$(A_1 \otimes A_2 + B_1 \otimes B_2)^{-1} = (V_1 \otimes V_2)(\Lambda_1 \otimes \Lambda_2 + I \otimes I)^{-1} (V_1^{-1}B_1^{-1} \otimes V_2^{-1}B_2^{-1}),$$ -$$(A_1 \otimes B_2 + B_1 \otimes A_2)^{-1} = (V_1 \otimes V_2)(\Lambda_1 \otimes I + I \otimes \Lambda_2)^{-1} (V_1^{-1}B_1^{-1} \otimes V_2^{-1}B_2^{-1}),$$ -provided none of the eigenvalues are zero. This simplifies further if one or both of the matrix pencils has more structure. E.g., if $A_2$ is symmetric and $B_2$ symmetric positive definite, then we can take $V_2$ such that $V_2^T B_2 V_2 = I$ so that $V_2^{-1} B_2^{-1} = V_2^T$. This is the case for the matrix in the question with $A_2 = J$, $B_2 = I$. -In the context of numerical computation, if the matrices are not symmetric and either matrix of eigenvectors $V_i$ is ill-conditioned, it may be better to work with the QZ decomposition (generalizing the Schur decomposition) instead, which has a similar tensor product structure. If $A_i, B_i$ have QZ decompositions -$$ A_i = Q_i S_i Z_i^*, \quad B_i = Q_i T_i Z_i^*, $$ -where $Q_i, Z_i$ are unitary and $S_i, T_i$ triangular, then we also have the QZ decomposition -$$ (A_1 \otimes A_2 + B_1 \otimes B_2) = (Q_1 \otimes Q_2) (S_1 \otimes S_1 + T_1 \otimes T_2) (Z_1 \otimes Z_2)^*$$ -$$ B_1 \otimes B_2 = (Q_1 \otimes Q_2) (T_1 \otimes T_2) (Z_1 \otimes Z_2)^*$$ -since the Kronecker product of triangular (unitary) matrices is triangular (unitary). Thus -$$ (A_1 \otimes A_2 + B_1 \otimes B_2)^{-1} = (Z_1 \otimes Z_2) (S_1 \otimes S_2 + T_1 \otimes T_2)^{-1} (Q_1 \otimes Q_2)^*,$$ -requiring the inverse of a large triangular instead of diagonal matrix. This approach works even when one or both $B_i$ are singular, requiring only that the pencils $A_i - \lambda B_i$ are regular (and again that no eigenvalues, which appear along the diagonal of $S_1 \otimes S_2 + T_1 \otimes T_2$, are zero).<|endoftext|> -TITLE: Good forcings with bad squares -QUESTION [15 upvotes]: Consistently with $\mathsf{ZFC}$ there is a forcing which preserves cardinals but whose square does not always preserve cardinals - that is, some $\mathbb{P}$ such that for every $\mathbb{P}$-generic $G$ we have $\mathrm{Card}^{V}=\mathrm{Card}^{V[G]}$ but for some $\mathbb{P}^2$-generic $H=\langle H_0,H_1\rangle$ we have $\mathrm{Card}^{V}\not=\mathrm{Card}^{V[H]}$. -However, the only way I know how to get this is via a bit of a cheat: find two different forcings which are individually "good" but have "bad" product, and then look at their lottery sum. This construction has the drawback that the square of the resulting forcing doesn't always collapse cardinals - we're only guaranteed "bad" behavior in the extension if the two coordinates of our generic lie on different "sides" of the original lottery sum. -I recall$^*$ seeing a stronger example of this phenomenon, but I can't track it down or reconstruct it on my own: - -Is it consistent with $\mathsf{ZFC}$ that there is a forcing $\mathbb{P}$ such that $\mathbb{P}$ preserves cardinals but $\Vdash_{\mathbb{P}^2}\mathrm{Card}^V\not=\mathrm{Card}^{V[\langle H_0,H_1\rangle]}$? - - -$^*$Actually my original memory was that even the weaker phenomenon can't happen, but after it was pointed out to me that it can, I now remember differently. I'm sure eventually my memory will have been right. :P - -REPLY [9 votes]: One can indeed prove more: starting with $V=L$, there exists a tame and cardinal preserving class forcing notion $\mathbb{P}$ such that forcing with $\mathbb{P} \times \mathbb{P}$ collapses -all uncountable cardinals. See - -Adam Figura, Collapsing algebras and Suslin trees, Fundamenta Mathematicae 114 (1981) 141-147, doi:10.4064/fm-114-2-141-147.<|endoftext|> -TITLE: Learning mathematics in an "independent and idiosyncratic" way -QUESTION [15 upvotes]: This is a question about learning mathematics outside of the standard undergraduate/graduate education. -The following is a quote from Thurston's On Proof and Progress in Mathematics: - -My mathematical education was rather independent and idiosyncratic, where for a -number of years I learned things on my own, developing personal mental models -for how to think about mathematics. This has often been a big advantage for me in -thinking about mathematics, because it’s easy to pick up later the standard mental -models shared by groups of mathematicians. This means that some concepts that -I use freely and naturally in my personal thinking are foreign to most mathematicians I talk to. My personal mental models and structures are similar in character -to the kinds of models groups of mathematicians share—but they are often different models - -The above quote suggests that Thurston's non-traditional mathematics education was key to his unique insight and contribution to mathematics. I know other examples of very successful mathematicians who benefited from learning outside the standard mathematical canon. -I am trying to incorporate this into my own learning, but I am struggling to get started. For one thing, the standard canons of mathematics are often written by masters of the respective fields who are also masterful writers (e.g. Thurston, Milnor, Serre, Stein). On the other hand, there are many lesser known texts which are poorly written and offer much less insight compared to the canons (even if they are idiosyncratic). Thus, one obstruction to pursuing the above "independent and idiosyncratic" approach to mathematics seems to be in judging what books are truly useful. -My Question. How specifically does one learn mathematics in an "independent and idiosyncratic" way? -Subquestions that may help answer the above: -1.) What should one look for when choosing a textbook? Again, I am looking for nonstandard ways of learning mathematics. -2.) Are there thing to keep in mind when one thinks about problems, theorems, etc.? -3.) What are the sociological obstructions to achieving the above goal? (e.g. as an undergraduate, there is an obvious push to learn mathematics in a certain way. Can this be an obstruction to achieving independence and idiosyncracy?) -I think the answer to my question depends on the audience, so for this question, let's restrict our attention to upper level undergraduates or beginning graduate students. Also, I am asking this question on MathOverflow (as opposed to stackexchange) because my goal is to become a successful research mathematician (as opposed to, say getting better grades in a math class). - -REPLY [2 votes]: To effectively learn "independently" (and perhaps "idiosyncratically"), you need to first learn to critically question the material being presented. Every instructor or author will teach/write in their own "idiosyncratic" method. Learn to differentiate the standard material with the idiosyncratic, then reorganize what you learned into your own "idiosyncratic" style. This isn't (nor shouldn't) be about taking shortcuts. It is actually harder. But if you are not satisfied with the "standard" presentation, you need to ask "why?" you are not satisfied. -In other words, don't tell someone to turn off their flashlight unless you know your flashlight is better.<|endoftext|> -TITLE: Isomorphisms of Pin groups -QUESTION [7 upvotes]: My goal is to identify the $Pin$ group -$$ -1 \to Spin(n) \to Pin^{\pm}(n) \to \mathbb{Z}_2 \to 1 -$$ -such that $Pin^{\pm}(n)$ are isomorphisms to other more familiar groups. -My trick is that to look at the center $Z$ of $Pin$ group, say $Z(Pin(n))$. -We see that, for $k \in \mathbb{Z}^+$, -when $n=4k+1$, -$$ -Z(Pin^+(4k+1))=\mathbb{Z}_2 \times \mathbb{Z}_2, -$$ -$$ -Z(Pin^-(4k+1))=\mathbb{Z}_4. -$$ -when $n=4k+3$, -$$ -Z(Pin^+(4k+3))=\mathbb{Z}_4, -$$ -$$ -Z(Pin^-(4k+3))=\mathbb{Z}_2 \times \mathbb{Z}_2. -$$ -when $n=4k$ or when $n=4k+2$, -$$ -Z(Pin^+(n))=Z(Pin^-(n))=\mathbb{Z}_2. -$$ - -So my naive attempt is that we have the following $Pin$ group isomorphisms: - - -$$Pin^+(1)=\mathbb{Z}_2 \times \mathbb{Z}_2, \quad Pin^-(1)=\mathbb{Z}_4.$$ - - -$$Pin^+(3)=SO(3) \times \mathbb{Z}_ 4 \text{ or } \frac{Spin(3) \times \mathbb{Z}_4}{\mathbb{Z}_2}, -\quad Pin^-(3)=Spin(3) \times \mathbb{Z}_2.$$ - - -$$Pin^+(5)=Spin(5) \times \mathbb{Z}_2,\quad Pin^-(5)=SO(5) \times \mathbb{Z}_4 \text{ or } \frac{Spin(5) \times \mathbb{Z}_4}{\mathbb{Z}_2}.$$ - - -Question 1: Am I correct about the above statements? More generally, do we have* - - -$$Pin^+(4k+1)\cong Spin(4k+1) \times \mathbb{Z}_2,\quad Pin^-(4k+1)\cong SO(4k+1) \times \mathbb{Z}_4.$$ - - -$$Pin^+(4k+3)\cong SO(4k+3) \times \mathbb{Z}_ 4, \quad Pin^-(4k+3)\cong Spin(4k+3) \times \mathbb{Z}_2.$$ - -I am not so sure, -$Pin^+(3)=SO(3) \times \mathbb{Z}_ 4$, could it be that instead -$Pin^+(3)=\frac{Spin(3) \times \mathbb{Z}_4}{\mathbb{Z}_2}?$ -(It looks that mine agrees with Wikipedia, but I doubt Wikipedia may be wrong about $Pin^+(3)=SO(3) \times \mathbb{Z}_ 4$...) possibly only for the Lie algebra, not for the Lie group. -Could it be that: - -$$Pin^+(4k+1)\cong Spin(4k+1) \times \mathbb{Z}_2,\quad Pin^-(4k+1)\cong \frac{Spin(4k+1) \times \mathbb{Z}_4}{\mathbb{Z}_2}.$$ - - -$$Pin^+(4k+3)\cong \frac{Spin(4k+3) \times \mathbb{Z}_ 4}{\mathbb{Z}_ 2}, \quad Pin^-(4k+3)\cong Spin(4k+3) \times \mathbb{Z}_2.$$ - - -Question 2: Are there other $Pin$ group isomorphisms for $n=4k$, or $4k+2$, say $$Pin^{+/-}(4k)\cong ?$$ and -$$Pin^{+/-}(4k+2)\cong?$$ - -Thanks for your comments in advance. - -REPLY [3 votes]: The second version given in the question is correct: -\begin{align} - \mathrm{Pin}^+(4k+1) &\cong \mathrm{Spin}(4k+1) \times \mathbb{Z}_2 \\ - \mathrm{Pin}^-(4k+1) &\cong (\mathrm{Spin}(4k+1) \times \mathbb{Z}_4)/\mathbb{Z}_2 \\ - \mathrm{Pin}^+(4k+3) &\cong (\mathrm{Spin}(4k+3) \times \mathbb{Z}_ 4)/\mathbb{Z}_2 \\ - \mathrm{Pin}^-(4k+3) &\cong \mathrm{Spin}(4k+3) \times \mathbb{Z}_2. -\end{align} -When $n$ is odd, the element $e_1 e_2 \cdots e_n$ of the Clifford algebra commutes with all elements. Every element of the pin group is either an element of the spin group, or is $e_1 \cdots e_n$ times an element of the spin group. If $(e_1 \cdots e_n)^2 = 1$, then $\mathrm{Pin}$ is the direct product $\mathrm{Spin} \times \langle e_1 \cdots e_n \rangle \cong \mathrm{Spin} \times \mathbb{Z}_2$. If $(e_1 \cdots e_n)^2 = -1$, then instead of a direct product, we could say $\mathrm{Pin} \cong (\mathrm{Spin} \times \mathbb{Z}_4)/\mathbb{Z}_2$. This can potentially be described neatly in other ways. For $\mathrm{Pin}^+(3)$, Harvey states that -$$\mathrm{Pin}^+(3) \cong \{ A \in U(2) : \det A = \pm 1 \}$$ -which makes sense along the same lines as $\mathrm{Spin}(3) \cong SU(2)$.<|endoftext|> -TITLE: Constructing a group of order $2187=3^7$ -QUESTION [8 upvotes]: I am trying to look for the $2$-generated groups of order $3^7$ and class $4$ all whose upper central series quotients are elementary abelian of order 9 except the center which has order $3$. -A small check through GAP reveals there is a unique one which is a semi-direct product of $C_{81}$ and $C_{27}$, namely SmallGroup($2187,194$). I am trying to get a structural argument for this. -The second center is $2$-generated abelian of order $27$. Is it possible to construct the third center using the cohomology argument from this without knowing how the generators of the group behave? -Apologies, if the question is too easy. -${\mathrm{\bf{Revised~notes}}}$: The group SmallGroup($2187,194$) turns out to be powerful (Thanks to Derek!). - -REPLY [4 votes]: I think I can see how to prove this now under the assumption that $G$ is powerful. I think the same approach would work without that assumption, but would involve eliminating more cases. -We are given that $G$ is a $2$-generated group, and that the upper central series of $G$ is $1=Z_0 < Z_1 < Z_2 < Z_3 < Z_4 = G$ with $|Z_1|=3$, $|Z_2|=27$, $|Z_3| = 243$, and $|Z_4|=|G|=2187$, with $Z_2/Z_1$, $Z_3/Z_2$ and $Z_4/Z_3$ elementary abelian -My approach is to identify the quotients $G/Z_i$ for $i=3,2,1,0$. I will just sketch the proof for now, and I can fill in details later if necessary. -We know that $G/Z_3 = C_p^2$ is elementary abelian. and it is not hard to see that $$G/Z_2 \cong \langle a,b \mid a^9=1, [[b,a],a] = [[b,a],b] = 1, b^3 = 1\ {\rm or}\ [b,a] \rangle.$$ -Since the group with $b^3=1$ is not powerful, we can assume that $b^3 = [b,a]$ and in fact $G/Z_2 \cong \langle a,b \mid a^9=b^9=1, a^{-1}ba=b^4 \rangle.$ -The hardest part is to identify $G/Z_1$, but using the facts that it is a powerful 2-generated group with centre is $Z_2/Z_1 \cong C_p^2$, it can be shown that $$G/Z_1 \cong \langle a,b \mid a^{27}=b^{27}=1, a^{-1}ba=b^4 \rangle.$$ -The final step, showing that $G \cong \langle a,b \mid a^{27}=b^{81}=1, a^{-1}ba=b^4 \rangle$ is similar but easier. -${\bf Edit\!:}$ Since we would prefer to prove this without assuming that $G$ is powerful, we need to eliminate the possibility that $b^3=1$ in $G/Z_2$. So assume that $b^3=1$ in $G/Z_2$ and now replace $a,b$ by inverse images in $G/Z_1$. -Suppose that $a^{-1}ba = bt$ in $G/Z_1$. Then $a^{-1}b^3a = (bt)^3 = b^3t^3[b,t]^3 = b^3t^3$, because $[b,t] \in Z_2/Z_1$. So, since $b^3 \in Z_2/Z_1$, we have $t^3=1$. -But then $b^{-1}a^{-1}b = ta^{-1}$, so $b^{-1}a^{-3}b = t^3 a^{-3} = a^{-3}$, and so $a^3 \in Z_2/Z_1$, contrary to assumption, because $a$ has order 9 in $G/Z_2$. -So we have determined $G/Z_2$ up to isomorphism without assuming that $G$ is powerful. -A similar, but more complicated calculation, reveals that there is one other non-powerful (powerless?) possibility for $G/Z_1$ other than the one above, which is -$$\langle a,b,t \mid a^{9} = b^{27} = 1, a^{-1}ba=b^4t, t^3=[a,t]=[b,t]=1 \rangle,$$ -where the generator $t$ is redundant. (This is $\mathtt{SmallGroup}(739,32)$, and the other (correct) option above is $\mathtt{SmallGroup}(739,22)$. -A similar argument to the one above shows that that this other option does not extend further to a group $G$ of order $2187$ with the required properties. -Although it is helpful to check things by computer, this can all be done by hand, and I can add further details if necessary. Of course whether you really gain much additional insight by doing it by hand rather than computer is an interesting question!<|endoftext|> -TITLE: Finite flat group schemes over $\mathbb{Z}$ that are extensions of $\mu_p$ by $\mathbb{Z}/p\mathbb{Z}$ -QUESTION [6 upvotes]: Suppose $X$ is a finite flat group scheme over $\mathbb Z$, killed by a prime number $p$ and such that there exists an extension as finite flat group schemes defined over $\mathbb Z$: -$$0\to \mathbb{Z}/p\mathbb{Z}\to X \to \mu_p \to 1.$$ - -Question: Can we conclude that $X\cong \mathbb{Z}/p\mathbb{Z}\times \mu_p$ over $\mathbb{Z}$? - -I know that the answer to this question is negative in general if you consider it over $\mathbb Q$, since you can take $X=E[7]$, the group scheme of $p$-torsion points of an elliptic curve with a $7$-torsion point defined over $\mathbb Q$, since we have such elliptic curves but no such curve with $E[7]\cong \mathbb{Z}/7\mathbb{Z}\times \mu_7$. Of course you can find examples easily for $p=2,3,5$, and probably for infinitely many prime numbers. -On the other hand, over the finite field $\mathbb{F}_p$ the answer is positive, since $\mu_p$ is connected and $\mathbb{Z}/p\mathbb{Z}$ étale, and one could use the connected-étale exact sequence of $X$ to get an splitting of the exact sequence above. -If the answer to the question is affirmative, I will be also interested for what other ring of integers the result is true. I suspect it should be related to the fact that $\mathbb{Q}$ has no unramified extensions. - -REPLY [3 votes]: It is proved in step 3 and 4 of section 3.4.3 in J-M. Fontaine. Il n’y a pas de variété abélienne sur Z. Invent. Math., 81(3):515– 538, 198 (using the ramification bound in that paper) that: -For $E=\mathbb Q$ and $\mathbb Q(\sqrt{-1})$, $\mathbb Q(\sqrt{-3})$, in the category of finite flat group schemes over $O_E$ killed by $p$, there is no non-trivial extension of $\mu_p$ by $\mathbb Z / p \mathbb Z$.<|endoftext|> -TITLE: Smooth complex projective surface as the total space of a Serre fibration -QUESTION [9 upvotes]: Let $M$ be the underlying topological manifold of a smooth complex projective surface. Assume $\pi_1(M)=\{0\}$ and $\pi_2(M)\neq \mathbb{Z}^2$. -Is there a Serre fibration $M\to B$ where $B$ is a CW complex of dimension $0 -TITLE: Inductively computing Mersenne primes / perfect numbers? -QUESTION [5 upvotes]: For two sets $A,B$ set $A+B = \{a +b | a \in A,b \in B\}$. -Let $(x_n)_{n \in \mathbb{N}}$ be independent variables. Let $\sigma(n)$ be the sum of divisors of $n$. -Set $\hat{\phi}(1) = \{x_1\}$ and then inductively: -$$\hat{\phi}(n) = \{x_n\}$$ -if $\sigma(k) \neq n$ for all $k \in \mathbb{N}$ -and -$$\hat{\phi}(n) = \bigcup_{ \sigma(m) = n} \left\{ \sum_{d|m} \hat{\phi}(d) \right\}$$ -otherwise. -Examples: -1 [x1] -2 [x2] -3 [x1 + x2] -4 [2*x1 + x2] -5 [x5] -6 [x1 + x5] -7 [3*x1 + 2*x2] -8 [4*x1 + 2*x2] -9 [x9] -10 [x10] -11 [x11] -12 [x1 + x11, 3*x1 + 2*x2 + x5] -13 [2*x1 + x2 + x9] -14 [3*x1 + x2 + x9] -15 [7*x1 + 4*x2] -16 [x16] -17 [x17] -18 [x1 + x10 + x2 + x5, x1 + x17] -19 [x19] -20 [x1 + x19] -21 [x21] -22 [x22] -23 [x23] -24 [9*x1 + 5*x2 + x5, x1 + x23, 7*x1 + 4*x2 + x9] -25 [x25] -26 [x26] -27 [x27] -28 [8*x1 + 5*x2 + 2*x5, 6*x1 + x11 + 3*x2 + x5] - -Conjecture: -Let $n=2^{p-1}(2^p-1)$ be an even perfect number. Then there exists a $x \in \hat{\phi}(2n)$, such that, if we plug in $1$ for all free variables in $x$ we get: -$$ y = x(1,\cdots,1)$$ -such that: -$$ y+1 = 2^q-1$$ -is a Mersenne prime. Furthermore we have: -$$N=2^{q-1}(2^q-1)$$ -is the next even perfect number after $n$. -I have tested this for $n=6,28,496$. -Unfortunately, my naive computation method hits the limit for the next perfect number. -So I am asking if somebody can check if the next perfect number satisfies this conjecture or not? -It would be nice if you share also your method to check the number. -Thanks for your help! -Here is some SAGE code I used. -And here is some output: -n y -6 2 -6 6 -28 30 -496 2 -496 126 - -Edit: -Here are the trees corresponding to $2\cdot 6$: -[ ( ) ] -[ ( (12, 6, 1), (12, 6, 2), ____(12, 6, 3) ____(12, 6, 6) ) ] -[ ( / / / / ) ( ) ] -[ ( (3, 2, 1) (3, 2, 2) , (6, 5, 1) (6, 5, 5) ), ( (12, 11, 1), (12, 11, 11) ) ] - -And here are the corresponding trees to $2\cdot 28$: -[ ( -[ ( (56, 28, 1), (56, 28, 2), _______(56, 28, 4) _____________(56, 28, 7)_____ _____________(56, 28, 14)____ -[ ( / / / / / / / -[ ( (4, 3, 1) ____(4, 3, 3) (7, 4, 1) (7, 4, 2) _______(7, 4, 4) (14, 13, 1) _______________(14, 13, 13)_____ -[ ( / / / / / / / -[ ( (3, 2, 1) (3, 2, 2) , (4, 3, 1) ____(4, 3, 3) (13, 9, 1) ____(13, 9, 3) (13, 9, 9) -[ ( / / / / -[ ( (3, 2, 1) (3, 2, 2) , (3, 2, 1) (3, 2, 2) , - - ) ( - ____________________________________________________________(56, 28, 28)___________________________________________________ ) ( (56, 28, 1), (56, 28, 2), _______(56, 28, 4) - / / / / / / ) ( / / -(28, 12, 1) (28, 12, 2) ____(28, 12, 3) _______(28, 12, 4) ____(28, 12, 6) _______________________(28, 12, 12)______________ ) ( (4, 3, 1) ____(4, 3, 3) - / / / / / / / / / / ) ( / / - (3, 2, 1) (3, 2, 2) (4, 3, 1) ____(4, 3, 3) (6, 5, 1) (6, 5, 5) (12, 6, 1) (12, 6, 2) ____(12, 6, 3) ____(12, 6, 6) ) ( (3, 2, 1) (3, 2, 2) , - / / / / / / ) ( - (3, 2, 1) (3, 2, 2) (3, 2, 1) (3, 2, 2) (6, 5, 1) (6, 5, 5) ), ( - - ) - _____________(56, 28, 7)_____ _____________(56, 28, 14)____ ___________________________________________________(56, 28, 28)__________________________________________ ) - / / / / / / / / / / / ) ( -(7, 4, 1) (7, 4, 2) _______(7, 4, 4) (14, 13, 1) _______________(14, 13, 13)_____ (28, 12, 1) (28, 12, 2) ____(28, 12, 3) _______(28, 12, 4) ____(28, 12, 6) _____(28, 12, 12) ) ( (56, 39, 1), - / / / / / / / / / / / / / ) ( - (4, 3, 1) ____(4, 3, 3) (13, 9, 1) ____(13, 9, 3) (13, 9, 9) (3, 2, 1) (3, 2, 2) (4, 3, 1) ____(4, 3, 3) (6, 5, 1) (6, 5, 5) (12, 11, 1) (12, 11, 11) ) ( - / / / / / / ) ( - (3, 2, 1) (3, 2, 2) , (3, 2, 1) (3, 2, 2) , (3, 2, 1) (3, 2, 2) ), ( - - - - ) ( - ____(56, 39, 3) _______________(56, 39, 13)_____ _______________________________________________(56, 39, 39)______________________________________ ) ( (56, 39, 1), ____(56, 39, 3) - / / / / / / / / / / / ) ( / / -(3, 2, 1) (3, 2, 2) , (13, 9, 1) ____(13, 9, 3) (13, 9, 9) (39, 18, 1) (39, 18, 2) ____(39, 18, 3) ____(39, 18, 6) (39, 18, 9) _________________(39, 18, 18)_______ ) ( (3, 2, 1) (3, 2, 2) , - / / / / / / / / / / ) ( - (3, 2, 1) (3, 2, 2) , (3, 2, 1) (3, 2, 2) (6, 5, 1) (6, 5, 5) (18, 10, 1) (18, 10, 2) (18, 10, 5) (18, 10, 10) ), ( - - ] - ] - ) ] - _______________(56, 39, 13)_____ _________________________________________(56, 39, 39)________________________________ ) ] - / / / / / / / / / ) ] -(13, 9, 1) ____(13, 9, 3) (13, 9, 9) (39, 18, 1) (39, 18, 2) ____(39, 18, 3) ____(39, 18, 6) (39, 18, 9) _____(39, 18, 18) ) ] - / / / / / / / / ) ] - (3, 2, 1) (3, 2, 2) , (3, 2, 1) (3, 2, 2) (6, 5, 1) (6, 5, 5) (18, 17, 1) (18, 17, 17) ) ] - - -The trees were constructed with the following code: -def inductiveGraph(n): - from itertools import product - ll = [i for i in range(1,n) if sigma(i)==n] - #print(ll) - if len(ll)==0: - return [LabelledOrderedTree([],label=n)] - else: - phll = [] - for m in ll: - ml = list(product(*[ [ LabelledOrderedTree(ph,label=(n,m)) for ph in inductiveGraph(d)] for d in divisors(m)])) - #print(n,ml) - for x in ml: - phll.append( x) - return phll - -print(ascii_art(inductiveGraph(2*28))) - -Furthermore, plugging in $1$ for all variables in $x(1,\cdots,1)$ counts the number of leafs in the tree which corresponds to $x \in \hat{\phi}(2n)$. -Related question: -Additive number theory, Hilbert spaces and polynomial rings? - -REPLY [2 votes]: The conjecture fails for $n=8128$, which can be verified in matter of seconds as explained below. I used PARI/GP for my verification. -First, since the conjecture concerns only values of at $x$'s being all ones, there is no need to compute explicitly $\hat\phi(n)$ but only its evaluation $f_n:=\hat\phi(n)(1,1,\dots,1)$. -Clearly, we have $f_1=\{1\}$ and for $n>1$, $f_n=\{1\}$ if $\sigma^{-1}(n)=\emptyset$, otherwise -$$f_n = \bigcup_{m\in\sigma^{-1}(n)} \left(\sum_{d|m} f_d\right).$$ -Second, to verify the conjecture for $n=8128$, we need to compute $f_{16256}$ -- let's round the index bound to $20000$. Since we will need to quickly get values for $\sigma^{-1}(x)$ for $x\leq 20000$, it's better to get them precomputed in vector is (although for larger values one also can use my invsigma() routine): -is = vector(20000,i,[]); for(i=1,#is, s=sigma(i); if(s<=#is,is[s]=concat(is[s],[i])) ); - -Third, we will need function sumset(S) that computes the sum of sets given as elements of the vector $S$: -setsum(S) = if(#S==0,return([])); my(r=S[1]); for(i=2,#S, r=Set(concat( apply(z->apply(t->t+z,S[i]),r) )) ); r; - -Finally, we are ready to compute $f_n$ for $n\leq 20000$ storing them in a vector, and print the exponents (i.e. $q$) of all $y+2$ that are powers of $2$ for $y\in f_{2n}$ when $n$ is a perfect number: -f=vector(20000); f[1]=[1]; for(n=2,#f, r=is[n]; f[n]=if(!r,[1],Set(concat(apply(t->setsum(apply(z->f[z],divisors(t))),r)))); if(n%2==0 && sigma(n/2)==n, print(n/2," ",apply(t->valuation(t+2,2),select(t->t+2==1< -TITLE: Proving an identity used in general relativity -QUESTION [5 upvotes]: I need to prove the following identity for scalar field ($\phi:M\rightarrow R$) in curved spacetime without torsion called $M$ -$\nabla_{\mu}[\Box \phi \nabla^{\mu}\phi-\frac{1}{2}\nabla^{\mu}(\nabla \phi)^{2}]=(\Box \phi)^{2}- -R_{\mu\nu}\nabla^{\mu}\phi\nabla^{\nu}\phi- \nabla^{\mu}\nabla^{\nu}\phi \nabla_{\mu}\nabla_{\nu}\phi$. -I opened up the total derivative and the term $(\Box \phi)^{2}$ appears but the others I cannot get the right combination. I tried to use the following identity to make appears the terms $R_{\mu\nu}\nabla^{\mu}\phi\nabla^{\nu}\phi$, namely, $[\nabla_\mu, \nabla_\nu]X_{\alpha}=-R^{\kappa}_{\alpha\mu\nu}X_{\kappa}$, but I got stuck at some point, and the third term does not appear at all. -Any help would be much appreciated. - -REPLY [6 votes]: This is the differential form of the Reilly formula. It holds for a function on any pseudo-Riemannian manifold. (Robert C. Reilly. Applications of the Hessian operator in a Riemannian manifold, Indiana Univ. Math. J. 26 (1977), no. 3, 459–472, doi:10.1512/iumj.1977.26.26036) -Use the product rule to say -$$(\Delta f)^2=\operatorname{div}(\Delta f\cdot\nabla f)-\langle\nabla f,\nabla\Delta f\rangle.$$ -Use the commutation formula for covariant derivatives to replace the last term by -$$\langle\nabla f,\nabla\Delta f\rangle=\langle\nabla f,\Delta\nabla f\rangle-\operatorname{Ric}(\nabla f,\nabla f).$$ -Use the product rule to replace the second to last term by -$$\langle\nabla f,\Delta\nabla f\rangle=\operatorname{div}\big(\nabla^2f(\nabla f,\cdot)\big)-|\nabla\nabla f|^2.$$ -Finally $\nabla^2f(\nabla f,\cdot)=\frac{1}{2}\nabla|\nabla f|^2$. This gives your formula. -Edit. As pointed out by Jeffrey Case below, this also follows from the Bochner formula -$$\frac{1}{2}\Delta|\nabla f|^2=|\nabla\nabla f|^2+\langle\nabla\Delta f,\nabla f\rangle+\operatorname{Ric}(\nabla f,\nabla f),$$ -where you just need to use the very first line above to replace the middle term on the RHS. The proof of the Bochner formula is by the other lines above<|endoftext|> -TITLE: How many Lie and associative algebras over a finite field are there? -QUESTION [10 upvotes]: This question is related to the following general question: -Given a variety of (non-associative) algebras $\mathcal V$, a finite field $\mathbb{F}_q$, with $q$ elements, and a positive integer $n$, how many $n$-dimensional $\mathbb F_q$-algebras in $\mathcal V$ are there? -It is well known that if $A$ is an $n$-dimensional algebra over a field $F$, with basis $\{e_1, \dots, e_n\}$ then its algebra structure is uniquely determined by the $n^3$-tuple $(\alpha_{ij}^{(k)})\in F^{n^3}$, defined by $e_i e_j=\sum_{k=1}^n \alpha_{ij}^{(k)} e_k$. -So the general question above can be reformulated as: -How many of these $n^3$-tuples of elements of $\mathbb F_q$ define algebras in $\mathcal V$? -Or in another language: given an arbitrary $n$-dimensional algebra, what is the probability that it lies in $\mathcal V$? -Let us denote such number by $N_{q,n}(\mathcal V)$. -Some examples are simple to compute. -For example, if $q$ is odd, one can easily show that if $\mathcal V$ is the variety of anticommutative algebras (i.e., the class of all algebras satisfying the identity $xy+yx=0$), then $N_{q,n}(\mathcal V)=q^{n^2(n-1)/2}$ and if $\mathcal C$ is the variety of commutative algebras, then $N_{q,n}(\mathcal C)=q^{n^2(n+1)/2}$. -But other examples seem to be much more difficult, for example for the varieties of Lie and associative algebras. -So my main questions (for now) are the following: - - -How many Lie algebras of dimension $n$ over a field with $q$ elements are there? -How many associative algebras of dimension $n$ over a field with $q$ elements are there? - - -I'd like to stress that I am not interested in isomorphism classes, but in the number of such algebras only (that is to say this is a problem of combinatorics and not of algebra). -Finally, I would like to remark that I have considered the possibility to write a computer program to compute some cases (example for $q=3$ and $n \leq 6$), so I could have a guess of the general answer, but in a first look I realized that this is would take too much time. -EDIT: -The answer I am expecting is an explicit formula for $N_{q,n}(\mathcal V)$ when $\mathcal V$ is the variety of associative or Lie algebras. - -REPLY [15 votes]: Bjorn Poonen addresses this question for commutative (associative, unital) algebras in The moduli space of commutative algebras of finite rank; asymptotically we have -$$q^{\frac{2}{27} n^3 + O(n^{8/3})}$$ -such algebras (Theorem 10.9). Bjorn also gives a more precise lower bound on the dimension of the corresponding affine scheme in Theorem 9.2 which is a collection of three polynomials with leading term $\frac{2}{27} n^3$ depending on the value of $n \bmod 3$. The $\frac{2}{27}$ may seem familiar from a corresponding count of the number of finite $p$-groups and it happens for very similar reasons as he discusses in Section 10: - -The approach towards both those results is to adapt the proof (begun in [Hig60] and completed in [Sim65]) that the number of $p$-groups of order $p^n$ is $p^{ \frac{2}{27} n^3 + O(n^{8/3})}$. As suggested to us by Hendrik Lenstra, there is an analogy between the powers of the maximal ideal of a local finite-rank $k$-algebra and the descending $p$-central series of a $p$-group. Although there seems to be no direct connection between finite-rank $k$-algebras and finite $p$-groups, the combinatorial structure in the two enumeration proofs are nearly identical. - -He also cites An estimate of the number of parameters defining an $e$-dimensional algebra by Yuri Neretin (which is in Russian, sadly for me) as addressing the Lie and associative cases; I'm not sure if the estimates immediately carry through to a finite field but if they do the answer is the same for Lie algebras and for associative algebras it is -$$q^{ \frac{4}{27} n^3 + O(n^{8/3}) }.$$ -Presumably the analogous structure for Lie algebras making the answer similar is the descending central series for a nilpotent Lie algebra. For the associative case maybe it's something like powers of the Jacobson radical? -Note also that because $\frac{8}{3} > 2$ the error term in the exponent absorbs multiplicative factors as large as $q^{O(n^2)}$ so these asymptotics hold regardless of whether or not we quotient by the action of $GL_n(\mathbb{F}_q)$ (which is equivalent to asking for the isomorphism classification), which you might see as unsatisfactorily lenient but I think these are state of the art. - -Edit: The lower bound for Lie algebras is easy enough to give here; it's very similar to the argument for finite $p$-groups and for commutative algebras but, I think, simpler. We consider only 2-step nilpotent Lie algebras $L$ of some dimension $n$, which arise as a central extension -$$0 \to [L, L] \to L \to A \to 0$$ -of an abelian Lie algebra $A$ (the abelianization) by another abelian Lie algebra $[L, L]$ (the commutator; I am not using fraktur here to save typing). Explicitly, the Lie bracket $[-, -]$ factors through $A$ and lands in $[L, L]$, and so the only constraint on it is that it's a surjective alternating map $\wedge^2(A) \to [L, L]$; given any such map we can construct a Lie bracket which trivially satisfies the Jacobi identity because all triple commutators vanish by 2-step nilpotence. This is a mild generalization of the construction of the Heisenberg algebra where $\dim [L, L] = 1$. -So, fixing the vector space $L$, we put a 2-step nilpotent Lie algebra structure on $L$ by first choosing a subspace $[L, L]$ we want to be the commutator and then choosing a surjection $\wedge^2(L/[L, L]) \to [L, L]$. In general the space of surjections from a f.d. vector space $V$ to a f.d. vector space $W$ admits a free action by $GL(W)$ and the quotient by this action is the Grassmannian of codimension $\dim W$ subspaces of $V$. So, setting $b = \dim [L, L]$, the space of choices we have available is the triple of choices of - -a $b$-dimensional subspace $[L, L]$ of $L$, -a $b$-codimensional subspace of $\wedge^2(L/[L, L])$, and -an isomorphism between the first choice and the quotient by the second choice. - -Write $a = n - b = \dim L/[L, L] = \dim A$, so that $a + b = n$. Over $\mathbb{F}_q$ there are exactly -$${n \choose b}_q { {a \choose 2} \choose b}_q |GL_b(\mathbb{F}_q)|$$ -ways to make the above choices. Now our job is to find $a, b$ which maximizes this, or at least which makes it quite big since we're aiming for a lower bound. The leading term in $q$ is $q$ to the power of -$$ab + \left( {a \choose 2} - b \right) b + b^2 = \frac{a(a+1)b}{2}.$$ -Subject to the constraint that $a + b = n$ this is maximized when $a \approx \frac{2n}{3}, b \approx \frac{n}{3}$, and we could be more careful depending on the value of $n \bmod 3$ if desired. Let's instead restrict to the case that $3 \mid n$ so that we can divide by $3$ exactly, and also take the liberty of dividing by $(q - 1)^b$ so that what remains is a polynomial in $q$ with nonnegative coefficients and so the leading term is a true lower bound. We get that there are at least -$$q^{ \frac{2}{27} n^3 + \frac{n^2}{9} - \frac{n}{3}}$$ -2-step nilpotent Lie brackets on $\mathbb{F}_q^n$ when $3 \mid n$. -To get a lower bound on the number of isomorphism classes we quotient badly by the action of $GL_n(\mathbb{F}_q)$. At this point we can actually restore the factor of $(q - 1)^b$ we lost above (although it doesn't matter too much either way); it's not hard to show that $\frac{|GL_n(\mathbb{F}_q)|}{|GL_b(\mathbb{F}_q)|} \le q^{n^2 - b^2}$, so we can then divide by $|GL_b(\mathbb{F}_q)|$ and then by $q^{n^2 - b^2}$ to get a lower bound, whereupon what remains is a polynomial in $q$ with non-negative coefficients which can be bounded from below by its leading term again. We get that there are at least -$$q^{ \frac{2}{27} n^3 - \frac{8n^2}{9}}$$ -isomorphism classes of 2-step nilpotent Lie algebras of dimension $3 \mid n$ over $\mathbb{F}_q$. It is maybe surprising that it's possible to prove a matching upper bound, at least up to leading order in the exponent; I don't know what that argument looks like in detail. -For small values of $n$ it would be feasible to not only maximize but sum over all $a + b = n$ above and so compute the exact number of 2-step nilpotent Lie brackets. The truly brave who wanted to compute the exact number of isomorphism classes could hope to apply Burnside's lemma... -The lower bound for finite groups is very similar, using 2-step nilpotent $p$-groups arising as the central extension of a f.d. $\mathbb{F}_p$-vector space by another one, etc.<|endoftext|> -TITLE: Value of an integral -QUESTION [5 upvotes]: I need to verify the value of the following integral -$$ 4n(n-1)\int_0^1 \frac{1}{8t^3}\left[\frac{(2t-t^2)^{n+1}}{(n+1)}-\frac{t^{2n+2}}{n+1}-t^4\{\frac{(2t-t^2)^{n-1}}{n-1}-\frac{t^{2n-2}}{n-1} \} \right] dt.$$ -The integrand (factor of $4n(n-1)$) included) is the pdf of certain random variable for $n\geq 3$ and hence I expect it be 1. If somebody can kindly put it into some computer algebra system like MATHEMATICA, I would be most obliged. I do not have access to any CAS software. -PS:-I do not know of any free CAS software. If there is any somebody may please share - -REPLY [23 votes]: The integral can be rewritten as -\begin{align*} -I&=\frac{n(n-1)}{2}\int_0^1\frac{t^{n-2}(2-t)^{n+1}-t^{2n-1}}{n+1}-\frac{t^n(2-t)^{n-1}-t^{2n-1}}{n-1}\,dt\\[6pt] -&=\frac{1}{2n+2}+\frac{n(n-1)}{2}\int_0^1\frac{t^{n-2}(2-t)^{n+1}}{n+1}-\frac{t^n(2-t)^{n-1}}{n-1}\,dt. -\end{align*} -Integrating by parts, we obtain -$$\int_0^1\frac{t^{n-2}(2-t)^{n+1}}{n+1}\,dt=\frac{1}{n^2-1}+\int_0^1\frac{t^{n-1}(2-t)^n}{n-1}\,dt.$$ -Therefore, -\begin{align*} -I&=\frac{1}{2}+\frac{n}{2}\int_0^1t^{n-1}(2-t)^n-t^n(2-t)^{n-1}\,dt\\[6pt] -&=\frac{1}{2}+\frac{1}{2}\int_0^1(t^n(2-t)^n)'\,dt=\frac{1}{2}+\frac{1}{2}=1. -\end{align*} -P.S. You can use SageMath and WolframAlpha for symbolic calculations. Both are free. - -REPLY [6 votes]: It seems your conjecture is true. Mathematica gives the result -$$ -(1 + 4^n (-1 + n) n \mbox{Beta} [1/2, -1 + n, 2 + n] - - 4^n n (1 + n) \mbox{Beta} [1/2, 1 + n, n])/(2 (1 + n)) -$$ -in terms of the incomplete Beta function, and putting in random integers $\geq 3$ always yields 1 (I haven't managed to get Mathematica to spit that out as a general result for arbitrary $n$). - -REPLY [4 votes]: You can use CoCalc. -For instance, type -integral(x^2,x) -and get -1/3*x^3 -It also permits symbolic parameters. -Input: -f(x,n)=x^2+n -integral(f(x,n),x) -Output: -1/3*x^3+n*x<|endoftext|> -TITLE: Growth of (integral of) Laplace transform of a function of compact support as $Re \to -\infty$ -QUESTION [6 upvotes]: Let $f:[0,\infty)\to \mathbb{R}$ be supported on $[0,1]$, with $\int_0^1 f(x) dx = 1$. Let $\mathcal{L} f$ be its Laplace transform. How slowly may -$$\int_{-\infty}^\infty |\mathcal{L} f(\sigma+i t)| dt$$ -grow as $\sigma\to -\infty$? It is clear it could be $\ll e^{\epsilon |\sigma|}$ (just let $f$ be supported on $[0,\epsilon]$). Could it grow polynomially on $|\sigma|$? Linearly on $|\sigma|$? - -REPLY [6 votes]: It has exponential growth: if your integral is $O(e^{-\sigma\epsilon}), \sigma\to -\infty,$ then the -support of your function $f$ is contained in $[0,\epsilon]$. This follows -from the inversion formula for the Laplace transform: -$$f(x)=\lim_{r\to\infty}\frac{1}{2\pi i}\int_{b-ir}^{b+ir}Lf(s)e^{sx}ds,$$ -where the integration is on any vertical line $\Re s=b$. -Under your assumptions, -$$|f(x)\leq\frac{1}{2\pi} e^{-b\epsilon}e^{bx},$$ -for every $b<0$, therefore, by letting $b\to-\infty$ we obtain $f(x)=0$ when $x>\epsilon$. - -REPLY [6 votes]: The Fourier inversion formula says that $$f(x)e^{-\sigma x} = (2\pi)^{-1} \int \mathcal{L}f(\sigma+it) e^{itx} dt$$ -so if your integral is $I(\sigma)$, one sees that -$$|f(x)| \le (2\pi)^{-1} e^{\sigma x} I(\sigma)$$ -It follows that if $I(\sigma)$ is $O(e^{\varepsilon\sigma})$, then f is supported on $[0,\varepsilon]$. So that can't hold for ALL $\varepsilon$. Thus the behavior that you mention is best possible.<|endoftext|> -TITLE: A unique equilibrium state which does not have Gibbs property -QUESTION [5 upvotes]: Let $T:\Sigma \rightarrow \Sigma$ be a topologically mixing subshift of finite type and let $f:\Sigma \rightarrow \mathbb{R}$ be a continuous functions over $(T, \Sigma)$. Assume that there is a unique equilibrium measure $\mu$ for $f$ because of some reason. -$\textit{Question}:$ Does $\mu$ necessarily have Gibbs property? -I guess the answer is no, but I can't find a reference. - -REPLY [4 votes]: The measure $\mu$ does not necessarily have the Gibbs property. In fact, it has the Gibbs property if and only if $f$ has the Bowen property: $\sup_n \sup \{ |S_n f(x) - S_n f(y)| : x_1 \dots x_n = y_1 \dots y_n \} < \infty$. Every such $f$ has a unique equilibrium measure, but there are some potentials without the Bowen property that still have unique equilibrium measures. -$\mu$ Gibbs iff $f$ Bowen. The Gibbs property requires that there be $K>0$ such that for every $x\in \Sigma$ we have -$$ -K^{-1}\leq \frac{\mu[x_1\dots x_n]}{e^{-nP(f) + S_nf(x)}} \leq K. -$$ -Given $x,y \in \Sigma$ with $x_1\dots x_n = y_1 \dots y_n$, the only quantity in the corresponding inequalities that can vary is $S_n f$, and comparing them gives -$$ -K^{-2} \leq e^{S_n f(x) - S_n f(y)} \leq K^2. -$$ -Thus $|S_n f(x) - S_n f(y)| \leq 2\log K$, which proves the Bowen property. The other direction is classical; see -Bowen, Rufus, Some systems with unique equilibrium states, Math. Syst. Theory 8(1974), 193-202 (1975). ZBL0299.54031. -which gives a more general result (expansive systems with specification, which includes mixing SFTs). -An example of a non-Bowen potential that has a unique equilibrium state. -Hofbauer, Franz, Examples for the nonuniqueness of the equilibrium state, Trans. Am. Math. Soc. 228, 223-241 (1977). ZBL0355.28010. -The example there is the full shift on two symbols 0,1, and the potential is $f(x) = a_k$ whenever $x = 1^k 0\dots$, where $a_k$ is a sequence of real numbers converging to $0$. (Also $f(1^\infty) = 0$.) Writing $s_k = a_0 + \cdots + a_k$, the table on page 239 of that paper is useful. The potential $f$ has the Bowen property iff $\sum a_k$ converges, but there are examples where $\sum a_k$ diverges and $f$ still has a unique equilibrium measure. -It is often the case that unique equilibrium measures, including the ones in Hofbauer's paper, satisfy a "non-uniform" Gibbs property: see -Climenhaga, Vaughn; Thompson, Daniel J., Equilibrium states beyond specification and the Bowen property, J. Lond. Math. Soc., II. Ser. 87, No. 2, 401-427 (2013). ZBL1276.37023.<|endoftext|> -TITLE: If the universal cover of a manifold is spin, must it admit a finite cover which is spin? -QUESTION [19 upvotes]: If $M$ is non-orientable, then it has a finite cover which is orientable (in particular, the orientable double cover). -If $M$ is non-spin, then it does not necessarily have a finite cover which is spin, e.g. $M = \mathbb{CP}^2$. As a cover of a spin manifold is spin, a necessary condition for $M$ to admit such a finite cover is that its universal cover is spin (which is not the case in the previous example). In analogy with the first sentence, note that the universal cover is always orientable. - -Let $M$ be a closed smooth manifold whose universal cover is spin. Is there a finite cover of $M$ which is spin? - -This question is partially motivated by the study of positive scalar curvature. The Dirac operator on a spin manifold can be used to obtain obstructions to positive scalar curvature à la Lichnerowicz, Hitchin, Gromov & Lawson, Rosenberg, etc. More generally, these techniques can be applied to manifolds which admit a spin cover. Things are generally more difficult in the non-compact case than in the compact case, so if the answer to my question were 'yes', we could just pass to a compact cover which is spin and apply the techniques there (as opposed to passing to the potentially non-compact universal cover). - -REPLY [3 votes]: As promised, here is my solution based on the Davis trick. First, there is a very general construction of PL aspherical 4-manifolds (it also works in higher dimensions). Start with a finite aspherical 2-dimensional CW complex $W$. Up to homotopy, $W$ always embeds in the Euclidean 4-space $E^4$ (I think, this is due to Stallings). Take such an embedding and let $N=N(W)$ denote a regular neighborhood of $W$ in $E^4$. Now, apply "Davis trick" to $N$: Introduce a reflection orbifold structure on the boundary of $N$ such that the corresponding stratification of the boundary is dual to a triangulation of $\partial N$. The resulting orbifold ${\mathcal O}$ is very good (admits a finite orientable manifold-covering $M\to {\mathcal O}$) and its universal covering (same for $M$ and for ${\mathcal O}$) is contractible. As a bonus, $\pi_1(W)$ embeds in $\pi_1(M)< \pi_1({\mathcal O})$. For details see -Mess, Geoffrey, Examples of Poincaré duality groups, Proc. Am. Math. Soc. 110, No. 4, 1145-1146 (1990). ZBL0709.57025. -and, of course, the original paper by Mike Davis from 1983. (Actually, it was Bill Thurston who came up with this trick in the context of 3-manifolds: He used it for the proof of his hyperbolization theorem.) This construction allows one to embed 2-dimensional finitely presented groups with "exotic properties" in fundamental groups of closed aspherical PL manifolds. -I will use a relative version of this construction. Start with a closed connected oriented surface of genus $\ge 1$; I'll take the torus $T^2$. Let $E\to T^2$ be the 2-disk bundle over $T^2$ with the Euler number $\pm 1$. The boundary of the 4-manifold $E$ is a 3-dimensional Nil-manifold: The total space of a nontrivial circle bundle over the torus. The group $\pi_1(\partial E)$ has two generators $a, b$, and $\pi_1(\partial E)$ has the presentation -$$ -\langle a, b| [a,b]=t, [a,t]=1, [b,t]=1\rangle. -$$ -Represent $a, b$ by simple disjoint loops $\alpha, \beta$ in $\partial E$. Now, take your favorite finite 2-dimensional aspherical complex $W$ whose fundamental group is nontrivial and has no proper finite index subgroups (I care only about the homotopy type of $W$). The standard example is the presentation complex of Higman group. But there are many other examples. As before, embed $W$ in $E^4$, take a regular neighborhood $N$ of $W$ in $E^4$. Then $\pi_1(\partial N)$ maps nontrivially to $\pi_1(W)$. Pick two simple loops $\alpha', \beta'\subset \partial N$ which map nontrivially to $\pi_1(W)$ (you can take the same loop). -Now, take two copies $N_a, N_b$ of $N$ and attach them to $E$ by identifying a regular neighborhood of $\alpha'$ to that of $\alpha$ for $N_a$ and identifying a regular neighborhood of $\beta'$ to that of $\beta$ for $N_b$. The result is a compact PL aspherical 4-manifold with boundary $Z$. The $\pi_1(Z)$ is an amalgam of $\pi_1(E)\cong {\mathbb Z}^2$ with two copies of $\pi_1(N)$ (along infinite cyclic subgroups). -For each homomorphism to a finite group -$$ -\phi: \pi_1(Z)\to \Phi -$$ -the subgroups $\pi_1(N_a), \pi_1(N_b)$ will have to map trivially. Hence, $a$ and $b$ will have to map trivially as well. Since $a, b$ generate $\pi_1(E)$, $\pi_1(Z)$ has no nontrivial homomorphisms to finite groups. Now, apply Davis trick to $Z$. The result is an orbifold ${\mathcal O}$. Since $Z$ was aspherical, so is ${\mathcal O}$ (i.e. it has contractible universal covering space). -Take a finite orientable manifold-covering $M\to {\mathcal O}$. Then $M$, of course, has contractible (hence, spin) universal covering. I claim that $M$ has no finite spin-covering spaces. Indeed, for each finite-sheeted covering $p: M'\to M$, the manifold $int(Z)\subset M$ has to lift trivially; more precisely, $p$ restricts to a trivial covering -$$ -p^{-1}(int Z)\to int Z.$$ -This is because $\pi_1(Z)$ has no nontrivial homomorphisms to finite groups. Thus, $M'$ contains a copy of $E$. In particular, $M'$ contains a 2-torus with odd self-intersection, i.e. the intersection form of $M'$ is not even, i.e. $M'$ is not spin. -I was working in the PL category but in dimension 4, PL is the same DIFF, so you get a smooth example as well. - -Edit. Lemma. Let $M$ be a triangulated manifold, $W\subset M$ is a subcomplex and $N=N(W)$ is the regular neighborhood of $W$ in $M$. Then the inclusion map $W\to N$ is a homotopy-equivalence; if $W$ is connected and has codimension $\ge 2$ in $M$ then $\partial N$ is connected and the induced map $\pi_1(\partial N)\to \pi_1(W)$ is surjective. -Proof. The homotopy-equivalence part is standard and holds for general simplicial complexes $M$, not just for manifolds. Moreover, the inclusion map $\partial N\to (N \setminus W)$ is also a homotopy-equivalence. (Both are proven using "straight-line homotopy.") -I will prove the second part. Take an arc $\alpha$ in $N$ connecting two points $x, y\in \partial N$. Since $W$ has codimension $\ge 2$, taking $\alpha$ in general position, we see that it will be disjoint from $W$, hence, is homotopic relative to $\{x, y\}$ to an arc in $\partial N$. -(I am using here and below the h.e. $\partial N\to N-W$.) Thus, $\partial N$ is connected. Next, let $\alpha$ be a loop in $N$ based at $x\in \partial N$. By the same argument, $\alpha$ is homotopic to a loop based at $x$ and contained in $N-W$, hence, to a loop in $\partial N$.<|endoftext|> -TITLE: Infinite dimensional irreducible representations of a tensor product -QUESTION [12 upvotes]: The second part of Theorem 3.10.2 of "Introduction to representation -theory" by Etingof, Golberg, Hensel, Liu, Schwender, Vaintrob and Yudovina states that -if $A$ and $B$ are $k$-algebras ($k$ an algebraically closed field) and $M$ is an irreducible finite dimensional representation of $A\otimes_k B,$ then $M\cong V\otimes_k W$ where $V$ and $W$ are finite dimensional irreducible representations of $A$ and $B$ respectively. -My question is about the first part of the remark following this theorem. This remark states that the previous proposition fails for infinite dimensional representations, "e.g. it fails when A is the Weyl algebra -in characteristic zero." I don't see how to construct an irreducible infinite dimensional representation $M$ of $A\otimes B,$ -where $A$ is the Weyl algebra, such that $M\ncong V\otimes_k W$. -(I asked the same question on Math.SE more than one year ago without receiving answers, also after starting bounties) - -REPLY [8 votes]: Nate's suggestion on math.SE works. We'll show that if $A = k[x, \partial_x]$ and $B = k[y, \partial_y]$ are both taken to be the Weyl algebra, then the module over $A_2 = A \otimes B \cong k[x, \partial_x, y, \partial_y]$ generated by $e^{xy}$ is 1) simple and 2) not a tensor product of simple modules of $A$ and $B$. -Explicitly this module $M$ consists of elements of the form $f(x, y) e^{xy}$ where $f$ is a polynomial, with the obvious action by multiplication and differentiation. Abstractly it is the quotient of $k[x, \partial_x, y, \partial_y]$ by the left ideal $(x - \partial_y, y - \partial_x)$. We can show very straightforwardly that every nonzero element of $M$ generates it, by computing that -$$(\partial_x - y) x^i y^j e^{xy} = ix^{i-1} y e^{xy}$$ -and similarly that -$$(\partial_y - x) x^i y^j e^{xy} = j x^i y^{j-1} e^{xy}.$$ -In other words, this module is the pullback of the usual module $k[x, y]$ under the (edit: inverse of the) automorphism $A_2 \to A_2$ sending $x \mapsto x, y \mapsto y, \partial_x \mapsto \partial_x - y, \partial_y \mapsto \partial_y - x$. Now $k[x, y]$ is irreducible which means so is this module (explicitly, every nonzero element is a generator because we can repeatedly differentiate resp. apply the above maps to get to $1$ resp. $e^{xy}$), and $k[x, y]$ is a tensor product $k[x] \otimes k[y]$. -But we can show that $M$ is not such a tensor product (this property is not invariant under twisting by automorphisms). If it were such a tensor product $V \otimes W$, then a pure vector $v \otimes w$ would have the property that there is some differential operator $a(x, \partial_x) \in k[x, \partial_x]$ such that $(a \otimes 1)(v \otimes w) = av \otimes w = 0$; in words, it would satisfy a polynomial differential equation involving $x$ only. -No nonzero element of $M$ has this property. The key point is that every nonzero element of $k[x, \partial_x]$ makes elements bigger in the lex order: we have -$$x \left( x^i y^j e^{xy} \right) = x^{i+1} y^j e^{xy}$$ -$$\partial_x \left( x^i y^j e^{xy} \right) = ix^{i-1} y^j e^{xy} + x^i y^{j+1} e^{xy}$$ -and so formally, writing an arbitrary element $a \in k[x, \partial_x]$ as a sum $\sum a_{k, \ell} \partial_x^k x^{\ell}$, we see that if $\partial_x^k x^{\ell}$ is the largest monomial in $a$ in the lex order where $x > \partial_x$, then $a(x^i y^j e^{xy})$ has largest monomial $x^{i + \ell} y^{j + k}$, and in particular it does not vanish, so the same is true if $x^i y^j e^{xy}$ is replaced by any other element with the same largest monomial.<|endoftext|> -TITLE: Separating a lattice simplex from a lattice polytope -QUESTION [8 upvotes]: Let $P\subset\mathbb{R}^n$ be a convex lattice polytope. -Do there always exist a lattice simplex $\Delta\subset P$ and an affine hyperplane $H\subset\mathbb{R}^n$ separating $\Delta$ from the convex hull of the integer points of $P\setminus \Delta$? -This is equivalent to say that there exist a degree one polynomial $h:\mathbb{R}^n\rightarrow\mathbb{R}$ that is positive on all the integer points of $\Delta$ and negative on all the integer points of $P\setminus \Delta$. - -REPLY [9 votes]: This is possible and here is how to do this. We will use an inductive argument, assume that the statement holds for polytops of dimension $ -TITLE: Are there axioms satisfied in commutative rings and distributive lattices but not satisfied in commutative semirings? -QUESTION [23 upvotes]: Consider the language of rigs (also called semirings): it has constants $0$ and $1$ and binary operations $+$ and $\times$. The theory of commutative rigs is generated by the usual axioms: $+$ is associative, commutative, and has unit $0$; $\times$ is associative, commutative, and has unit $1$; $\times$ distributes over $+$; and $0$ is absorbing for $\times$. -Every commutative ring is a commutative rig (of course), and every distributive lattice as well (interpreting $\bot$ as $0$, $\top$ as $1$, $\vee$ as $+$, and $\wedge$ as $\times$). In fact, the category of commutative rings is a full reflective subcategory of the category of commutative rigs, as is the category of distributive lattices. The intersection of the two is trivial, in the sense that only the trivial algebra is both a ring and a lattice. (In a lattice, $\top \vee \top = \top$; but in a ring $1 + 1 = 1$ implies $0 = 1$.) What I am wondering is how close do these two subcategories come to capturing "all" the possible behaviour of commutative rigs. More precisely: -Question 1. Is there a Horn clause in the language of rigs that is true in every commutative ring and every distributive lattice but false in some commutative rig? -Since commutative rings are not axiomatisable in the language of rigs using only Horn clauses, I would also be interested to hear about, say, cartesian sequents instead of Horn clauses. This can be phrased category theoretically: -Question 2. Is there a full reflective subcategory $\mathcal{C}$ of the category of commutative rigs that is closed under filtered colimits and contains the subcategories of commutative rings and distributive lattices but is not the whole category? (Furthermore, can we choose such a $\mathcal{C}$ so that the reflection of $\mathbb{N} [x]$ (= the free commutative rig on one generator) represents a monadic functor $\mathcal{C} \to \textbf{Set}$?) -I don't want to be too permissive, however – since commutative rings and distributive lattices can both be axiomatised by a single first order sentence in the language of rigs, taking their disjunction yields a sentence that is true in only commutative rings and distributive lattices but false in general commutative rigs. -Here is an example of a first-order axiom that is true in commutative rings and distributive lattices that is false in some commutative rig: - -For all $a$ and $b$, there exist $c$ and $d$ such that $(c + d) a + d b = b$. - -This axiom amounts to saying that every ideal is subtractive (which is a second-order axiom prima facie); it is the case that every ideal in a commutative ring or distributive lattice is subtractive. The way I prefer to think about it is that in a commutative ring, ideals are automatically subtractive because $-1$ exists, and in a distributive lattice, ideals are automatically subtractive because they are downward closed. As it turns out, this can be expressed as a first-order sentence, albeit not a cartesian sequent. - -REPLY [26 votes]: Following François's suggestion, I ran alg to find a unital commutative semiring which fails to satisfy -$$ - \forall x\, y\, z,\; x + z = y + z \land x \times z = y \times z \Rightarrow x = y. -\tag{1} -$$ -The smallest one has size 3. Here is the output of the program, cut off after the first example. - Theory unital_commutative_semiring. - - Constant 0 1. - Binary + *. - - Axiom: 0 + x = x. - Axiom: x + 0 = x. - Axiom: x + (y + z) = (x + y) + z. - Axiom: x + y = y + x. - - Axiom: 1 * x = x. - Axiom: x * 1 = x. - Axiom: x * (y * z) = (x * y) * z. - Axiom: x * y = y * x. - - Axiom: (x + y) * z = x * z + y * z. - Axiom: x * (y + z) = x * y + x * z. - - Axiom: 0 * x = 0. - Axiom: x * 0 = 0. - - # Extra command-line axioms - Axiom: not forall x y z, x + z = y + z /\ x * z = y * z => x = y. - -# Size 3 - -### unital_commutative_semiring_with_extras_3_1 - - - + | 0 1 a - --+--------- - 0 | 0 1 a - 1 | 1 1 1 - a | a 1 a - - - * | 0 1 a - --+--------- - 0 | 0 0 0 - 1 | 0 1 a - a | 0 a 0 - -We can also count the structures. For commutative unital semirings the counts are - size | count - -----|------ - 2 | 2 - 3 | 6 - 4 | 36 - 5 | 228 - 6 | 2075 - -The counts for commutative unital semirings that also satisfy (1): - size | count - -----|------ - 2 | 2 - 3 | 3 - 4 | 9 - 5 | 12 - 6 | 23 - -So clearly there will be some, but it looks like not too many for small sizes. -It is a bit weird that $2, 6, 36, 228, 2075$ is not in OEIS.<|endoftext|> -TITLE: Singular cardinal $\kappa$ with projective plane such that all edges have cardinality $<\kappa$ -QUESTION [6 upvotes]: Is there an infinite singular cardinal $\kappa$ such that there is a set $E\subseteq{\cal P}(\kappa)$ with the following properties? - -$|e| < \kappa$ for all $e\in E$, -whenever $\alpha\neq\beta\in \kappa$ there is $e\in E$ with $\{\alpha,\beta\} \subseteq e$, and -if $e_1\neq e_2\in E$ then $|e_1\cap e_2| = 1$. - -(There can be no infinite regular cardinal with this property.) - -REPLY [11 votes]: The answer is no. Let $\kappa$ be any infinite cardinal, regular or singular, and assume for a contradiction that there is a set $E\subseteq\mathcal P(\kappa)$ satisfying your conditions. I will call the elements of $\kappa$ points and the elements of $E$ lines. -The lines do not all go through one point: Given a point $\alpha$, choose a point $\beta\ne\alpha$ and a point $\gamma$ not on the line through $\alpha$ and $\beta$; the line through $\beta$ and $\gamma$ does not go through $\alpha$. -There are $\lt\kappa$ lines through any point: Consider any point $\alpha$ and let $\lambda$ be the number of lines through $\alpha$. Choose a line $e$ which does not go through $\alpha$. Since each line through $\alpha$ meets $e$ in a different point, $\lambda\le|e|\lt\kappa$. -Now choose two distinct points $\alpha$ and $\beta$. Say there are $\lambda$ lines through $\alpha$ and $\mu$ lines through $\beta$. Let $e$ be the line through $\alpha$ and $\beta$. Now every point which is not on the line $e$ is the point of intersection of a line through $\alpha$ and a line through $\beta$. Hence $\kappa\le|e|+\lambda\mu\lt\kappa$ which is absurd. -This argument is adapted from the proof that a finite projective plane of order $n$ has $n^2+n+1$ points. In that case we have $|e|=\lambda=\mu=n+1$ and the number of points is exactly $|e|+(\lambda-1)(\mu-1)=n^2+n+1$. -P.S. The answer is still no if condition (2) is weakened to "for every $\alpha\in\kappa$ we have $|\{e\in E:\alpha\in e\}|\gt1$". This more general result was proved in my answer to this old question.<|endoftext|> -TITLE: Diffeomorphisms pushing forward vector field to any non-zero multiple -QUESTION [7 upvotes]: Is there a closed smooth manifold $M$ such that for each real $x\neq 0$ there is a nowhere vanishing vector field $v$ on $M$ and a diffeomorphism $\phi:M\to M$ such that $\phi_*v=xv$? - -REPLY [6 votes]: Such a manifold exists. First let's construct a non-compact example. -Take $PSL(2,\mathbb R)$ and take two $1$-parameter subgroups, given by $$\begin{pmatrix} -e^{t} & 0 \\ -0 & e^{-t} -\end{pmatrix}, \begin{pmatrix} -1 & t \\ -0 & 1 -\end{pmatrix}$$ -Consider actions on $PSL(2,\mathbb R)$ of these two groups by multiplication on the left. Then $v$ is the vector field tangent to the second flow, while the first flow will give you a $1$-parameter family of diffeos that will dilate $v$ by any positive constant. -Now, to get the compact example, quotient $PSL(2,\mathbb R)$ from the right by a cocompact action of the fundamental group $\Gamma$ of a compact hyperbolic surface. -It remains to understand how to reverse the sign of $v$. For this recall, that we can identify $PSL(2,\mathbb R)/\Gamma$ with the unit tangent bundle of a hyperbolic surface (whose $\pi_1$ is equal to $\Gamma$). Now, the flow given by $v$ is the horocyclic flow. In order to reverse it, we can take a hyperbolic surface that admits and orientation reversing isometric involution. Such an involution clearly lifts to the unit tangent bundle and it sends the horocyclic flow to its inverse.<|endoftext|> -TITLE: Curve with no embedding in a toric surface -QUESTION [11 upvotes]: I am looking for a smooth proper curve $C$ such that there does not exist any closed embedding $C \to S$ where $S$ is a (normal projective) toric surface. -Since $C$ is smooth I believe it suffices to consider smooth projective toric surfaces $S$ since we may always perform a toric resolution of singularities and the strict transform of $C$ will be isomorphic to $C$ since $C$ is smooth. -Using the result on p.25 of Harris Mumford, On the kodaira dimension of the moduli space of curves, I can conclude that a very general curve cannot have any such embedding. -However, I am not able to write down an explicit example. Does anyone know such an example or what sort of obstruction might work to check this in particular examples. - -REPLY [13 votes]: A generic curve of genus $5$ is not a hypersurface in a toric surface. This argument is going to use conceptual ideas from Haase and Schicho's paper "Lattice polygons and the number $2i+7$", plus a bunch of case analysis. -Let's start with generalities about a curve $C$ in a toric surface $S$, other than one of the boundary divisors. The divisor $C$ gives a nef line bundle $\mathcal{O}(C)$ on $S$, which gives a lattice polygon $P$ in $\mathbb{Z}^2$. (See, for example, Lecture 4 here.) The number of interior lattice points of $P$ will be the genus of $C$. -Let $Q$ be the convex hull of the interior lattice points. So, in our genus $5$ case, $Q$ will be a lattice polytope with $5$ lattice points. Such a polytope either (Case 0) has no interior lattice points (Case 1) has one interior lattice point (Case 2) is a triangle with two interior lattice points or (Case 3) is a line segment of length $4$. -Case 0 Polytopes with no interior lattice points are trapezoids with vertices of the form $(0,0)$, $(a,0)$, $(0,1)$, $(b,1)$. If we want $5$ lattice points, our options are -$$\mbox{polygon 0a} = \begin{matrix} -\bullet&&& \\ \bullet&\bullet&\bullet&\bullet \\ -\end{matrix} \qquad -\mbox{polygon 0b} = \begin{matrix} -\bullet&\bullet& \\ \bullet&\bullet&\bullet \\ -\end{matrix}$$ -Case 1 There are $12$ polytopes with $1$ interior lattice point, which you can see in Figure 6 of Haase and Schicho. There are $3$ of these with $5$ lattice points: -$$ -\mbox{polygon 1a} =\begin{matrix} &\bullet& \\ \bullet&\bullet&\bullet \\ &\bullet& \\ \end{matrix} \qquad -\mbox{polygon 1b} =\begin{matrix} \bullet&& \\ \bullet&\bullet&\bullet \\ &\bullet& \\ \end{matrix} \qquad -\mbox{polygon 1c} =\begin{matrix} \bullet&& \\ \bullet&\bullet&\bullet \\ \bullet&& \\ \end{matrix}.$$ -Case 2 Wei and Ding, "Lattice polygons with two interior lattice points", list all lattice polytopes with $2$ interior lattice points. Only one of them has $5$ lattice points: -$$ -\mbox{polygon 2} = \begin{matrix} &&\bullet& \\ \bullet&\bullet&\bullet& \\ &&&\bullet \\ \end{matrix} -$$ -Case 3 Finally, here is the line segment: -$$\mbox{polygon 3}=\begin{matrix} \bullet&\bullet&\bullet&\bullet&\bullet \end{matrix}$$ -So $Q$ must be one of the $7$ polygons above. -Haase and Schicho show that not every polytope can occur as $Q$. Hasse and Schicho make the following definition: Let $v$ be a vertex of a lattice polytope $Q$, and let $x$ and $y$ be the minimal lattice vectors along the edges incident to $v$. Hasse and Schicho say that $v$ is a "good corner" if $(\vec{x}, \vec{y})$ is $SL_2(\mathbb{Z})$ equivalent to $((1,0), (-1,k))$ for some $k$. Haase and Schicho, Lemma 9, show that $Q$ is always either the empty set, a line segment, or a polytope with good corners. Polygons 0a and 2 do not have good corners, so we can eliminate them now. -We now look at Haase and Schicho, Lemma 8. This lemma says: Let $a_1$, $a_2$ and $b$ be integers with $GCD(a_1, a_2)=1$. Suppose that $a_1 x_1 + a_2 x_2 \leq b$ on $Q$, and that equality occurs at at least two lattice points. Then $a_1 x_1 + a_2 x_2 \leq b+1$ on $P$. -We will apply this result to eliminate polygons 0b and 3. Take $(a_1, a_2) = (0,1)$, so $a_1 x_1 + a_2 x_2$ is the second coordinate. Then Hasse and Schicho's lemma show that the polytope P is contained in a horizontal strip of width $3$ when $Q$ is polygon 0b, and width $2$ in polygon 3. This means that the corresponding curve is trigonal or hyperelliptic, respectively. A generic genus $5$ curve is not trigonal or hyperelliptic. -This leaves cases 1a, 1b and 1c. Haase and Schicho define a polygon they call $Q^{(-1)}$, whose definition I will let you read in their paper, and show that $P \subseteq Q^{(-1)}$. In the figure below, I have redrawn polygons 1a, 1b and 1c with solid dots $\bullet$, and drawn the additional vertices of $Q^{(-1)}$ as hollow circles $\circ$. -$$ -\begin{matrix} &&\circ&& \\ &\circ&\bullet&\circ& \\ \circ&\bullet&\bullet&\bullet&\circ \\ &\circ&\bullet&\circ& \\ &&\circ&& \\ \end{matrix} \qquad -\begin{matrix} \circ&&&& \\ \circ&\bullet&\circ&&& \\ \circ&\bullet&\bullet&\bullet&\circ \\ &\circ&\bullet&\circ& \\ &&\circ&& \end{matrix} \qquad -\begin{matrix} \circ&&&& \\ \circ&\bullet&\circ&&& \\ \circ&\bullet&\bullet&\bullet&\circ \\ \circ&\bullet&\circ&&& \\ \circ&&&& \\ \end{matrix}.$$ -In each case, there are $13$ lattice points in $Q^{(-1)}$, and thus $\leq 13$ lattice points in $P$. The curve $C$ is the zero locus of a polynomial with this Newton polytope, so it depends on $13$ parameters. But rescaling either of the variables, or rescaling the whole polynomial, does not change the isomorphism class of the curve, so we really only have $10$ parameters. The moduli space of curves of genus $5$ has dimension $12$, so the generic curve of genus $5$ is not a zero locus of a polynomial of any of the above forms. This concludes our case analysis. - -Writing down an explicit genus $5$ curve which works still seems interesting. We can avoid cases 0b and 3 by just not making our curve hyperelliptic or trigonal. It isn't clear to me what explicit criterion avoids cases 1a, 1b or 1c.<|endoftext|> -TITLE: Existence of smooth function that characterizes boundary and interior of set -QUESTION [6 upvotes]: It is well known that every closed set $A \subset \mathbb{R}^{n}$ is the zero level set of some smooth function. It follows that every closed set is also the zero sublevel set of some smooth function, i.e. -\begin{align*} -A &= \{x \in \mathbb{R}^{n} : f(x) \le 0 \}. -\end{align*} -I am wondering if one can easily characterize closed sets for which there exist a smooth function $f$ such that the following stronger conditions hold -\begin{align*} -\partial A &= \{x \in \mathbb{R}^{n} : f(x) = 0 \}, \text{ and}\\ -A^{\circ} &= \{x \in \mathbb{R}^{n} : f(x) < 0 \}? -\end{align*} - -REPLY [5 votes]: I think every closed set $A \subset \mathbb{R}^{n}$ has this property. Let $\{\phi_k\}_{k\in\mathbb{N}}\subset C^\infty_c(\mathbb{R}^{n})$ a countable collection of non-negative smooth functions with compact support such that $A^\circ=\bigcup_{k\in\mathbb{N}}\{\phi_k>0\}$ (for instance, $\{\phi_k>0\}$ may be balls of some countable covering of $A^\circ$). Then the series -$$ \sum_{k=0}^\infty\, {2^{-k}\| \phi_k\|_{C^k}^{-1}}\,\, \phi_k$$ -normally converges with derivatives of any order to a smooth function $f_-$ with $\{f_->0\}=A^\circ$. Here $\displaystyle\| \phi \|_{C^k}=\max_{ \alpha\in\mathbb{N}^n\atop |\alpha|\le k }\|\partial^\alpha\phi \|_{\infty,\mathbb{R}^{n}}$ is the standard ${C^k}$-norm. The same construction for $A^c$ produces a smooth $f_+$ with $\{f_+>0\}=\mathbb{R}^{n}\setminus \overline{A}$; then $f:=f_+-f_-$ has the required properties, that is $\{f<0\}=A^\circ$, $\{f>0\}=\mathbb{R}^{n}\setminus \overline{A}$, and $\{f=0\}=\partial A$.<|endoftext|> -TITLE: Pushouts and products in categories -QUESTION [7 upvotes]: This has to do with the "pushout-product" construction. -In a category $\mathcal{C}$, suppose we have $C\gets A\to B$ with pushout $D$ -and $Y\gets W\to X$ with pushout $Z$. Then we can form -$$ -(C\times Z) \cup_{C\times Y} (D\times Y) -\gets -(A\times X) \cup_{A\times W} (B\times W) -\to -B\times X . -$$ -This diagram cones to $D\times Z$. -Questions: - -Is $D\times Z$ the pushout? -It is the pushout in set and in various topological contexts, so is $D\times Z$ the pushout for certain good-enough categories? -Is there a good reference for these answers? - -REPLY [7 votes]: As Simon says in the comments, it is sufficient that the product preserves pushouts in each variable, which is the case in Set and in any cartesian closed category of spaces. (Indeed, the product can be replaced by any two-variable functor that preserves pushouts in each variable.) -Unfortunately at the moment I can't find a good reference that says exactly this, although I would be more surprised than not if it doesn't exist somewhere in the literature. Simon sketched one proof method in the comments; here's a slightly more abstract one that is at least closer to some things in the literature. -First notice that the goal is equivalent to saying that the pushout product functor $\hat{\times} : \mathcal{C}^{\mathbf{2}} \times\mathcal{C}^{\mathbf{2}} \to \mathcal{C}^{\mathbf{2}} $ takes a pair of pushout squares (regarded as morphisms in the arrow category $\mathcal{C}^{\mathbf{2}} $) to a pushout square. Since pushout squares are closed under composition (again, as morphisms in $\mathcal{C}^{\mathbf{2}} $), it suffices to show that $\hat{\times}$ preserves pushout squares in each variable separately. Thus, we can reduce to the case where we have a pushout $D$ of $C\leftarrow A \to B$ and a morphism $W\to X$, and we want to show that $D\times X$ is the pushout of -$$ (C\times X) \cup_{C\times W} (D\times W) \leftarrow (A\times X) \cup_{A\times W} B\times W \to B\times X $$ -Now there is a commutative cube in which the top and bottom faces are the images of our given pushout square $D = C\cup_A B$ under the functors $(-)\times W$ and $(-)\times X$, while the vertical arrows are induced by the map $W\to X$. I trust you can draw this cube; let's orient it so that $A$ and $B$ appear on the back face and $C$ and $D$ appear on the front face. -The back and front faces of this cube are not pushouts. But if we take the pushouts of their underlying spans, the induced maps from these pushouts to the lower-right corners are the two pushout-product maps in question, and the induced square between them is the one we're interested in. A diagram of this sort of "pushouts in two faces of a cube" can be found, for instance, at the top of page 9 of this paper; it's not there in the situation of a pushout product, but the immediate goal is the same, namely to show that the relevant square is a pushout. This follows by repeated application of the pushout pasting lemma (in both directions) to the four squares that are known to be pushouts.<|endoftext|> -TITLE: What is the relationship between spinors and supermanifolds and fermions? -QUESTION [13 upvotes]: I have the following two impressions about fermions in physics. I'm confused about their accuracy, and their compatibility: - -To consider the behavior of a fermion, whose intrinsic spin is described by a representation $V$ of the group $Spin(p,q)$, on a pseudo-Riemannian manifold $M$ of signature $(p,q)$, you first introduce a spin structure on $M$. Then the fermion field is a section of the bundle associated to $V$. - -To consider the behavior of a fermion on a pseudo-Riemannian manifold $M$ of signature $(p,q)$, you first turn $M$ into a supermanifold. Then the fermion field is a superfunction on $M$ with some constraints coming from its intrinsic spin. - - -Question: Is either of (1) or (2) close to accurate? What major points or subtleties have I missed? If both are close to accurate, then how does one "translate" between the formalism of (1) and the formalism of (2)? - -REPLY [5 votes]: This is a bit of a repackaging of the same info in the other answer, but maybe it will be more clear. -The short answer is (almost) both: A fermion is a section of the parity shifted spinor bundle on a manifold. As such, you can't have a fermion without a spin structure. -Each aspect of this can be considered separately: there is no classical reason that an anti-commutative field has to be a section of the spinor bundle, and there is no reason that a section of the spinor bundle has to be anti-commutative. However, in physics the Spin-Statistics Theorem says that to have a consistent, Lorentz-invariant theory in >2 spatial dimensions, all anti-commutative fields must be spinors (have half-integral spin). -However, you only need the parity shifted bundle here. The full formalism of supermanifolds is for when you have supersymmetry, which is an odd (ie, anti-commuting) symmetry that relates bosons and fermions. -You can look at this in two ways. The first is as supersymmetric quantum mechanics, where you have maps from, say, the super manifold $\mathbb{R}^{1|1}$ to a Riemannian manifold. Here, the need for a spin-structure arises when you try to quantize theory in order to patch together the Clifford algebras that arise on each local chart. -The second way to look at this is to have your fields be functions on a supermanifold. Here, the supermanifold is modeled on super-Minkowski space, which is acted on by the super-Poincare group. In super-Minkowski space, the odd part is (some number of copies of) the parity shifted the spinor bundle, so the need for the spin-structure is part of the definition. -Dan Freed's notes Classical field theory and supersymmetry on this stuff are very good.<|endoftext|> -TITLE: Existence of equivariant triangulations -QUESTION [5 upvotes]: I'm looking for a compatibility result which links two types of structures that could be imposed on a topological space $X$: - -Call $X$ triangulable if there exists a finite simplicial complex $K$ whose geometric realization $|K|$ admits a homeomorphism to $X$. -Call $X$ involutive if it admits a nontrivial $\mathbb{Z}/2$-action, in the sense that there is a continuous $\iota:X \to X$ satisfying $\iota \circ \iota = \text{id} \neq \iota$. - - -Here is the question: let's say a given space $X$ admits a triangulation $K$ and an involution $\iota$ in the senses described above. Is it true that we can find another triangulation $K'$, ideally a subdivision of $K$, with respect to which $\iota$ is a simplicial map? - -There are more general versions of this question about which I'm curious, obtained by replacing $\mathbb{Z}/2$ by another finite group $G$ acting on a triangulable $X$. If the desired statement is true, I would love to be pointed to a reference; and if it is false, I wonder if there is a nice obstruction to the existence of $K'$. - -REPLY [3 votes]: There are involutions $\sigma$ of the 3-sphere, whose fixed-point sets are wild 2-spheres: The fixed-point set cannot be a subcomplex of any triangulation, hence, $\sigma$ cannot be PL in any triangulation. -Bing, R. H., A homeomorphism between the 3-sphere and the sum of two solid horned spheres, Ann. Math. (2) 56, 354-362 (1952). ZBL0049.40401. -See also here for Calegari's take on Bing's proof. -Edit. There is even (unique in some sense) free involution of the 4-sphere which cannot preserve a triangulation (this is due to Ruberman). Thus, a bad fixed point set is not the only obstruction.<|endoftext|> -TITLE: Trace of a matrix associated to posets -QUESTION [7 upvotes]: Let $P$ be a finite connected poset with $n$ elements. Let $C=(c_{x,y})$ be the $n \times n$ matrix with entry 1 in case $x \leq y$ and 0 else. -The Coxeter matrix of $P$ is defined as the matrix $M_P=-C^{-1}C^T$. -Let $u_P$ be defined as the trace of $M_P^2$. - -Question 1: Is it true that $u_P$ is an odd integer in case $P$ is a lattice? Does it have a nice interpretation in this case? - -I can prove question 1 for distributive lattices. - -Question 2: Is $u_P$ an odd integer in case $P$ is just a bounded poset? - -Questions 1 and 2 have a positive answer for posets with at most 9 points. In case this is true for general bounded posets, there surely is a nice reason. -For a general connected poset $P$ $u_p$ can be zero. -Here is how to get a poset (not bounded) with $u_p=0$ using Sage: -n=6 - -posets=[P for P in Posets(n) if P.is_connected()] - -U=[P for P in posets if ((P.coxeter_transformation())^2).trace()==0] - -P=U[0] - -display(P) - -REPLY [4 votes]: Here is a nice proof for posets $P$ with a global maximum $M$ (it works dually for posets with a global minimum, but not for general posets as the example in the question shows) suggested by comments of Darij Grinberg and Fedor Petrov. -Note first that the entries of the coxeter matrix $Co=Co_P$ (formerly known as $M_P$) are given by $Co_{x,y}=- \sum\limits_{z \in P: z \geq x}^{}{\mu(y,z)}$, where $\mu$ is the Moebius function of $P$. -Now for any square matrix $A$ over the ring of integers, we have $\operatorname{tr}(A^2) \equiv \operatorname{tr}(A)^2 \equiv \operatorname{tr}(A) \mod 2$. -Thus when we show that $\operatorname{tr}(A) \equiv 1 \mod 2$, we are done. -Now by Hall's theorem on chains we have $\mu(x,y)=-c_1+c_2-c_3+c_4-....$ when $c_i$ denotes the number of length $i$ chains starting at $x$ and ending at $y$. -Mod 2 this is equal to the number of chains from $x$ to $y$ (since there is difference between - and + in mod 2). -Thus $\operatorname{tr}(Co)=- \sum\limits_{x \in P} \sum\limits_{z \in P: z \geq x}^{}{\mu(x,z)}$ is equal mod 2 to the total number of non-empty chains in the poset. -Now for any chain $K$ we can add or remove the maximum $M$ of the poset $P$, so we can partition all non-empty chains except $\{ M \}$ into pairs. -Thus $\operatorname{tr}(Co) \equiv 1 \mod 2$.<|endoftext|> -TITLE: When are principal bundles supporting Cartan connections isomorphic? -QUESTION [5 upvotes]: Suppose I have two Cartan geometries $(\mathscr{G}_1,\omega_1)$ and $(\mathscr{G}_2,\omega_2)$ of type $(G,H)$ over the same manifold $M$. What conditions on $G$ and $H$ allow us to conclude that $\mathscr{G}_1$ and $\mathscr{G}_2$ are isomorphic as principal $H$-bundles? -It seems to be a common implicit assumption in the literature that $\mathscr{G}_1$ is always isomorphic to $\mathscr{G}_2$ in the cases we usually look at. In particular, for parabolic geometries, it seems to be folklore that this is true. -Previously, I had implicitly assumed that such an isomorphism always exists for Cartan geometries of all types, but I recently thought of the following example. If I have a Hermitian holomorphic line bundle, then I can construct a Cartan geometry of type $(\mathbb{C}^m\rtimes\mathrm{U}(1),\mathrm{U}(1))$ corresponding to the Chern connection. However, in general, there are too many line bundles over a given complex manifold for them to all be associated (in the sense that $L\cong\mathscr{G}\times_{\mathrm{U}(1)}\mathbb{C}$) to the same principal $\mathrm{U}(1)$-bundle, so there must be nonisomorphic principal $\mathrm{U}(1)$-bundles admitting Cartan connections of this type over the same manifold. -I’ve thought about this for a few days now, and I imagine there’s probably a nice general condition on $(G,H)$, but I’m not seeing what that condition might be. - -REPLY [3 votes]: This is not a complete answer, but I think that it might help to clear up some misunderstandings. It is not, in general, true that all the principal $H$-bundles over $M$ supporting a Cartan connection of type $(G,H)$ are isomorphic, though the OP's proposed example does not actually show this. I think that the following discussion may help. -To fix notation, let's recall what we mean by a "Cartan connection of type $(G,H)$": Here $G$ is a Lie group with Lie algebra $\frak{g}$ and $H$ is a Lie subgroup with Lie algebra ${\frak{h}}\subset{\frak{g}}$. The representation $\mathrm{Ad}:H\to\mathrm{Aut}({\frak{g}})$ preserves the subalgebra ${\frak{h}}$ and so induces a representation $\rho:H\to \mathrm{Aut}({\frak{g/h}})$. If $\pi:B\to M$ is a principal right $H$-bundle, let $X_v$ for $v\in\frak{h}$ be the vertical vector field on $B$ whose flow is right action by $\mathrm{exp}(tv)$. Then a Cartan connection of type $(G,H)$ on $\pi:B\to M$ is a $\frak{g}$-valued $1$-form $\gamma:TB\to \frak{g}$ with the following properties: - -$\gamma_u:T_uB\to{\frak{g}}$ is an isomorphism for all $u\in B$. -$\gamma\bigl(X_v(u)\bigr) = v$ for all $u\in B$ and all $v\in\frak{h}$. -$R^*_h(\gamma) = \mathrm{Ad}(h^{-1})(\gamma)$ for all $h\in H$. - -It is important to note that not every principal right $H$-bundle over $M$ supports a Cartan connection of type $(G,H)$. This is because such a Cartan connection $\gamma$ defines an isomorphism $\iota_\gamma:TM\to B\times_\rho {\frak{g/h}}$. To see this, let $\omega = \gamma\,\mathrm{mod}\,{\frak{h}}:TB\to {\frak{g/h}}$. The above axioms imply that $\omega_u:T_uB/V_uB\to {\frak{g/h}}$ is an isomorphism for all $u\in B$, where $V_uB\subset T_uB$ is tangent to the fiber of $\pi:B\to M$. Since we have a canonical isomorphism $T_uB/V_uB\to T_{\pi(u)}M$, it follows that we can regard $\omega$ as defining an isomorphism $\omega_u:T_{\pi(u)}M\to {\frak{g/h}}$ for all $u\in B$ that satisfies $\omega_{u\cdot h} = \rho(h^{-1})(\omega_u)$ for all $u\in B$ and all $h\in H$. By the very definition of $B\times_\rho{\frak g/h}$, this establishes the claimed isomorphism $\iota_\gamma:TM\to B\times_\rho{\frak g/h}$. -Conversely, if an isomorphism $\iota:TM\to B\times_\rho{\frak g/h}$ is given, then one can construct a Cartan connection of type $(G,H)$ on $B$. -Thus, one can see why the OPs construction starting with a line bundle $L$ endowed with a $\mathrm{U}(1)$-connection does not automatically imply that there is a Cartan connection of the desired type on $M$. For example, in this case, if a Cartan connection existed, then $TM$ would have to be isomorphic to $L\otimes \mathbb{C}^n = B\times_\rho {\frak g/h}$, and this is generally not the case. -However, there is a simpler example to demonstrate that not all $H$-bundles that admit Cartan connections of type $(G,H)$ are isomorphic: Here, let $n=3$, let $H=\mathrm{SO}(2)$ and let $G = \mathbb{R}^3\rtimes H$, where $H=\mathrm{SO}(2)$ acts on $\mathbb{R}^3$ by rotation in the second and third coordinates. An $H$-bundle $\pi:B\to M^3$ is just an $\mathrm{SO}(2)$-bundle, so it has an Euler class (which could be nonzero) and the associated bundle $B\times_\rho \mathbb{R}^3$ is a sum of a trivial bundle and a $2$-plane bundle. If there is a Cartan connection on $B$, then we get an isomorphism of $TM$ with the sum of a trivial bundle and a $2$-plane bundle. In particular, this means that $M$ is oriented and we have a nonvanishing vector field on $M$ together with a $2$-plane subbundle that has a well-defined Euler class. -Now, every oriented $3$-manifold has a trivial tangent bundle, but once one chooses a nonvanishing vector field, the Euler class of the complementary $2$-plane bundle is determined and may very well be nonzero. For example, let $M = S^1\times S^2$. If we choose the vector field tangent to the $S^1$-fibers, then the complementary $2$-plane field is nontrivial on each $S^2$-fiber. Meanwhile, if we choose a trivialization of the tangent bundle of $M$, then letting the vector field be one of the three trivializing vector fields, the complementary $2$-plane bundle will be trivial. -Thus, we can have two $H$-bundles over $M$ that are not isomorphic even though they both admit Cartan connections of type $(\mathbb{R}^3\rtimes H,\ H)$. -It follows that the very first criterion one needs to have in order for all the Cartan connections of type $(G,H)$ to have isomorphic underlying $H$-bundles is that all of the structure reductions of the canonical $\mathrm{GL}(n,\mathbb{R})$-structure on $TM$ to a $\rho(H)$-structure be isomorphic. This is a very strong condition on $\rho(H)$ and $M$, and whether it is met depends on both $\rho(H)$ and $M$. -Meanwhile, for most of the familiar examples in parabolic geometry, $\rho(H)$ is some large group such as $\mathrm{GL}(n,\mathbb{R})$, $\mathrm{SL}(n,\mathbb{R})$, $\mathrm{CO}(n)$, or $\mathrm{SO}(n)$, and it happens that this uniqueness is met trivially. This may account for the common (false) belief that prompted this question in the first place.<|endoftext|> -TITLE: Is the space of metric topologies over a given set dense (in the order sense)? -QUESTION [5 upvotes]: Suppose that $S$ is an infinite set and that $\alpha$ and $\beta$ are metrics over $S$ such that the topology induced by $\alpha$ is everywhere strictly finer than the metric induced by $\beta$, meaning that every open set $U$ in $\beta$ contains a set $V$ that is open in $\alpha$ but not in $\beta$. Suppose further that $S$ is dense (in the metric sense) with respect to both metrics, in the sense that, for any $x \in S$, for any $\epsilon > 0$ there exists $y \neq x$ within distance $\epsilon$ of $x$. Does there exist a metric $\gamma$ over $S$ that is everywhere strictly finer than $\beta$ but everywhere strictly coarser than $\alpha$? Or, contrarily, are there cases where it is known that no such $\gamma$ exists? - -REPLY [2 votes]: $\def\cl{\operatorname{cl}}$ -A large family of counterexamples can be constructed using the following -proposition: -Let $(S, T_1)$ be a topological space with two complementary dense -subspaces $A, B$. -Define $T_3 = \{ (A \cap U) \cup (B \cap V) \mid U, V \in T_1 \}$, -in other words $(S, T_3)$ is the topological sum of $A$ and $B$. -Let $T_2$ be a topology on $S$, finer than $T_1$, such that $T_3$ is -everywhere strictly finer than $T_2$. -If $T_2$ is semi-regular, then $T_2 = T_1$. -Sketch of a proof: - -Observe that $T_1|A = T_2|A = T_3|A$ and $T_1|B = T_2|B = T_3|B$. Moreover, -$\cl_3 V = \cl_A (V \cap A) \cup \cl_B (V \cap B)$ for every $V \subset S$. -From the assumption that $T_3$ is everywhere strictly finer than $T_2$ -it follows that both $A$ and $B$ are dense in $(S, T_2)$. Hence for -every $U \in T_2$ we have $\cl_2 (U \cap A) = \cl_2 (U \cap B) = \cl_2 U$. -Show that for any $U \in T_2$ we have $\cl_2 U = \cl_3 U = \cl_1 U$. -Deduce that if $U$ is a regular open set in $(S, T_2)$, then $U \in T_1$. -Therefore, if $T_2$ is semi-regular, $T_2 \subset T_1$. - -The application to your problem is easy. If $(S, T_1)$ is nonempty, -metrizable and dense in itself, there are many choices of $A,B$ and -$T_3$ will also be metrizable and dense in itself. Of course for any -$U \in T_1 \setminus \{\emptyset\}$ we have $U \cap A \in T_3 \setminus T_1$ -and if $T_2$ is to be metrizable it must certainly be semi-regular.<|endoftext|> -TITLE: Source of infection on chessboard -QUESTION [11 upvotes]: I am looking for the original source of the following well known problem. - -Seven unit cells of a 8×8-chessboard are infected. In one time unit, the cells with at least two infected neighbors (having a common side) become infected. Can the infection spread to the whole chessboard? - -(It follows since the perimeter of infected part cannot increase.) -This problem appears in "connoisseur's collection" of Peter Winkler, with the following note: - -This lovely problem appeared in the Soviet magazine KVANT around 1986, then migrated to Hungary. - -I am also interested about Hungary. -P.S. It is found: Moscow mathematical olimpiad 1986, (8-4). Indeed, it appeared in Квант 1986, № 8, с. 57. - -REPLY [5 votes]: The Hungarian connection is in - -Gábor Pete: Hogyan gyepesítsünk kockát? [How to make the cube weedy?] Polygon (Szeged) VII:1 (1997), 69-80. - -József Balogh and Gábor Pete, Random Disease on the Square Grid (1997)<|endoftext|> -TITLE: Why is it still common to not motivate results in publications? -QUESTION [36 upvotes]: This is a question about practice and publication of research mathematics. -On the Wikipedia Page for Experimental Mathematics, I found the following quote: - -Mathematicians have always practised experimental mathematics. Existing records of early mathematics, such as Babylonian mathematics, typically consist of lists of numerical examples illustrating algebraic identities. However, modern mathematics, beginning in the 17th century, developed a tradition of publishing results in a final, formal and abstract presentation. The numerical examples that may have led a mathematician to originally formulate a general theorem were not published, and were generally forgotten. - -My question concerns the last two sentences. I've heard several of my professors complaining about precisely this: in many mathematics papers, one can read the whole paper without ever understanding how the authors came up with the arguments in the first place. -Questions: -1.) (Historical:) Why did mathematicians stop publishing motivational steps when publishing mathematics? (For example, were there particular schools of mathematicians who actively advocated for this?) -2.) Is there any movement amongst mathematicians today to change this tradition? (By this, I do not mean movement on a personal level; I know that many mathematicians motivate results in their publications with preliminary calculations they have performed when they initially thought about the problem. Instead, what I am looking for is a movement on the community level to initiate a change) -3.) There seems to be a lot of disadvantages to the above practice; by communicating how one thinks about a problem, others would be able to copy the author's ways of thinking which will add to the knowledge of the greater mathematical community. Is there any practical benefit for not motivating results in papers? -Clarification. By "publication", I mean everything that one makes available to the greater mathematical community (so anything that one makes available on one's webpage is included (e.g. preprints) as well as anything uploaded to the ArXiv. -4.) Is the assertion made in the last two sentences in the quote accurate. (Thanks to YCor for pointing this out.) - -REPLY [3 votes]: This is not in any way a comprehensive answer to your question, but it may be worth pointing out that while there are certainly several disadvantages to "erasing the trail" up to a result, there is also (at least) one major advantage: it lessens the "path dependence" and encourages others to think about possible applications of the result that would not have occured to the original discoverer. -For example, many people discover a result by thinking very carefully about one specific example. But the beauty of abstract math is that the same general result often applies to a huge range of concrete examples that differ hugely in their details, but share the minimal mathematical structure necessary for the result. The original discoverer may be so used to thinking about the result in the context of one type of example that they miss applications to other classes of examples. Presenting the detailed example that they worked through to motivate the result risks similarly locking other readers into the same mode of though. Whereas if a reader approaches a new result with a completely different example/application in mind, then they may be able to more easily extend the result to new corollaries that the original discoverer didn't think of.<|endoftext|> -TITLE: Roadmap for studying Galois deformation theory/modularity theorems from a modern perspective -QUESTION [11 upvotes]: I am a graduate student with some background in Galois deformation theory. I am familiar with the basics (the existence of a universal deformation space with prescribed conditions) and with some examples in Galois deformation theory, as well as with the some of the conjectural relations with $p$-adic systems of Hecke eigenvalues (i.e. $R=T$ theorems). I have seen applications and arguments in the case of $\operatorname{GL}_2$. -I'd like to get myself to a spot where I am somewhat familiar with the more modern advancements in the theory, by which I mean, Taylor–Wiles–Kisin patching, Kisin's deformation rings, potential modularity theorems, the Calegari–Geraghty method, the 6-author "Patching and the $p$-adic local Langlands correspondence" paper, etc... -It seems not so easy to start reading such papers without an idea where you're going because frequently they approach 100 pages with many logical dependencies. On the other hand, the more introductory accounts like the Darmon-Diamond-Taylor papers appears to be slightly outdated (as far as I understand). -So, what would be a good roadmap for studying Galois def theory/modularity theorems from a modern perspective? Where should I begin? -Thanks! - -REPLY [8 votes]: A fantastic place to start would be Toby Gee's notes from the 2013 Arizona Winter School. This gives a nice overview of the theory as it then existed -- things have of course moved on further since then, but it's significantly more "modern" than Darmon--Diamond--Taylor, for instance. The course concludes with a proof of a GL(2) modularity lifting theorem over totally real fields, a result which goes back to Fujiwara in 2000 or so, but rather than presenting Fujiwara's proof, the notes give a "modernized" proof using more up-to-date machinery.<|endoftext|> -TITLE: A selection principle in measure theory -QUESTION [7 upvotes]: A Borel subset $B$ of the unit interval $\mathbb I=(0,1)$ is defined to be a density neighborhood of a set $A\subseteq\mathbb I$ if for every $a\in A$ we have $$\lim_{\varepsilon\to0}\frac{\lambda(B\cap[a-\varepsilon,a+\varepsilon])}{2\varepsilon}=1$$where $\lambda$ denotes the Lebesgue measure on $\mathbb I$. - -Problem. Let $A\subseteq\mathbb I$ be a set of Lebesgue measure zero and $(B_n)_{n\in\omega}$ be a sequence of Borel density neighborhoods of $A$. Is there a sequence of compact sets $(K_n)_{n\in\omega}$ such that $K_n\subseteq B_n$ for all $n\in\omega$ and the set $K=\bigcup_{n\in\omega}K_n$ is a density neighborhood of $A$? - -REPLY [2 votes]: Professor Wladyslaw Wilczynski kindly informed me that the answer to this problem is negative. -Take any Lebesgue null dense $G_\delta$-set $A$ in the real line $\mathbb R$. Choose a countable dense subset $\{x_n\}_{n\in\omega}$ in $\mathbb R\setminus A$. Since the density topology $\tau_d$ on the real line is Tychonoff, for every $n\in\omega$ there exist disjoint $\tau_d$-open Borel sets $D_n,E_n\subseteq\mathbb R$ such that $x_n\in D_n$ and $A\subseteq E_n\subseteq\mathbb R\setminus\{x_k\}_{k\in\omega}$. -For every $n\in\omega$ consider the Borel set $B_n=\bigcap_{k\le n}E_k$. It is clear that $B_n$ is a density neighborhood of $A$. It can be shown that the sequence $(B_n)_{n\in\omega}$ has the required property: for any sequence of compact sets $K_n\subseteq B_n$, the union $\bigcup_{n\in\omega}K_n$ is not a density neighborhood of $A$.<|endoftext|> -TITLE: Bilinear forms in compact/semisimple Lie group theory -QUESTION [8 upvotes]: If you look up the list of compact or semisimple Lie groups, you will see that three out of four infinite families (B, C and D) are defined in terms of a bilinear form on a vector space, either symmetric or skew-symmetric. -Are there any underlying reasons for this prominence of bilinear/quadratic forms in Lie group theory? Why do they, and not any other geometric objects, play such a fundamental role? - -REPLY [8 votes]: According to a theorem of Serre, all semi-simple Lie groups are linear algebraic groups. See https://en.wikipedia.org/wiki/Complex_Lie_group for the precise statement and a reference. -That shows why we should look at "algebraic functions" when looking for complex semi-simple Lie groups. Natural place to start is subgroups of $GL(n, \mathbb{C})$ that preserve some linear forms. But then we get something isomorphic to $GL(n, \mathbb{C})$ as such subgroup has to preserve the kernel. Bilinear forms are the next best things and it turns out they provide plenty of examples. But we don't have to stop there! The complex Lie group $F_4$ can be defined as the subgroup of $GL(26, \mathbb{C})$ fixing a symmetric trilinear form. And the complex simple Lie group $G_2$ can be defined as the stabilizer of a generci $3$-form on $\mathbb{C}^7.$ There are similar descriptions for $E$-series. See e.g. this answer by Robert Bryant https://mathoverflow.net/a/99795/6818 (I think he wrote about this more explicitly somewhere else on MO, but I have trouble finding it. Anyway... this description of $E$-series goes back to Elie Cartan.)<|endoftext|> -TITLE: On free lattices -QUESTION [5 upvotes]: Free distributive lattices on a finite set exist and are finite, while free modular lattices on a finite set exist but are not finite when the set has at least 4 elements. - -Question: Is there a class (presumably, a variety in the sense of universal algebra) $C$ of lattices larger than the class of distributive lattices such that free $C$-lattices on a finite set exist (for all finite sets) and are finite? - -If yes, is there a "largest" such class $C$? - -REPLY [7 votes]: A variety is called locally finite if it has the property that its finitely generated algebras are finite. This is equivalent to the property that its finitely generated free algebras are finite. So, the questions may be rewritten as: -(1) Is there a locally finite variety of lattices larger than the variety of distributive lattices? -(2) Is there a largest locally finite variety of lattices? -Every variety generated by a single finite algebra is locally finite, so there are lots of locally finite varieties of lattices. Any variety generated by a finite, nondistributive lattice will properly contain the variety of distributive lattices, hence will be an answer to Question (1). -[There do exist locally finite varieties of lattices that are not generated by a single finite lattice, like the variety generated by all lattices of height 2.] -If there were a largest locally finite variety of lattices, then by the answer to Question (1) it would have to contain every finite lattice. But the variety generated by all finite lattices is the variety of all lattices, and the variety of all lattices is not locally finite. Hence the answer to Question (2) is No. -A comment on the comments: -The class of semidistributive lattices forms a quasivariety, and therefore -has free objects over any set. But this quasivariety is not locally finite. The free semidistributive lattice on 3 generators is infinite.<|endoftext|> -TITLE: Showing integrability of a locally integrable function on a bounded domain under some additional assumptions -QUESTION [5 upvotes]: Suppose $\Omega\subset \mathbb{R}^3$ is a smooth and bounded domain, and $f:\Omega\to[0,\infty]$ is a given function which is finite almost everywhere and satisfies - -Assumption A: For all $g\in C_0^1(\Omega)$ we have the product $fg\in L^1(\Omega)$. (Here $C_0^1(\Omega)$ refers to functions which are continuously differentiable in $\Omega$ and extend continuously to $0$ on $\partial\Omega$). - -Question 1: Can we show that $f\in L^1(\Omega)$? -Question 2: Does the answer to Question 1 change if we include some or all of the following assumptions: - -Assumption B: $f$ possesses a weak derivative which is finite almost everywhere in $\Omega$; - -Assumption C: There exists a nonnegative function $f_0 \in H^2(\Omega)\cap C(\bar{\Omega})$ such that $f-f_0=0$ (in the sense of trace) on $\partial\Omega$; - -Assumption D: There exists a nonnegative function $h\in H^1(\Omega)$ such that $h$ is nonzero almost everywhere in $\Omega$ and $f=-\ln h$ in $\Omega$. - - -Note: Assumption D more or less implies Assumption B. I wrote them separately in the hopes of formulating the problem as simply as possible. -Notation: Here $H^k$ is the standard Sobolev space notation for $W^{k,2}$. -9/14/20 Edit: -Question 1 has been answered in the affirmative. I additionally pose the following -Question 3: Answer Questions 1 and 2 in the case that Assumption A is replaced by - -Assumption A': $f\in L^1_{\text{loc}}(\Omega)$. - -REPLY [7 votes]: Let $(g_k)_{k\ge0}$ be a sequence of smooth functions such that $g_k(x)=1$ if $\text{dist}(x,\partial\Omega)\ge 2^{-k}$, -$g_k (x)=0$ if $\text{dist}(x,\partial\Omega)\le 2^{-k-1}$ and $0\le g_k\le 1$ everywhere. -To prove the affirmative answer to Question 1 by contrapositive, let $f\not\in L^1(\Omega)$ be given: we want to find $g\in C^1_0(\Omega)$ such that $fg\not\in L^1(\Omega)$. We can assume $fg_k\in L^1(\Omega)$ for all $k$, otherwise we are done with $g=g_k$ for some $k$. Then $ \int_\Omega fg_k$ is an increasing sequence of positive real numbers, that diverges to $+\infty$, for if it were bounded, $f\in L^1(\Omega)$ by Beppo Levi's theorem. So for some subsequence $(g_{k_j})_j$ we have $ \int_\Omega fg_{k_{j+1}}\ge \int_\Omega fg_{k_j}+1 $, that is $ \int_\Omega f(g_{k_{j+1}}-g_{k_j})\ge 1$. For all $j$ the function $g_{k_{j+1}}-g_{k_j}$ is bounded between $0$ and $1$, and supported in the set $\big\{ 2^{-k_{j+1}-1}\le \text {dist}(x,\partial\Omega)\le 2^{-k_j}\big\}$. -But then $g:=\sum _{j\ge1} \frac{g_{k_{j+1}}-g_{k_j} }j$ is a locally finite sum of smooth functions, hence smooth in $\Omega$; clearly $g(x)\to0$ for $x\to\partial\Omega$, and, again by Beppo Levi's theorem, $\int_\Omega fg\ge \sum_{j\ge1}\frac1j=+\infty$.<|endoftext|> -TITLE: Untangling two simple closed curves on a surface -QUESTION [5 upvotes]: Let $S$ be a smooth surface and $\gamma_1, \gamma_2$ be two transversal simple closed curves on it. Suppose moreover that there exists a simple closed curve $\gamma_1'$ on $S$ isotopic to $\gamma_1$ and such that $\#(\gamma_1\cap \gamma_2)>\#(\gamma_1'\cap \gamma_2)$. -Question. Is it true that there is a disk on $S\setminus (\gamma_1\cup\gamma_2)$ whose boundary is composed of one arc of $\gamma_1$ and one arc of $\gamma_2$? -Note that in case such a disk exists, one can construct an isotopy of $\gamma_1$ that would decrease the number of intersections of $\gamma_1$ with $\gamma_2$ by two. - -REPLY [3 votes]: This is also proved as Lemma 3.1 in -Joel Hass and Peter Scott, -Intersections of curves on surfaces, Israel Journal of Mathematics 51 (1985), 90–120. https://doi.org/10.1007/BF02772960<|endoftext|> -TITLE: A question on $SK_1$ of rings -QUESTION [5 upvotes]: Let $B$ be a commutative ring with unity and $B/nil(B):=B_{red}$, where $nil(B)$ is the nilradical of $B$. Is $SK_1(B)=SK_1(B_{red}) ?$ In particular, is it true when $B$ is an affine algebra over an algebraically closed field ? - -REPLY [7 votes]: Yes. An element in the kernel of $SK_1(B)\rightarrow SK_1(B_{red})$ is represented by a matrix $M\in GL_n(B)$ for some $n$. Write $\overline{M}$ for the reduction of $M$ mod $nil(B)$. Then $\overline{M}$ is a product of elementary matrices, all of which lift to elementary matrices over $B$. Adjusting $M$ accordingly, we can assume that $\overline{M}$ is the identity. -It follows that the elements on the diagonal of $M$ are all $1$ mod $nil(B)$, hence all units in $B$. This allows us to use elementary operations to convert $M$ to a diagonal matrix, which therefore (by Whitehead's lemma) represents the zero element of $SK_1(B)$. -(The same argument works if $B_{red}$ is replaced by $B/I$, where $I$ is any ideal contained in the Jacobson radical.)<|endoftext|> -TITLE: Set Theoretic Geology II: The structure of the directed partial order of grounds -QUESTION [6 upvotes]: In my previous question Set-theoretic geology: controlled erosion? -and the great answer by Jonas Reitz, I have learned a few things, starting from the awareness that I understand the fine-grain structure of Set Theoretic Geology even less than I had assumed. -That is, of course, good news: more to learn! -The second thing I have learned is: -if I want to understand more, I have to start from the STRUCTURAL STANDPOINT, ie I have to grasp, given a transitive model M (I could do away with that, by starting from V, but I prefer concrete set models), the structure of the partial order of grounds of $M$. -To be more specific, Let us begin with $GROUNDS(M)$, and take a look at its structure: it is a partial order, and looks like that it is directed. -So, given two grounds, say $G_1$ and $G_2$, there is a third G which refines both. -Joel's Modal Logic of Forcing is $S4.2$ (please correct me if I am wrong!), which makes sense to me: this logic corresponds exactly to directed partial pre-orders. -But here is where things become quite hazy to me: what about actual meets? -QUESTIONS - -When $GROUNDS(M)$ has the structure of a meet-semilattice? -When is $GROUNDS(M)$ equipped with a full lattice structure? -When $GROUNDS(M)$, assuming 1 and 2, is a complete (sups, infs) lattice? - -More related questions: -$GROUNDS(M)$ is a subclass of $TM(M)$, ie the class (set) of transitive sub-models of $M$, so it makes sense to loosen the questions above by asking when the infs and sups asked for are not part of the directed order, but still exist in $TM(M)$. -Any answer to any or some of the questions is welcome. - -REPLY [5 votes]: Mirco, this is also a fantastic question - the structure of the grounds as a partial order seems to be a very basic aspect of forcing that is not entirely understood. Once again I don’t have a complete answer, but I can provide some background & a few observations. -Intersection of grounds. As pointed out in the comments, it is not the case that the intersection of grounds is (necessarily) a ground - the intersection may fail to satisfy ZFC. However, the intersection does contain a ground (see Directedness) -Directedness. Theorem (Usuba): The grounds are downward-set-directed (that is, the intersection of any collection of set-many grounds contains a further ground). -This fundamental result resolved a number of open questions in Set Theoretic Geology - not least of which is the lovely fact that the Mantle (the intersection of all grounds) is always a model of ZFC. -Meets. Since we can get below any set-indexed collection of grounds, it’s natural to ask whether there is a unique, largest such ground below such a collection. I believe the answer is no (I think the example linked in comments by @gabe-goldberg Intersection of two generic extensions may be a good candidate for a counterexample, but I haven’t thought it through). -The Ground Axiom. The ground axiom states “there are no grounds except V itself” -- if this is the case, then GROUND(V) is trivial. -Least element. The Mantle (the intersection of all grounds) is, by Usuba’s result, a model of ZFC. If GROUND(V) has a least element, it is equal to the Mantle - this will happen exactly when V is a set forcing extension of a model of the Ground Axiom. If GROUND(V) does not have a least element, it may be the case that moving from the Mantle to V can be accomplished by class forcing. Finally, it may be the case that V is not even a class forcing extension of the Mantle (by any class forcing definable in the Mantle) -- see http://jdh.hamkins.org/the-universe-need-not-be-a-class-forcing-extension-of-hod/ . -With regards to your final “more related questions”, I believe that if infs or sups exist then they must be ground models, by an argument similar to @asaf-karagila ‘s in the comments - we can use directedness to get below a collection of grounds, so if an inf exists it is an intermediate model of ZFC between a ground and an extension, hence it is also a ground. -None of the above resolves your three Questions - under what circumstances do we get a nice structure on GROUND(V)? For example, if we start in a model of the Ground Axiom and carry out some forcing, what is the relationship between properties of the forcing we choose and the structure of grounds in the resulting extension? I really like this line of thinking as an avenue for understanding structural properties of forcing.<|endoftext|> -TITLE: What are good mathematical models for spider webs? -QUESTION [74 upvotes]: Sometimes I see spider webs in very complex surroundings, like in the middle of twigs in a tree or in a bush. I keep thinking “if you understand the spider web, you understand the space around it”. What fascinates me, in some sense it gives a discrete view on the continuous space surrounding it. -I started to wonder what are good mathematical models for spider webs. Obvious candidates are geometric graphs embedded in surfaces, or rather in space. One could argue that Tutte’s Spring Theorem from 1963 is the base model: a planar geometric graph, given as the equilibrium position for a system of springs representing the edges of the graph. It is the minimum-energy configuration of the system of springs (see the picture for illustration). There are generalizations of such minimum-energy configurations for convex graph embeddings into space (Linial, Lovász, Wigderson 1988), where you place, for example, four vertices of the graph at the vertices of a simplex in $\mathbb R^3$. -I think such systems of springs are good models, because the threads of the spider web are elastic. However, when viewed as models for spider webs, I wonder whether these minimum-energy spring models are missing two aspects: -The purpose of spider webs is to catch prey, so I feel the ideal model should also consider -(A) maximizing the area covered (or the volume of the convex hull) and -(B) minimizing the distances between the edges. -To me, formalizing (A) and (B) and combining it with the minimum-energy principle for a system of springs would be the ideal mathematical model for spider webs. - -Now, it is not obvious to me whether the minimum-energy principle alone determines a geometric graph satisfying (A) and/or (B)? Asking differently, if you add conditions like (A) or (B) to the minimum-energy principle, will this lead to different geometric graphs? - - -My second, broader question: Are you aware of any mathematical models developed explicitely to model spider webs? I checked MO and MSE and searched on the internet, but could not find anything. Maybe I am looking in the wrong fields, I wonder. Any help would be greatly appreciated! - -References: -Tutte, W. T. (1963), "How to draw a graph", Proceedings of the London Mathematical Society, 13: 743–767, doi:10.1112/plms/s3-13.1.743 -Linial, N.; Lovász, L.; Wigderson, A. (1988), "Rubber bands, convex embeddings and graph connectivity", Combinatorica, 8(1): 91–102, doi:10.1007/BF02122557 -The picture is from Daniel Spielman’s lecture notes pdf on the web - -REPLY [10 votes]: A biologist friend told me about this question on MathOverflow, so I wanted to contribute a useful link to a related article that appeared in NATURE. - -Published: 01 February 2012 -Nonlinear material behaviour of spider silk yields robust webs -Steven W. Cranford, Anna Tarakanova, Nicola M. Pugno & Markus J. Buehler -Nature volume 482, pages72–76(2012) - -This is the link https://www.nature.com/articles/nature10739 -The mathematically interesting feature investigated here is the nonlinear response of silk threads to stress: -From the abstract of this article: Here we report web deformation experiments and simulations that identify the nonlinear response of silk threads to stress — involving softening at a yield point and substantial stiffening at large strain until failure — as being crucial to localize load-induced deformation and resulting in mechanically robust spider webs. Control simulations confirmed that a nonlinear stress response results in superior resistance to structural defects in the web compared to linear elastic or elastic–plastic (softening) material behaviour. (...) The superior performance of silk in webs is therefore not due merely to its exceptional ultimate strength and strain, but arises from the nonlinear response of silk threads to strain and their geometrical arrangement in a web.<|endoftext|> -TITLE: Expressing primes $p\equiv 1 \pmod 3$ in the form $p = x^2 + xy + y^2$ -QUESTION [7 upvotes]: Fermat famously showed that the only primes $p$ of the form $x^2 + y^2$ are the primes such that $p \equiv 1 \mod{4}$. Furthermore, we now know “effective” versions of Fermat's theorem, i.e. given a prime $p$ such that $p \equiv 1 \mod{4}$, we know how to find integers $x$, $y$ such that $x^2 + y^2 = p$ in time polynomial in $\log p$ (see, for example section 4.5 in [1]). I would like an analogous theory for primes of the form $x^2 + xy + y^2$. In other words, I would like a precise characterization of which primes $p$ can be expressed in this form (EDIT: The comments explain that these are the primes $\not\equiv 2\mod 3$), as well as an efficient algorithm to obtain such a factorization given $p$. - -Shoup, Victor, A computational introduction to number theory and algebra, Cambridge: Cambridge University Press (ISBN 978-0-521-51644-0/hbk). xvii, 580 p. (2009). ZBL1196.11002. - -REPLY [15 votes]: This is an elaboration of the answer that Noam Elkies provided in the comments. -Suppose that $p=x^2 + xy + y^2$. Then note that $x$ and $y$ are small relative to $p$ (at most half as many digits). Note also that if $\zeta \not\equiv 1\pmod p$ satisfies $\zeta^3 \equiv 1\pmod p$ then $\zeta^2 + \zeta + 1 \equiv 0 \pmod p$, so -$$(x - \zeta y)(x - \zeta^2 y) = x^2 - (\zeta+\zeta^2)xy + \zeta^3 y^2 \equiv x^2 + xy + y^2 \equiv p \equiv 0 \pmod p.$$ -Therefore either $x \equiv \zeta y \pmod p$ or $x \equiv \zeta^2 y \pmod p$; in the latter case we have $\zeta x \equiv y \pmod p$. This means that in the 2-dimensional integer lattice generated by the vectors $(1,\zeta)$ and $(0,p)$, there is an unusually short vector $(y,x)$ or $(x,y)$, which can be found by lattice-basis reduction as long as we have $\zeta$. -It remains to find $\zeta$. Formally, we can write -$$\zeta := {\sqrt{-3} - 1 \over 2},$$ -and it is easy to check that if we can find a square root of $-3$ modulo $p$ then this formula does indeed give us a cube root of unity modulo $p$. But computing the square root can be done using the Tonelli–Shanks algorithm or Schoof's algorithm.<|endoftext|> -TITLE: General conditions for normality of blow-up -QUESTION [7 upvotes]: Let $X$ be an integral, affine, normal complex surface. I am looking for conditions on zero-dimensional closed subschemes $Z$ in $X$ such that the reduced scheme associated to the blow-up of $X$ along $Z$ is once again normal. Any reference will be most welcome. - -REPLY [7 votes]: Let $X=Spec(R)$. Blowing-up $Z=V(I)$ is the same as to look at $Proj$ of the graded ring $R[It]=\oplus_{j\geqslant 0} I^jt^j\subset R[t]$, the Rees ring associated to $I$. -Assume $R$ is a domain, integrally closed inside its fraction field $K$. In this case the integral closure of $R[It]$ inside its fraction field is -$$\overline{R[It]}=\oplus_{j\geqslant 0} \overline{I^j}t^j$$. Here, if $J\subset R$ is an ideal of $R$ the notation $\overline{J}$ is the integral closure of the ideal $J$. This is again an ideal inside $R$. -If $R$ is of dimension $2$ and $I=\overline{I}$ then results due to Zariski (if $R$ is a rlr) show that $I^j=\overline{I^j}$ for all $j$ and therefore $R[It]$ is integrally closed and you have the condition that ensures normality. These results were later generalized by Michael Artin to normal local rings of dimension $2$. -I should also say that there are very explicit combinatorial criteria (in terms of Newton polygons) to determine whether a monomial ideal $I\subset\mathbb{C}[[x,y]]$ or $\mathcal{O}_{\mathbb{C}^2,0}$ is integrally closed. -For details let me recommend you to take a look at "Integral Closure of Ideals, Rings and Modules" by Swanson and Huneke, chapters 5 and 14.<|endoftext|> -TITLE: What's the relationship between a $E_2$-Hochschild Cohomology module and a D-module? -QUESTION [5 upvotes]: Let's say for simplicity $A$ is a smooth algebra over a field $k$ ($A$ and $k$ are discrete commutative rings but from now on we are fully derived), and we will consider the $E_2$ algebra $HH^{\bullet}(A)$ which I'll just call $H_A$. We know by HKR it carries $T_A$ the tangent module in cohomological degree 1 and wedges of it in higher degrees and that the $E_2$ structure gives $T_A$ the normal Lie bracket structure and allows it to act on $A$. -Now let's suppose we have $H_A$ acting on $M$ which is concentrated in degree $0$ as a $E_2$ module. As for the homotopy groups of an $E_2$ algebra, it seems to me that $M$ should inherit both a action of $H_A$ normally and a degree lowering (sort of Lie bracket) action. More precisely, as $E_2$ algebra's homotopy groups inherit Gerstenhaber structure I'm guessing that $M$ will inherit a similar thing. (One supporting evidence is that the space of maps $H_A \otimes M \to M$ should be parametrized by 2 marked 2-discs in a 2-disc so there's an $S^1$ worth of such maps giving a degree lowering map of spectra) -Now assuming the above is correct, $M$ should have a D-module structure. (Does this seem correct?) One example of a $M$ is $A$ itself, which should inherit the standard D-module structure. -The question, assuming the above, is what class of D-modules is recoverable from this procedure? - -REPLY [5 votes]: (Incorporated David Ben-Zvi's comments below.) -An $E_2$-$\mathrm{HH}^\bullet(A)$-module is the same as a left module over the algebra of Hochschild chains $\mathrm{HH}_\bullet(\mathrm{HH}^\bullet(A))$. -Now assume $k$ has characteristic zero. It was shown by Tamarkin and Tsygan (see Theorem 2.7.1 in "The ring of differential operators on forms in noncommutative calculus") that this algebra is $A_\infty$ equivalent to $D(\Omega^\bullet(A))$, differential operators on differential forms on $A$ (with the zero differential and homologically graded). However, there is no natural morphism $D(A)\rightarrow D(\Omega^\bullet(A))$. -Here is another way to compute it. Suppose $H$ is an $E_2$-algebra. Then the $\infty$-category $\mathrm{LMod}_H$ of left modules over $H$ (viewed as an $E_1$-algebra) is monoidal; this uses the Dunn--Lurie additivity theorem of $E_n$ algebras, see Corollary 5.1.2.6 in Higher Algebra. I claim that the Drinfeld center of $\mathrm{LMod}_H$ is exactly the category of $E_2$-$H$-modules. One way to see it is to consider the corresponding framed 2d TFT; the value on the circle is the Drinfeld center which can be computed using factorization homology of $H$. -Now, let's say $A$ is a commutative dg algebra concentrated in non-positive degrees and $X=\mathrm{Spec}\ A$ the corresponding derived affine scheme. The monoidal category $\mathrm{LMod}_{\mathrm{HH}^\bullet(A)}$ is equivalent to the monoidal category $\mathbb{H}(A)$ defined in https://arxiv.org/abs/1801.03752 (see Section 4.1 for the precise claim). Its Drinfeld center is computed in https://arxiv.org/abs/1709.07867 to be $\mathfrak{D}^{der}(LX)$, a variant of the derived category of $D$-modules on the derived loop space $LX = X\times_{X\times X} X$. -If $X$ is smooth, $LX$ is eventually coconnective; so, by Example 0.2.5 $\mathfrak{D}^{der}(LX) = \mathfrak{D}(LX)$ is the usual derived category of $D$-modules on $LX$. (Note that $\mathfrak{D}(LX) = \mathfrak{D}(T[-1] X)$ which agrees with the Tamarkin--Tsygan computation.) The latter satisfies Kashiwara's lemma, i.e. it is insensitive to the derived structure. In particular, the natural pushforward functor along constant loops $\mathfrak{D}(X)\rightarrow \mathfrak{D}(LX)$ is an equivalence.<|endoftext|> -TITLE: unlinking when relaxing the homeomorphism condition -QUESTION [5 upvotes]: Say that we have two knots $K_1$ and $K_2$ in $S^3$ linked together in $S^3$ and forming the Hopf link. Usually, we can prove that we cannot unlink them by using a link invariant that shows that the "two-component unlink" that consists of two separate circles in $S^3$ have a different value (with respect to the invariant) in comparison to its value on the Hopf link. This effectively shows that there is no homomorphism from $S^3$ to itself that separates the two links. I want to relax the condition of homomorphism a little bit and ask: is there a continuous function that separates the images of the two links? in other words, is there a continuous function $f:S^3\to S^3$ with $deg(f)=\pm 1$ such that $f(K_1)$ is contained in a closed disk $D_1$ and $f(K_2)$ is contained in another closed disk $D_2$ and $D_1$ and $D_2$ are disjoint? Any pointer is appreciated. - -REPLY [8 votes]: It's possible to give such a map with degree $1$. -The complement of a Hopf link $H=H_1\cup H_2$ is $T^2\times I$. So if -we take $S^3$ and crush each component of the Hopf link to a point, we get a -map to the suspension of the torus $T^2$, $S^3 \to S^3/H_1/H_2 \cong ST^2$. -Moreover, if we had such a degree 1 map $f:S^3\to S^3$ with $f(K_i) \subset D_i$, we could get a map -factoring through the suspension, since we -can homotope the image $f(K_i)$ to a point in $D_i$, -and then extend by homotopy extension to a map -factoring through $S^3/H_1/H_2$. -Now we take a degree 1 map from $T^2$ to the sphere $S^2$, -e.g. by crushing a wedge of circles in $T^2$ whose -complement is a disk. - -This degree -one map suspends to a degree 1 map $S^3 /H _1/H_2 \cong ST^2 \to S S^2$. -The composition of these maps has the desired property.<|endoftext|> -TITLE: Roadmap to learning the classification of finite simple groups -QUESTION [17 upvotes]: I want to learn the classification of finite simple groups. But it is often commented that it is a theorem spanning tens of thousands of pages of research papers. So it is quite intimidating to an outsider like me. -Can someone please point me where to start and trace out atleast the first few basic must reads in order to get started in this business? -My current background: Representation Theory (of finite groups, semisimple Lie algebras; only very basic stuff) - -REPLY [14 votes]: A natural place to start would be Volume 1 of the so-called GLS project. - -Daniel Gorenstein, Richard Lyons, and Ronald Solomon, The Classification of the Finite Simple Groups, Mathematical Surveys and Monographs, Volume 40, Number 1, American Mathematical Society, 1994. - -The GLS project is a series of books that is intended to provide a complete proof of the Classification Theorem, modulo a relatively short list of references (such as Aschbacher and Smith's work on quasithin groups) containing crucial pieces of the proof that the GLS authors felt were already satisfactorily written up in a self-contained manner. The GLS project is expected to comprise 12 volumes, of which 8 have already been published. Volume 1 above contains an outline of the entire proof, as it was envisaged in 1994. There have been some changes to the plan since then, but nothing too radical, so Volume 1 still serves as a good starting point. Also you will probably want to read Ron Solomon's progress report in the June/July 2018 issue of the Notices of the AMS, written shortly before Volume 8 was published.<|endoftext|> -TITLE: Are lattice operations in a Lipschitz space sequentially continuous in the weak* topology? -QUESTION [6 upvotes]: This is a follow-up on this (answered) question on math.SE, but involves a different topology. I think this time it is more appropriate for MO. I will repeat the background from the question cited above. -Denote by $Lip_0(X)$ the set of all Lipschitz functions on a metric space $X$ vanishing at some base point $e \in X$. The norm in $Lip_0$ is defined as follows -$$ -\|f\|_{Lip_0} := Lip(f), -$$ -where $Lip(f)$ denotes the Lipschitz constant. With pointwise operations $f \vee g := \max\{f,g\}$ and $f \wedge g := \min\{f,g\}$ the space $Lip_0$ becomes a Lipschitz lattice, in which the following condition holds -$$ -\|f \vee g\|_{Lip_0} \leq \max\{\|f\|_{Lip_0},\|g\|_{Lip_0}\}. -$$ -The Banach lattice condition $|f| \leq |g| \implies \|f\| \leq \|g\|$, however, fails. (Nik Weaver. Lipschitz Algebras, 2nd ed.) -For a large class of metric spaces $X$, the space $Lip_0(X)$ has a unique predual, which is called the Arens-Eels space or the Lipschitz-free space, depending on the community. It can be seen as the completion of the space of Radon measures with zero mean $\mathcal M_0(X)$ with respect to the dual Lipschitz norm -$$ -\|\mu\|_{Lip*} := \sup\{\langle \mu,f \rangle \colon \|f\|_{Lip_0} \leq 1\}. -$$ -What is added by this completion are limits as $d(x,y) \to 0$ of linear combinations of the so-called elementary molecules -$$ -m_{xy} := \frac{1}{d(x,y)}(\delta_x - \delta_y), -$$ -where $d(x,y)$ is the distance between $x,y \in X$ and $\delta_x, \delta_y$ are delta-functions placed at $x,y$. (Nik Weaver. Lipschitz Algebras, 2nd ed.) -As pointed out in the answer to the question I cited above, lattice operations $f_+ := f \vee 0$, $f_- := (-f) \vee 0$ and $|f| := f \vee (-f)$ are not continuous in the $Lip_0$ norm, i.e. -$$ -\|f_n - f\|_{Lip_0} \to 0 \quad \text{does not imply} \quad \|(f_n)_+ - f_+\|_{Lip_0} \to 0. -$$ -Question. Are operations $f_+ := f \vee 0$, $f_- := (-f) \vee 0$ and $|f| := f \vee (-f)$ sequentially continuous in the weak* topology, i.e. does -$$ -f_n \rightharpoonup^* f \implies (f_n)_+ \rightharpoonup^* f_+ -$$ -hold? -Any help will be much appreciated. - -REPLY [10 votes]: Yes. If $f_n \to f$ weak* then the sequence $(f_n)$ must be bounded in ${\rm Lip}_0(X)$ (Banach-Steinhaus), and for bounded nets weak* convergence is the same as pointwise convergence. So $f_n \to f$ boundedly pointwise, which easily implies the same of the positive parts. -Let me also correct a couple of inaccuracies in your post: first, we do know that ${\rm Lip}_0(X)$ has a unique predual for a large class of spaces $X$ (if it has finite diameter, or if it is convex), but this is not known for all $X$. Second, what is added in the completion is limits of linear combinations of elementary molecules.<|endoftext|> -TITLE: Self-contained formalization of random variables? -QUESTION [8 upvotes]: I have not been able to find any formalization of random variables that supports construction of new random variables dependent on previously constructed ones. In what I have found, a random variable $A$ is a measurable function from a fixed probability space $(Ω,F,P)$ to a measurable space $(X,E)$. But that means that we have to fix $Ω$ first before we can define $A$. If later we want to define another random variable $B$ that depends on $A$, we are stuck. I know that in many cases we can 'backtrack' and define $Ω$ to accommodate all the random variables that we want to have, but I want to know if it is possible to avoid that, and I would love to know of any references. I believe it may be possible to do it by defining a random variable to carry a set of probability spaces that it depends on, rather than just one probability space, but I am unable to find any reference for such a notion. -For example, what if I want a definable function $f$ on the ordinals such that $f(k)$ for each ordinal $k$ is a random variable that with probability $1/3$ is an independent uniformly random bit and otherwise is equal to the parity of the minimum ordinal $m$ such that there is an increasing function $g : k_{≥m}{→}k$ satisfying $∀i{∈}k_{≥m}\ ( \ g(i) -TITLE: Finite order automorpisms of affine Kac-Moody Lie algebras -QUESTION [5 upvotes]: It is known that for a finite order automorphism $\phi$ of a complex semisimple Lie algebra $L$, the fixed point subalgebra $L^{\phi}$ is a reductive Lie algebra and the centralizer of a Cartan subalgebra of $L^{\phi}$ is a Cartan subalgebra in $L$. Moreover, we can always find a $\phi$-stable Cartan and a $\phi$-stable Borel subalgebra of $L$. For a reference see Chapter 8 of Kac's Book Infinite-dimensional Lie algebras. -It is too much to expect that the above results are true for a symmetrizable Kac-Moody Lie algebra. At least do we have similar theorems for affine Kac-Moody Lie algebras ? Is there a good reference where the finite order automorphisms of affine Kac-Moody Lie algebras are studied ? - -REPLY [2 votes]: Heintze, Ernst; Groß, Christian. -Finite order automorphisms and real forms of affine Kac-Moody algebras in the smooth and algebraic category. (English summary) -Mem. Amer. Math. Soc. 219 (2012), no. 1030, viii+66 pp. ISBN: 978-0-8218-6918-5 -Edit: I recommend the above reference for the following reasons. -Through the work of F. Levstein, G. Rousseau, their collaborators and many others the classification of finite order automorphisms and real forms -of affine Kac-Moody algebras has been completed -(in the algebraic case), after 15 years of work. It is said to fill hundreds of pages. -The work above is supposed to present a much simpler approach which gives more complete results; it also works in the smooth category. In fact the interest of the authors in those questions came from differential geometry, -namely in their study of infinite dimensional symmetric spaces originating from involutions of "smooth" affine Kac-Moody algebras. -Instead of using the structure of Kac-Moody algebras, the authors reduce the problems as fast as they can to the finite-dimensional case. They consider -automorphisms of two kinds, according to whether -they preserve or reverse the orientation of loops, -and show that they have a normal form, which allows to describe them in terms of curves of -automorphisms in the underlying finite-dimensional simple Lie algebra. -In the case of involutive automorphisms, they carry the analysis in detail and deduce a complete classification; this turns out to be -an extension of É. Cartan's classification of symmetric spaces. -They also apply their results in the complex case to conjugate linear involutions and obtain the classification of real forms.<|endoftext|> -TITLE: Mednykh's formula for 2d manifold with boundaries? How to count principal G-bundles with prescribed holonomies? -QUESTION [13 upvotes]: Let S be a closed, orientable 2d manifold and G a finite group. Since a principal G-bundle over S is specified by maps $\phi : \pi_1(S) \rightarrow G$ modulo the adjoint action by G, the way to count how many such bundles exist is -\begin{align} -\dfrac{ | Hom(\pi_1(S),G) |}{|G|} = \sum_{\text{irreps V of }G} \bigg( \dfrac{d(V)}{|G|} \bigg)^{2-2g} \; , -\end{align} -where $d(V)$ is the dimension of the irrep V, and g is the genus of S. -My question is how the formula above generalizes when we want to count principal G-bundles over S, when S has n boundaries with prescribed holonomies for the bundle specified by conjugacy classes $\{ k_i : i=1,...,n \}$ of G. Morally we should have something like -\begin{align} -\dfrac{ | Hom(\pi_1(S_{g,k_i}),G) |}{|G|} = \sum_{\text{irreps V of }G} \bigg( \dfrac{d(V)}{|G|} \bigg)^{2-2g-n} \prod_{i=1}^n \chi_V(k_i) \; , -\end{align} -where $\chi_V(k_i)$ is the character of the representation V evaluated at any representative of the conjugacy class $k_i$. -So, is the above formula completely correct? Are there factors missing? I would really appreciate any references that prove what the generalization is. - -REPLY [13 votes]: I get a slightly different formula: -$$\sum_{V} \frac{d(V)^{2g-2-n}}{|G|^{2g-2}} \prod_{i=1}^n |k_i|\chi_V(k_i)$$ -Here $|k_i|$ denotes the size of the conjugacy class. -I prefer to express this in a slightly different way. For a conjugacy class $k$ and irreducible representation $V$ of $G$, let $f^V_k$ denote the scalar by which the indicator function $k \in Z(\mathbb C[G])$ acts on $V$. In these terms the formula is: -$$\sum_V \left(\frac{d(V)}{|G|}\right)^{2g-2} \prod_{i=1}^n f^V_{k_i}$$ -To obtain the first formula from the second, note that the numbers $f^V_k$ are related to characters as follows: -$$f^V_k = \frac{|k|}{d(V)} \chi_V(k)$$ - -For me, these formulas come from general properties of commutative Frobenius algebras, or equivalently 2d topological field theory (TFT). Namely, given a commutative Frobenius algebra $A$, a genus $g$, and a collection of elements $k_1, \ldots , k_n$ one obtains a number -$$Z_A(g; k_1, \ldots , k_n)$$ -recording the value of the TFT on an oriented surface of genus $g$ with $n$ punctures labelled by the elements $k_1, \ldots , k_n$. You can compute this number by first multiplying together the elements $k_1 \ldots k_n$ in $A$, then applying a sequence of $g$ comultipication followed by multiplication operations, finally followed by the Frobenius trace (see cartoon below). - -In the case when $A$ is semisimple, one can be more explicit and write everything in terms of a basis of orthogonal idempotents. -In our case we take $A=Z(\mathbb C[G])$, the center of the group algebra, equipped with the trace $t$ (which takes the value $1/|G|$ at the identity element of $G$ and zero on all other elements). The numbers $f^V_k$ is just the change of basis matrix between the conjugacy classes $k$ and the orthogonal idempotents $e_V$ labelled by irreps. -I can't think of a reference for this right now. In this paper I explain some of this stuff in a related context, see e.g. Prop 2.13.<|endoftext|> -TITLE: A geometric definition of the addition law on abelian surfaces -QUESTION [13 upvotes]: Most people will have see a geometric "explanation" of the addition law on elliptic curves given by embedding it as a cubic in the projective plane and cutting it with lines. -Is there a similar explicit, geometric definition of the addition law on (a family of?) abelian surfaces? -So the question is really: Give a nice embedding of abelian surfaces into projective space and then define the addition law using this embedding - if not for all abelian surfaces, at least for some non trivial family. In fact, it would be really nice if we could do this for the embedding that realizes the surface as a degree 10 variety using the Horrocks-Mumford bundle. - -REPLY [8 votes]: Jacobians of genus-2 curves - and abelian surfaces in general, I suppose - can be realized as the variety of lines on the intersection of two quadrics in $\mathbb{P}^5$ (once you've chosen a line to act as the neutral element). This is analogous to seeing an elliptic curve as the variety of 0-dimensional spaces (i.e. points) on the intersection of two quadrics in $\mathbb{P}^3$ (which is sometimes called the "Jacobi" model of an elliptic curve). The group law has a really nice geometric expression. -This is covered at length in Chapter 17 ("A neoclassical approach") of Cassels and Flynn's Prolegomena to a middlebrow arithmetic of curves of genus 2, and in even more length in Chapter 6 of Principles of algebraic geometry by Griffiths and Harris (specifically Section 6.3, "Lines on the quadric line complex"). -Edit (bonus): If you're interested in higher dimensions, then let $X$ be the intersection of two quadrics in $\mathbb{P}^{2g+1}$, and let $S$ be the variety of $(g-1)$-planes in $X$. Then $S$ is a homogeneous space under the Jacobian of a hyperelliptic curve $C$ of genus $g$. The relationship between $X$, $S$, and $C$ (and the action of $\mathrm{Jac}(C)$ on $S$) is very explicit. Chapter 4 of Miles Reid's PhD thesis (The complete intersection of two or more quadrics) has the details.<|endoftext|> -TITLE: Generalized linear models: What's the benefit of the underlying theory? -QUESTION [5 upvotes]: I was recently trying to understand generalized linear models (GLMs) and after investing quite a few days, it still hasn't dawned on me what the fundamental benefit of the framework is. Normally, I am used to results like guarantees of convergence, limits for error etc, but all that seems to be missing here. -There is a common framework with underlying distribution, regressors/predictors linear in the coefficients, link functions and finally MLE but it seems to be branching off very quickly into the various subclasses, which each need a separate algebraical and numerical treatment. -So can anyone point me towards what is "general" about the GLMs and what is the benefit of that? - -REPLY [8 votes]: What are the benefits of a unified framework? You are right that we are rapidly going into some much used special cases line logistic regression or Poisson regression, but there is still benefit in having a common framework. - -Technology transfer from the general linear model (not generalized!), that is with gaussian errors and identity link. A lot of what one has learned from there can be directly used with glm's, especially all the modeling trick constructing the model matrix $X$. - -A common estimation algorithm (IRLS, iteratively reweighted least squares, see https://stats.stackexchange.com/questions/236676/can-you-give-a-simple-intuitive-explanation-of-irls-method-to-find-the-mle-of-a/237384#237384), leads to a common implementation framework. This is maybe not a mathematical advantage, but software implementation and modeling advantage. It is also a teaching advantage! This program is not a 100% success, as for some important glm's this algorithm do not work very well, as witnessed by the paper logbin: An R Package for Relative Risk Regression Using the Log-Binomial Model. - -Many common concepts applied to all or most of the special cases, like link function, offsets, variance function, mean function, ... , quasi-likelihood - -One area with little advantage of the common framework is residual analysis, which really need to be studied for each case separately. See for instance family of glm represents the distribution of the response variable or residuals. But see here for an approach using simulated residuals. - -The Nelder & Wederburn paper introding the glm is here at JSTOR and the original motivations can be found there. - - -A lot of information can be found at Cross Validated, see as a start https://stats.stackexchange.com/questions/104399/why-do-we-use-glm<|endoftext|> -TITLE: A domination property for the Hardy space $H^1$ -QUESTION [5 upvotes]: In the theory of Hardy spaces of the unit disc, a fact that is implicitely used quite often is that if $f\in H^p, 1 -TITLE: Cardioid-looking curve, does it have a name? -QUESTION [22 upvotes]: The curve, given in polar coordinates as $r(\theta)=\sin(\theta)/\theta$ -is plotted below. - -This is similar to the classical cardioid, but it is not the same curve (the curve above is not even algebraic, I believe). Does this curve have a name? Does it show up somewhere? -This curve has the property that it solves $\mathrm{Im}(1/z+\log(z))=0$, if this perhaps rings a bell. -This particular curve arises in some research I am working on at the moment, and it would be great if it perhaps connects to some classical area. -Edit: Thanks for the great references! -As a reward, here is a more artistic rendering of the shape -using a type of complex dynamical systems. - -REPLY [2 votes]: From what LSpice and skbmoore shared https://mathcurve.com/courbes2d/cochleoid/cochleoid.shtml, -There's this illustration of a helix on a cylinder -The picture in the middle is the view when lined up with the side of the cylinder. If I understand the site correctly, the line traced by the helix in this perspective is a cochleoid.<|endoftext|> -TITLE: Is the tensor product of symmetric pseudomonoids their coproduct? -QUESTION [5 upvotes]: The category of commutative monoid objects in a symmetric monoidal category is cocartesian, with their tensor product serving as their coproduct. This sort of result seems to date back to here: - -Thomas Fox, Coalgebras and Cartesian categories, Commun. Algebra 4 (1976), 665–667. - -I'm working on a paper with Todd Trimble and Joe Moeller, and right now we need something similar one level up — that is, for symmetric pseudomonoids. (For example, a symmetric pseudomonoid in Cat is a symmetric monoidal category.) -The 2-category of symmetric pseudomonoids in a symmetric monoidal 2-category should be cocartesian, with their tensor product serving as their coproduct. I imagine the coproduct universal property will hold only up to 2-iso. -Has someone proved this already? This paper: - -Brendan Fong and David I, Spivak, Supplying bells and whistles in symmetric monoidal categories. - -proves the result in the special case where the symmetric monoidal 2-category is Cat. In fact they do more, in this special case: -Theorem 2.3. The 2-category SMC of symmetric monoidal categories, strong monoidal functors, and monoidal natural transformations has 2-categorical biproducts. -Unfortunately their proof is not purely 'formal', so it doesn't instantly generalize to other symmetric monoidal 2-categories. And I believe the fact that the coproducts in SMC are biproducts must rely on the fact that Cat is a cartesian 2-category. - -REPLY [4 votes]: The result I wanted is Theorem 5.2 here: -Daniel Schäppi, Ind-abelian categories and quasi-coherent sheaves, Mathematical Proceedings of the Cambridge Philosophical Society, 157 (2014), 391–423. doi:10.1017/S0305004114000401 -The proof appears in Appendix A. He proves the result for symmetric pseudomonoids in a symmetric monoidal bicategory.<|endoftext|> -TITLE: What is the correct notion of representation for abelian varieties? -QUESTION [12 upvotes]: Zeroth question - am I right that in the "ordinary" sense an abelian variety does not possess any representations at all? -More precisely, a representation of an algebraic group $G$ (over an algebraically closed field $K$, say, over complex numbers) is a homomorphism $G\to\operatorname{GL}(V)$ for some $K$-vector space $V$, which is a morphism of algebraic varieties. I have vague feeling that if $G$ is a projective variety, hence an abelian variety, then there are no nontrivial such homomorphisms. -If the answer to this zeroth question is negative, then my actual ("number one") question would be whether there is a classification of such representations. -(As @Wojowu explains in a comment below, this is indeed true) -If it is positive, then the question is whether there exists a modification of the notion of representation that would give some meaningful result - mostly, would allow studying an abelian variety $G$ through such representations. -Possible approaches would, maybe, include allowing "representations with singularities", say, instead of polynomial homomorphisms to allow rational homomorphisms $G\to\operatorname{GL}(V)$. Or, say, one might consider $G$-equivariant vector bundles over $G$ (whatever this means). Or, one might look at algebraic homomorphisms $G\to\operatorname{Aut}(A)$ where $A$ is some commutative "thing" in $K$-varieties such that the algebraic group $\operatorname{Aut}(A)$ admits nontrivial algebraic homomorphisms from abelian varieties to it. Subquestion: are there such $A$? Can it be, say, another abelian variety? -Maybe one more hopefully simpler subquestion. Let $\operatorname{Aut}(|G|)$ be the algebraic group of all algebraic automorphisms of the underlying algebraic variety $|G|$ of $G$. Then (presumably) the map assigning to $x\in G$ the multiplication-by-$x$ operator $G\to G$ is an injective algebraic homomorphism $G\to\operatorname{Aut}(|G|)$, so $\operatorname{Aut}(|G|)$ contains a copy of $G$ as a subgroup. What are, if any, subgroups in between? How does $\operatorname{Aut}(G)$ sit inside $\operatorname{Aut}(|G|)$? Is this $\operatorname{Aut}(|G|)$ studied somewhere? - -REPLY [6 votes]: I am also very far from an expert here, but I think there's a case to be made that the "correct notion" involves actions on categories of sheaves, as Donu says in the comments. -Consider the following toy model: if $A$ is, say, a finite abelian group then its Pontryagin dual $\widehat{A}$ can be defined as the group $\text{Hom}(A, \mathbb{G}_m)$ of homomorphisms from $A$ into the multiplicative group $\mathbb{G}_m$ (say over the complex numbers, so $\mathbb{G}_m(\mathbb{C}) \cong \mathbb{C}^{\times}$, but any algebraically closed field of characteristic $0$ would do, or we could think in terms of Cartier duality). There is then a canonical pairing -$$A \times \widehat{A} \to \mathbb{G}_m$$ -which is used, for example, to define the Fourier transform $L^2(A) \cong L^2(\widehat{A})$. In representation-theoretic terms these homomorphisms correspond to $1$-dimensional representations and give exactly the irreducible representations of $A$. -Abelian varieties $X$ also have duals $X^{\vee}$, but they're defined not in terms of maps into the multiplicative group $\mathbb{G}_m$ but in terms of line bundles, or equivalently in terms of maps into the classifying stack $B\mathbb{G}_m$ of line bundles (although we need to restrict to degree $0$ line bundles). There is again a canonical "pairing" -$$X \times X^{\vee} \to B \mathbb{G}_m,$$ -namely the Poincaré bundle over $X \times X^{\vee}$, and it can be used to define the Fourier-Mukai transform $D(X) \cong D(X^{\vee})$ between derived categories of coherent sheaves. Among other things, what this analogy suggests is that the analogue of the "regular representation" for an abelian variety is its action on its derived category $D(X)$ by translation.<|endoftext|> -TITLE: Complete geodesics on hyperbolic a pair of pants -QUESTION [7 upvotes]: I have asked this question on MSE. But I think Mo is a better place to ask my question. Here is the link to my question on MSE. I will rewrite it here: - -I am trying to understand the article by Maryam Mirzakhani about Simple geodesics and Weil-Petersson volumes. In the third section of this article, the following proposition is stated. And I want to know why this proposition is correct. I mention the proposition: -Proposition: Let $P$ be a hyperbolic pair of pants with (non-empty)geodesic boundary components $\beta_1 , \beta_2, \beta_3$ of lenghts $x_1, x_2, x_3$ respectively. Then $P$ contains $5$ complete geodesics disjoint from $\beta_2 , \beta_3$ and orthogonal to $\beta_1$. More precisley, two of these geodesics meet $\beta_1$ respectively at -$y_1,y_2$ and spiral to $\beta_3$, the other two meet $\beta_1$ respectively at $z_1,z_2$ and spiral to $\beta_2$. there is also a unique common geodesic perpendicular from $\beta_1$ to itself meeting $\beta_1$ perpendicularly at two points $w_1, w_2$. - -REPLY [7 votes]: I suggest that you look at what happens in the universal cover given by the Poincare disk model. The five geodesics are quite easy to see. In the picture below, the greyed zone is a fundamental domain for the pair of pants. The five sought geodesics are in bold. The book of Peter Buser, geometry and spectra of compact Riemann surfaces, is a good reference concerning the representation of surfaces as quotients of the Poincare disk.<|endoftext|> -TITLE: Rational slice knot that is not slice -QUESTION [7 upvotes]: Does there exists a knot $K\subset \mathbb{S}^3$ such that - -$K$ is not slice -$\exists W^4$, $\partial W = \mathbb{S}^3$ rational homology ball -$\exists $ properly embedded smooth disk $(D,\partial D)\to (W,K)$. ? - -In other words $K$ is not slice in $B^4$ but is slice in some rational homology ball. - -REPLY [3 votes]: Rationally slice knots in $S^3$ are in abundance due to the collection of Kawauchi's theorems and the recent result of Kim and Wu: -Theorem ([1] + [2]): Any hyperbolic amphicheiral knot $K$ in $S^3$ is rationally slice. -Theorem ([3]): Any fibered, negative amphicheiral knot $K$ in $S^3$ with irreducible Alexander polynomial (called Miyazaki knot) is rationally slice. -Using these theorems, you may find plenty of rationally slice knots that are not smoothly slice. -[1]: Kawauchi, Akio. "The invertibility problem on amphicheiral excellent knots." Proceedings of the Japan Academy, Series A, Mathematical Sciences 55.10 (1979): 399-402. -[2]: Kawauchi, Akio. "Rational-slice knots via strongly negative-amphicheiral knots." Commun. Math. Res. 25.2 (2009): 177-192. -[3]: Kim, Min Hoon, and Zhongtao Wu. "On rational sliceness of Miyazaki's fibered,− amphicheiral knots." Bulletin of the London Mathematical Society 50.3 (2018): 462-476.<|endoftext|> -TITLE: Russell's paradox as understood by current set theorists -QUESTION [8 upvotes]: Many mathematicians like to think of the set of natural numbers as existing as a completed object. But it is difficult to make set theory as concrete, because Russell's paradox, in conjunction with some type of separation principle, tells us that arbitrary "collections" cannot be collected into a completed object. I view this as telling us that the metaphysical idea of "collection" has some natural limitations that we might not have realized, a priori. -Now, in terms of the formal mathematics of collections---known as set theory---there seem to be two standard fixes to address the paradox. -Class and set distinction First is the idea of creating a new level of collection called "proper classes". In some set theories like ZFC, classes are an informal notion referring to the formulas of the language. Some mathematicians still view those classes as referring to meta-collections in the metatheory. They even use set-builder notation to refer to them. In other versions of set theory, like NBG or KM, classes are also formal objects. Sometimes they are of a different type than sets, and sometimes sets are classes with extra properties. -Those theories with classes can often be reinterpreted inside the theories without classes, and vice versa. Thus, it seems that Russell's paradox does not prescribe the existence, Platonically speaking, of two distinct types of collections---the set and the proper class. Yet this language has also become very useful to mathematicians. My question is somewhat philosophical in nature. Do modern set theories view proper classes as a necessary, true concept? Do they favor the view that proper classes are only informal, or are they formal? -I have a follow up question, for those set theorists that believe a "true Platonic set theory" exists. How do you view that completed set theory in light of Russell's paradox? It seems that a "true set theory" couldn't be like a collection itself (hence not like a set, nor like a proper class even). In particular, "true Platonic set theory" would be unlike any model of formal set theory, since the domain of a model is a collection. -Type theory Another solution, which I am much less familiar with, is using type theory to limit collection principles. Are there many modern set theorists who favor this resolution? Or has the proper class idea overriden this solution? - -REPLY [13 votes]: Let me begin quoting W. Tait (lectures on proof theory, pages 4 and 5): - -I believe that what further has to be understood, in order to make -sense of these 'paradoxes' is that the notion of a transfinite number or, equivalently, -of a set of transfinite numbers is an essentially open-ended notion: -no matter what principles we introduce to construct sets of numbers, providing -only that these principles are well-defined, we should be able to admit all -numbers obtained by these principles as forming a set, and then proceed on -to construct new numbers. So $\Omega$ cannot be regarded as a well-defined extension: -we can only reason about it intensionally, in terms of those principles -for constructing numbers that we have already admitted, leaving open in our reasoning the possibility - in fact, the necessity - of always new principles for -constructing numbers. When this is not understood and $\Omega$ -is counted as a -domain in the sense of a well-defined extension, then the so-called paradoxes -force on us a partitioning of well-defined extensions into two categories: sets -and proper classes; and the only explanation of why such an extension should -be a proper class rather than a set would seem to be simply that the assumption -that it is a set leads to contradiction. The paradoxes deserve the name -'paradox' on this view because it offers no explanation of why there should -be this dichotomy of well-defined extensions. - -Emphases are mine ($\Omega$ is a reference to "the greatest ordinal"). So, one proposed rough sketch of an answer in the direction given by Tait (of course, there are many other directions in philosophy) is this: - -The subject matter of set theory is open-ended, therefore set theory must be about an intension, the concept of set, not about a well-defined extension. This intension is open-ended (it is hard to make sense of the oxymoron "open-ended well-defined extension"), and it is the unifying criterion behind the plurality of set theoretical practices. The contemporary criterion can be more or less approximated by $ZFC$, but there can be no definite final stage on the progressive conceptual unification of the set-theoretical practices, as there is a neccessary open-endedness (incompleteness) in this intension. - -There are many things to address here, but I will not try to pursue them, not even in outline, as this would lead us to a more hardcore philosophical activity. -As a final remark, there are similar arguments in the history of philosophy which were given many years before Russell. One of the most relevant is Plato's third man argument, in Parmenides. -SPECULATIVE ADDENDA: -I think the question "should there be a dichotomy of well-defined extensions and how can we deal with it?", a natural outcome of this discussion, is very relevant for the foundations of set theory, and there are many hints about this in traditional philosophy, say, from Plato to Hegel. I think the answer is no, and I agree with Tait's direction. (A small digression: "Platonism", as the term appears in the original question, has probably a very weak connection to Plato. Plato is very subtle, he wrote dialogues, not theoretical treatises in philosophy, in which the dramatic elements interact with the philosophical elements, probably because he sees philosophy as the kind of argumentative activity he shows in the dialogues, not as a body of theory. See W. Tait, Truth an Proof: The Platonism of Mathematics. Anyway, I think, along with Tait, that the man deserves a better fate.) -I will not dare to say much more about our questions here, but I would like to remark on the idea that there can be no final conceptual unification, for any unification is synthetic, that is, made on the basis of a new conceptual synthesis which, as "new", cannot be among those very things now unified. If reason operates this way, progressively unifying its previous practices through conceptual synthesis, open-endendness is its fate, and I believe mathematics is the primary example of this.<|endoftext|> -TITLE: Schwänzl and Vogt, Cofibration and fibration structures in enriched categories -QUESTION [5 upvotes]: In [Schwänzl and Vogt, Strong cofibrations and fibrations in enriched categories], the authors refer to an earlier preprint, [Schwänzl and Vogt, Cofibration and fibration structures in enriched categories] but give a URL that no longer works. Is this preprint still available somewhere online? -(It seems the preprint deals with some of complications arising from the fact that there is no "good" cylinder object in the category of simplicial sets. The later paper omits these details.) - -REPLY [7 votes]: Yes, it is also available here: -https://www.math.uni-bielefeld.de/sfb343/preprints/pr97044.ps.gz<|endoftext|> -TITLE: Fibonacci-like sequences in $\mathbb{F}_q$ where each element only depends on the previous one -QUESTION [11 upvotes]: Given a prime power $q$, consider all sequences $(a_n)_{n\in\mathbb{Z}}$ in $\mathbb{F}_q$ for which $a_{n+1}=a_n+a_{n-1}$ for all $n\in\mathbb{Z}$. Call such a sequence simple if there exists a function $f:\mathbb{F}_q\to\mathbb{F}_q$ such that $a_{n+1}=f(a_n)$ for all $n\in\mathbb{Z}$. -There are some trivial simple sequences. The null sequence is simple, as is $(cr^n)_{n\in\mathbb{Z}}$ for $c\in\mathbb{F}_q^*$ and $r$ a root of $X^2-X-1$. My questions are about nontrivial simple sequences. -I've asked a more specific version of this question on Math.Stackexchange. There, computations by the user @Servaes show that nontrivial simple sequences exist in $\mathbb{F}_p$ for $p\in\{199,211,233,281,421,461,521,557,859,911\}$ -Questions: - -Are there 'easy' conditions on primes $p$ such that no nontrivial simple sequences exist in $\mathbb{F}_p$ when $p$ satisfies these conditions? (and there are a large number of primes satisfying these conditions) -Are there infinitely many primes $p$ such that nontrivial simple sequences exist in $\mathbb{F}_p$? -Given a prime $p$, does there always exist a positive integer $n$ such that nontrivial simple sequences exist in $\mathbb{F}_{p^n}$? -In case the answer to the previous question is affirmative, let $n(p)$ be the smallest such positive integer. Is $n(p)$ bounded? If not, do there exist integers $m$ such that $n(p)=m$ for infinitely many primes? - -REPLY [5 votes]: $\def\ord{\mathop{\mathrm{ord}}}$Let $q=p^s$ for a prime $p$. -Let $\phi$ and $\psi$ be the roots of $X^2-X-1$; they may lie either in $\mathbb F_p$ (when $\left(\frac p5\right)=1$, call this case simple) or in $\mathbb F_{p^2}$. The case $\phi=\psi$, i.e. $p=5$, is covered by @YCor in the comments (1 2), so let us assume $\psi\neq \phi$. Notice that $\phi\psi=-1$. -The general form of a linear recurrence is then $a_n=a\phi^n+b\psi^n$; where $a,b\in\mathbb F_q$ if $\sqrt5\in\mathbb F_q$, and $a$ and $b$ are two conjugate elements in $K=\mathbb F_q[\sqrt5]$, otherwise (here, conjugate means that they are swapped by the nontrivial automorphism of $K$ over $\mathbb F_q$). Surely, this sequence is periodic with period $T=\ord \phi=\ord\psi$ (where $\ord$ means the multiplicative order in $\mathbb F_{p^2}$ which does not depend on $s$); so we need the terms $a_1,a_2,\dotsc,a_T$ to be distinct, while $a$ and $b$ are nonzero. -If two such terms are equal, we have -$$ - a\phi^n+b\psi=a\phi^{n+k}+b\psi^{n+k} - \iff a\phi^n(\phi^k-1)=b\psi^n(\psi^k-1) - \iff \frac ba=\phi^{2n}(-1)^n\frac{\phi^k-1}{\psi^k-1}. -$$ -For every prime $p$, the right-hand part attains finitely many values ($\leq T^2 -TITLE: Discretizing a line segment with pixels which satisfies the Pythagorean theorem -QUESTION [7 upvotes]: There are plenty of line drawing algorithms to discretize line segments using pixels. -The Bresenham's algorithm gives a line where the number of pixels in the segment is the same as its width (in x-direction) or height (y-direction), whichever is largest. -One can also imagine an algorithm where one starts in one of the points, and -choose the lattice path between start and end point which minimizes total -distance squared of pixel centers to the true geometric line. -The number of pixels produced is the width+height, as we have a lattice path. -Note that the (geometric) length of the line segment is somewhere between the number of pixels produced by the two approaches above. -My question is, is there some (standard) algorithm where the number of pixels in the constructed line segment is equal to the the (rounded to nearest integer) length of the line segment? We want the line-segment to be connected, in the sense that every x-coordinate between the endpoints are covered by at least one pixel (and same for y-coordinates). -Of course, one can take the lattice path approach above, and iteratively remove pixels furthest from the true geometric line, but this seems inefficient, and might not guarantee connectednes. - -REPLY [3 votes]: I think this question makes sense if we extend the planar grid with a certain collection of diagonals, and require that the drawn segment between two points always use a shortest path in this graph. It was posed in this form by Pach, Pollack and Spencer 30 years ago, and it is still open.<|endoftext|> -TITLE: Covariant splittings of Hopf algebra projections -QUESTION [5 upvotes]: What is an example of a pair of Hopf algebras $(A,B)$ with a surjective Hopf algebra map $\phi:A \to B$ such that $\phi$ does not admit a $B$-bi-comodule splitting $s:B \to A$? To be clear, the right $B$-comodule structure on $A$ is given by -$$ -(\textrm{id} \otimes \phi) \circ \Delta_A: A \to A \otimes B, -$$ -where $\Delta_A$ is the coproduct of $A$, and the left coaction is defined similarily. - -REPLY [2 votes]: I'll give an example "occuring in nature." It's not the simplest possible, but you can get a simpler one by removing the generators of degrees 3, 5, and 7, which don't feature in the argument. -According to results of Borel from 1954, the mod-2 homology Hopf algebra $$H_9 = H_* (\mathrm{Spin}(9);\mathbb F_2)$$ is the exterior algebra on one generator each of degrees 3, 5, 6, 7, and 15. The standard inclusion $i$ of $\mathrm{Spin}(9)$ in $\mathrm{Spin}(10)$ preserves this exterior algebra structure, but $H_{10} = H_*(\mathrm{Spin}(10);\mathbb F_2)$ has a new generator $u_{9}$ of degree 9 such that $H_{10}$ is a free module of rank two over $i_* H_9$ with basis $\{1,u_9\}$, -following from the collapse of the Serre spectral sequence of the fiber bundle $\mathrm{Spin}(9) \to \mathrm{Spin}(10) \to S^9$. The new generator $u_9$ doesn't anticommute with the old ones as one might expect: giving the other generators the obvious names, one has -$$u_6 u_9 + u_9 u_6 = u_{15}$$ -in $H_{10}$. -Particularly, the injection of left $H_9$-modules $H_9 \to H_{10}$ does not split. For degree reasons, $u_9$ would have to go to $u_3 u_6$ or zero under the splitting, but the product $H_9 \otimes H_{10} \to H_{10}$ sends $$u_6 \otimes u_9 \mapsto u_{15} + u_9 u_6,$$ while the product $H_9 \otimes H_9 \to H_9$ sends $$u_6 \otimes 0 \mapsto 0 \neq u_{15} + 0 u_6$$ and also $$u_6 \otimes u_3 u_6 \mapsto 0 \neq u_{15} + (u_3 u_6) u_6.$$ -A $H^*(\mathrm{Spin}(9);\mathbb F_2)$-comodule splitting of the cohomological Hopf algebra map $$i^*\colon H^*(\mathrm{Spin}(10);\mathbb F_2) \to H^*(\mathrm{Spin}(9);\mathbb F_2)$$ would lead on dualization to a forbidden module splitting of the sort we ruled out in the previous paragraph, so $i^*$ is an example of the type you wanted.<|endoftext|> -TITLE: Is there a closed-form expression for these integrals? -QUESTION [5 upvotes]: I am computing the following integrals by numerical integration and this takes a lot of time, although I'm sure there is a general closed-form formula but I can't find it. -Let $t$ be a vector of $\mathbb R_{+}^{d}$. For any integer $d \ge 1$, define $K_d$ as a convex subset of $\mathbb R^{d}$ by : -$$x \in K_d \iff \forall\, j \in {1,..,d}, \;x_j \ge 0 \text{ and }\sum\limits_{j=1}^d t_j x_j \le 1$$ -I think $K_d$ is usually called a simplex, but I am not sure. -Let now $i$ be a vector of integers in $\mathbb{N}^{d}$, and consider the integral : -$$I_{i}^{d} = \int\limits_{K_{d}} \;\;\prod_{j=1}^{d} x_j^{i_j} \;\;\partial x_1,...,\partial x_d$$ -Can we compute an expression for $I_{i}^{d}$? Maybe some recursion on $d$ or on $i$ can be found? -Edit: -I founded a paper that solves the problem, Lasserre - Simple formula for integration of polynomials on a simplex. it gives a formula a little more general, that reduces to the following : -$$\text{If and only if t_j = 1 for all j, }I_{i}^{d} = \frac{\prod\limits_{j=1}^{d} i_j}{(d+\lvert i \lvert)!}$$ -Then, as a comment showed, generalisation can be done via -$$I_i^d(t) = \bigl(\prod_{j = 1}^d t_j^{-(i_j + 1)}\bigr)I_i^d(1)$$ - -REPLY [2 votes]: If $\Sigma_t=\{x_i \geq 0: x_1+\cdots +x_d=t\}$and $d\sigma_t$ is its surface measure, then $I_t=\int_{\Sigma_t}\prod_{i=1}^d x_i^{\alpha_i}\, d\sigma_t=t^{|\alpha|+d-1}I_1$ (change variable $x_i=ty_i$). Then for $\alpha_i>-1$ -$$ -\prod_{i=1}^d \Gamma (\alpha_i+1)=\int_{[0,\infty[^d} \prod_{i=1}^d x_i^{\alpha_i} e^{-(x_1+\cdots +x_d)}dx=\int_0^\infty \frac{e^{-t}}{\sqrt n} I_tdt=I_1\int_0^\infty t^{|\alpha|+d-1}\frac{e^{-t}}{\sqrt n} dt. -$$ -This gives $I_1=\frac{\sqrt{n} \prod_{i=1}^d \Gamma (\alpha_i+1)}{\Gamma (|\alpha|+d)}$ and $$\int_{\{x_i \geq 0, x_1+\cdots +x_d \leq 1\}} \prod_{i=1}^d x_i^{\alpha_i}dx=\int_0^1 \frac{dt}{\sqrt n}I_t=\frac{\prod_{i=1}^d \Gamma (\alpha_i+1)}{\Gamma (|\alpha|+d+1)}.$$<|endoftext|> -TITLE: Double-diagonalisation of nxn matrices? -QUESTION [6 upvotes]: I've come up with the following piece of Python code (using the library Sympy): -def double_diagonalize(m1, m2): - V, _ = (m1.T * m2).diagonalize() - U, _ = (m1 * m2.T).diagonalize() - return U, V - -What I've found is that given many (but not all) random pairs of $ -n \times n$ matrices $M_+$ and $M_-$, it produces $U$ and $V$ such that both $U^{-1} M_+ V^{-T}$ and $U^T M_- V$ evaluate to diagonal matrices. -My question is, is this result known? And by what name? It looks similar to Generalised Schur Decomposition except that the matrices $U$ and $V$ are not necessarily unitary, and the (pair of) "normal forms" $U^{-1} M_+ V^{-T}$ and $U^T M_- V$ are usually fully diagonal instead of merely upper triangular. -It might be related to this as well: https://en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix#Generalized_eigenvalue_problem - -REPLY [3 votes]: It's the solution of the product eigenvalue problem $M_1^T M_2$, once you transpose the second relation.<|endoftext|> -TITLE: Does every category with a subobject classifier embed into a topos? -QUESTION [14 upvotes]: I've never seen an example of a category with a subobject classifier which didn't embed nicely into a topos. Is there a good reason for this? -Question 1: Let $\mathcal C$ be a category with a subobject classifier $\Omega$ (and whatever finite limits this entails -- namely, a terminal object and pullbacks along monomorphisms). Does there exist a fully faithful functor $\mathcal C \to \mathcal E$, where $\mathcal E$ is an elementary topos, which preserves the subobject classifier and the aforementioned finite limits? -Question 2: Same as Question 1, but assuming that $\mathcal C$ has all finite limits, and requiring that $\mathcal C \to \mathcal E$ preserves them. -Question 3: Same as Question 2, but throwing in finite colimits as well. -Question 4: Now assume that $\mathcal C$ is locally presentable and has a subobject classifer $\Omega$. Does it follow that $\mathcal C$ is a (necessarily Grothendieck) topos? -Question 4 may be the most heavy-duty-sounding formulation, but it also gives me the most reason to think the answer might be "yes" -- after all, in order for a category $\mathcal C$ with finite limits and a subobject classifier to be a topos, it just needs to additionally be cartesian closed. And if $\mathcal C$ is locally presentable, then by the adjoint functor theorem, to verify this one just needs to check that the functors $X \times (-)$ preserve colimits. Plausibly, the subobject classifier might force this. As partial progress, I think I can show that in this case, coproducts are disjoint. - -REPLY [16 votes]: Ivan's example in the comment actually proves that all the questions have negative answers. -As observed by Ivan, in the category of pointed set, there is a subobject classifier given by $\{*\} \to \{*,\bot \}$, where $*$ is the special point. -Indeed, a subobject of $X$, is just a subset of $X$ containing $*$ so it is classified by a unique map $X \to \{* = \top,\bot\}$ : the usual classifier of the map in Set. -Now, in a topos, you always have at least two maps from the terminal object to the sub-object classifier: the map $\top$ and the map $\bot$. If they are equal, the topos is degenerated. But in pointed set, there is only one map from $\{*\} \to \{*,\bot \}$, so there can't by a fully faithful functor to an elementary topos that preserve the subobject classifier and its universal subobject.<|endoftext|> -TITLE: What is the Lipschitz constant of the differential of the matrix exponential $\mathfrak{so}(3)\to \mathrm{SO}(3)$ -QUESTION [9 upvotes]: I'm interested in numerical methods on $\mathrm{SO}(3)$ manifold, and working on a particular problem using the exponential coordinates: -$$ -R(u) := \exp(u_\times) -$$ -with $u\in \mathbb{R}^3$ and where $u_\times \in \mathfrak{so}(3)$ is the cross-product matrix of vector $u$. -The directional derivative of $R(u)$ in the direction $Y$ is: -$$ -[D_u R]Y = [T(u)Y]_\times R(u) -$$ -for any vector $Y\in \mathbb{R}^3$, where -$$ -T(u) := \int_0^1R(su)ds -$$ -Both $R$ and $T$ are Lipschitz continuous with constants $1$ and $\tfrac{1}{2}$ respectively: -$$ -\|R(u)-R(v)\| \le|u-v| \\ -\|T(u)-T(v)\| \le \tfrac{1}{2}|u-v| -$$ -for any $u$ and $v$, where I use the operator norm (subordinate norm) of the Euclidean norm. -To find the convergence bounds of Newton's iterations for the numerical method I'm using (conditions of Kantorovich) I need to estimate a bound on the second derivative (which is really hard to compute explicitly as far as I know), or the Lipschitz constant of the differential. Experimentally (using a program) I found that it is $1$. How to prove this? - -REPLY [2 votes]: Found the proof! It's done using the integral definition of $T$: -$$ -T(v) = \int_0^1 R(su) ds = \lim_{n\rightarrow \infty} \frac{1}{n}\sum_{i=1}^n R\left(\tfrac{i}{n}v\right) -$$ -So for any vectors $X$ and $Y$: -\begin{align*} -&\biggl|\left[\mathrm{D}_v \left(R(v)X\right)\right]Y - \left[\mathrm{D}_u \left(R(u)X\right)\right]Y\biggr| = \biggl|\left[R(u)X\right] \times \left[T(u)Y\right] - \left[R(v)X\right] \times \left[T(v)Y\right]\biggr| \\ -&\le \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=1}^n \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| -\end{align*} -and we only need to prove that each summand is less than $|u-v| |X| |Y|$: -\begin{align*} - & \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| \\ - &= \biggl|R\left(\tfrac{i}{n}u\right)\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right] - R\left(\tfrac{i}{n}v\right) \left[\left(R\left(\tfrac{n-i}{n}v\right)X\right) \times Y\right]\biggr| \\ -& \le \biggl|\left[ R\left(\tfrac{i}{n}u\right) - R\left(\tfrac{i}{n}v\right) \right]\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right]\biggr| \\ -&+ \biggl| R\left(\tfrac{i}{n}v\right) \left[\left(\left(R\left(\tfrac{n-i}{n}v\right) - R\left(\tfrac{n-i}{n}u\right)\right)X\right) \times Y\right]\biggr| \\ - &\le \bigl|\tfrac{i}{n}u - \tfrac{i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| + \bigl|\tfrac{n-i}{n}u - \tfrac{n-i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| \\ - &= \bigl|u - v\bigr|\bigl|X\bigr| \bigl|Y\bigr| -\end{align*} -where we used the invariance of vector products under rotations, the triangle inequality and that $R$ is $1$-Lipschitz (see this question). -Since $X$ and $Y$ are arbitrary, -$$ -\left\|\mathrm{D}_v R(v) - \mathrm{D}_u R(u)\right\| \le |u-v| -$$ -in the subordinate norm.<|endoftext|> -TITLE: Subset of $\mathbb R$ with equal Fourier, Hausdorff and Minkowski dimensions -QUESTION [6 upvotes]: It is a standard fact that for $0\leq s\le1$, there is a compact set $C\subseteq [0,1]$ with Hausdorff and Minkowski dimensions $s$ (by modifying the construction of a Cantor set). -It is also a standard fact that for $0\leq s\le1$, there is a compact set $S\subseteq [0,1]$ with Fourier and Hausdorff dimensions $s$. -My question is: for an arbitrary $0\leq s\le1$, can we find a subset of $\mathbb R$ so that all three dimensions are equal? - -REPLY [2 votes]: Yes. We just use a Baire Category argument (a similar technique also works in high dimensions). Consider the complete metric space $X$ of pairs $(E,\mu)$, where $\mu$ is a probability measure supported on $E$ such that -$$ \sup_{\xi \in \mathbf{Z}} |\widehat{\mu}(\xi)| |\xi|^{s/2} < \infty, $$ -and $E$ is a compact subset of $[0,1]$. We define a distance function -$$ d((E_1,\mu_1),(E_2,\mu_2)) = \max \left( d_H(E_1,E_2), \sup_{\xi \in \mathbf{Z}} |\widehat{\mu_1}(\xi) - \widehat{\mu_2}(\xi)| |\xi|^{s/2} \right) $$ -where $d_H$ is the Hausdorff metric between two sets. It is a useful heuristic that a generic set is as `thin as possible' with respect to the Hausdorff metric. It is simple to see that for any $(E,\mu)$ in $X$, the Fourier dimension of $E$ is at least equal to $s$, so we should expect quasi-all elements of $X$ have dimension $s$. -For each $t > s$, $\delta > 0$, and $\varepsilon > 0$, set -$$ A(t,\delta,s) = \{ (E,\mu) \in \mathcal{X} : |E_\delta| < \varepsilon \cdot \delta^s \} $$ -where $E_\delta$ is the $\delta$ thickening of $E$. Then $A(t,\delta,s)$ is an open subset of $X$, and -$$ \bigcap_{n = 1}^\infty \bigcap_{m = 1}^\infty \bigcap_{k = 1}^\infty A(s+1/n,1/m,1/k) $$ -is the set of all pairs $(E,\mu)$ in $X$ where $E$ has Minkowski dimension $s$. Thus it suffices to argue that $A(t,\delta,\varepsilon)$ is dense in $X$ for all required parameters. It is slightly technical to argue this, but the basic idea is to consider a random construction which, given a pair $(E_0,\mu_0)$, considers the random measure -$$ \mu = \mu_0 \cdot \sum_{k = 1}^K \phi_{\varepsilon_0}(x - x_k) $$ -where $x_1,\dots, x_K$ are uniformly distributed on $[0,1]$, $\varepsilon_0 = K^{-1/s}$, and $\phi_{\varepsilon_0}$ is a smooth bump function supported on a ball radius $\varepsilon_0$. One then shows that with high probability that -$$ \sup_{\xi \in \mathbf{Z}} |\widehat{\mu}(\xi) - \widehat{\mu_0}(\xi)| = o(1) $$ -as $K \to \infty$, and that $d_H(\text{supp}(\mu), \text{supp}(\mu_0)) \to 0$.<|endoftext|> -TITLE: $p$-completeness of the function spectrum $F(\Sigma^{\infty} BS, \Sigma^{\infty} BK)$ -QUESTION [9 upvotes]: Let $S$ be a finite $p$-group and $K$ a compact Lie group, in the paper A Segal conjecture for $p$-completed classifying spaces, it is said that the function spectrum $F(\Sigma^{\infty} BS, \Sigma^{\infty} BK)$ is $p$-complete, but I have not succeeded in proving it. I hope this remains true when, more generally, $S$ is a $p$-toral group (replacing $\Sigma^{\infty} BS$ by $(\Sigma^{\infty} BS)^{\wedge}_p$, because $\Sigma^{\infty} BS$ is no longer $p$-complete). Any suggestion or idea?, please. - -REPLY [6 votes]: Tim's argument is correct, and here's a different way to see this. -To say that $\Sigma^{\infty} BS$ is a finite $p$-group has trivial rationalization and is $p$-local is the same as to observe that it is $p$-torsion, i.e. has $p$-torsion homotopy groups. -To see this without involving the Serre spectral sequence, observe that both the rational and $\mathbb{Z}/l$ cohomology (for $l \neq p$) of $\Sigma^{\infty} BS$ vanishes by Maschke's theorem. For a connective spectrum, this forces it to be $p$-torsion, by Hurewicz (applied to $\Sigma^{\infty} BS \otimes M(l)$). -We also have that $F(T, U)$ is p-complete for any $p$-torsion $T$ and arbitrary $U$. -To see the latter, observe that the subcategory of $p$-torsion spectra is generated under colimits and desuspensions by $M(p)$ (this is the same as saying that any such non-zero spectrum admits a non-zero map from $M(p)$, which is clear). Since $F(-, U)$ takes colimits to limits, and $p$-complete spectra are closed under limits, we then just need to know $F(S^{0}/p, U)$ is $p$-complete, but this is the same as $\Sigma^{-1} U \otimes S^{0}/p$, so we're done.<|endoftext|> -TITLE: When is a function on symmetric positive definite matrices an expectation of Gaussian? -QUESTION [5 upvotes]: Is there some characterization of real-valued functions of the form $\Phi(C)=\mathbb{E}F(X)$, where $X$ has the Gaussian $N(0,C)$ distribution, on the space of symmetric positive semidefinite $n\times n$ matrices $C$? In other words, given $\Phi\colon \mathrm{SPD}\to \mathbb{R}$, is there a way to tell whether such $F$ exists? -Edit: For example, is it clear that $\det(C)$ is not of this form? -Edit 2: Answer below shows that $\det(C)$ is not of this form even for $n=2$. - -REPLY [2 votes]: Here's an approach to determine if $\det(C)$ is of this form with $F$ analytic and $n=2$. -By Sylvester's criterion a matrix $\begin{pmatrix} a& b \\ b & c\end{pmatrix}$ is symmetric positive semidefinite iff $a\ge 0$ and $c\ge 0$, and $ac-b^2\ge 0$. -Let $X=(X_1,X_2)$ be normal with mean 0 and $E(X_1X_2)=b$, $E(X_1^2)=a$, $E(X_2^2)=c$. -The Pearson correlation coefficient is $\rho=b/\sqrt{ac}$. -Suppose $F(x,y)=\sum c_{mn}x^my^n$. -Is there an $F$ with $E(F(X))=ac-b^2$? -We have $X=\sqrt{a}Z_1$, $Y=\sqrt{c}(\rho Z_1+\sqrt{1-\rho^2}Z_2)$ where $Z_i$ are independent standard normal, so -using -$$E\left[(\rho Z_1+\sqrt{1-\rho^2}Z_2)^{2k}\right]=E\sum_{t=0}^{2k}\binom{2k}{t}\rho^t(1-\rho^2)^{(2k-t)/2}Z_1^t Z_2^{2k-t}$$ -$$=\sum_{t=0}^{2k}\binom{2k}{t}\rho^t(1-\rho^2)^{(2k-t)/2}E[Z_1^t Z_2^{2k-t}] -=\sum_{u=0}^{k}\binom{2k}{2u}\rho^{2u}(1-\rho^2)^{(2k-2u)/2}E[Z_1^{2u} Z_2^{2k-2u}]$$ -$$=\sum_{u=0}^{k}\binom{2k}{2u}\rho^{2u}(1-\rho^2)^{(2k-2u)/2}(2u-1)!!(2(k-u)-1)!!$$ -(where $(-1)!!=1$) -we have -$$E(X^{2\ell}Y^{2k})=a^\ell c^k\sum_{u=0}^{k}\binom{2k}{2u}\rho^{2u}(1-\rho^2)^{k-u}(2(\ell+u)-1)!!(2(k-u)-1)!!$$ - -When $k=\ell=1$, it is $ac((1-\rho^2)+3\rho^2)=ac+2b^2$ -since $\rho^2=b^2/ac$. -When $k=0$ and $\ell=2$, it is $3a^2$. -When $k=2$ and $\ell=0$, it is $3c^2-4b^4/a^2$. -I guess we should also do the case $E(X^3Y^1)$ and a couple of others. - -A Taylor series in variables $a,b,c$ defines the zero function only if all coefficients are zero. -So now by calculating these expectations $E(X^pY^q)$ we can determine whether -$ac-b^2$ is obtainable.<|endoftext|> -TITLE: Is it possible to create a polynomial $p(x)$ with this relation between $p(0)$ and $p(c)$? -QUESTION [8 upvotes]: Given $b$ and $c$ with $b,c>1$, is it possible to construct a polynomial $p(x)$, whose degree is $n$ for all $c$ and $b$, such that: - -$|p|$ is strictly increasing on $[1,c]$ - -and $|b \cdot p(c)| < |p(0)|$? - - -This might be satisfied by an interpolating polynomial, but how to actually construct it is beyond me. - -REPLY [2 votes]: Let me present a more explicit version of Fedor’s argument. -Choose distinct $x_0,\dots,x_n\in[1,c]$. -By Lagrange’s interpolation formula, there exist constants $a_0,\dots,a_n$ such that -$$ - p(0)=\sum_{I=0}^n a_ip(x_i) -$$ -for each polynomial $p$ of degree not exceeding $n$. Therefore, -$$ - |p(0)|\leq \sum_{I=0}^n |a_i|\cdot |p(x_i)|\leq - |p(c)|\cdot \sum_{I=0}^n |a_i|. -$$<|endoftext|> -TITLE: For self-adjoint $A$ and $B$, when is $(A+iB)^*$ the closure of $A-iB$? -QUESTION [5 upvotes]: Suppose that I have two self-adjoint operators $A$ and $B$ such that $\mathcal{D}(A)\cap\mathcal{D}(B)$ is dense and $B$ positive. Then $A\pm iB$ (with domains $\mathcal{D}(A)\cap\mathcal{D}(B)$) are closable. What are generic conditions so that $(A+iB)^*$ is the closure of $(A-iB)|_{\mathcal{D}(A)\cap\mathcal{D}(B)}$? $A$ or $B$ bounded suffices, but does this hold in general? -NB: What I'm really interested in is whether $0$ is in the resolvent set of the closure of $A+iB$, which follows easily from the above condition. - -REPLY [7 votes]: In general this is not true. Let $\Omega$ be a smooth bounded domain, let $A$ be $-\Delta$ with Neumann conditions and let $B$ be $-\Delta$ with Dirichlet conditions. The $(A+iB)$ is $(-1-i)\Delta$ with domain $H^2_0(\Omega)$. Its adjoint is $(-1+i)\Delta$ with domain $H^2(\Omega)$, which is not the closure of $A-iB$.<|endoftext|> -TITLE: Akbulut's cork involution -QUESTION [6 upvotes]: Akbulut's cork is the Mazur manifold $W$ shown in the picture below, -This manifold carries an involution of it's boundary $f:\partial W\to \partial W$ that exchanges a meridian of the 0-framed curve with a meridian of the dotted curve. -In Akbulut - 4-manifolds, the author says that $f$ is related to performing surgeries to exchange the dot with the 0-framed knot. -I don't understand how this relates to $f$ and how $f$ is defined. -We can surely perform two surgeries (one along a circle and one along a 2-sphere) to get a new 4-manifold $\tilde{W}$ that has a Kirby diagram obtained by exchanging dot and zero in the above picture. However surgery does not affect the boundary so I do not see how this could induce a nontrivial map between the boundaries. - -Precisely how is $f$ defined? - -REPLY [7 votes]: The boundary of W may be described as 0-framed surgery on both components of the link you drew. The link can be drawn in a more symmetric fashion, so that it is clear that there is an involution interchanging the two components. See for instance Figure 4 of Akbulut's paper, A solution to a conjecture of Zeeman, Topology 30 (1991, 513-515. This involution induces an involution on the surgered manifold that interchanges the two meridians. -If you don't care that the diffeomorphism on the boundary is an involution, you can just do an isotopy interchanging the components.<|endoftext|> -TITLE: The space of skew-symmetric orthogonal matrices -QUESTION [7 upvotes]: Let $M_n \subseteq SO(2n)$ be the set of real $2n \times 2n$ matrices $J$ satisfying $J + J^{T} = 0$ and $J J^T = I$. Equivalently, these are the linear transformations such that, for all $x \in \mathbb{R}^{2n}$, we have $\langle Jx, Jx \rangle = \langle x, x \rangle$ and $\langle Jx, x \rangle = 0$. They can also be viewed as the linear complex structures on $\mathbb{R}^{2n}$ which preserve the inner product. -I'd like to understand $M_n$ better as a topological space, namely an $(n^2-n)$-manifold. -$M_1$ is just a discrete space consisting of two matrices: the anticlockwise and clockwise rotations by $\pi/2$. -For $n \geq 2$, we can see that $M_n$ is an $M_{n-1}$-bundle over $S^{2n-2}$. Specifically, given an arbitrary unit vector $x$, the image $y := Jx$ must lie in the intersection $S^{2n-2}$ of the orthogonal complement of $x$ with the unit sphere $S^{2n-1}$. Then the orthogonal complement of the space spanned by $x$ and $y$ is isomorphic to $\mathbb{R}^{2n-2}$, and the restriction of $J$ to this space can be any element of $M_{n-1}$. -Since the even-dimensional spheres are all simply-connected, it follows (by induction) that $M_n$ has two connected components for all $n \in \mathbb{N}$, each of which is simply-connected. For instance, $M_2$ is the union of two disjoint 2-spheres: the left- and right-isoclinic rotations by $\pi/2$. The two connected components of $M_n$ are two conjugacy classes in $SO(2n)$; they are interchanged by conjugating with an arbitrary reflection in $O(2n)$. -Is [each connected component of] $M_n$ homeomorphic to a known well-studied space? They're each an: - -$S^2$-bundle over an $S^4$-bundle over $\dots$ an $S^{2n-4}$-bundle over $S^{2n-2}$ - -but that's not really very much information; can we say anything more specific about their topology? - -REPLY [8 votes]: Your $M_n$ is (two copies of) the Riemannian symmetric space $\mathrm{SO}(2n)/\mathrm{U}(n)$ (which is $DI\!I\!I$ in Cartan's nomenclature). Its topology is well-studied from that point of view.<|endoftext|> -TITLE: Is there a CAS that can solve a given system of equations in a finite group algebra $kG$? -QUESTION [6 upvotes]: Let $k$ be a finite field with char$(k)=p>0$. Let $G$ be a finite group. -Consider the group algebra $kG$. -I would like to solve a given system of equations in $kG$. -Question: - -Is there a computer algebra system that can solve a given system of equations in $kG$? - -Any help is appreciated. -Thank you very much. - -REPLY [2 votes]: A partial answer in a very special case, but still possibly useful to record. -The search for idempotents in a group algebra can be simplified by using GAP's MeatAxe: https://www.gap-system.org/Manuals/doc/ref/chap69.html -For example, if $G$ is the dihedral group with eight elements, and $F$ the field with eight elements, the following code verifies that $kG$ is indecomposable as a $G$-module, which means that the only idempotents are zero and one. -G := DihedralGroup(8) -k := FiniteField(8); -cayley := Action(G, AsList(G), OnRight); -reg := PermutationGModule(cayley, k); -MTX.IsIndecomposable(reg); - -More generally, if $M$ is a MeatAxe module, you can compute a direct sum decomposition using -MTX.Indecomposition(M) - -Of course, this doesn't solve for all the idempotents---it only returns a few. You can also use -MTX.HomogeneousComponents(M) - -to break the search into smaller algebras before using a more brute-force method.<|endoftext|> -TITLE: Reference on internal categories and externalization -QUESTION [7 upvotes]: I'm looking for a reference on internal categories and externalization of internally defined notions. -The nlab has a stub on externalization (more details are available under small fibration) and the page on internal categories gives enough of an introduction that I can sketch most internal notions, but I could really use a concise introduction to internal categories and externalization, and if possible the relationship between internalization and externalization. Are they adjoint in some sense? -I'm fine assuming a background of $2$-category theory, so talking about the $2$-category of internal categories in a category with pullbacks etc. would make sense, but ideally the reference would assume no familiarity with internal category theory or externalization. Any assistance is appreciated. - -REPLY [7 votes]: For what it's worth, I should add my comment as an answer. Chapter 1 of Bart Jacobs' book Categorical logic and type theory (Studies in Logic and the Foundations of Mathematics 141 (1999), (author's page, publisher page, pdf)) is a good intro to fibred category theory, and chapter 7 has an intro to internal category theory, and it links the two. Chapter 9 does some more advanced fibred category theory.<|endoftext|> -TITLE: Determining if a quadratic form is non-negative if variables are non-negative -QUESTION [7 upvotes]: Let $f(x_1,\dots,x_n) = \sum_{1 \le i \le j \le n} c_{i,j}x_ix_j$ be a homogeneous quadratic form. Is there a quick-ish way to determine whether $f(x_1,\dots,x_n) \ge 0$ for all $x_1,\dots,x_n \ge 0$? -I have a specific homogeneous quadratic form, where $n=44$. I am wondering whether I have to use a super computer to prove that it is non-negative if all of the variables are. I prefer not to disclose my quadratic form. -In general, I know how to figure out whether a given quadratic form is non-negative if the variables are, in $2^n$ time, since $f$ has at most $2^n$ (quickly computable) local minima on the set $\{(x_1,\dots,x_n) \in \mathbb{R}^n : x_1+\dots+x_n = 1, x_1,\dots,x_n \ge 0\}$ (choose a certain subset of the variables to be $0$, and then we get a bunch of linear equations, from looking at derivatives, that determine the rest). But I'm wondering if there's a quicker way, in general. - -REPLY [2 votes]: You might consider the "sum-of-squares" approach. The idea is to find a set of polynomials so that your expression is the sum of squares of the elements in the region of interest. For your case, you could replace each $x_i$ with a new variable $z_i^2$; you are now asking if the corresponding 4-th degree unconstrained polynomial is non-negative. -This restatement may not sound like an improvement, but it turns out that SOS problems can be attacked using semidefinite programming techniques (e.g., see this page). You can use a freely available SDP solvers. -This is a sufficient approach, i.e., it may prove that your original quadratic form is positive, but it can't disprove it. Since you're trying to solve a specific problem, though, it may be worth the gamble.<|endoftext|> -TITLE: Perfect numbers, Galois groups and a polynomial -QUESTION [6 upvotes]: Let $f(n,t) = \sum_{k=0}^{r-1} d_k t^k$ where $D_n = \{d_0=1,d_1,\cdots,d_{r-1}\}$ are all divisors of $n$. -For instance -$$f(28,t) = 28 t^{5} + 14 t^{4} + 7 t^{3} + 4 t^{2} + 2 t + 1$$ -For even perfect numbers $n = 2^{p-1}(2^p-1)$ this polynomial always factors as: -$$f(n,t) = ((2^p-1)t^p+1)(1+2t +2^2t^2+\cdots+2^{p-1}t^{p-1})$$ -Furthermore the two irreducible (?) factors seem to have interesting Galois groups: -6 (2*t + 1) * (3*t^2 + 1) -6 (2*t + 1, 1) Galois group PARI group [1, 1, 1, "S1"] of degree 1 of the Number Field in t with defining polynomial 2*t + 1 -6 (3*t^2 + 1, 1) Galois group PARI group [2, -1, 1, "S2"] of degree 2 of the Number Field in t with defining polynomial 3*t^2 + 1 -28 (4*t^2 + 2*t + 1) * (7*t^3 + 1) -28 (4*t^2 + 2*t + 1, 1) Galois group PARI group [2, -1, 1, "S2"] of degree 2 of the Number Field in t with defining polynomial 4*t^2 + 2*t + 1 -28 (7*t^3 + 1, 1) Galois group PARI group [6, -1, 2, "S3"] of degree 3 of the Number Field in t with defining polynomial 7*t^3 + 1 -496 (16*t^4 + 8*t^3 + 4*t^2 + 2*t + 1) * (31*t^5 + 1) -496 (16*t^4 + 8*t^3 + 4*t^2 + 2*t + 1, 1) Galois group PARI group [4, -1, 1, "C(4) = 4"] of degree 4 of the Number Field in t with defining polynomial 16*t^4 + 8*t^3 + 4*t^2 + 2*t + 1 -496 (31*t^5 + 1, 1) Galois group PARI group [20, -1, 3, "F(5) = 5:4"] of degree 5 of the Number Field in t with defining polynomial 31*t^5 + 1 -8128 (64*t^6 + 32*t^5 + 16*t^4 + 8*t^3 + 4*t^2 + 2*t + 1) * (127*t^7 + 1) -8128 (64*t^6 + 32*t^5 + 16*t^4 + 8*t^3 + 4*t^2 + 2*t + 1, 1) Galois group PARI group [6, -1, 1, "C(6) = 6 = 3[x]2"] of degree 6 of the Number Field in t with defining polynomial 64*t^6 + 32*t^5 + 16*t^4 + 8*t^3 + 4*t^2 + 2*t + 1 -8128 (127*t^7 + 1, 1) Galois group PARI group [42, -1, 4, "F_42(7) = 7:6"] of degree 7 of the Number Field in t with defining polynomial 127*t^7 + 1 - - -What are the Galois groups of these irreducible (?) factors? -Whats is the Galois group of $f(n,t)$ for $n$ an even perfect numbers? - -REPLY [7 votes]: The second factor is $P(2t)$ where $P=X^{p-1}+\cdots+X+1$, the $p$-th cyclotomic polynomial. -Hence the Galois group of $P(2t)$ is the same as the Galois group of $P(t)$, which is simply $(\mathbb{Z}/p\mathbb{Z})^\times$, which is cyclic of order $p-1$. -The other factor has the form $t^p -a$. The splitting field is $\mathbb{Q}(\zeta_p, a^{1/p})$, and it is known that the Galois group is $(\mathbb{Z}/p\mathbb{Z})\rtimes -(\mathbb{Z}/p\mathbb{Z})^\times$ (with the obvious action).<|endoftext|> -TITLE: Generalising the union-closed sets conjecture from lattice to a larger class of posets -QUESTION [9 upvotes]: (edit: I decided to simplify the question and only pose it for bounded posets first) -The Union-closed sets conjecture is equivalent for lattices P to: - -There exists a join-irreducible element $a$ with $|[a,M]| \leq |P|/2$, when $M$ is the maximum of $P$. - -Recall that an element a of a poset is join-irreducible if there is no subset $X \subseteq P$ with $a\not\in X$ and $a=\bigvee X$. -Call a (finite) bounded poset $P$ lattice-like in case an element $x \in P$ is join-irreducible iff $x$ is covers a unique element. -Every lattice is lattice-like but not every bounded poset is lattice-like. - -Question 1: Is the above conjecture also true for lattice-like posets? - -This is true for all such posets with at most 8 points. I would think there is a counterexample but I have not found one yet. - -Question 2: Are there attempts in the literature already to generalise the Union-closed sets conjecture from lattices to a larger class of posets? - -REPLY [11 votes]: Here is a counterexample of size 23. -Let $m=6$ and let $$P=\{0,a_1,\dots,a_m,1\}\cup\{b_{ij}: 1\le i|P|/2$ as long as -$$2+\binom{m}2 - (m-1) > \frac12\left(m+2+\binom{m}2\right)$$ -$$4+\binom{m}2>3m$$ -which is true for $m=6$ but not for $m=5$.<|endoftext|> -TITLE: A possible characterization of the category of finite $p$-groups -QUESTION [6 upvotes]: Let $\mathcal{FG}$ be the category of finite groups. Let $S$ be a full subcategory of $\mathcal{FG}$. -Assume that $G\in \mathcal{FG}$ and $P\in S$ is a subgroup of $G$. We say that $P$ is $S$-maximal if there is no object $P'\in S$ with $P\subset P' \subset G$. Assume that $S$ satisfies the following conditions: - -The subcategory $S$ is closed under taking subgroups, taking extensions" and isomorphy. That is: $P\in S, Q\subset P$ implies that $Q \in S$. Moreover for every short exact sequence $1\to P\to Q\to R\to 1$ we have $Q\in S$ if $P,R \in S$. Moreover every group isomorphic to an object of $S$ lies in $S$. - -For every $G\in \mathcal{FG}$, every two $S$-maximal subgroups of $G$ are conjugate. Moreover for every maximal $S$-subgroup $P$ of $G\in \mathcal{FG}$ we have $N(N(P))=N(P)$ and $\lvert G\rvert/\lvert N(P)\rvert$ is coprime to $|P|$. - - -Does it imply that $S$ is the category of $p$-groups for some prime number $p$? - -REPLY [7 votes]: $\DeclareMathOperator\GL{GL}\newcommand\card[1]{\lvert#1\rvert}$Here is a proof of "yes," using Tim Campion's proposition below. -Let $p$ be the smallest prime in $P$. For any prime $q\ne p$, let $o_q(p)$ be the multiplicative order of $p$ modulo $q$, or equivalently the least $n$ such that $q$ divides $\card{\GL_n(p)}$. Assuming $P\ne\{p\}$, let $q\in P-\{p\}$ and $G=\GL(V)$ where $V$ is a vector space of order $p^n$ and $n=o_q(p)$. As $p$ is smallest, $n\ge2$. By assumption, $G$ contains a Hall $P$-subgroup $X$. Then $X$ contains a Sylow $p$-subgroup $U$ of $G$ as well as an element $x$ of order $q$. If $U$ is normal in $X$, then $\card X$ divides $\card U(p-1)^n$, the order of the full upper triangular group, which is not divisible by $q$ as $q>p$. So $U$ is not normal in $X$. The theory of $B$-$N$ pairs then implies that $X$ contains a copy of $\operatorname{SL}_2(p)$. Hence every prime divisor of $p-1$ lies in $P$, which forces $p=2$. Then $U$ is a Borel subgroup of $G$, so $X$ must be a parabolic subgroup of $G$. But because of $x$, $X$ stabilizes no proper subspace of $V$. The only such parabolic subgroup is $X=G$. Hence $P$ contains all prime divisors of $\card G$. -In particular, $3\in P$. -Now suppose that $P$ is not the set of all primes and choose a prime $r\not\in P$ to minimize $m=o_r(2)$. Since $3\in P$, $m\ge3$. Let $H=\GL(W)$, where $W$ is a vector space of order $2^m$. Then $H$ contains an element $y$ of order $r$. Let $U$ be a Sylow $2$-subgroup of $H$. Let $W_1$ and $W_{m-1}$ be $U$-invariant subspaces of $W$ of respective dimensions $1$ and $m-1$. Let $H_1$ and $H_{m-1}$ be their respective stabilizers in $H$. Then $H_1$ and $H_{m-1}$ are maximal parabolic subgroups of $H$ containing $U$, and each is an extension of an elementary abelian $2$-group by $\GL_{m-1}(2)$. Hence $H_1$ and $H_{m-1}$ are $P$-groups, by our choice of $r$. However, they are maximal subgroups of $H$ and they are not conjugate in $H$, being distinct parabolic subgroups containing $U$. -By assumption, $\langle H_1^g, H_{m-1}\rangle$ must be a $P$-group for some $g\in H$. Since $H_{m-1}$ is maximal and not equal to $H_1^g$, $H$ must be a $P$-group. But $y\in H$ has order $r\notin P$, contradiction.<|endoftext|> -TITLE: Resonance arising when harmonic oscillator is excited using sawtooth -QUESTION [7 upvotes]: Solutions to the differential equation $my'' + ky = F \sin \omega t$ show resonance when the driving frequency $\omega$ equals the natural frequency $\sqrt{k/m}$. That is, solutions are unbounded when $\omega = \sqrt{k/m}$ and periodic for all other frequencies. It seems that when the sine function is replaced by a sawtooth function, there are more resonant frequencies. Numerical experiments here with $m = k = F = 1$ seem to be resonant when $\omega = 1/n$ for any integer $n$. Are there any theoretical results that prove this conjecture? - -REPLY [18 votes]: The sawtooth function $f$ has Fourier decomposition -$$ -f(t) = \frac{1}{2}-\frac{1}{\pi}\sum_{n=1}^\infty \frac{1}{n} \sin(n\omega t) -$$ -Therefore, if $\omega=\frac{\omega_0}{n}$, the $n$-th harmonic of $f$ will have angular frequency $n\omega=\omega_0$, resulting in resonance.<|endoftext|> -TITLE: Is Lie group cohomology determined by restriction to finite subgroups? -QUESTION [30 upvotes]: Consider the restriction of the group cohomology $H^*(BG,\mathbb{Z})$, where $G$ is a compact Lie group and $BG$ is its classifying space, to finite subgroups $F \le G$. If we consider the product of all such restrictions -$$H^*(BG,\mathbb{Z}) \to \prod_F H^*(BF,\mathbb{Z}),$$ -is this map injective? -Note, according to McClure - Restriction maps in equivariant K-theory a similar result holds in equivariant K-theory. Perhaps there is a way to derive the above from McClure's theorem? -I asked this question on stackexchange and didn't manage to get a (complete) answer nor solve it myself. However, Qiaochu Yuan offered a proof for the non-torsion elements. - -REPLY [3 votes]: $\DeclareMathOperator\Image{Image}$Here's the most general result I think I can muster. I've split it out into a second answer in order to keep the answer to the original question more self-contained. - -Theorem 1: Let $G$ be a compact Lie group, let $X$ be a $G$-space, and let $E$ be a spectrum. Then the following hold, where $F$ ranges over finite subgroups of $G$: - -The image of $\bigoplus_F E_\ast(X_{hF}) \to E_\ast(X_{hG})$ contains all of the torsion; - -The kernel of $E^\ast(X_{hG}) \to \prod_F E^\ast(X_{hF})$ is contained in the subgroup of divisible elements. - - - -This follows from the following two more precise theorems: - -Theorem 2: Let $G$ be a compact Lie group, and let $X$ be a $G$-space. Let $N \subseteq G$ be the normalizer of a maximal torus $T \subseteq G$, and let $W = N / T$ be the Weyl group. Then $\Sigma^\infty_+ X_{hG}$ splits off of $\Sigma^\infty_+ X_{hN}$. -Proof: The splitting is given by the Becker-Gottlieb transfer: the fiber of $X_{hN} \to X_{hG}$ is $G/N$, the same as the fiber of $BN \to BG$, which has Euler characteristic 1. - -Theorem 3: -Let $N$ be an extension of a finite group $W$ by a torus $T$, and let $E$ be a spectrum and $m \in \mathbb N_{\geq 2}$. Then the following hold, where $F$ ranges over finite subgroups of $N$: - -$\varinjlim_F (E/m)_\ast(X_{hF}) \to (E/m)_\ast(X_{hN})$ is an isomorphism; - -$(E/m)^\ast(X_{hN}) \to \varprojlim_F (E/m)^\ast(X_{hF})$ is an isomorphism. - - - -Proof of Theorem 1 from Theorems 2 and 3: By Theorem 2, it will suffice to consider the case where $G = N$ is an extension of a finite group by a torus. Theorem 3 establishes Theorem 1 for $E/m$. Then (1) follows by considering the natural short exact sequence $0 \to E_{\ast}(-)/m \to (E/m)_\ast(-) \to E_{\ast-1}(-)^{\text{$m$-tor}} \to 0$, and the argument for (2) uses a similar exact sequence. - -The proof of Theorem 3 will follow from a series of lemmas. For the remainder, we let $U(1)^n \to N \to W$ be an extension of a finite group by a torus, and we let $(\mathbb Q/\mathbb Z)^n \to N_\ast \to W$ and $(C_q)^n \to N_q \to W$ be the subextensions which exist by the analysis in my other answer. We fix a spectrum $E$, $m \in \mathbb N_{\geq 2}$, and an $N$-space $X$. - -Lemma 4: The fiber of $X_{hN_\ast} \to X_{hN}$ is $B\mathbb Q^n$, and in particular this map is an $(E/m)_\ast$ and $(E/m)^\ast$ equivalence. -Proof: This comes via a diagram chase from the fiber sequence $B\mathbb Q^n \to B(\mathbb Q/\mathbb Z)^n \to BU(1)^n$. - -Lemma 5: We have $(E/m)_\ast(X_{hN_\ast}) \cong \varinjlim_q (E/m)_\ast(X_{hN_q})$ canonically, and a canonical short exact sequence $0 \to \varprojlim^1 (E/m)^{\ast+1}(X_{hN_q}) \to (E/m)^\ast(X_{hN_\ast}) \to \varprojlim (E/m)^\ast(X_{hN_q}) \to 0$. -Proof: By the analysis in my other answer, we have $N_\ast = \varinjlim N_q$. Therefore $BN_\ast = \varinjlim BN_q$, and it follows that $X_{hN_\ast} = \varinjlim X_{hN_q}$. The lemma follows by the usual formulas for homology and cohomology of a filtered colimit. - -Lemma 6: If $E$ is $m$-torsion, then $\varprojlim^1 E^\ast(X_{hN_q}) = 0$. - -Proof of Theorem 3: This follows from Lemmas 4, 5, and 6, once we note that $E/m$ is $m^2$-torsion. - -It remains to prove Lemma 6. - -Lemma 7: Let $q,r \in \mathbb Z$, and consider the inclusion $C_{qr}^n \xrightarrow i (\mathbb Q/\mathbb Z)^n$. Consider also the inclusion $C_q^n \xrightarrow j C_{qr}^n$ with quotient $C_r^n$. Let $A$ be an $r$-torsion and $q$-torsion abelian group. Then $H^\ast(ij;A)$ is injective and $\Image(H^\ast(ij;A)) = \Image(H^\ast(j;A))$. -Proof: Direct calculation. More precisely, $H^\ast(BC_q;A)$ and $H^\ast(BC_{qr};A)$ both have $A$ in all degrees; the inclusion of $H^\ast(B(\mathbb Q/Z);A)$ is an isomorphism onto the even degrees, and $H^\ast(j;A)$ kills the odd classes while being an isomorphism on even classes. Then one extends this analysis to $n > 1$. - -Lemma 8: Let $A \xrightarrow i B \xrightarrow j C$ be maps of chain complexes. Suppose that $ji$ is injective and $\Image(ji) = \Image(i)$. Then the sequence of homologies $H_\ast(A) \xrightarrow{i_\ast} H_\ast(B) \xrightarrow{j_\ast} H_\ast(C)$ has $i_\ast$ injective and $Image(j_\ast i_\ast) = Image(i_\ast)$. -Proof: Diagram chase. - -Corollary 9: Fix $s \in \mathbb Z$, and consider the maps $H^s(BN_\ast;A) \xrightarrow i H^s(BN_{qr};A) \xrightarrow j H^s(BN_q;A)$. For $q$, $r$ sufficiently divisible by $m$ and $A$ $m$-torsion, we have that $ji$ is injective and $\Image(ji) = \Image(j)$. -Proof: Using Lemma 7 as a base case, use Lemma 8 to induct through the pages of the Serre spectral sequences for the fibrations over $BW$. This is a first-quadrant spectral sequence, so for fixed $s$ it stabilizes at a finite page. The statement can be tested on associated gradeds, so there are no extension problems. - -Corollary 10: Assume that $E$ is bounded below and $m$-torsion, and fix $s \in \mathbb Z$. Consider the maps $E^s(X_{hN_\ast}) \xrightarrow i E^s(X_{hN_{qr}}) \xrightarrow j E^s(X_{hN_q})$. For $q,r$ sufficiently divisible by $m$, we have that $ji$ is injective and $\Image(ji) = \Image(j)$. -Proof: Using Corollary 9 as a base case, use Lemma 8 inductively to walk through the Atiyah–Hirzebruch spectral sequences for the fibrations over $BN_\ast$, $BN_{qr}$, and $BN_q$ respectively (which all have fiber $X$). Since we are assuming that $E$ is bounded below, this is essentially a first-quadrant spectral sequence so the argument goes in the same way as before. - -Proof of Lemma 6: That this follows from Corollary 10 in the case where $E$ is bounded below can be seen in two ways — either from the eventual injectivity of $E^\ast(X_{hN_\ast}) \to E^\ast(X_{hN_q})$, or from the fact that sequence is Mittag–Leffler. When $E$ is not bounded below, we simply pass to a suitable connective cover of $E$, since we are always taking the cohomology of a suspension spectrum, which is bounded below.<|endoftext|> -TITLE: Truncation of infinity-categories -QUESTION [5 upvotes]: If we have a category $\mathcal{C}$, then we can see it as an $\infty$-category. Furthermore, we can truncate and $\infty$-category $\mathcal{X}$ to get a category $\mathcal{X}_{\leq 1}$. My question is if these functors are adjoint, i.e. if we have -$$\text{Hom}_{\mathfrak{Cat}}(\mathcal{X}_{\leq 1}, \mathcal{Y})\cong \text{Hom}_{\infty\text{-}\mathfrak{Cat}}(\mathcal{X},\mathcal{Y})$$ -where $\mathcal{Y}$ is a category (which on the right is seen as an $\infty$-category). - -REPLY [12 votes]: There is a bit of notation to be careful about here: -$\mathcal{X}_{\leqslant 1}$ is often used to denote the full subcategory of $\mathcal{X}$ of set-truncated object. For example if $\mathcal{X}$ is an $\infty$-topos, then $\mathcal{X}_{\leqslant 1}$ is its $1$-topos reflection. -with this definition, $\mathcal{X}_{\leqslant 1}$ is a $1$-category, but it is not the one that will have the property you want (it will be a right adjoint instead of a left adjoint, and only when restricted to finite limit preserving functor). -The $1$-category you want to consider is the homotopy category $h \mathcal{X}$ of $\mathcal{X}$, sometimes also denoted $\tau \mathcal{X}$, which is the category with the same objects as $\mathcal{X}$, and with the morphism sets -$$ h\mathcal{X}(a,b) \simeq \pi_0 ( \mathcal{X}(a,b) ) $$ -Which does satisfies the property you ask. -A rigorous proof of this of course depends on what model of $\infty$-category you use, but if you use quasi-categories, this follows from points 1.2 and 1.8 in Joyal notes on quasi-categories.<|endoftext|> -TITLE: Packing in uniform domains -QUESTION [5 upvotes]: Given $N$ points $X:=(x_i)_{i \in \{1,..,N\}}$, we now define a score function $S:X \rightarrow \mathbb{N}$ that is $S(X)= \sum_{i=1}^N S(x_i)$ where the score of $S(x_i)$ is -$$S(x_i) = 2* \vert \{x_j; \vert x_i-x_j \vert \in [1,2]\} \vert+ \vert \{x_j; \vert x_i-x_j \vert \in [2,3]\} \vert$$ -where $\vert \bullet \vert$ denotes the cardinality of the set. Moreover, we require that for all $i\neq j$ we have $\vert x_i-x_j \vert \ge \frac{1}{2}.$ -Question: Is it true that any configuration of $N$ points with maximal possible score is in a domain of diameter $c\sqrt{N}$ for some fixed c? - -REPLY [2 votes]: What follows below is the new answer to the modified question, where we assume in addition $|x_i-x_j|\ge 1$ (probably one can ask $|x_i-x_j|\ge 1-\varepsilon$ for sufficiently small $\varepsilon$). I want to propose a positive solution of this problem modulo the following guess, which, I hope, is correct. -Guess. Consider the equilateral triangular lattice $E$ with distance $1$ between neighbouring points. Then there are exactly $18$ points on distance at most $2$ from a given one, and $36$ on distance at most $3$. So the score of each point is $54=2*18+(36-18)$. I guess, that for any set $X$ such that any two points are on distance at least $1$, in the punctured $2$-neighbourhood of any point $x\in X$ there are at most $18$ points of $X$. I guess that the same holds for points on distance at most $3$. If this is true then we have the following corollary: for any set $X$ satisfying $|x_i-x_j|\ge 1$ we have $S(x_i)\le 54 $. -So from now on we assume that either the guess is correct, or we are working with a set $X$ such that each point of this set has score at most $54$. -I'll prove that under such condition the constant $c$ exists. -Proof. Note first of all that we can always construct a set $X$ with $N$ points, such that the score of $X$ is $54N-10^{10}\sqrt{N}$. Such a set can be given by intersecting $E$ with a disk of appropriate radius. (one can take a smaller constant than $10^{10}$, but it doesn't matter). -Assume by contradiction, that we constructed a set $X$ maximising the score and such that its diameter is more than $10^{10^{10}}\sqrt{N}$. Take the union of disks of radius $3$ around all points of $X$, and denote this set by $U_3$. It is easy to see that $U_3$ is connected. Indeed, if it is not, we can parallel translate its connected component by pushing one to another and increase this way the score of $X$. So, since the diameter of $X$ is at least $10^{10^{10}}\sqrt{N}$, the perimeter of the exterior boundary of $U_3$ is at least $10^{10^{10}}\sqrt{N}$. We will say that a point of $X$ contributes to the exterior boundary of $U_3$ if it is on distance $3$ from it. It is easy to see that the number of points of $X$ contributing to the exterior boundary is at least $10^{(10^{10}-2)}\sqrt{N}$ (because the length of a radius $3$ circle is $<100$). The final observation is that any point $x$ of $X$ that contributes to the boundary has score less than $54$. This is because the disk of radius $3$ around $x$ has a large sub-region, where points of $X$ can not lie (indeed, take a point $y\in \partial U_3$ on distance $3$ from $x$, then no point on distance less than $3$ from $y$ lies in $X$). Finally, taking into account the guess and the fact that the score of $X$ has to be at least $54N-10^{10}\sqrt{N}$, we get a contradiction. -Old answer -Let's consider two variations of this question. In both cases the answer is yes. In the first case $X$ is any subset of $\mathbb R^2$ in the second it is a subset of $\mathbb Z^2$. -1 We assume first that $X$ is any subset of $\mathbb R^2$. In such case the set with maximal possible score has diameter at most $6$. Let me prove this. Let's first construct a set with score approximatively $\frac{5}{3}N^2$. To do this we put $N/6$ points in each vertice of the regular hexagon with side of length $1$. -Now, suppose we have a set with maximal score and suppose its diameter is more than $6$. We will construct a set with larger score which will give us a contradiction. -So, suppose $X$ has two points $x_i, x_j$ such that $|x_i-x_j|>6$. Let's take two disks of radius $3$ around both points. One of them contains at most $N/2$ points, which means $S(x_i)$ or $S(x_j)$ is at most $N$. Without loss of generality assume $S(x_i)\le N$. On the other hand, we know that $S(X)\ge \frac{5 N^2}{3}$. So, there is a point $x_k$ such that $S(x_k)\ge\frac{5}{3}N$. Move $x_i$ at the place of $x_k$, this will increase the score $S(X)$. Contradiction. -2 What follows is just a sketch of proof. We assume $X\subset \mathbb Z^2$. In such case each point $x_i$ contributes at most $2*8+20=36$ to the sum $S(X)$. Indeed, there are $8$ integer points on distance at most $2$ from a given one, and $20$ on distance in $[2,3]$. From this one can deduce the answer applying the isomperietric inequality to the set that is the union of $2\times 2$ squares with centres at points of $X$. I can give more details, if you want.<|endoftext|> -TITLE: What is the cotangent complex good for? -QUESTION [36 upvotes]: The cotangent complex seems to be a pretty fundamental object in algebraic geometry, but if it's treated in Hartshorne then I missed it. It seems to be even more important in derived algebraic geometry, so I think I need to slow down and zoom out a bit. When first learning about object $X$, it's nice to have in mind some concrete applications of $X$ to structure one's thinking. -Question: Why study the cotangent complex? What problems is it intended to solve? -(Bonus points if there is something interesting to say about extending to the derived setting.) -I have the sense that the cotangent complex is such a fundamental object that it may be difficult to isolate its importance -- much like trying to articulate the significance of something like cohomology. In that case, it might be more appropriate to ask something like "what kinds of questions does the cotangent complex allow one to ask?". -EDIT: The answers so far are great, but I imagine there are a great many more examples which could be given (the more down-to-earth the better!). As suggested in the comments, it's probably appropriate to say a bit more about where I'm coming from. -I suppose the main ideas I have in my head right now are: - -The cotangent complex generalizes the Kahler differentials. - -The cotangent complex controls deformation theory. - - -This leaves me with a few difficulties: - -I'm not used to thinking of differential forms primarily as "things that control deformations". So it might be helpful to simply illustrate the use of the cotangent complex by describing some deformation problem and its solution in the smooth case using differntials -- it would then seem natural to want to generalize this situation to the non-smooth case. - -I'm not even sure why I should be interested in deformation theory as such. So it might be helpful to simply see an example of a problem which arises outside the context of deformation theory itself, see how it can be rephrased deformation-theoretically, and then see how its solution uses the cotangent complex. Bonus points if the story is geometric enough to see why the role of the cotangent complex here is really a generalization of the role of differentials. - -REPLY [32 votes]: Here is an example mentioned in passing by user ali's answer, but I think it is cute (and powerful) enough to be worth fleshing out the details. -Lifting from characteristic $p$ to characteristic zero -In short, studying a geometric object (say, a scheme) $X$ in characteristic $p$ often involves lifting it to characteristic zero. For example, if $X$ is a smooth projective variety over $\mathbf{F}_p$, we may try to find a (flat) lift $\mathcal{X}$ over the $p$-adic numbers $\mathbf{Z}_p$. Now, $\mathbf{Z}_p$ embeds into $\mathbf{C}$ (in some completely noncanonical way), and we can apply powerful methods such as Hodge theory to the complex manifold underlying $\mathcal{X}_\mathbf{C}$. -Now, recall that -$$ \mathbf{Z}_p = \varprojlim_n \mathbf{Z}/p^{n+1}. $$ -Thus lifting $X_0=X$ over $\mathbf{Z}_p$ involves finding compatible liftings $X_n$ over $\mathbf{Z}/p^{n+1}$ for all $n$. The system $\mathfrak{X} = \{X_n\}$ (or its inductive limit in locally ringed spaces) is a "$p$-adic formal scheme," and the next step involves checking that it is algebraizable, i.e. that it comes from an actual scheme $\mathcal{X}/\mathbf{Z}_p$ by the obvious "formal completion" functor. -Now the first step, finding the successive liftings $\{X_n\}$, is completely controlled by deformation theory. In our situation, it says the following: - -If $X_0$ is a scheme over $\mathbf{F}_p$, and $X_n$ is a flat lifting of $X_0$ over $\mathbf{Z}/p^{n+1}$, there there exists an obstruction class -$$ {\rm obs}(X_n, \mathbf{Z}/p^{n+2}) \in {\rm Ext}^2(\mathbf{L}_{X_0/\mathbf{F}_p}, \mathcal{O}_{X_0}) = {\rm Hom}_{D(X_0)}(\mathbf{L}_{X_0/\mathbf{F}_p}, \mathcal{O}_{X_0}[2]), $$ -which vanishes if and only if there exists a flat lifting $X_{n+1}$ of $X_n$ over $\mathbf{Z}/p^{n+2}$. It is functorial in the sense that for $f_n\colon X_n\to Y_n$ lifting $f_0\colon X_0\to Y_0$ we have a commutative square -$$\require{AMScd} \begin{CD} -\mathbf{L}_{Y_0/\mathbf{F}_p} @>>> \mathcal{O}_{Y_0}[2]\\ @VVV @VVV\\ -Rf_{0, *}\mathbf{L}_{X_0/\mathbf{F}_p} @>>> Rf_{0, *}\mathcal{O}_{X_0}[2] -\end{CD}$$ - -In case the obstruction class vanishes, the set of isomorphism classes of such liftings $X_{n+1}$ is in a natural way a torsor under -$$ {\rm Ext}^1(\mathbf{L}_{X_0/\mathbf{F}_p}, \mathcal{O}_{X_0}) = {\rm Hom}_{D(X_0)}(\mathbf{L}_{X_0/\mathbf{F}_p}, \mathcal{O}_{X_0}[1]). $$ - -The group of automorphisms of any lifting $X_{n+1}$ restricting to the identity on $X_n$ is naturally isomorphic to -$$ {\rm Hom}(\mathbf{L}_{X_0/\mathbf{F}_p}, \mathcal{O}_{X_0}). $$ - -There is a similar story for lifting morphisms $f_0\colon X_0\to Y_0$. - - -So if you can show that ${\rm Ext}^2(\mathbf{L}_{X_0/\mathbf{F}_p}, \mathcal{O}_{X_0})$ vanishes, then you know that $X_0$ admits a formal $p$-adic lifting $\mathfrak{X}$. For example, if $X_0$ is a K3 surface, then this group can be identified with the space of global vector fields on $X_0$, and its vanishing is a difficult theorem due to Rudakov and Shafarevich. (And the fact that there is an algebraizable formal lifting, i.e. that an ample line bundle can be lifted to all $X_n$'s for a good choice of $\mathfrak{X}$, was shown later by Deligne.) -Perfect schemes and Witt vectors -Recall that for every perfect field $k$ of characteristic $p>0$ there exists a unique complete discrete valuation ring $W(k)$ (its ring of Witt vectors) with residue field $k$ whose maximal ideal is generated by $p$. It is a functor of $k$, and we have $W(k) \simeq k^{\mathbf{N}}$ as functors into sets. The addition and multiplication laws on $k^{\mathbf{N}}$ obtained this way are given by complicated universal formulas, e.g. -$$ (x_0, x_1, \ldots) + (y_0, y_1, \ldots) = (x_0 + y_0, x_1 + y_1 - \sum_{0 -TITLE: Expected height of a poset? -QUESTION [23 upvotes]: I am interested in any known results/empirical studies done on the average height of a poset with $N$ elements. Obviously this would depend on how that poset relation was randomly defined, however, at this point, I'll take any reasonable result on the topic regardless of how the poset was formed. -In my case $N$ is a very large number (billions of elements), so I am interested whether the height $h \ll N$ asymptotically. - -REPLY [34 votes]: In Asymptotic Enumeration of Partial Orders on a Finite Set (1975), Kleitman and Rothschild showed that almost all partial orders on an $n$-element set have a simple description: they have three "layers" $L_1$, $L_2$, and $L_3$ of incomparable elements, of size $n/4$, $n/2$, and $n/4$ respectively. Each element of $L_1$ is covered by about half the elements of $L_2$. Likewise for $L_2$ and $L_3$, and in the reverse direction. -So almost all finite posets have height $3$. -(If you don't have the time/effort to read Kleitman and Rothschild's paper, I came across this reference via G. Brightwell, Linear extensions of random orders, Discrete Math. 125 (1994) pp. 87���96. If memory serves, Brightwell's paper presents a good summary of this one.)<|endoftext|> -TITLE: Taking points uniformly inside a general finite geometric domain -QUESTION [8 upvotes]: It is well known that if we want to take $n$ uniformly and randomly points inside a circle of radius $r$ and centered at the origin the following apparently correct approach for generating $x$ and $y$ -$$ x= U \cos(\theta), \;y= U \sin(\theta)$$ where $U$ is a uniform variate in $(0,1)$ and $\theta$ is uniform variate in $(0, 2 \pi)$, does not work. Rather the correct way to generate $x$ and $y$ coordinates is to use: -$$ x= \sqrt{U}\cos(\theta),\; y= \sqrt{U} \sin(\theta) $$ with $U$ and $\theta $ as mentioned above. I wonder is there a some way of generating a of $n$ points inside the bounded region defined by -$$\mathbf{D}=\{(x,y):f(x,y)=0\},$$ -given that $ \mathbf{D}$ forms a closed region with smooth boundary. -For instance let's say -$$\mathbf{D}=\{ (x,y):\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\},$$ -how can we create $n$ points that are uniformly distributed inside the ellipse. Thanks for any hints/responses in advance! - -REPLY [6 votes]: The approach (already alluded to in one of the answers) of sampling uniformly from a larger set and then throwing out the samples that you don't want is known as rejection sampling. You'll find all sorts of useful ideas in the linked Wikipedia article and references thereof.<|endoftext|> -TITLE: Set-theoretic geology III: inside the core -QUESTION [6 upvotes]: Thanks to Jonas, Asaf, and Gabe I understand a little more of grounds and the mantle (or mantles, because it looks like there may be more than one). -But, set-theoretic geology, or so it seems to me, should not be exclusively about grounds: the Earth has a core, and in fact perhaps several strata of cores, till it reaches the mantle (or some intermediate zone). - -So, how about doing the opposite? Rather than penetrating a model $M$ by erosion, what if we know that there is a solid center ( say $L^M$, assuming that the model is not $L[G]$ for some generic set), and grow that minimal core as far as we can go? -In other words, let us define a class of M-non grounds, -$$\mathrm{CORE}= \{ N\ \nsubseteqq M, N \vDash ZFC \land \nexists G M=N[G] \}$$ -and determine its structure. In some cases this class is empty, but suppose it is not. -Question: - -what can be said of the non grounds of $M$? Is this class a upward directed partial order? When do this class reach the mantle (in the sense that I cannot find anything below the mantle that is not in CORE)? - -Are there any layers for some models which lie between the outer core and the intersection of all mantles? -Sounds like Jules Verne's Voyage au Centre de la Terre $\dots$ - -REPLY [6 votes]: Here are a few observations about CORE. -Claim: It is consistent that CORE is not pairwise upwards directed. -Proof: -Let $\mathbb{P}_0$ be the class forcing for the Easton product of the Cohen forcing $\mathrm{Add}(\alpha^+, 1)$, over all cardinals $\alpha$, in $L$. Let $G$ be $L$-generic for $\mathbb{P}_0$. Then, we can pick a partition $A \cup B$ of the class of cardinals, with both $A, B$ being proper classes. Then $N_0 = L[G \restriction A], N_1 = L[G \restriction B]$ are both in CORE, and have no common upper bound in CORE. -On the other hand: -Claim: It is consistent that CORE is upwards directed: -Proof: Let $\mathbb{P}_1$ be the backwards Easton support iteration of $\mathrm{Add}(\alpha^+, 1)$ for $\alpha$ successor of a singular cardinal in $L$. Let $G$ be an $L$-generic and let $M=L[G]$. -Sub Claim: For every $N \in CORE^M$, there is an ordinal $\alpha$ such that $N \subseteq L[G \restriction \alpha]$. In particular, CORE is upwards directed. -Sketch of Proof: Let $\alpha$ be minimal (successor of singular) such that $L[G\restriction \alpha] \not\subseteq N$ and let $x\in N$ be a set of ordinals of minimal rank such that $x \notin L[G \restriction \alpha]$. Then $x$ has to be a fresh set over $L[G \restriction \alpha]$, and by minimality if has to be a subset of $\alpha^+$. By gap-type arguments, $x$ together with its name (which is in $L$) codes the set $G \restriction \alpha$, and thus $L[G \restriction \alpha] \subseteq N$. -Finally, large cardinals seems to have negative affect on the directedness of CORE: -Claim: Let $\kappa \in M$ measurable and $2^\kappa = \kappa^{+}$. Then, there are $N_0, N_1 \in CORE^M$, and $x \in N_1$, $y \in N_0$, such that $N_0[x] = N_1[y] = M$. -Proof: Let $\mathcal{U}$ be a normal ultrafilter on $\kappa$ and let $N$ the ultrapower by $\mathcal{U}$. Let us construct inside $M$ two $N$-generic filters $G_0, G_1 \subseteq \mathrm{Add}(\kappa^{+}, 1)$, such that their bitwise xor codes $\mathcal{U}$. This is possible, since $2^\kappa = \kappa^{+}$ (both in $M$ and in $N$). Let $N_0 = N[G_0], N_1 = N[G_1]$ and note that $N_0[G_1] = N_1[G_0] \supseteq N[\mathcal{U}] = M$. QED -On the other hand, if $A$ is a set of ordinals and $A^{\#}$ exists, then $L[A] \in CORE$, since no set forcing in $L[A]$ can introduce a sharp for $A$. -So under the large cardinal axiom "every set has a sharp" (which follows from the existence of class of measurable cardinals, for example), $\bigcup CORE = V$, so CORE can certainly contain sets which are not in the mantle.<|endoftext|> -TITLE: Separating spheres in $3$-manifolds of positive scalar curvature and mean convex boundary -QUESTION [8 upvotes]: Recently, A. Carlotto and C. Li proved a complete topological classification of those compact, connected and orientable $3$-manifolds with boundary which support Riemannian metrics of positive scalar curvature and mean-convex boundary. Namely, if $M^3$ is such a manifold, then there exist integers $A, B, C, D \geq 0$ such that $M$ is diffeomorphic to a connected sum of the form -\begin{align*} -P_{\gamma_1} \# \cdots \# P_{\gamma_A} \# \mathbb{S}^3/ {\Gamma_1} \# \cdots \# \mathbb{S}^3 / {\Gamma_B} \# \left( \#_{i=1}^C \mathbb{S}^2 \times \mathbb{S}^1 \right) \setminus \left( \sqcup_{i=1}^D B_i^3 \right), -\end{align*} -where $P_{\gamma_i}$, $i \leq A$, are genus $\gamma_i$ handebodies; $\Gamma_i$, $i \leq B$, are finite subgroups of $SO(4)$ acting freely on $\mathbb{S}^3$, $B_i^3$, $i \leq D$, are disjoint $3$-balls in the interior. -My question: Can we classify, in terms of $A,B, C, D$, the $3$-manifolds $M$ of the form above in which any smoothly embedded $2$-sphere in the interior separates $M$? -For instance, if $(A,B,C,D) = (1, 0, 0, 0)$, then this holds. Indeed, if $M = P_{\gamma_1}$, then $H_2(M) = 0$, so the connecting homomorphism $H_2(M, \partial M;\mathbb{Z}) \to H_1(\partial M; \mathbb{Z})$ is injective. Since an embedded $2$-sphere has no boundary, it lies in the kernel of this map, and thus equals to $0$ in $H_2(M, \partial M; \mathbb{Z})$. This means it separates $M$. - -REPLY [4 votes]: Every embedded 2-sphere will separate $M$ if and only if $C=0$. -Proof: Suppose $C=0$, and let $\Sigma$ be 2-sphere embedded in $M$. Let $\{S_j\}$ be a collection of 2-spheres which decompose $M$ into prime summands. Look at the intersection of $\Sigma$ with $\{S_j\}$. Let $\Delta$ be an innermost disk on some $S_k$. We can surger $\Sigma$ along $\Delta$, which will decompose $\Sigma$ into two 2-spheres. Repeating this process until there are no intersections with $\{S_j\}$, $\Sigma$ is decomposed into a collection of embedded $2$-spheres which we call $\Sigma'$. Then $\Sigma$ will be non-separating in $M$ if and only if some component of $\Sigma'$ is non-separating in $M$. -Each component of $\Sigma'$ is contained within a single prime summand of $M$. We then cut $M$ along $\{S_j\}$, and glue $3$-balls onto each $2$-sphere boundary component of the resulting $3$-manifolds. It is well-known that handlebodies and closed spherical $3$-manifolds are irreducible, which means that every embedded $2$-sphere bounds a $3$-ball. Thus each component of $\Sigma'$ is separating in its respective prime summand and hence in $M$. Removing the $D$ $3$-balls from $M$ which are disjoint from $\Sigma$ and $\Sigma'$ does not affect whether $\Sigma$ is separating. Therefore $\Sigma$ is separating in $M$. -Conversely, if $C\neq 0$, then we can find a non-separating 2-sphere $\Sigma''$ in some $S^2\times S^1$ component which is disjoint from each $B_i$ and each $S_j$. Furthermore the dual $1$-sphere to $\Sigma''$ is disjoint from each $B_i$ and each $S_j$. Therefore $\Sigma''$ is non-separating in $M$.<|endoftext|> -TITLE: A rather non-$F_\sigma$ Borel set -QUESTION [6 upvotes]: I asked this question at MSE a week ago, but received no answer, so I cross-post it here. -I obtained a negative answer to this MSE question provided each metric space $X$ such that $|X|=\frak c$ and density $d(X)<\frak c$, contains a Borel set $B$ such that $|B\setminus C|=\frak c$ for each $F_\sigma$-subset $C$ of $X$ with $C\subset B$. My question is whether the latter claim holds. I guess this is known (and true), but it is hard to find a reference. Thanks. -My try. I guess using Theorem 22.4 from [Kech] I can show the claim when $X$ is Polish. To prove the claim for a separable $X$, by Proposition 12.1 from [Kech], it suffices to prove it for subspaces of an arbitrary fixed Polish space. -References -[Kech] A. Kechris, Classical Descriptive Set Theory, Springer, 1995. - -REPLY [2 votes]: Here's an argument that the statement is false if the Continuum Hypothesis fails and the covering number for the null ideal is the same as the continuum. Wellorder the Borel sets of reals as $\langle B_{\alpha} : \alpha < \mathfrak{c} \rangle$. Choose for each $\alpha < \mathfrak{c}$ an $F_{\sigma}$ set $C_{\alpha} \subseteq B_{\alpha}$ such that $B_{\alpha} \setminus C_{\alpha}$ is null and a real $x_{\alpha}$ not in $B_{\beta} \setminus C_{\beta}$ for any $\beta < \alpha$. Let $X = \{ x_{\alpha} : \alpha < \mathfrak{c}\}$. Then any Borel subset of $X$ is $B_{\alpha} \cap X$ for some $\alpha$. Furthermore, $C_{\alpha} \cap X$ is an $F_{\sigma}$ subset of $B_{\alpha} \cap X$, and $(B_{\alpha} \cap X) \setminus (C_{\alpha} \cap X)$ is contained in $\{ x_{\beta} : \beta \leq \alpha\}$ which has cardinality less than $\mathfrak{c}$. -As for the consistency of the statement that CH is false and $\mathrm{cov}(\mathcal{N}) = \mathfrak{c}$, this is a standard consequence of MA + not-CH (see Theorem 26.39 of the 2003 edition of Jech's Set Theory). The consistency of MA and not-CH is Theorem 16.13 of Jech. -I have to revise my earlier claim that the statement in question follows from CH. This appears to be true for spaces $X$ having a Borel subset which is not $F_{\sigma}$, by the idea in my original comment : if $B$ has an $F_{\sigma}$ subset $C$ such that $B \setminus C$ is countable, then $B$ is $F_{\sigma}$. On the other hand, one can run the proof in the first paragraph of this answer under CH to produce a set of reals of cardinality $\mathfrak{c}$ such that every Borel set is $F_{\sigma}$. So the statement would fail for such a space.<|endoftext|> -TITLE: Reference request: the theory of currents -QUESTION [14 upvotes]: I am a graduate student and want to study the theory of currents. What is a good reference for a beginner? -I should be familiar with the theory of distributions or generalized functions on $\mathbb R^n$. - -REPLY [18 votes]: The theory of currents is a part of the geometric measure theory. Unfortunately, Federer made the subject completely inaccessible after he wrote his famous monograph: -H. Federer, Geometric measure theory. Die Grundlehren der mathematischen Wissenschaften, Band 153 Springer-Verlag New York Inc., New York 1969. -The problem is that the book contains `everything' (well, almost) and it is unreadable. After this book was published, people did not dare to write other books on the topic and only the bravest hearts dared to read Federer's Bible. -In my opinion the first accessible book on the subject is -L. Simon, Lectures on geometric measure theory. Proceedings of the Centre for Mathematical Analysis, Australian National University, 3. Australian National University, Centre for Mathematical Analysis, Canberra, 1983. -You can find it as a pdf file in the internet. Note that this book was written 14 years after Federer's book and there was nothing in between. -I would also suggest: -F. Lin, X. Yang, Geometric measure theory—an introduction. Advanced Mathematics (Beijing/Boston), 1. Science Press Beijing, Beijing; International Press, Boston, MA, 2002. -I haven't read it, but it looks relatively elementary (relatively, because by no means the subject is elementary). -The last, but not least is -F. Morgan, Geometric measure theory. A beginner's guide. Fifth edition. Illustrated by James F. Bredt. Elsevier/Academic Press, Amsterdam, 2016. -You will not learn anything form that book as it does not have detailed proofs, but you can read it rather quickly and after that you will have an idea about what it is all about. - -REPLY [15 votes]: A beginner friendly introduction can be found in chapter 7 of the book "Geometric Integration Theory " by Krantz and Parks. It is from 2008 and written in a modern and clear style and it starts nearly from "zero".<|endoftext|> -TITLE: Tools from other disciplines useful to mathematics research? -QUESTION [14 upvotes]: Obviously, mathematics provides essential tools for physicists, biologists, economists, engineers and many others to use in their research. Equally obviously, physics, biology, economy and engineering provide the inspiration for research topics and directions in mathematics. But this isn't what this question is about. Instead, I'm wondering about tools provided to mathematicians by other disciplines. -Are there any examples of tools (excluding software) from another discipline that you personally have used in your mathematical research? -E.g., I find dimensional analysis (as in assigning dimensionfull units to variables) a very useful tool from physics, which can greatly help in finding errors during algebraic manipulation of equations. - -REPLY [2 votes]: It's not clear how useful this would actually be, but the idea of using "citizen science" (a somewhat sociology-based notion which has been developed into a useful tool for — primarily observation-based ­— scientific research by astronomers, botanists and zoologists, among others) for mathematics research has been floated here on MO: Can pure mathematics harness citizen science?. It would be nice to know whether any of the projects proposed in answers to that question ever came to fruition.<|endoftext|> -TITLE: Enumerating subspaces of $\mathbb{F}_q^n$ in terms of words and inversions -QUESTION [6 upvotes]: When $q$ is a prime power, then on the one hand the $q$-binomial coefficient $\binom{n}{k}_q$ equals the number of $k$-dimensional subspaces of $\mathbb{F}_q^n$, and on the other hand it is the generating function of the sequence which sends $r$ to the number of words in two letters $X,Y$ of length $n$ with $k$ occurances of $X$ and with $r$ inversions (i.e. places where $Y$ comes before $X$). Therefore, there must a bijection -$$\{k\text{-dimensional subspaces of } \mathbb{F}_q^n\} \\ \downarrow{\small\cong}\\ \coprod_{r=0}^{k(n-k)} q^r \cdot \{\text{ words in X,Y of length } n \text{ with } k \text{ X's and } r \text{ inversions}\}$$ -My question is if you can write down a nice, explicit bijection. I require that its description is independent of the theory of $q$-binomial coefficients, thus doesn't use the calculation of the cardinalities, and that it is also without recursion (since you can easily transform the inductive proof of the equality of cardinalities into a recursive bijection; this doesn't count here). -For $k=1$ the bijection looks as follows: It sends $\langle a_1,\dotsc,a_n \rangle$ to $(a_1/a_{r+1},\dotsc,a_r/a_{r+1}) \cdot Y^r X Y^{n-r-1}$, where $r$ is maximal with $a_{r+1} \neq 0$. - -REPLY [6 votes]: You may act similarly as follows. -Let $V$ be a $k$-dimensional subspace of $\mathbb F_q^n$. Take any its base, put its elements into the rows of some matrix, and make it to the reduced row echelon form $B$ (which is unique). The rows of $B$ still form a base of $V$. -Let $r_1$, $r_2$, $\dots$, $r_k$ be the indices of leading elements in rows $1,2,\dots,k$. Then you put into correspondence to $V$ the tuple of all elements of $B$ which are non-fixed (i.e., those not in the leading columns, and not to the left of leading elements), along with the word $Y^{r_1-1}XY^{r_2-r_1-1}XY^{r_3-r_2-1}X\dots XY^{n-r_k}$ read backwards. The freedom of the coefficients provides the exponent of $q$ being -$$ - (n-r_1-(k-1))+(n-r_2-(k-2))+\dots+(n-r_k), -$$ -and the number of inversions (in the reversed word) is clearly -$$ - (n-r_k)+(n-r_{k-1}-1)+(n-r_{k-2}-2)+\dots+(n-r_1-(k-1)), -$$ -so they trivially coincide.<|endoftext|> -TITLE: 4-dimensional cohomology $\mathbb{CP}^2$'s -QUESTION [10 upvotes]: Let $M$ be a closed, smooth $4$-manifold with integral cohomology ring isomorphic to that of $\mathbb{CP}^2$, is it diffeomorphic to it? - -REPLY [5 votes]: This was originally going to be a comment to Marco Gallo's answer, but got too long. I figured since this provides an alternative solution to the problem, I'll post it as an answer. -An alternative answer to this question are the fake projective planes, these are algebraic surfaces which have the same cohomology as $\mathbb{CP}^2$, but nontrivial fundamental groups. They have universal cover the complex unit ball which topologically is just $\mathbb{R}^4$. -I offer this additional answer since they provide examples very different to Marco Galla's answer and answer a question of Anubhav Mukherjee's in the comment. Either they are irreducible or have an irreducible summand that is a cohomology $\mathbb{CP}^2$. -Suppose $X$ is one of the fake projective planes. If $X$ is irreducible, we're done. If $X$ is reducible, then it admits a connect sum decomposition $X = Y \# \Sigma$ where $Y$ is a cohomology $\mathbb{CP}^2$ and $\Sigma$ is a homology $S^4$ that is not a homotopy $S^4$. Again if $Y$ is irreducible and not homeomorphic to $\mathbb{CP}^2$, we're done. If $Y$ is reducible, repeat this process again until you get either an irreducible cohomology $\mathbb{CP}^2$. This process must terminate by compactness. -This gives a connect sum decomposition which we'll write as $X = Y \# \Sigma$ (slightly abusing notation here) where $Y$ is an irreducible cohomology $\mathbb{CP}^2$ and $\Sigma$ is a homology $S^4$ which is the connect sum of all of the homology $S^4$'s we found. If $Y$ is not $\mathbb{CP}^2$, we're done. -Now we see that $Y$ can not be $\mathbb{CP}^2$. Suppose $Y = \mathbb{CP}^2$ and consider the universal cover $f: \tilde{\Sigma} \rightarrow \Sigma$. Take a point $p \in \Sigma$ and it's inverse image $f^{-1}(p)$. We can now construct a cover $f^*: Y \# |f^{-1}(p)|\mathbb{CP}^2 \rightarrow \Sigma \# \mathbb{CP}^2 = X$ by connect summing a $\mathbb{CP}^2$ at each point in $f^{-1}(p)$. This covering space is simply connected since it's summands are simply connected and so it is the universal cover. This universal covering space has 2nd homology coming from the $\mathbb{CP}^2$ summands. This contradicts $X$ having universal cover $\mathbb{R}^4$. Thus $Y$ is not $\mathbb{CP}^2$ and is an irreducible cohomology $\mathbb{CP}^2$.<|endoftext|> -TITLE: Examples of "natural" finitely generated groups with an undecidable conjugacy problem -QUESTION [6 upvotes]: I am looking for natural groups with undecidable conjugacy problem. By natural, I mean that the word problem should be decidable, and the group should be given by some natural action. I know that $\mathbb{Z}^d \rtimes F_m$ (with a suitable action of $F_m$) has undecidable conjugacy problem. That's very nice, but I'd like to know other examples. I do not care about finite presentation, and I'm also fine with the group being a f.g. subgroup of something natural and geometric, which maybe simplifies things. A concrete case I was not able to resolve is whether all f.g. subgroups of right-angled Artin groups have decidable conjugacy problem. -Šunić, Zoran; Ventura, Enric, The conjugacy problem in automaton groups is not solvable., J. Algebra 364, 148-154 (2012). ZBL1261.20034. - -REPLY [4 votes]: Chuck Miller in [Miller, Charles F., III On group-theoretic decision problems and their classification. Annals of Mathematics Studies, No. 68. Princeton University Press, Princeton, N.J.; University of Tokyo Press, Tokyo, 1971] proves the following two rather nice and natural examples. -Theorem III.10. The free product of two free groups with finitely generated amalgamation can have unsolvable conjugacy problem. Further, the finitely presented HNN extension of a free group can have unsolvable conjugacy problem. -(Note that Miller calls HNN extensions 'Strong Britton extensions'). -Now by Bass-Serre theory, there is a natural action of an amalgamated free product/HNN on the associated Bass-Serre tree, which should satisfy your "natural action" criterion. - -Edit: The result mentioned by YCor can also be found in Miller's book. -Theorem III.23 The group $F_2 \times F_2$ has a finitely generated subgroup with undecidable conjugacy problem. -An important side remark, however, is that $F_2 \times F_2$ itself has decidable conjugacy problem, as do all RAAGs, in linear time. See [Crisp, John; Godelle, Eddy; Wiest, Bert; The conjugacy problem in subgroups of right-angled Artin groups. J. Topol. 2 (2009), no. 3, 442–460.].<|endoftext|> -TITLE: Functions which are periodic along every geodesic -QUESTION [5 upvotes]: In an effort to understand some geometric rigidity theorems, I am curious about the following: let $(M,g)$ be a complete Riemannian manifold and suppose there is a nonconstant real-valued function on $M$ which has a period of 1 when restricted to any unit-speed geodesic of $M$. Does this place restrictions on $(M,g)$? -My feeling is that the vast majority of manifolds cannot support such a function. The round sphere clearly can. -edit: if I understand correctly- according to the article arxiv.org/abs/1511.07852 of Radeschi and Wilking (Invent. Math. 2017) given by Igor Belegradek in the comments, a theorem of Wadsley (J. Diff. Geom. 1975) shows that, for every Riemannian manifold all of whose geodesics are closed, every function on $M$ satisfies the above condition with some constant period; the theorem is that all lengths of closed geodesics must be multiples of a single number, when $\pi_1(M)$ is finite - -REPLY [7 votes]: If $f\circ \gamma$ has period $1$ for all unit speed geodesics $\gamma:\mathbb{R} \to M$ then in particular $f(x) = f(y)$ whenever $x$ and $y$ are the endpoints of a geodesic segment of length $1$. -This implies that $f(x) = f(y)$ whenever there is a sequence $x_0 = x,x_1,\ldots,x_n = y$ such that $x_i$ and $x_{i+1}$ are the endpoints of a segment of length $1$ for $i = 0,\ldots,n-1$. Following Sunada call such a sequence a $1$-geodesic chain. -If any two points can be joined by a $1$-geodesic chain then $f$ would have to be constant. In a couple of papers of Sunada (e.g. Theorem C from "Mean value theorems and ergodicity of certain random walks" Compositio Mathematica 1983) conditions under which any two points can be joined by a $1$-geodesic chain are given. -His results imply that, in order for a non-constant $f$ to exist, there must be a point $o \in M$ such that for any unit speed geodesic with $\gamma(0) = x$ one has that $\gamma(n)$ is conjugate to $x$ along $\gamma$ for all integer $n$. -This implies in particular that $M$ is compact and its fundamental group is finite (since the condition above passes to the universal covering space as well). -For surfaces this shows (Theorem G in the aforementioned paper by Sunada) that $M$ would have to be either the sphere or the projective plane with a metric such that all geodesics from $o$ have period $2$.<|endoftext|> -TITLE: Formula expressing symmetric polynomials of eigenvalues as sum of determinants -QUESTION [20 upvotes]: The trace of a matrix is the sum of the eigenvalues and the determinant is the product of the eigenvalues. The fundamental theorem of symmetric polynomials says that we can write any symmetric polynomial of the roots of a polynomial as a polynomial of its coefficients. We can apply this to the characteristic polynomial of a matrix $A$ to write any symmetric polynomial of eigenvalues as a polynomial in the entries of $A$. -I stumbled upon an explicit formula for this. Let $A$ be an $n \times n$ matrix and $a_1, \dots, a_n$ be its eigenvalues. Then we have the following identity, provided the left hand side is a symmetric polynomial: -$$ -\sum_{i \in \mathbb{N}^n} p_i a_1^{i_1} \cdots a_n^{i_n} = \sum_{i \in \mathbb{N}^n} p_i \det(A_1^{i_1}, \dots, A_n^{i_n}) -$$ -The determinant $\det(A_1^{i_1}, \dots, A_n^{i_n})$ on the right hand side is the determinant of a matrix with those column vectors, where $A_i^k$ is the $i$-th column of the $k$-th power of $A$. The left hand side is a symmetric polynomial of the eigenvalues of $A$, and the right hand side is a polynomial of the entries of $A$. -Example: if $A$ is a $2\times 2$ matrix, then $$a_1 a_2^2 + a_1^2 a_2 = \det(A_1, A_2^2) + \det(A_1^2, A_2)$$ -Proof. Let $p(A) \in End(\bigwedge^n V^*)$ be given by $p(A)f(v_1,\dots,v_n) = \sum_{i\in \mathbb{N}^n}f(A^{i_1}v_1,\dots,A^{i_n}v_n)$. We have $End(\bigwedge^n V^*) \simeq \mathbb{R}$ and $p(A)$ is the right hand side of the identity under this isomorphism. Since $p(A)$ was defined basis independently, the right hand side is basis independent, and we get the left hand side in the eigenbasis. $\Box$ -Link to detailed proof and slight generalization to an identity on several commuting matrices. E.g. for commuting $2\times 2$ matrices $A,B$: -$$a_1 b_1 a_2^2 + a_1^2 a_2 b_2 = \det(AB_1, A_2^2) + \det(A_1^2, AB_2)$$ -This identity looks like it should be a few hundred years old, especially since the proof is quite simple, but I have not seen this in linear algebra courses. Is this a well known identity? Where should I look to learn more about these types of identities? Or, maybe I am mistaken and the identity is false? (though I have also empirically tested it with a computer program) I apologize if this question is too basic for mathoverflow; I am only doing pure mathematics for fun. I initially asked elsewhere but was advised to ask here. Thanks! - -REPLY [8 votes]: Concerning the reference request: -Several text books [1,2] give the theorem and proof for elementary symmetric polynomials $s_k=$ sum of all $k\times k$ principal minors of the $n\times n$ matrix. This also covers the trace ($s_1$) and the determinant ($s_n$). - -Matrix -Analysis and Applied Linear Algebra by Carl D. Meyer (Equation 7.1.6 on page 494, screenshot) -Matrix -Analysis by Roger A. Horn and Charles R. Johnson (Theorem -1.2.12 on page 42, screenshot). - - -Update: I have searched quite extensively for sources that give the generalized formula for complete homogeneous symmetric polynomials, but without success. The derivation could be analogous to the published derivation for the elementary symmetric polynomials, expanding ${\rm Det}\,(A+xI)^{-1}$ instead of ${\rm Det}\,(A+xI)$, but I have not seen it published.<|endoftext|> -TITLE: An example that the sum of two Borel sets which is not a Borel set in n-dimensional Euclidean space -QUESTION [5 upvotes]: By sum of two sets I mean $A+B := \{x+y:x \in A \quad y \in B\}$, and there is a tip in a book of real analysis by Zhou Minqiang which says: -“If $A,B$ are Borel sets in $\mathbb{R}^{n}$, $A+B$ may not be a Borel set.” -I want to know some specific examples.(Maybe $\mathbb{R}^{1}$ ?) -Any comments will be helpful. - -REPLY [8 votes]: This is a result of Erdos and Stone: https://www.ams.org/journals/proc/1970-025-02/S0002-9939-1970-0260958-1/S0002-9939-1970-0260958-1.pdf<|endoftext|> -TITLE: Is this lattice in the Tate module of an elliptic curve, coming from complex-analytic uniformization, stable under Frobenius? -QUESTION [7 upvotes]: Let $E$ be an elliptic curve over $\mathbb{Q}$, and let $p$ and $\ell$ be two distinct primes of good reduction. Let $T_\ell = T_\ell(E) = \varprojlim E[\ell^n](\overline{\mathbb{Q}})$ be the $\ell$-adic Tate module, and let $F_p \in \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ be a Frobenius element at $p$. Then $F_p$ acts $\mathbb{Z}_\ell$-linearly on $T_\ell$, and this action depends only up to conjugation on the choice of $F_p$. In particular, its characteristic polynomial is well-defined. A basic result is that the coefficients of this characteristic polynomial are integers. -This last fact is usually proved by considering the reduction of $E$ modulo $p$, which does not change the $\ell$-adic Tate module, and using that we can realize the $F_p$-action in characteristic $p$ as coming from an actual morphism of elliptic curves, namely the Frobenius morphism $E \to E^{[p]}$. But I was wondering if it is possible to give a more direct proof, namely by constructing a $\mathbb{Z}$-lattice $\Lambda \subset T_\ell$ (by which I mean a rank 2 free $\mathbb{Z}$-module such that the map $\Lambda \otimes \mathbb{Z}_\ell \to T_\ell$ is an isomorphism) which is preserved by $F_p$ in the sense that $F_p(\Lambda) \subset \Lambda$ (note that one cannot expect equality here since the determinant of $F_p$ acting on $T_\ell$ is $p$). Certainly, if you already know that $F_p$ has integral characteristic polynomial, then you can easily construct such lattices: take any $t \in T_\ell \setminus \ell T_\ell$ that is not an eigenvector for $F_p$, then $\Lambda = t \mathbb{Z} + F_p(t)\mathbb{Z} \subset T_\ell$ is an $F_p$-invariant lattice. So there should be plenty such lattices. But the goal is to construct an $F_p$-invariant lattice without using that we already know that $F_p$ has integral characteristic polynomial. -One potential lattice can be constructed as follows. We choose a complex-analytic uniformization $E(\mathbb{C}) = \mathbb{C}/\Lambda_0$ for some lattice $\Lambda_0 \subset \mathbb{C}$. Then we define a map $\Lambda_0 \to T_\ell$ by sending $\lambda \in \Lambda_0$ to the sequence $(\ell^{-1} \lambda, \ell^{-2} \lambda, \ell^{-3}\lambda, \ldots) \in T_\ell$, which is well-defined because $\ell^{-n}\lambda \in E(\mathbb{C})[\ell^n] = E(\overline{\mathbb{Q}})[\ell^n]$. Let $\Lambda_\ell \subset T_\ell$ be the image of this map. It is not hard to prove that $\Lambda_\ell$ is free of rank 2 and that $\Lambda_\ell \otimes \mathbb{Z}_\ell \to T_\ell$ is an isomorphism. Also note that $\Lambda_\ell$ does not depend on the choice of the uniformization. -Question: Does $F_p(\Lambda_\ell) \subset \Lambda_\ell$ hold? -P.S. I've tried searching for results in this direction in various places, but did not find much. If someone has suggestions for references or keywords to search for, I would be much obliged. - -REPLY [13 votes]: There is a subtle problem with this idea, that causes serious problems. You observed that $\Lambda_\ell \otimes \mathbb Z_\ell = T_\ell$ but didn't find any other information for it. There is a reason for that. -Let $K$ be the field generated by the coordinates of the $\ell$-power torsion points of $E$. Given an $\ell$-power torsion point defined over $F$, to make Frobenius act on it, we need to know its reduction mod $p$, so we need to embed $F$ into the maximal unramified extension $\mathbb Q_p^{ur}$ of $\mathbb Q_p$. -Given a point in the homology of $E_{\mathbb C}$, to find the corresponding point of $F$, we need to express the coordinates as complex numbers, so we need to embed $F$ into $\mathbb C$. -Are these embeddings canonical? Well, if we define $F$ as the field generated by the complex coordinates of $\ell$-power torsion points, then the second embedding is canonical but the first isn't. If we define $F$ as the field generated by the $p$-adic coefficients of torsion points, then the first embedding is canonical but the second isn't. So regardless, there is some ambiguity - we can translate one of our embeddings by an automorphism of $F$ and get one that looks equally reasonable. -How bad is that ambiguity? Fixing an automorphism $\sigma \in \operatorname{Gal}(F/\mathbb Q)$ of $F$, making this change of embeddings corresponds exactly to translating your lattice by the action of $\sigma$ on $T_\ell(E)$. So the set of lattices we obtain your construction is a $\operatorname{Gal}(F/\mathbb Q)$-orbit in $T_\ell(E)$. -For $E$ generic, we have $\operatorname{Gal}(F/\mathbb Q) \cong GL_2(\mathbb Z_\ell)$, so the orbit is quite large. In fact every single lattice $\Lambda$ with $\Lambda \otimes \mathbb Z_\ell = T_\ell$ lies in this orbit, because we can find a matrix in $GL_2$ taking the basis of one such lattice to another. So there is no more information available about these lattices than your initial observation that $\Lambda \otimes \mathbb Z_\ell = T_\ell$! -Of course, there are examples of such $\Lambda$ stable under $F$ and examples not stable under $F$. -For any $E$ non-CM, the situation is the same, because the Galois group is an open subgroup of $GL_2(\mathbb Z_\ell)$ and these act transitively on the set $GL_2(\mathbb Z_\ell)/GL_2(\mathbb Z)$ of lattices $\Lambda$, since $GL_2(\mathbb Z)$ is dense in $GL_2(\mathbb Z_\ell)$. -For $E$ CM, the situation is different, as the Galois group is much smaller. If $p$ is a supersingular prime, then David Speyer's argument shows $\Lambda_\ell$ is never stable under Frobenius. Conversely, if $p$ is an ordinary prime, then the endomorphism $V =p /F$ lifts to an endomorphism of the curve over the CM field and thus an endomorphism of the curve over $\mathbb Q$, thus always preserves $\Lambda_\ell$, and because its determinant is $p$, $F= p/V$ necessarily preserves $\Lambda_\ell$ as well. So for CM curves, Frobenius preserves this lattice if and only if $p$ is ordinary.<|endoftext|> -TITLE: 2-monads for categories with a class of (co)limits -QUESTION [7 upvotes]: This question concerns the strictness of (co)completions, at various levels of generality. -In Blackwell–Kelly–Power's Two-dimensional monad theory, the authors state - -For instance, the 2-category $\mathbf{Lex}$ of small finitely-complete categories, left-exact functors, and natural transformations is (see Subsection 6.4 below) $T\text{-}\mathbf{Alg}$ for a finitary 2-monad $T$ on Cat. - - -Where can be found an explicit proof that there is such a 2-monad, as opposed to a pseudomonad? This claim appears in several other papers, but I have only been able to find constructions of a pseudomonad for finite limit completion. -More generally, is there an explicit proof in the literature that, for a suitable class $\phi$ of (co)limits, there is a 2-monad on $\mathbf{Cat}$ for $\phi$-(co)completion, whose algebras and (pseudo)morphisms are $\phi$-(co)complete categories, $\phi$-(co)continuous functors, and natural transformations? -Similarly, is it known whether the small cocompletion of locally-small categories forms a 2-monad? -Most generally, are there any results that establish when a lax idempotent pseudomonad (i.e. KZ-doctrine) on a 2-category $\mathcal K$ may be replaced by a lax idempotent 2-monad on $\mathcal K$ (with isomorphic or equivalent 2-categories of algebras)? - -Power–Cattani–Winskel's A Representation Result for Free Cocompletions in particular seems promising, but the characterisation result there still assumes that such 2-monads exist in the first place. - -REPLY [7 votes]: Kelly and Lack's paper On the monadicity of categories with chosen colimits answers your questions (1),(2) and (3) affirmatively. The main theorems are Theorem 6.1, 6.2 and 7.1. Their main trick is Lemma 4.1, which allows them to modify a biadjunction (and so a pseudomonad) to a strict 2-adjunction (and so a 2-monad), assuming various hypothesis. I can imagine that something like this lemma might be helpful also in answering you question (4), but I am not aware of seeing any result of that nature. -Regarding your comment on presentations: if you have any presentation for a 2-monad (for instance, for categories with finite limits) and want to check that its pseudomorphisms are the expected ones (in this case, finite limit preserving functors), it is enough to know that those expected morphisms satisfy some natural properties such as doctrinal adjunction (so you don't have to work with the presentation at all). This is described in my paper Two-dimensional monadicity, with your example discussed in Section 7.2.<|endoftext|> -TITLE: Stacks for a string theory student -QUESTION [9 upvotes]: First, I'm a string theory student hoping to grasp some math involved in some physics developments. -I'm hoping to read the famous Kapustin-Witten Paper "Electric-magnetic duality and the geometric Langlands program" and the related "The Yang-Mills equations over Riemann surfaces". -The following statement serves to explain the origin of my trouble: Let $G$ be a simple complex Lie group and $C$ a Riemann surface. Geometric Langlands is a set of mathematical ideas relating the category of coherent sheaves over the moduli stack of flat $G^{L}$-bundles ($G^{L}$ is the dual Langlands group of $G$) over $C$ with the category of $\mathcal{D}$-modules on the moduli stack of holomorphic $G$-bundles over $C$. -My problem: I have working knowledge of representation theory, but I'm completely ignorant about the theory of mathematical stacks and the possible strategies to begin to learn it. What specifically worries me is how much previous knowledge of $2$-categories is needed to begin. -My background: I've read Hartshorne's book on algebraic geometry in great detail, specifically the chapters on varieties, schemes, sheaf cohomology and curves. My category theory and homological algebra knowledge is exactly that needed to read and solve the problems of the aforementioned book. I'm also familiar with the identification between the topological string $B$-model branes and sheaves at the level of Sharpe's lectures. -My questions: I'm asking for your kind help to find references to initiate me on the theory of stacks given my elementary background and orientation. I'm completely unfamiliar with the literature and the pedagogical routes to begin to learn about stacks. -Any suggestion will be extremely helpful to me. -Update: I have found the paper String Orbifolds and Quotient Stacks - very useful and explicit. - -REPLY [3 votes]: I still think that Vistoli's notes on descent, fibred categories, and stacks are the canonical non-infinity-categorical introduction: - -Angelo Vistoli, Notes on Grothendieck topologies, fibered categories and descent theory, arXiv:math/0412512.<|endoftext|> -TITLE: Numbers that don't start with (p-1) in base p for any p -QUESTION [13 upvotes]: Say that an integer $n$ is $p$-leading if its expansion in base $p$ starts with the digit $p-1$. My postdoc, Lifan Guan, asks: are there infinitely many positive integers $n$ that are not $p$-leading for any odd prime $p$? That is: is the set -$$S = \mathbb{N}\setminus \bigcup_{p\geq 3} \bigcup_{d\geq 1} [p^d - p^{d-1},p^d)$$ -infinite? -[My estimation is this is interesting, doable and non-trivial. I also suspect it must have been studied before!] - -REPLY [7 votes]: Following some discussion with Harald (offline)... It appears that the answer is YES, the set is indeed infinite. -First ingredient: -We take the idea of adding upper densities, but use logarithmic densities instead of natural densities (logarithmic density meaning one sums reciprocals to $x$, divides by $\log{x}$, and sends $x$ to infinity). A short computation bounds the upper logarithmic density of integers which start with $p-1$ in some base $p > C$ by $\sum_{p > C} \frac{1}{(p-1)\log{p}}$. Unfortunately, this sum is larger than $1$ when $C=2$, so this does not immediately imply a positive answer to the question. So we have to work harder. -2nd component: One shows that if $g_1,\dots,g_m \ge 2$ are integers with $1/\log{g_1}, \dots, 1/\log{g_m}$ being $\mathbb{Q}$-linearly independent, then the fractional parts $\{\log_{g_1} n\}, \dots, \{\log_{g_m} n\}$ are distributed like independent uniform random variables on $[0,1)$, with respect again to logarithmic density. This uses a Weyl-type criterion; the key point is that if $\gamma \ne 0$, then $e^{2\pi i \gamma \log{n}} = n^{2\pi i \gamma}$ has logarithmic mean value $0$. (The condition that $\gamma \ne 0$ explains the linear independence condition required on the numbers $1/\log g_i$.) -Now $1/\log{3}$ and $1/\log{5}$ are easily seen to be $\mathbb{Q}$-linearly independent, and so one gets from the second component that the integers which do not begin with $2$ in base $3$, and do not begin with $4$ in base $5$, comprise a set with logarithmic density -$$ (1-\log_3(3/2)) (1-\log_5(4/5)) \approx 0.543.$$ -Then one uses the first component with $C=5$. The sum of $\frac{1}{(p-1)\log{p}}$ for $p>5$ can be proved to be $< 0.53$ (say). It follows that the set in question is indeed infinite, with lower logarithmic density $> 1/100$. -If $\{1/\log{p}\}_{p\ge 3}$ is a linearly independent set (over $\mathbb{Q}$), modifying the above argument would show that the logarithmic density is exactly $\prod_{p \ge 3} (1-\log_p(p/(p-1))$. But I think this linear independence is open.<|endoftext|> -TITLE: Object classifiers in 1-toposes -QUESTION [9 upvotes]: In a Grothendieck $\infty$-topos, it is known that, for arbitrarily large regular cardinals $\kappa$, there is a classifier for the class of relatively $\kappa$-compact morphisms. It is also easy to show that this is not the case in 1-toposes, because we might have isomorphic, but not equal, such morphisms classified by the same map. However, we should be able to recover at least part of the definition of an object classifier. Namely, I need to know that in a Grothendieck 1-topos, for arbitrarily large regular cardinals $\kappa$, there is a map $t: U' \to U$ such that for every relatively $\kappa$-compact morphism $f: X \to Y$ there exists a pullback square -$\require{AMScd}$ -\begin{CD} -X @>>> U'\\ -@VfVV @VVtV\\ -Y @>>> U -\end{CD} -(not necessarily unique and such that the map $Y \to U$ doesn't necessarily only classify $f$). I feel that this should definitely be true, but I can't find it anywhere in the literature. It would be very nice to have a reference for it, or a confutation in the unfortunate case I'm wrong. Thanks! - -REPLY [2 votes]: If $\mathcal{X}$ is a topos, then there is an adjunction $\mathcal{X} \leftrightarrows \mathcal{P(C)}$ with a category of presheaves of sets, where the right adjoint is fully faithful and accessible and the left is left exact. -The idea is very similar to the one spelled out in https://www2.mathematik.tu-darmstadt.de/~streicher/NOTES/lift.pdf, except that where its authors first find a suitable class of relatively $\kappa$-compact morphisms in $\mathcal{P(C)}$ and then consider the class of all morphisms in $\mathcal{X}$ that become relatively $\kappa$-compact in $\mathcal{P(C)}$, my analysis allows to find some $\kappa$ such that the desired class is that of all relatively $\kappa$-compact morphisms in $\mathcal{X}$. A complete proof can be found in Appendix A of http://www.tac.mta.ca/tac/volumes/37/5/37-05.pdf -I'd like to point out that neither strategy is more general than the other. The reason of this is that a class of relatively $\kappa$-compact morphisms in $\mathcal{X}$ need not be sent precisely to a class of relatively $\lambda$-compct morphisms in $\mathcal{P(C)}$ and, conversely, the preimage of a class of relatively $\lambda$-compact morphisms in $\mathcal{P(C)}$ need not be precisely a class of relatively $\kappa$-compact morphisms in $\mathcal{X}$. -However, all relatively $\kappa$-compact morphisms found in the latter proof are indeed sent to relatively $\kappa$-compact morphisms in $\mathcal{P(C)}$.<|endoftext|> -TITLE: Can $1\ne H\cap H^g\lhd H$ happen if $G$ is a primitive permutation group with stabiliser $H$? -QUESTION [9 upvotes]: Assume everything is finite. -Let $G$ be a primitive permutation group with point stabiliser $G_\alpha$ for some $\alpha$. For $\beta\ne\alpha$, by an arc stabiliser we mean $G_{\alpha\beta}=G_\alpha\cap G_\beta$ and an edge stabiliser we mean $G_{\{\alpha,\beta\}}$, the stabiliser of the set $\{\alpha,\beta\}$. Note that $G_{\{\alpha,\beta\}}\ge G_{\alpha\beta}$. -My question is: can an arc stabiliser $G_{\alpha\beta}\ne 1$ be normal in $G_\alpha$? -This is impossible if $G_{\{\alpha,\beta\}}> G_{\alpha\beta}$. In this case, there exists $g\in G$ such that $\alpha^g=\beta$ and $\beta^g=\alpha$. It follows that $g\in N_G(G_{\alpha\beta})$. Note that $G_\alpha$ is maximal in $G$ and $G_{\alpha\beta}$ cannot be normal in $G$ (otherwise $G_{\alpha\beta}$ will be in the kernel of this action), the normaliser $N_G(G_{\alpha\beta})=G_\alpha$. This gives a contradiction: $g\in G_\alpha$ but $\alpha^g=\beta\ne \alpha$. -I think there might exist an example in the case when $G_{\{\alpha,\beta\}}=G_{\alpha\beta}$. However, by a quick check with Magma there is no such primitive permutation group $G$ with $|G|\le 300$. - -An equivalent statement is: Let $G$ be a group and $H$ a maximal and core-free subgroup of $G$. Is it possible that $1\ne H\cap H^g\lhd H$ for some $g\notin H$? -As shown in the comments there by Verret and Holt there is no such example for degree $\le 4095$. I also think the proof or an example, if any, will not be elementary (for example, applying O'Nan-Scott Theorem). - -This was initially a MSE question I asked. - -REPLY [6 votes]: An example was constructed by Pablo Spiga: -https://arxiv.org/abs/2102.13614 -"A generalization of Sims conjecture for finite primitive groups and two point stabilizers in primitive groups"<|endoftext|> -TITLE: First order PDE in complex variables? -QUESTION [5 upvotes]: Consider the equation -$$f'(x)+ g(x)f(x)=0$$ -This equation is an ODE and has a solution $$ f(x)=C e^{ \int_1^x g(x) \ dx}.$$ -Similarly, we can look at complex variables and consider the equation and Wirtinger derivatives -$$ (\partial_{\bar z} f)(z) +g(z) f(z)=0.$$ -Can one still write down an explicit solution? - -REPLY [6 votes]: You can start by looking at the chain rule for wirtinger derivatives, from which you deduce that -$$ -\partial_{\bar z} \exp(h(z)) = \exp(h(z)) \cdot \partial_{\bar z} h(z) -$$ -Therefore, if you find a function $h$ such that $\partial_{\bar z} h = - g(z)$ (I think you forgot a "$-$" sign in your solution for the real case!) taking $f(z) = \exp(h(z)) $ will solve your problem. In general, this is known as the d-bar problem (or $\bar\partial-$problem). As Daniele points out this Q&A is a god resource for the $\bar\partial-$problem in 1 dimension.<|endoftext|> -TITLE: Fast algorithms for calculating the distance between measures on finite ultrametric spaces -QUESTION [5 upvotes]: Let $X$ be a finite ultrametric space and $P(X)$ be the space of probability measures on $X$ endowed with the Wasserstein-Kantorovich-Rubinstein metric (briefly WKR-metric) defined by the formula -$$\rho(\mu,\eta)=\max\{|\int_X fd\mu-\int_X fd\eta|:f\in Lip_1(X)\}$$ where $Lip_1(X)$ is the set of non-expanding real-valued functions on $X$. - -Problem. Is there any fast algorithm for calculating this metric between two measures on a finite ultrametric space? Or at least for calculating some natural distance, which is not "very far" from the WKR-metric? - -Added in Edit. There is a simple upper bound $\hat \rho$ for the WKR-metric, defined by recursion on the cardinality of the set $d[X\times X]=\{d(x,y):x,y\in X\}$ of values of the ultrametric on $X$. If $d[X\times X]=\{0\}$, then for any measures $\mu,\eta\in P(X)$ on $X$ put $\hat\rho(\mu,\eta)=0$. Assume that for some natural number $n$ we have defined the metric $\hat\rho(\mu,\eta)$ for any probability measures $\mu,\eta\in P(X)$ on any ultrametric space $(X,d)$ with $|d[X\times X]|0\}$ and the probability measurees $\mu{\restriction} B$ and $\eta{\restriction}B$ assign to each subset $S$ of $B$ the numbers $\mu(S)/\mu(B)$ and $\eta(S)/\mu(B)$, respectively. -It can be shown that $\rho\le\hat\rho$. - -Question. Is $\rho=\hat\rho$? - -REPLY [3 votes]: This is a rather more fun problem than I thought. I must apologize though, as your question is a reference request and I have no references apart from pointing at any textbook on discrete optimization. It turns out, the key is that one can rewrite your problem into a flow problem on a tree, which then is almost trivial to solve. Thus, if I am not mistaken, not only is your upper bound $\hat{\rho}$ the correct value for $\rho$, but the same is true for many other heuristic ways to construct an upper bound. The ultrametric seems to try its best to actively prevent you from accidentally choosing bad solutions and you can use this to define some algorithms which should be almost optimal. -Preliminaries -I think the problem is easier to understand in the transport formulation (which is the dual of the one used in the question): -$$ \rho(\mu,\eta) := \min \left\{ \int_{X \times X} d(x,y) \,dT : T \in P(X\times X), T(.,X) = \mu,T(X,.)=\eta\right\} $$ -i.e. $T(A,B)$ tells us how much mass is transported from $A$ to $B$. I will mostly use this and some derived formulation, but it is good to have both around. In particular, if you have an $f$ for the formulation in the question and a $T$ for this formulation which both give you the same value, you know that both have to be optimal. -Futhermore, we can assume that $\operatorname{supp} \mu \cap \operatorname{supp} \eta = \emptyset$, as transporting from a point to itself is free. In fact, I will not assume that $\mu$ and $\eta$ are probability measures but only that $\mu(X) = \eta(X)$, which works equally well with all definitions and allows us to easily substract similar amounts from both without having to renormalize in every step. In fact in this context it can be useful to consider the signed measure $\nu = \mu -\eta$ instead, which sufficiently describes both of them. -The tree problem -As far as I can gather, any ultrametric can be written in form of a tree (rooted, as used in computer science), where the leaves correspond to the points of $X$ and each subtree to a set of balls containing precisely the points that are its leaves. One can then assign a distance $d_e$ to each edge $e \in E$ of the tree such that the distance between two points in $X$ corresponds to the length of their connecting path through the graph. -One can rewrite finding the WKR-metric into a flow problem on the tree: Extend $\mu$ to the interior nodes by $0$. Now we need to find a flow, i.e. an assignment of a direction and a value $p_e$ to each edge (It is simpler to assume a fixed direction, say upward in the tree and a signed $p_e$ instead) such that in each node $n$ the total of in and outgoing flow corresponds $\nu(n)$. The cost of such a flow then is given by $\sum_e d_e |p_e|$. -The interesting fact about this problem is that on a tree, such a flow is always unique. Also the cost of the unique flow is identical to the WKR-metric. In fact you can recover an $f$ with identical resulting value by assigning a fixed value to a given node $v$ and the recursively setting $f(w) = f(v) \pm d_{(v,w)}$ for all its neighbours, where the sign depends on the direction of the flow. Similarly, you can recover a $T$ by splitting the flow into a sum of weighted paths between leaves and setting $T(\{(x,y)\})$ to the weight of that path. If you take care to never have any cancellation (which is always possible), the corresponding value will again be the same as the cost of the flow. -A fast algorithm given a tree -There are fast algorithms to calculate an optimal flow in graphs, but as we only require the cost of the flow, there is an easy recursive algorithm to calculate it along the tree. For each subtree, we simultaneously construct the internal cost of the flow the flow that leads upwards from it. The total cost then is the internal cost of the whole tree. - -For each leaf $x$, the internal cost is 0 and the flow upwards is $\nu(x)$. - -For each subtree, we can recursively calculate internal cost and flow upwards of all of its child trees. -The internal cost of the subtree then is the sum of internal costs of its child trees plus the sum of the absolute values of the flows from each of those children multiplied by each respective distance. The flow upwards is simply the sum of all signed flows from the children. - - -This algorithm only visits each node in the tree once and does a rather simple calculation there, so I'd argue that it is next to optimal. In particular as there are always more children than internal nodes in a tree, it is of order $O(|X|)$. I also believe it is equivalent to the heuristic in the question. -A fast algorithm without a tree -If we do not have the tree structure but are instead only given the distance function, we do not need to calculate the tree. Instead there is a faster way to get to the same value by a simple greedy algorithm: - -Find the pair of nodes $x,y$ with $\mu(\{x\}) > 0$ and $\eta(\{y\}) > 0$ such that $d(x,y)$ is minimal. -Add $d(x,y)\min(\mu(\{x\}),\eta(\{y\}))$ to the total cost and reduce $\mu(\{x\})$ and $\eta(\{y\})$ by $\min(\mu(\{x\}),\eta(\{y\}))$ -Repeat until $\mu=\eta =0$ - -If initially one creates a binary heap of all distances this needs a runtime of order $O(|X|^2\log |X|)$. Then in each iteration this algorithm reduces $\operatorname{supp} \mu$ or $\operatorname{supp} \eta$ by a point, so it will run at most for $|X|$ iterations and in doing so remove all elements from the heap again in runtime $O(|X|^2\log |X|)$. As there are a potential $O(|X|^2)$ of distance values to check I'd argue that this again is close to optimal. -The reason why this algorithm returns the right result is evident if one considers the graph in parallel. In each iteration you can add the path between $x$ and $y$ with weight $\min(\mu(\{x\}),\eta(\{y\}))$. When the algorithm finishes, the sum of those paths then gives the flow and one can show that no cancellation occurs. The idea is that the tree is kind of filled from the bottom and a path of minimal distance starting can only ever leave a subtree, if either $\mu$ or $\eta$ is already zero on this subtree, so there will be no future path coming in the opposite direction. -Other distances -A fun observation I had while writing this: At least with Wasserstein-distances, one is generally interested in $d(x,y)^p$ for some $p \in [1,\infty)$ as a cost instead of just $d(x,y)$. But if $d$ is an ultrametric, so is $d^p$, so the whole argument gets adapted easily.<|endoftext|> -TITLE: Can we have stratified L? -QUESTION [6 upvotes]: Can one build a hierarchy of stratified constructible stages? That is a hierarchy that is built in a manner similar to Godel's constructible universe L, but with additionally requiring that the defining formulas must be also stratified! - -If that can be built, then my main question is whats the stance of choice in that hierarchy, would choice be satisfied internally in every limit stage of that hierarchy? Can choice fail? - -REPLY [7 votes]: Glad to see someone taking an interest in this stuff! -All but one of the Goedel operations are stratified, and the one that isn't is the existence of $\in$ "locally." Replace this operation by one that gives you the local version of $\in$ composed with singleton, so that you get -$A \cap \{\langle \{x\},y \rangle: x \in y \in B\}$. You can then prove an exact analogue of the original result that anything closed under the rudimentary functions and power set is a model of $\Delta_0$ separation - for the stratified operations. Any set closed under the stratrud functions and power set is a model of stratified $\Delta_0$ separation. This actually gives a finite axiomatisation of NF. Naturally one wonders about what happens if one tries to construct an analogue of $L$ in the fashion and the (annoying) answer is that what one gets depends very sensitively on how often one "sweeps up." The more often you sweep up the more of $L$ you get. I have some quite extensive notes on this, but they are from about 10 years ago and i would need to do some archaeology. I will if pressed.<|endoftext|> -TITLE: Representation of fundamental groupoid as $2$-sheaf -QUESTION [6 upvotes]: By https://arxiv.org/abs/1406.4419 (The fundamental groupoid as a terminal costack, Ilia Pirashvili), we know that for a topological space $X$, the $2$-functor -$$\text{Top}(X)\rightarrow \text{Gpd}, \quad (U\rightarrow X)\mapsto \Pi_1(U)$$ -is a $2$-cosheaf, in fact the terminal one. In particular, it follows that -$$\text{Top}(X)^{op}\rightarrow \text{Gpd}, \quad (U\rightarrow X)\mapsto \text{Hom}(\Pi_1(U),\mathcal{G})$$ -is a $2$-sheaf, where $\mathcal{G}$ is a groupoid. However, if we assume $X$ is a manifold (probably locally simply connected does the job, but I want to be cautious), then for any point $x\in X$, open simply connected sets $U\ni x$ are final in all open neighborhoods of $x$. Hence the stalk of this $2$-sheaf is -$$\text{colim}_{U\ni x}\text{Hom}(\Pi_1(U),\mathcal{G})\cong \text{colim}_{U\ni x,\text{ simp. conn}}\text{Hom}(\Pi_1(U),\mathcal{G})\cong \text{colim}_{U\ni x,\text{ simp. conn}}\text{Hom}(\{x\},\mathcal{G})$$ -where I denote by $\{x\}$ the groupoid whose objects are $\{x\}$ and has no automorphism (I think this is called the trivial groupoid). If we assume that $\mathcal{G}$ is a group $G$, then we get that the stalk consits only of the trival morphism. -However, having trivial stalks means that $\text{Hom}(\Pi_1(-),G)$ is trivial $2$-sheaf. But surely this can not be true, as that would mean -$$\text{Hom}(\Pi_1(X),\text{Gl}_n(\mathbb{C})$$ -is trivial, which is not true. Hence there is a gap/flaw in my reasoning, but I don't see it. -Basically, as $\Pi_1(-)$ is the terminal $2$-cosheaf, $\text{Hom}(\Pi_1(-),\mathcal{G})$ is the initial $2$-sheaf, but that would mean that there are almost no representations of the fundamental group, which seems wrong to me. - -REPLY [7 votes]: First of all, note that you haven't used the fact that $\Pi_1(-)$ was the terminal $2$-cosheaf, just that it was a ($2$-)cosheaf. -Then, as I pointed out in the comments, there's a question of whether you're considering the usual $\hom$ functor into $Set$, or the internal $\hom$ with values in $Gpd$. - -If you're considering the usual, external $\hom$, then the question is more about whether $\Pi_1(-)$ is a cosheaf than a $2$-cosheaf. But note that if you're considering $\hom$ into $Set$, then $\hom$ is not equivalence invariant. In particular, $\hom(\Pi_1(U),\mathcal G) \not\cong \hom(\{x\},\mathcal G)$ even when $U$ is simply-connected. For instance if you take for $\mathcal G$ an indiscrete groupoid on a set $S$ of objects, the former is $\hom_{Set}(U,S)$, while the latter is $\hom_{Set}(\{x\},S)$. So your stalk computation doesn't work anymore. - -(to perhaps get a better grasp on this, compare $\hom(2^{in},BG)$ and $\hom(\{x\},BG)$ where $2^{in}$ is the indiscrete groupoid on $2$ objects, and $BG$ the one-object groupoid associated to the group $G$) - -If you're considering the internal $\hom$, and looking at stacks/$2$-sheaves (so that you do have invariance under equivalence), then $\hom(\{x\},\mathcal G)\cong \mathcal G$, so it's definitely nontrivial. - -(moreover, note that in this context of stacks, or $2$-sheaves, you need to be more careful with colimits: this would probably be a $2$-colimit, so you'd need to check that the subcategory of simply-connected neighbourhoods of $x$ is final in the $2$-categorical sense in the category of neighbourhoods. This is a rather stronger condition than just finality in the $1$-categorical sense, and here I'm not sure whether it would be satisfied in general - I'm definitely not an expert on stacks and $2$-categories so I can't say much more)<|endoftext|> -TITLE: Parity of the multiplicative order of 2 modulo p -QUESTION [22 upvotes]: Let $\operatorname{ord}_p(2)$ be the order of 2 in the multiplicative group modulo $p$. Let $A$ be the subset of primes $p$ where $\operatorname{ord}_p(2)$ is odd, and let $B$ be the subset of primes $p$ where $\operatorname{ord}_p(2)$ is even. Then how large is $A$ compared to $B$? - -REPLY [34 votes]: This problem was asked by Sierpinski in 1958 and answered by Hasse in the 1960s. -For each nonzero rational number $a$ (take $a \in \mathbf Z$ if you wish) and each prime $\ell$, let $S_{a,\ell}$ be the set of primes $p$ not dividing the numerator or denominator of $a$ such that $a \bmod p$ has multiplicative order divisible by $\ell$. When $a = \pm 1$, $S_{a,\ell}$ is empty except that $S_{-1,2}$ is all odd primes. From now on, suppose $a \not= \pm 1$. -In Math. Ann. 162 (1965/66), 74–76 (the paper is at https://eudml.org/doc/161322 and on MathSciNet see MR0186653) Hasse treated the case $\ell \not= 2$. Let $e$ be the largest nonnegative integer such that $a$ in $\mathbf Q$ is an $\ell^e$-th power. (For example, if $a$ is squarefree then $e = 0$ for every $\ell$ not dividing $a$.) The density of $S_{a,\ell}$ is $\ell/(\ell^e(\ell^2-1))$. This is $\ell/(\ell^2-1)$ when $e = 0$ and $1/(\ell^2-1)$ when $e = 1$. -In Math. Ann. 166 (1966), 19–23 (the paper is at https://eudml.org/doc/161442 and on MathSciNet see MR0205975) Hasse treated the case $\ell = 2$. The general answer in this case is more complicated, as issues involving $\ell$-th roots of unity in the ground field (like $\pm 1$ in $\mathbf Q$ when $\ell = 2$) often are. The density of $S_{a,2}$ for "typical" $a$ is $1/3$, such as when $a \geq 3$ is squarefree. But $S_{2,2}$ has density 17/24, so the set of $p$ for which $2 \bmod p$ has even order has density $17/24$ and the set of $p$ for which $2 \bmod p$ has odd order has density $1 - 17/24 = 7/24$. -For example, there are $167$ odd primes up to $1000$, $1228$ odd primes up to $10000$, and $9591$ odd primes up to $100000$. There are $117$ odd primes $p \leq 1000$ such that $2 \bmod p$ has even order, $878$ odd primes $p \leq 10000$ such that $2 \bmod p$ has even order, and $6794$ odd primes $p \leq 100000$ such that $2 \bmod p$ has even order. The proportion of odd primes up $1000$, $10000$, and $100000$ for which $2 \bmod p$ has even order is $117/167 \approx .700059$, $878/1228 \approx .71498$, and $6794/9591 \approx .70837$, while $17/24 \approx .70833$. -The math.stackexchange page here treats $S_{7,2}$ in some detail and at the end mentions the case of $S_{2,2}$.<|endoftext|> -TITLE: Embedding of a derived category into another derived category -QUESTION [12 upvotes]: I am considering the following two cases: - -Assume that there is an embedding: $D^b(\mathcal{A})\xrightarrow{\Phi} D^b(\mathbb{P}^2)$and the homological dimension of $\mathcal{A}$ is equal to $1$($\mathcal{A}$ is an abelian category), for simplicity, maybe first I assume that $\mathcal{A}$ is a module category over a finite dimensional $A$, then $A$ is a hereditary algebra. Assume that $\Phi$ is a Fourier-Mukai functor, in addition, $A$ is $\textbf{not}$ fractional Calabi-Yau algebra. What kind of condition should I impose on $A$, to conclude that $A\cong KQ$(path algbera) such that $Q$ is a Kronecker quiver with three vertices and three arrows? - -Assume that there is an embedding: $D^b(\mathcal{A}')\xrightarrow{\Psi} D^b(J(\Gamma))$, where $\Gamma$ is a genus 2 degree 7 curve and $J(\Gamma)$ is its Jacobian, which is an abelian surface. Also $\mathcal{A}'$ has homological dimension 1 and $\Psi$ is also Fourier-Mukai functor. What condition should I impose to conclude that $\mathcal{A}'\cong\mathrm{Coh}(\Gamma)$? Note that in this case, $J(\Gamma)$ is an abelian surface and there isn't any non-trivial SOD for its derived category, which means that $\Psi(D^b(\mathcal{A}'))$ is not a left or right admissible subcategory of $D^b(J(\Gamma))$. - - -Motivation -I am considering $\mathbb{P}^2$ as certain moduli space of stable objects in $\mathcal{A}$ and $J(\Gamma)$ as certain moduli space of stable objects in $\mathcal{A}'$ and the embedding functor $\Phi$ and $\Psi$ are induced by Fourier-Mukai functor with the kernel given by universal family. - -REPLY [13 votes]: Any fully faithful functor from $D^b(\mathcal{A})$ has adjoints (because $D^b(\mathcal{A})$ is a smooth and proper category), so its image is an admissible subcategory. A recent result from Dmitrii Pirozhkov shows that any admissible subcategory in $D^b(\mathbb{P}^2)$ is generated by one or two exceptional objects obtained from the standard exceptional collection by mutations. Therefore, $D^b(\mathcal{A})$ must be generated by an exceptional pair of this sort. If $D^b(\mathcal{A})$ is a quiver with three arrows, the image must be the subcategory generated by the exceptional pair -$$ -\langle \mathcal{O}(i), \mathcal{O}(i+1) \rangle. -$$ -As for the second question—this never happens, because $J(\Gamma)$ is a Calabi-Yau variety and hence its derived category has no non-trivial admissible subcategories.<|endoftext|> -TITLE: Necessary and sufficient conditions for a holomorphic function defined in the unit disk to be univalent? -QUESTION [5 upvotes]: de Branges has proved de Branges's theorem (the famous Bieberbach conjecture) that if a holomorphic function $f(z) = z+\sum_{n=2}^{\infty} a_nz^n$ in the unit disk $D = \{z\in \mathbb{C},|z| \leq 1\}$ is univalent, then we have $|a_n| \leq n,\forall n\geq 2$. -Conversely, let's consider a holomorphic function $g(z) = z+\sum_{n=2}^{\infty} b_nz^n$ which is defined in $D$ and satifies $|b_n| \leq n$, then what are the general sufficient conditions(I've known some special conditions on this problems, such as Nehari's univalence criterion and other criterions, unfortunately, they are not in full generality) to ensure $g(z)$ is univalent. Any clues and facts are welcomed, best regards ! - -Updated question: necessary and sufficient conditions for a holomorphic function defined in the unit disk to be univalent (as far as I known, several conditions have be proposed, but all of them seem to be not practical), simple forms and only depend on function g(z) or its derivatives, integrals, their combinations, and so on. For example something like Milin's inequality. Unfortunately, I've tried several variants of this inequality (together with some additional conditons), but fails. - -REPLY [3 votes]: It is highly unlikely that anything "reasonable" can be said in terms of the coefficients. -Already the allowable region for $(a_2,a_3)$ (so they are the first non-trivial coefficients of $z+a_2z^2+a_3z^3+...$ univalent) is quite complicated as for example the sharpness (ie for each $0 \le \alpha \le 1$ there is an univalent functions for which the equality holds) of the Fekete-Szego inequalities $|a_3-\alpha a_2^2| \le 1+2e^{\frac{-2\alpha}{1-\alpha}}, 0 \le \alpha \le 1$ shows - here $\alpha =0$ corresponds to the highly non-trivial third coefficient bound $|a_3| \le 3$, while $\alpha=1$ corresponds to the easy $|a_3-a_2^2| \le 1$ -See also Schaeffer Spencer book on the coefficient regions for Schlicht functions https://www.ams.org/books/coll/035/<|endoftext|> -TITLE: Is every finite group the outer automorphism group of a finite group? -QUESTION [32 upvotes]: This question is essentially a reposting of this question from Math.SE, which has a partial answer. YCor suggested I repost it here. - -Our starting point is a theorem of Matumoto: every group $Q$ is the outer automorphism group of some group $G_Q$ [1]. It seems to be a research theme to place restrictions on the groups involved. For example, Bumagin and Wise proved that if we restrict $Q$ to be countable then we may take $G_Q$ to be finitely generated [2], and more recently Logan proved that if we restrict $Q$ to be finitely generated and residually finite group then we may take $G_Q$ to be residually finite [3, Corollary D] (this paper also cites quite a few other papers which play this game). -However, all the results I have found always produce infinite groups $G_Q$, even when the "input" groups $Q$ are finite. For example, Matumoto's groups $G_Q$ are fundamental groups of graphs of groups (so are always infinite), Bumagin and Wise use a variant of Rips' construction (so (as $Q$ is finite) their groups $G_Q$ have finite index in metric small cancellation groups, so are infinite), and Logan's groups $G_Q$ are HNN-extensions of hyperbolic triangle groups (so again are infinite). So we have a question: - -Does every finite group $Q$ occur as the outer automorphism group of some finite group $G_Q$? - -The answer is "yes" if we take $Q$ to be finite abelian or a symmetric group; this is what the answer to the original Math.SE question proves. -[1] Matumoto, Takao. "Any group is represented by an outerautomorphism group." Hiroshima Mathematical Journal 19.1 (1989): 209-219. (Project Euclid) -[2] Bumagin, Inna, and Daniel T. Wise. "Every group is an outer automorphism group of a finitely generated group." Journal of Pure and Applied Algebra 200.1-2 (2005): 137-147. (doi) -[3] Logan, Alan D. "Every group is the outer automorphism group of an HNN-extension of a fixed triangle group." Advances in Mathematics 353 (2019): 116-152. (doi, arXiv) - -REPLY [24 votes]: Yes. -For each finite group $Q$ I'll construct a finite group $H$ with $\mathrm{Out}(H)\simeq Q$, moreover $H$ will be constructed as a semidirect product $D\ltimes P$, with $P$ a $p$-group of exponent $p$ and nilpotency class $|Q|+1$) and $D$ abelian of order coprime to $p$ (actually $D$ being a power of a cyclic group of order $p-1$). - -I'll use Lie algebras which are convenient tools to encode $p$-groups when $p$ is less than the nilpotency class, taking advantage of linear algebra. -In Lie algebras, we denote $[x_1,\dots,x_m]=[x_1,[x_2,\dots,[x_{m-1},x_m]\cdots]]$. Also I choose the convention to let permutations act on the left. -The base field is $K=\mathbf{F}_p$, $p$ prime. Fix $n\ge 1$. Let $\mathfrak{f}_n$ be the free Lie $K$-algebra on generators $(e_1,\dots,e_n)$. It admits a unique grading in $\mathbf{Z}^n$ such that $e_i$ has degree $E_i$, where $(E_i)$ is the canonical basis of $\mathbf{Z}^n$, it is called multi-grading. For instance, $[e_3,[e_1,e_3]]$ has multi-degree $(1,0,2,0,\dots,0)$. -Let $I$ be a finite-codimensional multi-graded ideal contained in $[\mathfrak{f}_n,\mathfrak{f}_n]$: so the quotient $\mathfrak{g}=\mathfrak{f}_n/I$ is naturally multi-graded. There is a natural action of ${K^*}^n$ on $\mathfrak{g}$: namely $(t_1,\dots,t_n)$ acts on $\mathfrak{g}_{(m_1,\dots,m_n)}$ by multiplication by $\prod_{i=1}^n t_i^{m_i}$. Let $D\subset\mathrm{Aut}(\mathfrak{g})$ be the image of this action. Also denote by $c$ the nilpotency class of $\mathfrak{g}$: we assume $p>c$. -Using that $p>c$, we endow, à la Malcev–Lazard, $\mathfrak{g}$ with the group law given by the Baker-Campbell-Hausdorff formula: $xy=x+y+\frac12[x,y]+\dots$. We thus view $\mathfrak{g}$ as both a Lie algebra and a group; we denote it as $G$ when endowed with the group law (but feel free to refer to the Lie algebra law in $G$); this is a $p$-group of exponent $p$ and nilpotency class $cn+1$ there exist $n,p$ and $I$ finite-codimensional multi-graded ideal in $\mathfrak{f}_n(=\mathfrak{f}_n(\mathbf{F}_p)$, such that $\Gamma_I\simeq Q$. (And such that $\mathfrak{f}_n/I$ has nilpotency class $\le n+1$.) -This starts with the following, which provides for each finite group $Q$ a relation whose automorphism group is the group $L_Q\subset\mathfrak{S}(Q)$ of left translations of $Q$. -Lemma 4. Let $Q$ be a group, $n=|Q|$, and $q=(q_1,\dots,q_n)$ an injective $n$-tuple of $Q$. Define $X=Qq\subset Q^n$. Then $L_Q=\{\sigma\in\mathfrak{S}(Q):\sigma X=X\}$. -Proof. Clearly $L_Q$ preserves $X$. Conversely, if $\sigma$ preserves $X$, after composing with a left translation we can suppose that $\sigma(q_1)=q_1$, so $\sigma(q)\in\{q_1\}\times Q^{n-1}$; since $X\cap \{q_1\}\times Q^{n-1}=\{q\}$, we deduce $\sigma(q)=q$, which in turn implies $\sigma=\mathrm{Id}$. $\Box$. -Proof of the proposition. -Write $\mathfrak{f}_Q\simeq\mathfrak{f}_n$ as the free Lie algebra over the generating family $(e_q)_{q\in Q}$. It can be viewed as graded in $\mathbf{Z}^G$, with basis $(E_q)_{q\in Q}$. Write $E=\sum_q E_q$. -For $q\in Q^n$, define $\xi_q=[e_{q_n},e_{q_1},e_{q_2},\dots,e_{q_n}]$ (note that it is homogeneous of degree $E+E_{q_n}$; in particular $(\xi_h)_{h\in X}$ is linearly independent. Fix an injective $n$-tuple $q$ and define $X$ as in the proof of the lemma; for convenience suppose $q_n=1$. -Define $J$ as the $n$-dimensional subspace of $\mathfrak{f}_Q$ with basis $(\xi_h)_{h\in X}$. -Define $I=J\oplus \mathfrak{f}_Q^{n+2}$, where $\mathfrak{f}_Q^i$ is the $i$-th term in the lower central series. Hence $I$ is an ideal, and $\mathfrak{g}=\mathfrak{f}_Q/I$ is defined by killing all $i$-fold commutators for $i\ge n+1$ and certain particular $(n+1)$-fold commutators. (Since we assume $p>n+1$, we can view it as a group as previously.) -Claim. For $h=(h_1,\dots,h_n)\in Q^n$ with $h_{n-1}\neq h_n$, we have $\xi_h\in I$ if and only if $h\in X$. -By definition, $h\in X$ implies the condition. Now suppose that $h$ satisfies the condition. First, the condition $h_{n-1}\neq h_n$ ensures that $\xi_h\neq 0$; it is homogeneous in the multi-grading. If it belongs to $J$, its multi-degree is therefore some permute of $(2,1,\dots,1)$. This is the case if and only if the $h_i$ are pairwise distinct, so we now assume it; its degree is therefore equal to $E_{h_n}+E$. Now $J_{E+E_{h_n}}$ is 1-dimensional, and generated by $\xi_{h_nq}$. Hence $\xi_h$ is a scalar multiple of $\xi_{h_nq}$: -$$[e_{h_n},e_{h_1},\dots,e_{h_{n-1}},e_{h_n}]=\lambda -[e_{h_n},e_{h_nq_1},\dots,e_{h_nq_{n-1}},e_{h_n}].$$ -The next lemma implies that $h_i=h_nq_i$ for all $i\in\{1,\dots,n-1\}$. So $h\in X$. The claim is proved. -The claim implies that for every permutation $\sigma$ of $Q$, if the automorphism $u_\sigma$ of $\mathfrak{f}_Q$ preserves $I$, then $\sigma$ has to preserve $X$, and hence (Lemma 4) $\sigma$ is a left translation of $Q$. This finishes the proof. $\Box$. -Lemma 5 -Consider the free Lie algebra on $(e_1,\dots,e_n)$. If for some permutation $\sigma$ of $\{1,\dots,n-1\}$ and scalar $\lambda$ we have -$$[e_n,e_1,\dots,e_{n-1},e_n]=\lambda [e_n,e_{\sigma(1)},\dots,e_{\sigma(n-1)},e_n],$$ -then $\sigma$ is the identity and $\lambda=1$. -Proof. Use the representation $f$ in $\mathfrak{gl}_n$ mapping $e_i$ to the elementary matrix $\mathcal{E}_{i-1,i}$ (consider indices modulo $n$). Then -$[e_n,e_1,\dots,e_{n-1},e_n]=[e_n,e_1,[e_2,\dots,e_n]]$ maps to $$[\mathcal{E}_{n-1,n},\mathcal{E}_{n,1},\mathcal{E}_{1,n}]=[\mathcal{E}_{n-1,n},\mathcal{E}_{n,n}-,\mathcal{E}_{1,1}]=\mathcal{E}_{n-1,n}.$$ -Let by contradiction $j$ be maximal such that $\sigma(j)\neq j$; note that $2\le j\le n-1$ and $n\ge 3$. Then $[e_n,e_{\sigma(1)},\dots,e_{\sigma(n-1)},e_n]=[e_n,e_{\sigma(1)},\dots,e_{\sigma(j)},[e_{j+1},\dots,e_n]]$ maps to -$$w=[\mathcal{E}_{n,n-1},\mathcal{E}_{\sigma(1)-1,\sigma(1)},\dots,\mathcal{E}_{\sigma(j)-1,\sigma(j)},\mathcal{E}_{j,n}],$$ -which cannot be zero. Hence $[\mathcal{E}_{\sigma(j)-1,\sigma(j)},\mathcal{E}_{j,n}]\neq 0$. Since $\sigma(j) -TITLE: Geometric construction of the fourth intersection points of two conics -QUESTION [10 upvotes]: In general, two conics in the plane intersect at most 4 points. Suppose three of those points are given as $A,B,C$. Then let $c_1$ be the conic passing through those three points and $D_1,E_1$. Let $c_2$ be the conic passing through those three points and $D_2,E_2$. How can the fourth intersection point of these two conics be constructed geometrically, with ruler and compass? - -REPLY [5 votes]: Based on Projective conic sections - constructions, -the crux of the construction is this: - -let two conics intersect in $A,B,C,D$. -let any line through $A$ intersect the conics again in $M,M'$ -let any line through $B$ intersect the conics again in $N,N'$ -then $MN, M'N'$ and $CD$ are concurrent. - -To show this, consider the hexagons $MACDBN and M'ACDBN'.$ Let $P=MA\cdot DB=M'A\cdot DB$ and -$Q=AC\cdot BN=AC\cdot BN'$. By Pascal's Theorem $CD\cdot MN$ and $CD\cdot M'N'$ are on line $PQ$, and the concurrency follows. In particular, $F=MN\cdot M'N'$ lies on $CD.$ -For any $T\neq U$, let $UT\cdot UVWXY$ denote the other intersection $Z$ of the line $UT$ with the conic defined by the five points $U,V,W,X,Y$. There is a classic straightedge construction of $Z$ based on Pascal's Theorem which is described in Hatton's Projective Geometry (pg 240, 133.A.ii) -Putting it all together, the steps for the construction of $D$ are: - -$M=AT\cdot ABCD_1E_1$ -$M'=AT\cdot ABCD_2E_2$ -$N=BT\cdot BACD_1E_1$ -$N'=BT\cdot BACD_2E_2$ -$F=MN\cdot M'N'$ (as mentioned above, $F$ will lie on $CD$) -$D=CF\cdot CABD_1E_1$ - -Note that the construction can be done with a straightedge only - no compass required!<|endoftext|> -TITLE: Sustainability of ZBmath unrestricted access -QUESTION [27 upvotes]: ZBmath (formerly Zentralblatt für Mathematik) will become "Open access" in 2021 (see for instance at EMS site and at FIZ Karlsruhe site). It used to be under paywall, although some partial access was allowed. -My question (which was suggested in the comments of this answer to Is a free alternative to MathSciNet possible?): - -What are guarantees (technical and/or legal) that the reviews will remain in unrestricted universal access? - -(Concretely, I'd be reluctant to write reviews if ZBmath has the possibility to suddenly turn back to partial paywall in 2030 for whatever reasons, especially if this paywall affects searches for reviews made during the so-called open-access period.) - -REPLY [21 votes]: Thanks a lot for the question! On behalf of Klaus Hulek (as zbMATH Editor-in-Chief) we can confirm that it is correct that we will go Open Access as of 1st January 2021. This will mean that the database is freely accessible by everybody worldwide. It also means that most of the data will become open via a CC-BY-SA license. Such a license cannot legally be revoked. This applies in particular to the reviews written by our reviewers (only excluding the very rare case of reviews where the copyright is owned by a third party, which amounts to less than 0.5% of the reviews). The other main exception is third-party content such as author summaries, which may be provided under a different license. We will further have an API which will allow users to download parts (especially, the CC-BY-SA-licensed parts) of the database.<|endoftext|> -TITLE: Real forms of complex reductive groups -QUESTION [6 upvotes]: I have a collection of related (to me) questions, which stem from the fact that I feel like I have a bunch of pieces, but not a full clear picture. I'm curious about forms of reductive groups in general, so I'm only asking about $\mathbb{C}/\mathbb{R}$ for simplicity's sake and for explicit examples. -As a first fact, I know that $k$ forms of algebraic varieties $X_{k'}$ are classified by $H^1(\operatorname{Gal}(k'/k), \operatorname{Aut}_{k'}(X))$. There's an abstract (to me) way of producing the desired forms by twisting by cocycles. -However, the explicit ways I have of constructing different forms feel different to me. - -Tori. Here I immediately reach for $\operatorname{Res}_{k'/k}(T)$, or perhaps a norm torus $\operatorname{Res}_{k'/k}^{(1)}(T)$. For instance, two real forms of $\mathbb{G}_{m}(\mathbb{C})$ are precisely $\mathbb{R}^*$ and $\operatorname{Res}_{\mathbb{C}/\mathbb{R}}^{(1)}(\mathbb{G}_m(\mathbb{C})) = \mathbb{R}[x,y]/(x^2+y^2-1)$. - -I'm not clearly aware of how to view this second construction of a non-split (actually anisotropic?) torus as coming from twisting with a cocycle. - -Semisimple groups. Here the natural example is $\operatorname{SL}_2(\mathbb{C})$. The split real form is $SL_2(\mathbb{R})$, so I search for a way to construct $\operatorname{SU}_2(\mathbb{R})$. In my head, here I'm doing something much more cocycle-y, when I take the fixed points of $(x, (\overline{x}^{-1})^t)$ where $S_2$ is acting by exchanging coordinates: here I'm aware that I'm taking an automorphism of $\operatorname{SL}_2(\mathbb{C})$ given by inverse transpose, and composing it with the Galois action of complex conjugation, and taking fixed points. It should be clear that my understanding of this is pretty ad-hoc, but at least I'm aware that something of this sort is related to descent. - -So my questions are as follows: -A) How does restriction of scalars (and maybe taking norms) fit in with the more general cohomological machinery of constructing forms via twisting? -B) Let's say that I constructed the two real forms $\operatorname{SL}_2(\mathbb{R})$ and $\operatorname{SU}_2(\mathbb{R})$. Is there any way to predict or understand which forms of tori will appear? In $\operatorname{SL}_2(\mathbb{R})$ we get both forms, $\mathbb{R}^*$ embedded diagonally and $S^1$ embedded via $$\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}.$$ -In $\operatorname{SU}_2$, however, we only get the latter. Is there some more abstract way to parametrize which forms of tori will appear in a given form of a reductive group? I know that conjugacy classes of tori should be parametrized by $H^1(\operatorname{Gal}(k'/k), N_G(T))$ (at least I think this) but I'm not sure how to use this. -Sorry for the convoluted question, I just feel as though I have the pieces of the puzzle in hand... -I would also be delighted if anyone felt like there was a good reference (even if it only deals with $\mathbb{C}/\mathbb{R}$) for this material. - -REPLY [7 votes]: I answer Question 1. It is just a calculation. -Instead of a real torus, say ${\bf T}$, I consider a pair $(T,\sigma)$, -where $T$ is a complex torus and $\sigma\colon T\to T$ is an anti-holomorphic involution. -See this question and YCor's answer. -For a complex torus $T$, consider the cocharacter group -$${\sf X}_*(T)={\rm Hom}(T, {\Bbb G}_{m,{\Bbb C}}).$$ -To a real torus ${\bf T}=(T,\sigma)$ we associate a pair ${\sf X}_*({\bf T}):=({\sf X}_*(T),\sigma_*)$, -where $\sigma_*\in {\rm Aut\,}\,{\sf X}_*(T)$ is the induced automorphism. -It satisfies $\sigma_*^2=1$. -We denote $\Gamma={\rm Gal}({\Bbb C}/{\Bbb R})=\{1,\gamma\}$, where $\gamma$ is the complex conjugation. -We obtain an action of $\Gamma$ on ${\sf X}_*(T)$ (namely, $\gamma$ acts via $\sigma_*$). -In this way we obtain an equivalence between the category of ${\Bbb R}$-tori -and the category of $\Gamma$-lattices (finitely generated ${\Bbb Z}$-free $\Gamma$-modules): -$$ {\bf T}\rightsquigarrow {\sf X}_*({\bf T}). $$ -Moreover, this is an exact functor: a short exact sequence of real tori -$$ 1\to{\bf T}'\to{\bf T}\to{\bf T}''\to 1$$ -induces a short exact sequence of $\Gamma$-lattices -$$ 0\to {\sf X}_*({\bf T}') \to {\sf X}_*({\bf T}) \to {\sf X}_*({\bf T}'')\to 0.$$ -Now consider the torus ${\Bbb G}_{m,{\Bbb R}}=({\Bbb C}^\times,\,z\mapsto\bar z)$ and the corresponding $\Gamma$-lattice $({\Bbb Z},1)$. -Moreover, consider the torus -$$R_{{\Bbb C}/{\Bbb R}}{\Bbb G}_{m,{\Bbb C}}=(\,{\Bbb C}^{\times\,2},\, (z_1,z_2)\mapsto (\bar z_2,\bar z_1)\,)$$ -and the corresponding $\Gamma$-lattice $({\Bbb Z}^2,J)$, -where -$$ J=\begin{pmatrix}0&1\\1&0\end{pmatrix}. $$ -Consider the norm homomorphism -$$N\colon R_{{\Bbb C}/{\Bbb R}}{\Bbb G}_{m,{\Bbb C}}\to {\Bbb G}_{m,{\Bbb R}},\quad (z_1,z_2)\mapsto z_1z_2$$ -and the corresponding morphism of $\Gamma$-lattices -$$N_*\colon ({\Bbb Z}^2,J)\to ({\Bbb Z},1),\quad (x_1,x_2)\mapsto x_1+x_2.$$ -By definition, -$$R_{{\Bbb C}/{\Bbb R}}^{(1)}{\Bbb G}_{m,{\Bbb C}}=\ker N,$$ -and so its cocharacter group is $\ker N_*=\{(x, -x)\mid x\in{\Bbb Z}\}.$ -The complex conjugation $\gamma$ acts on $\ker N_*$ by $J$, that is, -$$(x,-x)\mapsto (-x, x).$$ -We see that $\ker N_*\simeq ({\Bbb Z},-1)$, and hence -$$R_{{\Bbb C}/{\Bbb R}}^{(1)}{\Bbb G}_{m,{\Bbb C}}\simeq ({\Bbb C}^\times, z\mapsto \bar z^{\,{-1}}).$$ -Since -$$ (z\mapsto \bar z^{\,{-1}})\,=\,(z\mapsto z^{-1})\,\circ\,(z\mapsto \bar z),$$ -we see that $R_{{\Bbb C}/{\Bbb R}}^{(1)}{\Bbb G}_{m,{\Bbb C}}$ can be obtained from ${\Bbb G}_{m,{\Bbb R}}=({\Bbb C}^\times,\,z\mapsto\bar z)$ -by twisting by the cocycle $\gamma\mapsto (z\mapsto z^{-1})$, as required. -Note that these three $\Gamma$-lattices -$({\Bbb Z},1),\ ({\Bbb Z}^2,J),$, and $({\Bbb Z},-1)$ are the only indecomposable $\Gamma$-lattices (up to isomorphism); -see this answer. -It follows that these three real tori ${\Bbb G}_{m,{\Bbb R}}$, -$R_{{\Bbb C}/{\Bbb R}}{\Bbb G}_{m,{\Bbb C}}$, and -$R_{{\Bbb C}/{\Bbb R}}^{(1)}{\Bbb G}_{m,{\Bbb C}}$ -are the only indecomposable real tori (again, up to isomorphism).<|endoftext|> -TITLE: Algebraic independence of shifts of the Riemann zeta function -QUESTION [17 upvotes]: Let $\zeta(s)$ denote the Riemann zeta function. Is the set $\{ -\zeta(s-j)\, \colon\, j\in\mathbb{Z}\}$, or even $\{\zeta(s-z)\, -\colon\, z\in\mathbb{C}\}$, algebraically independent over -$\mathbb{C}$? If not, then expanding a polynomial equation satisfied -by these functions into a Dirichlet series and taking the coefficient -of $n^{-s}$ would yield an "unlikely" number-theoretic identity. - -REPLY [20 votes]: Hmm, it was more difficult than I expected to leverage universality to establish the claim. But one can proceed by probabilistic reasoning instead, basically exploiting the phase transition in the limiting distribution of the zeta function at the critical line. The proof I found used an unexpectedly high amount of firepower; perhaps there is a more elementary argument. -Assume for contradiction that there is a non-trivial polynomial relation -$$ P( \zeta(s+z_1), \dots, \zeta(s+z_n) ) = 0$$ -for all $s$ (excluding poles if desired) and some distinct $z_1,\dots,z_n$ (it is slightly more convenient to reverse the sign conventions from the original formuation). We can assume $n$ to be minimal amongst all such relations. By translating we may normalize so that $z_1,\dots,z_m$ lie on the critical line $\{ \mathrm{Re}(s) = 1/2\}$ for some $1 \leq m < n$ and $z_{m+1},\dots,z_n$ lie to the right $\{ \mathrm{Re}(s) > 1/2 \}$ of the line. -Let $T$ be a large number, let $t$ be a random number in $[0,T]$, and define the random variables $Z_1,\dots,Z_n$ by $Z_j := \zeta(z_j+it)$. Then we have the identity -$$ P( Z_1,\dots,Z_n)=0$$ -with probability $1$. -Now we use the following form of Selberg's central limit theorem: the random variables -$$ (\frac{\log |Z_1|}{\sqrt{\frac{1}{2}\log\log T}}, \dots, \frac{\log |Z_m|}{\sqrt{\frac{1}{2}\log\log T}})$$ -and -$$ (Z_{m+1},\dots,Z_n)$$ -jointly converge to a limiting distribution as $T \to \infty$, with the limiting distribution of first tuple a standard gaussian that is independent of the limiting distribution of the second tuple (which will be some moderately complicated but explicit law). (The usual form of Selberg's theorem covers the case $m=n=1$, but the same machinery gives the general case, see e.g., Laurincikas' book. The intuition here is that the first tuple is largely controlled by the random variables $p^{it}$ for medium-sized primes $1 \lll p \ll T^\varepsilon$, while the second tuple is largely controlled by the random variables $p^{it}$ for small primes $p=O(1)$. The proof of this central limit theorem is unfortunately a bit complicated; the simplest proof I know of is by Radziwill and Soundararajan.) -Now expand $P$ as $\sum_{a_1,\dots,a_m} Z_1^{a_1} \dots Z_m^{a_m} Q_{a_1,\dots,a_m}(Z_{m+1},\dots,Z_n)$ for various polynomials $Q_{a_1,\dots,a_m}$. Extract out a leading term $Z_1^{a_1} \dots Z_m^{a_m} Q_{a_1,\dots,a_m}(Z_{m+1},\dots,Z_n)$ (using say lex ordering on $a_1,\dots,a_m$). The Selberg central limit theorem then shows that $Q_{a_1,\dots,a_m}(Z_{m+1},\dots,Z_n)$ must converge in distribution to zero as $T \to \infty$ (as otherwise there is an asymptotically positive probability event that this term dominates all the other terms put together). The random variable $Q_{a_1,\dots,a_m}(Z_{m+1},\dots,Z_n)$ is a Dirichlet series $\sum_n \frac{c_n}{n^{it}}$ with square-summable coefficients $c_n$ (indeed the coefficients decay like $O(n^{-\sigma+o(1)})$ for some $\sigma>1/2$ by the divisor bound), so by the $L^2$ mean value theorem for such series the variance of this series is asymptotic to $\sum_n |c_n|^2$ (and one can also check that the fourth moment is bounded, again by the divisor bound), so by the Paley-Zygmund inequality we must have $\sum_n |c_n|^2=0$, thus by analytic continuation we obtain a non-trivial polynomial relation $Q_{a_1,\dots,a_m}(s+z_{m+1},\dots,s+z_n)=0$ with fewer variables than the original relation, contradicting the minimality of $n$. - -REPLY [20 votes]: $\zeta(s - z)$ has an Euler product $\prod_p \frac{1}{1 - p^{z-s}}$, and so a monomial $\prod_i \zeta(s - z_i)$ (with the $z_i$ not necessarily distinct) has an Euler product -$$\prod_i \zeta(s - z_i) = \prod_p \prod_i \frac{1}{1 - p^{z_i - s}}.$$ -We want to show that these monomials are linearly independent. Now here's an observation: it's quite hard for Dirichlet series with Euler products to be linearly dependent. This is because any linear dependence must, by examining only the coefficients of $\frac{1}{p^{ks}}$ for each prime separately, be a linear dependence for every Euler factor separately, but also must be a linear dependence for all of the Euler factors multiplied together, and even for any subset of the Euler factors multiplied together. -In fact we can prove the following, passing from Dirichlet series to coefficients. If $S$ is a set of primes, write $\mathbb{N}_S$ for the set of positive integers divisible only by the primes in $S$, and write $\mathbb{N}_{-S}$ for the set of positive integers divisible only by the primes not in $S$. - -Lemma: Let $f_0, \dots f_k : \mathbb{N} \to \mathbb{C}$ be multiplicative arithmetic functions which are - - - -essentially nonzero in the sense that for any finite set of primes $S$, $f_i(n) \neq 0$ for some $n \in \mathbb{N}_{-S}$, and -essentially distinct in the sense that for any finite set of primes $S$, if $f_i(n) = f_j(n)$ for all $n \in \mathbb{N}_{-S}$ then $i = j$. - - - -Then the functions $f_i$ are essentially linearly independent in the sense that for any finite set of primes $S$ they are linearly independent over $\mathbb{C}$ when restricted to $\mathbb{N}_{-S}$. - -Proof. This ends up being a slight variant of the standard proof of linear independence of characters (which would apply directly if "multiplicative" were replaced by "completely multiplicative"). We induct on $k$. When $k = 0$ the result follows from the assumption that the $f_i$ are essentially nonzero. For general $k$, let $S$ be a finite set of primes and suppose by contradiction that we have a nontrivial linear dependence, which WLOG we take to be of the form -$$f_0(n) = \sum_{i=1}^k c_i f_i(n), n \in \mathbb{N}_{-S}.$$ -Since $f_0$ is essentially nonzero this requires that at least one of the $c_i$ also be nonzero. Now, if $m, n \in \mathbb{N}_{-S}$ are positive integers such that $\gcd(m, n) = 1$, then on the one hand -$$f_0(mn) = \sum_{i=1}^k c_i f_i(mn) = \sum_{i=1}^k c_i f_i(m) f_i(n)$$ -and on the other hand -$$f_0(mn) = f_0(m) f_0(n) = f_0(m) \sum_{i=1}^k c_i f_i(n) = \sum_{i=1}^k c_i f_0(m) f_i(n).$$ -Subtracting gives -$$\sum_{i=1}^k c_i (f_0(m) - f_i(m)) f_i(n) = 0.$$ -If $T$ is any finite set of primes, letting $m$ be any element of $\mathbb{N}_T \cap \mathbb{N}_{-S}$ (divisible by the primes in $T$ but not the primes in $S$) and letting $n$ range over $\mathbb{N}_{-(S \cup T)}$ gives, by the inductive hypothesis, that for each value of $m$ the above is a linear dependence of the $f_i$ which must be trivial, hence the coefficients $c_i (f_0(m) - f_i(m))$ must vanish for all $m \in \mathbb{N}_T \cap \mathbb{N}_{-S}$. (This bit of the argument is why we need the freedom to ignore finitely many primes.) -Since one of the $c_i$, say $c_j$, is nonzero, it follows that $f_0(m) = f_j(m)$ for all $m \in \mathbb{N}_T \cap \mathbb{N}_{-S}$, but since this is true independent of the choice of $T$, we in fact have $f_0(m) = f_j(m)$ for all $m \in \mathbb{N}_{-S}$, which contradicts essential distinctness. $\Box$ -Now it suffices to check that the monomials $\prod_i \zeta(s - z_i)$ are essentially nonzero and essentially distinct. Essential distinctness is a bit less straightforward than I thought, since deleting finitely many factors from the Euler product of $\zeta(s - z_i)$ produces zeroes at $s = z_i$ which may cancel some of the poles from other factors. But it doesn't affect the order of the pole at $s = z_i + 1$, which is further to the right, so we can still consider the rightmost $z_i$'s and the corresponding poles. We get that if two monomials are essentially equal then the rightmost $z_i$'s which occur in each must match (with matching multiplicities) so we can factor these out and inductively conclude that all of the $z_i$ must match. -We should also get algebraic independence for a broader class of Dirichlet series (anything for which it's clear that we can still show essential distinctness), e.g. shifts of Dirichlet L-functions.<|endoftext|> -TITLE: Deligne's letter to Bhargava from March 2004 -QUESTION [19 upvotes]: I am quite interested in moduli spaces for Rings and Ideals, a letter from Deligne to Bhargava is cited in Melanie Wood's thesis Moduli spaces for Rings and Ideals (pdf), studying the minimal free resolution of $n$ points in $\mathbb P^{n−2}$. It's also cited in the thesis of Kevin H. Wilson, Three perspectives on $n$ points in $\mathbb{P}^{n-2}$ (link) (after Theorem 135): - -Such strong canonicity in these multiplicative structures has been implicitly used by many authors. For instance, Wright and Yukie [116] implicitly used this structure to parameterize rank $n$ rings over $\mathbb{Q}$. And Bhargava also implicitly relied on a much stronger -canonicity (over $\mathbb{Z}$) in structuring his Higher Composition Laws [9, 10, 11, 14]. Perhaps the place where the idea of using such canonical multiplicative structures was first developed was in Deligne’s letter to Bhargava [41] which inspired Wood’s vast generalization in her thesis -[114, 113, 112] extending Bhargava’s quadratic, cubic, and quartic Higher Composition Laws to arbitrary base schemes. - -Where can I find a copy of this letter (or a short explanation of the idea)? Thank you very much. - -REPLY [24 votes]: The letter is here. Thanks to Will Sawin for alerting me to this request.<|endoftext|> -TITLE: What is the inverse in K-theory represented by Clifford module extensions? -QUESTION [8 upvotes]: I am working on a model for topological KO-theory which is represented by explicit spaces of orthogonal Clifford module extensions. That is, assuming $M$ compact, $KO^{-n+1}(M) := [M,X_n]$ where the spaces $X_n$ are defined as follows: -Fix the background inner product space $\mathbb{R}^{\infty}$ with the standard inner product, and a background Clifford module structure, i.e., orthogonal operators $e_1,\ldots,e_n$ such that $e_i^2 = -I$ and $e_i e_j = -e_j e_i$ for $i \neq j$. By Bott periodicity it suffices to do this only up to $n=8$. Then let $X_n = \{f \in O(\mathbb{R}^\infty) | f^2 = -I, fe_k = -e_kf \text{ for }k < n, \dim\ker(f-e_n)^\perp < \infty\}$. -This is the space of operators that play the role of $e_n$, i.e. define extensions of the module structure given by $e_1, \ldots, e_{n-1}$, additionally required to be skew-symmetric and differ from $e_n$ in only finite dimensions. This turns out to give the correct spaces in the $KO$-spectrum. -Given two maps $f,g:M \to X_n$ we define their sum as follows: let -$\rho: \mathbb{R}^\infty \oplus \mathbb{R}^\infty \to \mathbb{R}^\infty$ -be a module isomorphism. We can choose a particular $\rho$ so that it "shuffles" together irreducibles. Then $f \boxplus g = \rho(f \oplus g)\rho^{-1}$ -is the map from $M$ to $X_n$ that represents the sum in k-theory. Morally it is just the pointwise direct sum. -My question is this: what operation should play the role of the inverse? That is, given a map -$f:M \to X_n$ -what is $\operatorname{Inv}(f)$ such that $f \boxplus \operatorname{Inv}(f)$ is homotopic to the constant map to $e_n$? I suspect that something like $\operatorname{Inv}(f) := -f$ ought to do it, but I can't nail down a proof (and $-f$ on its own violates the finite-dimension condition). I know this is a fairly specific setup but any hints at all on how to define the inverse for k-theory represented by spaces, especially in the context of Clifford modules, would be a great help. Please feel free to make any simplifying assumptions or be imprecise. -Note that this definition is similar to Karoubi's definition using what he calls "gradations", but he works with triples $(E,\eta_1,\eta_2)$ where $E$ is the background module and $\eta_1,\eta_2$ are gradations (analogous to extensions). I would like to find out what swapping $\eta_1$ and $\eta_2$ corresponds to in terms of a single map into a colimit of spaces of gradations. - -REPLY [2 votes]: For what it's worth, I did eventually get the answer. The idea is to notice that if $\overline{W}$ denotes the background Clifford module $W$ (above $\mathbb{R}^n$) endowed with the opposite module structure, i.e., where the operators $e_i$ act instead by $-e_i$, then for any map $f:M \to X_n(W)$, the map $f \oplus -f: X(n,W \oplus \overline{W})$ is contractible by the homotopy -$$ \begin{bmatrix}f & 0 \\ 0 & -f\end{bmatrix} \cos\left(\frac{\pi}{2}t\right) + \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} \sin\left(\frac{\pi}{2}t\right). $$ -Then one notices that, for instance, the operator $e_{n+1}$ gives an isomorphism $W \to \overline{W}$ since $e_{n+1}e_k = (-e_k)e_{n+1}$ for $k=1,\ldots,n$. So the inverse of $f$ is given by conjugating $-f$ by $e_{n+1}$: -$$\mathrm{Inv}(f) = e_{n+1} f e_{n+1}.$$ -Since the shuffle sum induces a homeomorphism $X(n,W) \simeq X(n,W \oplus W)$, $f \boxplus e_{n+1}fe_{n+1}$ is contractible.<|endoftext|> -TITLE: Why should I look at the resolvent formalism and think it is a useful tool for spectral theory? -QUESTION [14 upvotes]: Wikipedia calls resolvent formalism a useful tool for relating complex analysis to studying the spectra of a linear operator on a Banach space. Sure, I believe you because I've seen results that use the resolvent in the proof. I've also read bits of Kato's Perturbation theory for linear operators where he says it simplifies proofs. -I also recognize that the resolvent of $T$ encodes $T$'s eigenvalues as poles. -Granting all that, when should I look at a problem and think "oh, of course, I should translate this problem into a resolvent formulation", and more generally, is it just because of the aforementioned fact that the resolvent is useful? Or is there some intuition that I'm missing? - -REPLY [21 votes]: Preliminary remark. As mentioned in the comments, I find the notion "resolvent formalism", as well as the description in the Wikipedia article, rather misleading - resolvents are not somekind of formalism, and they are certainly not a mere "technique for applying complex analysis to spectral theory" (as claimed in the Wikipedia article). On the contrary, resolvents lie at the very foundation of spectral theory and are an integral part of it. Even the most basic results in spectral theory on Banach spaces - say e.g. that the spectrum of an operator is always closed - rely on the properties of the resolvent. -Answer to the question. I think the best way to understand why resolvents are so useful and important in spectral theory is to discuss several situations where resolvents occur. I do this in the following list. -Notational remark. I am going to use the sign convention $R(\lambda,A) := (\lambda-A)^{-1}$, in contrast to the Wikipedia article that defines the resolvent at $\lambda$ as $(A-\lambda)^{-1}$. The convention used in the Wikipedia article is, for instance, used in Kato's book; but aside from this book I am under the impression that the other convention is much more common in the literature. -Setting. Let $X$ be a complex Banach space and let $A: X \supseteq D(A) \to X$ be a closed linear operator. I will call $A$ bounded if $D(A) = X$ (in which case $A$ is automatically a bounded linear operator $X \to X$). -Resolvents as solution operators. We often want to solve linear equations, i.e. equations of the type -$$ - (*) \qquad -Ax = y. -$$ -If $0$ is not in the spectrum of $A$, then this equation has a unique solution which is given by $x = R(0,A)y$; so the resolvent of $A$ at the point $0$ is the solution operator to the equation $(*)$. -In many cases, we want to study more general equations than $(*)$, for instance the equation -$$ - (**) \qquad (\lambda - A)x = y -$$ -for a scalar $\lambda$. There are various reasons why we could be interested in equations of the type $(**)$; for instance: - -It is a very simple generalisation of the equation $(*)$, thus raising $(*)$ to a slightly higher level of complexity. - -Equations of the type $(**)$ occur in various situations; for instance, if you consider an inhomogenious linear ODE of first order, you will end up with an equation of the type $(**)$ where $\lambda = 1$. - -If $0$ is in the spectrum of $A$, then the equation $(*)$ cannot be uniquely solved, in general. So we might be interested in slightly perturbing $(*)$ with a simple perturbation and solve the perturbed equation, instead. (This would lead to $(**)$ with small values of $\lambda$). - - -If $A$ is a differential operator, then people who work in PDE might often call the solution operator of $(**)$ (or rather, its integral kernel) the Green function of the PDE - so from a PDE perspective resolvent is an operator theoretic notion for Green function. -Locating the spectrum. The spectrum $\sigma(A)$ of $A$ consists of those complex scalars $\lambda$ for which $\lambda - A: D(A) \to X$ is not bijective. -Determining the spectrum of an operator is often an important and very often a very difficult task. But if we cannot compute the spectrum in detail we would at least like to know some information about its location. -Now an obvious way to show that a number $\lambda$ is not in the spectrum is to write down a formula for the resolvent $R(\lambda,A)$ and to show that it is indeed inverse to $\lambda - A$. This is less absurd than it may seem at first glance; here are two classical situations where one does precisely this: - -Assume that $A$ is bounded and that $|\lambda| > \|A\|$. Then it follows that the so-called Neumann series -$$ - \sum_{k=0}^\infty \frac{A^k}{\lambda^{k+1}} -$$ -converges absolutely with respect to the operator norm, and that this operator is inverse to $\lambda - A$ (i.e. the Neumann series equals the resolvent $R(\lambda,A)$). This proves that no spectral value of $A$ has modulus strictly larger then $\|A\|$; in particlar, the spectrum of $A$ is bounded. - -If $\lambda$ is not in the spectrum of $A$, then one can use a similar series expansion argument to show that numbers which are sufficiently close to $\lambda$ are not in the spectrum of $A$, either. Hence, the resolvent set of $A$ is open and thus the spectrum is closed. - - -Another way to use resolvents in order to obtain information about the location of the spectrum is as follows: -It is not difficult to show that the norm of the resolvent $R(\lambda,A)$ explodes as $\lambda$ approaches to spectrum of $A$. This can be useful to show that certain numbers are spectral values. -For instance, such an argument is typically used to show that the spectral radius of a positive operator on an ordered Banch space is - under certain assumptions on the space - always a spectral value. In finite dimensions, this can be used in a proof of the so-called Perron-Frobenius theorem (but there also exist other finite-dimensional proofs of this fact which do not use the resolvent). -Resolvents and functional calculi This is essentially the point of view where we consider the resolvent $R(\lambda,A)$ as the result that we obtain when we "substitute" the operator $A$ into the function $a \mapsto \frac{1}{\lambda - a}$. But why shoud we be interested in doing this? -Well, by a functional calculus we mean a method to substitute our operator $A$ into a "sufficiently nice" function $f: \mathbb{C} \supseteq \Omega \to \mathbb{C}$. For instance, if $f$ is a polynomial, this is straightforward. If $f$ is entire and $A$ is bounded, this can also be done in a straightforward way by using the Taylor series expansion of $f$. -For more general holomorphic functions, it is illuminating to recall the Cauchy integral formula: if $f: \mathbb{C} \supseteq \Omega \to \mathbb{C}$ is holomorphic, $a$ is a point in $\Omega$ and $\gamma$ is a cycle in $\Omega$ that encloses $a$ precisely once, but does not enclose any point outside of $\Omega$, then -$$ - (C) \qquad f(a) = \frac{1}{2\pi i} \int_\gamma f(\lambda) \frac{1}{\lambda-a} \, d\lambda. -$$ -Now if we "formally" replace $a$ with $A$ in this formula, this suggests that we define the operator $f(A)$ as -$$ - (C') \qquad f(A) := \frac{1}{2\pi i} \int_\gamma f(\lambda) R(\lambda,A) \, d\lambda. -$$ -In the Cauchy integral formula we required that the path $\gamma$ encloses the point $a$ - which is the only point where the function $\lambda \mapsto \frac{1}{\lambda - a}$ is not defined - exactly once. Similary, we shall assume for $(C')$ that the path $\gamma$ encloses each point $a$ where the function $\lambda \mapsto R(\lambda,A)$ is not defined - i.e., each spectral value of $A$ - exactly once. -Of course, as discussed here this is just heuristics. But it turns out that this definition of the holomorphic functional calculus works very well and has good properties. -Spectral projections. This can be seen as an application of the holomorphic functional calculus. For the sake of simplicity, assume that $A$ is bounded. -If $A$ is a finite-dimensional matrix, the underlying space $\mathbb{C}^d$ can be decomposed into generalized eigenspaces of $A$; this yields the well-known Jordan normal form of $A$. But for operators on infinite-dimensional spaces, we do not have such a normal form, in general. However, we can sometimes still decompose our operator $A$ into "simpler" operators: -To this end, assume that the spectrum of $A$ is not connected, i.e. it can be written as $\sigma(A) = \sigma_1 \cup \sigma_2$ for two non-empty and compact disjoint sets $\sigma_1$ and $\sigma_2$. Then we can find two open sets $\Omega_1$ und $\Omega_2$ which do not intersect and such that $\Omega_1$ contains $\sigma_1$ and $\Omega_2$ contains $\sigma_2$. Since our two open sets do not intersect, the indicator function $\mathbb{1}_{\Omega_1}$ is a holomorphic mapping on the union $\Omega := \Omega_1 \cup \Omega_2$. -Hence, we can use the functional calculus to define the operator $P := \mathbb{1}_{\Omega_1}(A)$, and similarly, $Q := \mathbb{1}_{\Omega_2}(A)$. -By using standard properties of the functional calculus one gets -$$ - P^2 = \mathbb{1}_{\Omega_1}(A) \mathbb{1}_{\Omega_1}(A) = \Big(\mathbb{1}_{\Omega_1} \mathbb{1}_{\Omega_1}\Big)(A) = \mathbb{1}_{\Omega_1}(A) = P, -$$ -and similarly, $Q^2 = Q$. Hence, $P$ and $Q$ are projections. Moreover, they sum up to -$$ - P + Q = \Big(\mathbb{1}_{\Omega_1} + \mathbb{1}_{\Omega_2}\Big)(A) = \mathbb{1}_{\Omega}(A) = \operatorname{id}_X, -$$ -so $P$ and $Q$ are even complementary projections. Both of them commute with $A$, so the decomposition -$$ - X = PX \oplus QX -$$ -reduces $A$. The projections $P$ and $Q$ are the so-called spectral projections of $A$ associated with $\sigma_1$ and $\sigma_2$. Their connection to resolvents is given by the functional calculus which we used to define $P$ and $Q$: for instance, -$$ - P = \frac{1}{2\pi i} \int_\gamma \mathbb{1}_{\Omega_1}(\lambda) R(\lambda,A) \, d\lambda -$$ -for an appropriately chosen path $\gamma$. -Evolutions equations: the Hille-Yosida theorem. Let $A$ be densely defined, by which we mean that the domain $D(A)$ is dense in $X$. Consider initial value problem -$$ - (IVP) \quad - \begin{cases} - \dot x(t) & = Ax(t) \quad \text{for } t \ge 0, \\ - x(0) & = x_0 - \end{cases} -$$ -for an intial vector $x_0 \in X$. The Hille-Yosida theorem say that the evolution problem (IVP) is well-posed (in a certain sense) for each $x_0 \in X$ if and only if the resolvent $R(\lambda,A)$ exists for all $\lambda$ of sufficiently large real part and its powers satisfy a certain growth estimate. -In other words: In order to show well-posedness of the evolution problem (IVP) we need to study the stationary problem $(\lambda - A)x = y$ and estimate its solutions. -So the resolvent occurs when we study well-posedness of evolution equations. -The resolvent as Laplace transform. This is related to the previous point. If $A$ is the generator of a $C_0$-semigroup $(T(t))_{t \ge 0}$ on $X$ (which means precisely that the evolution problem (IVP) is well-posed for each initial value), then for all $\lambda$ with sufficiently large real part the formula -$$ - R(\lambda,A)x = \int_0^\infty e^{-\lambda t} T(t)x \, dt -$$ -holds for each $x \in X$. This mean that the resolvent of $A$ is the Laplace transform of the semigroup. As a consequence of this observation, one can use so-called Tauberian theorems to obtain information about the long-term behaviour of the semigroup from information about the resolvent $R(\lambda,A)$. -An algebraich point of view: The resolvent identity. The resolvent of $A$ satisfies the so-called resolvent identity -$$ - (RI) \quad (\mu - \lambda) R(\lambda,A) R(\mu,A) = R(\mu,A) - R(\lambda,A) -$$ -for all scalars $\lambda, \mu$ outside the spectrum of $A$. -Now one can go one step further and call a mapping $R$ from a non-empty open subset of $\mathbb{C}$ into the space of bounded linear operators on $X$ a pseudo-resolvent if it satisfies the resolvent identity (RI). -If the operator $R(\lambda): X \to X$ is injective for one number $\lambda$ (in which case it is automatically injective for all further numbers $\lambda$, too), then one can show that there exists an operator $A$ such that $R(\lambda) = R(\lambda,A)$ for all $\lambda$. -If none of the operators $R(\lambda)$ is injective, one can still interpret $R$ as a resolvent - but of a so-called multi-valued operator rather than of a usual operator. -I do not discuss any specific applications of pseudo-resolvents here (there are applications of them!), but I thought it might be a good idea to mention them here in order to give yet one more perspective that one can take when looking at resolvents.<|endoftext|> -TITLE: Different ways to prove $L^p$-estimates for the heat equation -QUESTION [5 upvotes]: Let $p \in (1,\infty)$. We are interested in strong $L^p$-solutions to the heat equation in $\mathbb{R}^n$. -$$ -\begin{cases} - \partial_t u = \Delta u + f \\ -u(0) = 0. -\end{cases} -$$ -It is well-known that for all $f \in L^p((0,\infty;L^p(\mathbb{R}^n))$ there exists a unique function $u \in H^{1,p}((0,\infty);L^p(\mathbb{R}^n)) \cap L^p((0,\infty);H^{2,p}(\mathbb{R}^n))$, -which solves the heat equation and satisfies the estimate -$$ \| \partial_t u \|_{p} + \|\Delta u \|_{p}\le C \|f \|_{p} $$ -for some constant $C>0$. -I am interested in different ways to prove this. To start the discussion let me name two different methods. - -Using the theory of singular integrals applied to the solution formula given by means of the fundamental solution shows that the solutions operator is $L^p$-bounded. - -Fourier transformation in time and space gives $\hat{u} = \frac{{|\xi|}^2}{i \omega + |{\xi}|^2}\hat{f}$. Applying Mikhlin's multiplier theorem gives the desired estimate. - - -Do you know of any other methods to prove this? If so, feel free to extend the list. - -REPLY [3 votes]: I would say that methods 1) and 2) are very close each other but in both cases the proof is a bit harder than the elliptic counterpart since one has to use the Marcinkiewicz multiplier theorem instead of Mikhlin-Hormander. -There are a couple of similar approaches with some simplifications. One is in the book of N. Krylov: Elliptic and Parabolic equations in Sobolev spaces and relies on estimates of the sharp function of Fefferman-Stein. -The other one I know is based on an interpolation result originally due to Z. Shen, which can be found in the first of 4 papers by P. Auscher and J.M Martell. In few words if $Tf=D_t u$, where $D_t u-\Delta u=f$, it suffices to bound the $L^p$ means of $Tf$ over cubes with the corresponding $L^2$ means over the double cube, whenever $f$ vanishes on a bigger cube. This gives boundedness of $T$ in $L^p$, $p>2$, having at hands that in $L^2$. For PDE this is quite manageable since the criterion follows from interior estimates of homogenuous problems.<|endoftext|> -TITLE: Formal completion of a quotient stack -QUESTION [6 upvotes]: $\newcommand{\Rep}{\operatorname{Rep}}$ -$\newcommand{\mo}{\operatorname{-mod}}$ -$\renewcommand{\hat}{\widehat}$ -I apologize in advance if this is a naive question but my background in algebraic geometry is fairly superficial. I mostly care about global quotients $X/G$ where $X$ is an affine scheme over $\mathbb C$ and $G$ a complex connected affine algebraic (reductive if you like) group. My understanding of those is pretty much limited to the fact that we have an equivalence of symmetric monoidal categories -$$QC(X/G)\simeq O(X)\mo_{\Rep G}$$ -where $O(X)$ is the algebra of global functions, and $\Rep G$ the category of $O(G)$-comodules. -Let $x \in X$ be a fixed point of the $G$ action. In a nutshell my question is: - -What is the correct definition of the formal completion of $X/G$ at $x$ ? In particular what is its category of quasi-coherent sheaves thinking of it as an "ordinary" rather than formal stack (f that makes sense) ? - -A basic observation is that $\hat O(X)$, the completion of $O(X)$ by the ideal of functions vanishing at $x$, is not an object in $\Rep G$. Now it seems there are different things one can do: - -Look at the category $\hat O(X)\mo_{\Rep \mathfrak g}$, which I guess should be like quasi-coherent sheaves on $\hat X/\hat G$ -Think of $\hat O(X)$ as a topological algebra, hence as an object in a certain category of topological $G$-representation (say the pro-completion of the category of finite dimensional $G$-modules). -We can look at the coalgebra $C(X)$ of "distributions supported at $x$", i.e. the coalgebra which satisfies $C(X)^*=\hat O(X)$, which is a a coalgebra in $\Rep G$ so that you take take comodules over it.. This is the idea that formal affine scheme are the same as "cospectrum" of cocommutative coalgebras, and I think the category you get is equivalent to the one in 2 by taking duals. -Although $\hat O(X)$ is not an object in $\Rep G$, it still makes sense to look at modules for this algebra that happens to be in there, i.e. $\hat O(X)\mo_{\Rep G}$ do makes sense. - -Is any of those the, or a, correct definition ? Any insight or reference would be much appreciated. - -REPLY [5 votes]: I will assume $X$ is smooth for simplicity, but it is probably not needed. Given the stack $X/G$, there are two completions one may consider: - -Completing along $\mathrm{B}G\rightarrow X/G$ one obtains $\hat{X}/G$. -Completing along $\mathrm{pt}\rightarrow X/G$ one obtains $\hat{X}/\hat{G}$. - -Your next question is about quasi-coherent sheaves. I will assume your definition of quasi-coherent sheaves on a prestack is given by the right Kan extension from affines. In particular, $\mathrm{QCoh}$ sends colimits of prestacks to limits of categories. -Let me begin with $\mathrm{QCoh}(\hat{X})$. By definition, $\hat{X}$ is a colimit $\mathrm{colim} X_\alpha$ of affines. So, $\mathrm{QCoh}(\hat{X})=\lim \mathrm{QCoh}(X_\alpha)=\lim \mathrm{Mod}_{\mathcal{O}(X_\alpha)}$. If $\mathcal{O}(\hat{X})$ is the corresponding topological algebra, this limit may be identified with the category of complete $\mathcal{O}(\hat{X})$-modules. Also, since $\mathcal{O}(X_\alpha)$ are finite-dimensional, you can rewrite it as $\lim \mathrm{CoMod}_{\mathcal{O}(X_\alpha)^*}$. So, you can identify this category with the category of comodules over the coalgebra of distributions $\mathrm{Dist}(\hat{X})$. (The inclusion of the structure sheaf $p^*\colon\mathrm{Vect}\rightarrow \mathrm{QCoh}(\hat{X})$ admits a left adjoint $p_+\colon \mathrm{QCoh}(\hat{X})\rightarrow \mathrm{Vect}$ and $\mathrm{Dist}(\hat{X})=p_+\mathcal{O}_X$.) -Next, $\mathrm{QCoh}(\mathrm{B}\hat{G})$. Let $i\colon \mathrm{pt}\rightarrow\mathrm{B}\hat{G}$ be the inclusion of the basepoint. The pullback functor $i^*\colon \mathrm{QCoh}(\mathrm{B}\hat{G})\rightarrow \mathrm{Vect}$ does not have a colimit-preserving right adjoint. Instead, it has a left adjoint. One can show that $i^*\colon \mathrm{QCoh}(\mathrm{B}\hat{G})\rightarrow \mathrm{Vect}$ is monadic and identifies $\mathrm{QCoh}(\mathrm{B}\hat{G})\cong \mathrm{Mod}_{\mathrm{U}\mathfrak{g}}$. -(Here is a quick proof on the derived level. By Theorem 10.1.1 in https://arxiv.org/abs/1108.1738 $\Upsilon\colon\mathrm{QCoh}(\mathrm{B}\hat{G})\rightarrow \mathrm{IndCoh}(\mathrm{B}\hat{G})$ is an equivalence since $G$ is smooth. And $\mathrm{IndCoh}(\mathrm{B}\hat{G})=\mathrm{Mod}_{\mathrm{U}\mathfrak{g}}$ by Proposition 2.4.31 in https://www.math.ias.edu/~lurie/papers/DAG-X.pdf.) -Combining the two equivalences, you get -$$\mathrm{QCoh}(\hat{X}/\hat{G}) = \mathrm{CoMod}_{\mathrm{Dist}(\hat{X})}(\mathrm{Mod}_{\mathrm{U}\mathfrak{g}}),\qquad \mathrm{QCoh}(\hat{X}/G) = \mathrm{CoMod}_{\mathrm{Dist}(\hat{X})}(\mathrm{Rep}(G)).$$<|endoftext|> -TITLE: Enrichment as extra structure on a category -QUESTION [7 upvotes]: We will suppose, for the sake of simplicity, that everything is happening within a fixed 'metacategory' $\textbf{SET}$ of sets and functions. So, from now on, a 'category' just means a category object in $\textbf{SET}$ - i.e. a small category. -Let $\mathscr{V}$ be a monoidal category. A $\mathscr{V}$-enriched category $\mathscr{C}$ consists of: - -Objects: A set Ob($\mathscr{C}$). -Morphisms: For each pair of $\mathscr{C}$-objects $(X, Y)$, a $\mathscr{V}$-object Hom$(X, Y)$. -Composition: For each triple of $\mathscr{C}$-objects $(X, Y, Z)$, a $\mathscr{V}$-morphism $\circ$ : Hom$(X, Y)$ $\otimes$ Hom$(Y, Z)$ $\rightarrow$ Hom$(X, Z)$. -Identities: For each $\mathscr{C}$-object $X$, a $\mathscr{V}$-morphism id$_X$: $I$ $\rightarrow$ Hom$(X, X)$ (where $I \in \mathscr{V}$ is the unit of $\otimes$). - -This data is then subject to the usual associativity and unitality axioms which are expressed via the commutativity of certain diagrams in $\mathscr{V}$. -From this enriched category, we can extract an underlying category $\mathscr{C}_0$ by defining $\mathscr{C}(X, Y) = \mathscr{V}(I, \text{Hom}(X, Y))$. -My question is about if this is reversible - namely, can we define a $\mathscr{V}$-enriched category to be a category $\mathscr{C}$ equipped with a 'hom-functor' to $\mathscr{V}$? I'm having some trouble finding a reference for this but it seems like there should be a fairly obvious definiton. -A $\mathscr{V}$-atlas on a category $\mathscr{C}$ consists of: - -Morphisms: A functor Hom: $\mathscr{C}^{op} \times \mathscr{C} \rightarrow \mathscr{V}$. -Composition: For each triple of $\mathscr{C}$-objects $(X, Y, Z)$, a $\mathscr{V}$-morphism $\circ$ : Hom$(X, Y)$ $\otimes$ Hom$(Y, Z)$ $\rightarrow$ Hom$(X, Z)$. -Parametrisation: For each pair of $\mathscr{C}$-objects $(X, Y)$, an isomorphism $\eta: \mathscr{C}(X, Y) \xrightarrow{\sim} \mathscr{V}(I, \text{Hom}(X, Y))$ such that for all $X \xrightarrow{f} Y \xrightarrow{g} Z$ in $\mathscr{C}$, $\eta(g \circ f) = \eta(g)\circ\eta(f)$ (where on the left we have compositon in $\mathscr{C}$ and on the right we have composition in $\mathscr{V}$). - -I'm unsure though if this gives associativity and unitality as in the usual defintion of a $\mathscr{V}$-enriched category, or if we only get associativity and unitality for $I$-shaped elements of the hom-objects. Could this be remedied by just requiring the associativity and unitality laws to hold as in the usual definiton? Any help or references would be much appreciated. - -REPLY [7 votes]: When your enriched categories are bicomplete enough (specifically, tensored and -cotensored over $\mathscr{V}$), you can view the extra structure of -the enrichment as a kind of action of $\mathscr{V}$ on them: this is -called a closed $\mathscr{V}$-module in Definition 10.1.3 of Riehl's -Categorical homotopy -theory -(with comparison in Proposition 10.1.4). The point is that the -adjunction between the tensors and the internal hom (which is a better-behaved form of your "parametrisation") will allow you to -formulate the associativity and unitality for the hom composition very nicely in -terms of associativity of the action. -If you want to consider more general (not necessarily co/tensored) -$\mathscr{V}$-enriched categories, you can pass to a weaker form of -enrichment by relaxing the closed $\mathscr{V}$-module structure to a -simple (weak) $\mathscr{V}$-module structure (viewing $\mathscr{V}$ as -a weak monoid in the monoidal $2$-categories of categories); this -corresponds to enriching over the category of presheaves on -$\mathscr{V}$. Then the $\mathscr{V}$-enrichment is a condition of -representability of the action. -I do not know a reference for this story for $1$-categories, but it is -essentially what I understand of the construction in Definitions -4.2.1.25 and 4.2.1.28 of Lurie's Higher -algebra -and the explanations in the introduction of Heine's "An equivalence -between enriched $\infty$-categories and $\infty$-categories with weak -action" which compares these two -points of view on enriched $(\infty,1)$-categories.<|endoftext|> -TITLE: Maximum number of common zeros of n polynomials in n-1 variables -QUESTION [11 upvotes]: Given $n$ quadratic polynomials in $n-1$ variables over the complex field, what is the maximum number of common zeros? Can we have $2^{n-1}-1$ common zeros? I assume that a linear combination of the polynomials is always different from zero and the number of zeros is finite. -With $4$ polynomials, the maximum is not smaller than $6$. Using a projective space $(x_1,x_2,x_3,x_4)$, an example with $6$ roots is given by the polynomials. -$P_1=x_1 x_2$, -$P_2=x_1 x_3$, -$P_3=L_1 x_2+L_2 x_3$, -$P_4=(\text{a general quadratic polynomial})$, -$L_k$ being general linear polynomials. -Indeed, if $x_1=0$, then the first two polynomials are equal to zero and the remaining two polynomials in $x_2, x_3, x_4$ give four roots, which are distinct for a general choice of $L_k$ and $P_4$. If $x_2=x_3=0$, then the first three polynomials are equal to zero and the fourth polynomial in $x_1,x_4$ gives 2 additional roots. -This construction has a natural generalisation to $n$ polynomials, giving $2^{n-2}+2^{n-3}$ roots, which is about $3/4$ of the desired bound $2^{n-1}-1$. - -REPLY [14 votes]: There is a bound for the multiplicity of a (homogenous) almost complete intersection in Theorem 1 of this paper by Engheta. In case of $n$ quadrics in $n-1$ variables, that bound is $2^{n-1}-(n-2)$. So for $n\geq 4$, you can not get $2^{n-1}-1$. -(In Theorem 1 there was a condition that the first $n-1$ generators form a complete intersection, but I don't think that this is serious in your case, one could always make a linear change to turn the first $n-1$ generators into a regular sequence, as the whole ideal has maximal height) -There are further improvements and examples of Engheta's bound in this work and this very recent work.<|endoftext|> -TITLE: $\operatorname{SL}_2(k)$ invariant polynomials in $k[x_1,x_2,y_1,y_2]$ -QUESTION [8 upvotes]: Let $k$ be a field and let $\operatorname{SL}_2(k)$ act on $k[x_1,x_2]$ and $k[y_1,y_2]$ in the usual ways. These actions induce an action on the tensor product $k[x_1,x_2,y_1,y_2]$ that preserves the subspace $k[x_1,x_2,y_1,y_2]_{s,k}$ of polynomials that are homogeneous of degree $s+k$ with total $x_i$ degree $s$ and total $y_i$ degree $k$. I think these are sometimes said to be of bidegree $(s,k)$, but I'm not entirely sure that's standard terminology. -A computation I've performed in a seemingly unrelated mathematical field has led me to believe that for all $d \geq 0$, there should be a nonzero $\operatorname{SL}_2(k)$-invariant polynomial in $k[x_1,x_2,y_1,y_2]_{d,d}$ that is unique up to scaling. -Question: Assuming I'm right, how can I go about writing this polynomial down explicitly? - -REPLY [8 votes]: The polynomial you gave in the comments, $x_1y_2 - y_2 x_1$, after correcting the typo to $x_1 y_2 - x_2 y_1$, is invariant under $\operatorname{SL}_2$. -Proof: It's the determinant of -$$ \begin{pmatrix} x_1 & y_1 \\ x_2 & y_2 \end{pmatrix}$$ and determinants are invariant under left multiplication by matrices of determinant $1$. -It indeed generates the ring of invariants. You can check this using representation theory (bidegree $s, k$ polynomials form the representation $\operatorname{Sym}^s \otimes \operatorname{Sym}^k$ of $\operatorname{SL}_2$, and because $\operatorname{Sym}^j$ is irreducible this has one invariant if $s=k$ and $0$ otherwise) or by observing that any two nonzero matrices with the same determinant are equal up to the action of $\operatorname{SL}_2$. -The same idea can be used to find the $\operatorname{SL}_n$-invariants in the tensor product of $n$ copies of $k[x_1,\dotsc,x_n]$.<|endoftext|> -TITLE: Number of bounded Dyck paths with "negative length" -QUESTION [10 upvotes]: Let $c(n,k)$ denote the number of Dyck paths of semilength $n$ which are contained in the strip $0 \leq y \leq 2k + 1.$ -They satisfy the recursion $\sum_{j=0}^{k+1}(-1)^j \binom{2k+2-j}{j}c(n-j,k)=0$ for $n>k.$ -We can extend the sequence to negative $n$ such that this recursion holds for all $n \in \mathbb{Z}.$ -I am interested in the generating function of the sequence ${\left( {c( - n,k)} \right)_{n \geq 0}}.$ -It is well known that $\sum\limits_{n \geq 0} {c(n,k){x^n}} = \frac{{{F_{2k + 1}}( - x)}}{{{F_{2k + 2}}( - x)}}$ if by ${F_n}(x) = \sum\limits_{j = 0}^{\left\lfloor {\frac{n}{2}} \right\rfloor } \binom{n-j}{j} -x^j $ we denote the Fibonacci polynomials which satisfy ${F_n}(x) = {F_{n - 1}}(x) + x{F_{n - 2}}(x)$ with initial values $F_0(x)=F_1(x)=1.$ -Computations for small $k$ suggest that $\sum\limits_{n \geq 0} {c( - n,k){x^n}} = - \frac{1}{x}\frac{{{F_{2k}}( - \frac{1}{x})}}{{{F_{2k + 2}}( - \frac{1}{x})}}.$ -As mentioned in OEIS A080937 and A038213 for $n=2$ this result is due to Michael Somos. -These generating functions imply that $c(n,k)$ satisfies the recursion for $\left| n \right| > k.$ -But to show that $c(-n,k)$ is the looked for extension we need the recursion for all $n$. Any idea how to do this? - -REPLY [9 votes]: If $f(n)$ satisfies a linear recurrence with constant coefficients for all $n\in \mathbb{Z}$ and we set $F(x)=\sum_{n\geq 0} f(n)x^n$, then $\sum_{n\geq 1}f(-n)x^n = -F(1/x)$ (as rational functions). See Enumerative Combinatorics, vol. 1, second ed., Prop. 4.2.3. -Addendum. Using Exercise 3.66(d) in Enumerative Combinatorics, -vol. 1, second ed., it is not hard to show that $c(-n,k)$ is equal -to the number of sequences $(a_1,a_2,\dots,a_{2n-1})$ of positive -integers satisfying $1\leq a_i\leq k+1$ and $a_1\leq a_2 \geq a_3 \leq - a_4 \geq \cdots\geq a_{2n-1}$.<|endoftext|> -TITLE: Why the least action principle is always (?) used in this particular form? -QUESTION [9 upvotes]: The least action principle in (mathematical) physics says the following. Given a system, e.g. collection of particles, whose motion satisfies a known system of differential equations (of second order). Then there exists a so called action functional $S$ on the space of paths of all the particles such that solutions of the above differential equations are precisely the critical paths of $S$. Moreover in all known to me cases it is assumed that $S$ can be chosen in the form $$S=\int L(x,\dot x,t)dt,\,\,\,(1)$$ -where $L$ is called Lagrangian. -The least action principle is satisfied in this form for many problems of interest in physics, but not for arbitrary system of differential equations. - -Question. Why it is important that $S$ has the form (1) for some Lagrangian? Are there situations of interest in (mathematical) physics where the action functional is not given by any Lagrangian? - -Remark. If one takes the action functional in the form -$$S_1:=\exp(S)=\exp(\int L(x,\dot x,t)dt)$$ -then $S_1$ and $S$ have obviously the same critical paths. - -REPLY [11 votes]: In the form (1), if you compute the variation $\delta S / \delta x(t) = E(t)$, you find that $E(t) = E(x(t),\dot{x}(t), \ddot{x}(t) ,t)$ is a local/differential expression (the value of $E(t)$ does not depend on $x(t')$ or its derivatives at other times $t'\ne t$). This is no longer true if you use $\exp(S)$ instead of $S$. There is no dispute that $S$ and $\exp(S)$ have the same critical points (N.B.: first order variations cannot distinguish between critical points of different kinds like maxima, minima, or saddle points). But if like your $\delta S / \delta x(t)$ to be local (and some people do), then you are stuck with local action functionals, namely those in the form (1). -UPDATE: The proof that locality of $E(t)$ implies locality of $S$ is straightforward, essentially an application of the fundamental theorem of calculus. Morally, $E(x(t), \dot{x}(t), \ddot{x}(t), t)$ is the gradient of $S$ with respect to $x$. Conversely, $S$ is the primitive/anti-derivative of $E(t)$, and any two such primitives must differ by a constant. One primitive can be constructed by preserving locality: $$S = \int \left(\int_0^1 x(t) E(s x(t), s\dot{x}(t), s\ddot{x}(t), t) ds \right) dt,$$ where the expression in parentheses is known as the Vainberg-Tonti Lagrangian (Google the keywords for references). So all other primitives must differ by a constant. There may be some funny ways to express a constant which may not appear local in the way we have been discussing, but such non-locality can be dismissed as trivial. This discussion has obvious generalizations to more dependent and independent variables, as well as higher differential orders.<|endoftext|> -TITLE: Is the Chevalley-Eilenberg cohomology the only interesting cohomology for Lie algebra? -QUESTION [7 upvotes]: When talking about the cohomology space of a Lie algebras, it comes naturally to refer to the Chevalley-Eilenberg cohomology, is there other interesting type of cohomology for Lie algebra? - -REPLY [6 votes]: No, it's not the only one. For instance, Leibniz cohomology is interesting for Lie algebras themselves. See this answer of mine for references.<|endoftext|> -TITLE: Quaternionic and octonionic analogues of the Basel problem -QUESTION [33 upvotes]: I asked this question in MSE around 3 months ago but I have received no answer yet, so following the suggestion in the comments I decided to post it here. -It is a well-known fact that -$$\sum_{0\neq n\in\mathbb{Z}} \frac{1}{n^k} = r_k (2\pi)^k$$ -for any integer $k>1$, where $r_k$ are rational numbers which can be given explicitly in terms of Bernoulli numbers. For example, for $k=2$ the sum equals $\pi^2/3$ (this is essentially the Basel problem), and for $k=4$ it equals $\pi^4/45$. Note that for odd $k$ the sum vanishes. -The theory of elliptic curves with complex multiplication allows us to extend this result to systems of complex integers such as the Gaussian integers, or more generally the ring of integers in an imaginary quadratic number field of class number 1. Namely, for $k>2$ we have -$$\sum_{0\neq \lambda\in\mathbb{Z[\omega]}} \frac{1}{\lambda^k} = r_k \varpi^k,$$ -where again $r_k$ are rational constants and $\varpi \in \mathbb{R}$ (the "complex $2\pi$") depends only on the ring $\mathcal{O}=\mathbb{Z[\omega]}$ and is an algebraic multiple of a so-called Chowla–Selberg period, given by a product of powers of certain gamma factors (note that the sum is always a real number since it is invariant under conjugation). For example, for the Eisenstein ($\omega = (1+\sqrt{3} i)/2$), Gaussian ($\omega = i$) and Kleinian ($\omega = (1+\sqrt{7} i)/2$) integers, we have respectively -$$\varpi_3 = 3^{-1/4} \sqrt{2\pi} \left(\frac{\Gamma(1/3)}{\Gamma(2/3)}\right)^{3/2}, \quad \varpi_4 = 4^{-1/4} \sqrt{2\pi} \left(\frac{\Gamma(1/4)}{\Gamma(3/4)}\right), \quad \varpi_7 = 7^{-1/4} \sqrt{2\pi} \left(\frac{\Gamma(1/7)\Gamma(2/7)\Gamma(4/7)}{\Gamma(3/7)\Gamma(5/7)\Gamma(6/7)}\right)^{1/2}.$$ -For higher class numbers there is a similar formula, though in that case $r_k$ will in general not be rational but algebraic. A nice exposition of this result can be found in Section 6.3 of Zagier - Elliptic Modular Forms and Their Applications. - -My question is whether this is still true for hypercomplex number systems, such as the Hurwitz integers or the octonionic integers. Define $$S_k[\mathcal{O}] = \sum_{0\neq \lambda\in\mathcal{O}} \frac{1}{\lambda^k}$$ for $k>\operatorname{dim} \mathcal{O}$, where $\mathcal{O}$ is now an order in a totally definite rational quaternion/octonion algebra of class number 1. The restriction on $k$ is so that the sum converges absolutely. - -Subquestion 1: Do we have $S_k[\mathcal{O}] = r_k \varpi^k$ for some rational sequence $r_k$ and some real number $\varpi$ depending only on $\mathcal{O}$ (a "quaternionic/octonionic $2\pi$")? - -Obviously $\varpi$ will only be defined up to a nonzero rational factor. An equivalent question is whether $(S_m[\mathcal{O}])^n/(S_n[\mathcal{O}])^m$ is rational for any $m, n$ such that $S_n[\mathcal{O}]\neq 0$. - -Subquestion 2: If so, can (some fixed choice of) $\varpi$ be expressed in terms of -known constants such as $\zeta'(-1)$ or $\zeta'(-3)$? - -The reason I'm mentioning these particular constants is that in the previous cases (real and complex) the period $\varpi$ turns out to be equal to $e^{-\zeta'(\mathcal{O},0)/\zeta(\mathcal{O},0)}$ up to an algebraic factor, where the zeta function attached to the ring of integers $\mathcal{O}=\mathbb{Z}$ or $\mathbb{Z[\omega]}$ is defined as -$$\zeta(\mathcal{O},s) = \sum_{0\neq \lambda\in\mathcal{O}} |\lambda|^{-s}.$$ -(This is in general not the same as the previous sums, note the absolute value.) In the case that $\mathcal{O}$ is instead a quaternionic or octonionic order, the logarithmic derivative of this zeta function at $s=0$ can be expressed in terms of $\zeta'(-1)$ or $\zeta'(-3)$ respectively, where $\zeta(s)$ is the ordinary Riemann zeta function. - -I calculated a few sums numerically for the ring of Hurwitz quaternions. The result is $$S_6[\mathcal{O}] \approx 10.76,\quad S_8[\mathcal{O}] \approx 1.196,\quad S_{12}[\mathcal{O}] \approx 23.9905.$$ -Unfortunately the calculations take a lot of time, and the precision is not enough to determine whether e.g. $S_{12}[\mathcal{O}]/(S_6[\mathcal{O}])^2$ is rational to any degree of confidence. -I also found the recent paper Period Relations for Quaternionic Elliptic Functions by Z. Amir-Khosravi that refers to previous works by R. Fueter (Über die Quaternionenmultiplikation regulärer vierfachperiodischer Funktionen) and R. Krausshar (Generalized Analytic Automorphic Forms in Hypercomplex Spaces). A certain $3$-parameter family of quaternionic Eisenstein-like functions associated to a lattice in $\mathbb{R}^4$ is introduced, and shown to enjoy period-like relations resembling those in the complex case. Unfortunately, the form of these functions is restricted by quaternionic regularity to contain factors of the quaternionic norm (c.f. equations (2.5)-(2.7) in the paper), and as far as I can see they aren't directly related to the sums of pure powers I'm interested in. - -REPLY [22 votes]: This isn't really a full answer, but it's too long for a comment, and perhaps it's informative all the same. -Your sum $S_k[\mathcal{O}]$ can be written as the value at $s = k$ of the sum -$$\sum_{0 \ne \lambda \in \mathcal{O}} \frac{\lambda^k}{Nm(\lambda)^s} = \sum_{n \ge 1} a^{(k)}_n n^{-s},$$ -where $a^{(k)}_n := \sum_{N(\lambda) = n} \lambda^k$. -Now, I claim that $\sum_{n \ge 1} a^{(k)}_n q^n$ is the $q$-expansion of a modular form -- or something slightly more general, namely a quasi-modular form [*] -- of weight $k + 2$ and some level depending on $\mathcal{O}$; for the Hurwitz integers the level is $\Gamma_0(2)$. This should follow from thinking about Brandt matrices, which are a way of computing modular forms using quaternion algebras; see e.g. this article by Kimball Martin. -Anyway, once you know what to look for, it's now quite easy to recognise the sequences $(a^{(k)}_n)_{n \ge 1}$ for small $k$. For example, when $k = 6$, what you get is exactly the $q$-expansion of $12f_8$, where $f_8$ is the unique normalised modular cusp form of weight 8 and level 2. So $S_6[\mathcal{O}]$ is a value of the $L$-series of a modular form. In fact, we have $S_6[\mathcal{O}] = 12 L(f_8, 6) = 10.758540419274832757072...$, which agrees with your computations above. Similarly, unless I've slipped up in my computations, we have -$$S_8[\mathcal{O}] = 12 \big( L(f_{10}, 8) - L(f_8, 7) \big) = 1.18636076594110...$$ -where $f_{10}$ is the cusp form of weight 10. Since the periods of $f_{10}$ and $f_{8}$ have essentially nothing to do with each other, this strongly suggests that there is no tidy algebraic relation between $S_6[\mathcal{O}]$ and $S_8[\mathcal{O}]$. -[*] Quasi-modular forms are not too scary: they're exactly the ring of functions you get by starting with genuine modular forms and throwing in the function $E_2 = 1 - 24\sum \sigma(n) q^n$. - -EDIT. Further numerical experiments suggest the following explicit formula: if $\mathfrak{S}(m)$ denotes the set of normalised newforms of level 2 and weight $m$, then for every $k \ge 6$ we seem to have -$$S_k[\mathcal{O}] = 12\left( \sum_{f \in \mathfrak{S}(k+2)} L(f, k) - \sum_{f \in \mathfrak{S}(k)} L(f, k-1) \right).$$<|endoftext|> -TITLE: What's the dimension of the Lie algebra generated by transpositions on $n$ objects? -QUESTION [12 upvotes]: Define a Lie bracket on the group algebra of the permutation group $S_n$ in the following way: -$$[\sigma, \tau] = \sigma\circ\tau - \tau\circ\sigma,$$ -where $\sigma, \tau \in S_n$, and the multiplication on permutations is defined as composition. My question is, what is the dimension of the Lie subalgebra generated by transpositions, i.e. $(ij)$? My conjecture is that the dimension is given by $C_n - \lfloor \frac{n}{2} \rfloor$, where $C_n$ is the Catalan number. Is this correct and what is the proof? -For example, when $n=3$, using the cycle notation, we have -$$ -[(12),(23)] = (132) - (123) \\ -[(23),(31)] = (132) - (123) \\ -[(31),(12)] = (132) - (123) \\ -$$ -and -$$ -[(12), (132) - (123)] = 2((23) - (13)), \text{etc.} -$$ -Therefore this algebra is $4 = C_3 - 1$ dimensional. -If this conjecture is correct, the result should not be hard to generalize to the Lie subalgebras generated by $S_k$ for $k -TITLE: What are these recursively defined sequences called? -QUESTION [6 upvotes]: Let $F(x,y)$ be a function of two variables, defined for all positive integers $x$ and $y$. Define a sequence $a_n$ recursively by setting $a_1 = 1$ and -$$a_n = \sum_{k=1}^{n-1} F(k, n-k) \cdot a_k a_{n-k}.$$ -Is there a common name for such recursively defined sequences? A study for some general classes of $F$? I'll probably "show my cards" more in another post, but for now I'd like to search the literature a bit more... unfortunately I can't figure out the right term to search for! - -REPLY [9 votes]: The evaluation of your sequence is equivalent to the evaluation of certain weighted sums over binary trees. The resulting identities are often called hook length formulas. -Suppose $\mathcal B_n$ denotes the set of full binary trees with $n$ internal vertices. For some tree $T\in \mathcal B_n$ and vertex $v\in T$ we define the $F$-hook length of $v$ to be $H(v)=F(p+1,q+1)$ if the left tree below $v$ is in $\mathcal B_p$ and the right tree below $v$ is in $\mathcal B_{q}$. The elements of your sequence satisfy -$$a_{n}=\sum_{T\in \mathcal B_{n-1}}\prod_{v\in T}H(v).$$ -A particularly cool example due to Postnikov is given by the hook function $F(p,q)=1+\frac{1}{p+q-1}$ which leads to the strikingly simple -$$a_n=n^{n-2}\frac{2^{n-1}}{(n-1)!}.$$ -This sparked some curiosity about which functions $F$ give rise to simple evaluations for $a_n$. You can find more examples in the paper "Hook Length Formulas for Trees by Han's Expansion" by W. Chen, O. Gao, P. Guo but there are more papers out there on the topic. -All the investigated examples that I've seen use hook functions $F(p,q)$ that depend only on $p+q$. If we think in analogy with hook length formulas for partitions (where the analogue of Postnikov's formula is the Nekrasov-Okounkov formula) this is analogous to hook lengths being a sum $a+\ell+1$ where $a,\ell$ are the arm and leg of a box. The classical hook length formula, or Nekrasov-Okounkov formula use hook functions that depend only on $a+\ell$ but their $q,t$ generalizations, as well as the theory of Macdonald polynomials show that there are interesting formulas where the weight for each box depends on $a$ and $\ell$ separately. This makes me hopeful that the same can happen for trees, so I expect there to be hook length formulas for more general $F(p,q)$ that doesn't just depend on $p+q$. -Now, for those that are curious, such hook length formulas have been investigated for other classes of trees, and there is a unifying Hopf Algebraic perspective behind all such calculations. This is explained in "Tree hook length formulae, Feynman rules and B-series" by B. Jones, K. Yeats.<|endoftext|> -TITLE: Group action with unique word -QUESTION [6 upvotes]: This must be known or easy for some of you, but here goes: - -Suppose $f_0,f_1:[n]\to [n]$ are invertible functions, where $[n]=\{0,\dots,n-1\}$ is a set of $n$ elements. -For a word $w=w_1\dots w_m\in\{0,1\}^m$ we define $f_w=f_{w_m}\circ f_{w_{m-1}}\circ\dots\circ f_{w_1}$ (or make it the opposite order, if you prefer). -Suppose $c, d$ and $m$ are such that there is exactly one word $w\in\{0,1\}^m$ with $f_w(c)=d$. -Does it follow that $n\ge m+1$? - -I've checked that it does follow for all $m\le 11$ (eleven). - -REPLY [7 votes]: Fix $c\in [n]$. Let $\mathcal R_m(c)$ be $\{f_w(c)\colon |w|=m\}$ and $r_m(c)=|\mathcal R_m(c)|$. We define $\mathcal R_0(c)$ to be $\{c\}$. -Claim: Let $m\ge 0$. If $r_m(c)=r_{m+1}(c)$, then for all $M>m$ and for all $d\in\mathcal R_M(c)$, there are two $w$'s in $\{0,1\}^M$ with $f_w(c)=d$. -Proof: Let $S=\mathcal R_m(c)$. Then $\mathcal R_{m+1}(c)=f_0(S)\cup f_1(S)$. By invertibility if the $f_i$, $f_0(S)$ and $f_1(S)$ each have cardinality $r_m(c)=r_{m+1}(c)$, and their union also has cardinality $r_{m+1}(c)$. It follows that $f_0(S)=f_1(S)$. Now let $w\in \{0,1\}^M$ and let $f_w(c)=d$. Let $w$ be the concatenation of $u$ (of length $m+1$) and $v$ of length $M-(m+1)\ge 0$. By the above, there exists a $u'$ also of length $m+1$ with the opposite $(m+1)$st symbol such that $f_{u'}(c)=f_{u}(c)$. Now $f_v\circ f_{u'}(c)=f_v\circ f_u(c)$. $\square$ -It follows that if $c$ and $d$ are such that there is a unique $w$ of length $m$ with $f_w(c)=d$, then $r_{j+1}(c)>r_j(c)$ for each $j -TITLE: Mazur and contractible manifolds -QUESTION [6 upvotes]: A Mazur manifold is a contractible, compact, smooth $4$-manifold with boundary a homology $3$-sphere. -It is built from a single $0$-handle, a single $1$-handle and single $2$-handle. It is equivalent that the $4$-manifold must be of the form $\displaystyle S^{1}\times D^{3}$ union a $2$-handle. (Handles are all $4$-dimensional.) The following picture is from Akbulut and Durusoy's paper: - -Here, $W$ is a Mazur manifold with the boundary Brieskorn sphere $\Sigma(2,5,7)$. The dark blacked dotted circle shows the $1$-handle which can be drawn as $0$-framing unknot. -Q1. Are we free how we attach $2$-handles to $S^1 \times D^3$? -Q2. For example, the following picture describes a Mazur manifold? - -There are contractible $4$-manifolds built with a $0$-handle, two $1$-handles, and two $2$-handles. They are the examples of Stern. -Q3. Do we know the classification of contractible $4$-manifolds in terms of their handle numbers? - -REPLY [5 votes]: About terminology: wikipedia defines a Mazur manifold as a contractible compact smooth 4-manifold that is not diffeomorphic to the 4-ball. (It follows from this definition that the boundary of such a manifold is automatically an integral homology sphere.) It also says that frequently the definition is restricted to manifolds constructed with only one handles of each index 0, 1, and 2. I will stick to this latter definition, for consistency with your question(s). -What Anubhav refers to in his comment is a further restriction, but I wouldn't put it in the definition of a Mazur manifold (but rather speak of a Mazur cork if you have such an involution). -Now that we all agree on the objects, let's get to the questions. -Q1. No, we're not free. Suppose you have a presentation of a 4-manifold $W$ with one 1-handle and one 2-handle. The 1-handle gives you a generator in the presentation of the fundamental group of the 4-manifold, and the attaching circle of the 2-handle gives you a relation (which is the only relation, since you only have one 2-handle). If you want the fundamental group of $W$ to be trivial, you better have that the relation kills the generator, which translates to the attaching circle generating the homology of $S^1\times S^2$. Diagrammatically, you're asking for the linking number between the dotted circle and the framed knot to be ±1. -This is almost the only restriction, except that you need to check that the 4-manifold is not the 4-ball, but it follows from the property R that there is only one such knot. -This is actually a good segue for question 2. -Q2. No, these are not Mazur manifolds. Even interpreting the 0-framed 2-handle as a 1-handle (otherwise you don't even have the right homology groups), this is just $B^4$: the 2-handle geometrically cancels the 1-handle. -Q3. I highly doubt it, and I don't think the question can have a "nice answer". These questions about 4-manifolds are usually incredibly hard.<|endoftext|> -TITLE: Generalized Collatz sequences -QUESTION [6 upvotes]: Let $\mathbb{N}$ denote the set of positive integers. For $k\in\mathbb{N}$ let $c_k:\mathbb{N}\to\mathbb{N}$ be defined by $x\mapsto x/2$ for $x$ even and $x\mapsto kx+1$ otherwise. The Collatz sequence of $x\in \mathbb{N}$ with respect to $k$, denoted by $\text{Coll}_{x,k}:\mathbb{N}\to\mathbb{N}$ is defined by $1\mapsto x$ and $n\mapsto c_k(\text{Coll}_{x,k}(n-1))$ for $n\in\mathbb{N}\setminus\{1\}$. -The famous Collatz conjecture states that $$1\in\text{im(Coll}_{x,3})$$ for every $\in\mathbb{N}$. -For $k$ even, the behavior of $\text{Coll}_{x,k}$ is uninteresting, and it is easy to see that for every $x\in\mathbb{N}$, the sequence $\text{Coll}_{x,1}$ eventually periodic. Moreover, if $k>1$ and $k=4a+1$ for some $a\in\mathbb{N}$, we get that no member of $\text{im}(\text{Coll}_{k,k})$ is divisible by $4$ ... (Edit: apologies, this last statement is false as pointed out by user @wojowu! So I erroneously thought only $k=4a+1$ is uninteresting, so the questions below focus on $k=4a+3$.) -Questions. - -Is there $a\in\mathbb{N}$ such that there is a positive integer $x$ such that $\text{Coll}_{x,4a+3}$ is unbounded? (The smallest known value of $a$ satisfying this would be of interest.) - -Is there $a\in\mathbb{N}$ such that there is a positive integer $x$ such that $\text{Coll}_{x,4a+3}$ is bounded, but $1\notin \text{im}(\text{Coll}_{x,4a+3})$, or in other words, $\text{Coll}_{x,4a+3}$ is eventually periodic, but $1$ is not involved in the period? - - -Edit. I corrected the inductive definition of $\text{Coll}_{x,k}$. Thanks to user @wojowu for spotting my error. - -REPLY [6 votes]: This long comment might be helpful: -I think the answer is negative as I pointed out in the comments because almost all Collatz orbits attain almost bounded values, the result which is shown by Terras and was proven by Allouch which states that ${\mathrm{Col}_{\min}(N) < N}$ For almost all $N$ (in the sense of natural density), and there is an improvement here by Terry Tao in the sense of logarithmic density (see Theorem 2).<|endoftext|> -TITLE: Frobenius' article and the Markoff number unicity conjecture -QUESTION [16 upvotes]: A Markoff triple $(a,b,c)$ is a solution in positive integers to the equation -$$ a^2+b^2+c^2=3abc. $$ -Frobenius famously conjectured that a given integer $c$ may appear at most once as the largest coordinate of a Markoff triple $(a,b,c)$, a conjecture often referred to as the Unicity Conjecture. This conjecture appears (I've been told) in the paper -G. Frobenius, Über die Markoffschen Zahlen, S. B. Preuss Akad. Wiss., Berlin, 1913, pp. 458–487. -I was wondering if anyone knows where to find a copy of this paper on the web; or if you have a copy, would it be possible for you to send me a PDF? It's long out of copyright, so that shouldn't be an issue. I tried a general web search, and Google Scholar, but no success. [I'd be even more appreciative of a translation into English, since my German reading skills are rusty; but probably that's too much to hope for.] - -REPLY [18 votes]: I found Frobenius's 1913 publication in the Biodiversity Heritage Library. (Somehow Google Scholar does not index it.) You can view the paper online and download the pdf by submitting an email address.<|endoftext|> -TITLE: Weil cohomologies with given field of definition and coefficient field -QUESTION [6 upvotes]: Fix a perfect field $k$. Fix a field $K$ of characteristic $0$. -A Weil cohomology induces a functor from the category of smooth projective geometrically connected $k$-schemes to the category of $\mathbb{Z}_{\geq 0}$-graded finite-dimensional commutative associative unital $K$-algebras. -For any two Weil cohomologies are the corresponding functors naturally equivalent? - -REPLY [8 votes]: If $k$ is a number field, each embedding $\sigma:k\hookrightarrow\mathbb{C}$ determines a Weil cohomology theory $H^*_{B,\sigma}$ on smooth projective $k$-varieties given by taking the topological cohomology of the complex manifold obtained form $X$ via $\sigma$. In [1], Charles gives an example of $k$, $X$, and two complex embeddings $\sigma_1$, $\sigma_2$ such that $H^*_{B,\sigma_1}(X)$ and $H^*_{B,\sigma_2}(X)$ are not isomorphic as algebras with real coefficients. This gives a counterexample to your question with $K=\mathbb{R}$, since the functors $H^*_{B,\sigma_1}$ and $H^*_{B,\sigma_2}$ are not even pointwise-isomorphic. -[1] Charles, François, Conjugate varieties with distinct real cohomology algebras, J. Reine Angew. Math. 630, 125-139 (2009). ZBL1222.14122.<|endoftext|> -TITLE: SOS polynomials with rational coefficients -QUESTION [10 upvotes]: Suppose we are given a univariate polynomial with rational coefficients, $p \in \Bbb Q [x]$, and are told that $p$ can be expressed as the sum of $k$ squares of polynomials with rational coefficients. It is well-known that every univariate sum of squares (SOS) polynomial can be expressed as a sum of two squares. -Can we efficiently find an SOS decomposition $p = f^2 + g^2$, where both $f, g \in \Bbb Q [x]$? -Just to be clear: I want an efficient algorithm that takes as input a polynomial $p(x)$, which is guaranteed to have a representation as the sum of $k$ squares of polynomials with rational coefficients, and outputs two polynomials $f(x), g(x)$ with rational coefficients such that -$$p(x) = f^2(x) + g^2(x)$$ - -REPLY [20 votes]: In general you can't write $p = f^2 + g^2$ in ${\bf Q}[x]$ at all, -let alone do so efficiently. -For example, $2 x^2 + 3$ is positive for all $x$ -(and is the sum of three squares, $(x+1)^2 + (x-1)^2 + 1^2$); -but if $2 x^2 + 3 = f(x)^2 + g(x)^2$ then $3 = f(0)^2 + g(0)^2$, -which is impossible because $3$ is not a sum of two rational squares. -(Cf. the comment of Olivier Bégassat.) -A positive quadratic polynomial can still be written as $a f(x)^2 + b g(x)^2$ -for rational $a,b > 0$; but in degree $4$ and beyond even that is not usually -true, for Galois-theoretic reasons, using the factorization -$a f^2 + b g^2 = a (f+cg) (f-cg)$ with $c^2 = -b/a$. -For example, if $p$ has degree $n$ and Galois group $S_n$ -(which is the usual case) then $p$ cannot be written as $a f^2 + b g^2$. -Already $p = x^4 + x + 1$ is an example.<|endoftext|> -TITLE: Bound the number of the minimal generating set of group G by its abelianization -QUESTION [5 upvotes]: This is probably already well-known or too big to answer. Let $G$ be a finite group and $G^{ab}$ be the abelianization of the group G. Is there any bound on $d(G)=\min\{\#S\mid G=\langle S\rangle\}$ by using $d(G^{ab})$ without considering the order of $G$? -Thanks for any answer and comments. - -REPLY [9 votes]: Okay, so let's fill in the details on Ville's nice argument in the comments: there is no such bound, and to prove this it suffices to exhibit a sequence of finite perfect groups whose ranks are unbounded. We'll take the sequence $A_5^n$ to be concrete although the argument applies to powers of any finite perfect group. If we take $k$ elements $\{ g_1, \dots g_k \}$ of $G_n$, their projections to each copy of $A_5$ can take at most $60^k$ possible values, so by pigeonhole there's at least one subset of the indices $S \subseteq \{ 1, 2, \dots n \}$ of size at least $\left\lfloor \frac{n}{60^k} \right\rfloor$ such that the projections of the $g_i$ to each copy of $A_5$ indexed by each $i \in S$ are the same. If $|S| \ge 2$ it follows that $\{ g_1, \dots g_k \}$ can't generate $A_5^n$, hence -$$\text{rank}(A_5^n) > \log_{60} \frac{n}{2}.$$ -In the positive direction, Burnside's basis theorem implies that $\text{rank}(G) = \text{rank}(G^{ab})$ if $G$ is a finite $p$-group. -Edit: It's maybe also worth mentioning that we needed to do this construction because it doesn't suffice to just take, say, arbitrarily large finite simple groups; it's known that all nonabelian finite simple groups have rank exactly $2$.<|endoftext|> -TITLE: Is there a similar formula like Ramanunjan's Eisenstein series identity for $\sum_{k=1}^{n-1}k^2 \sigma(k)\sigma(n-k)$? -QUESTION [5 upvotes]: This question is related to the last question about van der Pol's identity for the sum of divisors. -In Touchard (1953) it is mentioned that the sum of divisors $\sigma(n)$ satisfies the following recurrence relation ($n>1$): -$$n^2(n-1) = \frac{6}{\sigma(n)} \sum_{k=1}^{n-1}(3n^2-10k^2)\sigma(k)\sigma(n-k)$$ -We can evaluate the convolution part with Ramanujan's identity: -$$\sum_{k=0}^n\sigma(k)\sigma(n-k)=\tfrac5{12}\sigma_3(n)-\tfrac12n\sigma(n)$$ -which for our case reads like this: -$$\sum_{k=1}^{n-1}\sigma(k)\sigma(n-k)=\tfrac5{12}\sigma_3(n)-\tfrac12n\sigma(n)+\tfrac{\sigma(n)}{12}$$ -Substituting in van der Pol's equation a perfect number $n = \sigma(n)/2$ and making use of Ramanujan's identity, we find that the perfect number $n$ satisfies the following quartic equation: -$$ -8n^4-2n^3+3 \sigma_3(n)n^2+24A_2 =0 -$$ -where -$$A_2 = \sum_{k=1}^{n-1}k^2 \sigma(k)\sigma(n-k)$$ -I asked an expert of convolution identities for $\sigma(n)$ if $A_2$ can be evaluated and he said, that one could prove a similar formula, like the one of Ramanujan, "simply by considering the first and the second derivative of suitable identities between Eisenstein series". -However I am not very confident with Eisenstein series, so I am asking the experts for help to help evaluate $A_2$. -Thanks for your help! - -REPLY [9 votes]: All these identities can indeed be proved essentially trivially using modular forms and quasi-modular forms (those involving $E_2$), and the fact that the dimension -of such spaces is $1$ for weight 4,6,8,10,14, and $2$ for weight 12, in which case the identities involve also the Ramanujan $\tau$ function. Explicitly, -sums $\sum_{1\le k\le n-1}k^a\sigma_b(k)\sigma_c(n-k)$ with $b$ and $c$ odd positive integers ($\sigma_b(k)=\sum_{d\mid k}d^b$) have weight -$w=b+c+2+2a$, so if $w=4,6,8,10,14$ you will obtain identities involving only $\sigma$, and if $w=12$ also $\tau(n)$.<|endoftext|> -TITLE: An inequality on some pairs of orthogonal vectors -QUESTION [13 upvotes]: Let $n,k\geq 1$. Suppose that -$a_1, \ldots, a_n\in \mathbb{R}^k$, $b_1, \ldots, b_n\in \mathbb{R}^k$ -and $a_i^T b_i = 0$ for $i=1,\dots, n$. Is it true that -$$ -\sum_{i=1}^n \|a_i\|_2^2 + \sum_{i=1}^n \|b_i\|_2^2 \geq \frac3n \sum_{i,j=1}^n | a_i^T b_j | -$$ - -A matrix reformulation of the problem: -Let $A$ be a matrix, we have (e.g. see here) $\|A\|_{(1)} = \frac 12 \min_{A=U^TV} (\|U\|_F^2 + \|V\|_F^2)$ -where $\|A\|_{(1)}$ is the sum of singular values of $A$ (known as the trace/nuclear norm). Now, the above problem could be stated as follows - -Let $A = [a_{ij}]$ be an $n \times n$ matrix with zero diagonal. Is -it true $$ \|A\|_{(1)} \geq \frac12\frac{3}{n} \sum_{i,j}|a_{ij}| $$ - -REPLY [9 votes]: $\def\Tr{\mathrm{Tr}}\def\Mat{\mathrm{Mat}}$I've been thinking about this problem a bunch, and I think the correct bound is -$$ \sum_{i,j} |A_{ij}| \leq \left( \cot \frac{\pi}{2n} \right)|A|_{(1)}. $$ -As $n \to \infty$, we have $\cot \tfrac{\pi}{2n} \sim \tfrac{2n}{\pi}$, so this matches the $\pi$ bound that fedja proved for $k=2$. -In particular, I will prove that this bound is correct for skew-symmetric $A$; almost all the work is not due to me but to a paper of Grzesik, Kral, Lovasz and Volec which was pointed out in a deleted answer by another user. -I'll write $\sigma_1(A) \geq \sigma_2(A) \geq \cdots$ for the singular values of $A$. Note that we have -$$\sum |A_{ij}| = \max_{P \in \mathrm{Mat}_n(\pm 1)} \Tr(AP) $$ -and -$$|A|_{(1)} = \max_{Q \in O(n)} \Tr(AQ). $$ -Here $P$ is ranging over $\pm 1$ matrices, and $Q$ is ranging over the orthogonal group. -We may replace the orthogonal group by its convex hull without changing the max. The convex hull of $O(n)$ is the set of matrices of operator norm $\leq 1$; call that $B_1$. -So -$$|A|_{(1)} = \max_{R \in B_1} \Tr(AR). $$ -As a warm up, let's consider the best inequality we can prove of the form $\sum |A_{ij}| \leq C |A|_{(1)}$ without imposing that the diagonal is $0$. The answer is that the best is $C = n$, and that is easy to prove by elementary means, but I want to demonstrate my approach instead. So we want to find a $C$ such that, for every $\pm 1$ matrix $P$ and for every matrix $A$, we have $\Tr(AP) \leq C \max_{R \in B_1} \Tr(AR)$. Since $B_1$ is convex, this is the same as asking for $C$ such that $P \in C B_1$. In other words, we want to bound $\sigma_1(P)$ for $P$ in $\Mat_n(\pm 1)$. It wouldn't be hard to obtain the bound $n$ from here, but we move on. -Let's leave the warm up and get to the real problem. What we actually want is that $\Tr(AP) \leq C \max_{R \in B_1} \Tr(AR)$ for $A$ having zero diagonal. Thus, we only need $\pi(P)$ to lie in $\pi(C B_1)$, where $\pi$ is orthogonal projection onto matrices of diagonal $0$. In other words, we want $P$ to lie in $CB_1 + \Delta$ where $\Delta$ is the vector space of diagonal matrices. So we come to the following problem: -Problem 1: Find the best constant $C_1$ such that, for every $\pm 1$ matrix $P$, there is a diagonal matrix $D$ with $\sigma_1(P+D) \leq C_1$. -Unfortunately, it seems hard to even guess a rule for choosing the optimal $D$. For example, if $P$ is identically $1$, the best choice of $D$ is $-\frac{n}{2} \mathrm{Id}_n$. -Having no success here, I move on to the case of $A$ skew symmetric. We now can consider only skew symmetric $P$ (which are $0$ on the diagonal and $\pm 1$ off the diagonal.) For such a $P$, we now want to solve the problem: -Problem 2: Find the best constant $C_1$ such that, for every skew-symmetric $\pm 1$ matrix $P$, there is a symmetric matrix $H$ with $\sigma_1(P+H) \leq C_1$. -Fortunately, here I can make a little progress. It turns out that the symmetric matrix is irrelevant! -Lemma: Let $P$ be a skew symmetric matrix and $H$ a symmetric matrix. Then $\sigma_1(P+H) \geq \sigma_1(P)$. -Proof: Since $P$ is skew symmetric, it is diagonalizable over $\mathbb{C}$ with purely imaginary eigenvalues, and the largest such is $i \sigma_1(P)$. Let $v$ be an eigenvector with $P v = i \sigma_1 v$. Writing $\dagger$ for the conjugate transpose, normalize $v^{\dagger} v =1$. Then $\sigma_1(P+H) \geq | v^{\dagger} (P+H) v | = |i \sigma_1 + v^{\dagger} H v|$. But $v^{\dagger} H v$ is real, so $|i \sigma_1 + v^{\dagger} H v| \geq \sigma_1$. $\square$. -Thus, we have reduced to the problem: -Problem 3: Find the largest operator norm of any skew-symmetric $\pm 1$ matrix. -Another poster answered and then deleted his answer to point out that this problem is solved in Lemma 11 of Cycles of a given length in tournaments! (On reflection, I have removed this poster's name since they choose to self-delete, but I hope they will identify themselves and claim the credit; this is useful!) The largest operator norm is always achieved by the matrix which is $1$'s above the diagonal and $-1$'s below it. (As well as by the many other matrices which are conjugate to this one by signed permutation matrices.) -This matrix can be explicitly diagonalized: -The eigenvectors are of the form $(1, \zeta, \zeta^2, \ldots, \zeta^{n-1})$ where $\zeta = \exp(\pi i (2j+1)/(2n))$. The corresponding eigenvalues are $i \cot \tfrac{(2j+1) \pi}{2n}$. In particular, the largest singular value is $\cot \tfrac{\pi}{2n}$, thus explaining my guess. -I am guessing this is optimal for Problem 1 as well as Problem 2, but this is based on a very weak intuition that skew symmetric choices are good, plus fedja's answer.<|endoftext|> -TITLE: How to deal with an advisor that offers you nearly no advising at all? -QUESTION [40 upvotes]: I am a young PhD student (24) at a Germany university and I am not sure whether this is the right place to ask this kind of question. If not feel free to move it elsewhere or delete it completely. -Currently, I a have a half time position in Analysis and my doctoral advisor more and more turns out to be not very involved in my PhD. I started 1 1/2 years ago at the age of 22 and my PhD advisor was at least somehow involved as I wrote my Master thesis but gave me much freedom. He gave this thesis the best grade possible and I felt I also deserved it to an extent. The topic was one that I had chosen myself and I learned much new things writing it; combining different areas that I did not know before. This time was quite stressful for me personally as I tend to pressure myself too hard if I have to perform like this. After that I was quite exhausted but wanted to pursue a PhD at my university in the field I wrote my Master thesis in because I felt like the right thing to do; I really like the people and also the topic I wrote my thesis on. -However, things changed rapidly as I became officially a PhD student. First thing my advisor told me was that he had no time to spend on me as his oldest PhD student had to finish after over four years. I did not wonder and at that time I had the luck to write a paper with a guy, which still is like some mathematical godfather to me (and much more capable than me), who I met on a conference. The paper that we wrote had quite nice new ideas in it, however I felt my part in creating it was minor. But I also did some good work I think. -At the start of the year I investigated some other question on my own and managed to produce a positive result with the tools I learned from writing the first paper. I also extended the question quite a bit to write a paper on my own about the topic; without any advising at all. The only thing I currently do, is to speak with some colleagues of mine over very specific questions. I sent this "paper" to my advisor and the only thing he told me is that he would be too busy to read it in near future. -Currently, I have another collaboration going with the guy I mentioned beforehand and several others who advise me more than my own advisor, although they work on completely different universities. So currently I am quite lucky to have some advision and a perspective in research. -Finally, my PhD advisor didn't give me a question to work one. He just mentioned very vaguely that one maybe could extend some of the concepts used in my Master thesis but he couldn't tell me any possible applications for these abstractions. So I did not feel like this would be promising to work on. He also does not meet up with me on a weakly basis to discuss. Furthermore, my advisor also holds a record on suggesting topics to this PhD students that are completely inept to work on at this stage of their mathematical career. My older "PhD brothers", for example, spend to years working on a big conjecture in one specific field without making any progress, whatsoever. My PhD advisor had also no new idea how to approach that problem; he basically just told them to try it without giving much help. -So I frequently ask myself the following question: "Do I feel it is worth to pursue a PhD under this circumstances?" -I saw how other advisors work with their PhD students and I feel their advisors have a clear initial idea on the "what" and on the "how". Moreover, they meet up and discuss the current problems that arise while pursuing the question. All this I do not have at the moment. I really enjoy teaching courses but I do not have the impression that I move forward in research to much and that really pulls me down. And I also feel that this whole situation damages me mentally to a point where I frequently get anxiety attacks. On the other hand, I know that I could earn good money in the economy with my qualities and my intellectual capacity. -So my question is: Would you advise me to quit my PhD and to try it at another university in my field? Or should I stay and fight? Or should I just skip the PhD and do something that earns my money and gives me more structure? -I know that the "right" answer to this question is not determined in any way and that it might fit in the category "vague question" that we usually try to avoid on this platform. But I do not know where to ask it elsewhere and I would like to get answers from people with more academic experience than I have. I really cannot really pigeonhole my whole situation and do not know what to do at the moment. - -REPLY [8 votes]: I've gotten here late, but I'd like to emphasize one point that I don't think has been made: I don't know what it is like in Germany, but from a US perspective, you have already done enough for a Ph.D. Many of us graduate with no papers yet published, their dissertations containing their only published results. You are already doing substantially better than many recent American Ph.D. graduates. The only requirement for a Ph.D. is having done independent research (beyond the level of a master's degree, whatever that means), and you have already done that. I know of people whose dissertations were a few published papers tied together with some interstitial wording—this is obviously not ideal, but I state it to point out it is an option: it has been done. -The fact your advisor has been too uninvolved to propose a thesis problem actually grants you a great deal of freedom, because you don't need to pursue some impossible white whale of a problem to please him. He can't fault you for choosing and answering your own problem. All you need to do to graduate is decide one of these independent projects, ideally a harder one, is the one you want to be your dissertaion, write some additional background chapters you wouldn't put in a paper, and submit that, and it sounds like you have plenty of time to do that. So by no means quit. -Unfortunately, unlike the Ph.D. itself, there are no standards required to be an advisor, and many of us get rather little advice. There's not much you can do to fix this problem, which is on his end, but it sounds like you're already doing the right thing, pursuing collaborations with people whose research is of interest to you and your own projects. -As far as career-related advice goes, many of us have had advisors who aren't terribly involved or aren't tremendously worldly, and what one usually does is to ask advice of postdocs you've made friends with, other established people in the department you trust, your peers, to compare, some of your older coauthors who you feel you have some rapport with, etc. It's unfortunate your advisor is unwilling to advise, but you have coauthors who want to see you stay in mathematics, as well as surely friends who have your best interest at heart, and it is always OK to ask. (And if all else fails, there's the Academia StackExchange.)<|endoftext|> -TITLE: Lowenheim-Skolem numbers for SOL + correctness quantifiers -QUESTION [7 upvotes]: For a logic $\mathcal{L}$, say that a cardinal $\kappa$ is $\mathcal{L}$-correct iff every satisfiable $\mathcal{L}$-theory of size $<\kappa$ has a model of size $<\kappa$. First-order correctness is of course boring, but quite quickly we enter the realm of strong large cardinal properties (see e.g. here). -I'm interested in a kind of "iterated correctness:" repeatedly add to a given logic the ability to quantify over cardinals which are correct for it (or rather, for the previous iteration of this process). This doesn't make obvious sense in general, but for logics with reasonably nice syntax things are better. In particular, I'm interested in what happens when we iteratively "add correctness quantifiers" to second-order logic, as follows: -Let $\mathcal{L}^2_0$ be usual second-order logic, and let $\mathcal{L}^2_{n+1}$ be $\mathcal{L}_n^2$ augmented with a quantifier $$\mathsf{C}_nx\varphi(x)\equiv\mbox{$\vert\{x: \varphi(x)\}\vert$ is $\mathcal{L}^2_n$-correct.}$$ So $\mathcal{L}^2_k$, in addition to the usual logical symbols, has $k+2$ different types of quantifier: the first-order quantifiers, the second-order quantifiers, and $k$-many correctness quantifiers. These quantifiers can alternate however is desired. -My question is: - -Do any of the "usual" large cardinal properties imply $\mathcal{L}^2_n$-correctness for every $n$? - -This is not at all obvious to me. On the other hand, I can't pin down a concrete obstacle to e.g. supercompactness having the above property. - -REPLY [4 votes]: Assume ZFC. Let $n$ be a meta-integer. Then $\kappa$ is $\mathcal{L}^2_n$-correct iff $\kappa$ is $\Sigma_{n+2}$-reflecting, i.e. $V_\kappa\preccurlyeq_{n+2}V$. -Proof: For $n=0$, i.e. 2nd order logic, we have: If $V_\kappa\preccurlyeq_2 V$ then easily $\kappa$ is $\mathcal{L}^2_0$-correct. Suppose now that $\kappa$ is $\mathcal{L}^2_0$-correct. Now for each $\alpha<\kappa$, if $|V_\alpha|<\kappa$ then $|V_{\alpha+1}|<\kappa$, as is easily seen using $\mathcal{L}^2_0$-correctness. So there is a limit ordinal $\eta\leq\kappa$ such that $|V_\eta|=\kappa$. Now observe that $V_\eta\preccurlyeq_2 V$ by using $\mathcal{L}^2_0$-correctness. It follows that $\eta=\kappa$ (if $\eta<\kappa$, take $\alpha<\eta$ such that $\eta\leq\beta=|V_\alpha|<\kappa$, and using $V_\eta\preccurlyeq_2 V$, show $\beta<\eta$ for a contradiction). So $V_\kappa\preccurlyeq_2 V$ as desired. -For $n=1$: Suppose $V_\kappa\preccurlyeq_3V$. Let $T\in V_\kappa$ be a satisfiable $\mathcal{L}^2_1$-theory. Then there is an ordinal $\eta$ such that $V_\eta\preccurlyeq_2 V$ and $M\in V_\eta$ such that $V_\eta\models$"$M\models T$", where the latter statement uses $V_\eta$'s own notions to compute $\Sigma_2$-elementarity, and hence the $C_0$ quantifier. But this is a $\Sigma_3$ statement, so $V_\kappa$ models it, and note that this really gives a true model of $T\in V_\kappa$. -Conversely, suppose $\kappa$ is $\mathcal{L}^2_1$-correct. So in particular it is $\mathcal{L}^2_0$-correct, and hence $V_\kappa\preccurlyeq_2V$. We want to see $V_\kappa\preccurlyeq_3V$. Let $x\in V_\kappa$ and suppose $V\models\exists w\varphi(x,w)$ where $\varphi$ is $\Pi_2$. Let $\alpha<\kappa$ with $x\in V_\alpha$. Consider the $\mathscr{L}^2_0$-theory theory $T$ in parameters in $V_\alpha$, which describes a rank segment $V_\eta$ of $V$ such that $\eta$ is a cardinal and $V_\eta\preccurlyeq_2V$ (by using the $C_0$-quantifier to require $\eta$ to be $\mathcal{L}^2_0$-correct), and $V_\eta\models\exists w\varphi(w,x)$. Note this is satisfiable, and hence satisfiable in cardinality ${<\kappa}$, and hence $V_\kappa\models\exists w\varphi(w,x)$. -Now proceed in this manner. -So in particular, least $\mathcal{L}^2_0$-correct cardinal is $<$ least strong -and least supercompact, these are $<$ -least $\mathcal{L}^2_1$-correct $<$ least extendible -< least $\mathcal{L}^2_2$-correct.<|endoftext|> -TITLE: What is difference between working with small and large category of spaces? -QUESTION [7 upvotes]: The following construction have always bugged me. This is p328, Remark 5.1.6.1 in Lurie's Higher Topos Theory. Lurie begins with the following: - -Construction: Let $C$ be a simplicial set. $S$ denote the $\infty$-category of small spaces. $s$ a vertex. Composing the Yoneda embedding $$j:C \rightarrow P(C)$$ by the evaluation map $$e_s: P(C)=Fun(C^{op}, S) \rightarrow Fun(\{ s \}, S)\simeq S$$ -We obtain the map $j_s:=e_s\circ j:C \rightarrow S$. Where $j_s$ is referred to as the functor correpresented by $s$. - -The remark states that we should replace $S$ by $\widehat{S}$ the large $\infty$-category of spaces. - -I agree this is what we do 1-categorically when $S$ is replaced with sets. -Q1. But provided this construction makes sense, what's wrong with not replacing it? -Q2. Now suppose we are to work with it $\widehat{S}$. Is it safe to regard $Fun(C,S)$ as a fullsubcategory $Fun(C,\widehat{S})$? What do we know about the inclusion $S \hookrightarrow \widehat{S}$ i.e. if this inclusion is - -limit/colimit preserving -conservative - -? - -(Unfortuantely I am unable to work through the material after this - which may answer the above questions) - -REPLY [6 votes]: The Yoneda embedding $y: S \rightarrow P(S)$ is informally given by $y(s)(s^{\prime}) = map_{S}(s^{\prime}, s)$ (some care must be taken if $S$ is not an $\infty$-category in interpreting the mapping space, but Lurie gives a precise definition). If $S$ is a small simplicial set, then $map_{S}(s, s^{\prime})$ is a small space, so the functor does in fact land in presheaves of small spaces, as claimed. This is not necessarily true if $S$ is a large simplicial set, and this is what I believe Lurie's remark is about. What is perhaps slightly confusing is that you only really need $S$ to be locally small (ie. to have small mapping spaces), rather than to be small itself, and that's actually a rather common occurrence. - -The inclusion $S \hookrightarrow \widehat{S}$ preserves all small limits and colimits. The same will be true for $Fun(C, S) \hookrightarrow Fun(C, \widehat{S})$, because (co)limits in functor $\infty$-categories are pointwise.<|endoftext|> -TITLE: Is there a standard way of defining the integral of an extended real function with respect to a finitely additive probability measure? -QUESTION [5 upvotes]: Let $X$ be a set, and let $\mu$ be a finitely additive probability measure defined on $2^X$. Let $\Phi$ be the set of functions from $X$ to $\mathbb R \cup \{-\infty, \infty\}$. - -Is there a standard way of defining a integrable function that allows us to define $\int f d\mu$ for some unbounded $f \in \Phi$? - -To be clear, the integral should satisfy the following properties: -(1) $\int (af+bg) d\mu = a\int f d\mu + b\int g \mu$ for $f,g \in \Phi$ and $a,b \in \mathbb R$, provided all the integrals exist and $\infty - \infty$ doesn't occur anywhere. -(2) If $f \geq 0$, then $\int f d\mu \geq 0$. -(3) If $1_A$ is the indicator of $A \subset X$, then $\int 1_A d\mu = \mu(A)$. -The standard way of defining the integral for bounded, real-valued functions is the familiar one: first define the integral for simple functions, then uniformly approximate any bounded function by a sequence of simple ones. I'm wondering to what extent this can be pushed beyond bounded functions. Obviously the integral will be badly behaved in the sense that it won't satisfy the convergence theorems that we take for granted when $\mu$ is countably additive, but as (1)-(3) indicate, I don't mind that. - -REPLY [3 votes]: Once the integral over measurable bounded functions is defined, a “standard” way is this: For a nonnegative (measurable) function $f$ one takes the supremum of the integral over all bounded measurable functions $g$ with $0\le g\le f$. For general measurable $f$ one splits into integrals over the positive and negative part and takes the difference, requiring that at least one of the integrals (according to the previous definition) is finite. (For functions for which both integrals are infinite, there is no “standard” way.) -For functions with values in a Banach space, one cannot deal with infinite integrals, and one cannot use the supremum-argument. A “standard” procedure in this case is carried out in the “bible” of functional analysis, Dunford N., Schwarz Jacob T., Linear Operators I, Int. Publ./Wiley-Interscience 1966, but be aware that there are some oversights (IIRC, e.g. in the proof of the monotone convergence theorem; not sure whether they were corrected in later editions).<|endoftext|> -TITLE: Topological factors of complex projective manifolds -QUESTION [10 upvotes]: Let $M$ be a closed orientable smooth 4-manifold. Assume $\pi_1(M)=\{0\}$ and $b_2(M)>0$. -Let $S$ be a closed orientable surface. Denote $P=M\times S$. -Can it so happen that there is no complex projective manifold homotopy equivalent to $P$? -Is it possible to rule out the existence of a closed symplectic 6-manifold homotopy equivalent to $P$? It seems unlikely, see this preprint. -Note that - -$P$ is formal -the Betti numbers are even in odd degree -there is a class $c\in H^2(P, \mathbb{R})$ satisfying hard Lefschetz -$\pi_1(P)$ is Kähler -$P$ admits an almost complex structure. - -A related example is constructed here but I don't think it decomposes as a direct product. It may be relevant that Hsueh-Yung Lin claims that every closed Kähler threefold is deformation equivalent to a complex projective manifold. - -REPLY [2 votes]: Let $M=\mathbb CP^2\#\mathbb CP^2$ and let $S=T^2$ be the $2$-dimensional torus. I think this gives an example for the original question. As for the symplectic version of the question, I am sure it is an open problem. -Proof. Suppose by contradiction $P=M\times S=\mathbb CP^2\#\mathbb CP^2\times T^2$ is homotopic to a complex projective manifold. The contradiction will be derived from the Hodge index theorem, applied to $H^{1,1}(P)$. -Let us calculate the cubic intersection form on $H^2(P,\mathbb Z)$. First we choose the basis $e_1,e_2, e_3$ in $H^2(P,\mathbb Z)$ as follows. Let $S_1\subset \mathbb CP^2\#\mathbb CP^2$ be the sphere generating $H_2$ of the first summand and $S_1$ of the second summand. Then we set $e_1$ to be Poincare dual to $S_1\times T^2$, $e_2$ dual to $S_2\times T^2$, and $e_3$ dual to the fibre $M$ in $M\times T^2$. It is easy to see then, that -$$(a_1e_1+a_2e_2+a_3e_3)^3=3(a_1^2+a_2^2)a_3=Q$$ -Let us now choose the any positive class $h$ in $H^{1,1}(P)$ with $h^3=3$. It is easy to see that applying a linear transformation to $\mathbb R^3$ that preserves $Q$ we can send $h$ to the vector $(1,0,1)$. Let us now apply Hodge index theorem to $H^{1,1}$. First, the class $h$ induces a quadratic from on $H^{1,1}$ and this form should be definite on the orthogonal $h^{\perp}$ to $h$. The quadratic form is $(a_1e_1+a_2e_2+a_3e_3)^2h=$ -$$=(a_1e_1+a_2e_2+a_3e_3)^2(e_1+e_3)=a_1^2+a_2^2+2a_1a_3=(a_1+a_3)^2+a_2^2-a_3^2.$$ -So, we see, its signature is $(2,1)$. The vector $h$ is positive by definition, so the orthogonal to it has signature $(1,1)$. This contradicts the Hodge index theorem. QED. -By the way, would $P$ be complex projective, one would be able to prove that $h^{1,1}(P)=3$. I'll give the argument, even if it is not needed. Indeed, since $H^1(P,\mathbb Z)\cong\mathbb Z^2$, we have the Albanese map $A:P\to Alb(P)$, where $Alb(P)$ is a 1-dimensional abelian variety, i.e. an elliptic curve. Take a regular point $x\in Alb(P)$, then the fiber $A^{-1}(x)$ is smooth $2$-dimensional divisor on $P$ with zero square. However, since $P$ is projective, there is a class in $H^{1,1}(P)$ with positive cube. I follows $h^{1,1}\ge 2$, and so $h^{2,0}=h^{0,2}=0$.<|endoftext|> -TITLE: Smooth structure on direct product -QUESTION [11 upvotes]: Let $M$ be the $E_8$ manifold. Is there a closed manifold $N$ such that $M\times N$ is smoothable? What is the smallest possible dimension of $N$? - -REPLY [11 votes]: Extending Michael Albanese's answer above, $M \times N^k$ will never be smoothable. For if it were then choose a point $p\in N$ and a chart U around $p$. Then $M \times U$ is an open subset of $M \times N$, and hence is smoothable. But as argued in Scorpan (p. 219, Lemma), $M \times \mathbb{R}^k$ is not smoothable. -Scorpan's argument uses the Kirby-Siebenmann product structure theorem to get down to dimension $5$. But I believe (it's been a while) that you could construct an argument based on Novikov's work on signatures.<|endoftext|> -TITLE: Complex manifold defined over $\mathbb{Q}$ -QUESTION [9 upvotes]: If we consider complex projective varieties, to be defined over $\mathbb{Q}$ means that there is a projective embedding whose image is the vanishing locus of a polynomial system with coefficients in $\mathbb{Q}$. -If we consider closed complex manifolds there is no obvious ambient space. -However, we could require that there be a positive integer $n$ and a choice of - -for any $1\leq i\leq n$, open sets $U_i\subset M$ -for any $1\leq i -TITLE: Arzelà-Ascoli theorem and Hölder spaces -QUESTION [5 upvotes]: Let $B\subset \mathbb{R}^n$ be a open ball. Let $\{f_i\}$ be a sequence of functions bounded in the Hölder norm $C^{k,\alpha}(B)$ for a given integer $k\geq 0$ and $\alpha\in (0,1)$. -Does there exist a subsequence which converges to a function $f$ (necessarily $f\in C^{k,\alpha}(B)$) in the norm $C^{k,\alpha/2}(\bar B')$ for any closed ball $\bar B'\subset B$? -A reference would be very helpful. - -REPLY [12 votes]: For completeness, let's mention a simpler and more general statement: For $\Omega\subset\mathbb{R}^n$ a bounded open set, $k\in\mathbb{N}$ and $0<\beta<\alpha\le1$ there is a compact embedding -$$ C^{k,\alpha}(\Omega) \to C^{k,\beta}(\Omega) . $$ -Some details: -1. For a compact metric space $(E,d)$, and for $0<\beta<\alpha\le1$ we have a compact embedding of the space of the $\alpha$-Hölder functions into the space of $\beta$-Hölder functions: -$$\big( C^\alpha(E),\|\cdot\|_{\alpha,E}\big)\to\big( C^\beta(E),\|\cdot\|_{\beta,E}\big).$$ -Here $\|u\|_{\alpha,E}:= \|u \|_\infty+|u|_{\alpha,E}$ and -$$|u|_{\alpha,E}:=\sup_{x\neq y\in E} \frac{|u(x)-u(y)|}{d(x,y)^\alpha} . $$ -Indeed, let $(u_k)_{k\in\mathbb{N}}\subset C^\alpha(E)$ be a $\|\cdot\|_{\alpha,E}$-bounded sequence, that is, it is uniformly bounded and equicontinuous w.r.to a common modulus of continuity $Ct^\alpha$. By Ascoli-Arzelà, some subsequence $(u_{k_j})$ converges uniformly to some $u$ with the same modulus of continuity, so that $u\in C^\alpha(E)$. We may assume w.l.o.g. that $u$ is the null function (for we just replace $(u_{k_j} )$ with $(u_{k_j}-u)$). The thesis then follows since for $j\to\infty$ we have $\|u_{k_j}\|_\infty= o(1)$ and -$$\left|\frac{u_{k_j}(x)-u_{k_j}(y)}{d(x,y)^\beta}\right|=\left| \frac{u_{k_j}(x)-u_{k_j}(y)}{d(x,y)^\alpha}\right|^{\frac{\beta}{\alpha}} - \left|u_{k_j}(x)-u_{k_j}(y)\right|^{1-\frac{\beta}{\alpha}} $$whence also -$$ |u_{k_j}|_\beta\le |u_{k_j}|_\alpha^{\frac{\beta}{\alpha}}\left(2\|u_{k_j}\|_\infty \right)^{1-\frac{\beta}{\alpha}}=o(1).$$ -2. The same compact embedding holds true if $(E,d)$ is only assumed totally bounded: its completion $(\tilde E,\tilde d)$ is compact, and the map "extension by density of uniformly continuous functions" gives an isometry (whose inverse map is the restriction to $E$) -$$C^\alpha(E)\to C^\alpha(\tilde E).$$ -3. For $\Omega\subset\mathbb{R}^n$ a bounded open set, $k\in\mathbb{N}$ and $0<\beta<\alpha\le1$ the analogous compact embedding -$$\big( C^{k,\alpha}(\Omega),\|\cdot\|_{k,\alpha}\big)\to\big( C^{k,\beta}(\Omega),\|\cdot\|_{k,\beta}\big) $$ -follows from the case $k=0$, because of the usual closed-range embedding -$$ C^{k,\alpha}(\Omega) \to C^{0,\alpha}(\Omega)^N$$ -given by $u\mapsto \big( \partial^\nu u \big)_{\nu\in\mathbb{N^n},|\nu|\le k}$, for $N:=\big({k+n-1\atop k}\big)$.<|endoftext|> -TITLE: Advice for researchers outside academia -QUESTION [34 upvotes]: Perhaps some personal background is relevant to this question. A couple of years ago, I graduated with a master's degree in Applied Mathematics from a good Dutch university. Even though I obtained somewhat decent grades for the courses and the thesis (a 7/10 on average, maybe a B in the American system), I sensed that I probably wasn't good enough to pursue a PhD in a relevant mathematical research direction. -Therefore, I opted for a career in a different area. However, I keep coming back to mathematics over and over again. I am still drawn to some mathematical problems, and I work on them in my spare time. One day, I'd like to be a (co)author of (at least) one publication. -I've now come at a point that I think I have some interesting ideas for a certain problem. In order to flesh them out, I think it would be a good idea to collaborate with someone who has more expertise than me in the relevant domain. I've found a postdoc at my former institution, whom I suspect has the appropriate background to work on this problem with me. -I'm not quite sure, though, how to approach this person. I've never met him/her, and it seems somewhat odd or inappropriate to send him/her an e-mail out of the blue. I've thought of proposing to give a short presentation of my ideas to this person, after which s/he can decide whether the approach seems worthwhile enough to delve into more deeply. -This -- finding and approaching trustworthy fellow mathematicians who have the time and willingness to collaborate -- is one particular problem I've encountered while working as a researcher outside academia. As an “independent researcher”, if you will, I encounter all sorts of other problems, including: - -Getting access to some journals, books and research papers; -Obtaining licenses for expensive proprietary software; - -I wonder whether you have any advice to independent researchers on these matters, and whether you can perhaps give some general guidelines or nuggets of wisdom to those working outside academia. - -REPLY [3 votes]: One point (mentioned in parenthesis by Alexandre Eremenko, but worth highlighting) is that since this postdoc is at your former institution, there's a chance there are professors there who remember you. (I myself is a professor who is terrible with names of people I meet in real life, but there are those who seem to remember everyone.) If so, that could be another way "in". -One way to play it could simply be to add a professor as a Cc: in your email to the postdoc, as a way of saying "BTW Prof, this is what I'm up to these days" without requesting a response on that part. Furthermore that does give the postdoc a chance to ask the professor about you before responding, should (s)he feel that advisable. -Another way to go ahead, especially if the postdoc you found ends up not being interested in your ideas (this happens; what research one wants to engage in is as much a matter of personal taste as one's choice of music), is to explore the possibility of you giving a talk at your old institution; maybe there is someone else there who would be interested. How feasible this is depends on the character of the institution; in a large department with research seminars every day it might be hard for a hobbyist to squeeze in, but in a smaller department (or one which for some reason doesn't have that many seminars going on) it can be easier to find an opening. Being an alumnus (even if at a low level) is a plus for you — many professors find it somewhat novel to hear of a former student who isn't himself a math professor doing math anyway. Talks by alumni addressing current students is also its own genre in academia, which can be given better resources than research activities. -One matter that can be tricky is knowing where to put the emphasis in your talk, when you do not know your audience. (Some things they all know and do not need to be reminded of, other things could be elementary to you but novelties to them.) If you get as far as being invited to give a talk then you may however be able to get the host to comment on an outline for your planned talk, if you ask nicely. They usually want the talk to be interesting as well.<|endoftext|> -TITLE: About the dual of the cube lemma in homotopy theory -QUESTION [6 upvotes]: Consider the Reedy category $2\rightarrow 1 \leftarrow 0$. Consider a map of diagrams of topological spaces $D\to E$ over this Reedy category: - -The maps which are fibrations are depicted with the symbol $\twoheadrightarrow$: the map of diagrams $D\to E$ is a pointwise fibration, and moreover the maps $D_2\to D_1$ and $E_2\to E_1$ are fibrations as well. This implies that the two diagrams $D$ and $E$ are Reedy fibrant. We suppose moreover that the map $D_2 \twoheadrightarrow E_2 \times_{E_1} D_1$ (not depicted in the image) is a fibration as well. Therefore the map of diagrams $D\to E$ is a Reedy fibration. For the Reedy model structure, the inverse limit is a right Quillen functor. This implies that the map $f$ from the inverse limit of D to the inverse limit of E is a fibration. -Here is now the question. - -With these hypotheses, is it sufficient to conclude that -$$\varprojlim D \longrightarrow \varprojlim E \times_{E_0} D_0$$ -is a fibration as well ? If not, what is missing ? - -REPLY [8 votes]: Yes, $D_0 \times_{D_1} D_2 \to (E_2 \times_{E_1} E_0) \times_{E_0} D_0$ is a fibration. -First, observe that -$$(E_2 \times_{E_1} E_0) \times_{E_0} D_0 \cong E_2 \times_{E_1} D_0 \cong (E_2 \times_{E_1} D_1) \times_{D_1} D_0$$ -by the pullback pasting lemma. Also, -$$D_2 \times_{D_1} D_0 \cong D_2 \times_{E_2 \times_{E_1} D_1} ((E_2 \times_{E_1} D_1) \times_{D_1} D_0)$$ -but we assumed that $D_2 \to E_2 \times_{E_1} D_1$ is a fibration, so -$$D_2 \times_{E_2 \times_{E_1} D_1} ((E_2 \times_{E_1} D_1) \times_{D_1} D_0) \to (E_2 \times_{E_1} D_1) \times_{D_1} D_0$$ -is also a fibration. But this is isomorphic to $D_0 \times_{D_1} D_2 \to (E_2 \times_{E_1} E_0) \times_{E_0} D_0$ so we are done. (So, in fact, we don't need to assume anything else is a fibration.) -I apologise for the lack of diagrams. I think it is easier to understand if you draw them yourself.<|endoftext|> -TITLE: Why does the number of permutations of $n-1$ adjacent transpositions where the outputs are different equal $2^{n-2}$? -QUESTION [6 upvotes]: Maybe I'm wrong, but I just noticed that the different permutations of $(1,2)(2,3)(3,4),\dots,(n-1,n)$ seem to be $2^{n-2}$ and I don't know why this is true. Can someone help if I'm right about this and explain a little bit? -e.g.: $n=4$, $(1,2)(3,4)(2,3) = (3,4)(1,2)(2,3)$ and $(2,3)(1,2)(3,4) = (2,3)(3,4)(1,2)$ but $(1,2)(2,3)(3,4)$ and $(3,4)(2,3)(1,2)$ gives unique outputs. So the number of different permutations is $2^{4-2}=4$, I checked it for 5, 6, 7 and gives the same pattern. - -REPLY [14 votes]: These are the Coxeter elements of the symmetric group, and they correspond to orientations of the Type A Dynkin diagram, of which there are $2^{n-2}$.<|endoftext|> -TITLE: Harmonic functions on knot complements -QUESTION [7 upvotes]: In Axler's Harmonic Function Theory, he and his coauthors develop the theory of harmonic functions on spheres and discs by considering the restrictions of arbitrary polynomials on the sphere $S^{n-1} = \{x \in \mathbb{R}^n : ||x||^2 = 1 \}$ and taking the Poisson integral to get a harmonic polynomial in the interior ball. One can then take the Kelvin transform to get a harmonic function on the exterior of the sphere. This process yields a canonical projection $\mathscr{P}(\mathbb{R}^n) \to \mathscr{H}(\mathbb{R}^n)$, from the space of polynomials to the space of harmonic functions, factoring through the restriction map to $L^2(S^{n-1})$. -Does this theory generalize to knot complements? Say we have a knot $K \subseteq \mathbb{R}^3$, and we take a small tubular neighborhood $V$ around $K$, whose boundary is topologically a torus $T$. Given a function on the knot complement, one could restrict to $T$ and then solve the Dirichlet problem on the knot complement to get a projection like the one above. However, in the sphere case, there are many nice properties of the harmonic function theory; namely it comes with an efficient algorithm for computation of a harmonic polynomial basis of $L^2(S^{n-1})$ which involves repeatedly differentiating the function $f(x) = |x|^{2-n}$. -Is anyone aware of any theory along this vein? Are there any obstacles to generalizing what happens in the sphere case? - -REPLY [2 votes]: This is more of a comment, but way too long. First, two remarks on the initial part of the question: - -The Kelvin transform of a harmonic polynomial is of course harmonic, but it is not a polynomial. For example, the constant $1$ gets transformed into $|x|^{2-n}$. - -The extension of the projection $\pi : \mathscr P(\mathbb R^n) \mapsto \mathscr H(\mathbb R^n)$ is not clear to me. Let $\pi = \pi_2 \circ \pi_1$ be the factorisation mentioned in the question: $\pi_1$ maps polynomials to their restrictions to the unit sphere $\mathbb S^{n-1}$, and $\pi_2$ extends that harmonically to the unit ball $\mathbb B^n$. Then $\pi_2$ clearly extends to the usual extension from $L^2(\mathbb S^{n-1})$ to the harmonic Hardy space $\mathscr H^2(\mathbb B^n)$ in the unit ball, given by the Poisson integral. And for $\pi_1$ all that we need is to be able to restrict our function to the unit sphere and get something square-integrable (so for example the Sobolev space $H^{1/2}(\mathbb B^n)$ will do). However, if we require our projections to be in $\mathscr H(\mathbb R^n)$, the class of entire harmonic functions, then their power series converge everywhere, which is a severe restriction. I am not aware of any intrinsic characterisation of the inverse image of $\mathscr H(\mathbb R^n)$ through (the extension of) $\pi_1$, let alone $\pi = \pi_2 \circ \pi_1$. - - -When it comes to the main question, I have troubles understanding the proposed construction. Of course any (reasonable — say, integrable with respect to the surface measure) boundary values on $T$ correspond to a harmonic function $h$ in the complement of $V$, again given by a Poisson integral (with kernel which is no longer known explicitly). This $h$ is given uniquely if we assume, say, that $h$ is bounded at infinity. If we are lucky, this function $h$ might extend to the complement of $K$, but I am not aware of any reasonable conditions for such an extension to exist, even in the simplest possible setting when $K$ were a point and $T$ a sphere (this is then essentially what I was trying to describe in the first part of this comment, after a Kelvin transform). -So it looks like I got something completely wrong...<|endoftext|> -TITLE: Picard-surjectivity and Morita-equivalence -QUESTION [9 upvotes]: Let us say that an algebra $A$ over a field $k$ is Picard-surjective if the canonical map -$$ \mathrm{Aut}(A) \rightarrow \mathrm{Pic}(A)$$ -is surjective. Here $\mathrm{Pic}(A)$ denotes the group of isomorphism classes of invertible $A$-$A$-bimodules and the map sends an automorphism $\alpha$ to the $A$-$A$-bimodule $A_\alpha$, where the left action is the usual one and the right action is via $\alpha$. -Q: For any given finite-dimensional $k$-algebra $A$, does there exist a Morita-equivalent one that is Picard-surjective? -If not, I am interested in conditions under which this is true. I am mainly interested in the case $k=\mathbb{R}$ or $\mathbb{C}$, and for all the examples that I came up with so far, this seems to be correct, as far as I can tell. - -REPLY [8 votes]: Yes, the basic algebra of $A$ will be Picard-surjective. -The basic algebra is the endomorphism algebra $\operatorname{End}_A(\bigoplus_{i=1}^{n}P_i)$ of the direct sum of indecomposable projective (right) modules, one from each isomorphism class. It is Morita equivalent to $A$. -Suppose $A$ is basic. Then as a left or right module, $A$ is the direct sum of indecomposable projective modules, one from each isomorphism class. Suppose $M$ is an invertible bimodule. Since $S\otimes_AM$ is nonzero for every simple module $S$, a direct sum decomposition of $M$ as a left module must contain at least one copy of each indecomposable projective. Let $X=\bigoplus_{i=1}^nS_i$ be the direct sum of simple (right) $A$-modules, one from each isomorphism class. Since $X\otimes_AM\cong X\cong X\otimes_AA$, as a left module $M$ must contain exactly one copy of each indecomposable projective in a direct sum decomposition. So $M\cong A$ as left modules. The right $A$-module structure of $M$ is then given by an injective algebra homomorphism $A^{op}\to\operatorname{End}_A(_AA)\cong A^{op}$, which is an isomorphism by finite dimensionality: i.e., the right action is induced by an automorphism of $A$.<|endoftext|> -TITLE: Hodge diamonds of complex threefolds -QUESTION [7 upvotes]: There is no closed complex curve or surface with $h^{1, 0}-h^{0, 1}=1$. -Now consider threefolds. Can this condition be satisfied? -Is Serre duality in fact the only restriction on the Hodge diamond? - -REPLY [10 votes]: Let $R$ be a commutative ring with identity. The three-dimensional Heisenberg group over $R$ is -$$\mathbb{H}(3, R) = \left\{\begin{bmatrix} 1 & z^1 & z^3\\ 0 & 1 & z^2\\ 0 & 0 & 1\end{bmatrix} : z^1, z^2, z^3 \in R\right\}.$$ -The Iwasawa manifold $\mathbb{I}_3$ is the quotient of $\mathbb{H}(3, \mathbb{C})$ by the discrete subgroup $\mathbb{H}(3, \mathbb{Z}[i])$ acting on the left, i.e. $\mathbb{I}_3 := \mathbb{H}(3, \mathbb{Z}[i])\setminus\mathbb{H}(3, \mathbb{C})$. It is a compact complex threefold with $h^{1,0}(\mathbb{I}_3) = 3$ and $h^{0,1}(\mathbb{I}_3) = 2$. The holomorphic one-forms $dz^1$, $dz^2$, and $dz^3 - z^1dz^2$ on $\mathbb{H}(3, \mathbb{C})$ descend to $\mathbb{I}_3$ and form a basis for $H^{1,0}_{\bar{\partial}}(\mathbb{I}_3)$, while the $(0,1)$-forms $d\bar{z}^1$ and $d\bar{z}^2$ descend to a basis of $H^{0,1}_{\bar{\partial}}(\mathbb{I}_3)$. -As for your second question, consider the following potential Hodge diamond (which satisfies the symmetries imposed by Serre duality): -$$\begin{array}{ccccccc}&&&1&&&\\&&0&&0&&\\&0&&0&&0&\\0&&0&&0&&0\\&0&&0&&0&\\&&0&&0&&\\&&&1&&&\end{array}$$ -There does not exist a compact complex threefold with this Hodge diamond. In particular, Serre duality is not the only restriction on the Hodge diamond of a compact complex threefold. -To see this, first note that on a compact complex manifold, it follows from the Frölicher spectral sequence that $b_k \leq \sum_{p+q=k}h^{p,q}$. So if $X$ were to have the Hodge diamond above, it would satisfy $b_0(X) = b_6(X) = 1$ and $b_i(X) = 0$ otherwise; that is, $X$ is a six-dimensional rational homology sphere. It follows that $c_1(X) \in H^2(X; \mathbb{Z})$ and $c_2(X) \in H^4(X; \mathbb{Z})$ are torsion, so -$$\chi(X, \mathcal{O}) = \int_X\operatorname{Td}(X) = \int_X\frac{1}{24}c_1(X)c_2(X) = 0,$$ -but $\chi(X, \mathcal{O}) = h^{0,0}(X) - h^{0,1}(X) + h^{0,2}(X) - h^{0,3}(X) = 1 \neq 0$. Therefore, no such $X$ exists. -This fact can be generalised in two ways. - -The above argument can be used to show that if a compact complex threefold $X$ has $h^{1,1}(X) = 0$, then $1 + h^{0,2}(X) = h^{0,1}(X) + h^{0,3}(X)$ which gives a restriction on the Hodge diamond which does not follow from Serre duality. - -Consider the potential Hodge diamond with $h^{0,0} = h^{n,n} = 1$ and all other numbers zero (the case $n = 3$ gives the Hodge diamond above). This cannot be realised unless $n = 1$. As before, we see that if $X$ has the given Hodge diamond, then $X$ is a $2n$-dimensional rational homology sphere. In this paper, Aleksandar Milivojevic and I showed that a rational homology sphere which admits an almost complex structure must have dimension $2$ or $6$, see Theorem 2.2, so the Hodge diamond cannot be realised for $n \neq 1, 3$. For $n = 1$, the Hodge diamond is realised by $\mathbb{CP}^1$, while the argument above shows that it cannot be realised for $n = 3$.<|endoftext|> -TITLE: Subgroups of Sp(2g,Z) that map onto all Sp(2g,Z/m) -QUESTION [8 upvotes]: I stumbled into the following problem. I apologize for being a bit naive. - -For $g\geq 3$, consider the group $\mathrm{Sp}(2g,\mathbb{Z})$ of symplectic square matrices of order $2g$ with integral entries. -For every integer $m\geq 2$ let $p_m:\mathrm{Sp}(2g,\mathbb{Z})\rightarrow\mathrm{Sp}(2g,\mathbb{Z}/m)$ be the natural projection. -Let $G$ be a finitely-generated subgroup of $\mathrm{Sp}(2g,\mathbb{Z})$ such that $p_m(G)=\mathrm{Sp}(2g,\mathbb{Z}/m)$ for all $m\geq 2$ -(really "all $m$" and not just "all but finitely many $m$"). - -(1) Can I say that $G$ is the whole $\mathrm{Sp}(2g,\mathbb{Z})$? -If the answer to (1) is no, then: -(2) what are typical counterexamples? -(3) is there some further (non-tautological) hypothesis that would ensure that $G= \mathrm{Sp}(2g,\mathbb{Z})$? - -REPLY [12 votes]: The answer to your question as to whether $G=Sp_{2g}({\mathbb Z})$ is NO. $Sp_{2g}({\mathbb Z})$ contains a pro-finitely dense FREE subgroup (hence has infinite index) which is also finitely generated. In fact the number of generators may be taken to be four: -https://arxiv.org/pdf/1205.1140. -The same conclusion holds for any arithmetic subgroup of $G({\mathbb Q})$ for a ${\mathbb Q}$-simple semi simple algebraic group defined over ${\mathbb Q}$ which satisfies the congruence subgroup property.<|endoftext|> -TITLE: Regarding a positive Lebesgue measure set in $\mathbb{R}^2$ -QUESTION [12 upvotes]: Let $P\subset \mathbb{R}^2$ be a positive Lebesgue measure set. Then $P$ does not necessarily contain a subset of the form $A\times B$ where $A,B\subset \mathbb{R}$ are of positive Lebesgue measure. -For example consider $P=\{(x,y)\in [0,1]\times[0,1]:x-y\notin \mathbb{Q}\}.$ -This example leads me to ask: -Given any $P\subset \mathbb{R}^2,$ a positive Lebesgue measure set, does there exists a measure zero set $U\subset \mathbb{R}^2$ such that $P\cup U$ contains a subset of the form $A\times B$ where $A,B\subset \mathbb{R}$ are of positive Lebesgue measure? - -REPLY [7 votes]: The question was answered by Robert Israel 1995 on Usenet, essentially by the set mentioned in fedja's comment. The proof that this set has the required property is carried out in detail in Example 4.3.1 of M. Väth, Ideal Spaces, Springer 1997. -Here is a sketch of the proof: Let $F\subseteq[-1,1]$ be closed with empty interior and positive measure and $P=\{(t,s)\in[0,1]\times[-1,2]:t-s\in F\}=\{(t,t+s):\text{$t\in[0,1]$, $s\in F$}\}$. Then $P$ has positive measure by the Cavalieri principle. Assume by contradiction that there are sets $A,B$ of positive measure such that $N=(A\times B)\setminus P$ is a null set. Consider $$x(s)=\int_A\chi_B(s+t)dt.$$ Then $x$ is continuous (translation is $L_1$-continuous) and vanishes a.e. on the complement $C$ of $P$, because by Fubini-Tonelli $$\int_Cx(s)ds=\text{mes}N=0.$$ Since $C$ is dense, it follows that $x$ is the null function. But again by Fubini-Tonelli $$\int_{-\infty}^{\infty}x(s)ds=\text{mes}A\,\text{mes}B,$$ a contradiction. -Robert Israel originally formulated the proof in terms of a convolution (IIRC $\chi_A*\chi_B$); the argument above is a variant of that.<|endoftext|> -TITLE: Contractible Rips complex from non-hyperbolic group -QUESTION [8 upvotes]: I heard that the Rips complexes associated to the Cayley graphs of hyperbolic groups are contractible for a sufficiently large radius. Is the converse true? Namely, if a group is non-hyperbolic, then is its Rips complex never ``asymptotically" contractible? -For example, we can ask if the non-hyperbolic group $\mathbb{Z}^2$ has contractible Rips complex associated to its Cayley graph for large radius. -I don't know much geometric group theory, and quick Google searches for this question weren't enough to fetch me an easy answer for this. This seemed like a natural question that would have been addressed at some point, so if anyone has a standard answer for this, it would be nice to know :) - -REPLY [8 votes]: Another source of Cayley graphs with contractible Rips complexes comes from Helly graphs. -Proposition: Rips complexes of uniformly locally finite Helly graphs are contractible. -See Lemma 5.28 and Theorem 4.2(v) from the preprint arXiv:2002.06895. -One construction of Helly graphs is the following: Given a CAT(0) cube complex $X$, the graph obtained from $X^{(1)}$ by adding an edge between any two vertices which belong to a common cube is a Helly graph. And there exist many groups admitting one-skeleta of CAT(0) cube complexes as Cayley graphs. For instance: -Corollary: Let $\Gamma$ be a finite simplicial graph. Let $G$ denote the Cayley graph of the right-angled Artin group $A(\Gamma)$ constructed from the generating set $\{ u_1\cdots u_n \mid u_1, \ldots, u_n \in V(\Gamma) \text{ pairwise adjacent} \}$. Then all the Rips complexes of $G$ are contractible. -Notice that $A(\Gamma)$ is hyperbolic if and only if $\Gamma$ has no edges, so most of these examples are not hyperbolic. For instance, the corollary includes $\mathbb{Z}^2$ with the generating set $\{(1,0), (1,1), (0,1)\}$. Other groups having one-skeleta of CAT(0) cube complexes as Cayley graphs include right-angled Coxeter groups and right-angled mock reflection groups. -Edit: I should also mention arxiv:1904.09060, which proves that Garside groups (such as braid groups) and Artin groups of type FC also admit Cayley graphs that are Helly, and so whose Rips complexes are contractible.<|endoftext|> -TITLE: Symplectic structure on the square of a 3-manifold -QUESTION [9 upvotes]: Let $M$ be a connected closed orientable 3-manifold. Assume $M$ is not the direct product of a surface and the circle. -Can there be a symplectic or Kähler manifold homeomorphic to $M\times M$? I think this might work if $M$ is a non-trivial circle bundle over the torus. - -REPLY [12 votes]: Let $M$ be a 3-manifold fibering over $S^1$, so there exists a fibration $\Sigma \to M \to S^1$. -Then $M\times M$ will admit a symplectic structure. -There is a symplectic structure on $M\times S^1$, associated to the fibration $\Sigma \to M\times S^1 \to S^1\times S^1=T^2$ which is trivial in the second factor, by a result of Thurston (the converse also holds). -Similarly, there is a fibration $\Sigma \to M\times M \to S^1\times M$ which is trivial in the second factor. Hence by the result in the fourth paragraph of Thurston's paper, $M\times M$ admits a symplectic structure. -These manifolds cannot be Kähler usually. See Theorem 1.2 of Biswas-Mj-Seshadri, which implies that if $M\times M$ is Kähler, then either $M$ is a manifold admitting Nil geometry (the fundamental group is commensurable with the Heisenberg group $H$), or $M$ is finitely covered by $\Sigma \times S^1$ for some surface $\Sigma$ of positive genus. See Dmitri Panov's answer for a non-trivial example. Question 4.4 in the paper leaves open the possibility of whether $H\times H$ can be Kähler. - -REPLY [9 votes]: Here is a Kahler example. Consider a hyper-elliptic curve $C$ of positive genus with involution $\sigma$. Take $C\times S^1$ and quotient $C\times S^1$ by $\mathbb Z_2$ that rotates $S^1$ by a half-turn and acts by $\sigma$ on $C$. Now, to get a Kahler structure on $M\times M$ take $C\times C\times E$, where $E$ is an elliptic curve and quotient by an obvious action of $(\mathbb Z_2)^2$.<|endoftext|> -TITLE: Reciprocity for fans of bounded Dyck paths -QUESTION [7 upvotes]: This is a continuation of some questions asked by Johann Cigler: Number of bounded Dyck paths with "negative length" and Number of bounded Dyck paths with negative length as Hankel determinants. -Let $\mathcal{D}(k,n)$ denote the following planar directed graph: - -It has $k+1$ vertices in the leftmost column and $n+1$ vertices in the bottom row. It always has an odd number of columns, and even number of rows. Also, all edges are directed from left to right. -For $0\leq i \leq k+1$, let $C(k,i;n)$ denote the number of $i$-tuples of nonintersecting lattice paths in $\mathcal{D}(k,n)$ which connect the bottom $i$ vertices of the leftmost column to the bottom $i$ vertices of the rightmost column. -Note that these tuples of nonintersecting lattice paths could also be called $i$-fans of $(2k+1-2(i-1))$-bounded Dyck paths of semilength $n$. -There is of course a Lindström-Gessel-Viennot determinantal expression for $C(k,i;n)$. -Conjecture/Proposition: As a function of $n$, $C(k,i;n)$ satisfies a linear recurrence with constant coefficients. -The reason this should be true is via a "transfer matrix"-style argument. We can make $\mathcal{D}(k,n+1)$ from $\mathcal{D}(k,n)$ by adding two columns on the right; and if we consider $i$-tuples of nonintersecting lattice paths in $\mathcal{D}(k,n)$ that start at the bottom $i$ vertices of the leftmost column, there are finitely many patterns of sinks at which they could terminate at; and in turn there are a fixed number of ways to continue these patterns for the two additional columns. -If that is indeed so, then we can define $C(k,i;-n)$ at negative values via the recurrence. -Question: Do we have the "reciprocity" result that $C(k,i;-n)=C(k,k+1-i;n+1)$? -The resolution of the previous questions implies that this is true for $i=1$ (and it is trivially true for $i=0$). - -REPLY [6 votes]: Let's say we have $n+1$ sets of vertices $V_t$, and for each $0\le t\le n$ we have $|V_t|=k+1$. The subsets of $V_t$ will often be identified with subsets of $\{1,2,\dots,k+1\}$. -Given some directed graph $G$ with $k+1$ sources and $k+1$ sinks satisfying the conditions of Lindström–Gessel–Viennot, we can form a graph $\widehat{G}_n$ by gluing together $n$ copies of $G$ as follows: for all $t$, the $t$-th copy has its sources identified with $V_{t-1}$ and its sinks identified with $V_t$. -Let $A$ be the $(k+1)\times (k+1)$ matrix whose $(i,j)$ entry counts the number of paths from source $i$ to sink $j$ in $G$. Let's denote by $A_s$ the matrix of $s\times s$-minors of $A$. The Lindström–Gessel–Viennot lemma tells us that the number of non-intersecting paths connecting $s$ sinks to $s$ sources in $G$ is the appropriate entry in $A_s$. Therefore the generating function for non-intersecting s-tuples of paths in graphs $\widehat{G}_n$ is given by -$$\sum_{n\geq 0} C(k,s,n)x^n=(I-xA_s)^{-1}$$ -where $C(k,s,n)$ denotes the $\binom{k+1}{s}\times\binom{k+1}{s}$ matrix where each entry counts the number of nonintersecting paths connecting the appropriate subsets of sinks and sources. This is just a rephrasing of the transfer matrix argument, and we see that each entry of $C(k,s,n)$ satisfies a linear recurrence. For the negative extension we obtain the generating function -$$\sum_{n\geq 1} C(k,s,-n)x^n=-(I-x^{-1}A_s)^{-1}=xA_s^{-1}(I-xA_s^{-1})^{-1}$$ -therefore $\sum_{n\geq 0} C(k,s,-n)=(I-xA_s^{-1})^{-1}$. Now up to a factor of $\det A$ (which for your original graph is 1) the inverse of the s-compound matrix is the s-adjugate matrix. When you unpack what this means for our situation it says that -$$C(k,s,-n)_{I,J}=(-1)^{\sigma(I)+\sigma(J)}C(k,k+1-s,n)_{J^{c}, I^{c}}$$ -where $I,J$ are subsets of size $s$ that index the sources/sinks and $\sigma(I)$ is the sum of the elements in $I$. This reciprocity is true for all graphs $G$ that have $\det A=1$. -Now returning to your graph, we have another symmetry at our disposal. Choosing $I$ to be be the lowest $s$ vertices we obtain from the argument above that $C_{I,I}(k,s,-n)=C_{I^c,I^c}(k,k+1-s,n)$. We observe that there is a very easy bijection between the non-intersecting family of paths joining the lowermost $k+1-s$ sources/sinks in $\mathcal D(k,n+1)$ and the non-intersecting family of paths joining the uppermost $k+1-s$ sources/sinks in $\mathcal D(k,n)$ (erase the first and last column and flip everything upside down). This proves the statement in your question.<|endoftext|> -TITLE: Can American Math. Monthly be used to publish hard research? -QUESTION [7 upvotes]: My question pertains to the journal "American Mathematical monthly" published by the MAA. -I wish to ask whether a paper as a part of a PhD thesis (subject: Combinatorics ) can be submitted to the AMM. Like, how does the community take publications in AMM to be? Most articles there are like extensions to Putnam/ IMO or such matter which can be called advanced undergraduate (according to my valuation). But, will a publication in AMM be valuable for PhD and post-doctoral work (like, will it be considered in the same vein as a publication in a specialized reputed journal), or will it be taken as a minor work? Thanks beforehand. - -REPLY [12 votes]: Definitely they publish serious research. For example, this paper -MR0379852 L. Zalcman, A heuristic principle in complex function theory. Amer. Math. Monthly 82 (1975), no. 8, 813–817 -has 175 citations as recorded on Mathscinet. (And 351 on Google Scholar). -And many other such examples can be given. As I understand their criterion is -that the paper is a) of sufficient interest to the broad audience, and b) does not require much background. There are plenty of serious research papers satisfying both criteria.<|endoftext|> -TITLE: Reference requests : Presentation of the braided dual of $U_q(\frak{sl_2})$ -QUESTION [5 upvotes]: I am interested in the braided dual of the quantum group $U_q(\frak{sl_2})$. This is the algebra generated by the matrix coefficients but where the multiplication is twisted by an action of the $R$-matrix. I have seen (for example in https://arxiv.org/pdf/1908.05233.pdf example 1.23) that it is isomorphic to the algebra generated by elements $a^1_1, a^1_2, a^2_1$ and $a^2_2$ together with the relations : -\begin{align*} -a^1_2 a^1_1 &= a^1_1 a^1_2 + ( 1-q^{-2})a^1_2a^2_2\\ -a^2_1 a^1_1 &= a^1_1 a^2_1 - ( 1-q^{-2})a^2_2a^2_1\\ -a^2_1 a^1_2 &= a^1_2 a^2_1 + ( 1-q^{-2})(a^1_1a^2_2 -a^2_2a^2_2)\\ -a^2_2a^1_1 &= a^1_1a^2_2\\ -a^2_2a^1_2 &= q^2 a^1_2a^2_2 \\ -a^2_2a^2_1 &= q^{-2} a^2_1a^2_2\\ -a^1_1a^2_2 &= 1 -q^{-2}a^1_2a^2_1 -\end{align*} -If $V$ is the standard representation of $U_q(\frak{sl_2})$ and we set $a^i_j := v^i \otimes v_j$, I can see that those elements indeed generate the whole algebra, but I don't know if there are more relations needed. According to the literature this is enough, but I cannot find a proof of this. - -REPLY [2 votes]: There are a few different ways to see that these relations are sufficient. - -One can appeal to the fact that the braided dual is a flat PBW deformation of the algebra O(SL_2), so that a basis is given by ordered monomials in the generators $a^i_j$, as has been proved in many places (and in which one may assume $a^1_2a^2_1$ does not appear, using the q-determinant relation. One can then confirm that these expressions satisfy the criteria of Bergman's diamond lemma, so that ordered expressions in the $a^i_j$'s form a basis of the algebra thusly presented, hence if there were any additional relations, it would break the flatness. This kind of computation is done in Juliet Cooke's paper (in more complicated examples) https://arxiv.org/abs/1811.09293, though of course this particular result you're asking about is much much older, appearing in papers in the 90's which I won't try and dig up. -This is essentially a variation on the above, or one way of proving the PBW claim being made there. Since Repq(SL_2) is semisimple, its braided dual has a Peter-Weyl type decomposition as the direct sum of $C(\lambda) = V_\lambda^* \otimes V_\lambda$, and one can see that the degree $\leq k$ elements in the filtration on the algebra above map isomorphically onto the subspace $C(0) + C(1) + ... + C(k)$. Here's a filtration because the q-determinant relation is not homogeneous. One can then see that if there were more relations than those listed, it would not define an injective map.<|endoftext|> -TITLE: Perfect sphere packings (as opposed to perfect ball packings) -QUESTION [8 upvotes]: I came across this question when I was discussing the rather wonderful Devil's Chessboard Problem with my colleague, Francis Hunt. -We realised that there is a nice connection to a packing question in $(\mathbb{F}_p)^n$ and I want to ask what is known about this. -Construction: Let $d$ be a positive integer and let $n=2^d$. Now construct the $d$-by-$2^d$ matrix $M$ over $\mathbb{F}_2$ which has vectors in $(\mathbb{F}_2)^d$ as columns (in some order). So, for instance, when $d=2$, we might have -$$ -M= \begin{pmatrix} -0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \end{pmatrix}.$$ -Now let $U$ be the solution set in $V=(\mathbb{F}_2)^{2^d}$ for the homogeneous system with coefficient matrix $M$. (In coding terminology, $U$ is the code with parity-check matrix $M$.) So, in the example above, we have -$$ -U=\left\{\begin{pmatrix}0\\0\\0\\0\end{pmatrix}, \begin{pmatrix}0\\1\\1\\1\end{pmatrix}, \begin{pmatrix}1\\0\\0\\0\end{pmatrix}, \begin{pmatrix}1\\1\\1\\1\end{pmatrix}\right\}. -$$ -In general $U$ has $2^{2^d-d}$ vectors $v_i$. For each $i$, let $S_i$ be the sphere in $V$ of radius $1$ with centre $v_i$. Then $|S_i|=2^d$ and it is easy to verify that the spheres $S_i$ partition the vectors in $V$. In other words these spheres are a perfect packing of the vector space. - -Question: Is this the only way to construct a perfect sphere packing in a finite vector space? - -Some comments: - -There's obviously loads of stuff in the literature on perfect packings but, so far as I can tell, they normally involve packing balls rather than spheres. Packing balls is the sensible thing to do when working with codes. -The subspace $U$ is clearly some kind of extension of the Hamming code... But not the extension that goes by the name of "the extended Hamming code"! In coding theory adding that column of 0's to the parity-check matrix is a dumb thing to do, but it works if you think about spheres instead of balls. -I have had some very preliminary thoughts about the numbers involved here. Suppose we are in a vector space $V=(\mathbb{F}_p)^n$ and, for $r=1,\dots, n-1$, we let $S_r$ be a sphere of radius $r$. Observe that $|S_r|=\binom{n}{r}(p-1)^r$. For a perfect packing to exist we need $|S_r|$ to divide $p^n$. Thus we must have $p=2$. We must also have $\binom{n}{r}$ equal to a power of $2$. I'm thinking that this can only happen if $r\in\{1,n-1\}$ but haven't been able to write down a proof. So... - - -Question. Is it true that $\binom{n}{r}$ is only equal to a prime power when $r\in\{1,n-1\}$? - - -Observe that in $(\mathbb{F}_2)^n$ any sphere of radius $r$ is also a sphere of radius $n-r$ (take the centre $v$ of the first sphere, change all the entries so that you get the unique vector at distance $n$ from $v$ and this will be the centre of the second sphere). Thus the construction given above can be thought of as a partition by $1$-spheres or by $(n-1)$-spheres. - -REPLY [5 votes]: Question: Is this the only way to construct a perfect sphere packing in a finite vector space? - -No. Take the linear space $V$ generated by the following vectors in $\mathbb{F}_2^8$: -$(0,0,0,1,1,1,1,0)$ -$(0,0,1,0,1,1,0,1)$ -$(0,1,0,0,1,0,1,1)$ -$(1,0,0,0,0,1,1,1)$. -One can see that any two different elements in $V$ differs in at least $4$ places. Projecting $V$ to its first $7$ coordinates generates the Hamming(7,4) code, so any two different elements differs in at least $3$ places. Furthermore, all the elements of $V$ have even weight, so no two elements could differ in exactly $3$ places. -By a counting argument, there are $2^4 \times 8= 128$ elements in $\mathbb{F}_2^8$ that differ from some element of $V$ in exactly $1$ place, and they are exactly the elements with odd weight. -One can take a copy of $V$ and permute its coordinates to get a different linear space $V'$. This is possible because $|V|=16$, and there are $70$ weight-4 vectors in $\mathbb F_2^8$. Displace $V'$ by a odd-weight vector $\alpha$ and call it $U$. The elements that differ from some element of $U$ in exactly $1$ place are exactly those with even weight. Thus $V \cup U $ is a perfect sphere packing. -The only thing left is to check that $V \cup U $ is not a linear space. To see this, let $v\in V \text{\\} V'$. Then $0$, $\alpha$, $v$ are elements of $V \cup U$, but $v+\alpha$ is not, so $V \cup U$ is not a linear space.<|endoftext|> -TITLE: Deformation equivalent vs diffeomorphic to projective manifold -QUESTION [9 upvotes]: Let $M$ be a closed complex manifold that is not deformation equivalent to a complex projective manifold. -Can $M$ be orientedly diffeomorphic to a complex projective manifold? What if $M$ is moreover Kähler? - -REPLY [10 votes]: I believe the answer is yes and follows from the combination of Theorem 4.6 here https://arxiv.org/pdf/math/0111245.pdf -and Theorem 1.3 here https://arxiv.org/pdf/math/0111245.pdf -The first result shows that deformations of standard complex tori are complex tori (i.e. $\mathbb C^n/\Gamma$ where $\Gamma\cong \mathbb Z^{2n}$). The second result shows that on $T^6$ there is an infinite dimensional family of complex structures. -PS. As for the second version of the question, where $M$ is asked additionally to be Kahler, I would guess that it can be safely counted as an open problem. Remember in https://link.springer.com/article/10.1007/s00222-003-0352-1 Voisin solved negatively Kodaira problem by constructing the first ever example of a Kahler manifold that is not deformation equivalent to a projective one. It seems to me that since then no new examples were found of such a phenomenon. And as you pointed out in a different post, it was proven recently that in dimension $3$ every Kahler manifold can be deformed to a projective one, but this is hard work (strongly relying on dim $3$). So in order to advance in your question one has to advance in one of these two directions - trying to extend the $3$-dimensional result to dimension $4$ and trying to look for new constructions of Kahler manifolds... - -REPLY [8 votes]: Yes. -As everyone knows, flat complex structures on $\Bbb R^4$ are parametrised by two copies of $\Bbb CP^1$. Take your favourite elliptic curve $E$; map $e \mapsto -e$ gives two-sheeted ramified covering $p: E \to \Bbb CP^1$. Consider a complex structure on $E \times T^4$ where complex structure on the fiber over $e \in E$ corresponds to the point $p(e)$. This almost complex structure is integrable (somewhat easy exercise) and non-Kahler (if it was, then corresponding Kahler metric would be flat and globally pluriclosed, but computation shows that such metric could not be compatible). -There is a paper by Catanese from early 00s where he shows that any deformation of a flat complex 3-dim torus is again a flat torus (and same result for products of curves of higer genus with torus $\Sigma \times T^4$). (UPD: it's the first link in Dmitri's answer) -I'm not aware of any Kahler examples.<|endoftext|> -TITLE: A question about $p$-adic monodromy of abelian varieties -QUESTION [9 upvotes]: Let $S_0$ be a smooth (projective?) and (geometrically) connected scheme over a finite field of characteristic $p$ and let $S$ be its base change to an algebraic closure of the finite field. Let $\pi:A \to S_0$ be an abelian scheme of relative dimension $g$ such that the Newton polygon of $A[p^{\infty}]$ is constant. For all $\ell \not=p$ we can consider the local systems $R^1 \pi_{\ast} \mathbb{Q}_{\ell}$ as a representation $\rho_{\ell}$ of $\pi_1^{\text{ét}}(S)$, and deep results of Deligne tell us that this is a semisimple representation; in fact the Zariski closure of the image of $\rho$ is "independent of $\ell$". Similarly there is a a $p$-adic variant $\mathcal{E}$ of $R^1 \pi_{\ast} \mathbb{Q}_{\ell}$, which is an (overconvergent) isocrystal, it also has the same monodromy group (see https://arxiv.org/abs/1711.06669) as its $\ell$-adic cousins. -Let $\mathbb{X}_b$ be a $p$-divisible group over $\overline{\mathbb{F}}_p$ with the same Newton polygon as $A[p^{\infty}]$ and let $J_b(\mathbb{Q}_p) \subset G(\breve{\mathbb{Q}}_p)$ be the automorphism group of the isocrystal associated to $\mathbb{X}_b$. Then Proposition 4.3.13 of Caraini-Scholze (https://arxiv.org/abs/1511.02418) gives us a pro-étale $J_b(\mathbb{Q}_p$)-torsor which roughly speaking parametrises quasi-isogenies $A[p^{\infty}] \to \mathbb{X}_b$. -Question: Is there any relation between the Zariski closure of the image of $\rho':\pi_1^{\text{ét}}(S) \to J_b(\mathbb{Q}_p)$ and the geometric monodromy groups of $\rho_{\ell}$ and $\mathcal{E}$? -Question: Is the Zariski closure of the image of $\rho':\pi_1^{\text{ét}}(S) \to J_b(\mathbb{Q}_p)$ reductive? -Example: If $S_0=Y_1(N)^{\text{ord}}$ is the mod $p$ fiber of the ordinary locus of the modular curve, then the $\ell$-adic monodromy group is equal to $\operatorname{SL}_2$ and the Zariski closure of the image of $\rho'$ is $\mathbb{G}_m=J_b$. - -REPLY [4 votes]: Let me briefly answer your question. There are two $p$-adic analogues of $R^1\pi_*\mathbb{Q}_\ell/S_0$: a (convergent) $F$-isocrystal $\mathcal E$ and an overconvergent $F$-isocrystal $\mathcal{E}^\dagger$. These two objects define algebraic monodromy groups $G_F(\mathcal E)$ and $G_F(\mathcal E^\dagger)$, where the first algebraic group is naturally a subgroup of the other. The algebraic group $G_F(\mathcal E^\dagger)$ (which for simplicity I assume connected) is “the same” as the arithmetic monodromy group of $R^1\pi_*\mathbb{Q}_\ell$ (as proven in the article you quoted). Moreover, one can attach to $G_F(\mathcal E^\dagger)$ a cocharacter $\lambda$ induced by the slope filtration of $\mathcal E$. What I have also proven is that $G_F(\mathcal E)$ is the parabolic subgroup $P(\lambda)\subseteq G_F(\mathcal E^\dagger)$ attached to $\lambda$. As a consequence, if $G(\rho')$ is the Zariski closure of your $\rho'$ and $Z(\lambda)$ is the centraliser of $\lambda$, there is an exact sequence $$1\to G(\rho')\to Z(\lambda)\to T\to 1$$ where $T$ is a torus. In particular, $G(\rho')$ is a reductive group. I gave two talks here in Bonn about a generalisation of this result. You can find some extended notes at http://guests.mpim-bonn.mpg.de/daddezio/MSPC-BN19.pdf. If you have any further questions you can send me an email. -EDIT: Here is the article about the results of this post: https://arxiv.org/abs/2012.12879 (see Proposition 5.1.4).<|endoftext|> -TITLE: $(\infty,2)$-categories: current applications and future prospects -QUESTION [17 upvotes]: Lately there has been a lot of progress on the foundations of $(\infty,2)$-categories (for example, all currently-known models for them were shown to be equivalent and finally we have a construction of the Gray tensor product). -While I'm aware they are used in Gaitsgory–Rozenblyum, I don't know of any other applications (out of ignorance, not because there aren't other ones)... - -Q$_1$: What are some current applications of $(\infty,2)$-categories (say in homotopy theory or DAG)? -Q$_2$: What are some expected future applications of them, say as the theory gets in a better overall shape and starts to be more widely-adopted? - -REPLY [15 votes]: Topological field theory (TFT) is a major client of higher-dimensional category theory. For $(\infty, -2)$-categories specifically, this specializes to two-dimensional TFT. One significant research area in this field -is taking physics ideas, making them mathematically rigorous, and then using the resulting mathematical objects to -prove interesting theorems internal to mathematics. -For example, string theorists originally predicted a form of duality now known as “mirror symmetry,” -coming out of a more general physics phenomenon called T-duality. The context is that given $X$ a Calabi-Yau -$3$-fold, physicists studied a two-dimensional supersymmetric quantum field theory called "the sigma model with -target $X$" and found that it admits two twists, called the A and B models, which are topological field theories -(in the physicists' sense). Mirror symmetry predicts that given $X$, there is another CY3 called the mirror -of $X$ and denoted $X^\vee$, such that the A model with target $X$ is equivalent to the B model with target -$X^\vee$, and vice versa. -Now, because these are topological field theories, it should be possible to make sense of them mathematically: -the A and B model should admit descriptions as symmetric monoidal functors from an $(\infty, 2)$-category of -bordisms in dimension 2 to some target $(\infty, 2)$-category. This has amounted to a lot of hard work by many -teams of researchers, but large parts of this story have been translated into mathematics, and the -higher-categorical language is important here!<|endoftext|> -TITLE: Combinatorial meaning of Kazhdan-Lusztig-Stanley polynomial -QUESTION [7 upvotes]: This question is motivated by - -Why do combinatorial abstractions of geometric objects behave so well? -The algebraic geometry of Kazhdan-Lusztig-Stanley polynomials - - -Kazhdan-Lusztig-Stanley polynomials (KLS) are vast generalizations of the classical Kazhdan-Lusztig polynomials, whose special values have deep meaning in representation theory [1]. They also include the matroid analogue studied in the past ten years. (Aside: they also include general zeta functions.) -KLS bridge combinatorics and algebraic geometry. While the nonnegative coefficients of the KLS can be interpreted as the dimension of suitable cohomologies of certain perverse sheaves [2], it seems to be less transparent in pure combinatorial settings. -I hope to understand KLS more from its combinatorial perspective, without any interference by the geometrical side. However, the definition of KLS [2] is done algebraically on the deformed dual of the underlying poset, making its meaning less transparent. -Question - -Why did combinatorialists consider KLS in their point of view? -Any baby examples of posets whose KLS shows rich combinatorial information right away? - -Remark: I have no background in combinatorics. Being aware of that KLS also generalizes (in some sense) many combinatorial invariants (h-vector, g-polynomials).. I'd hope the answer can be pedagogical, and show the easiest nontrivial example. -Related - -Twisted Incidence Algebras and Kazhdan-Lusztig-Stanley Functions-[Brenti], in which a nonassociative algebra is naturally given. - -The Hodge theory of Soergel bimodules, hinting its relation with higher category theory. - -The Kazhdan-Lusztig polynomial of a matroid -, defining matroid analogue of KL polynomials. - -The algebraic geometry of Kazhdan-Lusztig-Stanley polynomials -. As Sam pointed out in the comment, this paper does a great job collecting many different examples. - -REPLY [2 votes]: This is maybe more addressed at the comments and the discussion with Timothy Chow, but I just wanted to point out that, at least in a certain context, there is a very very "concrete" description of the $h$-vector. Say $\mathcal{P}$ is a simple (convex, full-dimensional) polytope in $\mathbb{R}^n$. Then let $\phi$ be a generic enough linear functional on $\mathbb{R}^n$. Use $\phi$ to orient the $1$-skeleton of $\mathcal{P}$: orient an edge $uv$ from $u$ to $v$ if $\phi(u) < \phi(v)$ (since $\phi$ is generic there will not be ties). Then if $h=(h_0,h_1,\ldots,h_n)$ is the $h$-vector of $\mathcal{P}$ (defined in the usual way as a transform of the $f$-vector), we have that -$$ h_i = \# (\textrm{vertices $v$ with indegree $=i$})$$ -according to our orientation of the $1$-skeleton. So for instance this explains that the $h_i$ are positive, that $h_0+h_1+\cdots+h_n$ is the number of vertices; also we will have a $h_i=h_{n-i}$ symmetry which swaps indegree according to $\phi$ for outdegree according to $-\phi$, etc. -Incidentally, I don't know who to attribute this simple but nice perspective on the $h$-vector to; to me it is folklore. -EDIT: As Richard notes in the comments this perspective is the same as the idea of a line shelling for a simplicial polytope, which I guess was assumed by Schläfli in his proof of the Euler-Poincaré formula and formally established by Bruggesser and Mani.<|endoftext|> -TITLE: Proof of existence of Joyal model structure via Cisinski theory? -QUESTION [10 upvotes]: I'm looking for a proof of the existence of the Joyal model structure -- with its usual description -- which uses Cisinski theory directly. The closest thing I know of is Theorem 5.26 of Ara's Higher quasi-categories vs higher Rezk spaces, but even there it seems he needs to assume the existence of the Joyal model structure before he can compare it to a certain model structure obtained from Cisinski theory. -More precisely, let - -$\mathsf{IH} = \{\Lambda^k[n] \to \Delta[n] \mid 0 < k < n\}$ be the set of inner horns; - -$\mathsf{J} = \{\Delta[0] \to J\}$ where $J$ is the walking isomorphism. - - -Then we have the following -Theorem: (Existence of the Joyal model structure) There exists a Cisinski model structure on $sSet$ such that - -$(\ast(\mathsf{IH} \cup \mathsf{J})):$ The fibrant objects (resp. fibrations between fibrant objects) are those objects (resp. morphisms between fibrant objects) which have the right lifting property with respect to $\mathsf{IH} \cup \mathsf{J}$. - -What I'm looking for is a write-up of the following proof outline of the above theorem: -Proof Sketch: By Cisinski theory, there exists a Cisinski model structure on $sSet$ such that $\ast(\Lambda(\mathsf{IH} \cup \mathsf{J}))$ holds, where for any $S$ we define - -$\Lambda^0(S) = S \cup \{\Delta[n] \cup \partial \Delta[n] \times J \to \Delta[n] \times J\}$ - -$\Lambda^{n+1}(S) = \{B \times \partial \Delta[1] \cup A \times J \to B \times J \mid A \to B \in \Lambda^n(S)\}$ - -$\Lambda(S) = \cup_n \Lambda^n(S)$ - - -We now verify (and these verifications are what I'd like to see!) that the morphisms of $\Lambda(\mathsf{IH} \cup \mathsf{J})$ can be constructed as retracts of transfinite composites of cobase changes of morphisms of $\mathsf{IH} \cup \mathsf{J}$. Therefore $\ast(\mathsf{IH} \cup \mathsf{J})$ and $\ast(\Lambda(\mathsf{IH} \cup \mathsf{J}))$ are equivalent, and we are done. - -I'm pretty sure that all the necessary combinatorics has been done somewhere, but I'd like to see it strung together. -I'd also be interested in an analogous approach to the Kan-Quillen model structure. - -REPLY [10 votes]: Such a proof is given in Chapter 3 of Cisinski's book Higher categories and homotopical algebra, see Definition 3.3.7 and Theorem 3.6.1. (Note that Cisinski's proof uses as the interval object not the nerve of the free-living isomorphism, but the simplicial set freely generated by a "left and right invertible" 1-simplex, see Definition 3.3.3.) -I attempted to give a geodesic such proof in the Appendix of my paper Joyal's cylinder conjecture, see Theorem A.7 (take $B = \Delta[0]$). I take the nerve $J$ of the free-living isomorphism as the interval object. This choice has the advantage that the necessary combinatorics can be boiled down to the following lemma (and the standard fact that the pushout-product of an inner horn inclusion with a monomorphism is inner anodyne), see Lemma A.5: -Lemma. Let $j \colon S \to T$ be a bijective-on-$0$-simplices monomorphism of simplicial sets, and suppose that $T$ is a quasi-category. Then the pushout-product of $j \colon S \to T$ with the inclusion $\{0\} \to J$ is inner anodyne. -For the Kan--Quillen model structure, see Section 3.1 of Cisinski op. cit.<|endoftext|> -TITLE: Generalization of symmetric functions -QUESTION [10 upvotes]: A $n$-variable function $f$ is a symmetric function if -$$f(x_1,x_2, \ldots, x_n) = f(x_{\sigma(1)}, x_{\sigma(2)}, \ldots, x_{\sigma(n)})$$ -for every permutation $\sigma \in S_n$. -In particular, if $f$ is a polynomial, then $f$ is a symmetric polynomial. -These objects have been studied extensively. -I wonder if the following generalization has been studied. -A $n^2$-variable function $f$ is an $S_n$-symmetric function if -$$f(x_{11}, x_{12}, \ldots, x_{1n}, \ldots, x_{nn}) = f(x_{\sigma(1)\sigma(1)}, x_{\sigma(1)\sigma(2)}, \ldots, x_{\sigma(1)\sigma(n)}, \ldots, x_{\sigma(n)\sigma(n)})$$ -for every permutation $\sigma \in S_n$. -I think such objects must have been studied as they are so natural. -But I don't know the keywords and couldn't find the literature. -I'm very grateful if anyone could provide information on them. -Thanks in advance. - -REPLY [9 votes]: Let $w\in S_n$ (the symmetric group) have cycle type -$\lambda =(\lambda_1,\dots, \lambda_\ell)\vdash n$, where -$\ell=\ell(\lambda)$ is the length (number of nonzero parts) -of $\lambda$. Then the induced action of $w$ on $[n]\times -[n]$ (where $[n]=\{1,2,\dots,n\}$) has cycle enumerator -$$ \prod_{i=1}^{\ell(\lambda)} \prod_{j=1}^{\ell(\lambda)} - z_{\mathrm{lcm}(\lambda_i,\lambda_j)}^{\mathrm{gcd}(\lambda_i,\lambda_j)}. $$ -Let $f_n(d)$ be the dimension of the space of complex polynomials in -the variables $x_{ij}$, $1\leq i,j\leq n$, that are -homogeneous of degree $d$ and $G$-invariant. Then by -Molien's theorem, -$$ F_n(x):=\sum_{d\geq 0} f_n(d)x^d $$ -$$ \ \ = \frac{1}{n!}\sum_{\lambda\vdash - n} \frac{n!}{z_\lambda}\prod_{i=1}^{\ell(\lambda)} \prod_{j=1}^{\ell(\lambda)} - \frac{1}{\left(1-x^{\mathrm{lcm}(\lambda_i,\lambda_j)}\right)^ - {\mathrm{gcd}(\lambda_i,\lambda_j)}}. $$ -I am using standard symmetric function notation, so -$n!/z_\lambda$ is the number of permutations in $S_n$ of -cycle type $\lambda$. For instance, -$$ F_1(x) = \frac{1}{1-x} $$ -$$ F_2(x) = \frac{1+x^2}{(1-x)^4(1+x)^2} $$ -$$ F_3(x) = \frac{1+3x^2+10x^3+16x^4+12x^5+16x^6+10x^7+3x^8+x^{10}} - {(1-x)^9(1+x)^4(1+x+x^2)^3}. $$ -Addendum. The invariant theory of finite groups, such as can be -found here, -can be used to obtain some further information about the ring $R$ of -invariant polynomials. For instance, if $S$ is the subring of all -symmetric functions in the $x_{ij}$'s, then $R$ is a -finitely-generated free $S$-module of rank $n^2!/n!$.<|endoftext|> -TITLE: Is there Matrix-Tree theorem for counting the bases of a connected matroid? -QUESTION [11 upvotes]: The famous Kirchhoff's Matrix-Tree theorem counts the number of spanning trees of a connected graph, that is, the number of bases of its cycle matroid. But it appeals to vertices, that's why I do not see how to generalize it to general matroids. Also the graphs with isomorphic cycle matroids may have quite different Laplacian matrices. However, possibly there exists a known generalization? - -REPLY [12 votes]: A broader class of matroids for which you have a Matrix Tree theorem are the regular matroids (those representable over every field): see, e.g., https://arxiv.org/abs/1404.3876. -EDIT: Let me actually try to give a very simple explanation of what's going on here. -Let $\mathbf{M}$ be an $n\times m$ matrix representing (i.e., its columns represent) our rank $n$ regular matroid $M$. In the case of a graph, this would be the vertex-edge incidence matrix (well, almost- we have to delete the first row of the incidence matrix to get a full rank matrix). Then the analog of the (reduced) Laplacian is given by $\mathbf{L}:=\mathbf{M}\mathbf{M}^T$. Now, the key (in fact, by a result of Tutte, equivalent) property of regular matroids is that we can choose $\mathbf{M}$ to be totally unimodular, meaning every minor is $=0, \pm 1$. Let's say we've done that. Then a routine application of the Cauchy-Binet formula shows that $\mathrm{det}(\mathbf{L})$ is equal to the number of $n$-tuples of linearly independent columns of $\mathbf{M}$, i.e., the number of bases of $M$. -Note that here we only used that the maximal minors of $\mathbf{M}$ are $0,\pm 1$, so maybe you can get away with slightly less than a regular matroid, I'm not sure.<|endoftext|> -TITLE: The contractivity of the heat semigroup in $L^p$ spaces -QUESTION [5 upvotes]: Let $M$ be a Riemannian manifold. By functional calculus, it is immediate to show that the heat semigroup is a contraction in $L^2(M)$. I can also show that it is a contraction in any $L^p(M)$ with $p \in [1, \infty]$, but this requires a bit of work. The thing is that I need to use this contractivity property in a work of mine, and I don't want to waste journal pages by providing a (not so short) proof of a result that I believe is already known and quite a classic. I already know of theorem X.55 in volume 2 of Reed & Simon, but this only treats the case of $M$ of finite measure. -Edit: Crucially, I do not want to assume the existence of the heat kernel (this must be obtained as a by-product, in fact), and in general I am willing to assume only those facts about the heat semigroup that are immediate consequences of its construction through functional calculus. - -Could you please point me to a citeable reference that proves the $L^p$-contractivity of the heat semigroup (or any other more general result from which this contractivity can be immediately obtained)? - -Once again: I am not looking for a proof, I already have one. I need a reference to include in an article, instead of my own proof, and for arbitrary manifolds (not just of finite measure). - -REPLY [6 votes]: For the record, the result requested in the question is given in Theorem 1.3.3 in: E.B. Davies, Heat Kernels and Spectral Theory, DOI:10.1017/CBO9780511566158. -The assumptions are: - -$L$ is a positive definite self-adjoint operator given by a quadratic form $Q$ on the Hilbert space $L^2(\Omega)$, where $\Omega$ is a $\sigma$-finite measure space; -if $u$ is in the domain of the form, then $|u|$ is in the domain of the form, too, and $Q(|u|) \leqslant Q(u)$ (this is one of the equivalent conditions of Theorem 1.3.2 in the same book); -if $u$ is in the domain of the form and $u \geqslant 0$, then $v = \min\{u,1\}$ is in the domain of the form, too, and $Q(v) \leqslant Q(u)$ (this is one of the equivalent conditions of Theorem 1.3.3 therein). - -The proof proceeds by proving first that $\exp(-t L)$ is contractive on $L^\infty$, using duality to get the result for $L^1$, and applying Riesz–Thorin interpolation to conclude.<|endoftext|> -TITLE: Can every simple polytope be inscribed in a sphere? -QUESTION [17 upvotes]: It is known that not every convex polytope (even polyhedron, e.g. this one) can be made inscribed, that is, we cannot always move its vertices so that - -all vertices end up on a common sphere, and -the polytope has not changed its combinatorial type in the process. - -Is there anything known about whether this is possible if we instead ask for simple polytopes, i.e., $d$-dimensional polytopes of vertex-degree $d$? - -REPLY [21 votes]: Not all simple polytopes are incribable, e.g. the dual of the cyclic polytope $C_4(8)$ is simple and not inscribable, as shown recently in Combinatorial Inscribability Obstructions for Higher-Dimensional Polytopes by Doolittle, Labbé, Lange, Sinn, Spreer and Ziegler -In dimension $3$, there is a combinatorial criterion by Rivin describing inscribabilty completely. I think already a cube with corner cut, which is simple, will be a non-inscribable $3$-polytope. -This can be checked with the following two lines of sage: -sage: C = polytopes.cube().intersection(Polyhedron(ieqs = [[15/8,1,1,1]])) -....: C.graph().is_inscribable() -False - -sage: C.is_simple() -True - -It's nice that Rivin's criterion is implemented in sage... -Here's an image of the graph of the "cube without one corner" 3-polytope, which is non-inscribable and simple: - -I just checked that this is the smallest non-inscribable simple 3-polytope: all other simple 3-polytopes with up to 10 vertices are inscribable.<|endoftext|> -TITLE: Chern classes of a mapping torus vector bundle in terms of the construction data -QUESTION [10 upvotes]: Let $\pi:E\to X$ be a complex vector bundle*, and $f:E\to E$ a bundle isomorphism. -Consider the mapping torus -$$E(f) := \frac{E\times [0,1]}{E \times \{0\}\sim_f E \times \{1\}}$$ -where the identification is the obvious one: $(x,0)\sim_f (f(x),1)$. -$E(f)$ is also a complex vector bundle over $ X\times \mathbb{S}^1$, the fibration is given by the map $[x,t]\in E(f) \mapsto (\pi(x), t)\in X\times \mathbb{S}^1$. -The rank of $E\to X$ and $E(f)\to X\times \mathbb{S}^1$ is the same. -Problem: - -Express the Chern classes of $E(f)$ in terms of the Chern classes of $E$ and the automorphism $f$. - -This naively should be possible since all the data used to build $E(f)$ is in $E$ and $f$. -Of course if $f=id$ then the characteristic classes are the same. Probably also the case when $f$ is of finite order may be carried out. -$E(f)$ depends only on the isotopy class of $f$ therefore, the characteristic classes of $E(f)$ should tell us (something) about whether $f$ lies in the identity component of the gauge group $\Gamma(Aut(E))$. -If this is correct however the problem might not be that easy, since I believe that $\pi_0 (\Gamma(Aut(E)))$ for generic $X$ (any dimension, any rank of $E$) is not known. But maybe I am missing something, please correct me if I'm wrong. -Since characteristic classes are related to sections, it is natural try to understand them. -If we have a section $u: X\to E$, we can construct a section $\mu:X\times \mathbb{S}^1\to E(f)$ by putting $\mu(x,t) = [u(x)(1-t) + tf (x)(u(x)),t]$. -Notice $\mu(x,t)$ is zero when $u(x)$ meets a $\lambda$-eigenspace of $f(x)\in Aut(E|_x)$ with $\lambda<0$ (or when u(x) is zero)). -However eigenspaces of $f$ do not define a subbundle (dimension can jump). I expect these subspaces to show up in a possible formula (if there is one). -Hopefully experts in homotopy theory will tell me more. -*Everything is smooth/manifold. - -REPLY [5 votes]: In the case that $E$ is trivial, there is a "universal" example of the construction you describe, which is the vector bundle on $S^1\times U(n)$ formed the "canonical" automorphism (each point acts on the fiber over it). For a general $X$ you take the pullback along the map to $S^1\times U(n)$ defined by the given automorphism. You can reduce the general case to this, assuming $X$ is compact, by adding to $E$ some bundle $F$ such that $E\oplus F$ is trivial and declaring the automorphism to be the identity on $F$. Note that $H^*(S^1\times U(n))$ is an exterior algebra generated by $x_1, y_1, y_3,...,y_{2n-1}$ (where the subscript of each generator indicates its degree) and the $k$-th chern class of our "universal" bundle on $S^1\times U(n)$ is $x_1y_{2k-1}$. In particular it seems that the problem you are posing is equivalent to the following: given a map $X\rightarrow U(n)$, determine its behavior on cohomology. -By the way, for a nice application of ideas similar to yours (study large automorphism groups using characteristic classes) check out this paper by Paul Seidel.<|endoftext|> -TITLE: Fano manifold becoming general type upon conjugation -QUESTION [5 upvotes]: Let $(M, J)$ be a Fano projective manifold. Can $(M, -J)$ be general type? -For complex curves and surfaces Kodaira dimension is diffeomorphism invariant so this cannot happen. - -REPLY [8 votes]: No. $(M,J)$ and $(M,-J)$ have conjugate pluri-canonical rings, hence have same Kodaira dimension. -Proof. Take a section $\mu$ of $K_{(M, J)}^{\otimes n}$, then $\bar \mu$ is a holomorphic section of $K_{(M,-J)}^{\otimes n}$. And vice versa.<|endoftext|> -TITLE: An averaging game on finite multisets of integers -QUESTION [9 upvotes]: The following procedure is a variant of one suggested by -Patrek Ragnarsson (age 10). Let $M$ be a finite multiset of -integers. A move consists of choosing two elements -$a\neq b$ of $M$ of the same parity and replacing them with the -pair $\frac 12(a+b)$, $\frac 12(a+b)$. If we continue to -make moves whenever possible, the procedure must eventually -terminate since the sum of the squares of the elements will -decrease at each move. What is the least and the most number -of moves to termination, in particular, if $M=\{1,2,\dots, -n\}$? If $M=\{a_1,\dots,a_n\}$, then an upper bound on on -the maximum number of moves is $\frac 12\sum (a_i-k)^2$, -where $k$ is the integer which minimizes this sum. (In fact, -$k$ is the nearest integer to $\frac 1n(a_1+\cdots+a_n)$.) -We can turn this procedure into a game by having Alice and -Bob move alternately, with Alice moving first. The last -player to move wins. (We could also consider the misère -version, where the last player to move loses.) Which -multisets are winning for Alice, especially -$M=\{1,2,\dots,n\}$? The game is impartial, so it has a -Sprague-Grundy number. However, it doesn't seem to be useful -for analyzing the game since a position $M$ never breaks up -into a disjoint union (or sum) of smaller independent -positions. Nevertheless we can ask for the Sprague-Grundy -number of a position $M$. - -REPLY [4 votes]: This doesn't address the whole question, but symmetry considerations show that when $M = \{1,2,\ldots, 2m\}$, the second player has a winning strategy. Details are below... -Let's say that a multiset $M$ is symmetric about $c$ if the multiplicity of an element $x$ in $M$ is equal to the multiplicity of $2c-x$. By taking the sum of the elements, we see that $M$ can be symmetric about at most one element $c$; $c$ is forced to be the arithmetic mean of $M$. During the game, $M$ may cease to be symmetric, or may become symmetric, but the point of symmetry is determined. (Since $M$ consists of integers, such a $c$ would be forced to be in $\frac{1}{2}\mathbb{Z}$, so this doesn't happen for most multisets of integers.) -In the case where $M = \{1,2,\ldots, 2m\}$, $M$ is symmetric about $c=m+\frac{1}{2}$. Consider the following strategy for the second player, Bob. In the previous turn, Alice chose two numbers $a_1, a_2$ of the same parity. Bob chooses $b_1 = 2c-a_1, b_2 = 2c-a_2$. If $M$ was symmetric before Alice's turn, then the fact that $a_1, a_2 \in M$ implies $b_1, b_2 \in M$. Bob's move is then guaranteed to be valid because $b_1, b_2$ have the same parity, which is different to the parity of the elements $a_1, a_2$ chosen by Alice (so Alice cannot have removed either of $b_1, b_2$ prior to Bob's turn, because $a_1, a_2$ have different parity). Moreover, it is also easy to see that Bob's move returns $M$ to a state that is symmetric about $c$. So Bob will always be able to play, and therefore will win. -This argument doesn't extend to the odd case. Suppose $M = \{1,2,3,4,5\}$. Alice could remove $1, 3$, and the symmetric entries, $3$ and $5$, are not a valid move for Bob. Alternatively, Alice could remove $2, 4$ which gives Bob a symmetric board state.<|endoftext|> -TITLE: Non-commutative duality I: Which C*-algebras are (isomorphic to a) convolution algebra? -QUESTION [14 upvotes]: Many interesting C*-algebras can be realized as convolution algebras over a groupoid, an idea introduced in 1980 by Jean Renault (this entry in nLab provides plenty of context to the general approach of attaching an algebra to a groupoid). -Perhaps due to my incompetence in this formidable field, I was not able to identify some results in the way of characterizing which C*-algebras are in fact convolution algebras. -So, here is my question: - -Is there a duality theorem stating that the sub-category of C*-algebras satisfying (add here some properties) is dual to the category of locally compact groupoids ? - -ADDENDUM: I have changed the title to make clear what I said in my comments below: though this is a very specific question (and by no means trivial, see the excellent responses below), it is just a tassel toward a broad goal, namely get some kind of duality theory for non commutative spaces. I was (happily) surprised not only with the answers I have received, but also by the views and the likes this question has gotten. That means that numerous folks share this interest. There will be further questions under the same rubric, and I hope they will attract other members of MO from a broad variety of backgrounds (operator theory, higher category theory, mathematical physicists, and generally researchers with a passion for duality theory) - -REPLY [3 votes]: There are already excellent answer so I hope my small remarks here will not seem to be too trivial. They concern the geometric origin, in a way, of the Fourier morphism mentioned by Simon Henry in a very specific contect. -When quantizing Poisson manifolds into NC $C^*$-algebras on possible way is through what is called groupoid quantization. The procedure is, here, to build the symplectic groupoid integrating the original Poisson manifold (thus a very specific kind of Lie groupoid) and then performing geometric quantization on this symplectic manifold keeping in mind, in a way, its groupoid structure. -The relevant part is the choice of a polarization compatible with the groupoid structure: what is called a multiplicative polarization. Such polarizations induce a groupoid fibration from the original symplectic groupoid to a quotient groupoid (which is not Lie but only topological) and a cocycle on this quotient groupoid coming from the symplectic structure. The (twisted by the cocycle) groupoid $C^*$ algebra thus resulting is the outcome of the groupoid quantization procedure. -Different choice of polarizations may result in quotient groupoids much differing one from the other and in the quotient cocylce being trivial or not. The typical example is an invariant symplectic structure $\omega_\theta$ ($\theta\not\in \mathbb Q$) on the torus which, depending on the choice of polarization, may be either quantized by the irrational rotation algebra (groupoid $\mathbb Z\ltimes_\theta \mathbb S^1$ with trivial cocycle) or by a trivial groupoid based on $\mathbb R^2$ (0-isotropy) but with non trivial cocycle depending on $\theta$. That the two are isomorphic may be considered a quantization does not depend on polarization type of result. The relation between the two is a partial Fourier transform applied to one of the two variables, in a way, and in general whenever the symplectic groupoid can be identified with a cotangent bundle as a manifold (which happens for a wide class of Poisson manifolds) different choices of polarization defines some Fourier-type relation between different (twisted) groupoid $C^*$-algebras. -One of the subtle points here is that in some sense groupoid $C^*$-algebras behave in a contravariant functorial way with respect to the base and in a covariant functorial way with respect to the isotropy so that all relations (like this choice of polarization) that result in some interchange of variables between base and isotropy have a quite complicated functorial description. - -Eli Hawkins, A groupoid approach to quantization, J. Symplectic Geom. 6 Number 1 (2008) 61–125. Project Euclid, arXiv:math/0612363.<|endoftext|> -TITLE: BCT equivalent to DC -QUESTION [7 upvotes]: Do you know where I can find proof of equivalence Baire Category Theorem and DC (Axiom of Dependent Choice)? It is well known fact but I can't find appropriate literature with the proof. - -REPLY [5 votes]: Yet another formulation of Blair's proof is in M. Väth, Topological Analysis, DeGruyter 2012.<|endoftext|> -TITLE: Number theory in symmetric cryptography -QUESTION [10 upvotes]: One of the most famous application of number theory is the RSA cryptosystem, which essentially initiated asymmetric cryptography. -I wonder if there are applications of number theory also in symmetric cryptography. -Thank you in advance for any comment / reference. -NOTE: Since RSA is based on Euler's theorem, I'm looking for applications of number theory to symmetric cryptography that involve number-theoretic theorems at least as "complex" as Euler's theorem. For example, I do not consider Caesar cipher as an application of number theory to symmetric cryptography, because it uses only the most basic definition of modular arithmetic. - -REPLY [11 votes]: Here are a few interesting examples of symmetric primitives whose claimed security is/was based on number-theoretic problems: - -From the 1980s: the famous Blum-Blum-Shub deterministic random bit generator is a classic example. Let $N = pq$ be the product of two large safe primes, and consider the sequence defined by $x_{i+1} = x_i^2 \pmod{N}$, where $x_0$ is the random seed (which can be any value in $(\mathbb{Z}/N\mathbb{Z})^\times\setminus\{1\}$). After each squaring, you extract some of the bits of $x_i$ to form the pseudorandom stream. The security of the bit generator - that is, the indistinguishability from a uniform random stream - can be reduced to number-theoretic problems. The idea is that if you only take the least significant bit of $x_i$ (or up to $O(\log\log N)$) at each iteration, then breaking this generator reduces to solving the Quadratic Rediduosity Problem $\bmod N$. - -A second classic example (this time from the 1990s): the KN cipher (Knudsen-Nyberg) was a number-theoretic block cipher designed specifically to resist differential cryptanalysis. The cipher was applied to 64-bit blocks, and the round function was defined as follows: choose a basis of $\mathbb{F}_{2^{37}}$ where the operation $x \mapsto x^3$ is particularly efficient. Let $E: \mathbb{F}_{2}^{32}: \to \mathbb{F}_{2^{37}}$ be some affine map, and let $F: \mathbb{F}_{2^{37}} \to \mathbb{F}_{2}^{32}$ be the map defined by cubing in $\mathbb{F}_{2^{37}}$, followed by throwing away five coefficients of the polynomial representation (w.r.t. the "nice cubing" basis). Now, dividing the 64-bit cipher state into two 32-bit values $L$ and $R$ in $\mathbb{F}_2^{32}$, the round function is $(L,R) \mapsto (R,L+F(E(R)+K))$, where $K \in \mathbb{F}_{2^{37}}$ is the secret key. The nonlinearity of the cubing permutation is important. The KN-cipher was subsequently broken using higher-order differential cryptanalysis, but its ideas have proven influential: the more recent MiMC cipher, for example, revisits the KN-cipher targeting applications in multi-party computation and zero-knowledge proofs. - -An example from the 2000s using "deeper" results in number theory: the Charles-Goren-Lauter hash function. Here we consider the $2$-isogeny graph of supersingular $j$-invariants over a suitably large $\mathbb{F}_{p^2}$: this is an important example of a Ramanujan graph, and this is key to the construction. The bits of the message $(m_0,m_1,\ldots,m_n)$ drive a non-backtracking walk of length $n$ in the isogeny graph (which is $(2+1)$-regular, so at each step you have $2$ choices: "low" or "high" w.r.t. some ordering on $\mathbb{F}_{p^2}$, and you go "low" if $m_i = 0$ and "high" if $m_i = 1$). The final hash value is a projection of the ending point $j_n$ of your walk into $\mathbb{F}_p$. The security of the hash function reduces to problems connected with finding cycles in the isogeny graph, which are provably large. - -Edit (I forgot one of my favourites): Wegman-Carter authenticators, which give high-performance MACs (message authentication codes) with information-theoretic security. Here, take a $\ge k$-bit finite field $\mathbb{F}_q$ and fix an inclusion $\iota: \{0,1\}^k \to \mathbb{F}_q$ (everything will operate on $k$-bit chunks of data) and a mapping $\pi: \mathbb{F}_{q} \to \{0,1\}^t$ (this will produce a $t$-bit MAC). For each $n > 0$, we can define a map $(\{0,1\}^k)^n \to \mathbb{F}_q[X]$ by $$M = (M_1,\ldots,M_n) \mapsto f_M(X) := \iota(M_n)X^n + \cdots + \iota(M_1).$$ Now to produce (and verify) an authenticator for a message $M$ given a shared secret $(R \in \{0,1\}^k, S \in \{0,1\}^t)$, we compute $T = f_M(R)\oplus S$ (where $\oplus$ denotes XOR in $\{0,1\}^t$). A crucial part of the security argument depends on the distribution of evaluations of polynomials over finite fields (see e.g. Bernstein 2005 for an up-to-date description and analysis of this). - - -In all four examples, number-theoretic arguments are used to give strong justifications for the security of the primitive. But the last example is important because it is also used in practice: the Wegman-Carter construction can be seen in GHASH, which is used in AES-GCM (in this case, $q$ is a power of $2$), and it is also the basis of Poly1305, a high-speed software authenticator. AES-GCM and ChaCha20-Poly1305 are two state-of-the-art algorithms for Authenticated Encryption that are widely used on the internet today.<|endoftext|> -TITLE: Reference of J.L. Waldspurger's paper on Shimura correspondence -QUESTION [6 upvotes]: I want to find reference of Waldspurger's paper referred at "Sur les coefficients de Fourier des formes modulaires de poids demi-entier" J. Math. Pures Appl. (9) 60 (1981), no. 4, 375–484 (available here at J. Voight's web page). -The name of ref. is [W] J.L Waldspurger, Correspondance de Shimura -(J. Math. pures et appl., n 60, 1980, p.1-132). -Is there anyone who has the latter ref.? -I don't know the definition of Hecke algebra of -metaplectic group $\mathrm{SL}(2,A)\times\{1,-1\}$. -There are many senses of Hecke algebra but I don't know in the context of that paper ("Sur les coefficients de Fourier des formes modulaires de poids demi-entier" ) - -REPLY [4 votes]: Here is Sur les coefficients de Fourier des formes modulaires de poids demi-entier -and here is Correspondances de Shimura (not the 1980 paper, but a summary of that paper from 1983)<|endoftext|> -TITLE: What is known about sums of the form $\sum_{n=2}^{\infty}[\zeta(n)-1]^{p} $? -QUESTION [10 upvotes]: A fair bit is known about rational zeta series. This includes identities like $$ \sum_{n=2}^{\infty} [\zeta(n) -1] = 1 . $$ -Many more identities can be found in articles by e.g. Borwein and Adamchik & Srivastava (here). -So far, I have not been able to find identities for series involving powers of zeta values. For instance, I wonder what the collection of series $$ R(p) := \sum_{n=2}^{\infty}[\zeta(n)-1]^{p} $$ amounts to, for some positive integer $p$. -For $p=2$, we can use the first identity to establish: -\begin{align} \sum_{n=2}^{\infty} [\zeta(n)-1]^{2} &= \sum_{n=2}^{\infty} [\zeta(n)^{2} - 2 \zeta(n) + 1] \\ -&= \sum_{n=2}^{\infty} (\zeta(n)^{2} - 1) -2 \sum_{n=2}^{\infty} (\zeta(n)-1) \\ -&= \sum_{n=2}^{\infty}(\zeta(n)^{2} -1) -2 .\end{align} -In order to proceed with the sum on the left, we can plug in the definition of the Riemann zeta function, use the multinomial theorem and interchange the order of summation to obtain: -\begin{align} \sum_{n=2}^{\infty}(\zeta(n)^{2} -1) &= \frac{7}{4} - \zeta(2) + 2\sum_{m=2}^{\infty} \frac{H_{m-1-\frac{1}{m}} - H_{-\frac{1}{m}} - H_{m-1}}{m} \\ -\end{align} -Here, $H_{m}$ is the $m$'th Harmonic number. -Let $$S := \sum_{m=2}^{\infty} \frac{H_{m-1-\frac{1}{m}} - H_{-\frac{1}{m}} - H_{m-1}}{m} . $$ -I've considered using the following generalization of the Harmonic numbers for real and complex values $x$: $$H_{x} = \sum_{k=1}^{\infty} \binom{x}{k} \frac{(-1)^{k}}{k} $$ at $x=-\frac{1}{m}$, but I'm somewhat stuck at finding a useful expression for $\binom{-\frac{1}{m}}{k} $. -Questions: - -Can the sum $S$ be evaluated? -What is known about the series $R(p)$ when $p \in \mathbb{Z}_{\geq 2}$? -Are there any results regarding rational sums of powers of zeta values in the literature? - -Note: A copy of this question with fewer details can be found here - -REPLY [7 votes]: An integral representation:$$\sum (\zeta(n)^2-1)=\sum_{n\geqslant 2, (a,b)\ne (1,1)}\frac1{a^nb^n}=\sum_{(a,b)\ne (1,1)}\frac1{ab(ab-1)}\\= -\sum_{(a,b)\ne (1,1)}\int_0^1 x^{ab-2}(1-x)dx=\int_0^1 (1-x)\left(\sum_{b\geqslant 2}x^{b-2}+x^{-2}\sum_{a=2}^\infty \sum_{b=1}^\infty x^{ab}\right)\\=1+ -\int_0^1 \sum_{a=2}^\infty(1-x)\frac{x^{a-2}}{1-x^a}dx= -1+ -\int_0^1 \sum_{k=0}^\infty\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx.$$<|endoftext|> -TITLE: What is the natural motivation for smooth/étale/unramified morphisms restricting from formally smooth/étale/unramified morphisms? -QUESTION [5 upvotes]: (I asked it first in MathStackExchange but I haven't get an answer yet) -Smooth (resp. étale) morphisms are just locally finitely presented + formally smooth (resp. étale) morphisms. -For unramified morphisms, it is originally defined in EGA as locally finitely presented + formally unramified morphisms, but now they are widely accepted as locally of finite type + formally unramified morphisms. -My question is, why do we need to add the "locally finitely presented" or "locally of finite type" conditions in the "true definition" of smooth/étale/unramified morphisms? -According to vakil's discussion and this note about motivations of unramified morphisms, we can see that the three morphisms are analogues of some important notions in differential geometry: - -Smooth-Submersions: surjections on tangent space, e.g. $\mathbb{A}^9\to \mathbb{A}^5$ -Étale-Covering Spaces: bijections on tangent space, e.g. $\mathbb{A}^5\to \mathbb{A}^5$ -Unramified-Immersions: injections on tangent space, e.g. $\mathbb{A}^2\to \mathbb{A}^5$ - -From my point of view, given a morphism of schemes $f:X\to Y$, the natural analogue of surjection (resp. bijection, resp. injection) on tangent spaces is perfectly described by surjection (resp. bijection, resp. injection) of -$$\DeclareMathOperator{\Spec}{Spec}\DeclareMathOperator{\Hom}{Hom}\Hom_Y(\Spec A,X)\to \Hom_Y(\Spec A/I,X)$$ -where $\Spec A$ is any afine $Y$-scheme with $I^2=0$. -In the language of this note about motivations of unramified morphisms, they are all the "differential like data", and tangent vectors can be thought as differentals. So I would be happy to accept the above definitions as the defitions of smooth (resp. étale, unramified) morphisms. -Is there any natural motivations that we include these finiteness conditions? The idea "we need the fibres of smooth morphisms to be smooth varieties" is not enough to convince me, because there are still the case étale morphisms and unramified morphisms, also why do we need that naturally? -e.g. - -Is there any morphisms of schemes that are not expected to be smooth/étale/unramified intuitively but they fall into the cateogy of formally smooth/étale/unramified? So to exclude them we need to introduce finiteness condition. -Is there any big theorems that have to include finiteness conditions? -Maybe the true analogue indeed contains finiteness conditions from the begining? - -REPLY [6 votes]: The main difference between formally smooth/smooth and formally étale/étale is that the formal versions don't need to be open maps. Chevalley's theorem on constructible topology says that a locally finite presentation map is an open map for the constructible topology, and if in addition you include flatness, you can show that it is an open map for the specialization topology. A map is open for the Zariski topology if and only if it is constructibly open and specialization-open. -As a consequence, smooth/étale maps (being flat and locally of finite presentation) are (universally) open maps. -There are other reasons involving Noetherian approximation to include finite presentation as well. Formally smooth/étale maps don't descend along inverse limits, while smooth and étale maps do. (In the qcqs case, that is).<|endoftext|> -TITLE: Toroidal Heegaard splittings -QUESTION [9 upvotes]: Suppose I have a Heegaard splitting of a closed oriented irreducible 3-manifold $M$, defined by the Heegaard diagram $(\Sigma_{g},\{\alpha_{1},\dots,\alpha_{g}\},\{\beta_{1},\dots,\beta_{g}\})$. Are there any obvious sufficient or necessary conditions for the attaching curves for when $M$ is toroidal (or atoroidal)? -Any sort of lead would be helpful. - -REPLY [3 votes]: I have found the paper `The disjoint curve property and genus 2 manifolds' by Abigail Thompson, where she proves that if a Heegaard splitting for a 3-manifold $M$ does NOT have the disjoint curve property, then $M$ is atoroidal. -Let $M_{1}\cup_{\Sigma}M_{2}$ be a Heegaard splitting for $M$. Then the Heegaard splitting has the disjoint curve property if there exist essential simple closed curves $c, a$ and $b$ on $\Sigma$ where $c$ is disjoint from $a$ and $b$ on $\Sigma$, and $a$ bounds a disk in $M_{1}$, and $b$ bounds a disk in $M_{2}$.<|endoftext|> -TITLE: Integer decomposition property with a partial order -QUESTION [9 upvotes]: Let $\mathcal{P}$ be a convex lattice polytope in $\mathbb{R}^n$. We say that $\mathcal{P}$ has the integer decomposition property (or "is IDP") if for all $k\in \mathbb{N}$ and $\alpha \in k\mathcal{P}\cap\mathbb{Z}^n$, there are $\alpha_1,\ldots,\alpha_k \in \mathcal{P}\cap\mathbb{Z}^n$ such that $\alpha=\alpha_1+\cdots+\alpha_k$. IDP polytopes are a well-studied class with connections to commutative algebra, algebraic geometry, etc. -Here's a new/nonstandard notion. Let's say $\mathcal{P}$ is IDP$\leq$ if there exists a partial order $\leq$ on $\mathcal{P}\cap\mathbb{Z}^n$ such that for all $k\in \mathbb{N}$ and $\alpha \in k\mathcal{P}\cap\mathbb{Z}^n$, there is a unique multichain $\alpha_1 \leq \cdots \leq \alpha_k \in \mathcal{P}\cap\mathbb{Z}^n$ with $\alpha=\alpha_1+\cdots+\alpha_k$. -For example, the order polytope $\mathcal{O}(P)$ and the chain polytope $\mathcal{C}(P)$ of a poset $P$ are IDP$\leq$ where $\leq$ is the natural distributive lattice order (on order filters/antichains). -Question: Are there other interesting families of IDP$\leq$ polytopes? -Motivation: this property gives a canonical/algorithmic choice of decomposition for all latice points of dilates. Also then the zeta polynomial of $(\mathcal{P}\cap\mathbb{Z}^n,\leq)$ is the Ehrhart polynomial of $\mathcal{P}$. - -REPLY [2 votes]: Have you checked the family of marked order polytopes? These include the classical Gelfand-Tsetlin polytopes, and I think I can construct such a partial order in case of GT-polytopes. -Let $T \in kP_\lambda$, where $T$ is a GT-pattern, which is in bijection with some SSYT in $SSYT(k \lambda,m)$. We want to find a partial order, such that -$T=T_1 + \dotsb + T_k$ in a unique manner. -Well, the $T_i$ are elements in $SSYT(\lambda,m)$, -and $+$ is interpreted as concatenate-and-sort rows. So, if we take -the partial order on $SSYT(\lambda,m)$ where $T \leq T'$ iff -for each $i \leq j$, column $i$ from $T$ together with column $j$ from $T'$ -form a valid SSYT when placed next to each other. -We can then find the unique decomposition $T=T_1 + \dotsb + T_k$ -by letting $T_i$ be every $k$th column from $T$, starting with the $i$th one. -I think one can extend this construction to all marked order polytopes.<|endoftext|> -TITLE: Does Publ. Math. Institute Hung. have a new name? -QUESTION [9 upvotes]: Narrow question: Did Publ. Math. Institute Hung. (Publications of the Mathematical Institute of the Hungarian Academy of Sciences) change its name? I am finalizing the bibliography of an article in progress that refers to an article that supposedly appeared in Publ. Math. Institute Hung. (see e.g. this MathSE question), but my go-to webpage for official journal abbreviations, https://mathscinet.ams.org/msnhtml/serials.pdf, appears to list no such journal. Perhaps it's now Acta Math. Hungar. or Period. Math. Hungar. or Studia Sci. Math. Hungar.? -Broader question: What is the best resource for investigating such questions? When a journal changes its name, how is a working mathematician to know? - -REPLY [9 votes]: According to Mathscinet, the name of this journal was -A Magyar Tudományos Akadémia. Matematikai Kutató Intézetének Közleményei. -Abbreviation: Magyar Tud. Akad. Mat. Kutató Int. Közl. -It changed the name in 1977 and now is called Alkalmazott Matematikai Lapok. -But the paper you refer to is listed under the old name.<|endoftext|> -TITLE: Realizing mapping classes as isometries? -QUESTION [8 upvotes]: Let $\phi : M \to M$ be a diffeomorphism. Is there a metric $g$ on $M$ and a diffeomorphism $\psi$ isotopic to $\phi$ so that $\psi$ is an isometry with respect to $g$? I'm guessing the answer is no, since there are manifolds such that all metrics admit no nontrivial isometries and maybe some of these manifolds have a nontrivial mapping class group. -I'm actually particularly interested in the case of the 3-torus which has mapping class group $GL_3(\mathbb{Z})$. Which of these mapping classes can be realized as isometries for some metric? - -REPLY [6 votes]: To complement Dmitri Panov's answer: - -Let $\phi$ be a (smooth) diffeomorphism of a closed manifold. Then $\phi$ preserves a Riemannian metric if and only if $\phi$ is isotopic to a diffeomorphism of finite order. - -$\Leftarrow$ is clear. Conversely, suppose there's such a metric and let $G$ be its isometry group, which is a compact Lie group. Then the closure $H$ of $\langle\phi\rangle$ in $G$ is a compact abelian Lie group, hence isomorphic to $F\times T$ for some finite group $F$ and some torus $T$. In particular, every connected component of $H$ contains an element of finite order, and this applies in particular to the connected component of $\phi$. - -As a corollary, when it holds, the image of $\phi$ in the mapping class group ( = group of diffeomorphisms modulo smooth isotopy, I guess? hence also modulo continuous isotopy) has finite order (this is the contents of Dmitri Panov's answer). I don't know if the converse holds. In other words, I don't know if a mapping class of finite order always has a lift of finite order (comments welcome if you know more!).<|endoftext|> -TITLE: Axiomatic QFT, the reconstruction theorem and functional integrals -QUESTION [11 upvotes]: Before posting my question, let me make some remarks: -[MS] Salmhofer's book on renormalization begins with a nice discussion on Feynman's path integral. At some point, the author states the following: - -In quantum field theory, one is not dealing with a single particle, but with infinitely many particles, because one has to account for the creation and annihilation of particles. One can formally write down a Hamiltonian, but it becomes very difficult to give a mathematical definition of it. We shall simply define the theory by the functional integral. - -[AA] I think in the same spirit of the above statement, Abdelmalek's paper on QFT for mathematicians states that, from the mathematical point of view, the fundamental problem is to give meaning to and study the properties of integrals of the form: -\begin{eqnarray} -\mathbb{E}[\mathcal{O}_{A_{1}}(x_{1})\cdots \mathcal{O}_{A_{n}}(x_{n})] = \frac{\int_{\mathcal{F}}\mathcal{O}_{A_{1}}(x_{1})\cdots \mathcal{O}_{A_{n}}(x_{n})e^{-S(\phi)}D\phi}{\int_{\mathcal{F}}e^{-S(\phi)}D\phi}\tag{1}\label{1}. -\end{eqnarray} -Now, I'd like to understand both these statements, in particular the one in bold typed in [MS]. -[EW; BS] As discussed in Edson de Faria and Wellington de Melo's book and Reed & Simon's book, a first mathematical description of QFT was given by Garding and Wightman. They proposed a set of axioms, known today as Wightman axioms, which defines mathematically what we mean by a quantum field theory. This is called Axiomatic QFT. Also, there is a famous result called the Wightman reconstruction theorem which states that one can completely recover a QFT from its Wightman correlations functions. -QFT and Euclidean QFT are related by a Wick rotation to imaginary time. As a consequence, Wightman correlation functions become Schwinger functions and a set of axioms for Schwinger functions can also be defined. These axioms are called Osterwalder-Schrader axioms. As before, there is a reconstruction theorem which states that the Euclidean QFT can be fully recovered from its Schwinger functions. -Concerning the above discussion, I have two questions: -Q1: Are these reconstruction theorems the reason for both statements [MS] and [AA]? In other words, is Euclidean QFT's mathematical description basically a study of Schwinger functions (and, thus, functional integrals) because the underlying QFT can be recovered from them? -Q2: I've been told once that, although axiomatic QFT is very precise mathematically, it is still very limited in its ability to produce results in terms of the physics behind it. I was even told that axiomatic QFT is "more like a mathematical theory than a physics theory". I'm very inexperienced and I don't know if this is accurate or not, but with the above discussion, it seems to me that axiomatic QFT is not necessarily trying to produce results in terms of physics, but rather it is trying to produce a solid mathematical ground which will certainly contribute to finding results in physics at some point. Is this accurate? Moreover, is axiomatic QFT even limited? - -REPLY [6 votes]: Q1: This is basically correct. For a discussion, cf. the discussion in the first chapter of the book by Montvay and Münster and the references given therein. -Q2: This is quite correct. Axiomatic QFT can rigorously prove results like the CPT theorem or the spin-statistics theorem, but it is of very limited (not to say no) use in calculating physical observables. For a discussion cf. the book by Streater and Wightman, from whence I lift the (somewhat tongue-in-cheek) quotation about axiomatic QFT that - -[c]ynical observers have compared -them to the Shakers, a religious sect of New England who built solid barns and led -celibate lives, a non-scientific equivalent of proving rigorous theorems and calculating -no cross sections.<|endoftext|> -TITLE: $0$-surgeries on trefoil and figure-eight -QUESTION [11 upvotes]: Let $M$ and $N$ be $3$-manifolds obtained by zero-surgery on (left-handed) trefoil and figure-eight knot respectively. -What is the easy way to prove that $M$ and $N$ are not homeomorphic? -Note: When they are knot homology spheres (they are both homology $S^1 \times S^2$'s), I cannot use the classical invariants. - -REPLY [11 votes]: As long as we're collectively throwing the kitchen sink at this, note that the Alexander polynomial of the knot is an invariant of the 0-surgered manifold. So since the figure eight and trefoil knots have different polynomials, the 0-surgeries are not homeomorphic.<|endoftext|> -TITLE: Is every $A \in \mathrm{SL}_n(\mathbb C)$ a product of four unipotent matrices? -QUESTION [11 upvotes]: Is every matrix $A \in \mathrm{SL}_n(\mathbb C)$ a product of four unipotent matrices? -I have verified that this is true if $n = 2$, and I believe I have came across this result before. However, I cannot find a reference to this. - -REPLY [11 votes]: If you only need four factors then you can have much more than that. Namely, you can take as the unipotent matrices upper and lower uni-triangular matrices. It was noted by G. Strang in "Every unit matrix is a LULU" that for any field $K$ anfy $A\in M(n,K)$ can be presented as $A=U_1L_1U_2L_2$, where $U_i$ are upper uni-triangular and $L_i$ are lower uni-triangular. -In fact, this is true not only for fields, but also for rings of stable rank $1$ (in particular, for semilocal rings, for boolean rings, for the ring of all entire functions and for the ring of algebraic integers), see this paper (the result was known to the experts long before that, but mostly in terms of triangular decompositions). -As noted in the answer by Gjergji Zaimi, there is a paper by A. Sourour, where he proved that over a field $F$ there is a presentation as a product of three unipotent matrices (two for non-central elements). This can not be carried to the decompositions in terms of lower and upper uni-tringular marices. Namely, every matrix $A\in \operatorname{SL}(n, R)$ is a product $U_1LU_2$ if and only if $R$ is a boolean ring (reference). However, T. Toffoli proved that almost all (w.r.t. Lebesgue measure) elements of $\operatorname{SL}(n,\mathbb{R})$ are ULU, and that for any $A$ there exists a permutation matrix $P$ such that either $PA$ or $-PA$ is ULU.<|endoftext|> -TITLE: Deleting triangles in a graph -QUESTION [5 upvotes]: I'm sure it is well-known how many edges you must delete in a (highly linked) graph to destroy all cycles. Is it also known how many edges you must delete to destroy only all triangles? And even, how many you need if additionally, you may not use an edge to destroy more than one triangle? (Note the latter doesn't mean you can't touch a common edge, but after deleting this edge only one triangle "counts" as destroyed. Thus you need 4 edges for the graph $K_4$, and $K_6$ has no solution.) -A reference would be welcome, especially if the problem would be in NP (but a non-NP lower bound around $O(n^2)$ would also be fine). - -REPLY [2 votes]: Given a graph $G$, the problem of determining the minimum size $\tau(G)$ of a set of edges $X$ such that $G-X$ is triangle-free is indeed NP-complete. This was proved by Yannakakis. Regarding bounds for $\tau(G)$, we can consider the following dual problem. Let $\nu(G)$ be the maximum number of edge-disjoint triangles of $G$. Clearly $\nu(G) \leq \tau(G) \leq 3\nu(G)$. A famous conjecture of Tuza asserts: - -$\tau(G) \leq 2\nu(G)$, for all graphs $G$. - -The complete graph $K_4$ shows that this bound is best possible. Tuza's conjecture is still open, but it has been shown to hold for many restricted graph classes. For example it holds for threshold graphs by this recent paper of Bonamy, Bożyk, Grzesik, Hatzel, Masařík, Novotná, and Okrasa. The best general bound is by Haxell, who proved that $\tau(G) \leq \frac{66}{23} \nu(G)$ for all graphs $G$.<|endoftext|> -TITLE: When does $\left\Vert f(\mathbf{N}) - f(\mathbf{M})\right\Vert_{\mathrm{op}} \leq k\left\Vert \mathbf{N} - \mathbf{M}\right\Vert_{\mathrm{op}}$ hold? -QUESTION [9 upvotes]: Define the Frobenius norm of a matrix as $\left\Vert A \right\Vert_{\mathrm{F}}=\sqrt{\sum_{i,j} A_{ij}^2}$ and the operator norm as $\left\Vert A \right\Vert_{\mathrm{op}}=\sup_{x \not = 0} \frac{\left\Vert Ax\right\Vert_2}{\left\Vert x \right\Vert_2}$ where the the norm in the numerator and denominator are the standard Euclidean norm. -If $\mathbf{N}$ and $\mathbf{M}$ are normal matrices on a separable complex Hilbert space $H$, and $f$ is a Lipschitz function defined on the spectrum of both matrices $\Omega = \sigma(\mathbf{N}) \cup \sigma(\mathbf{M}) $ with Lipschitz constant $k$ then $\left\Vert f(\mathbf{N}) - f(\mathbf{M})\right\Vert_\mathrm{F} \leq k\left\Vert \mathbf{N} - \mathbf{M}\right\Vert_\mathrm{F}$. This is a result from Kittaneh (1985). Note that in this paper they use the notation $\left\Vert \cdot \right\Vert_2$ to be the Hilbert–Schmidt operator which I believe is the Frobenius norm in the finite dimensional case. -I found this recent survey paper on operator Lipschitz functions which states similar results. From what I can tell the norm isn't specified but I think the results are also for the Frobenius norm. -I would like to know - -Is the term "operator Lipschitz function" reserved for functions which have the property under the Frobenius norm, or is it a more general concept defined for any norms? -Does the result from Kittaneh (1985) hold for operator norms and if so what is the reference? (The proof in this paper seems specific to the Frobenius norm). The review paper says for equation 3.1.2 "It follows easily from the spectral theory for pairs commuting normal operators", I'm not quite sure what this is and if it holds for norms generally (I haven't had much much formal training in functional analysis or measure theory) - -Specifically for my research I have real symmetric matrices $\mathbf{N}$ and $\mathbf{M}$ with eigenvalues lying in the interval $[-1, 1]$. If I have a Lipschitz continuous function $f:[-1,1] \rightarrow \mathbb{R}$ (we can also add differentiability or infinitely differentiability as an assumption if it helps) does it hold that $\left\Vert f(\mathbf{N}) - f(\mathbf{M})\right\Vert_{\mathrm{op}} \leq k\left\Vert \mathbf{N} - \mathbf{M}\right\Vert_{\mathrm{op}}$ where $k$ is the Lipschitz constant of $f$? -I hope this question isn't too basic for MO, I asked on the mathematics SE but didn't get a response. - -REPLY [9 votes]: The term "operator Lipschitz function" is definitely not reserved to the Hilbert-Schmidt norm. On the opposite, I would say that it is mostly used for the operator norm (but not only, see for example https://arxiv.org/abs/0904.4095 ). In particular, the survey that you are citing is using the operator norm. -It is known that not every Lipschitz function is operator Lipschitz (for the operator norm). So the answer to you question 2) is negative. I think that this was a conjecture of Krein, and that the first countexample was produced by Farforovsakaya (1972). A very simple example (Davies and Kato) is given by the absolute value map, which is $1$-Lipschitz but only $\simeq \log(n)$-Lipschitz on $M_n(\mathbf{C})$ (for the operator norm). This is very closely related to the fact that the triangular truncation has norm $\simeq \log(n)$ on $M_n(\mathbf{C})$. In fact, any $1$-Lipschitz function is $O(\log n)$-Lipschitz on $M_n(\mathbf{C})$. -In another direction, sufficient conditions are known to ensure that a function is operator-Lipschitz (for the operator norm), in terms of Besov spaces (Peller, see the survey you cite). It is also known that Lipschitz functions are $O(\max(p,1/(p-1))$-Lipschitz on $S^p$, the Schatten $p$-class. This is the paper I cite above by Potapov and Sukochev. -Note: I am not sure, but I would expect that the fact that Lipschitz functions remain Lipschitz for the Hilbert-Schmidt norm was already known to Krein in the 1960's. - -REPLY [8 votes]: There are several inaccuracies or mistaken impressions in the OP's original question – I say this not to be denigrating, since it is always tricky reading the literature in a different mathematical community from one's own. However it seems worth writing an answer rather than "sniping in comments". - -First of all, Aleksandrov and Peller have written many papers on this theme, but in their setting they almost always mean the operator norm not the Frobenius norm. Hence, when you link to their paper https://arxiv.org/abs/1611.01593 you cannot use it to say things about the Kittaneh paper. -Regarding -Q1. My impression is that among the infinite-dimensional functional analysis community, "operator Lipschitz" is interpreted with respect to the operator norm. This is for instance what is meant by the Aleksandrov—Peller paper that you linked to, contrary to your impression. -Q2. Note that the equation in A+P that you refer to is for commuting operators only! When two matrices commute and are normal in the technical sense of that word, they can be simultaneously diagonalized (over the complex numbers) and hence estimates of the form you seek can be obtained directly. The difficulty, in both the Kittaneh paper and the series of papers by A+P, lies in the fact that the given matrices might not commute with each other. -Finally, there is an example going back to Kato, and refined in work of Davies I think, which shows that the absolute value function is not "operator Lipschitz". So in full generality the answer to Q2 is no. (I don't have the references to hand but will try to update this later)<|endoftext|> -TITLE: Toposes with only preorders of points -QUESTION [16 upvotes]: For a Grothendieck topos $\mathcal{E}$, are the following assertions equivalent? -$(i)$ $\mathcal{E}$ is localic. -$(ii)$ The diagonal geometric morphism $\mathcal{E} \to \mathcal{E} \times \mathcal{E}$ is an embedding. (Here $\mathcal{E} \times \mathcal{E}$ is the product topos, not the product category.) -$(iii)$ For every Grothendieck topos $\mathcal{E}'$, $\mathrm{Geom}(\mathcal{E}', \mathcal{E})$ is a preorder (no parallel geometric transformations). -The implications $(i) \Rightarrow (ii)$ and $(ii) \Rightarrow (iii)$ do hold: - -$(i) \Rightarrow (ii)$: Any diagonal morphism $X \to X \times X$ (in any category) is a split mono and a split mono of locales is an embedding. The (forgetful) functor from locales to toposes preserves the product and turns embeddings of locales into geometric embeddings. -$(ii) \Rightarrow (iii)$: If $\mathcal{E} \to \mathcal{E} \times \mathcal{E}$ is an embedding, then the diagonal functor $\mathrm{Geom}(\mathcal{E}', \mathcal{E}) \to \mathrm{Geom}(\mathcal{E}', \mathcal{E} \times \mathcal{E}) \simeq \mathrm{Geom}(\mathcal{E}', \mathcal{E}) \times \mathrm{Geom}(\mathcal{E}', \mathcal{E})$ must be fully faithful. But this means precisely that $\mathrm{Geom}(\mathcal{E}', \mathcal{E})$ is a preorder. - -So in summary, is a topos with only a preorder of $\mathcal{E}'$-based points for every $\mathcal{E}'$ already localic? - -REPLY [18 votes]: $(i) \Leftrightarrow (ii)$ is true and is Proposition C.2.4.14 in Peter Johnstone's Sketches of an elephant. More generally he shows that a bounded geometric morphism $f: \mathcal{E} \to \mathcal{S}$ is localic if and only if $\mathcal{E} \to \mathcal{E} \times_{\mathcal{S}} \mathcal{E}$ is an embedding. -$(ii)$ and $(iii)$ are not equivalent: there is a large gap between "the diagonal is a monomorphism" and "the diagonal is an embedding" -For a typical example, take a free but non-proper action of a group $G$ on a locale (or space) $X$. To fix the idea, take $G = \mathbb{Z}$ acting on $X=S^1$ the unit circle by rotation by an irrational angle. -The topos of equivariant sheaves $X//G$ classifies "orbits" for the action of $G$ on $X$, that is a $G$-torsor $T$ (a principale $G$-bundle) together with a $G$-equivariant map $T \to X$. Because the action is free, the category of point in any topos will have no non-trivial morphisms. -But that topos is not localic at all: its subterminal objects are the $G$-invariant open subset so in our concrete example its only $\emptyset$ and $1$. -One can also compute the diagonal map. $\mathcal{T} \times \mathcal{T}$ can be shown to be the topos corresponding to the action of $G \times G$ on $X \times X$. Subtopos of this would corresponds to $G \times G$-equivariant sublocales of $X \times X$ and the diagonal is not $G \times G$-equivariant. -To make more explicit construction, we can use that for a discrete group $G$, a topos localic over $BG$ (the topos of $G$-set) is the same as a local with a $G$-action. $\mathcal{T}$ corresponds to $X$ with its $G$ action. $\mathcal{T} \times \mathcal{T} \to BG \times BG$ is also localic (product of localic map), and the corresponding locale is obtained by pulling back along the point $* \to BG \times BG$, which allows to see that the corresponding locale is indeed $X \times X$. Now if I see $\mathcal{T}$ over $BG \times BG$ as $\mathcal{T} \to \mathcal{T} \times \mathcal{T} \to BG \times BG$, then it corresponds to the local $X \times G$ where $G \times G$ acts on $X$ and $G$ separately (to be more symetric it is the the locale of triplets $(x,x',g)$ where $x'=gx$). -As locale with $G \times G$ action, the diagonal map of $\mathcal{T}$ hence corresponds to the map $X \times G \to X \times X$ that sends $(g,x)$ to $(x,gx)$. Which is a mono because $G$ acts freely, but is not en embedding. -Of course some of the claim I made above would require a proof... but that might be a bit too long for MO.<|endoftext|> -TITLE: Treating citation numbers as objects -QUESTION [11 upvotes]: The following item appears in the $\LaTeX$ style-guide for the Journal of Integer Sequences: - -To me this seems correct and reasonable, but it in most articles I read, authors tend toward the first example of "wrong" usage. Is this simply journal-specific, or is it abusive to use "In [1]..."? - -REPLY [26 votes]: I don't think this kind of thing is "abusive." I'm certain I've done it in my own writing before, and I can think of several scenarios where it's better than the alternative: -(1) If a citation has no author, like a collaboratively edited industry standard for some software. -(2) If a citation has many authors and it's awkward to write "Smith, et. al. proved ... [1]" -(3) If a citation is an edited volume and it's unclear who contributed the thing you need to cite. -Generally, I think the guiding principle should be to do whatever will make your writing easier for the reader. It seems to me that there are many times when the reader just needs to know the result and which citation they should look it up in, rather than who was the first author on the paper where that result appears. Much more important to me than the question of whether I write "In [1], the authors prove..." instead of "So and so proved ... [1]" is that the citation contain a more precise reference, like Theorem 5.3.2 instead of just citing a 1000 page book and leaving the reader to go hunting for the result.<|endoftext|> -TITLE: Antiholomorphic involution with a fixed point -QUESTION [6 upvotes]: Let $M$ be a connected closed complex manifold. Assume it has an antiholomorphic involution. Must it have an antiholomorphic involution with a fixed point? - -REPLY [10 votes]: No. There exist both non-algebraic and projective counterexamples. -1 Non-algebraic example. Take a flat Euclidean torus $T^4=M$ and let $Z$ be its twistor space. It has an antiholomorphic involution without fixed points which is central symmetry in all the fibres. I claim that $Z$ doesn't have an anti-holomorphic involution that has a fixed point. -Suppose by contradiction that such $\sigma$ exists. Let $\tilde Z$ be the universal cover of $Z$. Recall that $\tilde Z$ is the complement to a line in $\mathbb CP^3$. Now, $\sigma$ induces an anti-holomorphic involution $\tilde \sigma$ on $\tilde Z=\mathbb CP^3\setminus \mathbb CP^1$. I claim that $\tilde \sigma$ extends a holomporphic self-map of $\mathbb CP^3$. The point is that $\tilde \sigma $ sends any complex line in $\mathbb CP^3\setminus \mathbb CP^1$ to a complex line in $\mathbb CP^3\setminus \mathbb CP^1$ . One can deduce from this that the map $\tilde \sigma$ is induced by a linear (i.e. degree $1$) map from $\mathbb CP^3$ to itself. It remains to check that the standard real involution on $\mathbb CP^3$ that fixes an $\mathbb RP^3$ doesn't commute with the action of $\mathbb Z^4=\pi_1(T^4)$ on $\mathbb CP^3$, which is not super hard. -2 Projective example. To get a projective example one should take a generic quartic curve in $\mathbb CP^2$ defined by a real equation but without real points. A generic such curve doesn't have a real involution that has a fixed point. Otherwise, by taking a composition of such an involution with the real involution we get a non-trivial holomorphic automorphism (which doesn't exist on a generic quartic)<|endoftext|> -TITLE: Skolem's method for checking truth-value assignments - a "cut-free proof procedure" for first-order logic? -QUESTION [5 upvotes]: In his intro to ( Skolem 1923a), Van Heijenoort (From Frege to Godel, p. 509) describes Skolem as giving “an alternative to the axiomatic approach” to proving a first-order formula. This is referring to the effective procedure Skolem gives for checking whether or not a first-order formula U has a solution of level n. A solution of level n is an assignment of truth-values to the atomic propositions of the nth level expansion. The nth level expansion of U is the conjunction of instances of U formed by dropping the quantifiers, letting the universal variables range over the domain of level n-1, and introducing new integers for the existential variables. -Skolem's procedure is as follow: - -Form the expansions of U up the nth level. -At each level, write down all possible truth-value assignments to the atomic propositions. -If a truth-value assignment at level m has no continuations at level m+1, reject it. - -U has a solution of level n if and only if there are truth-value assignments remaining when this procedure is carried out up to level n. If for some n there is no solution of level n, then we have shown that U is truth-functionally unsatisfiable. -VH says that this procedure “provides proofs that are cut free and have the subformula property”. I know what those properties are in the context of the sequent calculus, but I don’t understand what he means in this context. -References: -SKOLEM, THORALF -[1923a] Einige Bemerkungen zur axiomatischen Begründung der Mengenlehre. -Matematikerkongressen i Helsingfords den 4–7 Juli 1922, Den -femte skandinaviska matematikerkongressen, Redogörelse . Helsinki: -Akademiska Bokhandeln, 1923, pp. 217–232. English translation in van -Heijenoort (ed.) [1967], pp. 290–231. -VAN HEIJENOORT, JEAN -[1967a] From Frege to Gödel; a source book in mathematical logic, 1879-1931. Cambridge, Harvard University Press. - -REPLY [5 votes]: The answer to the question can be found in Section I (especially p.11) of this source (it is a newly typeset version of Joseph E. Quinsey's 1980 Oxford doctoral thesis Some Problems in Logic: APPLICATIONS OF KRIPKE’S NOTION OF FULFILMENT).<|endoftext|> -TITLE: Relative homology of free loop space with respect to constant loops -QUESTION [6 upvotes]: Let $Q$ be a closed manifold with $\dim Q\geq2$ and let $\Lambda_0Q$ be the connected component of the free loop space of $Q$ whose elements are contractible loops. I am looking for conditions on the homotopy groups $\pi_k(Q)$ ensuring that the relative homology $H(\Lambda_0Q,Q)$ with respect to the space of constant loops $Q$ is non-zero. -Using the relative Hurewicz theorem, I can see that $H(\Lambda_0Q,Q)\neq0$ if $Q$ is $2$-connected and I was interested in a sharper result. Is it true that it is enough to assume that $Q$ is $1$-connected? What about a non-simply connected manifold $Q$ having $\pi_k(Q)\neq0$ for some $k\geq2$? - -REPLY [8 votes]: Let me tackle the case that $Q$ is $1$-connected, but not $2$-connected. Because $Q$ is $1$-connected we have that $\Lambda_0Q=\Lambda Q$, as all loops are contractible. -Two sequences are relevant here: The long exact sequence for homotopy groups of the free loop space fibration $\Omega Q\rightarrow \Lambda Q\xrightarrow{\mathrm{ev}} Q$, and the long exact sequence in homology of the pair $(\Lambda Q,Q)$. -The evaluation of the basepoint $\mathrm{ev}$ has a section: It sends a point in $q\in Q$ to the constant loop at $q$. This means that the long exact sequence in homotopy groups splits into split short exact sequences: -$$ -0\rightarrow \pi_k(\Omega Q)\rightarrow \pi_k(\Lambda Q)\rightarrow \pi_k(Q)\rightarrow 0 -$$ -Now $\pi_k(\Omega Q)=\pi_{k+1}(Q)$, and all groups in sight are abelian because $Q$ is connected. The splitting lemma then shows that $\pi_k(\Lambda Q)\cong\pi_{k+1}(Q)\oplus \pi_k(Q)$. Similarly, the long exact sequence in homology breaks down and -\begin{equation} -H_k(\Lambda Q)\cong H_k(Q)\oplus H_k(\Lambda Q,Q) -\end{equation} -By assumption $\pi_2(Q)\not=0$. By Hurewicz (note that $\pi_1(\Lambda Q)$ is abelian) we have $H_1(\Lambda Q)\cong \pi_1(\Lambda Q)\cong \pi_2(Q)\not=0$. As $H_1(Q)\cong\pi_1(Q)=0$, the last displayed equation gives that $H_1(\Lambda Q,Q)\not=0$.<|endoftext|> -TITLE: Penrose’s singularity theorem -QUESTION [29 upvotes]: Roger Penrose won today the Nobel Prize in Physics for the singularity theorem, which at first glance seems to be a result in pure mathematics. -Questions about the theorem: - -What kind of mathematical technology was used to prove this? -What ideas did it require that were new at the time? -Has its interest since then been mainly in physics or has it also led to new mathematics? - -There is a related question on Penrose's broader contributions. - -REPLY [7 votes]: Wille's answer is technically true, but he doesn't talk about the historical context of the result. I think that is important for understanding why such a "simple" result is deserving of a Nobel prize. -Sections below are answers to the numbered questions in your question. - -The tech here is differential topology as applied to Lorentzian geometry. Penrose wrote a book, "Techniques of differential topology in Relativity". It's an amazingly well written account of the maths needed to prove his original singularity theorem. Good luck getting your hands on a copy. In no way do I endorse searching for the book on libgen. - -The actual results of this book are not like your Riemannian differential topology. He wasn't the originator of the basics, but he did put them together in a novel way. There are many many results that use the same kinds of techniques that have amazing consequences (see Krolak's cosmic censorship series for example, or Minguzzi's recent work about evaporating black holes). -What separates this kind of study of relativity from the "more normal approaches" is that it is purely kinematic. There are no dynamics involved. In particular Einstein's field equations arn't used. The only physical assumption is an "Energy Condition" which is expressed as an inequality to do with Ricci curvature. Penrose doesn't do a great job of justifying the inequality on physical grounds. For that I suggest you consult Hawking and Ellis which very clearly demonstrate that the inequality follows from some assumptions about how classical matter behaves. The inequality is used in the Raychaudhuri equation to ensure that solutions diverge in finite affine parameter. The inequality is just there to ensure a technicality is true. - -As mentioned the differential topology techniques had been (or were being) developed by a bunch of people. Geroch, Hawking, Ellis, Penrose and a little later Clarke, Kronheimer (and a few others) are the main names. Penrose's insight was to combine the various existing results in an interesting way. - -Fundamentally the singularity theorem works like this: Assume that the manifold is maximally extendend, assume a condition that implies that no conjugate points along geodesics can exist, and assume a condition that ensures that if a curve is complete then it has a pair of conjugate points. The resulting contradiction is used to justify that the manifold has an incomplete geodesic. -Beem, Erhlich and Easley presented a super refined version of this argument in Theorem 12.43 of their book. You should check this theorem out. If you read it I guarantee you will first say "WTF" then "but that follows directly from the assumptions". Penrose has been accused of this BTW. Some people have historically claimed that Penrose assumed what he proved. As a philosophical point this is true of all mathematical knowledge, but some how Penrose's singularity theorem is a little "too on the nose". -So what was the reason for the Nobel prize? Where is his original contribution? It comes from the context of physical research back when Penrose published. I know a paper from Senovilla has already been mentioned, but you should read an earlier and much better paper by him: https://arxiv.org/abs/1801.04912 "Singularity theorems and their consequences". -So... there's these three Russians. In 1963 Lifschitz and Khalatnikov publish a paper, (https://www.tandfonline.com/doi/abs/10.1080/00018736300101283) in which they explicitly state, "An attempt is made to provide an answer to one of the principal questions of modern cosmology: ‘does the general solution of the gravitational equations have a singularity?’ The authors give a negative answer to this question.' (That's a quote from the abstract). In 1965 Penrose publishes his paper which proves that Lifschitz and Khalatnikov's paper is bullshit. In 1970 Belinskii, Lifschitz and Khalatnikov (usually write BKL) publish a paper claiming to have demonstrated that generic solutions of Einstein's field equations do have singularities. -Lifschitz and Khalatikov are / were big names. Their second paper birthed the dynamical systems approach to cosmology and one of the more important conjectures in General Relativity (check out https://en.wikipedia.org/wiki/BKL_singularity). These guys work was important. -Their first paper is super complicated analysis of a certain class of solutions to Einstein's field equations that they claimed were generic. -Penrose's theorem is even more generic. As in so generic it has been claimed that the result is self evident. Yet... big wig in the field didn't "believe" in generic gravitational singularities. -In fact, just as Senovilla states in the paper above, Penrose's contribution was convincing astronomers that they should take black holes seriously. Because in in the 60's they didn't. They believed that black holes were a failure of GR and demonstrates that the Einstein's model for gravity was flawed. -That's why Penrose's Nobel prize is jointly awarded with the people who produced the physical evidence for black holes. That's his contribution. It's not actually about the math, it's about a result that changed the whole direction of an area of research. That's why it's a deserving award. - -Well yeah. Absolutely piles of new research and ides. I did my PhD on this stuff, and that wasn't that long ago. I could chat about this literally forever. The big big big question is: Given only the (super minimal) assumptions of the singularity theorems is it possible to provide bounds on curvature divergence? I.e. do the kinematics of gravity imply behaviour as an observer falls into a singularity or do you need dynamics too. I literally mean this: EVERY THING IN CURRENT RESEARCH IN GENERAL RELATIVITY IS ABOUT SINGULARITIES. Or at least can be traced back to it. Penrose's theorem gave GR a reason for existing beyond bragging rights for solving a difficult set of differential equations. For independent justification of this: https://en.wikipedia.org/wiki/History_of_general_relativity#Golden_age. - -So yeah... Beyond a more specific question about research I think yelling is ok right? -Edit: Sorry I left of some commentary about mathematical innovation. All singularities theorem look suspiciously like Riemannian rigidity results. So one mathmatical example of the influence of Penrose's work is "The Lorentzian Splitting Theorem", see Chapter 14 of BEE. More generally Penrose's theorem shows that geodesic completeness is a big issue in Lorentzian manifolds. More so since Hopf-Rinow is false. There's been a lot of work on understanding geodesic completeness as a result. There's also been a lot of work on geometrically inspired compactifications of Lorentzian manifolds (unlike the Riemannian case there is no canonical distance). There is an ongoing conference series on Lorentzian geometry. Here is a link to next years one: http://www.uco.es/gelocor/. You could have a look through the list of speakers and their topics to get more of an idea. -Lorentzian geometry is very difficult (due to the failure of basic results - like Hopf-Rinow) and is rife with unresolved issues of completeness. There is also not a great deal of interest in the area among mathematicians.<|endoftext|> -TITLE: Shannon entropy and doubly stochastic matrices -QUESTION [6 upvotes]: Suppose that $A$ is a stochastic matrix. We know that if $A$ is doubly stochastic, then $H(Ap)\geq H(p)$ where $H$ is Shannon entropy and $p$ is a probability vector. Is the converse true? i.e., if $H(Ap)\geq H(p)$ then $A$ is doubly stochastic. - -REPLY [4 votes]: Consider any $n \times n$ left stochastic matrix $A$, i.e. each column sums up to $1$. We argue that if $H(Ap) \geq H(p)$ for all probability distributions $p$, then $A$ is doubly stochastic. -Take $p$ to be the uniform distribution. Then $H(Ap) \geq H(p)$ implies that $Ap = p$, since the uniform distribution on $[n]$ is the unique maximizer of Shannon entropy among all probability distributions on $[n]$. -Since $p = \frac{1}{n} 1$ (the all ones vector), we have that $A1 = 1$ -- all rows sum up to $1$. Thus, $A$ is doubly stochastic.<|endoftext|> -TITLE: What are the main contributions to the mathematics of general relativity by Sir Roger Penrose, winner of the 2020 Nobel prize? -QUESTION [50 upvotes]: I received an email today about the award of the 2020 Nobel Prize in Physics to Roger Penrose, Reinhard Genzel and Andrea Ghez. Roger Penrose receives one-half of the prize "for the discovery that black hole formation is a robust prediction of the general theory of relativity." Genzel and Ghez share one-half "for the discovery of a supermassive compact object at the centre of our galaxy". Roger Penrose is an English mathematical physicist who has made contributions to the mathematical physics of general relativity and cosmology. I have checked some of his works which relate to mathematics, and I have found the paper - -M. Ko, E. T. Newman, R. Penrose, The Kähler structure of asymptotic twistor space, Journal of Mathematical Physics 18 (1977) 58–64, doi:10.1063/1.523151, - -which seems to indicate Penrose has widely contributed generally to the mathematics of general relativity like tensors and manifolds. Now my question here is: - -Question -What are contributions of Sir Roger Penrose, the winner of the 2020 Nobel prize in physics, to the mathematics of general relativity, like tensors and manifolds? - -We may motivate this question by adding a nice question which is pointed out in the comment by Alexandre Eremenko below where he asks: Is Sir Roger Penrose the first true mathematician to receive a Nobel prize in physics? If the answer is yes, then Sir Roger Penrose would say to us "before being a physicist you should be a mathematician". On the other hand, in my opinion the first mathematician to be awarded several physics prizes is the American mathematical and theoretical physicist Edward Witten. This seems to meet Sir Roger Penrose in his research such as cosmology and research in modern physics (Einstein general relativity). -Related question: Penrose’s singularity theorem - -REPLY [25 votes]: I answered about the incompleteness theorem in the other thread. Let's talk about some of his other contributions here. (This list is definitely incomplete*, but just some stuff off the top of my head.) -1 -The "black hole" theorem (incompleteness theorem) is closely related to, yet subtly different from, the Hawking-Penrose Singularity Theorems. The Hawking Penrose theorems again prove the geodesic incompleteness of spacetime under certain cosmologically reasonable assumptions. The difference is in the interpretation. The Penrose theorem proves the genericity of black hole formation; the Hawking-Penrose Theorem guarantees, in some sense, the genericity of the Big Bang. -2 -Penrose made significant contributions to how we understand causal geometry of space-times. A particularly interesting paper is Kronheimer and Penrose, "On the structure of causal spaces" (Proc. Camb. Phil. Soc. (1967)). In this paper they abstracted the relation between two space-time events (as being time like or light like) into a partial order. From this one is naturally led to study the ideals and filters, and their principality. This leads to a beautiful description of what the idealized "boundary at infinity" should look like for space-times. -3 -The GHP Calculus (named after the authors Geroch, Held, and Penrose of the 1973 paper "A space-time calculus based on pairs of null directions" (Journal of Mathematical Physics)) and the more general Newman-Penrose formalism ((1962) "An Approach to Gravitational Radiation by a Method of Spin Coefficients" (Journal of Mathematical Physics)) are some of the most common ways to perform symbolic computations in GR. -The calculus is a version of the Cartan formalism (or a special way of looking at Ricci rotation coefficients), but taking special advantage of the four dimensionality of space-time and the Lorentzian structure of spacetime. -4 -The Penrose inequality is a conjectured (and partially proven in many special cases) relation between the area of the apparent/event horizon of a black hole space-time with the mass (as observed at infinity) of the corresponding black holes. -This inequality actually lead to a lot of interesting recent works in Riemannian geometry. -5 -Also, he formulated and named the Strong and Weak Cosmic Censorship Conjectures. -6 -Penrose is also responsible for suggesting his namesake process for extracting energy from rotating black holes through backscattering. The process, combined with some putative nonlinear feedback mechanism, gained popular fascination under the martial name of the Black Hole Bomb. In the literature this is called the superradiant instability and has been proven to work in certain linearized matter models around rotating black holes (such as the Klein-Gordon model for massive scalar waves). -An interesting modern mathematical discovery is that the superradiant instability does not apply to massless scalar fields. Understanding how this works for tensor fields, especially for those solving the linearized Einstein equations, is a massive undertaking and crucial in the current effort to demonstrate nonlinear dynamical stability of the Kerr black hole. -7 -One way to probe the nonlinear effects of gravity is by understanding how gravitational waves can interact. Our experience from Fourier theory suggests that it can be useful to start with the interaction with plane wave pulses. This was treated first in Khan and Penrose "Scattering of Two Impulsive Gravitational Plane Waves" (Nature, 1971). The impact of this collision still reverberates to this day. (The state of the art, as I understood it, is that we can now understand a bit about what happens when we collide three waves. Four is still somewhat out of reach.) -8 -Finally, something a bit more whimsical, since I don't know anyone who actually uses it: the Penrose notation for tensor computations. I tried to use it for a few weeks when I was in graduate school, but gave up mostly because they are impossible to type up. - -* Pun very much intended.<|endoftext|> -TITLE: Does asymptotic behavior of $\left|\sum_{d|n}f(d)\right|$ imply asymptotic properties of $f(d)$? -QUESTION [6 upvotes]: The classic example of a function that has a drastic cancelation when summed over divisors is $\mu(n)$, with complete cancellation for every number other than $1$. Another such function is the Liouville function $\lambda(n)$. Both of these functions have have the property that $\sum_{n=1}^{\infty}\frac{f(n)}{n}=0$. Is this a general pattern? I.e do the conditions -\begin{equation} -\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}\left|\sum_{d|n}f(d)\right|=0\tag{i} -\end{equation} -\begin{equation} -f(n)=O(1)\tag{ii} -\end{equation} -imply -\begin{equation} -\sum_{n=1}^{\infty}\frac{f(n)}{n}=0\tag{iii} -\end{equation} -The interesting question here is about the $\mathit{convergence}$ of the sum. If we know that the sum in (iii) converges then we simply note that -$$\lim_{\epsilon\to0^+}\sum_{n=1}^{\infty}\frac{f(n)}{n^{1+\epsilon}}=\lim_{\epsilon\to0^+}\frac{1}{\zeta(1+\epsilon)}\sum_{n=1}^{\infty}\frac{\sum_{d|n}f(d)}{n^{1+\epsilon}}$$ -Now since the terms $\sum_{d|n}f(d)$ tend to be small, the sum $\frac{\sum_{d|n}f(d)}{n^{1+\epsilon}}$ will go to infinity slower than $\zeta(1+\epsilon)$ and so -$$\lim_{\epsilon\to0^+}\sum_{n=1}^{\infty}\frac{f(n)}{n^{1+\epsilon}}=0$$ -which implies that the value of the sum must be zero. A rigorous proof of the above statement is not much harder than the handwaving done above. - -REPLY [3 votes]: Interestingly enough, it is actually enough to know -\begin{equation} -\frac{1}{N}\sum_{n=1}^{N}\left|\sum_{d|n}f(d)\right|=o(1)\tag{1} -\end{equation} -to deduce -\begin{equation} -\sum_{n=1}^{\infty}\frac{f(n)}{n}=0\tag{2} -\end{equation} -to do this, we work instead under the change of variables $g(n):=\sum_{d|n}f(d)$ which shows that the problem $"(1)\implies(2)"$ is equivalent to saying that -\begin{equation} -\frac{1}{N}\sum_{n=1}^{N}\left|g(n)\right|=o(1)\tag{3} -\end{equation} -implies that -\begin{equation} -\sum_{n=1}^{\infty}\frac{\sum_{d|n}\mu\left(\frac{n}{d}\right)g(d)}{n}=0\tag{4} -\end{equation} -by Mobius inversion. Working with this sum, we get that -\begin{align*} -\sum_{n=1}^{N}\frac{\sum_{d|n}\mu\left(\frac{n}{d}\right)g(d)}{n}&=\sum_{n=1}^{N}\frac{\sum_{d|n}\mu\left(\frac{n}{d}\right)g(d)}{n}\\ -&=\sum_{d=1}^{N}\sum_{n=1}^{N/d}\frac{\mu\left(n\right)g(d)}{nd}\\ -&=\sum_{d=1}^{N}\frac{g\left(d\right)}{d}M\left(N/d\right) -\end{align*} -Where here -$$M(x):=\sum_{n0$ -\begin{align*} -\sum_{n=1}^{N}\frac{\sum_{d|n}\mu\left(\frac{n}{d}\right)g(d)}{n}&=\sum_{d=1}^{N}\frac{g(d)}{d}M(N/d)\\ -&=\sum_{d=N/L}^{N}\frac{g(d)}{d}M(N/d)+\sum_{d -TITLE: Group of surface homeomorphisms is locally path-connected -QUESTION [9 upvotes]: I think the following is true and I need a reference for the proof. (Given a closed surface $S$, i.e. a compact 2-dimensional topological manifold (without boundary), we endow $S$ with a distance generating its topology, and endow the set of self-homeomorphisms of $S$ with the distance max(uniform distance between two maps, uniform distance between their inverses)). -For any $\varepsilon>0$ there exists $\eta>0$ such that for any self-homeomorphism $f$ of $S$ whose distance to the identity is $<\eta$, there is an isotopy $t\mapsto f_t$ from $f$ to the identity that stays at distance $<\varepsilon$ to the identity. -I'd be interested in a reference and/or a hint of the proof. -Note: Unless I'm too tired and got it wrong, this result implies (and is implied by the fact) that the group of self-homeomorphisms of $S$ is locally path-connected (in the sense that every point has a path-connected neighbourhood, not necessarily open), whence the title of this question. - -REPLY [4 votes]: In the particular case of surfaces, I found the following reference which includes a proof that is not too complicated: Regular Mappings and the Space of Homeomorphisms on 2-Manifolds by Hamstrom and Dyer. They prove local contractibility, which is more than I asked. It works for surfaces with or without boundaries and includes a slight generalization with fixed points on the boundary. This is Theorem 1 in this article. The proof fits in 6 pages, is a bit heavy on notations but this remains managable. Unfortunately there is no figure. The proof uses conformal maps for a couple of lemmas, Alexander's trick, and a technique due to J.H. Roberts (Local arcwise connectivity in the space $H^n$ of homeomorphisms of $S^n$ onto itself, Summary of Lectures, Summer Institute on Set Theoretic Topology, Madison, Wisconsin, 1955, p. 100) but I cannot find the corresponding reference. They also cite a German article of Kneser (Die Deformationssätze der einfach zusammenhägenden Flächen, Mathematische Zeitschrift, Vol. 25(1926), pp. 362-372) as a source of inspiration, but my knowledge of German is very basic so reading it would represent quite an investment.<|endoftext|> -TITLE: Prove that there are no composite integers $n=am+1$ such that $m \ | \ \phi(n)$ -QUESTION [9 upvotes]: Let $n=am+1$ where $a $ and $m>1$ are positive integers and let $p$ be the least prime divisor of $m$. Prove that if $a1 $? ( I have not found any counter - examples). -Note that if $n=am+1$ is prime, we have $\phi(n)= n-1=am$. We see that $m \ | \ \phi(n) $. Its the converse of this statement that we want to prove i.e. If $m \ | \ \phi(n) $ then $n$ is prime. -If this conjecture is true, then we have the following theorem which is a generalisation ( an extension) of Lucas's converse of Fermat's little theorem. -$\textbf {Theorem} \ \ 1.$$ \ \ \ $ Let $n=am+1$, where $a$ and $m>1$ are positive integers and let $p$ be the least prime divisor of $m$ with $a 1$ then this gcd gives a non-trivial divisor of $n$. -Correspondingly, the given proof of Theorem 1 is easy to make work: instead of concluding that $m\mid\phi(n)$ and relying on the unproved claim, one can show that $m\mid (r-1)$ for every prime divisor $r\mid n$, implying that $n$ does not have prime divisors below $\sqrt{n}$ and thus it must be prime.<|endoftext|> -TITLE: Unique almost complex structure up to diffeomorphism -QUESTION [12 upvotes]: For which closed smooth manifolds does the action of the diffeomorphism group on the set of almost complex structures have exactly one orbit? -For example it is true for $S^2$. - -REPLY [14 votes]: Dmitri's answer is fine, but there is a different argument that is purely local that is worth bearing in mind as well: -On a $2n$-manifold $M$, the set of almost complex structures on $M$ are the sections of a smooth bundle $\mathscr{J}(M)\to M$ whose fibers are diffeomorphic to $\mathrm{GL}(2n,\mathbb{R})/\mathrm{GL}(n,\mathbb{C})$, a space of real dimension $4n^2 - 2n^2 = 2n^2$. -Thus, almost complex structures in dimension $2n$ depend locally on $2n^2$ functions of $2n$ variables, while diffeomorphisms of $M$ depend locally on $2n$ functions of $2n$ variables. Since $2n^2>2n$ when $n>1$, it follows that, when $n>1$, almost complex structures have local invariants, i.e., the diffeomorphism group cannot act transitively on the space of $k$-jets of almost complex structures for $k$ sufficiently large. Hence, not all almost complex structures can be equivalent under diffeomorphism when $n>1$, even locally.<|endoftext|> -TITLE: Homotopy groups of Diff(X) and Homeo(X) -QUESTION [15 upvotes]: For a compact closed smooth manifold $X$, the group Diff(X) has a natural homomorphism $\Phi$ to the homeomorphism group Homeo(X). If $X$ has dimension at least $5$, I'm looking for some general information about the map(s) $\Phi_*$ induced on the homotopy groups of these spaces. I’m mainly interested in the simply connected case but of course would be interested in a more general result. -In particular, are the kernel and cokernel of $\Phi_*$ always finite for every homotopy group? This would be my guess, based naively on the statement that a higher-dimensional simply connected manifold has finitely many smoothings. I tried to look in some older literature (eg papers by Burghelea-Lashof) but am a bit overwhelmed by the notation and technicalities. Such a finiteness statement would contrast sharply with the situation in dimension $4$. - -REPLY [18 votes]: No, this is not true, not even for spheres. Consider the following commutative diagram: -$\require{AMScd}$ -\begin{CD} -\text{Diff}_{\partial}(D^d) @>>> \text{Homeo}_{\partial}(D^d) \sim *\\ -@V f V V @VV V\\ -\text{Diff}(S^d) @>>g> \text{Homeo}(S^d) -\end{CD} -We have that $\text{Homeo}_{\partial}(D^d)$ is contractible by the famous Alexander trick. Now if $d \geq 5$ is odd, it is known that $\pi_{\ast} \text{Diff}_{\partial}(D^d) \otimes \mathbb Q$ is often non-trivial. In degrees $4i-1$ below some range growing with the dimension this was discovered by Farrell and Hsiang; a recent improvement was found by Krannich. But there are also other classes, see Watanbe: On Kontsevich's characteristic classes for higher‐dimensional sphere bundles II: Higher classes, for instance. -In any case, there are many non-trivial classes in rational homotopy that die under $g \circ f$. As $f$ can be seen as the fiber of $t\colon \text{Diff}(S^d) \to Fr(S^d) \sim O(d+1)$ where $Fr(S^d)$ is the frame bundle of $S^d$ and $t$ sends a diffeomorphism to the differential of the north pole of $S^d$ and there is the obvious section coming from the action of $O(d+1)$ and $S^d$, we see that these classes all survive under $f_{\ast}$. Hence they lie in the kernel of $g_{\ast}$. - -REPLY [15 votes]: No, the statement about the kernel and cokernel being finite is not true. -For a closed $d$-manifold, $d \neq 4$, smoothing theory identifies the homotopy fibre of -$$B\mathrm{Diff}(M) \longrightarrow B\mathrm{Homeo}(M)$$ -with (certain path components of) the space of sections of a bundle -$$Top(d)/O(d) \longrightarrow E \longrightarrow M$$ -constructed from te tangent bundle of $M$. Supposing for simplicity that $M$ is parallelisable, this space of sections is equivalent to -$$\mathrm{map}(M, Top(d)/O(d)).$$ -So your question is more or less equivalent to asking about the homotopy groups of $Top(d)/O(d)$. Until recently it was not even known whether the homotopy groups of $Top(d)/O(d)$ are finitely-generated, but that was proved by Kupers, in Some finiteness results for groups of automorphisms of manifolds. These days quite a bit is known about the rational homotopy groups of these spaces. Using the above formulation of smoothing theory (which goes through unchanged for manifolds with boundary) for the disc $D^d$, you will find most results are stated for -$$B\mathrm{Diff}_\partial(D^d) \simeq \Omega^d_0(Top(d)/O(d)).$$ -(Here I have used that $B\mathrm{Homeo}_\partial(D^d) \simeq *$ by the Alexander trick.) Any result which shows that $B\mathrm{Diff}_\partial(D^d)$ has a rationally nontrivial homotopy group provides a counterexample to your finiteness question. -For example: - -$\pi_i(Top(2n)/O(2n)) \otimes \mathbb{Q}=0$ for $i< 4n-2$, but -$$\pi_{4n-2}(Top(2n)/O(2n)) \otimes \mathbb{Q}= \mathbb{Q}.$$ -(See Kupers--Randal-Williams On diffeomorphisms of even-dimensional discs.) - -$\pi_i(Top(2n+1)/O(2n+1)) \otimes \mathbb{Q}=0$ for $i< 2n+5$, but -$$\pi_{2n+5}(Top(2n+1)/O(2n+1)) \otimes \mathbb{Q}= \mathbb{Q}.$$ -(This follows from Farrell--Hsiang On the rational homotopy groups of the diffeomorphism groups of discs, spheres and aspherical manifolds.) - - -There are some slides from a talk I recently gave at https://www.dpmms.cam.ac.uk/~or257/MIT2020.pdf, which give some more detail about the state of the art.<|endoftext|> -TITLE: Endomorphism ring of trivial source modules for abelian p-groups -QUESTION [6 upvotes]: Bernhard Böhmler  (who is also on MO) and myself had the following idea: -Let $G$ be a finite group and $k$ a field of characteristic $p$ (algebraically closed when it is needed) such that $p$ divides the order of $G$. -Let $A=kG$ be the group algebra of $G$ and $M$ the direct sum of indecomposable all trivial source modules (that are modules which are indecomposable direct summands of modules of the form ${k\!\uparrow}_H^G$ for some $p$-subgroup $H$ of $G$). -One might ask what properties $B:=End_{kG}(M)$ has. - -Quesion 1: Is $B$ studied already in the literature? - -The simplest case is when $G$ is abelian and then we can also assume that $G$ is an abelian $p$-group. Then any indecomposable direct summand of $M$ is of the form $k(G/H_i)$ for some subgroup $H_i$ of $G$. - -Question 2: When $G$ is an (elementary) abelian $p$-Group, is $B$ a Gorenstein ring? - -It might also be interesting whether the relations of $B$ have an easy description, since the Hom-spaces can in principle be described purely combinatorially. We can show that $B$ has dominant dimension equal to $2$. -Our question has a positive answer when $G$ is cyclic and then $B$ has Gorenstein dimension $2$. When $G$ is the Klein four group it is also true and $B$ has Gorenstein dimension 3. One can show that the quiver of $B$ is given by doubling the Hasse quiver of the poset of subgroups of $G$ (that is for every arrow in the Hasse quiver we add the opposite arrow). -For non-abelian groups it is not true, the quaternion group gives a counterexample. - -REPLY [9 votes]: Representations of $B$ (or at least an equivalent category) are studied in the literature under the name of "cohomological Mackey functors". -Theorem 1.1 of -Bouc, Serge; Stancu, Radu; Webb, Peter, On the projective dimensions of Mackey functors, Algebr. Represent. Theory 20, No. 6, 1467-1481 (2017). ZBL1422.20005 -implies that $B$ is Gorenstein if and only if the Sylow $p$-subgroups of $G$ are cyclic or dihedral. (In the latter case $p$ must be $2$, of course.)<|endoftext|> -TITLE: Which homotopy 2-types are H-spaces? -QUESTION [23 upvotes]: Let $X$ be a connected CW-complex with $\pi_k(X)$ trivial for $k >2$. Is it known under which circumstances $X$ is an $H$-group? -I have been able only to deduce the necessary condition that $\pi_1(X)$ has to be abelian and act trivially on $\pi_2(X)$. Furthermore, if the necessary condition holds, vanishing of the Postnikov invariant $\beta \in H^3( \pi_1(X), \pi_2(X))$ is a sufficient condition. Thus the interesting case is that of $\beta$ nonzero. - -REPLY [5 votes]: I just wanted to add to Tyler Lawson's answer that all the maps $\beta\colon K(G,1)\rightarrow K(A,3)$ ($G$ and $A$ abelian and no action of $G$ on $A$) satisfying his additivity condition are loop maps by Stasheff's Homotopy Associativity of $H$-Spaces II (Theorem 5.3). Hence for 2-types being an $H$-space is the same as being a loop space. See also Qiaochu Yuan's comment. -In terms of $k$-invariants, the condition is that the $k$-invariant of the 2-type is in the image of the so-called cohomology 'suspension' morphism -$$H^4(K(G,2),A)\longrightarrow H^3(K(G,1),A),$$ -which is induced by taking loops on the corresponding sets of homotopy classes of maps between Eilenberg-MacLane spaces. -The source is very well understood, it coincides with -$$\hom(\Gamma(G),A).$$ -Here $\Gamma(G)$ is the target of the universal quadratic map $G\rightarrow\Gamma(G)$. Recall that a map $\gamma\colon G\to B$ between abelian groups is quadratic if $G\times G\to B\colon (x,y)\mapsto\gamma(x+y)-\gamma(x)-\gamma(y)$ is bilinear. -An example with non-trivial $k$-invariant can be constructed as follows. It suffices to show that -$$H^4(K(\mathbb{Z}/2,2),\mathbb{Z}/4)\longrightarrow H^3(K(\mathbb{Z}/2,1),\mathbb{Z}/4)$$ -coincides with -$$\mathbb{Z}/4\twoheadrightarrow \mathbb{Z}/2.$$ -Indeed, $H^3(K(\mathbb{Z}/2,1),\mathbb{Z}/4)$ is well-known to be the elements annihilated by $2$ in $\mathbb{Z}/4$, which identifies with $\mathbb{Z}/2$. A normalised $3$-cocycle representing the generator is -$$f\colon \mathbb{Z}/2\times \mathbb{Z}/2\times \mathbb{Z}/2\longrightarrow \mathbb{Z}/4,\qquad f(1,1,1)=2.$$ -(I will omit notations for classes in quotients of $\mathbb{Z}$ since the meaning in each case will be clear from the context.) -The universal quadratic map for $\mathbb{Z}/2$ is $\gamma\colon \mathbb{Z}/2\rightarrow \mathbb{Z}/4$, $\gamma(0)=0$, $\gamma(1)=1$. Hence $\Gamma(\mathbb{Z}/2)=\mathbb{Z}/4$ and -$$\hom(\Gamma(\mathbb{Z}/2),\mathbb{Z}/4)=\hom(\mathbb{Z}/4,\mathbb{Z}/4)=\mathbb{Z}/4.$$ -In order to compute the suspension morphisms I'm going to use crossed modules and their semi-stable version called reduced quadratic modules by Baues (see his book on 4-dimensional complexes). It is well known that 3-dimensional group cohomology classifies crossed modules (Eilenberg-MacLane or one of them alone, I don't currently remember). Similarly $\hom(\Gamma(\mathbb{Z}/2),\mathbb{Z}/4)$ classifies reduced quadratic modules. The generator is represented by the reduced quadratic module -$$\mathbb{Z}\otimes \mathbb{Z} \stackrel{\langle-,-\rangle}\longrightarrow \mathbb{Z}\oplus \mathbb{Z}/4\stackrel{\partial}\longrightarrow \mathbb{Z}$$ -where $\partial(a,b)=2a$ and $\langle x,y\rangle =(0,{xy}).$ This is because $\ker \partial=\mathbb{Z}/4$, $\operatorname{coker}\partial=\mathbb{Z}/2$ and the map $\mathbb{Z}/2\mapsto \mathbb{Z}/4$ defined by ${x}\mapsto{\langle x,x\rangle}$ coincides with the aforementioned universal quadratic map. The loop crossed module of this reduced quadratic module is -$$\mathbb{Z}\oplus \mathbb{Z}/4\stackrel{\partial}\longrightarrow \mathbb{Z},$$ -where the source is equipped with an exponential action of the target defined by -$$x^a=x+\langle\partial(x),a\rangle.$$ -(Crossed modules usually consist of non-abelian groups and reduced quadratic modules too, but in this case everything is abelian because they are very small, this simplifies a lot the computations). A 3-cocycle -$$g\colon \mathbb{Z}/2\times \mathbb{Z}/2\times \mathbb{Z}/2\longrightarrow \mathbb{Z}/4$$ -representing the cohomology class of this crossed module is defined by the following choices: -We first need a set-theoretic splitting of $\partial$, that we define as -$$s\colon \mathbb{Z}/2\longrightarrow \mathbb{Z},\qquad s(0)=0,\quad s(1)=1.$$ -Now, for each pair of elements $x,y\in \mathbb{Z}/2$ we need $t(x,y)\in \mathbb{Z}\oplus \mathbb{Z}/4$ such that -$$\partial(t(x,y))=-s(y)-s(x)+s(x+y).$$ -This measures the failure of $s$ to be a morphism. -We take $t(0,-)=(0,0)$, $t(-,0)=(0,0)$, and $t(1,1)=(1,0)$. -With these choices $g$ is -$$g(x,y,z)=t(x,y)^{s(z)}+t(x+y,z)-t(x,y+z)-t(y,z)\in\ker\partial=\mathbb{Z}/4.$$ -Different choices produce cohomologous cocycles. -This cocycle is normalised because $t(0,-)$ and $t(-,0)$ vanish, so we only have to compute -$$\begin{array}{rcl} -g(1,1,1)&=&t(1,1)^{s(1)}+t(1+1,1)-t(1,1+1)-t(1,1)\\ -&=&t(1,1)+\langle \partial t(1,1), s(1)\rangle +t(0,1)-t(1,0)-t(1,1)\\ -&=&\langle \partial t(1,1), s(1)\rangle\\ -&=&\langle \partial (1,0), 1\rangle\\ -&=&\langle 2, 1\rangle\\ -&=&(0,2). -\end{array}$$ -The second coordinate (the kernel of $\partial$) is $2\in \mathbb{Z}/4$, therefore, $g=f$ above. This concludes the proof of the claim.<|endoftext|> -TITLE: Invertibility of specific function -QUESTION [8 upvotes]: This is my first post. I'm not a mathematician, just an electronics engineer who loves mathematics. In one of my projects, I arrived at the following function: -$$V\left(\varphi\right)=\frac{A\sqrt{\pi-\varphi+\sin{\varphi\cos{\varphi}}}}{\sqrt{2\pi}}$$ -The project requires $V\left(\varphi\right)$ to be inverted, to obtain angle $\varphi$ (unknown), from a voltage $V$ (known). $V\left(\varphi\right)$ is continuous and strictly monotonic (descending), so an inverse mapping should exist. I tried to invert it symbolically, but couldn't arrive at a closed-form solution for $\varphi\left(V\right)$. I ended up using MATLAB to compute it numerically, and the project was successfully completed. -Out of pure curiosity, I asked my cousin (a mathematician) to attempt to symbolically invert the above function, but he also couldn't do it, and couldn't even give me a rigorous answer as to the existence of such solution. So, my questions are the following: - -Does a closed-form expression for $\varphi\left(V\right)$ exist? -If the answer to (1) is YES, can someone provide that function, or point me to a method for deriving it? -If the answer to (1) is NO, what is the formal reason for it? Is there a way to show/prove that such solution does not exist? - -I apologise if this problem is too easy, too obvious, or even irrelevant to the MathOverflow community. I've already posted it in the Mathematics Stack Exchange community (for students and professionals), where I got some replies which loosely confirm my suspicion that a closed-form expression for $\varphi\left(V\right)$ does not exist. However, the replies were either too descriptive, or used Taylor series expansions to invert the function, which is not what I want. No reply provided rigorous answers to my questions. So, I thought of posting the problem here, where more advanced topics are discussed, in the hope that someone can provide me with some rigorous answers. This is not a homework exercise, and the associated practical problem has already been solved numerically. This post was made out of pure curiosity about the invertibility of functions of the form of $V\left(\varphi\right)$. Many thanks to all for your replies. - -Adding some graphics, in order to better illustrate the problem. -The following figure shows how the function we're looking for, $\varphi\left(V\right)$, looks like. Notice that there appears to be no symmetry in this function. The values of $\varphi$ lie in the interval $[0,\pi]$, while the values of $V\left(\varphi\right)$ lie in the interval $[0,\frac {A} {\sqrt2}]$. - -Based on the comment by @PietroMajer, the problem can be reduced to the inversion of function $k=x-\sin(x)$. In this case, the values of both $x$ and $k$ lie in the interval $[0,2\pi]$. The following figure shows a plot of the inverse of $k=x-\sin(x)$, together with a plot of the function itself (dashed line). Plotting them both on the same graph is useful, since they both have the same range for their independent and dependent variables. It can be observed that now there is a clear symmetry of this function at the point $(\pi,\pi)$, thanks to the removal of the square root term. This means that we only need to deal with the interval $[0,\pi]$, and use symmetry on that result to obtain the other half ($[\pi,2\pi]$). - - -Please see my accepted answer at Mathematics Stack Exchange, for some useful approximations of the inverse of $k=x-\sin(x)$, and of $\varphi\left(V\right)$. They may not be of much interest to pure mathematicians, but, being an electronics engineer, I consider them very useful in solving practical problems involving the inversion of $k=x-\sin(x)$ or similar functions. - -REPLY [12 votes]: The answer to the question of whether the inverse has a closed form depends of course on one's definition of "closed form." One plausible definition is that a closed-form function is a function that lies in a so-called Liouvillian extension of $\mathbb{C}(x)$, the field of rational functions of $x$ with complex coefficients. I won't give the exact definition of a Liouvillian extension, but suffice it to say that any function that you can get via a finite number of applications of addition, subtraction, multiplication, division, taking $n$th roots, exponentiation, and taking logarithms is going to be a closed-form function in this sense. Note that since we're working over the complex numbers, we get trig functions and their inverses as well. So this covers everything that most everyone would agree is "closed form." (Liouvillian extensions also include algebraic functions that aren't expressible using radicals; not everyone would consider such functions to be expressible in "closed form," but since we're going to show that a certain function is not expressible in closed form, it doesn't hurt to include extra functions in our class of "closed-form functions.") -Rigorous proofs that specific functions of interest are not Liouvillian go back, naturally, to Liouville, with later contributions by other authors (e.g., Ritt, as mentioned by Iosif Pinelis). Again, reviewing the general theory is beyond the scope of a MathOverflow answer, but fortunately, when it comes to finding inverses, there is a theorem of Rosenlicht ("On the explicit solvability of certain transcendental equations," Publications Mathématiques de l'IHÉS 36 (1969), 15–22) that can be used to handle many of the simple "transcendental equations" that arise in practice. Stated slightly informally, the relevant special case of Rosenlicht's theorem is the following. - -Theorem. Suppose that $y_1, \ldots, y_n$ and $z_1, \ldots, z_n$ are closed-form functions of $x$ satisfying -$$\frac{y_i'}{y_i} = z_i', \qquad i=1,\ldots,n.$$ -If $\mathbb{C}(x,y_1, \ldots,y_n,z_1,\ldots,z_n)$ is algebraic over both $\mathbb{C}(x,y_1,\ldots,y_n)$ and $\mathbb{C}(x,z_1,\ldots,z_n)$, then $y_1,\ldots,y_n$ and $z_1,\ldots z_n$ are all algebraic over $\mathbb{C}(x)$. - -In light of Pietro Majer's observation, let's use this theorem to show that the function $f$ defined implicitly by the equation $x = f - \sin f$ (Kepler's equation, as noted by Rob Corless) has no closed-form expression. We're only going to need the special case $n=1$ of the theorem. The first step is to write everything in terms of exponentials. Recall that $\sin f = (e^{if}-e^{-if})/2i$, so if we set $z := if$ then we have the equation -$$x = -iz - \frac{e^z - e^{-z}}{2i}.\qquad\qquad(*)$$ -The equation $y'\!/y = z'$ appearing in Rosenlicht's theorem is secretly the equation $y=e^z$ in disguise. So what we need to do is to introduce extra functions to represent the exponentials that appear, to turn our equations into polynomial equations. Here, all we need to do is to set $y=e^z$. Then $y'\!/y = z'$ and Equation $(*)$ becomes -$$x = -iz- \frac{y - 1/y}{2i}.\qquad\qquad (**)$$ -We're now ready to apply Rosenlicht's theorem with $n=1$. Certainly $\mathbb{C}(x,y,z)$ is algebraic over $\mathbb{C}(x,y)$ because $z$ is actually a rational function of $x$ and $y$. It's also true that $\mathbb{C}(x,y,z)$ is algebraic over $\mathbb{C}(x,z)$ because $y$ satisfies a quadratic equation with coefficients that are polynomial (in fact, linear) in $x$ and $z$. The hypothesis of the theorem is therefore satisfied. What does this tell us? -Well, if $f$ is a closed-form function of $x$, then so is $z=if$ as well as $y=e^z = e^{if}$. So if $f$ is a closed-form function, then Rosenlicht's theorem tells us that $y$ and $z$ must in fact be algebraic functions of $x$. -This isn't quite a contradiction yet, but it's not so hard to show that $y$ and $z$ can't be algebraic functions of $x$. We can use the argument given by Bronstein et al. in their paper showing that the Lambert $W$ function is not Liouvillian ("Algebraic properties of the Lambert $W$ function from a result of Rosenlicht and Liouville," Integral Transforms and Special Functions 19 (2008), 709–712). If $z$ has a pole of finite order (in the extended complex plane), then $y$ has an essential singularity, but this contradicts Equation $(**)$ since the left-hand side has no singularities. So $z$ must be constant, but this is absurd. -By the way, according to Rosenlicht, Liouville himself already knew that the solution to Kepler's equation is not Liouvillian, but I haven't checked Liouville's paper myself. -[I'm making this answer community wiki since I benefited from the observations of several other respondents.] - -REPLY [8 votes]: The change of variable given above by Pietro Majer shows that this is equivalent to Kepler's equation Wikepedia on Kepler's Equation which is believed not to have any closed form solution (let alone an elementary solution). I am actually not so sure that's true and I don't know of any proof.<|endoftext|> -TITLE: Etale fundamental group of an order in a number field -QUESTION [6 upvotes]: Let $\mathcal{O}$ be an order in a number field $K$, that is a subring of $K$ with rank as abelian group equal to $[K:\mathbb{Q}]$. What is known about the SGA3-étale fundamental group of $X=\mathrm{Spec}(\mathcal{O})$ ? Are there example where it is not profinite ? -My motivation for asking this question is for computing the étale cohomology group $H^1(X,\mathbb{Z})=\mathrm{Hom}_{cont}(\pi_1^{SGA3}(X),\mathbb{Z})$. -If X is geometrically unibranch then the SGA3 étale fundamental group equals the étale fundamental group, hence is profinite and the cohomology group vanishes. - -REPLY [2 votes]: I only vaguely know about étale fundamental groups at the moment so I went the other way and computed $H^1(X,\mathbb{Z})$. Let $\pi : Y=\mathrm{Spec}(\mathcal{O}_K) \to X$ denote the normalization. Denote $Z$ the singular locus and put $s_v=\#\pi^{-1}(v)-1$ for every $v\in Z$. - -Claim : $H^1(X,\mathbb{Z})$ is a free abelian group of rank $\sum_{v\in Z}s_v$. - -Proof : Consider the cokernel $F$ of the injection $0\to \mathbb{Z} \to \pi_\ast \mathbb{Z}$. Then $F$ is a skyscraper sheaf supported on singular points and we can compute by Galois cohomology that $H^0(X,F)\simeq \oplus_{v\in Z}\mathbb{Z}^{s_v}$ (this comes from the vanishing of the $H^1$ in Galois cohomology with coefficients in $\mathbb{Z}$). Because $H^1(Y,\mathbb{Z})=0$ since $Y$ is normal (see etale-cohomology-with-coefficients-in-the-integers), the long exact sequence in cohomology reads -$$ -0 \to \mathbb{Z} \xrightarrow{\mathrm{Id}} \mathbb{Z} \to H^0(X,F) \to H^1(X,\mathbb{Z}) \to 0 -$$ which proves the claim. $\blacksquare$ -In particular, as soon as there is a singular point with at least two primes above it in the normalization, the (SGA3) étale fundamental group of $X$ be not profinite.<|endoftext|> -TITLE: Example of a PID with a residue field of finite characteristic and a residue field of characteristic 0? -QUESTION [11 upvotes]: I understand that for any nonempty set $S$ of characteristics, there exists a PID $R$ such that the set of characteristics of residue fields of $R$ (i.e. quotients by of $R$ by maximal ideals -- I'm not including the residue field at the generic point. Thanks to Steven Landsburg for pointing out this terminological ambiguity in the comments below) is precisely $S$. I learned this from a paper of Heitmann, PID’s with specified residue fields (which proves much more), which I originally found at Exotic principal ideal domains. -Question: What is a "nice" example of a PID $R$ such that $R$ has a residue field of characteristic 0 and a residue field of finite characteristic? -By "nice", I'd ideally mean that $R$ is not just custom-built for the purpose of providing such an example, and might be a ring I'd meet on the street one day. Failing that, I'd settle for a streamlined description of such a ring $R$ (in order to understand Heitmann's example one must wade through several layers of extra generality related to his more ambitious aims). -If we only require $R$ to be Noetherian, then YCor gave a simple example in the comments (1 2 3) on If a PID has no nonzero divisible elements, then is the same true of its finitely-generated modules?: $R = \mathbb Z_p[t]$ has residue fields $\mathbb F_p$ and $\mathbb Q_p$ (the latter obtained by modding out by $(1-pt)$). Similarly, $\mathbb Z_{(p)}[t]$ has residue fields $\mathbb F_p$ and $\mathbb Q$. It would be nice if there were an example of a PID with this property just as "nice" as $\mathbb Z_p[t]$. - -REPLY [21 votes]: You can take the ring of fractions $\frac{a}{b}$ with $a,b \in \mathbb Z[x]$, where $b$ is nonzero mod $p$ and nonzero mod $px-1$. -Given any polynomial $a$, we can remove all factors of $p$ and remove all factors of $px-1$, obtaining a polynomial that is nonzero mod $p$ and nonzero mod $px-1$. So every polynomial is a power of $p^i (px-1)^j$ times a unit for natural numbers $i,j$. -Because the ideal generated by $p$ and $px-1$ contains $1$, the ideal generated by $p^{i_1} (px-1)^{j_1}$ and $p^{i_2} (px-1)^{j_2}$ is also generated by $p^{ \min(i_1,i_2)} (px-1)^{\min(j_1,j_2) } $. So every ideal is generated by a single element of the form $p^i (px-1)^j$. -There are two maximal ideals, $(p),$ and $(px-1)$, whose quotients $\mathbb F_p(x)$ and $\mathbb Q$ have characteristics $p$ and $0$ respectively.<|endoftext|> -TITLE: Very particular kind of 4-manifolds. Classification -QUESTION [14 upvotes]: Let $M$ be a smooth orientable compact connected (with boundary) manifold of dimension $4$. In addition $M$ is assumed to be aspherical and acyclic. - -Question: is there a "classification" of such manifolds? Or can they be classified in any effective way? - -REPLY [18 votes]: There are plenty of such manifolds, but as Danny indicates in his answer, there is not a known classification. -Take any acyclic group $G$ with a finite aspherical 2-complex $C$ with $\pi_1(C)=G$. Then one can create an aspherical 4-manifold with boundary having $G$ as fundamental group. We may assume that the 1-skeleton $C^{(1)}$ of $C$ is a wedge of $k$ circles. Then take a 4-dimensional handlebody $H$ with $k$ 1-handles, with a spine of $C^{(1)}$. There are $k$ disks attached to the 1-skeleton in $C$. Attach $2$-handles to $H$ in such a way that the core of the attaching map is homotopic to the attaching map in the 2-skeleton $C^{(1)}$ to get a manifold $W$ with handle structure so that $C$ is a deformation retract of $W$, and hence $\pi_1(W)\cong G$. By the Poincaré-Lefschetz theorem, $\partial W$ is a homology 3-sphere. But in general we may get many different boundaries depending on the choice of isotopy class and framing of the boundary of the cores of the 2-handles. -To get such groups $G$, one can choose a small-cancellation -$C'(\frac16)$ balanced presentation with $H_1(G)=0$. Then a presentation complex $C$ will be aspherical and acyclic. Added: See an explicit example due to Rylee Lyman in the comments. A simpler presentation of the Higman group is given (which perfect and has aspherical presentation complex). -The difficulty here is that one has no idea what 3-manifold the boundary of such a manifold will be. Moreover, it's not clear what the homeomorphism classification of such manifolds is, even if they have the same aspherical 2-skeleton spine and boundary. -Presumably there are also examples which do not have a 2-dimensional spine. The only obvious restriction I see is that the fundamental group must have cohomological dimension three.<|endoftext|> -TITLE: Uniformly approximating a function and its derivative using polynomials -QUESTION [6 upvotes]: I'm struggling either proving or disproving the following statement: - -Let $K\subset \mathbb{R}$ be compact, and $S = \mathrm{span}\{p_k, k = 0, 1, \ldots\}$, where $p_k$'s are polynomials over $K$. If $S$ is dense in $C^1(K)$ with respect to sup-norm, then for any $f\in C^1(K)$, there exists $\{g_n\} \subset S$ such that $g_n$ uniformly approximates $f$ with $f'$ uniformly approximated by $g_n'$. - -This statement seems to be true, although $S$ is only a subset of all polynomials over $K$. If the statement appears false, is there any prior assumption that makes it true? Any reference or direct answer would greatly help me. - -REPLY [9 votes]: The span of $\{x^{2k}\colon k=0,1,2,\dots\}$ is dense in $C([0,1])$. But all their derivatives vanish at $0$. - -REPLY [3 votes]: Any prior assumption that makes it true. A simple case where your statement is true is the case of an interval $K\subset\mathbb{R}_+ $, and $(p_k)_{k\ge0}$ are monomials, $p_k(x)=x^{n_k}$, of degrees $0=n_0 -TITLE: Scottish Book Problem 172 -QUESTION [9 upvotes]: The problem is formulated using old terminology and I want to understand what it actually says. -The problem reads: "A space $E$ of type (B) has the property (a) if the weak closure of an arbitrary set of linear functionals is weakly closed. [A sequence of linear functionals $f_n(x)$ -converges weakly to $f(x)$ if $f_n(x) \to f(x)$ for every $x$.] -The space $E$ of type (B) has the property (b) if every sequence of linear functionals weakly convergent converges weakly as a sequence of elements in the conjugate space $\bar{E}$. -Question: Does every separable space of type (B) which has property (a) also possess property (b)?" -I know that "space of type (B)" means "Banach space" in modern terminology. - -Q1: Does "linear functional" include the assumption of continuity? - - -Q2: What is the conjugate space? Does it mean the dual space of continuous linear functionals on $E$? - -If the answers to Q1 and Q2 are yes, then this would mean Banach spaces with property (b) are exactly what nowadays are called Grothendieck spaces. [A Grothendieck space is a Banach space $E$ such that every sequence in the dual space $E^*$ that converges weak* - that is, with respect to $\sigma(E^*,E)$ - also converges weakly - that is, with respect to $\sigma(E^*,E^{**})$.] -I am also unsure how to understand the condition "the weak closure of an arbitrary set of linear functionals is weakly closed", as this seems like a tautology. Is the point here that we consider a sequential closure? - -REPLY [4 votes]: The answer to both Q1 (in the context given) and Q2 is yes. -Regarding the weak closure of an arbitrary set of linear functional being linear that appears to me to be sequential weak-star convergence — that is, if they converge sequentially in the weak-star topology, then they converge in the weak topology on the dual.<|endoftext|> -TITLE: Arbitrary torsion in cohomology of Kähler manifolds -QUESTION [5 upvotes]: Fix an integer $i\geq 3$ and a finite abelian group $G$. -Is there a connected closed Kähler manifold $M$ such that $H^i(M, \mathbb{Z})\approx \mathbb{Z}^n\oplus G$ for some integer $n\geq 0$? - -REPLY [9 votes]: The answer is positive and can be deduced from Proposition 15 of "Sur la topologie des varietes algebriques en characteristique p" by Serre. According to this proposition for any finite group $G$ there exists a complete intersection $X$ on which $G$ is acting freely. Set $Y=X/G$. Then $\pi_1(Y)=G$. Let now $G$ be your abelian group. Then $\pi_1(Y)\cong G$ and $H_1(Y,\mathbb Z)\cong G$. It follows that $H^2(Y,\mathbb Z)\cong \mathbb Z\oplus G$. To get torsion $G$ is higher cohomologies take the product $Y\times \mathbb CP^n$.<|endoftext|> -TITLE: When does the loop functor $\Omega^\infty:Sp(C) \rightarrow C$ commute with filtered colimits? -QUESTION [6 upvotes]: Let $C$ be a pointed $\infty$-category which admits finite limits. - -Let $Sp(C)$ denote the $\infty$ category of spectrum objects. One way to define, i.e. 1.4.2.24, is by taking the homotopy limit in $Cat_\infty$, the $\infty$-category of categories. -$$Sp(C):= \varprojlim \left( \cdots \xrightarrow {\Omega} C \xrightarrow {\Omega} C \right) $$ -Let us denote $\Omega^\infty: Sp(C) \rightarrow C$ as the projection onto the last component. - -I would like to understand what categorical properties of $\Omega^\infty$ satisfy. My question is -If each $\Omega$ commute with $I$-indexed limit does this imply $\Omega^\infty$ does too? - -The reason I am concerned with this question: It is claimed in C.1.4.1, that - -if $C$ is a prestable and presentable $\infty$-category and $\Omega:C \rightarrow C$ commutes with filtered colimits then $\Omega^\infty$ commutes with filtered colimits. - -A prestable $\infty$-category by definition can be intriscally characterized, C.1.2.1 as a category which satisfies the following conditions - -pointed and admits finite colimits. -suspension is fully fiathful -every morphism $Y \rightarrow \Sigma Z$ lies in a pullback square with top right part $X \rightarrow Y \rightarrow \Sigma Z$ and bottom left $0$. Further, the sequence $X \rightarrow Y \rightarrow \Sigma Z$ is a cofiber sequence. - - -I have recorded my thoughts below, which one may safely ignore. -Both strategies I know do not really apply - these are based on the case $C=S_*$, the $\infty$-cat of pointed spaces. -Strategy 1. $\Omega^\infty: Sp(S) \rightarrow S_*$. $\Omega^\infty$ is corepresented by $\mathbb{S}=\Sigma^\infty S^0$, the sphere spectrum, where we $\Sigma^\infty$ is left adjoint of $\Omega^\infty$. Now by noting that that $S^0$ is a compact object in $S_*$ result follows. -Strategy 2. Consider the $\infty$-cat $Pr^\omega$ of compactly generated, in the sense of 5.5.7.1, $\infty$-categories with right adjoints. -We prove that $S\in Pr^w$ and that $Pr^w \hookrightarrow Cat_\infty$ reflects (filtered) limits. - -REPLY [5 votes]: The result is true, more generally, if you take a class of diagrams $\mathcal K$ and the $\infty$-category $\widehat{Cat_\infty}^\mathcal K$ of $\infty$-categories that have all $\mathcal K$-indexed colimits, and functors between them that preserve those, then the forgetful functor $\widehat{Cat_\infty}^\mathcal K\to \widehat{Cat_\infty}$ preserves all limits, in fact it has a left adjoint. -This is stated as Corollary 5.3.6.10. in Lurie's Higher Topos Theory (with his notations, $\mathcal K' =$ my $\mathcal K$, and his $\mathcal K= \emptyset$). -From this, your result follows, as if $\Omega$ preserves $I$-indexed colimits, then your diagram lives in $\widehat{Cat_\infty}^{\{I\}}$, so its limit does too, and the projection functors too, in particular $\Omega^\infty: Sp(C)\to C$ is one of those projection functors, so it preserves $I$-indexed colimits (this is, of course, assuming that $C$ has all $I$-indexed colimits - which is the case in the statement you refer to, as of course a presentable $\infty$-category has all filtered colimits) -Your strategy 1 is in this sense somehow misguided, as proving that $\mathbb S$ is compact essentially uses that $\Omega^\infty$ preserves filtered colimits. -Actually, a less general, but perhaps easier proof works in the special case of $Sp(C)$ and filtered colimits: $Sp(C)$ can be seen as a certain full subcategory of $Fun(\mathbb{Z\times Z},C)$ (such a functor is a grid, $Sp(C)$ is the full subcategory on those grids that are only $0$ objects off the diagonal, and such that certain squares are pullbacks), and $\Omega^\infty$ is then simply the restriction to this subcategory of the evaluation at $0$. -Now if $C$ has all filtered colimits, and $\Omega$ commutes with those, then $Sp(C)\subset Fun(\mathbb{Z\times Z},C)$ is closed under filtered colimits (the only pullbacks appearing in its definition are pullbacks defining $\Omega$), so that, as in functor categories in general, filtered colimits in $Sp(C)$ are computed pointwise; and so in particular $\Omega^\infty$ commutes with those. -This second proof is less general, but it's easier and gets you what you want- and perhaps it allows for a better understanding in this specific context ?<|endoftext|> -TITLE: Algebra with a certain abelian group as the multiplicative group -QUESTION [6 upvotes]: Let $A$ be an abelian group. Are there an algebra $\mathfrak{X}(A)$ s.t. the multiplication group is isomorphic to A ? i.e. -$$ -\mathfrak{X}(A)^{\times} \simeq A. -$$ -For example, for $A=\mathbb{Z}/4\mathbb{Z}$, $\mathfrak{X}(A)=\mathbb{F}_{5}$. I want to know the sufficient conditions of $A$ for existence of $\mathfrak{X}(A)$. Is it well-known ? - -REPLY [14 votes]: I am going to assume that by "algebra" you simply mean a ring. -The answer is "no", in general. For example $\mathbb{Z}/5\mathbb{Z}$ is not the unit group of a ring. Indeed, suppose it was the unit group of a ring $R$, let $u\in R$ denote a generator of the unit group. Now $-1$ is always a unit, and has order dividing $2$. But it is supposed to live in a cyclic group of order $5$, which forces its order to be $1$, i.e. $1=-1$ in $R$, so that $R$ has characteristic $2$. But then $u$ generates a subring of $R$ that is a quotient of $\mathbb{F}_2[\mathbb{Z}/5\mathbb{Z}]\cong \mathbb{F}_2[x]/(x^5-1)\cong \mathbb{F}_2\times \mathbb{F}_{16}$. Since that quotient has at least $5$ distinct elements, it must be at least all of $\mathbb{F}_{16}$, so that $R^\times$ contains $\mathbb{F}_{16}^\times\cong \mathbb{Z}/15\mathbb{Z}$ — contradiction. -In general, this argument shows that a group of prime order $p$ is isomorphic to the unit group of a ring if and only if $p=2$ or of the form $2^n-1$ for some $n$.<|endoftext|> -TITLE: Is Thompson's group $T$ co-Hopfian? -QUESTION [17 upvotes]: A group $G$ is co-Hopfian if every injective homomorphism $G\to G$ is bijective, i.e., if $G$ contains no proper subgroups isomorphic to $G$. My question is whether Thompson's group $T$ is co-Hopfian. -For context, Thompson's groups $F$ and $V$ are very much not co-Hopfian, roughly due to the fact that there are many copies of the unit interval inside the unit interval, and many copies of the Cantor set inside the Cantor set. However, there are no copies of the circle properly inside the circle, so the same intuition does not apply to $T$. - -REPLY [9 votes]: The answer to the question is no, $T$ is not co-Hopfian, i.e., it does contain proper subgroups isomorphic to itself. Nicolás Matte Bon explained this to me over email (he doesn't use Mathoverflow, but someone showed him this question). -Matte Bon's strategy is to look at $T$ acting on the Cantor set $C=\{0,1\}^{\mathbb{N}}$ and define a new (faithful) action of $T$ on $C$ that yields an (injective) endomorphism $T\to T$ (and then argue that it is not surjective). For non-surjectivity, the idea is that under the new action there is a proper non-empty invariant open subset (so the action is not minimal), unlike for the usual action of $T$. Constructing such an action is a bit too complicated to explain here, but it uses ideas from Section 11 of Matte Bon's paper [1]. What I can do here is explain one concrete example of a proper endomorphism $T\to T$, which I sorted out after understanding Matte Bon's general construction. -To describe the endomorphism $T\to T$, I will use the strand diagram model for elements of $T$ (and $F$ and $V$), see, e.g., Definition 2.7 of Belk-Matucci [2]. For each split or merge vertex (Definition 2.1(2)), draw a small neighborhood around the vertex, not meeting any other vertices. For a split this neighborhood has one incoming strand and two outgoing strands, and for a merge this neighborhood has two incoming strands and one outgoing strand. Now to define $T\to T$ we replace the picture inside each such neighborhood with a more complicated picture. For a split, replace the strand going from the split vertex to the right exit with the picture in Figure 2 of Belk-Matucci (the usual "$x_0$" generator). For a merge, do this same thing but flipped upside-down. (Don't change any cyclic permutations.) This defines a well defined injective endomorphism $T\to T$ (if we only allow cyclic permutations that is; if we allow all permutations then it's $V\to V$, and if we allow no permutations then it's $F\to F$). To see it's non-surjective, one can use the strand diagram method of analyzing dynamics (see, e.g., Figure 19 of Belk-Matucci) to check that if $f$ is in the image of this endomorphism and $c\in C$ starts with $11$ then $f(c)$ has a "$11$" somewhere in it. (Since $T$ certainly contains elements violating this rule, this gives non-surjectivity.) -[1] Nicolás Matte Bon, Rigidity properties of full groups of pseudogroups over the Cantor set. arXiv link -[2] James Belk, Francesco Matucci, Conjugacy and dynamics in Thompson's groups. Geometriae Dedicata 169.1 (2014) 239-261. arXiv link<|endoftext|> -TITLE: Maximal in-degree in directed voting graph -QUESTION [10 upvotes]: Real-life motivation. Our team has $n$ members. For the next in-team presentation session, everyone had 1 talk prepared that he or she would be able to present. Now everyone could cast $1$ vote about whose talk they would like to hear. (Everyone is modest, so no-one voted for their own talk.) Now, unfortunately, there were two talks receiving the maximum vote number, so a coin had to be tossed. - Informally speaking, I was wondering whether this dilemma situation would be less likely as $n$ grows. See formal version below. -Formal version. In this context, we define a directed graph $G=(V,E)$ to consist of a finite set $V$ and of a set $$E \subseteq (V\times V)\setminus\{(v,v):v\in V\}.$$ -Given $v\in V$, we define the in-degree by $\text{deg}^-(v) = |\{a\in V:(a,v)\in E\}|$, and the out-degree is defined in a dual fashion. Let $\Delta^-(G)$ be the maximal in-degree. -In this question we only consider directed graphs $G=(V,E)$ with out-degree $\text{deg}^+(v) = 1$ for all $v\in V$ (everyone casts exactly 1 vote). Let ${\cal M}(G) = |\{v\in V: \deg^-(v) = \Delta^-(G)\}|$ be the number of vertices having maximal in-degree. (Trivially, ${\cal M}(G) \geq 1$.) Moreover, let $M_n$ be the expected value of ${\cal M}(G)$ where $G$ is an arbitrary graph with $V(G) = \{1, \ldots, n\}$ satisfying $\text{deg}^+(v) = 1$ for all $v\in V$. -Do we have $\lim\sup_{n\to\infty}M_n = 1$? - -REPLY [4 votes]: Let's do a back of the envelope computation for a slightly simpler problem (when voting for one's own talk is allowed). It doesn't seem like it matters too much, but formally it is a bit different. There will be no formal proof, just an "educated guess" anyway. We have $n$ numbered urns (talks) and $n$ balls (votes) and the question is what approximately is the probability that the maximum is unique when $n$ is very large. We can think instead of the independent Poisson distribution in each urn (a leap of faith occurs here), i.e., the probability that each urn has exactly $k$ balls is $\frac 1{ek!}$. -Suppose that the maximum is $K$ and it is unique. It means that there is an urn with exactly $K$ balls and every other urn has fewer balls. The probability of that is -$$ -\frac n{eK!}\left[\sum_{k=0}^{K-1}\frac 1{ek!}\right]^{n-1} -$$ -after which we have to sum over $K$ from $1$ to $\infty$. -Notice that it is the left Riemann sum for $nx^{n-1}$ on $[0,1]$ with fixed partition refining near $1$. The issue is that the partition interval lengths decay very fast near $1$ while $nx^{n-1}$ is pretty much the same as $ne^{-n(1-x)}$, so, reflecting about $1$, we arrive at the following situation: -We have the function $e^{-x}$ and take the right Riemann sum over some partition in which each next (to the right) interval is huge compared to the previous one. Then there is at most one interval of length comparable to constant (perhaps none) and it determines the probability up to $o(1)$. Thus, the upper limit should be $\max_{x>0} xe^{-x}=1/e$ and the lower limit should be $0$, so the tie is more likely than not in every case. -It is worth mentioning, however, that the current Earth population ($\approx 7.6\times 10^9$) is way too low to see that effect, so for any reasonably sized group of humans, you can safely say that the probability of tie is $50\%\pm 10\%$.<|endoftext|> -TITLE: Fixed point for a map from $\{0,1\}^N$ to itself -QUESTION [7 upvotes]: Let $N\geq2.$ Let $F$ be a function from $\left\{ 0,1\right\} ^{N}$ to itself -dreceasing for the product order defined by $$ (x_1,x_2,\ldots,x_N)\leq (y_1,\ldots,y_N)\ \text{ if and only if for all }i,\ x_i\leq y_i $$ -Here, $F$ being decreasing means $$x\leq y \Rightarrow F(y)\leq F(x)$$ -Suppose moreover that the $i^{th}$ component of $F$ -does not depend on the $i^{th}$ variable. -Is it true that $F$ has a unique -fixed point ? - -REPLY [6 votes]: For $N = 3$ there are exactly $58$ counterexamples. -$54$ of them have two fixed points. E.g.: -\begin{eqnarray} -000 &\mapsto& 110\\ -100 &\mapsto& 100\\ -010 &\mapsto& 010\\ -110 &\mapsto& 000\\ -001 &\mapsto& 000\\ -101 &\mapsto& 000\\ -011 &\mapsto& 000\\ -111 &\mapsto& 000\\ -\end{eqnarray} -$2$ of them have three fixed points. E.g.: -\begin{eqnarray} -000 &\mapsto& 111\\ -100 &\mapsto& 100\\ -010 &\mapsto& 010\\ -110 &\mapsto& 000\\ -001 &\mapsto& 001\\ -101 &\mapsto& 000\\ -011 &\mapsto& 000\\ -111 &\mapsto& 000\\ -\end{eqnarray} -2 of them have zero fixed point. E.g.: -\begin{eqnarray} -000 &\mapsto& 111\\ -100 &\mapsto& 101\\ -010 &\mapsto& 110\\ -110 &\mapsto& 100\\ -001 &\mapsto& 011\\ -101 &\mapsto& 001\\ -011 &\mapsto& 010\\ -111 &\mapsto& 000\\ -\end{eqnarray}<|endoftext|> -TITLE: Inverse Hodge and inverse Betti problems for Kähler manifolds -QUESTION [7 upvotes]: A Betti sequence is a map $\mathbb{Z}_{\geq 0}\to \mathbb{Z}_{\geq 0}$. -A Betti sequence $b$ is realizable if there is a connected closed Kähler manifold $M$ such that $b(k)=b_k(M)$. -A Hodge diamond is a map $\mathbb{Z}_{\geq 0}\times \mathbb{Z}_{\geq 0}\to \mathbb{Z}_{\geq 0}$. To any Hodge diamond $h$ we associate a Betti sequence $b$ given by $b(k)=\sum_{i=0}^k h(i, k-i)$. -A Hodge diamond $h$ is realizable if there is a connected closed Kähler manifold $M$ such that $h(i, j)=h^{i, j}(M)$. Realizable Hodge diamonds have realizable Betti sequences. -A Hodge diamond is naively realizable if $h(0, 0)=1$ and there is an integer $n\geq 0$ such that - -$h(i, j)=h(j, i)=h(n-i, n-j)$ if $0\leq i, j\leq n$ -$h(i, j)\geq h(i-1, j-1)$ if $i, j\geq 1$ and $i+j\leq n$ -$h(i, j)=0$ if $i, j\geq 0$ and $\mathrm{min}(i, j)\geq n+1$ - -Is there a naively realizable, non-realizable Hodge diamond with a realizable Betti sequence? -For example, the naively realizable Hodge diamond with $n=2$ and $h(0, 1)=h(1, 1)=1$, $h(0, 2)=0$ is not realizable but its Betti sequence is not realizable either. - -REPLY [11 votes]: The Hodge diamond \begin{array}{ccccc}&&1&&\\&0&&0&\\a&&1&&a\\&0&&0&\\&&1&&\end{array} is naively realisable. -Suppose $M$ is a compact Kähler surface with the given Hodge diamond with $a \geq 2$. As $h^{2,0}(M) > 1$, the Kodaira dimension of $M$ is either $1$ or $2$. Note that $$c_1(M)^2 = 2\chi(M) + 3\sigma(M) = 2(2a + 3) + 3(2a + 1) = 10a + 9.$$ It follows from the Kodaira-Enriques classification that $c_1(M)^2 \leq 0$ for a surface of Kodaira dimension $1$, so we must have $\kappa(M) = 2$. Note however that $$c_1(M)^2 = 10a + 9 > 6a + 9 = 3(2a + 3) = 3c_2(M)$$ which contradicts the Bogomolov-Miyaoka-Yau inequality. Therefore, the above Hodge diamond with $a \geq 2$ is not realisable. -Finally, the corresponding Betti sequence is realisable as the blowup of $\mathbb{CP}^2$ at $2a$ points demonstrates. - -The same arguments apply to the Hodge diamond \begin{array}{ccccc}&&1&&\\&0&&0&\\a&&b&&a\\&0&&0&\\&&1&&\end{array} with $a \geq b \geq 1$ and $a \geq 2$. -The case $a = b = 1$ provides another example. The same arguments as above can be used to rule out Kodaira dimensions $-\infty$, $1$, and $2$. To rule out $\kappa(M) = 0$, note that $M$ must be minimal and hence must be finitely covered by a K3 surface or a complex torus. As $\chi(M) = 5$ and the Euler characteristic is multiplicative under finite covers, this is impossible.<|endoftext|> -TITLE: How to characterize $E_n$ morphisms from $\mathrm{Mod}(A)$ to $\mathrm{Mod}(B)$? -QUESTION [9 upvotes]: Suppose $A$ and $B$ are $E_{\infty}$ rings, then $\mathrm{Mod}(A)$ and $\mathrm{Mod}(B)$ are $E_{\infty}$ monoidal categories (left modules over those rings). We can ask about $E_n$ colimit-preserving morphisms between these two categories. How do we characterize them? If $n=\infty$ we just have maps from $A$ to $B$ and if we don't ask for any monoidal compatibility I guess we just get bimodules over $A$ and $B$, but what are the intermediates? Maybe we can also ask for a characterization of $E_n$ maps of the category of $E_n$ modules over $E_n$ rings? -Edit: Thanks to Dylan's answer below the first question is solved. However I believe the second question is still not answered, which is about studying $E_n$ maps between categories of $E_n$ modules over $E_n$ rings. - -REPLY [2 votes]: To clarify my comment: suppose you're interested in lax $E_n$-monoidal functor (so that indeed, you get coherent maps $F(M)\otimes F(N)\to F(M\otimes N)$ which aren't necessarily equivalences, similarly for $1\to F(1)$) that preserve colimits, then the category of such functors is essentially $Alg_{E_n}(Fun^{ex}(Mod_A^\omega,Mod_B))$ where $Fun^{ex}(Mod_A^\omega, Mod_B)$ has a symmetric monoidal structure induced by the Day convolution structure on $Fun(Mod_A^\omega, Mod_B)$. -Then you get a symmetric monoidal identification $Mod_{A^{op}\otimes B} \simeq Fun(Mod_A^\omega,Mod_B)$ which, since $A$ is $E_\infty$, can be thought of as $Mod_{A\otimes B}$. -Therefore lax $E_n$-monoidal colimit preserving functors $Mod_A\to Mod_B$ can be thought of as $E_n-A\otimes B$-algebras. -Of course, these rarely classify strict monoidal functors (those where $F(M)\otimes F(N)\to F(M\otimes N)$ is an equivalence, and $1\to F(1)$ is as well). In fact they do if and only if this $A\otimes B$-module is $B$ with structure induced from an $E_{n+1}$-map $A\to B$: Tyler's answer shows this. -I made a silly comment below Tyler's answer (then deleted, as I realized my mistake), which is perhaps noteworthy: if you try to take this approach and check when, for an $A\otimes B$-module $R$, the associated lax $E_n$-monoidal functor is actually strict, you'll get two conditions: one on the map $F(M)\otimes F(N)\to F(M\otimes N)$ and one on the unit. The second one is easy to forget (which is what happened in my silly comment), and if you do, you'll get something like an $E_n-B$-algebra such that the structure map $B\to R$ is an epimorphism. -This is why for instance $Mod_\mathbb Q\to Sp$ is almost symmetric monoidal, although it is only lax: you get the condition on tensor products but not the one on the unit.<|endoftext|> -TITLE: Does such a function exist? -QUESTION [5 upvotes]: I am looking for a function with the following property: -Let $v_1,v_2$ be two linearly independent vectors in $\mathbb{R}^2.$ -I am given a smooth function $g:(0,1) \rightarrow (0,\infty).$ -I am trying to understand if there exists a smooth (non-constant) function $f:(0,1) \times \mathbb R^2 \rightarrow \mathbb R^2$ with the property that for all $n,m \in \mathbb Z$ such that $g(t)=n/m$, where $n/m$ is a reduced fraction, we have for all $x \in \mathbb R^2$ -$$f(t,mv_1+x)=f(t,x)=f(t,nv_2+x)$$ and $n,m$ are the minimal periods of the function $f$, i.e. for all natural numbers $1\le n_0 -TITLE: Wayback Machine for mathematics? -QUESTION [31 upvotes]: I have had a couple of experiences recently which have made me wonder whether the mathematics community should try to establish and maintain something like the Wayback Machine, but specifically focused on mathematics. -In a paper that I wrote back in 1999, I cited a webpage with the title "Is elementary?" and provided the following URL: -http://www.apmaths.uwo.ca/~rcorless/AM563/NOTES/Nov_23_95/Nov_23_95.html -As you might guess, that URL is no longer functioning. I tried the Wayback Machine and I was lucky that it had indeed archived a copy of the relevant part of the page. Unfortunately, crucial parts of the page were not preserved; the original page used LaTeX2HTML, whose results were not satisfactorily archived. -The above example is not so important from a mathematical point of view because there is now a much better reference available for the mathematical fact in question, but it is a good illustration of the sort of thing I am concerned about—there's a considerable amount of material on the web that is of mathematical value but that is disappearing because it is not being formally published or archived. -My second example is more (ahem) consequential, and is something that the mathematical community might be able to do something about if we act now. For years there has been a useful web resource at Purdue for researching Consequences of the Axiom of Choice. Unfortunately, the page is no longer functioning, as you will quickly discover if you try submitting a form number. The URLs have changed. I suspect that Purdue redesigned its website at some point, changing the URLs, and that since Herman Rubin died a couple of years ago, there is now nobody responsible for maintaining the Axiom of Choice page. I tried emailing a couple of random people in the Purdue mathematics department to find out if something could be done to revive the page, but have received no response. -The ideal solution for the Axiom of Choice page may be for some researchers with an active interest in the area to create a wiki, whose survival will not depend on the survival of a single person. That is the route that the OEIS took and it seems to have worked out well. One would like to have not just a snapshot of the contents frozen at a single point in time, but a dynamic resource that is continually updated. Failing that, though, a snapshot would be better than nothing. However, according to my understanding, the Wayback Machine is not well designed for something like this where you're supposed to access the content by querying a form. -These two examples are of course only the tip of an iceberg. Scattered across the Internet are all kinds of lecture notes, computer code, databases, blog posts, etc., that are of long-term mathematical interest but that are at risk of disappearing when people retire or die. Even something like MathOverflow should perhaps be archived from time to time in some independent location in case something goes awry with the corporation in charge of it. - -Would it be feasible to set up something like the Wayback Machine but specifically targeted at mathematics, so that we could ensure higher quality preservation than the actual Wayback Machine is able to provide? If so, which organizations are best equipped to create and maintain such a resource? - -REPLY [9 votes]: In my previous answer, I focused on unintended consequences of creating a Wayback Machine for math. But I'm worried that answer is preventing us from answering the actual question. - -Would it be feasible to set up something like the Wayback Machine but specifically targeted at mathematics, so that we could ensure higher quality preservation than the actual Wayback Machine is able to provide? If so, which organizations are best equipped to create and maintain such a resource? - -The Wayback Machine works via web crawlers, and it would be somewhat difficult to replicate what they do just for math. You'd have to help the algorithm decide which pages to archive, and I don't see any clean way to do that. For example, if you only crawled pages with a .edu endpoint, and with math in the URL, you'd miss personal web pages. You could crawl pages where the index page has mathematician in the text, but you'd miss pages about software, paper repositories not tied to a personal web page, etc. -I think a better solution, in the short term, would be to get journals to keep a copy of any web page that is cited when a paper is published, just in case that page goes down in the future. In case the journal can't be convinced, authors could also keep copies of web pages they cite. For every paper I've published, I have a folder with the tex file. I can just make a subfolder where I keep a copy of web pages I cited as of the day I submitted the paper. This would also be a good idea for citing online lecture notes (e.g., Stefan Schwede's notes on symmetric spectra or equivariant homotopy theory) which have many different version and where the theorem numbering might eventually change from what it was when you cited it. -Keeping these folders is very little extra work for us (or for the journals) and feels important to the enterprise of mathematics. To me, it's analogous to how other scientific fields are combating the Replication Crisis, by keeping track of what they did to their data, keeping a copy of the data as it was when the paper was written, and keeping their R code, so that a future researcher could replicate what they had done. I want people to be able to see the same documents I saw when I wrote my paper, so for any that lack a permanent home, I should keep a copy just in case they vanish. If they do vanish, at least I'll have a copy I can share in case someone asks, or potentially host if I have permission (like the Wayback Machine does).<|endoftext|> -TITLE: holomorphy in infinite dimensions (holomorphic families of operators) -QUESTION [6 upvotes]: Let $X$ be a Banach space (over $\mathbb C$), and let $\mathcal L(X)$ be its algebra of bounded linear operators. -Let $U\subset \mathbb C^N$ be an open subset, and $f:U\to \mathcal L(X)$ a function that is locally bounded (with respect to the operator norm on $\mathcal L(X)$), and holomorphic when $\mathcal L(X)$ is equipped with the topology of pointwise convergence (the strong operator topology). - -Does it then automatically follow that $f$ is holomorphic when one equips $\mathcal L(X)$ with respect to the topology of uniform convergence on bounded sets (the norm topology)? - - -Let $X$ be a locally convex topological vector space, and let $\mathcal L(X)$ be its algebra of continuous linear operators. -Let $U\subset \mathbb C^N$ be an open subset, and $f:U\to \mathcal L(X)$ a function that is locally bounded, and holomorphic when $\mathcal L(X)$ is equipped with the topology of pointwise convergence. Here, $f$ being locally bounded means that for every compact $K\subset U$ and every bounded $B\subset X$, the set $\{f(z)(x): z\in U, x\in B\}$ is again bounded in $X$. - -Does it then automatically follow that $f$ is holomorphic when one equips $\mathcal L(X)$ with respect to the topology of uniform convergence on bounded sets? - -REPLY [3 votes]: In addition to the information given by user bathalf15320, I think that a bit more information on the Banach space case could be useful: -Here is a very general theorem about vector valued functions: -Theorem 1. Let $Y$ be a complex Banach space and let $f: U \to Y$ be locallly bounded. Let $W \subseteq Y'$ be a subset which is norming for $Y$ (or more generally, almost norming as defined in [1, p. 779]). If $z \mapsto \langle y', f(z) \rangle$ is holomorphic for each $y' \in W$, then $f$ is holomorphic with respect to the norm on $Y$. -Reference: [1, Theorem 1.3]. -Corollary 2. Let $X$ be a complex Banach space and let $f: U \to \mathcal{L}(X)$ be such that $z \mapsto \langle x', f(z) x\rangle$ is holomorphic for each $x \in X$ and each $x' \in X'$. Then $f$ is holomorphic with respect to the operator norm. -Proof: (a) Note that $f$ is automatically locally bounded as a consequence of the uniform boundedness theorem. -(b) Now apply Theorem 1 to $Y = \mathcal{L}(X)$, where $W$ is to the linear span of the set of all functionals on $\mathcal{L}(X)$ of the form -$$ - \mathcal{L}(X) \ni T \mapsto \langle x', Tx \rangle \in \mathbb{C}, -$$ -where $x \in X$ and $x' \in X'$. qed -What is, however, probably more surprising is the fact that, in Theorem 1, we can replace the assumption that $W$ be (almost) norming with the assumption that $W$ merely separates the points of $X$. This result can be found in [1, Theorem 3.1]. -Further information on the topic can for instance be found in [1] and [2]. -References: -[1] W. Arendt, N. Nikolski: Vector-valued holomorphic functions revisited (Math. Z., 2000) -[2] W. Arendt, N. Nikolski: Addendum to 'Vector-valued holomorphic functions revisited' (Math. Z., 2006)<|endoftext|> -TITLE: An explicit expression for the naturality of the Serre automorphism in the bicategory of algebras -QUESTION [7 upvotes]: By the cobordism hypothesis, there is an $O(2)$-action on the maximal subgroupoid $\hat{\mathcal{C}}$ of the subcategory of fully dualizable objects in a bicategory $\mathcal{C}$. The $SO(2)$-part of this action can equivalently be described by a natural transformation $id_{\hat{\mathcal{C}}} \to id_{\hat{\mathcal{C}}}$ which maps an object $C$ to the Serre automorphism $S_C$ (see Chris Schommer-Pries' lecture notes "Dualizability in Low-Dimensional Higher Category Theory)". As we have a natural isomorphism between 2-functors, given a 1-morphism $f$, we also expect a 2-morphism $S_f$ satisfying certain properties. In Section 4.1.1. of Jan Hesse's thesis the existence of $S_f$ is proven, but no explicit expression is given. -Consider the 2-category $Alg$ in which objects are algebras over $\mathbb{C}$, 1-morphisms are bimodules and 2-morphisms are intertwiners. The subcategory of fully dualizable objects consists of finite-dimensional semisimple algebras, finite-dimensional bimodules and intertwiners (ref: lemma 3.2.1 and 3.2.3 of Orit Davidovich' thesis). In $Alg$, the Serre automorphism is given by the $\mathbb{C}$-linear dual $S_A = A^*$ as an $(A,A)$-bimodule (Lemma 4.18 of Jan Hesse's thesis). -Now let $M$ be an invertible $(A,B)$-bimodule, where $A,B$ are finite-dimensional semi-simple. The 2-morphism $S_M$ expresses a canonical filling of the diagram -$\require{AMScd}$ -\begin{CD} -A @>M>> B\\ -@V A^* V V @VV B^* V\\ -A @>>M> B -\end{CD} -In other words, it is an $(A,B)$-bimodule isomorphism $S_M: A^* \otimes_A M \to M \otimes_B B^*$. -Question: Is there an explicit expression for $S_M$, preferably one that does not depend on a lot of choices, such as bases and direct sums into simples? - -REPLY [4 votes]: We will use the fact that $M$ is invertible. Let ${}_BN_A$ be an inverse to $M$. Thus we have isomorphisms -$${}_AM \otimes_B N_A \cong {}_AA_A$$ -and -$${}_BN \otimes_A M_B \cong {}_BB_B$$ -If we make this data part of an adjoint equivalence (as we should, and as I will assume) then the construction I am about to explain won't depend on these choices. -Rather than construct the map you ask for, I will construct an equivalent map: -$$S_A: {}_B N \otimes_A A^* \otimes_A M_B \to {}_BB^*_B$$ -This is easier to express since we are not mapping into a tensor product. -Given an element $b \in B$ we can write it as $\sum_i n_i \otimes m_i$ in $N \otimes_A M$. -Given $n \otimes f \otimes m$ in $N \otimes_A A^* \otimes_A M$, the map $S_A$ sends it to the following linear map on $B$: -$$b = \sum_i n_i \otimes m_i \mapsto \sum_if(mn_i \cdot m_in)$$ -Here $m n_i$ and $m_i n$ are taken as elements in $M \otimes_B N = A$, which are multiplied together before applying the linear functional $f$. It is not too hard to check that this map is well-defined (doesn't depend on the choice of representation $b =\sum_i n_i \otimes m_i$) and also that it is a $B$-$B$-bimodule map. -It is a little harder to see that this is an isomorphism and I don't have time to write it out just now, but notice that the same construction gives a map the other way: -$$M \otimes_B B^* \otimes_B N \to A^*$$ -I claim that you can use this to show $S_A$ is an isomorphism.<|endoftext|> -TITLE: Non-calibrated area-minimising surface -QUESTION [8 upvotes]: Let $(M^{n+k},g)$ be a Riemannian manifold. Call a surface $\Sigma^n \subset M$ calibrated if there is a closed $n$-form $\omega$ defined on a neighbourhood $U \subset M$ of $\Sigma$ so that $\omega \lvert \Sigma = \mathrm{vol}_\Sigma$ and for any $p \in U$ and $n$-tuples $(X_1,\dots,X_n) \in T_p M$ of orthonormal vectors $\omega(X_1,\dots,X_n) \leq 1$. (This is slightly different from the usual definition, where usually $\omega$ is defined on $M$.) A simple argument shows that a calibrated surface $\Sigma$ is area-minimising in the neighbourhood $U$, and a small perturbation of $\Sigma'$ of $\Sigma$ will have $\mathrm{Area}(\Sigma') \geq \mathrm{Area}(\Sigma)$. In particular a calibrated surface is minimal, that is stationary for the area functional, and has mean curvature $H_\Sigma = 0$. -There are many examples of calibrated area-minimising surfaces: - -linear subspaces of $\mathbf{R}^n$, -minimal graphs of $u: \Omega \subset \mathbf{R}^n \to \mathbf{R}$, where $\Omega$ is an open domain in $\mathbf{R}^n$, -special Lagrangian submanifolds $\Sigma \subset M$ in Calabi-Yau manifolds, that is Lagrangian submanifolds so that $\mathrm{Im} \, \Omega \lvert \Sigma = 0$ where $\Omega$ is the holomorphic volume form, -holomorphic subvarieties of $\mathbf{C}^n$, -area-minimising cones with an isolated singularity at the origin, for example the Simons cone $\mathbf{C}_S = \{ (X,Y) \in \mathbf{R}^n \times \mathbf{R}^n \mid \lvert X \rvert = \lvert Y \rvert \}$. (I believe these are calibrated because of the Hardt-Simon foliations.) - -However I cannot think of any examples of area-minimising surfaces which are not calibrated. -Question: -What are they? I am especially interested in the codimension one case, where $\Sigma^n \subset M^{n+1}$. In which settings, or under which hypotheses, is an area-minimising surface not be calibrated? -Remark: I can formulate a more technically precise question, at the price of using some terms from geometric measure theory. Let $B \subset \mathbf{R}^{n+k}$ be the unit ball, and $T \in \mathbf{I}_n(B)$ be an integral current with $\partial T = 0$ in $B$. Suppose that $T$ is area-minimising in the sense that for some $\epsilon > 0$ and all currents $S \in \mathbf{I}_{n+1}(B)$ with $\mathrm{spt} \, S \subset \subset B$ and $\mathrm{dist}(\mathrm{spt} \, S,\mathrm{spt} \, T) \leq \epsilon $, $\mathrm{Area} \, (T + \partial S) \geq \mathrm{Area} \, T$. Is there a neighbourhood of $T$ on which it admits a calibration? Here again I would be most interested in the case $k = 1$. - -REPLY [7 votes]: Actually, a better example along the lines Otis suggests would be the geodesic $\mathbb{RP}^1\subset\mathbb{RP}^2$. Of course, $\mathbb{RP}^1$ is orientable and it is homologically mass-minimizing, but it can't be calibrated on any open set $U\subset\mathbb{RP}^2$ containing $\mathbb{RP}^1$ because twice it is not even stable. -Of course, this also works for any $\mathbb{RP}^{2n-1}\subset\mathbb{RP}^{2n}$, and there are higher codimension examples of closed geodesics in (orientable) lens spaces that are homologically mass-minimizing but that cannot be calibrated on any open neighborhood of the geodesic. One can even foliate $\mathbb{RP}^3$ by homologically mass-minimizing geodesics that cannot be calibrated on any open neighborhood. -What one probably needs to assume, at least, is that every multiple of $\Sigma$ is homologically area-minimizing in some neighborhood before one could hope to construct a 'neighborhood' calibration. -Remark (10/12/20): I just recalled one example of possible interest for this question, since the OP is interested in what can happen in Euclidean space. A student of mine, Timothy Murdoch, in his PhD thesis "Twisted calibrations and the cone on the Veronese surface" (Rice University, 1988), showed that the $3$-dimensional cone in $\mathbb{R}^5$ on the Veronese surface in $S^4$ is area-minimizing, but, of course, it's not orientable. However, its 'double cover' is a cone on the $2$-sphere and so is orientable. I don't know whether this double cover is area-minimizing in $\mathbb{R}^5$ or not. It obviously cannot be calibrated, even if it is area-minimizing. -Explicitly, here is the example: Think of $\mathbb{R}^5$ as $S^2_0(\mathbb{R}^3)$, the traceless $3$-by-$3$ matrices with real entries endowed with the quadratic form $\langle a,b\rangle = \mathrm{tr}(ab)$, which is invariant under $\mathrm{SO}(3)$ with the irreducible action $A\cdot a = AaA^{-1}$ for $A\in\mathrm{SO}(3)$ and $a\in S^2_0(\mathbb{R}^3)$. Then the Veronese cone $C\subset S^2_0(\mathbb{R}^3)$ is the set of matrices $a$ with eigenvalues $t^2,t^2, -2t^2$ for some $t\ge0$. It is a cone on an $\mathrm{SO}(3)$-homogeneous minimal surface $\mathbb{RP}^2\subset S^4$ known as the Veronese surface. (Note that $C$ and $-C$ intersect only at the origin.) $C$ is smooth except at the origin, and, if you define the 'double cover' by counting each smooth point as two points with different orientations, then the double cover is homeomorphic to $\mathbb{R}^3$, parametrized by the quadratic map $s:\mathbb{R}^3\to S^2_0(\mathbb{R}^3)$ defined by -$$ -s(x) = |x|^2\, I_3 - 3\,x\,x^T\quad\text{for}\ x\in\mathbb{R}^3. -$$ -Tim showed that, if you take the (literal) Riemannian double cover of $S^2_0(\mathbb{R}^3)\setminus (-C)$, then the double cover of $C\setminus\{0\}\simeq \mathbb{R}^3\setminus\{0\}$ can be calibrated in the ambient double cover as a Riemannian manifold.<|endoftext|> -TITLE: Simple quotients of a triple tensor product -QUESTION [5 upvotes]: Let $\mathcal{H}$ be a Hopf algebra over $\mathbb{C}$. Let also $V_1, V_2, V_3$ finite-dimensional simple modules over $\mathcal{H}$ and $Q$ be a simple quotient of $V_1\otimes V_2\otimes V_3$. Is it possible to show that one of the following statements is true? Is there any counterexample? -i) $Q$ is a quotient of $N\otimes V_3$, for some simple quotient $N$ of $V_1\otimes V_2$; -ii) $Q$ is a quotient of $V_1\otimes P$, for some simple quotient $P$ of $V_2\otimes V_3$. - -REPLY [7 votes]: Both of these statements are true (at least if $H$ is semisimple). It suffices to prove the first one. By hypothesis there is a nonzero map $V_1 \otimes V_2 \otimes V_3 \to Q$. It dualizes to a nonzero map $V_1 \otimes V_2 \to Q \otimes V_3^{\ast}$ (I don't know if I need to distinguish between left and right duals here if $H$ isn't cocommutative but I don't think it matters, just whichever dual makes this true), which factors through its image $P \to Q \otimes V_3^{\ast}$. Dualizing again we get a nonzero map $P \otimes V_3 \to Q$.<|endoftext|> -TITLE: An induction formula for spectral Mackey functors, and a fake proof -QUESTION [5 upvotes]: I'm trying to get a grasp of Barwick's model for genuine $G$-spectra, that is, spectral Mackey functors 1. There's a classical formula about induction, that should be easy to prove, that I was trying to prove in this model; but I failed, and it's worse than that because it's not just that my proof isn't conclusive, it proves that the formula fails. -So I'm trying to understand where my proof went wrong: the rest of my question is devoted to presenting said proof, and the question is : - -Where is the mistake ? - -I apologize in advance for the long proof, but I wanted to make sure all the details were there - of course if I say "one easily sees that" too much, then the mistake is bound to be in those moments. -So here goes: -$\newcommand{\ind}{\mathrm{Ind}} \newcommand{\Hom}{\mathrm{Hom}} \newcommand{\map}{\mathrm{map}} \newcommand{\Map}{\mathrm{Map}} \newcommand{\Mack}{\mathbf{Mack}} \newcommand{\res}{\mathrm{res}}$ -My conventions are as follows: $\map$ denotes the mapping space, $\Map$ denotes the mapping spectrum between two objects in a stable $\infty$-category, and in the category $\Mack_G$ (of spectral Mackey functors), $\Hom_G$ will denote the internal hom, defined by adjunction with the Green tensor product (defined in 2). $G$ will be a fixed finite group. -$A^{eff}(G)$ is the effective Burnside category, defined in 1. -I will say "$G$-spectrum" to mean "spectral Mackey functor on $A^{eff}(G)$". -The claim in question is that for any subgroup $H$, $H$-spectrum $Y$, and $G$-spectrum $X$, $\ind_H^GY\otimes X\simeq \ind_H^G(Y\otimes \res_H^GX$). This is a pretty basic claim, analogous to the situation in $1$-category theory, for representations over a commutative ring, so it should be true. -If the proof is correct, then maybe the mistake is in my assumption that the Green tensor product corresponds the usual smash product of genuine $G$-spectra ? - -Is that the case ? - -Note that $\ind_H^G$ is defined to be left adjoint to $\res_H^G$, while $\res_H^G$ is defined by restriction along $A^{eff}(H)\to A^{eff}(G)$ given informally by $L\mapsto G\times_H L$ (this is a functor $F_H\to F_G$ which preserves pullbacks and coproducts, so it induces an additive functor on the effective Burnside categories) -Then, $$\map(\ind_H^GY\otimes X,Z) \simeq \map(\ind_H^GY, \Hom_G(X,Z))\simeq \map(Y,\res_H^G\Hom_G(X,Z))$$ -where all the equivalences are natural in $X,Y,Z$, and also -$$\map(\ind_H^G(Y\otimes \res_H^GX),Z)\simeq \map(Y\otimes \res_H^GX,\res_H^GZ)\simeq \map(Y, \Hom_H(\res_H^GX,\res_H^GZ))$$ -so the claim is equivalent to $\res_H^G\Hom_G(X,Y)\simeq \Hom_H(\res_H^GX,\res_H^GZ)$, which is a reasonable claim, and again, valid in the analogous situation in $1$-category theory. -So to prove this claim I tried to compute the value of $\res_H^G\Hom_G(X,Z)$ on $H/K$ for $K\leq H$. This is the same as $\Hom_G(X,Z)(G/K)$ by definition, and so I wanted to compute the values of $\Hom_G(X,Z)$. -Note that $ev_{G/K}\circ \Hom_G(X,-)$ is right adjoint to $X\otimes i_!(-)$, where $i : \{G/K\}\to A^{eff}(G)$ is the inclusion, and $i_!$ is left Kan extension along $i$ followed by the left adjoint to the inclusion $\Mack_G\to Fun(A^{eff}(G),Sp)$. -So, naturally in $X,Y,Z$, $\map_{\Mack_G}(X\otimes i_!(Y), Z) \simeq \map_{Fun(A^{eff}(G),Sp)}(X\otimes_{Day} i_! Y,Z)$ by definition of the Green tensor product; then by definition of the Day tensor product, this is $\simeq \map_{Fun(A^{eff}(G)\times A^{eff}(G),Sp)}(X(-) \boxtimes i_!Y(\bullet),Z(-\times \bullet)$ where $(X\boxtimes i_!Y)((L,L')) := X(L)\otimes (i_!Y)(L'))$. -So now we get $\map_{Fun(A^{eff}(G),Sp)}(i_!Y(\bullet), \Map(X(-), Z(-\times \bullet))$. -Let me specify what I mean here, as the notations are annoying: $\bullet$ being fixed, $Z(-\times \bullet)$ is a functor of $-$ so $\Map(X(-),Z(-\times \bullet))$, the mapping spectrum in the stable $\infty$-category $Fun(A^{eff}(G),Sp)$ is well-defined, and it's a functor of $\bullet$, which makes this meaningful. -Now finally, using the definition of $i_!$, and given that $\Map(X(-),Z(-\otimes \bullet))$ is an additive functor of $\bullet$, this mapping space is $\map_{Sp}(Y, \Map(X(-), Z(-\times G/K))$. -It follows that $ev_{G/K}\circ \Hom_G(X,Z)\simeq \Map(X,Z(-\times G/K))$. -Therefore, on the one hand, -$$\res_H^G\Hom_G(X,Y)(H/K)\simeq \Map(X,Z(-\times G/K))$$ -and on the other hand, -$$\Hom_H(\res_H^GX,\res_H^GZ)(H/K) \simeq \Map(\res_H^GX, (\res_H^GZ)(-\times G/K)) \simeq \Map(\res_H^GX, Z(G\times_H(-)\times G/K)) \\ -\simeq \Map(X,\mathrm{CoInd}_H^G(Z(G\times_H(-)\times G/K)))$$ the $\res_H^G\dashv \mathrm{CoInd}_H^G$ adjunction being automatically $Sp$-enriched. -To conclude, we must compute $\mathrm{CoInd}_H^G M$, for a $G$-spectrum $M$. But note that the forgetful functor $A^{eff}(G)\to A^{eff}(H)$ (induced by the forgetful $F_G\to F_H$) is left adjoint to $G\times_H -$ (as it is a right adjoint to it, and we have compatible equivalences $A^{eff}(G)^{op}\simeq A^{eff}(G))$), so that precomposing with it is right adjoint $Fun(A^{eff}(H),Sp)\to Fun(A^{eff}(G),Sp)$. But now both precompositions preserve the full subcategories of Mackey functors, so they induce an adjunction between $\Mack_G$ and $\Mack_H$, the left adjoint of which is precomposition with $G\times_H-$, i.e. $\res_H^G$. -So precomposition with the forgetful $U: A^{eff}(G)\to A^{eff}(H)$ is exactly coinduction (one can make a reality check of this by looking at what this means when we evaluate in $G=G/e$, which should yield the underlying spectrum); so that, in the end -$$\Hom_H(\res_H^GX,\res_H^GZ)(H/K)\simeq \Map(X, Z((G\times_H U(-)) \times G/K))$$ -and those two spectra are simply not the same, as $G\times_H U(-)$ is not equivalent to the identity. Worse, one can make it pretty explicit : $G\times_H U(-)\simeq G/H\times -$, as one checks on $F_G$, so that the latter is $\Map(X,Z(G/H\times G/K\times -)$. Taking, e.g., $H=K=e$, by a usual trick you can make on of the $G$'s have a trivial action, and so this is $\bigoplus_{g\in G}\Map(X,Z(G\times -))$ which is, in general, different from $\Map(X,Z(G\times -))$. -But note that the reduction from the beginning was an equivalence, not only a sufficient condition, so this computation seems to not only mean I can't prove the claim, but actually that the claim is wrong ! However this is seemingly a well-known fact, and analogous to an easy $1$-categorical fact, so I must be making a mistake. -1: Barwick, C. (2017). Spectral Mackey functors and equivariant algebraic K-theory (I). Advances in Mathematics, 304, 646-727. -2: Barwick, C., Glasman, S., & Shah, J. (2019). Spectral Mackey functors and equivariant algebraic K-theory, II. Tunisian Journal of Mathematics, 2(1), 97-146. - -REPLY [3 votes]: $\newcommand{\Hom}{\mathrm{Hom}} \newcommand{\res}{\mathrm{res}}$ -Ah, well, I found the mistake (at a surprising time: I'm more tired now than I was when I looked for it earlier) : $A^{eff}(H)\to A^{eff}(G)$ given by $G\times_H-$ preserves pullbacks, not products ! In particular, in my computation for $\Hom_H(\res_H^GX,\res_H^GZ)(H/K)$, I used $G\times_H(-\times H/K) \simeq G\times_H - \times G/K$, whereas it should be (apologies for the notation) $(G\times_H -) \times_{G/H} G/K$ (where of course $\times_H$ is not a pullback, but $\times_{G/H}$ is one) -This precisely solves the problem, as $G\times_H U(-) \simeq G/H\times -$ so this $G/H$ gets cancelled with the $\times_{G/H}$ and you get the same result, which makes the proof work. -As Dylan pointed out in the comments, there's a more direct, and perhaps more elegant proof using a reduction to $F_G$ and $F_H$ (which, I think, requires a lemma of the type $\Sigma^\infty_+\res_H^G\simeq \res_H^G\Sigma^\infty_+$, but this is not hard to prove). Of course, "my" proof is still interesting as it provides an explicit formula for $\Hom_G(X,Z)(G/K)$, which is perhaps interesting (although it was of course known in general for the internal hom in a Day convolutional symmetric monoidal structure)<|endoftext|> -TITLE: Is the theory Flow actually consistent? -QUESTION [34 upvotes]: Recently the paper - -Adonai S. Sant'Anna, Otavio Bueno, Marcio P. P. de França, Renato Brodzinski, Flow: the Axiom of Choice is independent from the Partition Principle, arXiv:2010.03664 - -appeared on the arXiv, which claims that the authors' new theory, Flow, can prove the Partition Principle ("if $A$ is nonempty and surjects onto $B$, $B$ injects into $A$") but not the Axiom of Choice. If correct, this would resolve a long-open question in Set Theory (whether PP implies AC). -However, nowhere in the paper is the mentioned theory proved consistent (even relative to some LCA, which would be necessary as the paper claims their theory can interpret "ZF + there is a Grothendieck Universe") as far as I can tell. The axioms/principles that make up the theory seems to be dispersed throughout the paper. Definitions 7 and 10 (p13/p14) also seems to be self-referential, and no way of resolving self-referential definitions is discussed. -Therefore, I wonder if anyone on MO knows is the mentioned theory Flow consistent (relative to some LCA)? - -REPLY [25 votes]: I am one of the authors of this preprint. Concerning the Grothendieck Universe, my first answer is this. Some terms of Flow are called ZF-sets. Through the use of our restriction axiom we are able to define a term u which acts on all ZF-sets. That means x belongs (membership relation is defined from the concept of action in Flow) to u iff x is a ZF-set. Next we prove u satisfies all conditions of a Grothendieck Universe. Concerning self-referential definitions, that is something we intend to fix for a next version of this work. Somehow we missed that important issue. Sorry about that. Concerning consistency of Flow, we should be very careful: if Flow is consistent, then we have a model of ZF where PP holds but AC is not valid. We are still considering questions regarding relative consistency. Anyway, it has been a great pleasure to collect all those first criticisms to our work. - -In response to a question in comments: -The idea of Theorem 55 is something like this. AC is equivalent to the statement that every set can be well-ordered. So, if we want to prove AC does not hold, we need to exhibit a ZF-set which cannot be well-ordered. We chose a hyperfunction psi. But an analogous proof can be done to certain well-founded ZF-sets (unless, obviously, we made some mistake - that is why this is just a preprint). We suppose psi can be well-ordered. That means it is possible to fix (to define) an injection from psi into some ordinal o. And that is supposed to be recursively definable in a similar way how Zermelo's theorem is proven. Zermelo's theorem takes into account AC to prove every set can be well-ordered. Instead of AC we have our F-Choice axiom. By using recursively F-Choice, every time we define an injection from a proper restriction of psi into some ordinal, in the next step of such a transfinite recursion we "loose information" about some terms which belong to psi that were previously fixed in the previous step. Some terms previously fixed are no longer fixed, and we simply do not know who are those terms. That happens because F-Choice axiom demands the existence of an injection g whose domain is a restriction of a given surjection f but such that g itself is no restriction of f. That is why the concept of lurking functions is quite important within our proposal. In Zermelo's theorem that is not an issue, since AC is used. All terms which were previously assigned to an ordinal by means of AC, keep their assignment during all next steps of transfinite recursion. Our motivation is our focus on functions rather than sets. We hope our F-Choice axiom is strong enough to grant PP but too weak to grant AC.<|endoftext|> -TITLE: Which abelian groups are $\varprojlim^1$ groups? -QUESTION [6 upvotes]: Question 1: Let $\mathcal A$ be an abelian group. Does there exist an inverse system $(A^n)_{n \in \mathbb N} = (\cdots \to A^n \to A^{n-1} \to \cdots \to A^0)$ such that $\varprojlim^1 A^\bullet \cong \mathcal A$? If not, can we characterize the abelian groups which are $\varprojlim^1$ groups or at least say anything interesting about their isomorphism types? -The remaining questions are meant to be refinements of Question 1. -Question 2: Let $\mathcal B^0,\mathcal B^1$ be abelian groups. Does there exist an inverse system $(B^n)_{n \in \mathbb N}$ such that $\varprojlim^i B^\bullet \cong \mathcal B^i$ for $i=0,1$? -If $(C^n)_{n \in \mathbb N}$ is an inverse system, there is a canonical two-term chain complex which we'll call $\mathbf{Lim} (C^\bullet) = (\prod_{n \in \mathbb N} C^n \to \prod_{n \in \mathbb N} C^n)$, where the differential is $(c^0,c^1,\dots) \mapsto (c^0 - \gamma c^1,c^1-\gamma c^2,\dots)$ where $\gamma$ ambiguously denotes any of the linking maps for the inverse system $C^\bullet$. The point, of course, is that $H^i(\mathbf{Lim} (C^\bullet)) = \varprojlim^i(C^\bullet)$ for $i=0,1$. -Question 3: Let $\mathcal C^\ast = (\mathcal C^0 \to \mathcal C^1)$ be a two-term chain complex of abelian groups. Does there exist an inverse system $(C^n)_{n \in \mathbb N}$ of abelian groups such that $\mathbf{Lim}(C^\bullet)$ is quasi-isomorphic to $\mathcal C^\ast$? -If $(D^{n,\ast})_{n \in \mathbb N}$ is an inverse system of chain complexes of abelian groups, then define $\mathbf{Lim}(D^{\bullet,\ast})$ by by applying $\mathbf{Lim}$ levelwise to obtain a double complex, and then taking the diagonal. -Question 4: Let $\mathcal D^\ast$ be a chain complex of abelian groups. Does there exist an inverse system $(D^{n,\ast})_{n \in \mathbb N}$ of chain complexes of abelian groups such that $\mathbf{Lim}(D^{\bullet,\ast})$ is quasi-isomorphic to $\mathcal D^\ast$? - -REPLY [14 votes]: Abelian group $A$ is cotorsion if $\rm{Ext}(F, A) = 0$ for every flat $F$, or, equivalently, $\rm{Ext}(\Bbb Q, A) = 0$ -Every $\varprojlim^1$ of an inverse system of abelian group is cotorsion, and, conversely, every cotorsion group is a $\varprojlim^1$. Proof can be found in Warfield, Huber. On the values of the functor $\varprojlim^1$. -If every group in the inverse system $G_i$ is f. g., then $\varprojlim^1$ is $\rm{Ext}(A, \Bbb Z)$ for $A$ torsion free countable: take $A$ equal to direct limit of $\rm{Hom}(G_i, \Bbb Z)$. In particular, that limit is always a divisible group. -Converse is also true, but version of proof I know is messy and uses structure teorem of alg. compact groups. Probably you can find something on that in Fuchs or Jensen books.<|endoftext|> -TITLE: Dimension of classifying space of a group -QUESTION [7 upvotes]: If $N$ is a normal subgroup of a group $G$ such that $G/N= \mathbb{Z}$. Suppose that the classifying space of $G$ is a finite CW-complex of dimension $n$. Does it follow that the classifying space of $N$ is a finite CW-complex of dimension $n-1$ ? - -REPLY [4 votes]: No, not even in the case when $G$ has a 1-dimensional classifying space: if $G$ is free of rank at least two, then for any $N$ with $G/N\cong \mathbb{Z}$, $N$ will be free of infinite rank and so $N$ will need a 1-dimensional classifying space too. -What you get easily is that if $G$ has cohomological dimension $n$ (modulo Eilenberg-Ganea issues for $n=2$ this is the same as the minimal dimension of a classifying space for $G$) then $N$ with $G/N\cong \mathbb{Z}$ cannot have a classifying space whose dimension is lower than $n-1$. Also easily $N$ does have a classifying space of dimension $n$ provided that $G$ does. But both of these possibilities can and do occur. -Concerning the question of whether $N$ can have a finite classifying space, the Euler characteristic gives a simple obstruction. If the Euler characteristic of the classifying space for $G$, $\chi(G)$ is non-zero, then $N$ cannot have a finite classifying space, because Euler characteristics multiply for group extensions and $\chi(\mathbb{Z})=0$, -giving $\chi(G)=\chi(N)\chi(\mathbb{Z})=0$ whenever $\chi(N)$ is well-defined.<|endoftext|> -TITLE: Plateau's Problem from an annulus -QUESTION [6 upvotes]: Let $(M^n,g)$ be a complete Riemannian manifold with bounded geometry, that is, it has bounded curvature and positive injectivity radius. Given two disjoint smoothly embedded homotopically trivial closed curves $\gamma_1$ and $\gamma_2$, we consider the problem of minimizing annulus $\Sigma$ with $\partial \Sigma=\gamma_2\cup \gamma_2$. -More precisely, we consider all maps $u\in W^{1,2}(A,M)\cap C^0(\bar A,M)$, where $A:=\{z\in \mathbb C \,\mid 1 < |z| <2\}$, such that $u$ restricted to $\partial A$ are reparametrizations of $\gamma_1$ and $\gamma_2$. Here, we assume at least one such map $u$ exists. -Can we prove the existence of such $u$ whose image has the least area? Is there any reference for such existence? - -REPLY [10 votes]: Such an annulus need not exist. For example, consider two circles in $\mathbb{R}^3$ defined by $x^2+y^2 = 1$ and $z = \pm R$. If $R$ is sufficiently large, then there cannot be a minimizing annulus (or, indeed, any minimizing connected surface) with these two circles as boundary. -The reason is the following: First, one can certainly find an annulus with these two surfaces as boundary whose area is as close to $2\pi$ as desired: Just take the two flat disks with these circles as boundary and join them by a very thin tube $x^2+y^2 = \epsilon^2>0$ and $|z|\le R$ for $\epsilon$ very small and smooth the result to get an annulus. -Meanwhile, if there were a connected minimal surface $A$ with these circles as boundary, then $A$ would have to pass through the plane $z=0$ at some point $p = (x_0,y_0,0)$, and hence the ball $B$ of radius $r2$, then no connected minimal surface with these two circles as boundary can achieve the lower bound of $2\pi$.<|endoftext|> -TITLE: Spanning trees: the last darn $1/4$ -QUESTION [16 upvotes]: Let $\Gamma$ be a connected graph. By (Kleitman-West, 1991), -if every vertex of $\Gamma$ has degree $\geq 3$, then $\Gamma$ has a spanning -tree with $\geq n/4+2$ leaves, where $n$ is the number of vertices of $\Gamma$. -It is relatively forward (though not completely trivial) -to deduce that, if every vertex of $\Gamma$ has -degree $\geq 2$, then $\Gamma$ has a spanning -tree with $\geq n/4+2$ leaves, where $n$ is the number of vertices of -$\Gamma$ of degree $\geq 3$. -Question: can the assumption on the degree of all vertices be dropped -altogether? That is, is it true that every connected graph $\Gamma$ -with $n$ vertices of degree $\geq 3$ has a spanning tree with -$\geq n/4+2$ leaves? If not, can you give a counterexample? - -Note 1: -The one case in doubt is that where there is exactly one vertex of degree $1$. -All other cases follow from (Bankevich-Karpov, 2011), which gives the lower -bound $\geq m/4+3/2$, where $m$ is the number of vertices of $\Gamma$ of degree -not $2$. Alternatively, one may reduce the general problem to the case -where exactly one vertex has degree $1$ as follows: given two vertices -$v_1$, $v_2$ of degree $1$, we may identify them (not changing the number of -vertices -of degree $\geq 3$ thereby) and apply the bound we are proving, recursively -(since the number of vertices of degree $1$ has decreased); if the spanning -tree contains the new vertex $v$ as a leaf, it is valid as a spanning tree -of the original graph; if it contains $v$ as an internal vertex, we -separate $v$ again into $v_1$ and $v_2$ (thus increasing the number of leaves -by $2$), and find that we have two trees, covering all vertices of $\Gamma$; -there is some edge of $\Gamma$ connecting them, and we may add it at a cost -of at most $2$ leaves. -Note 2: It obviously follows from Bankevich-Karpov that, when there is exactly one vertex of degree $1$, the bound $\geq n/4+7/4$ holds. It then follows -from (Karpov, 2012) that a counterexample to $\geq n/4 + 2$ would need -to have no vertices of degree $>3$. - -REPLY [14 votes]: Consider connected $G$ with $n$ vertices of degree $\ge 3$ and exactly one vertex $v$ of degree 1. Take an extra copy $G'$ of $G$ with $v'$ being its vertex of degree 1. -Now identify $v$ and $v'$ to make a new graph $H$ which has $2n$ vertices of degree $\ge 3$ and no vertices of degree 1. The identified $v=v'$ has become a cut-vertex of degree 2. By the previous theorems, $H$ has a spanning tree with at least $2n/4+2$ leaves, and so at least $n/4+1$ leaves on one side of the cut. $v=v'$ isn't one of these leaves since it is a cut-vertex. Now take this spanning tree back to $G$ and $G'$. The side, say $G$, which had $n/4+1$ leaves in $H$ now has the extra leaf $v$, making $n/4+2$ leaves.<|endoftext|> -TITLE: What determines the maximal dimension of the irreps of a (finite) group? -QUESTION [28 upvotes]: I am chemist and ask for apologies for all my mathematical inabilities when asking this question in advance, but after quite a bit of searching I found that this problem could be "open" or at least hard enough to find addressed in the literature and also advanced enough that it's possibly suitable to be asked here. -I work over a subset of the finite groups called point groups these are all the ((essentially) finite) subgroups of $O(3)$. The "degeneracies" arising in those groups are of importance in Chemistry so I started to investigate them. With "degeneracy" the dimensions of the irreps over $\mathbb R$, (this is of essential importance) occurring for the group under consideration is meant. Whenever a group is represented by an irrep of dimension $n>1$ we speak about $n$-fold degeneracy. -The term degeneracy in this context relates to the fact that the quantum mechanical eigenstates of (symmetric) systems/molecules form such $n$-dimensional sub spaces of the Hilbert space. Since the Hamilton operator is self-adjunct it makes sense to regard representations over $\mathbb R$ instead of over the usually more elegant $\mathbb C$. -My first question is, under which conditions in terms of group elements degeneracy can occur in a group? And the second question is how is the maximal dimension $n_{\max}$ of real irreps, over all irreps of the group, determined by the structure of the group? -My primary observation is that groups that contain exactly one generator $y$ of order $m=3$, like a group $\langle x,y,...| x^2=y^m = 1 = ...\rangle $ have $n_{\max}=2$ -Then there are just a very few point groups with $n_{\max} > 2$. We call them "high symmetry groups". Basically its the symmetry groups of the tetrahedron, the octahedron and the icosahedron (with 2 or 3 certain subgroups of theirs), where the former two have $n_{\max}=3$, the latter $n_{\max}=5$. -These high symmetry groups all have two generators, the tetrahedral groups one of order $m=3$, the octahedral groups one of order $m=4$ and the icosahedral groups one of order $5$. -So I would assume that there is a connection between the order of the generators and the degeneracies that can occur in a group. What it is exactly remains very obscure to me. So I would be very grateful about any hints also to the literature. - -Edit -Since the audience is so fantastically knowledgeable, I can't resist to make a small comment on the motivation of my research in the hope it might ring some bell and give rise to more inspiring comments: -The motivation of my question is, that you can see certain interesting physical properties in the states (of physical systems) if they are degenerate. "States" are some manifestations of irreps where we have direct numerical access to and a good intuition about their visual representations. It happens that there are certain, it seems, deep connections between the angular momentum operator, which is essentially an infinitesimal rotation in the (physical space) $\mathbb R^3$, and the occurrence of degeneracy (at least if it's 2-fold). At the same time we see that these states which are instances of degenerate representations are transformed into each others by (finite) rotations. Such that the question was arising if all such degeneracies are related to rotations, or if there is at least something in the structure of rotations that is general in $d>1$ dimensional representations. -There was recently a result that suggested that there is a "hidden" anti-unitary symmetry (state transforming operator) at the basis of any $2$-fold degeneracy of the form -$$ \mathcal{O} = i \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \mathcal{K}$$ (with $\mathcal{K}$ as complex conjugation operator) that transforms between the two degenerate irreps. Such an operator can be easily constructed from the angular momentum operator, for example, but not only, constructions of operators like {\mathcal{O}} based on spin or time reversal and others are possible if one considers more general "parametrizations" of physical systems then only $\mathbb R^3$. Anyway, as far as symmetries of the real space $\mathbb R^3$ are concerned it seems that rotation (and infintesimal rotation) are crucial and I aim to understand what "crucial" here exactly means. -What I find in particular intriguing in the light of the answer from @QiaochuYuan is the connection with "non-Abelianess", because the defining relation of $J$, the angular momentum or also spin operator is the so called "angular momentum algebra", that is a commutator relation between its components -$$ [J_i,J_j] = i \varepsilon_{ijk} J_k$$ -(with the Levi-Civita symbol $\varepsilon$). So this seems to suggest that angular momentum is an essential source of non-commutativity somehow. I like then to understand what types of sources else there might be, for this non-commutativity in quantum mechanical systems, especially if only representations of states in $\mathbf{R}^3$ are considered. One thing which complicates the question is that in most systems rotational symmetry "is broken" but degeneracy can nevertheless occur. Then my question would be, what is the explicit form of the $\mathcal{O}$ operator and if it can be continuously related to angular momentum if one views the "symmetry breaking" as a continuous process. -(Sorry for the post-answer edit, I hope it complies with the MO rules!) - -REPLY [21 votes]: A simple bound on the largest dimension of a complex irreducible representation (which is either equal to or half of the largest dimension of a real irreducible representation) is the following: we know that - -$|G| = \sum d_i^2$ where $d_i$ are the dimensions of the irreducibles, -the number of (complex) irreducibles is the number $c(G)$ of conjugacy classes, and -the size $a(G) = |G^{ab}|$ of the abelianization is the number of $1$-dimensional irreducibles (so the number of $d_i$ terms equal to $1$). - -It follows that the largest dimension $d_{max}$ satisfies $a(G) + d_{max}^2 \le |G| \le a(G) + (c(G) - a(G)) d_{max}^2$, and rearranging these gives -$$\sqrt{ \frac{|G| - a(G)}{c(G) - a(G)} } \le d_{max} \le \sqrt{|G| - a(G)}.$$ -$c(G)$ is a measure of "how abelian" $G$ is; it's a nice exercise to show that $\frac{c(G)}{|G|}$ is the probability that two random elements of $G$ commute. Roughly speaking this means that $d_{max}$ is a measure of "how nonabelian" $G$ is. For example, if $G = A_5$ is the icosahedral group then $|G| = 60, a(G) = 1, c(G) = 5$ gives -$$ \sqrt{ \frac{59}{4} } \approx 3.84 \dots \le d_{max} \le \sqrt{59} \approx 7.68 \dots $$ -so $4 \le d_{max} \le 7$, and since we also know that the dimensions $d_i$ divide $|G|$ we have $4 \le d_{max} \le 6$, and the true value $d_{max} = 5$ is right in the middle. Loosely speaking this says that $A_5$ is "more nonabelian" than, say, a dihedral group, which satisfies $d_{max} = 2$. -This bound is most useful when the abelianization is large. A different bound useful when the center $Z$ is large is the following: we know that - -by Schur's lemma every irreducible representation has a central character, and if $\lambda : Z \to \mathbb{C}^{\times}$ is a central character, then the irreducibles with central character $\lambda$ can be identified with simple modules over the twisted group algebra obtained by quotienting $\mathbb{C}[G]$ by the relations $z = \lambda(z)$ for $z \in Z(G)$, -every twisted group algebra as above has dimension $|G/Z|$, so the dimensions $d_i(\lambda)$ of the irreducibles with central character $\lambda$ satisfy $|G/Z| = \sum d_i(\lambda)^2$, -the number of irreducibles with a fixed central character is the number of conjugacy classes of $G/Z$ satisfying a certain condition, and in particular is at most the number of conjugacy classes of $G/Z$. - -Now it follows that the largest dimension $d_{max}$ satisfies $d_{max}^2 \le |G/Z| \le c(G/Z) d_{max}^2$, which gives -$$\sqrt{ \frac{|G/Z|}{c(G/Z)} } \le d_{max} \le \sqrt{|G/Z|}.$$ -For example, the upper bound is tight for a finite Heisenberg group $H_3(\mathbb{F}_p)$, which satisfies $|G/Z| = p^2$ and has $p^2$ one-dimensional characters and $p - 1$ irreducibles of dimension $p$. The lower bound actually produces $1$ here which shows that it can be worse than the previous lower bound (which applied here gives $\sqrt{ \frac{p^3 - p^2}{p^2 + p - 1} } \approx \sqrt{p}$). The size of the center is another measure of "how abelian" $G$ is so this gives another sense in which $d_{max}$ measures "how nonabelian" $G$ is.<|endoftext|> -TITLE: Wiki for consequences of axiom of choice? -QUESTION [31 upvotes]: I raised the following question as part of another MO question, but I am following the suggestion of Nate Eldredge to make it a question in its own right. -For many years, there has a been a valuable web resource, hosted by Purdue, on the -Consequences of the Axiom of Choice. Unfortunately, the page is no longer functioning, as you will quickly discover if you try submitting a form number. The URLs have changed. I suspect that Purdue redesigned its website at some point, changing the URLs, and that since Herman Rubin died a couple of years ago, there is now nobody responsible for maintaining the Axiom of Choice page. I tried emailing a couple of random people in the Purdue mathematics department to find out if something could be done to revive the page, but have received no response. -I am wondering if there is a way to revive this resource, ideally in a way that will prevent it from suffering a similar extinction risk a few years down the line. Perhaps some people can turn the page into a wiki, much in the way the OEIS evolved from a personal project of Neil Sloane's into a wiki? Also, maybe someone reading this knows more than I do about the situation at Purdue and can comment on what would be involved in making the data publicly available again. - -REPLY [7 votes]: Paul Howard is a colleague of mine here at EMU. Several years ago I asked him if I could try to put his book into an online format (with a database underneath and Ruby on Rails on top). -My progress has stalled on the last 2 years, but you can see what we were able to do at http://104.237.130.142/consequences. - -In response to comments below: -The links at the top of the page correspond to different aspects of Paul's book. -In order, the - -articles, authors, journals, books, and excerpts links give information about every one of them listed in the book. if you click on the authors link, for example, you'll see every author listed in the book. click on an author, you'll see information about papers by that author that were cited in the book -forms link: goes to a page that lists every equivalence class of forms of the axiom of choice. if you click on a specific form, you'll see basic information about that equivalence class. -models link: goes to a page that has information about each of the models of set theory; click on one of the models, and you'll see, for example, a list of all the forms of the axiom of choice that are known to be true in that model, (each with its own link), a list of all the forms of the axiom of choice that are known to be false in that model), along with a lot more information. -notes link: Paul and Jean included a lot of little independent proofs of things that they did not find in the literature. This link lists all of the notes, and clicking on any note takes you to the statement and proof -implications link: Paul and Jean classified the different implications by 6 different types; this page lists all of the implications by type. Each type of implications has its own page; going to that page, and click on an implication, and you'll see information about the forms involved and how that implication was derived -tools: this was something that Paul had on his site that I reproduced here. You can put in versions of the axiom of choice by number and make little implication arrays, etc. - -I hope this helps; I apologize for the length.<|endoftext|> -TITLE: Are any of these complex surfaces ever projective? -QUESTION [13 upvotes]: Let $C$ and $T$ be compact connected Riemann surfaces (or: smooth projective connected curves over $\mathbb{C}$) of genus at least two and let $X:=C\times T$. Let $(c,t)$ be a point of $X$, and let $X'\to X$ be the blow-up of $X$ in $(c,t)$. By Grauert's contraction theorem, we may contract the strict transform of $\{c\}\times T$ on $X'$ and obtain a normal complex-analytic surface $X'\to S$. - - -Under what conditions (if any) is $S$ projective? - - -Note that $S$ contains a unique rational curve (given by the image of the exceptional curve $E$ of $X'\to X$), and that $S$ has a unique singular point $\sigma$ in $S$. -My interest in this surface is related to Lang's conjectures, and I first learned about this surface from Frederic Campana. Indeed, the surface $S$ has the peculiar property that, for any point $s$ which does not lie on the rational curve and any pointed curve $(D,d)$, the set of pointed maps $(D,d)\to (S,s)$ is finite. However, for the pointed curve $(C,c)$ and the singular point $\sigma$, the space of pointed maps $(C,c)\to (S,\sigma)$ covers $S$. -I was not able to prove projectivity of $S$, not even whilst assuming it is proper (so that one might appeal to https://arxiv.org/abs/1112.0975 ) - -REPLY [8 votes]: Here is a simple method for constructing projective examples: -Assume there exist maps $f:C \to \mathbb{P}^1$ and $g:T \to \mathbb{P}^1$ of the same degree which are totally ramified at $c$ and $t$. Let $X = C \times T$, $Y = \mathbb{P}^1 \times \mathbb{P}^1$, and consider the map $p:=(f,g): X \to Y$. Let $X'$ be the blowup of $X$ at $(c,t)$ and $Y'$ the blowup of $Y$ at $(f(c), g(t))$. An easy local computation shows that $p$ induces a morphism $p': X' \to Y'$. -Now let $Y_1$ be the blow down of the strict transform of $\{f(c)\} \times \mathbb{P}^1 $ in $Y'$. The surface $Y_1$ is projective, so its normalisation $X_1$ in the function field of $X'$ is also projective. By using that the map $X' \to Y_1$ factors through $X_1$, it follows easily that $X_1$ is equal to the surface $S$, so $S$ is projective.<|endoftext|> -TITLE: Is there an infinitary sentence which is absolutely not second-order expressible? -QUESTION [13 upvotes]: This is a "forcing-absolute" followup to this question, whose answer was largely unsatisfying. The question is: - -Suppose $V=L$. Is there an $\mathcal{L}_{\infty,\omega}$-sentence $\varphi$ such that in no forcing extension is $\varphi$ equivalent to a second-order sentence? - -Throughout, by "forcing" I mean "set forcing," although the (tame) class forcing version also seems potentially interesting. -EDIT: I forgot to add that I'm restricting attention to infinite structures here (which is key to my comment below that every projectively-definable infinitary sentence is second-order expressible). As Fedor Pakhomov commented below, without this restriction the problem is trivial since second-order theories of finite structures can't be changed by forcing. I do not, however, want to restrict attention to countable structures. - -I've decided to focus on models of $\mathsf{V=L}$ since that hypothesis seems to add interesting flavor to the question in a few ways: - -It implies that such a $\varphi$ must not be in $\mathcal{L}_{\omega_1,\omega}$. This is because $(1)$ every constructible real is projective in some forcing extension and $(2)$ every projectively definable $\mathcal{L}_{\omega_1,\omega}$-sentence is equivalent to a second-order sentence. So a positive answer would have to crucially rely on uncountable Boolean combinations - which seems a bit odd, because each specific $\mathcal{L}_{\infty,\omega}$-sentence is equivalent to some $(\mathcal{L}_{\omega_1,\omega})^{V[G]}$-sentence in an appropriate forcing extension $V[G]$ (just collapse the size of the sentence), but isn't an obvious contradiction since the "potential projectivity" fact about $L$ doesn't seem to lift to arbitrary forcing extensions of $L$. - -It rules out the "silly" solution provided by large cardinals. If large cardinals exist - specifically, enough to guarantee projective absoluteness - then any infinitary sentence which is not equivalent to a second-order sentence in $V$ remains so in all forcing extensions. (Note that this would give us an example in $\mathcal{L}_{\omega_1,\omega}$ for that matter.) But $\mathsf{V=L}$ breaks this "hammer," so that we seem to be forced to do some actual work. - -If the answer is yes, there is in fact a definable example, namely the least such sentence with respect to the $L$-ordering. Of course this is silly, but it suggests that there might be canonical examples in a more interesting sense. By contrast I could imagine models of $\mathsf{ZFC+V\not=L}$ where the existence of such a sentence is guaranteed nonconstructively (e.g. by a more intricate counting argument), and so no canonical example need exist. - - -That said, since it seems plausible that the $\mathsf{V=L}$-situation is more difficult to attack than I'd hoped, I'm also interested in results for other extensions of $\mathsf{ZFC}$. - -REPLY [5 votes]: For any given finite signature $\Omega$ there is a second-order sentence $\varphi$ of the signature $\Omega$ such that $\mathsf{ZFC}+V=L$ proves that for any $\mathcal{L}_{\infty,\omega}$-formula $\psi$ of the signature $\Omega$ there is a poset $P$ for which it is $\Vdash_P$-forced that for any infinite model $\mathfrak{M}$ we have $\mathfrak{M}\models \varphi$ iff $\mathfrak{M}\models \psi$. -$\varphi$ should be a second-order sentence expressing in all infinite models $\mathfrak{M}$ that the following holds (see below for a more explicit construction of $\varphi$): - -there exists the greatest ordinal $\alpha$ such that the value $\aleph_\alpha^{L_\beta}$ is the same for all large enough countable limit ordinals $\beta$; -$\alpha$ is the position in $<_L$ of some $\mathcal{L}_{\omega_1,\omega}$-sentence $\psi$ -$\psi$ is true in $\mathfrak{M}$. - -Given $\mathcal{L}_{\infty,\omega}$-formula $\psi$ we consider $\alpha_0$ that is the position of $\psi$ in $<_L$. Let $P$ be the Levy collapse of $\aleph_{\alpha_0}$ onto $\omega$. Let us check that for this $\varphi,\psi$, and $P$ we have the desired equivalence. -Indeed let us reason inside $\Vdash_P$. Clearly, $\psi$ became an $\mathcal{L}_{\omega_1,\omega}$-formula. -Observe that $\alpha$ from 1. will always be equal to $\alpha_0$. Indeed, let us choose countable limit $\beta_0>\alpha_0$ such that all ordinals $\gamma<\alpha_0$ that aren't $L$-cardinals aren't cardinals in $L_{\beta_0}$. Clearly, for any limit $\beta>\beta_0$, we have $\aleph_{\alpha}^{L_\beta}=\aleph_\alpha^L$. For any $\alpha'>\alpha_0$ since $\aleph_{\alpha'}^L$ isn't countable and any countable $\gamma$ we could find limit countable $\beta'$ such that $\aleph_{\alpha'}^{L_{\beta'}}$ is either undefined or has the value $>\gamma$. Hence $\alpha_0$ is the ordinal $\alpha$ from 1. -Using this the proof of semantical equivalence of $\varphi$ and $\psi$ in infinite models is trivial. - -Now let me give more detailed description of $\varphi$. -Let $\mathsf{KPUL}_2(\in,\in_u)$ be a second-order formula depending on element variable $\alpha$ and binary predicate symbols $\in,\in_u$ asserting that we have $(\mathsf{KPU}-\mathsf{Foundation})+\textsf{Second-Order-Foundation}+L[\mathfrak{M}]=V$ for the structure where each element of the underlying domain simultaniously represent itself (as an urelement) and some set, $\in$ gives membership of sets in sets, $\in_u$ gives membership of urelements in sets. Formula $\mathsf{Emb}(R,\in,\in_u,\in',\in_u')$ expresses that $\mathsf{KPUL}_2(\in,\in_u)\land \mathsf{KPUL}_2(\in',\in_u')$ and the unary function $f$ gives an end embedding of the admissible set given by $(\in,\in_u)$ into the admissible set $(\in',\in_u')$. Formula $\mathsf{St}(\alpha,\beta,\kappa,\in,\in_u)$ expresses that $\mathsf{KPUL}_2(\in,\in_u)$ and in the corresponding admissible set $\alpha$ is a countable ordinal, $\beta$ is a countable limit ordinal, $\kappa$ is the ordinal that is $\alpha$-th cardinal according to $L_\beta$, and for any $f,\in',\in_u',\beta'$ if $\mathsf{Emb}(f,\in,\in_u,\in',\in_u')$ and $\beta'>f(\beta)$ is a countable limit ordinal according to the admissible set given by $(\in',\in_u')$, then there is $f(\alpha)$-th cardinal according to $L_{\beta'}$ and it is equal to $f(\kappa)$. Formula $\mathsf{MSt}(\alpha,\in,\in_u)$ expresses that $\mathsf{St}(\alpha,\in,\in_u)$ and we don't have $\mathsf{St}(\alpha+1,\in,\in_u)$. Formula $\mathsf{LC}(\alpha,\psi,\in,\in_u)$ expresses that in the admissible set given by $(\in,\in_u)$, $\alpha$ is an ordinal, $\psi$ is a constructible $\mathcal{L}_{\omega_1,\omega}$-sentence of the signature $\Omega$, and its place in $<_L$-order is $\alpha$. Formula $\mathsf{Tr}(\psi,\in,\in_u)$ expresses that in the admissible set given by $(\in,\in_u)$, $\psi$ is an $\mathcal{L}_{\omega_1,\omega}$-sentence of the signature $\Omega$ that is true in the underlying model $\mathfrak{M}$. -We put $\varphi$ to be: -$$\exists \in,\in_u,\alpha,\psi(\mathsf{KPUL}_2(\in,\in')\land \mathsf{MSt}(\alpha,\in,\in')\land \mathsf{LC}(\alpha,\psi,\in,\in')\land \mathsf{Tr}(\psi,\in,\in')).$$<|endoftext|> -TITLE: Elliptic factors in the Jacobian and zeta function -QUESTION [7 upvotes]: Consider a hyperelliptic curve $\mathcal{C}$ over $\mathbb{Q}$ and its Jacobian $J(\mathcal{C})$. Assume that $J(\mathcal{C})$ admits an elliptic factor $\mathcal{E}$. -For almost all primes, we can reduce $\mathcal{C}$ modulo $p$, and consider its zeta function $Z_p$. Can the elliptic factor property be seen as a special property of $Z_p$? -Conversly, if such property is satisfied for almost all primes, can we prove that a hyperelliptic curve possess an elliptic factor in its Jacobian? - -REPLY [5 votes]: This is an incomplete answer to the second question (Jackson has indicated the answer to the first question in the comments). -If you ask the same question over a general number field $F$, then you will find counterexamples by considering "false elliptic curves" i.e. principally polarized abelian surfaces with quaternionic multiplication. These are automatically Jacobians of hyperelliptic curves of genus two. They have the property that for almost all primes their reduction is a product of elliptic curves, but over $F$ (and even over its algebraic closure) the Jacobian may very well be simple (hence no elliptic factors). -Over $\mathbb{Q}$, you might look at abelian surfaces with "potential" quaternion multiplication, but actually I would guess that these don't give counterexamples-- just a guess. In genus greater than 2, it's even less clear to me.<|endoftext|> -TITLE: First Chern class and field extensions -QUESTION [6 upvotes]: Let $X$ be a smooth, complex projective algebraic variety defined over a number field $K$. -Let $D$ be a divisor of $X$ defined over $K$ with the following property: - -For any curve $C$ defined over $K$, we have $\operatorname{deg (D_{|C})=0}$ - -Is it then true that $c_1(D)=0$? -In general, in order to have $c_1(D)=0$, I should check that $\operatorname{deg (D_{|C})=0}$ -for any curve (not just the ones defined over $K$). I'm asking if in this particular setting, the curves defined over $K$ are enough. - -REPLY [5 votes]: Every curve on $X$ is algebraically equivalent to a curve defined over a finite extension of $K$, and then a union of Galois conjugates will be defined over $K$. So, if you allow reducible curves, then the answer is yes. -Added: The intersection product is Galois invariant. -For a nonperfect field $k$ and a divisor $D$ defined over a purely inseparable extension of $k$ of degree $p^m$, the divisor $p^m D$ is defined over $k$. -Regard $D$ as the -Cartier divisor defined by a family of pairs $(f_{i},U_{i}^{\prime})$, -$f_{i}\in k^{\prime}(X)$, and let $U_{i}$ be the image of $U_{i}^{\prime}$ in -$X$; then $k^{\prime}(X)^{p^{m}}\subset k(X)$, and so the pairs $(f_{i}% -^{p^{m}},U_{i})$ define a divisor on $X$ whose inverse image on $X_{k^{\prime}}$ is $p^{m}D$.<|endoftext|> -TITLE: Pants decomposition and moduli space of $\Sigma_g$ for $g>1$ -QUESTION [6 upvotes]: By Section 8.3.1 of the book: A primer on mapping class groups by Farb and Margalit, a pair of pants is a compact surface of genus 0 with three boundary components. Let $S$ be a compact surface with $\chi(S)<0$. A pair of pants decomposition of $S$, or a pants decomposition of $S$, is a collection of disjoint simple closed curves in $S$ with the property that when we cut $S$ along these curves, we obtain a disjoint union of pairs of pants. Equivalently, a pants decomposition of $S$ is a maximal collection of disjoint, essential simple closed curves in $S$ with the property that no two of these curves are isotopic. In particular, a pants decomposition of $\Sigma_g$ (the orientable closed surface of genus $g$) for $g>1$ has $3g-3$ curves, cutting $\Sigma_g$ into $2g-2$ pairs of pants. -On the other hand, according to Riemann's computation, the dimension of the moduli space $\mathcal{M}_g$ of algebraic curves of genus $g>1$ is $3g-3$. See, for example, the book: Algebraic curves towards moduli spaces. -My question: why are the number of curves in the pants decomposition of $\Sigma_g$ for $g>1$ and the dimension of the moduli space of $\Sigma_g$ for $g>1$ the same number $3g-3$? Is there any deep relation? -Thank you! - -REPLY [2 votes]: I am unsure if this is what you are looking for, but the moduli space $\mathcal{M}_g$ can be viewed as an orbifold quotient of the Teichmüller space by the mapping class group. I cannot offer more details, but this is explained in a paper by John H. Hubbard and Sarah Koch. -EDIT -I notice that this has been addressed in another question as well.<|endoftext|> -TITLE: Formal definition of homotopy type theory -QUESTION [15 upvotes]: The HoTT community is quite friendly, and produces many motivational introductions to HoTT. The blog and the HoTT book are quite helpful. However, I want to get my hands directly onto that, and am looking for a formal treatment of HoTT. Therefore this question: what's the formal definitions of the gadgets in homotopy type theory (HoTT)? -I expect an answer with absolutely no motivations and no explanations. It should define all terminologies, such as type, term, :, for all, exist, identity, equivalence.. etc. It should also clarify which words are undefined. I feel like I'm almost asking for a computer program.. so if that's your answer, I'm happy to study it. - -REPLY [18 votes]: Andrej’s answer gives an excellent list of resources. But I think it’s also worth saying a little upfront to ward off common misunderstandings. HoTT is a foundational system — analogous to ZFC set theory — and the way in which the basic vocabulary of foundational systems is “defined” is a bit different from how most mathematical objects are defined. And for newcomers to type theory, especially experienced mathematicians, this is a frequent source of misaligned expectations. -Whenever you’re setting up some logical formal system, you are typically “defining” things in several different senses: - -you’re defining the formal system itself — its syntax, its rules of inference, its axioms… - -you’re introducing terminology to talk about the basic notions of the formal system - -as you work within the formal system, you’re then defining further objects within it. - - -With ZFC, for instance: - -This is the sense in which you define what “first-order logic” and “ZFC” formally are, for studying them from the outside as a logician. - -This is the sense in which you define set, implies, for all, and so on. You don’t define “a set is a such-and-such satisfying the so-condition”. We just introduce set as a word for the primitive objects that the ZFC axioms axiomatise. Similarly, implies is just a certain primitive symbol of our language, and first-order logic gives us rules for reasoning with it. - -This is the sense in which you define power sets, the rank of sets, and generally, definitions made while doing mathematics within ZFC. - - -So (1) and (3) are definitions in the usual mathematical sense, but happening at two different levels — external to the formal system under consideration, or internal to it. Meanwhile, (2) is a bit more different: you are “defining” things by taking them as primitive notions and axiomatising them. -I say all this because most of the things you mention in your question are “defined” in HoTT in sense (2) — so you will not find definitions of them in the sense you may expect! They are “defined” in the same way that sets are defined by ZFC. Following through the senses above, for HoTT: - -This is the sense in which you define what “HoTT” is, as a formal system from the outside. (HoTT and other type theories don’t have the same kind of separation between the “logical language” and the “theory in it” that you have with e.g. FOL+ZFC.) - -This is the sense in which you define type, term, for all, and so on: They’re the words we use for the various primitive objects and symbols of the formal system; and the formal system gives us rules for reasoning with them. - -This is the sense in which you define contractible, eqivalence, univalent, and so on, when developing mathematics within type theory.<|endoftext|> -TITLE: Set of quadratic forms that represents all primes -QUESTION [6 upvotes]: A SPECIFIC CASE: -Any prime number can be classified as either $p \equiv 1 \pmod 3$ or $p \equiv 2 \pmod 3$. -If $p = 3$ or $p = 1 \pmod 3$, then the prime $p$ can be represented by the quadratic form $ x^2 + 3y^2, x,y \in \mathbb Z.$ -But what if $p \equiv 2 \pmod 3$? -Is there a quadratic form $ax^2+bxy+cy^2$ such that $p= ax^2+bxy+cy^2, $ when $p \equiv 2 \pmod 3$ where $x,y, a, b,c \in \mathbb Z$? -GENERAL CASE: -The general question is, is there a set of quadratic forms which represent all prime numbers? -We will classify the prime numbers, say, by $m$. Any prime is defined by $p \equiv i \pmod m$ where $1 \leq i\leq m-1$. -In above example, $i \in \{1, 2\}, m=3$. Let, the set of quadratic forms is $A$, then the number of elements in $A$ is at-least $(m-1)$. -QUESTION: -For a given $m$ can we find a set $A$ such that any prime $p$ can be represented by one of the quadratic form of $A$ ? -If it is possible then how? If there is a condition on $m$, what is it? -Does the question has any relation to the following theorem ? - -One can answer only the specific case, if they wish to do so. -EDIT: -Is there a finite set of (preferably irreducible) binary quadratic forms such that every prime is represented by at least one of the forms in the set? - -REPLY [14 votes]: Every prime $p$ is represented by at least one of the following quadratic forms: $x^2+y^2$, $x^2+3y^2$, $3x^2-y^2$: - -if $p=2$ or $p\equiv 1\pmod{4}$, then $p$ is represented by $x^2+y^2$; -if $p=3$ or $p\equiv 1\pmod{3}$, then $p$ is represented by $x^2+3y^2$; -if $p\equiv 11\pmod{12}$, then $p$ is represented by $3x^2-y^2$. - -This follows from Lemma 2.5, Corollary 2.6, (page 26) in Cox: Primes of the form $x^2+ny^2$ coupled with the fact that $x^2+y^2$, $x^2+3y^2$, $3x^2-y^2$, $x^2-3y^2$ represent all integral binary quadratic forms of discriminant lying in $\{-4,\pm 12\}$. -Added. More generally, if an odd number of discriminants multiply to a square, then the quadratic forms of those discriminants together represent all primes coprime to those discriminants. In the example above, the discriminants were the elements of $\{-4,\pm 12\}$, and we could do without the form $x^2-3y^2$. See also this related post.<|endoftext|> -TITLE: Can an "almost injective'' function exist between compact connected metric spaces? -QUESTION [10 upvotes]: Let $\pi: X\to Y$ be a surjective continuous function between the compact, metric and connected spaces $X$, $Y$, and $Y_0 = \{y\in Y: \#\pi^{-1}(y)>1\}$. Suppose that: - -$Y_0$ is dense in $Y$, -$Y\setminus Y_0$ is a dense $G_\delta$ in $Y$, and -for some constant $N<\infty$, we have $\#\pi^{-1}(y) \leq N$ for all $y\in Y$. - -My question is: can a function $\pi$ like this exist? Maybe I need more hypothesis on the topology of $X$ and $Y$. -Observe that the connectedness is crucial: Sturmian codings of irrational rotations of the circle are surjective continuous functions $\pi:K\to S^1$ from the Cantor set (a totally disconnected space) and the circle such that $\#\pi^{-1}(y) = 2$ for $y$ in a countable dense subset $Y_0 \subseteq S^1$ and $\#\pi^{-1}(y) = 1$ for $y \in S^1\setminus Y_0$. -Condition (3) is also needed: if $f\colon[0,1]\to[0,1]$ is the Thomae's function, $X = \{(x,y)\in[0,1]^2 : 0\leq y\leq f(x)\}$ is the subgraph of $f$, $\pi: X\to[0,1]$ is the projection onto the first coordinate, and $Y_0 := \{y\in Y: \#\pi^{-1}(y)>1\} = [0,1]\cap\mathbb{Q}$, then $\pi$ is continuous, $X$ is connected and (1),(2) hold, but $\pi(y)$ is an uncountable set when $y \in Y_0$. -After building these examples I am more convinced than at the beginning that $\pi$ must be injective. -I would appreciate any comment. - -REPLY [6 votes]: A simple example is the Peano curve $f:[0,1]\to[0,1]^2$. -Specifically, here $Y_0=([0,1]\times T)\cup (T\times [0,1])$ where $T$ is the set of triadic rationals in $[0,1]$, a dense $F_\sigma$ with empty interior, so $Y\setminus Y_0=T^c\times T^c$ is a dense $G_\delta$. Check here for details. Note that here $N=4$ and, as per Anton Petrunin's remark in a comment to the linked answer, one could modify the construction to make $N=3$.<|endoftext|> -TITLE: Forcing notions adding minimal reals -QUESTION [12 upvotes]: I am looking for a comprehensive list of known forcing notions which add a minimal real into the ground model. I know some of them like the Sacks forcing, or the Judah-Shelah's example of a c.c.c. forcing which adds a minimal real (and a few more). - -Question. What are other known (or less known) examples of forcing notions which add a minimal generic real? - -REPLY [11 votes]: Splitting forcing is a little-known forcing that adds a splitting real and creates a minimal extension. It consists of splitting trees ordered by inclusion. A splitting tree is a perfect tree $T \subseteq 2^{<\omega}$ where for every $s \in T$ there is $m \in \omega$ so that for all $n \geq m$, $i \in 2$, there is $t \in T$, $s \subseteq t$ with $t(n)= i$. -In my recent preprint (arxiv link), I show that splitting forcing adds a minimal real (Corollary 4.20) and provide part of an argument for the minimality of the forcing extension (Corrolary 3.21). -Other than that, splitting forcing is $\omega^\omega$-bounding but as far as I know it is unknown whether it has the Sacks-property (see also Question 6.4 in On splitting trees by Laguzzi, Mildenberger and Stuber-Rousselle).<|endoftext|> -TITLE: Trees in chain complexes -QUESTION [7 upvotes]: $\DeclareMathOperator{\Ch}{\mathit{Ch}}$Let $\Ch_\mathbb{Q}$ denote the model category of chain complexes over rational numbers. Let $T_\ast$ be a tree in $\Ch_{\mathbb{Q}}$ with $n$ vertices. -How to classify trees with respect to weak equivalences i.e., chain homotopies? Is it true that the classification can be recovered from the $\mathit{ho}(\Ch_{\mathbb{Q}})$? -I think the key factor is that any chain complex $C_\ast\cong \oplus_n V\langle n\rangle_\ast$, here $V\langle n\rangle_\ast$ is the chain complex concentrated in degree $n$ and $V\langle n\rangle_n= H_n(C_\ast)$ -For example if we take a path with length 2, $f_\ast : C_\ast \to C_\ast'$ then it is equivalent to maps $H_n(f_\ast): H_n(C_\ast) \to H_n(C_\ast')$, that is maps between vector spaces. We know that any map between vector spaces is completely describe by the dim(Ker). In this case, any path of length 2 is fully describe by $\dim(\mathrm{ker}(H_n(f_\ast)))_n$. -I really appreciate it if someone could say something for trees. - -REPLY [6 votes]: Let $T$ be a well-founded poset and $k$ a field. Let $H: Ch_k \to Ho(Ch_k) = Gr_k$ be the homology functor and $\iota: Gr_k \to Ch_k$ be the canonical section which sends a graded vector space to the corresponding chain complex with zero differential. -Proposition 1: For any $F: T \to Ch_k$, $F$ and $\iota H(F)$ are quasi-isomorphic. -In particular, if $H(F) \cong H(G)$, then $F \simeq \iota H(F) \cong \iota H(G) \simeq G$. -Proof: In the Reedy = injective model structure on $Fun(T,Ch_k)$, fibrations and weak equivalences are levelwise, and an object is cofibrant (and thus bifibrant) if and only if each morphism of $T$ is carried to a monomorphism. We may assume that $F \in Fun(T,Ch_k)$ is cofibrant and prove the more specific claim that there is a quasi-isomorphism $F \to \iota H(F)$. By induction on the structure of $T$, we are reduced to the following: -Lemma 2: Let $X \to Y$ be a monomorphism in $Ch_k$. If $X \to \iota H(X)$ is a quasi-isomorphism, then there exists a quasi-isomorphism $Y \to \iota H(Y)$ forming a commutative square. -Proof: We may decompose $X \to Y$ into a series of cell attachments, and construct the map $Y \to \iota H(Y)$ by induction, one cell at a time. There are two cases. In the first case we have $Y = X \oplus k[n]$, in which case we have $H(Y) = H(X) \oplus k[n]$, and we extend using the identity on $k[n]$. In the second case we have $Y = X \cup_z w$ where $z$ is a cycle representing a nonzero homology class of $X$ and $w$ is a cell we are attaching to kill it. In this case we have $H(Y) = H(X)/v$, and we can extend via the zero map on $w$. - -Added in response to comment: It's worth noting that although the the proposition only applies directly to tree-shaped diagrams, we can leverage it to understand certain more complicated diagrams. For example: -Let $\mathcal C$ be the "marked category" -$$\require{AMScd} \begin{CD} a @>>> b\\ @VVV @VVV\\ 0 @>>> c \end{CD}$$ -with the understanding that $Fun(\mathcal C,Ch_k)$ means functors $\mathcal C \to Ch_k$ carrying $0$ to an acyclic chain complex. Then $Fun(\mathcal C,Ch_k)$ with levelwise weak equivalences models "maps $A \xrightarrow f B \xrightarrow g C$ equipped with a nullhomotopy of $gf$". -Lemma 3: If $\mathcal D$ is/presents a stable $\infty$-category, then the $\infty$-category (presented by) $Fun(\mathcal C, \mathcal D)$ is equivalent to the $\infty$-category (presented by) $Fun(T,\mathcal D)$ where $T$ is the tree $D \leftarrow F \to B$. -Proof: Consider the following diagram, where the rows and columns are fiber sequences: -$$\require{AMScd} \begin{CD} A @>>> F @>>> D \\ @V{=}VV @VVV @VVV \\ A @>>> B @>>> Q \\ @VVV @VVV @VVV \\ 0 @>>> C @>{=}>> C \end{CD}$$ -The diagram can be uniquely reconstructed from either the bottom left square, or the top right square subject to the requirement that it be an exact square, or from the part $D \leftarrow F \to B$ subject to no requirements. -Proposition 4: In $Fun(\mathcal C, Ch_k)$, a diagram is quasi-isomorphic to its homology. Consequently, if the homologies of two such diagrams are isomorphic, then the diagrams are quasi-isomorphic. -Proof: By Proposition 1, this is true for $Fun(T,Ch_k)$. Start with a model of the top right square such that $F \to D$, $F \to B$, and $B \oplus D \to Q$ are monomorphisms. By Proposition 1, we have a commuting square of quasi-isomorphisms from $F \to D \oplus B$ to the corresponding map of homologies. By Lemma 2, we may extend this to a quasi-isomorphism from $Q$ to its homology, in a natural way. We think of this as a natural transformation from the top-right square of the above grid to the corresponding grid of homologies, and we seek to fill out the remaining components of a natural transformation between grids. -Since $D \to Q$ is a monomorphism, we may use Lemma 2 again to extend this natural family of quasi-isomorphisms to $C$, and hence to the lower-right square. Modeling $A$ by the dual mapping cone of $F \to D$, we obtain a model of $A \to F$ which is epic. By the dual of Lemma 2, we may extend our natural family of quasi-isomorphisms to $A$ and hence to the upper-left square. To model the lower-right square, take an arbitrary factorization through an acyclic complex, which comes with a unique quasi-isomorphism to its homology. The remaining naturality squares all necessarily commute, and the naturality of the quasi-isomorphisms out of the lower-left square is precisely what we are looking for.<|endoftext|> -TITLE: Chain rule for the superderivative -QUESTION [6 upvotes]: A one dimensional complex supermanifold $X$ is locally described by an ordinary complex coordinate $z$ and an anticommuting coordinate $\theta$, $\theta^2 = 0$. -The superderivative is the square root of the derivative in the following sense: -Take the vector field $D_{\theta} = \partial_{\theta} + \theta\partial_z$ so that $\frac12[D_{\theta},D_{\theta}]= \partial_z$, where $[ \ , \ ]$ denotes the super-commutator. -It is stated (see for example pg. 4 Equation 2.9 in https://arxiv.org/pdf/1209.2459.pdf) that under a superanalytic change of coordinates $(z, \theta) \mapsto (\overline{z}(z, \theta), \overline{\theta}(z, \theta))$ the superderivative $D_{\theta}$ transforms as -$$ D_{\theta} = (D_{\theta} \overline{z} + \overline{\theta} D_{\theta} \overline{\theta} )\frac{\partial}{\partial_{\overline{z}}} + D_{\theta} \overline{\theta} D_{\overline{\theta}} $$ -My Question: When I use the chain rule I find that: -$$D_{\theta} = (D_{\theta} \overline{z})\frac{\partial}{\partial_{\overline{z}}} + D_{\theta} \overline{\theta} D_{\overline{\theta}}. $$ -In particular, I am missing the term $\overline{\theta} D_{\theta} \overline{\theta} $ term. I cannot figure out how to reproduce the given transformation $D_{\theta}$ using the ordinary chain rule from calculus. -How does one use the chain rule on $D_{\theta}$ to produce the term $\overline{\theta} D_{\theta} \overline{\theta} $? - -REPLY [2 votes]: Direct application of definitions and chain rule gives -$$ -D_\theta = (D_\theta \hat{\theta}) \partial_{\hat{\theta}}+(D_\theta \hat{z}) \partial_{\hat{z}}. -$$ -Then eliminate $\partial_{\hat{\theta}}$ in favour of $D_{\hat{\theta}}$: -$$ -D_\theta = (D_\theta \hat{\theta}) (D_{\hat{\theta}}-\hat{\theta}\partial_{\hat{z}})+(D_\theta \hat{z}) \partial_{\hat{z}}. -$$ -Rearranging terms, we get -$$ D_\theta =(D_\theta \hat{\theta})D_{\hat{\theta}} +(D_\theta \hat{z}-\hat{\theta}D_\theta \hat{\theta}) \partial_{\hat{z}} -$$ -as claimed in (2.9) of the given reference (note the sign).<|endoftext|> -TITLE: Complete stable minimal hypersurface in positively curved manifolds -QUESTION [5 upvotes]: Let $(M^n,g)$ be a complete noncompact orientable Riemannian manifold with positive sectional curvature. Can we find an orientable stable minimal hypersurface $N$ in $M$? -It follows from R. Schoen's work that if $n=3$, no such hypersurface $N$ exists. Moreover, if $N$ is compact, this is also impossible by the stability inequality. Are there any result for the general case? - -REPLY [2 votes]: You can construct a positively curved $(M^n,g)$ for any $n\geq 4$ that admits a stable minimal hypersurface. This is described in Example 1.2 here. (That paper also contains some non-existence results for $n=4$ under additional curvature assumptions.)<|endoftext|> -TITLE: reference for: no finite set of positive (integer) binary quadratic forms represents all primes -QUESTION [14 upvotes]: This recent question asks for a set of forms (binary quadratic) representing all primes. -Set of quadratic forms that represents all primes -When the question was asked on MSE last month -https://math.stackexchange.com/questions/3820129/non-linear-forms-for-all-prime-numbers -I made the claim that no finite set of positive binary forms would suffice. This still seems right to me, but I lack a proof or any reference. The subject is traditional, I would guess there is a mention in, say Dickson's History, which I do have. I will check. -Let's see, this will take some time, but there is no problem writing a Manjul Bhargava style "truant" program, begin with $x^2 + y^2,$ prime $3$ missing says add $x^2 + 2 y^2,$ then $7$ missing says add $x^2 + xy + 2 y^2,$ and so on. Eventually I would expect to see some non-principal forms as the smallest absolute discriminant form. - -REPLY [20 votes]: This is indeed correct; I don't know a reference, but here's a proof. Let ${\mathcal D}$ be a finite set of $K$ negative fundamental discriminants. We want to show that the set of primes not represented by any binary quadratic form with discriminants in ${\mathcal D}$ has density at least $2^{-K}$. -Let $X$ be large. We are interested in -$$ -\sum_{p\le X} \log p \prod_{d\in {\mathcal D}} \frac{(1 -\chi_d(p))}{2} -= \frac{1}{2^K} \sum_{j=0}^{K} (-1)^j \sum_{\substack{ d_1, \ldots, d_j \in {\mathcal D} \\ \text{distinct} }} \sum_{p\le X} \Big(\frac{d_1\cdots d_j}{p}\Big) \log p. -$$ -The term $j=0$ gives a contribution $\sim X/2^K$. If $j$ is odd, then $d_1 \cdots d_j$ is negative, and therefore the character $(\frac{d_1 \cdots d_j}{\cdot})$ is non-principal, and the sum over primes cancels out. If $j$ is even, then it is conceivable that $d_1\cdots d_j$ is a square when the character $(\frac{d_1\cdots d_j}{\cdot})$ is principal, but here $(-1)^j =1$ and so these terms only increase our density of primes. That completes the proof. -Obviously the argument breaks if indefinite forms (positive discriminants) are allowed, since one can have an odd number of terms in the product equalling a square (e.g. take discriminants $5$, $17$, $85$ and all the binary quadratic forms with these discriminants).<|endoftext|> -TITLE: The Hopf Invariant 1 Problem through L-polynomials -QUESTION [12 upvotes]: Adams and Atiyah give a wonderfully simple proof of the Hopf invariant 1 problem that uses the Adams operations on K-theory to reduce the Hopf Invariant 1 question to an elementary number theory question. In this theme, I think we should also be able to reduce the Hopf Invariant 1 problem to a number theoretical question about the L-polynomials. -Recall the Hopf Invariant 1 problem asks for which $n$ there is $f: S^{2n-1} \rightarrow S^n$, $\operatorname{cofiber}(f)=X$ has its middle dimensional cohomology generator square to a generator of its top dimension cohomology. Of course, this implies that $X$ has Poincare duality, i.e. it is a Poincare duality space. -It is not difficult to show that for $f$ to have Hopf invariant 1, $n$ must be a power of 2, so let us assume such an $n$ and that $n>2$. Thus $X$ has cohomology concentrated in even dimensions. We might ask when is $X$ actually the homotopy type of a manifold. The first obstruction is a lift of the Spivak normal fibration (that $X$ has as a result of being a PD space) to $BTop$. Recall $G$ is used to denote the space classifying stable spherical fibrations -Since the homotopy groups of $G/Top$ are the surgery obstruction groups of the trivial group, the homotopy groups are trivial in odd dimensions. So all obstructions to lifting must be trivial because the cohomology of $X$ is in even dimensions. Hence, we have a lift from $BG$ to $BTop$. -This means we have a surgery problem in dimension divisible by four, so if the surgery obstruction vanishes $X$ has the homotopy type of a $8k$-manifold that is $4k-1$ connected with signature 1. Perhaps it is helfpul to mention here that the surgery obstruction will just be the difference of signatures. -$\bf Question:$ Suppose I have a 8k-manifold $M$ so (1) the rank of $H^{4k}(M)$ is 1, (2) Hirzebruch L-polynomials for $M$ have contributions only from $p_{2k}$ and $p_{k}^2$, (3) $p_k$ is some multiple $n$ of the generator $x$ of $H^{4k}(M)$, and (4) that $x^2$ generates $H^{8k}(M)$, can we deduce a contradiction? -Of course, the solution of the Hopf invariant 1 problem implies no such manifold can exist (in dimensions greater than 16, but I am wondering if this can be proven only from the $L$ polynomials. - -REPLY [13 votes]: If I have understood your question correctly, the answer is no. -A rational projective plane is a closed $2n$-dimensional manifold $M$ with $H^*(M; \mathbb{Q}) \cong \mathbb{Q}[\alpha]/(\alpha^3)$ where $\deg\alpha = n$. Such manifolds were studied by Su in her paper Rational Analogs of Projective Planes. Note that if $n = 4k$, a rational projective plane satisfies the criteria (1) - (4). However, in that same paper, Su used the Barge-Sullivan theorem to show the existence of a 32-dimensional rational projective plane. In particular, the conditions you state are insufficient to obtain a solution of the Hopf invariant one problem.<|endoftext|> -TITLE: General Fourier inversion formula (Gil-Pelaez) -QUESTION [7 upvotes]: Gil-Pelaez (1951) proves the Fourier inversion formula -\begin{align*} -F(x) &= \frac{1}{2} + \frac{1}{2\pi} \int_0^\infty \frac{e^{itx}\phi(-t)-e^{-itx}\phi(t)}{it}dt \\ -&= \frac{1}{2} - \frac{1}{\pi} \int_0^\infty \Im\left(\frac{e^{-itx}\phi(t)}{t}\right)dt, -\end{align*} -where $F$ is the cdf of a random variable and $\phi$ the characteristic function, $\phi(t)=\int_\mathbb{R}e^{itx}dF(x)$. -Is it possible to relate the integral (or its principal value) $$\int_0^\infty \Im\left(\frac{e^{-itx}\phi(t)}{t^n}\right)dt$$ for $n\in\mathbb{N}$ to the cdf $F$? -Gil-Pelaez' original proof wouldn't work because it would then involve integrals like $\int_0^\infty \frac{\sin(x)}{x^n}dx$ and $\int_0^\infty \frac{\cos(x)}{x^n}dx$ which don't necessarily exist but perhaps somebody has seen a different proof that is adaptable for powers in the denominator? - -REPLY [7 votes]: Whenever the distribution with characteristic function $\phi$ has a finite mean $a$, we have $\phi(t)=1+iat+o(t)$ (as $t\downarrow0$). So, for any real $x\ne a$, the integrand in your integral is $\sim (a-x)t^{1-n}$ and hence for any $n\ge2$ the integral diverges to $\pm\infty$ in a right neighborhood of $0$. So, your integral does not exist for any $n\ge2$ and any real $x\ne a$, even as a principal value. -However, in this paper or its arXiv version, one can find many formulas of the same flavor as the Gil--Pelaez one, with $t^p$ for however large $p$ in the denominator of the integrand.<|endoftext|> -TITLE: A density question for the Hilbert transform -QUESTION [11 upvotes]: Let $\mathscr Hf$ denote the Hilbert transform of a function $f$ defined on the real-line $\mathbb R$. Are the set of functions -$$ \{(f+\mathscr Hf)_{|_{(0,1)}}\,:\, f \in C^{\infty}(\mathbb R)\quad \text{and}\quad \textrm{supp} f \Subset (0,\infty)\}$$ -dense in $L^2((0,1))$? - -REPLY [11 votes]: Yes, it is dense. -Indeed, if $g$ is an $L^2$ function supported on $[0,1]$ such that $g$ is orthogonal to every $f+\mathscr Hf$ with $f$ compactly supported on $(0,+\infty)$, then $g-\mathscr Hg=0$ on $(0,+\infty)$. However, $\mathscr H$ is an isometry in $L^2(\mathbb R)$, so this would imply that $\mathscr Hg=g$ on $(0,1)$ and, hence, $\mathscr Hg=0$ a.e. outside $[0,1]$, i.e., that $\mathscr Hg=g$ in $L^2(\mathbb R)$, which is impossible unless $g\equiv 0$.<|endoftext|> -TITLE: Is $\Gamma(s, x=s-1)/\Gamma(s)$ decreasing for real $s>1$? Is $\Gamma(s, x=s)/\Gamma(s)$ increasing? -QUESTION [5 upvotes]: This has received no full solution at StackExchange. -As per https://dlmf.nist.gov/8.10#E13 we have -$$\frac{\Gamma\left(n,n\right)}{\Gamma\left(n\right)}<\frac{1}{2}<\frac{\Gamma% -\left(n,n-1\right)}{\Gamma\left(n\right)}$$ for $n=1,2, \ldots$. -My question is: 1) show that both of these are monotone in $n$ AND 2) replacing $n$ by real $s>1$, show that these are monotone in $s$ (of course 1 follows from 2, but perhaps 1 is easier). -Note: For large $n$ both of these converge to $\frac{1}{2}$: rewriting via repeated integration by parts we get $\frac{\Gamma% -\left(n,n-1\right)}{\Gamma\left(n\right)}=\exp\{-(n-1)\}\sum_{i=0}^{n-1} \frac{(n-1)^i}{i!}$ which is the CDF of Poisson distribution with $\lambda=n-1$ evaluated at $x=\lambda=n-1$; as $n$ increases this approaches CDF of normal with mean and variance $\lambda$, which at $x=\lambda$ is $\frac{1}{2}$. (I would be interested in alternative proofs of this fact as well.) - -REPLY [3 votes]: $\newcommand\Ga\Gamma -\newcommand\tp{\tilde p} -\newcommand\tq{\tilde q} -\newcommand\tr{\tilde r}$First, let us show that $\dfrac{\Ga(s,s-1)}{\Ga(s)}$ decreases in real $s>1$. This is equivalent to $\dfrac{\Ga(s)-\Ga(s,s-1)}{\Ga(s,s-1)}$ being increasing in real $s>1$. Note that -\begin{align*} - \Ga(s)-\Ga(s,s-1)&=\int_0^{s-1} t^{s-1}e^{-t}\,dt=(s-1)^s\int_0^1 x^{s-1}e^{-(s-1)x}\,dx, \\ - \Ga(s,s-1)&=\int_{s-1}^\infty t^{s-1}e^{-t}\,dt=(s-1)^s\int_1^\infty x^{s-1}e^{-(s-1)x}\,dx, \\ -\end{align*} -So, what we need to show is that -\begin{equation} - R(u):=\frac{I(u)}{J(u)} -\end{equation} -is increasing in real $u>0$, where -\begin{align*} - I(u)&:=\int_0^1 f(x)^u\,dx=\int_0^1 z^u\,p(z)\,dz, \\ - J(u)&:=\int_1^\infty f(x)^u\,dx=\int_0^1 z^u\,q(z)\,dz, -\end{align*} -where $f(x):=xe^{1-x}$, -$p(z):=x_1'(z)>0$, $q(z):=-x_2'(z)>0$, $x_1(z)$ is the only root $x\in(0,1)$ of the equation $f(x)=z$ for $z\in(0,1)$, and $x_2(z)$ is the only root $x\in(1,\infty)$ of the equation $f(x)=z$ for $z\in(0,1)$. -To show this, note that -\begin{align*} - 2J(u)^2R'(u)&=2\int_0^1\int_0^1 dx\,dy\,(xy)^u p(x)q(y)(\ln x-\ln y) \\ - &=2\int_0^1\int_0^1 dy\,dx\,(yx)^u p(y)q(x)(\ln y-\ln x) \\ - &=\int_0^1\int_0^1 dy\,dx\,(xy)^u [p(x)q(y)-p(y)q(x)] (\ln x-\ln y)\\ - &=\int_0^1\int_0^1 dy\,dx\,(xy)^u\,p(y)q(y) [r(x)-r(y)](\ln x-\ln y), -\end{align*} -where -\begin{equation} - r:=p/q. -\end{equation} -So, it remains to show that $r$ is increasing on $(0,1)$. -We have -\begin{equation} - p=\frac{x_1}{(1-x_1) z},\quad q=\frac{x_2}{(x_2-1) z}, -\end{equation} -\begin{equation} - r'=\frac{x_1(x_2-x_1)(x_1+x_2-2)}{(1-x_1)^3 (x_2-1) x_2 z}. -\end{equation} -So, it remains to show that $x_1+x_2-2>0$ or, equivalently, $x_2>2-x_1$ or, equivalently, $f(x_2)0$. This proof is similar to the one above. Here in place of $p,q,r=p/q$ we get -\begin{equation} - \tp:=pe^{-x_1},\quad \tq:=qe^{-x_2},\quad \tr:=\tp/\tq, -\end{equation} -with -\begin{equation} - \tr'=\frac{(x_2-x_1) (x_1 x_2-1)}{(1-x_1)^3 (x_2-1) z} -\end{equation} -So, it remains to show that $x_1x_2-1<0$ or, equivalently, $x_2<1/x_1$ or, equivalently, $f(x_2)>f(1/x_1)$ or, equivalently, $f(x_1)>f(1/x_1)$. -So, it remains to show that $f(x)>f(1/x)$ for $x\in(0,1)$ or, equivalently, $10$ and $x_1x_2-1<0$, used in the proofs above, is to note that the condition $f(x_1)=f(x_2)$ means that the logarithmic mean of $x_1,x_2$ is $1$, and then use the arithmetic-logarithmic-geometric mean inequality. - -The same method can be used to establish monotonicity of the ratio $\dfrac{\Ga(s,s+a)}{\Ga(s)}$ for real $a\notin\{-1,0\}$. For any given real $a$, it will follow that (i) $\dfrac{\Ga(s,s+a)}{\Ga(s)}$ is increasing in real $s>\max(0,-a)$ if the function -$$m_a:=(a+1) (x_1 x_2-1)-a (x_1+x_2-2)$$ -is $<0$ on $(0,1)$ and (ii) $\dfrac{\Ga(s,s+a)}{\Ga(s)}$ is decreasing in real $s>\max(0,-a)$ if -$m_a>0$ on $(0,1)$. So, it follows from the inequalities $x_1 x_2-1<00$ for each real $a\ge0$ and (ii) $\dfrac{\Ga(s,s+a)}{\Ga(s)}$ is decreasing in real $s>-a$ for each real $a\le-1$. -Moreover, one can show that $m_a>0$ on $(0,1)$ iff $a\le-1/3$. -So, $\dfrac{\Ga(s,s+a)}{\Ga(s)}$ is decreasing in real $s>-a$ for each real $a\le-1/3$. Furthermore, one can show that $\dfrac{\Ga(s,s+a)}{\Ga(s)}$ is neither decreasing nor increasing in real $s>-a$ for each real $a\in(-1/3,0)$. -We conclude that $\dfrac{\Ga(s,s+a)}{\Ga(s)}$ is -(i) increasing in real $s>0$ for each real $a\ge0$; -(ii) decreasing in real $s>-a$ for each real $a\le-1/3$; -(iii) non-monotonic in real $s>-a$ for each $a\in(-1/3,0)$.<|endoftext|> -TITLE: Integral representation of product of two Whittaker functions -QUESTION [11 upvotes]: Does anyone know anything about the following formula involving special functions: -$$\begin{multline*} -W_{\kappa,\mu}(z)W_{\lambda,\mu}(w)=\frac{e^{-(z+w)/2}(zw)^{\mu+1/2}}{\Gamma(1-\kappa-\lambda)}\int_0^\infty e^{-t}t^{-\kappa-\lambda}(z+t)^{\kappa-\mu-1/2}(w+t)^{\lambda-\mu-1/2} \\ -\qquad \qquad \times {}_2F_1\left(\mu-\kappa+1/2,\mu-\lambda+1/2,1-\kappa-\lambda;\frac{t(z+w+t)}{(z+t)(w+t)}\right)\mathrm{d} t \ .\\ -\text{for }\qquad \mathrm{Re}(\kappa+\lambda)<1\ ,\qquad z,w\neq 0\ . -\end{multline*}$$ -This formula says that the product of two Whittaker functions $W_{\kappa,\mu}$ is equivalent to an integral of a hypergeometric function ${}_2F_1$ against some weight. -I came across this formula on page 74 of Iwanami mathematical formulas 3 (written in Japanese). You can also find this in equation 7.526.3 of Table of Integrals, Series, and Products, on page 401 of Tables of Integral Transforms volume 2, and in equation 6.15.3.21 in higher transcendental functions. vol. i But I could not find the original paper where this formula is derived. Could you tell me any reference books for this formula? - -REPLY [6 votes]: I finally found the right paper written by A. Erdélyi in which the formula is introduced. They derived this formula by writing Whittaker functions as Laplace type integral forms and using convolutional properties of Laplace transformation.<|endoftext|> -TITLE: Integral monoid rings and Ore conditions -QUESTION [5 upvotes]: Consider a cancellative monoid $S$ satisfying the left Ore condition, so it embeds in a group $G=S^{-1}S$. Consider also the integral monoid rings $\mathbb Z[S]$ and $\mathbb Z[G]$. -I have two, probably trivial, questions: - -Can one prove that $S$ satisfies the left Ore condition (for rings) as a multiplicative set in $\mathbb Z[S]$? I have tried both to find references and to prove it myself, but without success. - -Even if the answer to (1) is negative, one can still consider the map $\mathbb Z[S]\to \mathbb Z[G]$ and use it to make a left $\mathbb Z[S]$-module $M$ into a left $\mathbb Z[G]$-module -$$ -\bar M:=\mathbb Z[G]\otimes_{\mathbb Z[S]}M. -$$ -Furthermore, there is a canonical map $\varphi \colon M\to \bar M$, defined by $\varphi(m):=1\otimes m$, for all $m\in M$. Is there an easy description for $\ker(\varphi)$? For example, is $\ker(\varphi)=\{m\in M:\exists s\in S, sm=0\}$? - -REPLY [5 votes]: Let $s\in S$ and $r\in \mathbb{Z}[S]$. We can then write $r=\sum_{i=1}^{n}\alpha_i t_i$ for some $\alpha_{i}\in \mathbb{Z}$ and some $t_i\in S$. -The left Ore condition on $S$ implies that there exists some $u_1$ and $v_1$ (in $S$) with $u_1t_1 = v_1 s$. Similarly, the left Ore condition gives us $u_2$ and $v_2$ with $u_2(u_1t_2) = v_2s$. Repeating, we see that $$(u_n u_{n-1}\cdots u_2 u_1)r\in \mathbb{Z}[S]s.$$ This shows that $S$ is a left Ore set inside the domain $\mathbb{Z}[S]$.<|endoftext|> -TITLE: Are algebras with invertible linear duals always Frobenius? -QUESTION [11 upvotes]: Let $A$ be a finite dimensional algebra over a ground field $k$. The linear dual $A^* = Hom_k(A,k)$ is naturally an $A$-$A$ bimodule. I am interested in those algebras such that $A^*$ is an invertible $A$-$A$ bimodule. That is, there is another $A$-$A$ bimodule $L$ and $A$-$A$ bimodule isomorphisms $L \otimes_A A^* \cong A \cong A^* \otimes_A L$. -One class of algebras which has this property are the Frobenious algebras. One of the classical definitions of a Frobenius algebra is that it is an algebra with an isomorphism of right $A$-modules ${A^*}_A \cong A_A$. If this is an isomorphism of bimodules then this is a symmetric Frobenius algebra. More generally we have ${}_A{A^*}_A \cong {}_A{}^\sigma A_A$, where the right-hand side is simply $A$ as a bimodule but where the left action is twisted by the Nakayama isomorphism $\sigma$. In particular since the Nakayama isomorphism is an isomorphism, $A^*$ is an invertible bimodule. -Question: If $A$ is an algebra such that $A^*$ is an invertible bimodule, does $A$ admit the structure of a Frobenius algebra? -Upon reviewing some old notes to myself, apparently at one time I believed that the answer to the above question is yes. However I don't remember the reasoning and didn't record a reference. Further, I am suspicious of my old self because in general there are certainly invertible bimodules which do not come from twisting the left action of the trivial bimodule. I would be happy to understand a counterexample or to find out that my old self was right. -One motivation for studying these algebras is that they arise naturally in extended topological field theory. There is a certain variant of 2D framed tqfts (the "non-compact" variant) and these algebras are in bijection with those tqfts with values in the Morita 2-category. So I would also be interested in anything further that could be said about these algebras, even with further assumptions like $k$ being characteristic zero. - -REPLY [6 votes]: For a finite dimensional algebra $A$, $A^{\ast}$ being an invertible -bimodule is equivalent to $A$ being self-injective (which is the same -as quasi-Frobenius for finite dimensional algebras). -One implication has already been covered in comments. If $A^{\ast}$ is -invertible, then $-\otimes_{A}A^{\ast}$ is a self-equivalence of the -right module category, and so sends projectives to projectives. So -$A^{\ast}$ is projective. -For the other implication, assume $A$ is self-injective. Then -$-\otimes_{A}A^{\ast}$ is left adjoint to -$\operatorname{Hom}_{A}(A^{\ast},-)$, and it is easy to check that -the unit -$$A\to \operatorname{Hom}_{A}(A^{\ast},A\otimes_{A}A^{\ast}),$$ -which is given by $a\mapsto[\varphi\mapsto a\otimes\varphi]$ for -$a\in A$, $\varphi\in A^{\ast}$, is an isomorphism. -But $\operatorname{Hom}_{A}(A^{\ast},-)$ is exact and therefore -isomorphic to $-\otimes_{A}L$, where -$L=\operatorname{Hom}_{A}(A^{\ast},A)$, by the Eilenberg-Watts -theorem. So $A^{\ast}\otimes_{A}L\cong A$ as $A$-bimodules. -The same argument with left modules shows that $A^{\ast}$ has a left -inverse, and so $A^{\ast}$ is invertible. -For a typical example of a self-injective algebra that is not -Frobenius, start with a Frobenius algebra $A$ with an indecomposable -projective right module $P$ such that $P\otimes_{A}A^{\ast}\not\cong -P$, and take a Morita equivalent algebra $B$ that is the endomorphism -algebra of a progenerator that contains $P$ and $P\otimes_{A}A^{\ast}$ -as direct summands with different multiplicities. -The simplest example is where $A$ is the path algebra of a quiver with -two vertices $v_{1}$ and $v_{2}$, with an arrow $a$ from $v_{1}$ to -$v_{2}$ and an arrow $b$ from $v_{2}$ to $v_{1}$, modulo the relations -$ab=0=ba$. Let $e_{i}$ be the idempotent corresponding to vertex -$v_{i}$, and $P_{i}=e_{i}A$ the corresponding indecomposable -projective right module. -Then $B=\operatorname{End}_{A}(P_{1}^{2}\oplus P_{2})$ is -self-injective (since it's Morita equivalent to $A$) but not -Frobenius. The indecomposable projective corresponding to $P_{1}$ -under the Morita equivalence occurs with multiplicity two as a summand -of $B$, but with multiplicity one as a summand of $B^{\ast}$.<|endoftext|> -TITLE: Are the Morse inequalities sharp for 5-manifolds -QUESTION [19 upvotes]: Given a compact smooth manifold $M$ denote by $b_i(M)$ the $i$-th Betti number and denote by $q_i(M)$ the minimal number of generators for $H_i(M)$. Let $f$ be a Morse function on $M$. The Morse inequalities say that the number of critical points of index $k$ equals at least $b_k(M)+q_k(M)+q_{k-1}(M)$. -If $M$ is simply connected one can ask whether the Morse inequality is sharp, i.e. whether exists always a Morse function such that the inequalities become an equality for every $k$. Smale showed that if $\dim(M)\geq 6$, then one can always find such $f$. By the resolution of the Poincare conjecture the Morse inequality is sharp for simply connected 3-manifolds. In the 4-dimensional setting it's a famous open problem whether the Morse inequalities are sharp. Doing some literature search I could not find anything though on the status of 5-manifolds. Is it know that the Morse inequalities are (not) sharp? Or is that an open problem? - -REPLY [10 votes]: Presumably you mean closed. Otherwise a non-trivial h-cobordism would not have the minimal number of critical points. -For closed simply connected manifolds, it seems to me that this is Corollary 2.2.2 of Barden (Simply Connected Five-Manifolds, Annals of Mathematics Vol. 82, No. 3 (Nov., 1965), pp. 365-385)<|endoftext|> -TITLE: Can this result in cardinal arithmetic be established without using pcf theory? -QUESTION [11 upvotes]: Suppose $\kappa\leq\mu$ are infinite cardinals. Let us agree to call a family $\mathcal{P}\subseteq[\mu]^{<\mu}$ a countably generating family for $[\mu]^\kappa$ if every member of $[\mu]^\kappa$ can be written as a union of countably many elements of $\mathcal{P}$. Note that we can extend this in the obvious way to structures of the form $[\mu]^{<\kappa}$ as well. - -Theorem: If $\mu$ is a strong limit singular cardinal of uncountable cofinality, then the minimum cardinality of a countably -generating family for $[\mu]^{{\rm cf}(\mu)}$ is the same as the -minimum cardinality of a countably generating family for -$[\mu]^{<\mu}$ (and hence, the same as the minimum cardinality of a -countably generating family for $[\mu]^\kappa$ for any $\kappa$ with -${\rm cf}(\mu)\leq\kappa<\mu$). - -The only proof I have relies on heavy machinery from pcf theory, but I do not believe this should be required, and I think that a more direct proof would shed light on some related questions. The following vague question asks for such a proof in a simple special case: - -Question : Suppose $\mu$ is a strong limit singular cardinal of cofinality $\omega_1$, and let $\mathcal{P}$ be a countably generating -family for $[\mu]^{\omega_1}$ (WLOG closed under subsets). Is there a reasonably constructive way -to build a countably generating family for $[\mu]^{\omega_2}$ (of the -same cardinality) from $\mathcal{P}$? - -I don't have the language of Galois-Tukey connections -available in this context to formulate this question more precisely, but I want to avoid tricks like "let $M$ be the Skolem hull of $\mathcal{P}$ and the other parameters, and then $M\cap [\mu]^{<\mu}$ works because the theorem is true". -Roughly speaking, the pcf-theoretic proof relies on Shelah's "cov vs. pp Theorem" to convert things into an equivalent question about pseudopowers, and then solves the associated pseudopower question using other results from his book Cardinal Arithmetic. Can we do better? I want to make sure I'm not just suffering from a blind spot and missing something easy. - -REPLY [6 votes]: There IS an easy proof of this, but I just had to reframe the way I was thinking of the problem. The cardinals in question (and many of their relatives) turn out to be $2^{\mu}$ if $\mu$ is strong limit because we can do suitable coding: -Suppose $\mu$ is a singular strong limit cardinal of cofinality $\kappa$. Since $\mu$ is a strong limit, we can enumerate the bounded subsets of $\mu$ as $\langle X_\alpha:\alpha<\mu\rangle$ in such a way that every such set is indexed cofinally often. -We also let $\langle \mu_\xi:\xi<\kappa\rangle$ be increasing and cofinal in $\mu$. -Given $X\subseteq\mu$, we code $X$ as an element of $[\mu]^\kappa$ in the natural way: for each $\xi<\kappa$ we choose $\alpha(\xi)<\kappa$ such that $X\cap\mu_{\xi}= X_{\alpha(\xi)}$, and define the code of $X$, $cd(X)$ to be $\{\alpha(\xi):\xi<\kappa\}$. -Note that we can assume the sequence $\langle \alpha(\xi):\xi<\kappa\rangle$ is strictly increasing, so that without loss of generality $cd(X)$ has cardinality $\kappa$. -Now since the sequence $X\cap \mu_{\xi}$ increases with $\xi$, we can recover $X$ from any subset of $cd(X)$ of cardinality $\kappa$ as $X\cap\mu_\xi$ is indexed (in the enumeration of the bounded subsets of $\mu$) by the $\xi$th element in the increasing enumeration of $cd(X)$. Thus, given $Y\subseteq cd(X)$ of cardinality $\kappa$, -$$ X = \bigcup_{\alpha\in Y}X_\alpha.$$ -Now if $\mathcal{D}$ is any dense subset of $[\mu]^{\kappa}$ (that is, $\mathcal{D}\subseteq [\mu]^{\kappa}$ and every set in $[\mu]^{\kappa}$ has a subset in $\mathcal{D}$, then the ``decoding map'' -$$d:\mathcal{D}\rightarrow \mathcal{P}(\mu)$$ -defined by -$$d(A) = \bigcup_{\alpha\in A} X_\alpha$$ -maps $\mathcal{D}$ onto $\mathcal{P}(\mu)$. -Now turning to the actual question, if $\mu$ is a strong limit singular cardinal of cofinality $\kappa$, and $\mathcal{P}\subseteq[\mu]^{<\mu}$ has the property that every member of $[\mu]^{\kappa}$ is a union of fewer than $\kappa$ members of $\mathcal{P}$, then the set -$$\mathcal{D}=\mathcal{P}\cap [\mu]^{\kappa}$$ -is dense in $[\mu]^\kappa$, and therefore $|\mathcal{P}|= 2^{\mu}$. -The pcf stuff I made reference to is then properly viewed as a relative of the above argument that holds true even when $\mu$ is a strong limit in a very weak sense.<|endoftext|> -TITLE: Are the Galois actions on automorphisms of twists isomorphic? -QUESTION [6 upvotes]: This might be a trivial question and I might be overlooking something: -Suppose $k$ is a field with algebraic closure $\overline k$ and absolute Galois group $\Gamma$. Let $X,Y$ be two distinct varieties over $k$ that are isomorphic over $\overline k$. Consider their automorphisms groups $Aut_{\overline k}(X)$ and $Aut_{\overline k}(Y)$. These groups are isomorphic as abstract groups but they each also have a Galois action of $\Gamma$ acting on the groups by conjugation. -Are these two automorphism groups isomorphic as "groups with a $\Gamma$ action"? Note that $H^1(\Gamma,Aut_{\overline k}(X)) \cong H^1(\Gamma,Aut_{\overline k}(X))$ because both groups classify twists of $X$ (or equivalently of $Y$). -If these two "groups with $\Gamma$ action" are distinct, are their higher cohomologies distinct too in any examples? How are these two related otherwise? - -REPLY [12 votes]: They are not. $H^0(\Gamma, \text{Aut}(X_{\bar{k}}))$ computes the subgroup of automorphisms which are Galois-invariant, which is equivalently the automorphism group of $X$, and similarly for $Y$, so to find a counterexample it suffices to find varieties $X, Y$ which are isomorphic over $\bar{k}$ but which have non-isomorphic automorphism groups over $k$. -Let $k = \mathbb{R}$, let $X = \mathbb{P}^1$, and let $Y$ be the conic $\{ X^2 + Y^2 + Z^2 = 0 \}$ in $\mathbb{P}^2$. Then $\text{Aut}(X) \cong PGL_2(\mathbb{R})$ but $\text{Aut}(Y) \cong PO(3) \cong SO(3)$ (actually I'm not entirely confident I know how to prove this, at least not without passing to the complexification). These two groups can be distinguished abstractly by their abelianizations: I believe (but haven't checked carefully) that the abelianization of $PGL_2(\mathbb{R})$ is $\{ \pm 1 \}$ (given by the sign of the determinant) but the abelianization of $SO(3)$ is trivial.<|endoftext|> -TITLE: Is there a non-degenerate quadratic form on every finite abelian group? -QUESTION [9 upvotes]: Let $G$ be a finite abelian group. A quadratic form on $G$ is a map $q: G \to \mathbb{C}^*$ such that $q(g) = q(g^{-1})$ and the symmetric function $b(g,h):= \frac{q(gh)}{q(g)q(h)}$ is a bicharacter, i.e. $b(g_1g_2, h) = b(g_1, h)b(g_2, h)$ for all $g, g_1, g_2, h \in G.$ -The quadratic form $q$ is called non-degenerate if the corresponding bicharacter $b$ is non-degenerate. -Question: Is there a non-degenerate quadratic form on every finite abelian group? -Motivation: it is used to make pointed braided/modular tensor categories, see Chapter 8 of this book (in particular Sections 8.4, 8.13 and 8.14). - -REPLY [5 votes]: Yes. It's necessary and sufficient to show that every finite abelian group admits a nondegenerate quadratic form valued in a finite cyclic group. The following slightly stronger statement is true: every finite abelian $p$-group admits a nondegenerate quadratic form valued in $C_{p^k}$ for some $k$ (this suffices by the Chinese remainder theorem). So let's prove this. -If $A$ is a finite abelian $p$-group for $p$ an odd prime, we can pick any isomorphism $A \cong A^{\ast}$ from $A$ to its Pontryagin dual $A^{\ast} = \text{Hom}(A, \mathbb{Q}/\mathbb{Z}) \cong \text{Hom}(A, \mathbb{Q}_p/\mathbb{Z}_p)$ and we'll get a nondegenerate bilinear form $B : A \times A \to C_{p^k}$ whose associated quadratic form $Q : A \to C_{p^k}$ is nondegenerate. As you say in the comments, this almost but doesn't quite work when $p = 2$. -When $p = 2$ the following slight modification works. Again pick an isomorphism $A \cong A^{\ast}$ to the Pontryagin dual and get a nondegenerate bilinear form $B : A \times A \to C_{2^k}$. Now we do something a bit funny. Consider the inclusion (not a group homomorphism!) $C_{2^k} \to C_{2^{k+1}}$ given by $k \mapsto k$, thinking of elements of $C_n$ as elements of $\mathbb{Z}/n$. Composing this inclusion with $B$ gives a map (not a bilinear map!) $B' : A \times A \to C_{2^{k+1}}$. Now I claim that the diagonal $Q(a) = B'(a, a)$ of this map is a nondegenerate quadratic form. We clearly have $Q(-a) = B'(-a, -a) = B'(a, a)$, and the associated bilinear form $Q(a + b) - Q(a) - Q(b)$ recovers $B$, now taking values in $2 C_{2^{k+1}} \cong C_{2^k}$. -In particular, when $A = C_2$ we get the quadratic form $Q : C_2 \to C_4$ given by $Q(0) = 0, Q(1) = 1$. This quadratic form can be interpreted as a cohomology class in $H^4(B^2 C_2, C_4)$ and so in turn a cohomology operation $H^2(-, C_2) \to H^4(-, C_4)$ which I believe is exactly the Pontryagin square.<|endoftext|> -TITLE: Proof that $x^2 + y^2 - z^2$ is universal -QUESTION [5 upvotes]: The (ternary) quadratic form $x^2 + y^2 - z^2$ is universal, meaning that any integer $n$ can be represented as $n = x^2 + y^2 - z^2$ for some integers $x, y, z$. -My question is this: who proved this fact first? I want to know to whom I should credit this fact. The oldest literature I can find is Dickson's 1929 paper "The forms $ax^2+by^2+cz^2$ which represent all integers" in Bulletin of AMS (ProjectEuclid link to paper), where he gives quite a general theorem on universality of all diagonal forms. And I would think that the universality of this specific form can go back further. - -REPLY [2 votes]: As the others point out, there is no knowing about the earliest this was written down. For example, the notion of regularity of a ternary form is due to Dickson, but universality is an easier concept and could have gone without any name for quite some time. -Worth pointing out that all universal ternaries can be described. Three out of four types are in Dickson's 1939 book, page 161 in Modern Elementary Theory of Numbers; the one type with odd "mixed" coefficients was proved by A. Oppenheim in 1930. Sir Alexander Oppenheim was a student of Dickson and got his Ph.D. in 1930. The dissertation was titled The Minima of Indefinite Quaternary Quadratic Forms -http://www.numbertheory.org/obituaries/OTHERS/oppenheim.html -I am having trouble finding this: Quarterly Journal of Mathematics (1930) 179-185. Evidently this is where Oppenheim published a few items. His obituary goes back as far as 1941 .. -The following are representative forms under the action of $SL_3 \mathbb Z$ as follows: given the Hessian matrix $H$ of a quadratic form, a new representative is $P^T H P.$ -Taking $N$ odd, while $M$ is any integer, we have universal -$$ xy-Mz^2 $$ -$$ 2xy - N z^2 $$ -$$ 2xy + y^2 - N z^2 $$ -$$ 2xy + y^2 - 2N z^2 $$<|endoftext|> -TITLE: What are the "correct" references for the Vassiliev invariant? -QUESTION [7 upvotes]: Is there a good survey paper which describes the general ideas of -Vassiliev's invariant? I am not an expert on knot theory, many references are too technical for me. -Could Vassiliev's invariants be defined for general embeddings (rather than knot theory)? What are the useful references? -Thanks. - -REPLY [2 votes]: Along with the references provided already, I think many people like New points of view in knot theory by Joan Birman. She gives a good explanation of the connections between these finite type invariants and knot polynomials.<|endoftext|> -TITLE: Strict transform of a tangent curve under blow-up -QUESTION [7 upvotes]: $\DeclareMathOperator{\Bl}{\operatorname{Bl}}$It is known that if we have a projective variety $X$ and a projective smooth subvariety $Y$ then the exceptional divisor $E \subset \Bl_{Y}X$ of the blow-up of $X$ along $Y$ is the projectivization of the normal bundle $N_{Y|X}$. In particular points in $E$ parametrizes lines (directions) normal to $Y$. -My question now is the following: suppose for simplicity that $X= \mathbb{P}^3$ and $Y=\ell$ is a line. If we have a point $p \in \ell$ and a smooth curve $C \subset \mathbb{P}^3$ such that $\mathbb{T}_pC=\ell$, then if $$\nu:\Bl_{\ell}\mathbb{P}^3 \rightarrow \mathbb{P}^3$$ what is the intersection $\widetilde{C} \cap E$, where $\widetilde{C}$ is the strict transform of $C$ under $\nu$? -In general, if I have a curve tangent to the locus that I'm blowing up, where does its "direction" go if the exceptional locus parametrize only normal directions? -Thanks in advance. - -REPLY [9 votes]: If $C \subset X$ is a smooth curve and $p \in C$ there is a unique plane (so-called osculating plane) -$$ -T^2_pC \subset T_pX -$$ -in the tangent space $T_pX$ to $X$ at $p$ such that $C$ any element of $T_p^\vee X = \mathfrak{m}_p/\mathfrak{m}^2_p$ vanishing on $T^2_pC$ vanishes to order 2 on $C$ at $p$. Of course, it contains the tangent line $T_pC$. So, if $T_pC = \ell$ then this plane defines a normal direction to $\ell$ at $p$; the corresponding point of $E$ is the intersection point of $\tilde{C}$ with $E$.<|endoftext|> -TITLE: Irrationality measure of powers -QUESTION [8 upvotes]: Let $\alpha$ be an irrational number. Denote by $\mu(\alpha)$ its irrationality measure. Can one say anything about $\mu(\alpha^n)$ for every $n\in\mathbb N$? -Even more, one knows that $\mu(e)=2$. Can one say anything for $\mu(e^{p/q})$ for $\frac pq\in\mathbb Q^*$? - -REPLY [3 votes]: Let $\alpha$ be irrational. There are two cases: $\alpha$ can be either algebraic or transcendental. Products of algebraic numbers are algebraic, while rational powers of transcendental numbers are transcendental. Hence, for all positive integer $n$, $\alpha^n$ is algebraic in the first case, and transcendental in the second one. -Roth proved that algebraic irrational numbers all have irrationality measure $2$. Instead, little is known about transcendental numbers. We can only say, in general, that the irrationality measure is $\geq 2$. Thus, by the previous discussion, we can conclude that: -$\alpha$ irrational algebraic $\Rightarrow$ $\mu(\alpha^n)=2$ for all integers $n \geq 1$ (in fact, this also holds for all nonzero rationals $n$, since roots of algebraic numbers are algebraic). -$\alpha$ transcendental $\Rightarrow$ $\mu(\alpha^n) \geq 2$ for all integers $n \geq 1$ (in fact, as before, this also holds for all nonzero rationals $n$). -I think that the actual value of the irrationality measure of a power of a transcendental number highly depends on the particular case. However, it is worth recalling that almost (in the sense of Lebesgue measure) all irrational numbers have irrationality measure $2$.<|endoftext|> -TITLE: Incompleteness theorems for theories with omega-rule -QUESTION [7 upvotes]: Recall that the omega-rule is an infinitary rule of inference that allows one to deduce $\forall x A(x)$ from $A(0), A(1), \dots$. It's known that adjoining PA (or even Q) with the omega-rule results in a complete theory (true arithmetic). I'm curious what happens to stronger theories when we allow the omega-rule as the sole infinitary rule of inference (and all the axioms must be recursively enumerable, as I will hereafter assume). For example, can there be a complete theory of analysis or set theory if we allow ourselves the omega-rule? I suspect the answer is no, but I'm not sure how to prove it. -We can also generalize the omega-rule to allow deducing a universal statement from the set of all true instances of size at most, say, $2^{\aleph_0}$ (so for example, deduce $\forall x \in \mathbb{R} B(x)$ from the $2^{\aleph_0}$ instances of $B(x)$ for each real number $x$). Again, I suspect but cannot prove that there will be a sufficiently strong theory (something stronger than analysis) that must be incomplete even if one allows this generalized omega-rule (and similarly for generalized omega-rules for each cardinality). - -REPLY [6 votes]: Footnote to Emil Jeřábek's answer: -(1) Rosser (Journal of Symbolic Logic, 1937) was the first to show that there is a true $\Sigma^1_1$-statement that is unprovable in (second order arithmetic + the $\omega$-rule) with the essentially the same proof outlined by Emil. -(2) In contrast, as shown in a 1961-paper of Grzegorczyk, Mostowski, and Ryll-Nardzewski, every true $\Pi^1_1$-statement is provable in (second order arithmetic + the $\omega$-rule). -I learned the above facts as a graduate student from Barwise' article "The role of the Omitting Types Theorem in infinitary logic" (see p.57), published in Arch. math. Logik 21 (1981),55-68.<|endoftext|> -TITLE: hereditary C*-subalgebra of a non-elementary simple C*-algebra -QUESTION [5 upvotes]: A is said to be elementary if A is isomorphic to some $K(H)$ or $M_n$. -A C*-subalgebra $B$ is said to be hereditary if for every $0≤a≤b∈B$ we have $a∈B$. -I wanted to know that is this statement true? -every hereditary C-subalgebra of a non-elementary simple C-algebra has infinite dimensions? -If so, could you help me to prove it? - -REPLY [6 votes]: This is true in the separable case (and more generally) and a consequence of Larry Brown's stable isomorphism theorem (1977 Pacific Journal of Math). A special case of his theorem states: If $A$ is a separable, simple C*-algebra and $B$ is a hereditary subalgebra of $A$, then $A\otimes K(H)\cong B\otimes K(H).$ One could probably answer your question in the general case from the separable case. -For your question: Suppose $A$ is (separable) non-elementary and simple and $B$ is a hereditary subalgebra. Brown's theorem implies that $A\otimes K(H)\cong B\otimes K(H).$ If $B$ were finite dimensional it would have to be isomorphic to $M_n$ (otherwise $A$ wouldn't be simple). This would imply that $A$ is stably isomorphic to $K(H)$ which is impossible (for example it would imply that $A$ is Type I, which it can't be because the only separable simple Type I C*-algebras are the elementary ones).<|endoftext|> -TITLE: If all transitive models have the same height, are they all "simple"? -QUESTION [10 upvotes]: Suppose that $\alpha$ is the unique ordinal for which $L_\alpha$ is a model of $\sf ZFC$. In other words, there is no transitive model of $\sf ZFC$ in which there is a transitive model of $\sf ZFC$. -We know, of course, that there are many different models of $\sf ZFC$ of height $\alpha$. Starting with $L_\alpha$ itself, it is a countable model, so we can do a lot of different forcings over it. In fact, also class forcings can be used to extend $L_\alpha$. So we get models which are class-generic extensions, which may not have any set which is set-generic over $L_\alpha$ (e.g. a minimal coding real). - -Is it true/consistent that if all transitive models have the same height, then all transitive models are class-generic extensions of $L_\alpha$? -(Yes, I'm including here things like "hyperclass" generic extensions, it's just the question of whether there is some relatively "tame" operation that generates all models from the minimal model; relative constructibility is not tame.) - -If the answer is somehow positive, how much can this be pushed up to include other heights of transitive models? Can it include "there are 2/3/infinitely many different heights of transitive models"? What about "every real is in a transitive model"? What about uncountable heights? - -REPLY [6 votes]: I think the following result is related: - -Theorem 1(Mack Stanley). Let $L_α$ be a minimal countable standard transitive model of ZFC. There -exists a real $x_{nwg}$ having the following three properties: - -$x_{nwg}\notin L_α$. - -$L_α[x_{nwg}] \models ZFC$. - -$x_{nwg}$ is not definably generic over any outer model of $L_α$ that does not already -contain $x_{nwg}$. - - - -Indeed Stanley proves something stronger. See his paper - -Stanley, M. C., A non-generic real incompatible with $0^\#$, Ann. Pure Appl. Logic 85, No. 2, 157-192 (1997). ZBL0877.03025. (Also on Stanley's homepage.) - -But on the other hand, we have partial positive answers as well. For example see Stanley's paper - -Stanley, M. C., Invisible genericity and $0^\#$, J. Symb. Log. 63, No. 4, 1297-1318 (1998). ZBL0924.03097. (Also on Stanley's homepage.) - -What is proved in this paper is simply that some instances of any type of non-constructible object are class generic over $L$. See also the paper ''$^*$forcing'' by Garvin Melles.<|endoftext|> -TITLE: Pontrjagin-Thom model for units of the sphere spectrum? -QUESTION [15 upvotes]: Is there a framed bordism model for the units of the sphere spectrum, $gl_1(S)$? -At the level of individual homotopy groups, the Pontrjagin-Thom construction identifies the group of bordism classes of stably framed $k$-manifolds with the stable homotopy group $\pi_k(S)$, and even captures the product. But for $k > 0$ this group agrees with $\pi_k(gl_1(S))$. Is there an operation on framed manifolds corresponding to the action of $\pi_*(S)$ on $\pi_*(gl_1(S))$? What if we just asked for multiplication by $\eta$? - -REPLY [3 votes]: You should definitely have a look at Bokstedt and Waldhausen's "The map $BG \to A(*) \to QS^0$": -MR0921487 Bökstedt, Marcel ; Waldhausen, Friedhelm . The map BSG→A(∗)→QS0. -Algebraic topology and algebraic K-theory (Princeton, N.J., 1983), -418--431, Ann. of Math. Stud., 113, Princeton Univ. Press, Princeton, NJ, 1987. -They study a geometrically defined transfer map $BG \to G$, and show it is multiplication by $\eta$, and that this agrees with the usual action of $\eta$ under the iso to $\pi_*(S)$ in dimensions 3 and up, but not on $\pi_2$ as I mentioned above. This appearance of $\eta$ reminds me of the theorem of Blumberg, Cohen and Schlichtkrull on THH of Thom spectra, which might also be worth looking at in this connection. -MR2651551 -Blumberg, Andrew J.; Cohen, Ralph L.; Schlichtkrull, Christian. -Topological Hochschild homology of Thom spectra and the free loop space. -Geom. Topol. 14 (2010), no. 2, 1165--1242. -Sorry if this turns out to be irrelevant, but these are fun papers to read in any case.<|endoftext|> -TITLE: Does there exist a complete implementation of the Risch algorithm? -QUESTION [49 upvotes]: Is there a generally available (commercial or not) complete implementation of the Risch algorithm for determining whether an elementary function has an elementary antiderivative? -The Wikipedia article on symbolic integration claims that the general case of the Risch algorithm was solved and implemented in Axiom by Manuel Bronstein, and an answer to another MO question says the same thing. However, I have some doubts, based on the following comment by Manuel Bronstein himself on the USENET newsgroup sci.math.symbolic on September 5, 2003: - -If Axiom returns an unevaluated integral, -then -it has proven that no elementary antiderivative exists. There are -however -some cases where Axiom can return an error message saying that you've -hit -an unimplemented branch of the algorithm, in which case it cannot -conclude. -So Richard was right in pointing out that the Risch algorithm is not -fully implemented there either. Axiom is unique in making the difference -between unimplemented branches and proofs of non-integrability, and also -in actually proving the algebraic independence of the building blocks -of the integrand before concluding nonintegrability (others typically -assume this independence after performing some heuristic dependence -checking). - -Bronstein unfortunately passed away on June 6, 2005. It is possible that he completed the implementation before he died, but I haven't been able to confirm that. I do know that Bronstein never managed to finish his intended book on the integration of algebraic functions. [EDIT: As a further -check, I emailed Barry Trager. He confirmed that the implementation that -he and Bronstein worked on was not complete. He did not know much about -other implementations but was not aware of any complete implementations.] -I have access to Maple 2018, and it doesn't seem to have a complete implementation either. A useful test case is the following integral, taken from the (apparently unpublished) paper Trager's algorithm for the integration of algebraic functions revisited by Daniel Schultz: -$$\int \frac{29x^2+18x-3}{\sqrt{x^6+4x^5+6x^4-12x^3+33x^2-16x}}\,dx$$ -Schultz explicitly provides an elementary antiderivative in his paper, but Maple 2018 returns the integral unevaluated. - -REPLY [16 votes]: Fricas, an open-source clone of Axiom, implements a considerable chunk of Risch, see -http://fricas-wiki.math.uni.wroc.pl/RischImplementationStatus -Fricas is also available as a optional package of SageMath open-source system. -Edit: here how it goes in SageMath with Fricas as backend. - -sage: r=integrate((29*x^2+18*x-3)/(x^6+4*x^5+6*x^4-12*x^3+33*x^2-16*x)^(1/2),x,algorithm="fricas") -sage: r -log(x^29 + 40*x^28 + 776*x^27 + 9648*x^26 + 85820*x^25 + 578480*x^24 + 3058536*x^23 + 12979632*x^22 + 45004902*x^21 + 129708992*x^20 + 317208072*x^19 + 675607056*x^18 + 1288213884*x^17 + 2238714832*x^16 + 3548250712*x^15 + 5097069328*x^14 + 6677210721*x^13 + 8106250392*x^12 + 9056612528*x^11 + 8991685504*x^10 + 7944578304*x^9 + 6614046720*x^8 + 4834279424*x^7 + 2374631424*x^6 + 916848640*x^5 + 638582784*x^4 - 279969792*x^3 - 528482304*x^2 + (x^26 + 38*x^25 + 699*x^24 + 8220*x^23 + 68953*x^22 + 436794*x^21 + 2161755*x^20 + 8550024*x^19 + 27506475*x^18 + 73265978*x^17 + 165196041*x^16 + 324386076*x^15 + 570906027*x^14 + 914354726*x^13 + 1326830817*x^12 + 1731692416*x^11 + 2055647184*x^10 + 2257532160*x^9 + 2246693120*x^8 + 1939619840*x^7 + 1494073344*x^6 + 1097859072*x^5 + 640024576*x^4 + 207618048*x^3 + 95420416*x^2 + 50331648*x - 50331648)*sqrt(x^6 + 4*x^5 + 6*x^4 - 12*x^3 + 33*x^2 - 16*x) + 150994944*x - 134217728)<|endoftext|> -TITLE: Why do we associate a graph to a ring? -QUESTION [21 upvotes]: I don't know if it is suitable for MathOverflow, if not please direct it to suitable sites. -I don't understand the following: -I find that there are many ways a graph is associated with an algebraic structure, namely Zero divisor graph (Anderson and Livingston - The zero-divisor graph of a commutative ring), Non-Commuting Graph (Abdollahi, Akbari, and Maimani - Non-commuting graph of a group) and many others. -All these papers receive hundreds of citations which means many people work in this field. -I read the papers, it basically tries to find the properties of the associated graph from the algebraic structure namely when it is connected, complete, planar, girth etc. -My questions are: - -We already have a list of many unsolved problems in Abstract Algebra and Graph Theory, why do we mix the two topics in order to get more problems? - -It is evident that if we just associate a graph with an algebraic structure then it is going to give us new problems like finding the structure of the graph because we just have a new graph. -Are we able to solve any existing problems in group theory or ring theory by associating a suitable graph structure? Unfortunately I could not find that in any of the papers. - -Can someone show me by giving an example of a problem in group theory or ring theory which can be solved by associating a suitable graph structure? - - -For example suppose I take the ring $(\mathbb Z_n,+,.)$, i.e. the ring of integers modulo $n$. What unsolved problems about $\mathbb Z_n$ can we solve by associating the zero divisor graph to it? -NOTE: I got some answers/comments where people said that we study those graphs because we are curious and find them interesting. -I am not sure if this is how mathematics works. -Every subject developed because it had certain motivation. -So I don't think this reason that "Mathematicians are curious about it, so they study it" stands. -As a matter of fact, I am looking for that reason why people study this field of Algebraic Graph Theory. - -REPLY [2 votes]: The Fischer graph is one of examples; see page 569 of -Suzuki, Michio. Group theory. II. Translated from the Japanese. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 248. Springer-Verlag, New York, 1986. -An special important case of the Fischer graph is the induced subgraph of the commuting graph of a group on a certain conjugacy class of involution so-called 3-traspositions. -The title of Section 10 of Chapter 6 of the above book of Suzuki is ``Graphs and Simple groups".<|endoftext|> -TITLE: An inequality involving the beta distribution -QUESTION [7 upvotes]: Let $F$ be the CDF for a Beta distribution, $$F(x)=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\int_{0}^{x}t^{a-1}(1-t)^{b-1}\,dt$$ with $a,b\geq 1$. Is it true that $$\frac{b}{a+b} \leq\int_0^1\sqrt{F(x)}dx \leq \frac{2b}{a+b}\,?$$ Numerical simulations very strongly suggest that both inequalities hold, but I have no idea how to prove it. - -REPLY [8 votes]: You understand that it may be a good time to quit when you start forgetting what used to be your favorite tricks. -Let $u=\frac ab>0$. Let $F_s$ be the CDF corresponding to $a'=u(1+s),b'=1+s$, so $F_s'(t)=c_s t^{u-1}[t^u(1-t)]^s$. Notice that $F_s(0)=0$, $F_s(1)=1$ and $\int_0^1 F_s=\frac{b}{a+b}$ for all $s$. Now, when $s>0$ (we are interested in $s=b-1$), we have $F_sF_0$ slightly to the left of $1$ (the order of tangency is greater). Thus, the graphs of $F_s$ and $F_0$ intersect at least once on $(0,1)$. -If they intersect more than once on $(0,1)$, then, by Rolle, the difference of the derivatives $F_s'-F_0'$ has at least three zeroes on $(0,1)$ but that difference is $t^{u-1}\{c_s[t^u(1-t)]^s-c_0\}$, which, clearly, has at most two roots on $(0,1)$ (the function $t\mapsto t^u(1-t)$ is unimodal). Thus we have $F_s\le F_0$ on $[0,x_s]$, $F_s\ge F_0$ on $[x_s,1]$ and $F_s(x_s)=F_0(x_s)=Y$ for some $x_s\in(0,1)$. -Then, since both $F_s,F_0$ are increasing and $\Phi(v)=\sqrt v$ is concave, we get -$$ -\Phi(F_s)-\Phi(F_0)\le \Phi'(Y)(F_s-F_0) -$$ -on $[0,1]$ (consider $[0,x_s]$ and $[x_s,1]$ separately), -so, since $F_s$ and $F_0$ have the same integral, we get $\int_0^1\Phi(F_s)\le\int_0^1 \Phi(F_0)$. -Now it remains to evaluate $\int_0^1\sqrt{F_0}=\int_0^1 x^{u/2}\,dx=\frac{2}{u+2}=\frac{2b}{a+2b}$.<|endoftext|> -TITLE: Is the Alexander horned sphere a cofibration? -QUESTION [18 upvotes]: The Alexander horned sphere is a closed embedding of $S^2$ into $S^3$ which is not flat because otherwise the Schoenflies Theorem would be true for it. However, not being flat is not the same as not being able to find a neighborhood deformation retract. I suspect the answer to the question is no, but I have no valid argument. To be precise, let $\iota:S^2\hookrightarrow S^3$ be the embedding whose image is the Alexander horned sphere. Is $\iota$ a cofibration? - -REPLY [19 votes]: We'll make use of the following. - -If $X$ is an ANR and $j:A\subseteq X$ is a closed subspace, then $A$ is an ANR if and only if the inclusion $j$ is a cofibration. - -I don't know a good reference for this, although I suspect it may be in Borsuk's book in some form or another. A reference I do have which will cover the case at hand is found in chapter 3 of Daverman's book Decompositions of Manifolds, where Theorem 6 states that - -If $A$ is a closed subset of a metric space $X$, then $A$ has HEP in $X$ with respect to every ANR $Y$. - -The point is that if $A$ is itself an ANR, then so is $A\times I$, and in particular also the mapping cyclinder $M_j=A\times I\cup X\times 0$. In this case we see that $j:A\subseteq X$ is a cofibration by noticing that it satisfies the HEP with respect to $M_j$ (i.e. we show that $M_j$ is a retract of $X\times I$). -The point of course is that $S^3$ is an ANR, as is any closed subspace $A\subseteq S^3$ which is homeomorphic to $S^2$.<|endoftext|> -TITLE: Does the union of the subcomplexes of $X$ that contain a given subcomplex and whose inclusion in $X$ is trivial on $\pi_1$, have trivial $\pi_1$? -QUESTION [5 upvotes]: Let $X$ be a simplicial complex and let $A \subset X$ be a contractible subcomplex on the same set of vertices as $X$. Is it true that the union $$\bigcup C$$ taken over all complexes $A \subset C \subset X$ whose inclusion in $X$ induces the trivial map on fundamental groups, has trivial fundamental group? - -REPLY [6 votes]: I think it does not. Begin with $A$ a path $\{1,2\},\{2,3\},\{3,4\},\{4,5\}$. To construct $X$, add an edge $\{1,3\}$ and triangles $\{1,2,5\},\{1,3,5\},\{2,3,5\}$ to $A$. -$A$ is contractible. If $C$ contains $A$ and its image in $X$ has trivial fundamental group, it does not contain any of the edges $\{1,5\},\{2,5\},\{3,5\}$: for instance, if it contains $\{3,5\}$, then it contains the closed loop $3\to 4 \to 5 \to 3$ (a triangle) but there is no triangle containing the vertex $4$, so there is no nullhomotopy. -The path $1\to 2\to 3\to 1$ is contractible in $X$: the union of triangles we added is a disk which bounds it. Thus $\bigcup C$ of the question is just $A \cup \{1,3\}$. It contains none of the triangles of the complex: they each contain one of the forbidden edges $\{i,5\}$ where $i\in \{1,2,3\}$. So the path $1\to 2\to 3$ is not contractible in $\bigcup C$.<|endoftext|> -TITLE: Function of $(x_1,x_2,x_3,x_4)$ that factors in two ways as $\phi_1 (x_1 ,x_2 )\phi_2(x_3 ,x_4 )=\psi_1 (x_1,x_3)\psi_2(x_2,x_4)$ -QUESTION [26 upvotes]: Suppose we have a function $f(x_1 ,x_2 ,x_3 ,x_4).$ We know that we can factor it in two ways as $f(x_1 ,x_2 ,x_3 ,x_4)=\phi_1 (x_1 ,x_2 )\phi_2(x_3 ,x_4 )=\psi_1 (x_1,x_3)\psi_2(x_2,x_4)$ -Show that we can completely factor the function as: $f(x_1 ,x_2 ,x_3 ,x_4)=\varphi_1(x_1)\varphi_2(x_2)\varphi_3(x_3)\varphi_4(x_4).$ -I stumbled a little bit on this elementary problem as the proof is not as immediate as I think. But eventually I can prove this. -Here the overlap of partition {{1,2} {3,4}} and {{1,3},{2,4}} is {{1},{2},{3},{4}} and indeed satisfying the first two partition implies that we can factor by the overlap of both partitions. -I wonder if there is a general statement/theory of this. - -REPLY [6 votes]: Your problem can be recast into the language of factor graphs as follows: you have two factor graphs $G_1$, $G_2$ for the same function that are both comprised of two isolated edges, with vertex sets the partitions you indicate. These two factor graphs must have a common refinement, i.e., there must be a common factor graph $G_{12}$ and graph morphisms $g_j : G_{12} \rightarrow G_j$. -In your example, the only way that can happen is if the vertex set of $G_{12}$ is $\{1,2,3,4\}$, i.e., if $f$ factorizes completely.<|endoftext|> -TITLE: What is the automorphism group of the projective line minus $n$ points? -QUESTION [11 upvotes]: $\DeclareMathOperator{\AGL}{\operatorname{AGL}}\DeclareMathOperator{\PGL}{\operatorname{PGL}}$What is the automorphism group of $\mathbb P^1$ minus $n$ points (let's say over an algebraically closed field of characteristic $0$ if it matters). I want to consider the removed points without order. I can do small cases by hand but it seems hard in general and it seems to depend on which points are removed. -Here's what I have thought about so far: - -$n = 0$: The automorphism group of $\mathbb P^1$ is $\PGL_2(k)$ -$n = 1$: The automorphism group of $\mathbb A^1$ is $\AGL(1)$. -$n = 2$: The automorphism group of $\mathbb G_m$ is $\mathbb Z/2 \ltimes k^\times$. -$n = 3$: Since $\PGL_2$ acts three transitively, it doesn't matter which points we remove. Any automorphism of $\mathbb P^1 - \{0,1,\infty\}$ will extend to an automorphism of $\mathbb P^1$ fixing $\{0,1,\infty\}$ as a set and is determined by what it does to this set. We get all of $S_3$ in this case. -$n = 4$: Any automorphism has to preserve the cross ratio and every permutation that does so is obtainable. So we get the Klein $4$ group - $\mathbb Z/2 \times \mathbb Z/2$ in the generic case. - -I don't know what happens for $n \geq 5$. We can get non trivial automorphisms for large $n$ by doing the following: Pick a finite subgroup of $\PGL_2(k)$ (these are classified) and pick any finite subset of $\mathbb P^1$ and remove the entire orbit of this set by the finite subgroup. -Also generically, I believe there are no automorphisms for $n\geq 5$ by the following argument: We require all $4$ element subsets to have distinct cross ratios and since automorphisms have to preserve the cross ratio, this means that all $4$ element subsets are preserved. But this implies that the automorphism is trivial. - -REPLY [11 votes]: For $n \geq 5$, we can describe the locus of configurations that have nontrivial automorphisms. To do this, note that if there is any nontrivial automorphism, there is an automorphism of order $p$ for some prime $p$. Such an automorphism acts on $\mathbb P^1$ with two fixed points and the remaining points orbits of size $p$. -So the automorphism restricts to $\mathbb P^1$ minus $n$ points if and only if $n=ap+b$ for some $a \in \mathbb N$ and $b \in \{0,1,2\}$, and those $n$ points consist of $b$ of the fixed points as well as $a$ orbits of size $p$. -The space of such configurations has dimension $a$ for any given automorphism, hence $a+2$ in total, which is at most $\frac{n}{p}+ 2 \leq \frac{n}{2} + 2 -TITLE: Canonical multiplication representation of self-adjoint operator in quantum chemistry and coding theory research -QUESTION [5 upvotes]: In my applied math research group, we are studying and going over functional analysis results from papers and theses from our institution to generalize their results and apply them in our discrete dynamics in quantum chemistry and coding theory research. Right now, we are dealing with self-adjoint operators in the context of the spectral theorem's many forms. One form, the multiplication operator form, says - -Let $A$ be a bounded, self-adjoint operator on a separable Hilbert space $H$. Then, there exist measures $\{\mu_n\}_{n=1}^{N}$ (where $N$ is a natural number or infinity) on $\sigma(A)$ and a unitary operator -$$ -U : H \rightarrow \bigoplus_{n=1}^N L^2(\mathbb{R},d\mu_n) -$$ -and we have -$$ -(UAU^{-1}\psi)_n(\lambda) = \lambda \psi_n(\lambda) -$$ -where we write an element $ \psi \in \bigoplus_{n=1}^N L^2(\mathbb{R},d\mu_n) $ is written as an N-tuple $(\psi_1(\lambda),\psi_2(\lambda),...,\psi_N(\lambda))$. If we do not insist on the function we multiply by to be $f(\lambda)=\lambda$, we have that $A$ is unitarily equivalent to the multiplication operator $ M_F$ on $L^2(M,d\mu) $ which multiplies by the function $F(\lambda)$. That is the background we are using. - -Here is my problem. We consider the operator $A=L+R$ on $\ell^2(\mathbb{Z})$ which is the sum of the left and right shift operators on square-summable sequences indexed by all integers. I know, via simple Fourier series, that $A$ is unitarily equivalent to $M_{2\cos(x)}$ on $L^2([0,2\pi),dx/2\pi)$. Here is what I am missing and need. We take $(Bf)(x)=xf(x) \; \; \text{on} \; \; L^2([-2,2],dx)$. The old Russian thesis I am working on says that $A$ is unitarily equivalent to $ B \oplus B $ on $L^2([-2,2],dx)=H_1 \oplus H_2$, and it defers the proof of this to the appendix which, as luck would have it, seems to be missing (I checked all databases I know of it is not there, it is quite old so maybe not digitized and uploaded to any database). This result is really important to my research in dynamics in quantum chemistry and coding theory where $A$ shows up a lot, and having the decomposition of $L^2([-2,2],dx)$ as stated above into two disjoint invariant subspaces of $B$ and actually finding the unitary $U$ that takes $ A $ to $ B \oplus B $ (or vice versa) can really boost my research, but I am just now entering functional analysis as a user/researcher whose background is mainly in applied math. I was hoping that someone here can help me find $U$ such that $UAU^*=B \oplus B$. Unfortunately, I am not so proficient in functional analysis and spectral theory to do anything that is not mostly intuitive. I would also love to share more about the context of our research if people are curious. I thank all helpers. - -REPLY [4 votes]: If you think about it, then on $[0,2\pi)$, the function $2\cos(t)$ takes all the values in $[-2,2]$, with multiplicity $2$ (except for $\pm 2$ which have multiplicity $1$). So the claim seems very plausible. The exercise now is to pick the correct unitary, which is basically a "change of variables" problem. -The unitary you want (assuming my rusty Calculus is okay) is -$$ U : L^2([0,2\pi), dt/2\pi) \rightarrow L^2([-2,2],dx) \oplus L^2([-2,2],dx); -\quad \xi\mapsto (f,g), $$ -where -\begin{align*} -f(2\cos(t)) &= \xi(t) / \sqrt{4\pi\sin(t)} \quad (0\leq t\leq \pi), \\ -g(2\cos(t)) &= \xi(t) / \sqrt{-4\pi\sin(t)} \quad (\pi < t < 2\pi) -\end{align*} -Notice that $g(\pm 2)$ is not defined, but we are in an $L^2$ space, so it's okay to not define a function at some isolated points. Similarly, where $\sin(t)=0$ the functions are not defined. -Similarly let $U(\eta) = (h,k)$. Then -\begin{align*} -(U^*(B\oplus B)U\xi|\eta) -= \int_{-2}^2 x f(x) \overline{h(x)} \ dx -+ \int_{-2}^2 x g(x) \overline{k(x)} \ dx -\end{align*} -Change variables to see that this equals -\begin{align*} -\int_{\pi}^0 2\cos(t) f(2\cos(t)) \overline{h(2\cos(t))} (-2)\sin(t) \ dt \\ -+ \int_\pi^{2\pi} 2\cos(t) g(2\cos(t)) \overline{k(2\cos(t))} (-2)\sin(t) \ dt. -\end{align*} -Now substitute in, and see why the strange $\sin$ parts occurred before, -\begin{align*} -\int_0^\pi 2\cos(t) \frac{\xi(t)}{\sqrt{4\pi\sin(t)}} \frac{\overline{\eta(t)}}{\sqrt{4\pi\sin(t)}} - 2\sin(t) \ dt \\ -+ \int_\pi^{2\pi} 2\cos(t) \frac{\xi(t)}{\sqrt{-4\pi\sin(t)}} \frac{\overline{\eta(t)}}{\sqrt{-4\pi\sin(t)}} - (-2)\sin(t) \ dt -\end{align*} -This cancels down to -$$ \int_0^{2\pi} 2\cos(t) \xi(t) \overline{\eta(t)} \ \frac{dt}{2\pi}, $$ -as we want. -Edit: Here I have written $(\cdot|\cdot)$ for the inner-product. (A sort of mix between maths and physics notation; but I often write papers which need to consider both bilinear, and sesquilinear, pairings, and it's nice to have a notational difference between these). $\xi,\eta$ are members of $L^2([0,2\pi), dt/2\pi)$ (so functions, or equivalence classes thereof). -A similar calculation shows that $U$ is a unitary. More accurately, repeat the calculation without the operator $B\oplus B$ to show that $U^*U=1$. To see that $UU^*=1$, equivalently, $U$ has dense range, one could argue that "this is clear"; or you could compute $U^*$ (use change of variable again) which has a similar form.<|endoftext|> -TITLE: Is the factorization of $a_m-a_n$ affected by the fact that $\Sigma \frac{1}{a_k}<+\infty$? -QUESTION [5 upvotes]: I would like to ask the following. - -Let $(a_n)$ be a sequence of natural numbers such that -$\sum_{k=1}^{\infty}\frac{1}{a_k}$ converges. Is it true that for -infinitely many $m$, there is a $n -TITLE: From Vitali to Heine-Borel in reverse mathematics -QUESTION [5 upvotes]: The Vitali and Heine-Borel covering theorems are house-hold names of analysis, and rightly well-studied in reverse mathematics. As shown in Simpson's excellent monograph [1], for countable coverings of the unit interval, the Heine-Borel theorem is equivalent to WKL (weak Koenig's lemma), while the Vitali covering theorem is equivalent to WWKL (weak weak Koenig's lemma). The theorem numbers in [1] are IV.1.2 and X.1.13. -My question is then as follows: -Is there a natural statement X such that [WWKL +X ] $\leftrightarrow$ WKL, say over RCA$_0$? -Here, $X$ should be weaker than WKL, obviously. Results in related frameworks (computability theory, Weihrauch reducibility, constructive math, ...) are also welcome. -PS: I am asking this question because in the case of uncountable coverings, such an X does exist. -[1] Stephen G. Simpson, Subsystems of second order arithmetic, 2nd ed., Perspectives in Logic, Cambridge University Press, 2009. - -REPLY [2 votes]: In the Weihrauch reducibility framework, my hunch is that the answer is no. Of course, "natural statement" does not lend itself to easily disprove existence, so I can't rule out changing my mind in the future. -First, we are looking at principles below $\mathrm{WKL}$ which are incomparable with $\mathrm{WWKL}$. To my knowledge, the only such principles which were studied in the literature are variants of convex choice $\mathrm{XC}_{[0,1]^n}$. All these variants are themselves below the principle $\mathrm{XC}_{[0,1]}^\diamond$, which lets us invoke choice for subsets of the unit interval finitely many times (with subsequent queries depending on previous answers). Note that $\mathrm{XC}_{[0,1]}$ is equivalent to the intermediate value theorem. -However, $(\mathrm{XC}_{[0,1]} \sqcup \mathrm{WWKL})^\diamond <_{\mathrm{W}} \mathrm{WKL}$ (the left hand side is "make finitely many calls to both $\mathrm{XC}_{[0,1]}$ and to $\mathrm{WWKL}$). The reason for this is that on a computable input, $\mathrm{XC}_{[0,1]}$ can always return something computable, and $\mathrm{WWKL}$ can always return some ML-random. Thus, the left-hand side can always return something which is computable from a ML-random, but this doesn't hold for $\mathrm{WKL}$, as $\mathrm{PA}$-degrees are not computable from ML-randoms. -Another piece of evidence is that there is multivalued function $g$ at all such that $\mathrm{WKL} \leq_{\mathrm{W}} \mathrm{WWKL} \circ g$. On the right, we have that every input has a positive measure of potential solutions, and $\mathrm{WKL}$ has instances where the Turing-upper cone of solutions is measure $0$.<|endoftext|> -TITLE: Motivation for the Kazhdan-Lusztig involution -QUESTION [8 upvotes]: I would like to know about the motivation behind the Kazhdan–Lusztig involution on an Iwahori–Hecke algebra. -I'll borrow the conventions from Libedinsky's Gentle introduction to Soergel bimodules I: The basics. The Iwahori–Hecke algebra $\mathcal{H}$ of a Coxeter system $(W,S)$ is the $\mathbb{Z}[v,v^{-1}]$-algebra with generators $h_s$ for $s\in S$ and relations - -$h_s^2 = (v^{-1}-v)h_s +1$ for all $s\in S$ -$\underbrace{h_sh_rh_s\cdots}_{m_{rs}} = \underbrace{h_rh_sh_r\cdots}_{m_{rs}}$ for all $s,r\in S$. - -The Kazhdan–Lustig involution is the $\mathbb{Z}$-algebra involution $d\colon \mathcal{H}\ -\to\mathcal{H}$, defined by $d(h_s)=h_s^{-1}=h_s+v-v^{-1}$ and $d(v)=v^{-1}$. - -Question 1: What is the motivation for considering the Kazhdan–Lusztig involution? Is there a motivation that is intrinsic to Coxeter systems and Iwahori–Hecke algebras? - -Let me clarify my aim: I know that this involution leads to the definition of the Kazhdan–Lusztig basis, and that many representation-theoretical wonders ensue. But I am a topologist by nature and I can't claim to fully appreciate these applications. But I am familiar with Coxeter groups and Iwahori–Hecke algebras. So I am looking for some motivation, if it can be given, on the level that I do understand! Perhaps a sub-question might help: - -Question 2: Let $\mathcal{A}\subseteq \mathcal{H}$ denote the $\mathbb{Z}$-submodule fixed by the Kashdan-Lusztig involution. This is a $\mathbb{Z}[v+v^{-1}]$-algebra with basis given by the Kazhdan–Lusztig basis. What is known about $\mathcal{A}$? - -REPLY [2 votes]: I'm mostly a combinatorialist who doesn't completely understand this stuff, so I may have something slightly wrong, but... -When $W$ is a Weyl group, the Kazhdan--Lusztig involution is (the $K$-theoretic image of) Verdier duality on the bounded derived category of constructible $B$-equivariant sheaves on the flag variety $G/B$. -The references that will have this right are Springer's Quelques applications de la cohomologie d'intersection and Reitsch's An introduction to perverse sheaves.<|endoftext|> -TITLE: Spectrum of a ring (studied by Krull?) of rational functions -QUESTION [6 upvotes]: Let $k$ be an algebraically closed field and $\mathbb A^2_k=\operatorname {Spec}k[x,y]$ the affine plane over $k$. -Consider the ring $R \subset k(x,y)$ of the rational functions on the plane defined and constant on $V(x)$ (the $y$-axis $x=0$). -What is $\operatorname {Spec}R$ ? -(This is the geometric translation of an example due, I think, to Krull for which I unfortunately have no reference.) -Edit -Sorry,my definition of $R$ above is a bit ambiguous. -What I mean is that $R$ consists of those fractions $r(x,y)=\frac {p(x,y)}{q(x,y)}$ which can be written as the quotient of two polynomials $p(x,y),q(x,y)\in k[x,y]$ such that $q(0,y)\neq 0\in k[y]$ and $\frac {p(0,y)}{q(0,y)}\in k\subset k[y]$. -For example the rational function $\frac {y+x}{y-x}$ mentioned by @YCor in the comments does belong to $R$ since $y-0\neq0\in k[y]$ and $\frac {y+0}{y-0}=1\in k \subset k[y]$ - -REPLY [3 votes]: I will join the party if you don't mind. -Let $A=k[x,y]_{(x)}$. This is the ring of fractions $p/q$ such that $q$ is not divisible by $x$. Our ring $R$ can be written as $R=k+x A$. -Inverting $x$ produces $R[x^{-1}] = A[x^{-1}]=k(x,y)$. The latter is a field with a unique prime ideal $(0)$. Thus the only prime ideal not containing $x$ is $(0)$. -The radical of $(x)$ is $x A$ because $(x A)^2 \subset x ( x A) \subset x R$ implies $xA\subset \sqrt{(x)}$ and $x A$ is maximal. So the only prime ideal containing $x$ is $x A$. -Any prime ideal either contains $x$ or it doesn't, so $(0), xA$ is the complete list.<|endoftext|> -TITLE: Minimal number of (Dehn twists) generators of the mapping class group of a marked sphere -QUESTION [5 upvotes]: Let $\Gamma_{g,n}$ denote the mapping class group of an oriented surface of genus $g$ and with $n$ marked points. We assume that elements of $\Gamma_{g,n}$ are not allowed to permute the marked points. I am interested in the case $g=0$. -In Farb & Margalit, on page 114, it is claimed that $\Gamma_{g,n}$ can be generated by $2g+n$ Dehn twists along the curves drawn on Figure 4.10. I was wondering if this statement is also true in the case $g=0$. -In math/9912248, Wajnryb exhibits a family of generators for $\Gamma_{0,n}$ in Lemma 23. The curves $\alpha_{i,j}$ illustrated on Figure 12 are a family of $n(n-1)/2$ generators. -My questions are the following: - -Is the minimal number of Dehn twists generators of $\Gamma_{0,n}$ known ? -What would be the $n$ generators of $\Gamma_{0,n}$ if the construction of Farb-Margalit cited above applies in the case $g=0$ ? - -REPLY [8 votes]: The minimum number of Dehn twist generators (and in fact the minimum number of generators of any kind) for $\Gamma_{0,n}$ is ${n-1 \choose 2} - 1$. Here's why. -A presentation for $\Gamma_{0,n}$ is known, and can be found in Lemma 4.1 of this paper by Rebecca R. Winarski and myself. In the paper, $\operatorname{PMod}(\Sigma_0,\mathcal B(n))$ is the group $\Gamma_{0,n}$. -Number the $n$ marked points. The generators $A_{i,j}$ are Dehn twists about curves that surround only the $i$th and $j$th marked points (see Figure 3 from the paper). These are essentially the same curves in the paper by Wajnryb that is linked in the question. -The generating set used in our paper is the set $\{A_{i,j} \mid 1 \leq i < j \leq n-1\}$, and one of the relations (relation (5)) is $$(A_{1,2}A_{1,3} \cdots A_{1,n-1})\cdots(A_{n-3,n-2}A_{n-3,n-1})(A_{n-2,n-1}) = 1. $$ In this relation, each $A_{i,j}$ appears exactly once, so you can use a Tietze transformation to eliminate one of the generators. We are now left with a generating set consisting of ${{n-1}\choose{2}} - 1$ Dehn twists. -The other 4 relations are all commutation relations (that is, of the form $[W,X] = 1$), so we can conclude that the abelianization of $\Gamma_{0,n}$ is a free abelian group of rank ${n-1 \choose 2} - 1$. Therefore $\Gamma_{0,n}$ cannot be generated by less than ${n-1 \choose 2} - 1$ elements. -I guess this also answers your question 2, in the sense that the result stated in Farb & Margalit does not hold for genus 0.<|endoftext|> -TITLE: Was Cantor aware of Lebesgue theory of integration? -QUESTION [14 upvotes]: Georg Cantor died in 1919, more than ten years after appearance of the Lebesgue theory of measure and integration at the beginning of the twentieth century. Lebesgue theory has a deep connection with Cantor's theory of sets, for instance one of first Lebesgue's contributions after his thesis was about Fourier series, which is one of motivations of Cantor in developing theory of sets. It seems interesting to know about any (possible) reaction of Cantor to the measure and integration theory of Lebesgue. -Added in Edit: It seems that there are at-least some correspondence. The following quote is from a letter of Lebesgue to Borel, in February 17, 1904, where he talks about an unpublished publication of him, (see p. 52 in Lettres d’Henri Lebesgue à Émile Borel, Cahiers du séminaire d’histoire des mathématiques, tome 12 (1991), p. 1 -506), Lebesgue writes that - -Vous pouvez envoyer a Fatou et G. Cantor et vous savez qu' il m' en restera tres probablement pendant quel que temps si vous avez l' idee d' une ou deux personnes. - -I do not claim that this prove anything (I even don't know whether Cantor received such document). Two days later, in another letter (ibid, p. 54), Lebesgue writes - -Cantor existe-t-il? - -REPLY [8 votes]: The following quote is from Joseph Dauben (in his article Georg Cantor: The Personal Matrix of His Mathematics, Isis, Vol. 69, No. 4, Dec., 1978, pp. 534-550). - -When he suddenly suffered his first breakdown in May 1884, Cantor had just -returned from an apparently successful, quite enjoyable trip to Paris. He had met a -number of French mathematicians, including Hermite, Picard, and Appell, and was -delighted to report to Gosta Mittag-Leffler that he had liked Poincare very much and -was happy to see that the Frenchman understood transfinite set theory and its -applications in functional analysis. - -It appears that Cantor did not keep up with the work of the later generation of French analysts. Recall that Lebesgue wrote his dissertation under the supervision of Borel, and Borel was born in 1871, so Borel was only 13 years old at the time of Cantor's visit. One key reason for this might be that after his first breakdown (1884), Cantor's range of interests were widely expanded to many other domains, as indicated by the following excerpt from Dauben's aforementioned paper: - -He began to emphasize other interests. The amount of time he devoted to -various literary and historical problems steadily increased, and he read the history -and documents of the Elizabethans with great attentiveness in hopes of proving that -Francis Bacon was the true author of Shakespeare's plays. As time progressed, he also began to intensify his study of the Scriptures and of the church fathers, and he -developed new interests in Freemasonry, Rosicrucianism, and Theosophy.<|endoftext|> -TITLE: Is there a notion of Čech groupoid of a cover of an object in a Grothendieck site? -QUESTION [5 upvotes]: Given a topological space $X$, and a cover $\mathcal{U} :=\cup_{\alpha \in I}U_{\alpha}$ of $X$, one can define a groupoid called Čech groupoid $C(\mathcal{U})$ of the cover $\mathcal{U}$ by $\sqcup_{i,j \in I} U_i \cap U_j \rightrightarrows \sqcup_{i \in I} U_i$ whose structure maps are obvious to define. -Now given a site $(C,J)$ and an object $c \in C$, one has a cover $J_c$ of $c$ induced from $J$. -My question: -Is there an analogous notion of Čech Groupoid corresponding to $J_c$? Or the investigation in this direction may not be fruitful? -I will also be very grateful if someone can provide some literature references regarding these. -Thanks in advance. - -REPLY [9 votes]: Take $U=\coprod_{i∈I}Y(U_i)$, where $Y\colon C\to\mathop{\rm Presh}(C,{\rm Set})$ is the Yoneda embedding. -We have a canonical morphism $U→Y(X)$. -The Čech groupoid of $J_c$ can now be defined as -the groupoid with objects $U$ and morphisms $U⨯_{Y(X)}U$, -with source, target, composition, and identity maps defined in the usual manner. -In the case of a site coming from a topological space, this construction -recovers the usual Čech groupoid. -In fact, iterating fiber products produces a simplicial presheaf, -namely, the Čech nerve of $J_c$, which is used to define Čech descent -for simplicial presheaves.<|endoftext|> -TITLE: Shrinking subset and product -QUESTION [7 upvotes]: Given a segment and a value $c$ less than the segment length, let $A_1,\dots,A_n$ be finite unions of intervals on the segment. We choose a finite union of intervals $B$ with $|B|=c$ that maximizes $|B\cap A_1|\times\dots\times |B\cap A_n|$, where $|\cdot|$ denotes the length (i.e. Lebesgue measure). If there are many such $B$, we choose one arbitrarily. -Now, we shrink $A_1$ to $A_1'\subseteq A_1$, and choose $B'$ using the same procedure. Is it always true that $|B'\cap A_1'|\le |B\cap A_1|$? -If $A_1,\dots,A_n$ are disjoint finite unions, the answer is positive, as shown here. - -REPLY [2 votes]: OK, it seems that this is a counterexample. -Take 8 disjoint segments $I_1,\dots,I_8$ of length 1. Take 8 sets -$$ - S_1=\{1,2,3\}, \quad - S_1=\{4,5,6\}, \quad - S_1=\{1,2,4\}, \quad - S_1=\{1,2,5\}, \quad - S_1=\{1,2,6\}, \quad - S_1=\{1,3,4\}, \quad - S_1=\{1,3,5\}, \quad - S_1=\{1,3,6\}. -$$ -Say that $I_i$ lies in $A_j$ iff $j\in S_i$, otherwise $I_i$ and $A_j$ are disjoint. Finally, set $c=2$. -In this situation, the optimal $B$ is $I_1\cup I_2$. where the product equals $1$. This follows from the fact that $\sum_j |B\cap A_j|\leq 6$, since any point is covered by at most three of the $A_j$. -Now set $A_1’=A_1\setminus I_1$. Consider the quantities -$$ - x=\left|B’\cap\left(\bigcup_{i=3}^8 I_i\right)\right|, - \quad - y=|B’\cap I_2|. -$$ -Then, by AM—GM, -$$ - |B’\cap A_1’|\leq x, \quad - \prod_{j=2}^3 |B’\cap A_j|\leq (x/2+2-x-y)^2, \quad - \prod_{j=4}^6 |B’\cap A_j|\leq (x/3+y)^3, -$$ -and the equalities are achievable simultaneously. Hence, in the optimal case, we have -$$ - \prod_{j=1}^6|B’\cap A_j|=x(2-x/2-y)^2(x/3+y)^3 - =\frac{6x\cdot (36-9x-18y)^2\cdot (4x+12y)^3}{6\cdot 18^2\cdot 12^3}. -$$ -So we seek for a maximizer $(x_0,y_0)$ of -$$ - f(x,y)= 6x\cdot (36-9x-18y)^2\cdot (4x+12y)^3 -$$ -under the conditions $x.y\geq 0$, $x+y\leq 2$. We claim that such a maximizer has $x_0\geq 24/17$, which provides $|B’\cap A_1’|>|B\cap A_1|=1$, as desired. -Indeed, we have -$$ - f\left(\frac{24}{17},\frac{10}{17}\right) - =\frac{144}{17}\cdot \left(\frac{216}{17}\right)^5. -$$ -On the other hand, if $x\leq 24/17$, by AM—GM we have -$$ - f(x,y)\leq 6x\cdot\left(\frac{2(36-9x-18y)+3(4x+12y)}5\right)^5 - =6x\cdot\left(\frac{72-6x}5\right)^5; -$$ -the right hand part is an increasing function for $0\leq x\leq 2$, so -$$ - f(x,y)\leq 6\cdot \frac{24}{17}\cdot\left(\frac{72-6\cdot 24/17}5\right)^5 - = f\left(\frac{24}{17},\frac{10}{17}\right), -$$ -as desired.<|endoftext|> -TITLE: Maximal ideals of ultraproducts of full matrix algebras -QUESTION [12 upvotes]: Let $\mathscr U$ be a non-principal ultrafilter over the natural numbers. Let $M_{\mathscr U}$ be the ultraproduct of all full matrix algebras $M_n$ along $\mathscr U$. This is a C*-algebra that is not simple as it contains a non-zero proper ideal, for example $\{[(x_n)]\colon \lim_{n, \mathscr U} \|x_n\|_{\rm HS} = 0\}$, where $\|\cdot\|_{\rm HS}$ stands for the Hilbert–Schmidt norm. - -Is the cardinality of the set of maximal ideals of $M_{\mathscr U}$ known? -Does $M_{\mathscr U}$ have an ideal of finite-codimension? - -I anticipate that for Q1 the answer should be $2^{\mathfrak{c}}$ and for Q2 it should be no but I am somehow stuck. - -REPLY [17 votes]: I think Nik Weaver is right that the ideal mentioned is the unique maximal ideal. -This simultaneously answers both questions (since the quotient is clearly infinite dimensional). Let $\tau$ be the trace on $M_\mathcal{U}$ defined as $\tau(x_n)=\lim_{n\rightarrow \mathcal{U}}\tau_n(x_n)$ where $\tau_n$ is the normalized trace on $M_n.$ As Nik Weaver mentioned in the comments, the ideal $\{ x\in M_\mathcal{U}:\tau(x^*x)=0 \}$ is maximal. I claim this is the only maximal ideal. First we need a lemma from linear algebra -Claim: Let $a\in M_N$ be positive norm 1 and set $\varepsilon=\tau_N(A)>0.$ Then there are -$k=\frac{2}{\varepsilon}$ partial isometries $v_1,...,v_k\in M_N$ such that $\sum v_i^*av_i\geq \frac{\varepsilon}{2}I.$ -Proof of Claim: Order the eigenvalues of $a$ as $a_1\geq a_2\geq\cdots \geq a_N$ and if $a_{\frac{N\varepsilon}{2}+1}<\frac{\varepsilon}{2}$ the trace is strictly less than $\varepsilon$ hence $a_i\geq \frac{\varepsilon}{2}$ for $1\leq i\leq \frac{N\varepsilon}{2}.$ Let $v_1$ project onto the $\frac{N\varepsilon}{2}$-dimensional subspace spanned by the first $\frac{N\varepsilon}{2}$ eigenvectors (corresponding to the ordering of the eigenvalues $a_i$). Then twist this projection down the line with appropriate partial isometries to obtain the claim. -Back to the Answer: Let $I$ be an ideal that contains a positive, norm 1 element $x=(x_n)$ such that $\tau(x)>0.$ We will show that the ideal generated by $x$ contains the identity. By replacing $(x_n)$ with an equivalent sequence we can assume each $x_n$ is positive, norm 1 and has $\tau_n(x_n)\geq\varepsilon$ for some $\varepsilon>0.$ Now just apply the above claim coordinate wise to produce $k=\frac{2}{\varepsilon}$ partial isometries $w_1,...,w_k\in M_\mathcal{U}$ so $\sum w_ixw_i^*\geq \frac{\varepsilon}{2}I.$<|endoftext|> -TITLE: Is being simply connected very rare? -QUESTION [27 upvotes]: Essentially, my question is how strong a restriction it is to be simply connected. -Here is a way of making this precise: Let's say we want to count simplicial complexes (of dimension 2, though that does not matter much, any fixed dimension is fine) on N simplices that are subject to the following restrictions: -A: every vertex is contained only in a bounded number of simplices (say, 10000). -B: the complex is simply connected. -So properly: How many distinct complexes like this are there? In fact, I only want a rough answer: is it exponential in N, or is it superexponential. Note that if I remove either restriction, the answer is superexponential. - -REPLY [6 votes]: The following does not answer your question, but adding just in case it is helpful. -If you weaken "simply connected" to $H_1(\Delta, \mathbb{Q}) = 0$, and weaken "every vertex is in a bounded number of edges" to "the average number of triangles containing an edge is bounded", then there are super-exponentially many such complexes. This is an old result of Kalai, on enumerating $\mathbb{Q}$-acyclic complexes. https://link.springer.com/article/10.1007/BF02804017 -There are $\exp( cn^2)$ such complexes, where $c>0$ is some constant not depending on $n$. -For the latest on enumerating $\mathbb{Q}$-complexes, see Linial and Peled: -https://onlinelibrary.wiley.com/doi/abs/10.1002/rsa.20841 -Andrew Newman and I recently showed that a typical $\mathbb{Q}$-acyclic complex (according to a natural `determinantal measure') is not simply connected. With high probability, the fundamental group is a hyperbolic group of cohomological dimension 2. -https://arxiv.org/abs/2004.13572<|endoftext|> -TITLE: Square root of doubly positive symmetric matrices -QUESTION [15 upvotes]: I wonder whether the following property holds true: For every real symmetric matrix $S$, which is positive in both senses: -$$\forall x\in{\mathbb R}^n,\,x^TSx\ge0,\qquad\forall 1\le i,j\le n,\,s_{ij}\ge0,$$ -then $\sqrt S$ (the unique square root among positive semi-definite symmetric matrices) is positive in both senses too. In other words, it is entrywise non-negative. -At least, this is true if $n=2$. By continuity of $S\mapsto\sqrt S$, we may assume that $S$ is positive definite. Denoting -$$\sqrt S=\begin{pmatrix} a & b \\ b & c \end{pmatrix},$$ -we do have $a,c>0$. Because $s_{12}=b(a+c)$ is $\ge0$, we infer $b\ge0$. - -REPLY [31 votes]: Robert Bryant already showed via an example that the answer is "no". To come up with lots of counterexamples, recall that (under some mild assumptions) if $A$ has maximal eigenvalue $\lambda_{\text{max}}$ and corresponding unit eigenvector $\mathbf{v}$ then $(A/\lambda_{\text{max}})^k \rightarrow \mathbf{v}\mathbf{v}^*$ as $k \rightarrow \infty$. -So if you pick any matrix that is (a) positive semidefinite with a negative entry, and (b) has a unique maximal eigenvalue with corresponding entrywise positive eigenvector, then repeatedly squaring $A$ will eventually give a counterexample to the original question. - -REPLY [26 votes]: No. If $$A = \begin{pmatrix}10&-1&5\\-1&10&5\\5&5&10\end{pmatrix},$$ then $A$ is positive definite but does not have all entries positive, while -$$ -A^2 = \begin{pmatrix}126&5&95\\5&126&95\\95&95&150\end{pmatrix} -$$ -is positive in both senses.<|endoftext|> -TITLE: Blow-up of projective variety $P^1 \times P^1..... \times P^1$ ($n$ times) and blow-up of $P^n$ -QUESTION [9 upvotes]: It is known that the blow-up of $P^1 \times P^1$ at a point is isomorphic to the blow-up of $P^2$ at two points. I'm wondering if there is any general statement for the blow-up of $P^1 \times P^1 \times.... \times P^1$ ($n$ times) and of $P^n$. - -REPLY [3 votes]: The blow-up of $(\mathbb{P}^1)^n$ at one point is always pseudo-isomorphic to the blow-up of $\mathbb{P}^n$ at $n$ points in general position. -To see this, let us consider, for each $n\ge 2$, the toric birational map $\tau\colon (\mathbb{P}^1)^n\dashrightarrow (\mathbb{P}^n)$ given by $$([x_1:1],[x_2:1],\ldots,[x_n:1])\mapsto [x_1:\cdots:x_n:1]$$ -and which restricts then to an isomorphism on the affine space $\mathbb{A}^n$. -You can write $\tau$ as -$$ ([x_1:y_1],[x_2:y_2],\ldots,[x_n:y_n])\mapsto [x_1y_2\cdots y_n:y_1x_2y_3\cdots y_n:\cdots:y_1\cdots y_{n-2}x_{n-1}y_n:y_1\cdots y_{n-1}x_n:y_1y_2\cdots y_n].$$ -and observe that it is the blow-up of the point $([1:0],\ldots,[1:0])$ followed by a pseudo-isomorphism (isomorphism if $n=2$, flop of three curves if $n=3$, ...) and then the contraction of the strict transforms of the hyperplanes given by $y_1=0,\ldots,y_n=0$ onto the toric points of $\mathbb{P}^n$ that are contained in the hyperplane at infinity. -You can also see the pseudo-isomorphism as blow-ups and blow-downs if you prefer (for instance, in dimension $3$ it is only $3$ disjoint Atiyah flops: blow-ups of three lines and contractions of three lines).<|endoftext|> -TITLE: Many numbers with pairwise differences squares -QUESTION [9 upvotes]: How many different natural numbers are there such that the difference of any two is a perfect square? - -I could find that with 'sums' instead of 'differences' this has been asked by Erdos and L. Moser (see Guy: Unsolved Problems in Number Theory, 3rd ed., p. 268), and that is open, with six being the current record. - -REPLY [8 votes]: Four is possible, an example is $0,451584,462400,485809$. -This solution comes from the example of a cuboid with integer sides, space diagonal and two out of three face diagonals given here. -I don't know for $5$ or more but it is probably a very hard question.<|endoftext|> -TITLE: Applications of symplectic geometry to classical mechanics -QUESTION [16 upvotes]: It is claimed that classical mechanics motivates introduction of symplectic manifolds. This is due to the theorem that the Hamiltonian flow preserves the symplectic form on the phase space. - -I am wondering whether symplectic geometry has applications to classical mechanics. Was this connection useful for classical mechanics? Were methods of symplectic geometry relevant for it via, say, the above theorem? - -REPLY [15 votes]: The list will be long, very long indeed. But to start: - -Questions about dynamics of Hamiltonian systems are at the heart of symplectic topology, symplectic capacities are precisely introduced for that purpose to understand the difference between a mere volume-preserving flow and a Hamiltonian flow. This includes questions about closed orbits etc. Here the books of Hofer-Zehner or McDuff-Salomon are a good start. - -Even if interested only in mechanics in $\mathbb{R}^{2n}$: as soon as it comes to symmetries (and mechanics deals a lot with symmetries) one inevitably ends up with concepts of phase space reduction. The reduced phase space of the isotropic harmonic oscillator (could there be something more relevant for mechanics ?) is $\mathbb{CP}^n$ with Fubini-Study Kähler structure. Quite a complicated geometry already. In classical textbooks you discuss the Kepler problem by fixing the conserved quantities (angular momentum, etc) to certain values. This is just a phase space reduction in disguise. The geometry becomes less-dimensional but more complicated by doing so. Coadjoint orbits are symplectic and needed for descriptions of symmetries in a similar fashion. Without geometric insight, their structure is hard to grasp, I guess. The aforementioned textbook of Abraham and Marsden as well as many others provide here a good first reading. In fact, up to some mild topological assumptions any symplectic manifold arises as reduced phase space from $\mathbb{R}^{2n}$ according to a theorem of Gotay and Tuynman. From that perspective, symplectic geometry is mechanics with symmetries. - -If trying to understand Hamilton-Jacobi theory, it is pretty hard to get anywhere without the geometric notion of a Lagrangean submanifold. This was perhaps one of the main motivations for Weinstein's Lagrangean creed. - -Mechanical systems with constraints require a good understanding of the geometry of the constraints. This brings you into the realm of symplectic geometry where coisotropic submanifolds (aka first class constraints in mechanics) are at home. - -When restricting the configuration space of a mechanics system (think of the rigid body) then you are actually talking about the cotangent bundle of the config space as (momentum) phase space. This is perhaps one of the very starting points where symplectic geometry takes of. - -Going beyond classical mechanics, one perhaps is interested in quantum mechanics: here symplectic geometry provides a very suitable platform to ask all kind of questions. It is the starting point to try geometric quantization, deformation quantization and alike. - -Maybe more exotic, but I really like that: integrable systems can have quite subtle and non-trivial monodromies. There is a very nice book (and many papers) of Cushman and Bates on this. The mechanical systems are really simple in the sense that you find them in all physics textbooks. But the geometry is hidden and highly non-trivial as it involves really a global point of view to uncover it. - -From a more practical point of view, non-holonomic mechanics is of great importance to all kind of engineering problems (robotics, cars, whatevery). Here a geometry point off view really help and is a large area of research. Also mechanical control theory is not only about fiddling around with ode's but there is a lot of (symplectic) geometry necessary to fully understand things. The textbooks of Bloch as well as Bullo and Lewis might give you a first hint why this is so. - -As a last nice application of (mostly linear) symplectic geometry one should not forget optics! This is of course not mechanics, but optics has a very interesting symplectic core, beautifully outlined in the textbook by Guillemin and Sternberg. - - -Well, I could go on, but the margin is to small to contain all the information, as usual ;) Of course, for many things one can just keep working in local coordinates and ignore the true geometric features. But one will miss a lot of things on the way. - -REPLY [5 votes]: There is a "symplectic structure" on the set of body motions. -During the years 1960–1970, Jean-Marie Souriau, proved that under very general assumptions, the set of all possible solutions of a classical mechanical system, involving material points interacting by very general forces, has a smooth manifold structure (not always Hausdorff) and is endowed with a natural symplectic form. He called it the manifold of motions of the mechanic -J.-M. Souriau, Structure des systèmes dynamiques, Dunod, Paris, 1969. -J.-M. Souriau, La structure symplectique de la mécanique décrite par Lagrange en 1811, Mathématiques et sciences humaines, tome 94 (1986), pages 45–54. Numérisé par Numdam, http://www.numdam.org. -https://www.google.com/url?sa=t&source=web&rct=j&url=http://marle.perso.math.cnrs.fr/diaporamas/ManifoldMotionsMass.pdf&ved=2ahUKEwjjt5SWz8LsAhXKMewKHfg-D5wQFjAGegQIBxAB&usg=AOvVaw00m-2HERW5vWi10i1m3_hd<|endoftext|> -TITLE: Existence of eigenvalues in a k-linear abelian category -QUESTION [7 upvotes]: I cannot find any categorical definition of an eigenvalue, so I ask this question. Let $\mathbb{k}$ a be a field and $\mathcal{C}$ be a $\mathbb{k}$-linear abelian category. Let $f: X \rightarrow X \in \mathrm{End}_\mathcal{C}(X)$. To me, it makes sense to call $\lambda \in \mathbb{k}$ an eigenvalue of $f$ if $\ker(f - \lambda 1_X)$ is nonzero (and call this the corresponding eigenspace). By considering pullbacks one can show that these kernels do not "intersect" either for different $\lambda$. -If this indeed is the accepted definition, what are some reasonable set of conditions so that any such $f$ always has an eigenvalue (for instance, algebraic closedness of $\mathbb{k}$ will probably be necessary and some finiteness assumption)? -The greater context for such a question is from trying to prove categorical Schur's lemma for a tensor category, where any endomorphism of a simple object is a scalar multiple of the identity. And a similar statement about an endomorphism of an indecomposable being of the form $\lambda 1_X + n$, where $n$ is nilpotent. - -REPLY [8 votes]: Schur's lemma has the same proof in a $k$-linear abelian category $C$ as usual: if $T : M \to M$ is a nonzero endomorphism of a simple object, by simplicity it must have trivial kernel and cokernel, so is an isomorphism. Hence $\text{End}(M)$ is a division algebra over $k$. If furthermore $k$ is algebraically closed and $\text{End}(M)$ is finite-dimensional (e.g. if $C$ has finite-dimensional homsets) then $\text{End}(M) = k$. -Similarly if $k$ is algebraically closed and $\text{End}(M)$ is finite-dimensional then every endomorphism $T : M \to M$ has at least one eigenvalue (if $M$ is nonzero), because the natural map -$$k[x] \ni f(x) \mapsto f(T) \in \text{End}(M)$$ -has nontrivial kernel (generated by the minimal polynomial of $T$). Working a little more carefully to check that all the details still work as usual without elements: if $m(t) = \prod (t - \lambda_i)^{m_i}$ is the minimal polynomial of $T$, then $m(T) = 0$ implies that (if $M \neq 0$) at least one of the factors $(T - \lambda_i)^{m_i}$ is not a monomorphism, hence has nontrivial kernel. -As for the indecomposable case, with the same hypotheses as above $M$ is naturally a module over $k[x]/m(x) \cong \prod k[x]/(x - \lambda_i)^{m_i}$. The primitive idempotents of this product split $M$ into the direct sum of generalized eigenspaces of $T$ (this is a general feature of idempotent endomorphisms in abelian categories and also does not require elements), so if $M$ is indecomposable then $T$ has exactly one eigenvalue $\lambda$ and $T - \lambda$ is nilpotent as usual.<|endoftext|> -TITLE: Integrality of a binomial sum -QUESTION [11 upvotes]: The following sequence appears to be always an integer, experimentally. - -QUESTION. Let $n\in\mathbb{Z}^{+}$. Are these indeed integers? -$$\sum_{k=1}^n\frac{(4k - 1)4^{2k - 1}\binom{2n}n^2}{k^2\binom{2k}k^2}.$$ - -POSTSCRIPT. After Carlo's cute response and several useful comments, I like to ask this: is there a combinatorial proof? - -REPLY [3 votes]: There is a way to prove Zhi-Wei Sun's identity as well as Carlo Beenakker's identity. Of course, both can be treated in accord with Fedor Petrov's induction. Let's focus on Sun's identity. Divide through by $\binom{2n}n\binom{3n}n$ to write -$$A_n:=\sum_{k=1}^n\frac{(9k-2)27^{k-1}}{k^2\binom{2k}k\binom{3k}k}=\frac{27^n}{3\binom{2n}n\binom{3n}n}-\frac13. \tag1$$ -so that -$$A_n-A_{n-1}=\frac{(9n-2)27^{n-1}}{n^2\binom{2n}n\binom{3n}n}.$$ -Let $a_n=\binom{2n}n\binom{3n}nA_n$ (which is exactly Sun's LHS) to get the recursive equation -$$n^2a_n-3(3n-1)(3n-2)a_{n-1}=(9n-2)27^{n-1}.\tag2$$ -First, we find a solution to the homogeneous equation $n^2a_n-3(3n-1)(3n-2)a_{n-1}=0$ as follows -$$a_n^{(h)}=\binom{2n}n\binom{3n}n. \tag4$$ -A particular solution to the non-homogeneous equation (2) can be determined by mimicking the RHS as $a_n^{(p)}=(bn+c)27^n$. Now, plug this back in (2) to solve for $b$ and $c$: -\begin{align*} n^2(bn+c)27^n-3(3n-1)(3n-2)(bn-b+c)27^{n-1}&=(9n-2)27^{n-1} \\ -\iff 27n^2(bn+c)-3(3n-1)(3n-2)(bn-b+c)&=9n-2 \\ -\iff \qquad b=0 \qquad \text{and} \qquad c=\frac13. -\end{align*} -Therefore, the general solution takes the form -$$a_n=a_n^{(p)}+\beta\,a_n^{(h)}=\frac{27^n}3+\alpha\binom{2n}n\binom{3n}n.$$ -Since $a_0=A_0=0$, we compute $\beta=-\frac13$ and hence -$$a_n=\frac{27^n}3-\frac13\binom{2n}n\binom{3n}n=\frac{27^n}3-\binom{2n}n\binom{3n-1}{n-1}. \qquad \square$$<|endoftext|> -TITLE: Optimal constant in Sobolev embedding -QUESTION [6 upvotes]: It is well-known that the Sobolev space $H^1(0,s)$ embeds continuously in the space of continuous functions $C[0,s]$; in fact, Marti has found in 1983 that the optimal embedding constant is $\sqrt{\coth(s)}$, with -$$\|\cosh\|_\infty = \sqrt{\coth(s)} \|\cosh\|_{H^1}.$$ -Is the optimal embedding constant of $H^1_0(0,s)$ in $C[0,s]$ also known? - -REPLY [4 votes]: I do not know a reference but the following argument gives the best constant. Consider the interval $[0,a]$ and $G(t,s)$ the Green function of $I-D^2$ with zero boundary conditions at $0,a$. If $u \in H^2 \cap H^1_0$, then -$$u(t)=\int_0^a G(t,s)\left (u(s)-u''(s)\right )ds=\int_0^a \left(G(t,s)u(s)+G_s(t,s)u'(s) \right )ds -$$ so that -$$|u(t)| \le \left (\int_0^a G(t,s)^2ds \right )^{1/2}\|u\|_2+\left (\int_0^a G_s(t,s)^2ds \right )^{1/2}\|u'\|_2 \le M_t \|u\|_{H^1} -$$ -with $M_t^2=\int_0^a \left(G(t,s)^2+G_s(t,s)^2\right )ds$. This inequality extends to $H^1_0$, by density. Then the best constant $C$ satisfies $C \leq \sup_t M_t$. On the other hand, if $u(\cdot)=G(t, \cdot)$ then -$$G(t,t)=\int_0^a \left(G(t,s)^2+G_s(t,s)^2\right )ds=M_t \|u\|_{H^1} -$$ -and hence $C=\sup_t M_t$. At this point one has to compute patiently $G(t,s)$ and $M_t$ and maximize it over $[0,a]$. The maximum is in $a/2$ (as expected) and I got (if no mistake occurred) -$$C=\left(\frac{1}{2} \coth a-\frac{1}{4 \sinh a}\right )^{1/2} -$$ which tends to $1/\sqrt 2$ as $a \to \infty$, the best constant on the line.<|endoftext|> -TITLE: What is a map for the representation theory of reductive groups? -QUESTION [5 upvotes]: I have finished learning about linear algebraic groups (minus their representation theory) and the associated algebraic structures (root data, root systems, etc.), and will next attempt to summarize for myself the main components related to their representation theory. -It's quite confusing for the uninitiated! - -I want the beginning of the story to be "the easy case", by which I mean the case for which classification of irreducible representations is done via the Theorem of the Heighest Weight. Sources that I've glanced at discuss two types of cases: the semisimple Lie algebra case (which I choose not to care about), and the compact real Lie group case. I somehow care about neither one... I want to discuss (split) reductive groups over a general field. Over the reals, the reductive groups correspond to the real compact Lie groups... Is it correct to say that the Theorem of Heighest Weight applies in general to split reductive groups over a general field? And that this is the "easy case"? Would it apply to reductive or semisimple groups? - -I'm somewhat confused in general about at what point it is necessary to restrict to unitary representations. This is my understanding: for finite groups and for compact groups all group representations can be given an inner product in such a manner as to make them unitary, and this is essentially the proof that the category of representations in these cases are semisimple. So I guess the point is that for general reductive groups, even though their category of representations is semisimple, not all representations can be made unitary... Or am I confused, and somehow being reductive should be seen as a generalization of being compact? - -On the one hand, it appears that the classification of irreducible (unitary?) representations of reductive groups is classified using the Theorem of Highest Weight and is therefore "the easy case". But I guess the point is that once you look at $G(K)$ for some ring $K$ then this stops being the easy case? For example: $K=\mathbb{R}$, or the adeles, or $\mathbb{C}$. So let's start with an easy question: is the representation theory of $G(\mathbb{C})$ the same as the representation theory of $G$? - -Can you put into context for me the following phrases: cuspidal representations - is that a term that only applies to the representation theory of the adelic points of $G$? What about tempered representations? Smooth representations? Admissible representations? Are they only for $G(\mathbb{R})$? Are there several unrelated notions of admissible/smooth representations? I see them arise with very different definitions in different context, and I'm not sure if I need to think of them as specific examples of one phenomenon. What are these good for, and why is it not covered by the Theorem of the Highest Weight? Is it hopeless to classify unitary representations that are not smooth/admissible? - -The Langlands classification "is a description of the irreducible representations of a reductive Lie group G". Why was that not already covered by the Theorem of the Highest Weight? Is that point that here we are dealing with a reductive Lie group as opposed to a reductive linear algebraic group? Or is that point that we're looking at $G(\mathbb{R})$? It's very hard for me to draw the line between what is easy and what is difficult... - -REPLY [3 votes]: Probably someone will step in with a more detailed answer soon... but here are some comments. -I think the line you are looking for between easy and hard could be the following: - -The algebraic representations of a split reductive algebraic group $G$, and - -The representations of some associated Lie groups $G(\mathbb R)$ or $G(\mathbb C)$ (or indeed p-adic groups $G(\mathbb Q_p)$, or adelic groups $G(\mathbb A)$, ...). - - -In case 1), the algebraic representations (of a split reductive algebraic group over a field, say) are determined by highest weight theory. In this case all irreducible representations are finite dimensional. The classification of representations of compact Lie groups is the same (so, for example, the algebraic representations of $SL_2(\mathbb R)$ and $SL_2(\mathbb C)$ are the same as the Lie group representations of $SU(2)$). The book of Fulton and Harris covers this topic in some detail. -In case 2), say $G$ is defined and split over $\mathbb R$, then we have an associated Lie group $G(\mathbb R)$. This is a non-compact Lie group, and it will typically have infinite dimensional irreducible representations. This theory is much more intricate. For example, one must think about what kind of topologies you want to consider on the underlying vector space of the representation. -Amongst such representations, we have the class of admissible representations. A key point about admissible representations is that they they are determined by their Harish-Chandra $(\mathfrak g, K)$-module, which is a purely algebraic gadget. The Langlands classification for real reductive groups is about admissible representations (one version of which reduces the classification to so-called tempered representations). -Amongst admissible representations, unitary representations (those that can be represented by unitary operators on a Hilbert space) are of particular importance and interest. The classification of such is more subtle and less well-understood. -There are a bunch of textbooks and lecture notes (e.g. Knapp, Trapa). One approach is to focus on the case of $SL_2(\mathbb R)$. David Ben-Zvi taught a class in this direction at UT Austin some years ago - you can find notes here: https://web.ma.utexas.edu/users/benzvi/GRASP/lectures/benzvi/mylectures.html<|endoftext|> -TITLE: Nontrivial signed measure on Lebesgue measurable sets being trivial on Borel sets -QUESTION [15 upvotes]: Let $\mathfrak{L}(\mathbb{R})$ be the collection of Lebesgue measurable sets and $\mathfrak{B}(\mathbb{R})$ be the Borel sets. -Question: Is there a nontrivial signed measure on $\mathfrak{L}(\mathbb{R})$ that is trivial on $\mathfrak{B}(\mathbb{R})$? -Obviously, any positive measure that is trivial on $\mathfrak{B}(\mathbb{R})$ is also trivial on $\mathfrak{L}(\mathbb{R})$, since any Lebesgue measurable set is a subset of a Borel set. -For the signed case, I have tried doing Jordan decomposition but it doesn't seem work. It is hard (if ever possible) to show $(\mu|_{\mathfrak{B}(\mathbb{R})})^+ = \mu^+|_{\mathfrak{B}(\mathbb{R})}$ and $(\mu|_{\mathfrak{B}(\mathbb{R})})^- = \mu^-|_{\mathfrak{B}(\mathbb{R})}$. -In fact, If I can deal this problem by decomposition, there must be something special about Borel sets, since the above equalities do not hold in general. Let $\mathfrak{C} = \{\varnothing,\{0\},\{1\},\{0,1\}\}$, $\mathfrak{D} = \{\varnothing, \{0,1\}\}$. The signed measure $\lambda$ on $\mathfrak{C}$ is defined that $\lambda(\{0\})=1$ and $\lambda(\{1\})=-1$. Then $\lambda|_\mathfrak{D}$ is trivial and the equalities fail. -Background: I am trying to prove (or disprove) that if $\mu$ and $\lambda$ are signed measures on $\mathfrak{L}(\mathbb{R})$, then $\mu|_{\mathfrak{B}(\mathbb{R})} = \lambda|_{\mathfrak{B}(\mathbb{R})}$ implies $\mu = \lambda$. - -REPLY [6 votes]: So, promoting my answer to a comment, this is unprovable in ZFC (assuming ZFC is consistent). I claim that such a signed measure $\nu$ exists only if there exists a nontrivial, atomless, countably additive probability measure $\mu$ on the discrete $\sigma$-algebra of $\mathbb{R}$ (or equivalently $[0,1]$). As I understand it, the latter is equivalent to the existence of a real-valued measurable cardinal of size at most $\mathfrak{c}$, which is independent of ZFC. -Suppose such $\nu$ exists. Consider its Hahn decomposition $\mathbb{R} = H^+ \cup H^-$. Since $H^+ \in \mathfrak{L}(\mathbb{R})$, it can be written $H^+ = B^+ \cup N^+$ where $B^+$ is Borel and $N^+$ is Lebesgue-null. By assumption $\nu(B^+) = 0$ so we must have $\nu(N^+) > 0$, and $\nu$ is positive on $N^+$. Now every subset of $N^+$ is Lebesgue measurable, so $\nu$ is defined for every such subset. Thus define $\mu(A) = \nu(A \cap N^+)$ for any subset $A \subset \mathbb{R}$. This is a nontrivial, countably additive, finite, positive measure on $2^{\mathbb{R}}$, which we may rescale to a probability measure. And since singletons are Borel, and therefore have $\nu$-measure zero, $\mu$ is atomless. -Gerald's answer, with Michael's comments, seems to be establishing the converse, that the existence of a real-valued measurable cardinal implies the existence of a desired $\nu$. Combining these would show that the original statement is independent of ZFC.<|endoftext|> -TITLE: Counterexamples to the Penrose Conjecture -QUESTION [15 upvotes]: I have noticed that in the literature on causality in general relativity one sees apparent counterexamples to the cosmic censorship hypothesis (somehow you have models for gravitational collapse which assume spherical symmetry and things like this so that naked singularities can in fact arise). Hawking conceded that these were counterexamples, but then re-instated the hypothesis because these examples are in some sense unrealistic or unphysical. -I was wondering if the Penrose conjecture is also likely to have 'unphysical' or 'unrealistic' violations (so somehow make some special assumptions and then cook up a black hole spacetime which violates the Penrose inequality), or whether the conjecture is that one can simply never create a counterexample at all to the inequality? -Edit: I am aware of the counterexample of Carrasco and Mars to a stronger version of the conjecture. In that paper they find slices of the Kruskal spacetime for which the outermost generalized apparent horizon has area strictly greater than $16 \pi M^2$, and so this is not a counterexample to the true Penrose inequality as far as I am aware. -Jarosław Kopiński has mentioned to me in private communication that there is in fact already a counterexample to the Penrose inequality with 'apparent horizon': - -Ishai Ben-Dov, The Penrose inequality and apparent horizons, Phys.Rev. D 70 (2004) 124031, doi:10.1103/PhysRevD.70.124031, arXiv:gr-qc/0408066, - -and so that it is not that surprising that one can construct counterexamples when the inner boundary is even more general. - -REPLY [4 votes]: Having thought about this more and discussed it with others, the answer seems to be that there are likely no counterexamples to the Penrose inequality, even if one allows for unphysical violations. -For recent numerical evidence of this, a paper of Kulczycki and Malec investigates the Penrose inequality and other modifications, finding counterexamples to the modified versions but no violations of the original inequality.<|endoftext|> -TITLE: Does every topological group embed as a closed subgroup in an amenable group? -QUESTION [11 upvotes]: It is a standard result that closed subgroups of locally compact amenable groups are themselves amenable, so for example $F_2$, the free group on two generators, cannot be embedded as a closed subgroup of a locally compact amenable group. However, by a result of Pestov, $F_2$ embeds as a closed subgroup of $\mathrm{Aut}(\Bbb Q,\leq)$ and the latter group is (extremely) amenable. -Are there topological groups that cannot be embedded as a closed subgroup of any amenable group? - -REPLY [2 votes]: At least for Polish groups, which was the case I was most interested in, the answer is positive. -I mentioned this question to Ola Kwiatkowska yesterday and she immediately pointed out that one of the standard universal Polish groups, the group $\mathrm{Iso}(\Bbb U)$ of isometries of the Urysohn space, is in fact not only amenable, but even extremely amenable (and Polish subgroups of Polish groups are closed). -For arbitrary groups I still expect the answer to be positive, but I have no meaningful comments to make about that case, apart from the fact that if a group embeds as a closed subgroup into an amenable group $G$, then it also embeds as a closed subgroup into an extremely amenable group, namely $L^0(G,X,\mu)$.<|endoftext|> -TITLE: J.-P. Serre: Duality of regular differentials on singular curves -QUESTION [8 upvotes]: I already asked this on math.stackexchange.com, but didn't get any responses. I hope it is appropriate here. -Let $X'$ be an irreducible singular algebraic curve over an algebraically closed field $k$, let $X \to X'$ be its normalization, and consider a singular point $Q \in X'$. Let $K = Q(X)$ be the function field of $X$ and $X'$. -Let $\mathcal{O}_Q' = \mathcal{O}_{X', Q}$ be the stalk of the structure sheaf of $X'$ at $Q$, and let $\mathcal{O}_Q = \bigcap_{P \mapsto Q} \mathcal{O}_P$ be its normalization. Here $\mathcal{O}_P$ is the stalk of the structure sheaf of $X$ at $P \in X$, and the intersection is over all points mapping to $Q$. -In Algebraic Groups and Class Fields by J.-P. Serre, chapter IV §3, Serre introduces the module $\underline{\Omega}_Q'$ of regular differentials at $Q$. A differential $\omega \in D_k(K)$ is called regular, iff -\begin{equation}\sum_{P \mapsto Q} \operatorname{Res}_P(f \omega) = 0 \quad \text{for all} \ f\in \mathcal{O}_Q'.\end{equation} -Similarly to $\mathcal{O}_Q$, Serre defines -$$ \underline{\Omega}_Q = \bigcap_{P \mapsto Q} \Omega_P.$$ -Since every differential $\omega \in \underline{\Omega}_Q$ has no poles at any point $P \mapsto Q$, clearly $\operatorname{Res}_P(f \omega) = 0$ for $f \in \mathcal{O}_Q'$, so that $\underline{\Omega}_Q \subset \underline{\Omega}_Q'$. -Now to my question: The mapping -\begin{align} -\mathcal{O}_Q / \mathcal{O}_Q' \times \underline{\Omega}_Q' / \underline{\Omega}_Q & \to k \\ -(f, \omega) & \mapsto \sum_{P \mapsto Q} \operatorname{Res}_P(f \omega) -\end{align} -is clearly bilinear and well-defined. Serre claims, that it is a perfect pairing, but I don't know why. I think we have to show two things: - -If $f \in \mathcal{O}_Q$, with the property that for each $\omega \in \underline{\Omega}_Q'$, one has $\sum_P \operatorname{Res}_P(f \omega) = 0$, then in fact $f \in \mathcal{O}_Q'$. -If $\omega \in \underline{\Omega}_Q'$, such that for each $f \in \mathcal{O}_Q$, one has $\sum \operatorname{Res}_P(f \omega) = 0$, then $\omega \in \underline{\Omega}_Q$, i.e. $\omega$ is regular at every $P \mapsto Q$. - -Any help would be appreciated :) - -REPLY [2 votes]: I think I can show this in the analytic setting. The argument may possibly work in the algebraic setting if one considers completions, but I don't feel too familiar with completions, and my own interest in this is actually analytic. -Let $X'_1, \dotsc, X'_r$ be the irreducible components of $X'$ at $Q$. Then $X$ is the disjoint union of the normalizations $X_i \to X_i'$. Since we work in the analytic setting and the question is local in $X'$ we can assume that each $X_i$ contains exactly one point $P_i \mapsto Q$. Algebraically this means $\mathcal O_Q \cong \prod_i \mathcal O_{P_i}$. Since the $\mathcal O_{P_i}$ are regular rings, we can choose a local coordinate $\mathcal O_{P_i} \cong \mathbb C\{x_i\}$. - -Consider an element $f \in \mathcal O_Q \setminus \mathcal O_Q'$. We have to find a differential form $\omega \in \underline{\Omega}_Q'$ such that $\sum_P \operatorname{Res}_P(f \omega) \neq 0$. -1.1. Suppose $f(P_i) \neq f(P_j)$ -for some $i$ and $j$, consider the differential form $\omega$ which vanishes at all $P_k$ for $i \neq k \neq j$, and has simple poles at $P_i$ and $P_j$ with residues $$\operatorname{Res}_{P_i} = - \operatorname{Res}_{P_j} = 1.$$ -Since each $g \in \mathcal O_Q'$ can be developed locally at $P_i$ in a power series, $\operatorname{Res}_{P_i}(g \omega) = g(Q)$, and similarly $\operatorname{Res}_{P_j}(g \omega) = - g(Q)$. Hence $\omega \in \underline{\Omega}_Q'$, but -$$\sum_{P \mapsto Q} \operatorname{Res}_P(f \omega) = f(P_i) - f(P_j) \neq 0.$$ -1.2. If $f(P_i) = f(P_j)$ for all $i,j$, one irreducible component of $X'$ has to be singular. It is sufficient to show the existence of $\omega$ on the irreducible component, so we might suppose $X'$ itself is irreducible. This means we have an inclusion $\mathcal O_Q' \subset \mathbb C\{x\}$. Since the quotient -$$ 0 \to \mathcal O_Q' \to \mathbb C\{x\} \to \Bbb C^\delta \to 0$$ -is finite-dimensional one has $x^k \in \mathcal O_Q'$ for some $k > \delta$, and so $\mathcal O_Q'$ is given by the vanishing of $\delta$ linear equations on the coefficients $a_1, \dotsc, a_{k-1}$ of a power series $\sum_n a_n x^n$. Let $$l = \gamma_1 a_1 + \dotsb + \gamma_{k-1} a_{k-1}$$ be one of those linear equations with $l(f) \neq 0$. Then define -$$ \omega = \left( \frac{\gamma_1x^{k-2} + \dotsb + \gamma_{k-2} x + \gamma_{k-1}}{x^{k}} \right) dx$$ -such that for each power series $g \in \mathbb C\{x\}$ we have $\operatorname{Res}_0(g \omega) = l(g)$. Thus $\omega \in \underline{\Omega}_Q'$ and $\operatorname{Res}_0(f \omega) \neq 0$. - -Suppose $\omega \in \underline{\Omega}'_Q$, but $\omega \notin \underline{\Omega}_Q$. Then $\omega$ has a pole at some $P_i$, and we can write -$ \omega = h(x_i) dx_i$ for some Laurentseries -$$h(x_i) = \sum_{k \geq -n} h_k x^k_i, \quad h_{-n} \neq 0.$$ -Thus $\operatorname{Res}_{P_i}(x_i^{n-1} \omega) = h_{-n} \neq 0$. So if we define -$$ f = (0, \dotsc, 0, x_i^{n-1}, 0, \dotsc, 0) \in \prod_j \mathcal O_{P_j}$$ -then $\sum_j \operatorname{Res}_{P_j}(f \omega) = h_{-n} \neq 0$.<|endoftext|> -TITLE: A piecewise-linear or topological Fulton-MacPherson compactification -QUESTION [9 upvotes]: The Fulton-MacPherson compactifications of configuration spaces are smooth manifolds with corners which have the ordered configuration spaces of distinct points in a smooth manifold as their interior. You can construct them by iterated spherical blow-ups, or directly as the closure of the image of a certain map. See this paper by Dev Sinha for details and an overview of the rich history of this construction. - -Is it possible to construct the Fulton-MacPherson compactification up to (PL-)homeomorphism -without reference to a smooth structure. That is, can we give a construction that also works for piecewise-linear or topological manifolds? - -REPLY [4 votes]: In this note I used recent results of Chen and Mann rule out the existence of such a topological compactification in all dimensions $\geq 2$.<|endoftext|> -TITLE: Minimum number of independent pairs in a matroid -QUESTION [5 upvotes]: Given a matroid $M$ with ground set $E$ of size $2n$, suppose there exists $A\subseteq E$ of size $n$ such that both $A$ and $E\setminus A$ are independent. What is the minimum number of $B\subseteq E$ such that both $B$ and $E\setminus B$ are independent? - -With $n=2$, some casework shows that the answer is $4$: suppose $\{1,2\},\{3,4\}$ are independent. Using the augmentation property with $\{1\}$ and $\{3,4\}$, we get that wlog $\{1,3\}$ is independent. If $\{2,4\}$ is independent, we get four sets $B$, so using $\{2\}$ against $\{3,4\}$, it must be that $\{2,3\}$ is independent. But then using $\{4\}$ against $\{1,2\}$ gives us the claim. It is possible that the independent sets are $\emptyset,\{1\},\{2\},\{3\},\{4\},\{1,2\},\{3,4\},\{1,3\},\{2,4\}$, giving the answer of $4$. - -REPLY [7 votes]: As observed by Geva Yashfe, the answer is $2^n$. This can be achieved when each of $A$ and $\overline{A}:=E\setminus A$ are bases, with $A = \{a_1,\ldots,a_n\}$, $\overline{A} = \{b_1,\ldots,b_n\}$, and $a_i$ parallel to $b_i$ for all $i \in [n]$. -For the lowerbound, by truncation, we may assume that $A$ and $\overline{A}$ are both bases. It is well-known that every matroid actually satisfies the following stronger basis exchange axiom: for all distinct bases $B_1$ and $B_2$ and every $X \subseteq B_1 \setminus B_2$, there exists $Y \subseteq B_2 \setminus B_1$ such that both $(B_1 \setminus X) \cup Y$ and $(B_2 \cup X) \setminus Y$ are bases. Applying this axiom to the bases $A$ and $\overline{A}$ and every $X \subseteq A$, we get $2^n$ distinct bases $B$ such that $\overline{B}$ is also a basis. -As requested by TZM, here is a proof that the stronger exchange axiom always holds. The key idea is to use the Matroid Partition Theorem on two appropriately defined matroids. Given two matroids $M_1$ and $M_2$ on the same ground set $E$, we say that a set $X \subseteq E$ is $(M_1, M_2)$-partitionable if $X$ is the disjoint union of $I_1$ and $I_2$ where $I_i$ is independent in $M_i$. We denote the size of a largest $(M_1, M_2)$-partitionable set as $\pi(M_1, M_2)$. -Matroid Partition Theorem. Let $M_1$ and $M_2$ be matroids on the same ground set $E$ with rank functions $r_1$ and $r_2$. Then -$$\pi(M_1, M_2)=\min_{A \subseteq E} (|E-A|+r_1(A)+r_2(A)).$$ -We can now prove the stronger exchange axiom. -Lemma. Let $B_1$ and $B_2$ be distinct bases of $M$ and $X \subseteq B_1 \setminus B_2$. Then there exists $Y \subseteq B_2 \setminus B_1$ such that $(B_1 \setminus X) \cup Y$ and $(B_2 \cup X) \setminus Y$ are both bases. -Proof. Let $M_1$ be the restriction of $M / (B_1 \setminus X)$ to $B_2 \setminus B_1$ and $M_2$ be the restriction of $M / (X \cup (B_1 \cap B_2))$ to $B_2 \setminus B_1$. Let $r_1$ and $r_2$ be the rank functions of $M_1$ and $M_2$. A simple calculation (using submodularity) shows that $r_1(A)+r_2(A) \geq |A|$ for all $A \subseteq B_2 \setminus B_1$. Therefore, by the Matroid Partition Theorem, $\pi(M_1, M_2)=|B_2 \setminus B_1|$. That is, there exists a partition $Y \cup Z$ of $B_2 \setminus B_1$ such that $Y$ is independent in $M_1$ and $Z$ is independent in $M_2$. In other words, $(B_1 \setminus X) \cup Y$ is independent in $M$ and $X \cup (B_1 \cap B_2) \cup Z$ is independent in $M$. Note that this implies $|Y|=|X|$; otherwise one of these two sets has size more than $|B_1|$. Thus, $(B_1 \setminus X) \cup Y$ and $(B_2 \cup X) \setminus Y$ are both bases of $M$, as required.<|endoftext|> -TITLE: Must a path of compact sets in $X$ descend to a path in $X$? -QUESTION [14 upvotes]: (I am most interested in the case $X=\mathbb R^2$, but of course one could ask the same question for manifolds, or metric spaces in general.) -Let $\text{Com}(\mathbb R^2)$ denote the space of nonempty compact subsets of the plane, equipped with the Hausdorff metric. Let $S_\bullet:[0,1]\to\text{Com}(\mathbb R^2)$ be a continuous path, and let $p\in S_0$. Must there exist a path $\gamma:[0,1]\to\mathbb R^2$ such that $\gamma(0)=p$, and $\gamma(t)\in S_t$ for all $t\in[0,1]$? - -REPLY [13 votes]: $\DeclareMathOperator{\R}{\mathbf{R}}\DeclareMathOperator{\Z}{\mathbf{Z}}$The answer is no, even in the circle (and hence in the plane). -As coordinates, write the circle as the 1-point compactification $\bar{\R}$ of $\R$. -For $t\in\mathopen]0,1]$, write $$X_t=\{\infty\}\cup\big(t\Z+\sin(1/t)\big).$$ -For $t\to 0$, this tends to $X_0=\bar{\R}$, hence defines a continuous path on $[0,1]$. Any continuous lift, for $t>0$ has to have the form $x(t)=tn+\sin(1/t)$ for some fixed $n$. This does not converge when $t\to 0$ (it accumulates to all of $[-1,1]$). So there is no continuous lift. -(Note that the answer is clearly positive when $X=\mathbf{R}$, as $x\mapsto \max(x)$ is then a lift (not only for paths).)<|endoftext|> -TITLE: What's the localic reflection of a presheaf topos? -QUESTION [8 upvotes]: $\newcommand{\Psh}{\operatorname{Psh}} -\newcommand{\Sh}{\operatorname{Sh}} -\newcommand{\O}{{\mathcal{O}}}$ -Let $X$ be a locale, $\O(X)$ the corresponding frame. - -What's the localic reflection of $\Psh X$? - -We know that -$$ -\O(X) \cong \mathrm{Sub}_{\Sh X}(1) -$$ -Call $Y = \mathrm{Sub}_{\Psh X}(1)$ the localic reflection of $\Psh X$. -One has $\Psh X \simeq \Sh Y$. -Since $\Sh X$ is a subtopos of $\Psh X$, $X$ should be a sublocale of $Y$, i.e. there should be a nucleus $j$ on $\O(Y)$ such that $\O(X) = \O(Y) / j$. - -Who's $j$? - -REPLY [11 votes]: I'm writing $\mathcal{O}(X)$ for the frame corresponding to $X$. -Opens of $Y$ are sieves on $X$, i.e. the collection of open subsets $V \subset \mathcal{O}(X)$ such that $v \in V$ and $u \leqslant v \Rightarrow u \in V$. Thus $Y$ is the locale of downward closed subsets of $\mathcal O(X)$. -More generally if $P$ is a poset, a subterminal presheaf on $P$ is a functor $P \to \{ 0,1\}$ and the set of $p \in P$ sent to $1$ is a sieve on $P$. This induces a bijection between sieves and subterminal presheaves. -When you identify $X$ as a sublocale of $Y$, its opens correspond to principal sieves, i.e. the ones of the form $\downarrow\! v = \{ u \in \mathcal{O}(X) | u \leqslant v \}$. -Thus the inclusion of locales $X \hookrightarrow Y$ is given by the quotient of frames $\mathcal{O}(Y) \to \mathcal{O}(X)$ defined as $V \mapsto \bigvee V$ and whose right adjoint is $v \mapsto {\downarrow\! v}$. -The nucleus $j$ hence takes a general sieve $V \subset \mathcal{O}(X)$ and sends it to the smallest principal ideal containing $V$, that is, $\downarrow\! \bigvee V$.<|endoftext|> -TITLE: When do the lengths of simple closed curves determine a hyperbolic surface? -QUESTION [9 upvotes]: Consider hyperbolic metrics on $\Sigma_g$ a closed orientable surface of genus $g$. Let $[\gamma_1] , \cdots, [\gamma_n]$ be a finite collection of isotopy classes of simple closed curves on $\Sigma_g$. -At what conditions on $[\gamma_1] , \cdots, [\gamma_n]$ is any (isotopy class of) metric completely determined by the lengths of the geodesic representatives of respective classes $[\gamma_1] , \cdots, [\gamma_n]$ ? -I am not necessarily looking for a complete answer and partial answers/comments are very welcome. - -REPLY [5 votes]: This paper is probably relevant for your question: -MR0528966 -Wolpert, Scott -The length spectra as moduli for compact Riemann surfaces. -Ann. of Math. (2) 109 (1979), no. 2, 323–351. -And here is a more recent paper on the subject: -MR3770180 -Parlier, Hugo -Interrogating surface length spectra and quantifying isospectrality. (English summary) -Math. Ann. 370 (2018), no. 3-4, 1759–1787. -These papers suggest that finitely many geodesic lengths never determine the surface.<|endoftext|> -TITLE: History of Study's Lemma? -QUESTION [15 upvotes]: The following theorem is usually attributed to Eduard Study: - -Let $f(x,y)$ and $g(x,y)$ be polynomials in two variables over a field, with $f$ irreducible. If $f\nmid g$ then the curves $C_f:f=0$ and $C_g:g=0$ have finitely many points of intersection. Consequently, If the field is algebraically closed and $C_f\subseteq C_g$ (hence $C_f\cap C_g$ has infinitely many points) then $f|g$. - -However, I have not been able to track down any reference to this result outside of modern textbooks. -Questions: - -What is the original reference for this result? -What did Study actually prove? -What was the context? -Did this result directly influence later versions of the Nullstellensatz? - -Thanks. -Edit: I see that Study wrote a book on on the Theory of Ternary Forms (1889). I suppose the result must be in there somewhere. - -REPLY [17 votes]: I found the lemma on page 63 of Study's Einleitung in die Theorie der Invarianten linearer Transformationen auf Grund der Vektorenrechnung (1923). - -The source cited for the proof is page 202 of Study's Methoden zur Theorie der ternären Formen (1889), where it appears as an endnote and refers to a "less elementary" proof by - -O. Hölder, Zum Invariantenbegriff, Math. Naturwiss. Mitteilungen -1, 59–65 (1884). - -Volume 1 of the journal is not online, but there is a review at zbMath [thanks, @spin] that explicitly mentions the lemma: - -Sind die ganzen rationalen Functionen $F(x_1,x_2,\ldots,x_m)$, -$G(x_1,x_2,\ldots,x_m)$ so beschaffen, dass für alle (reellen oder -complexen) Wertsysteme $x_1,x_2,x_m$, wofür $G=0$ ist, auch die -Gleichung $F=0$ gilt, und ist G irreduzibel, so ist $G$ ein Teiler -von $F$. - Let the entire rational functions $F(x_1,x_2,\ldots,x_m)$ and -$G(x_1,x_2,\ldots,x_m)$ be such, that for any (real or complex) values $x_1,x_2,\ldots x_m$ where $G=0$ also $F=0$, and let $G$ be irreducible, then $G$ divides $F$. - -It's too late (and not productive) to change common practice, but priority wise I would conclude that Study's lemma is Hölder's lemma.<|endoftext|> -TITLE: Abandoned notions in mathematics? -QUESTION [7 upvotes]: I'm looking for examples of abandoned or demised notions/concepts in mathematics, preferably (but not necessarily) after the age of foundations. To be clear: I'm not looking for abandoned ideas or theories (as opposed to notions or concepts); neither am I looking for revised (as opposed to abandoned) notions -- I'm aware that there is plenty of both kinds. To illustrate what I'm looking for, consider the notion of phlogiston in natural sciences. Do we have anything like that in math? -Thank you very much! - -REPLY [5 votes]: How about consulting Euclid's Elements? - -The three main classes of irrational magnitudes are the medial, binomial, and apotome. ... It was Euclid who generalized the idea of commensurable and incommensurable to squares, and also ordered the binomial and apotome irrational lines into six distinct classes each (Knorr, 1983).<|endoftext|> -TITLE: What are the references of the form [D blah] in Giraud's cohomologie nonabelienne? -QUESTION [5 upvotes]: In Giraud's book "Cohomologie non-abelienne", the author repeatedly cites sources using something like [D blah]. E.g., Chapter 1, section 1, first line: "Nous renvoyons à [D 1] et à [SGA 1 VI]..." -The problem is there's nothing in the bibliography labelled [D]. The closest thing are two papers of Dedecker's. The first paper does not seem to be the right one, and the second one (published in "Lille"), my university doesn't seem to have access to. -What reference is he citing? - -REPLY [13 votes]: In the bibliography you can find the following item: - - -Giraud,J.: Méthode de la descente. Mémoires Soc. Math. Fr. 2 (1964) (cité [D]). - - -My understanding then is that citationd like [D 1] refer to specific sections or results in the reference [D], which is above. Note something similar is done with references to EGA or specific volumes of SGA - abbreviated name followed by a section/result number for a citation.<|endoftext|> -TITLE: Discovered 240 new circles associated with Pascal's line -QUESTION [5 upvotes]: I am looking for a proof or a reference request for a problem as follows: -Problem: Let a cyclic hexagon with sidelines $l_1$, $l_2$, $l_3$, $l_4$, $l_5$, $l_6$ and $l_1 \cap l_4 =A$, $l_3 \cap l_6 = B$, $l_5 \cap l_2 = C$. Let $l’_1$ is the line through $A$ and parallel to $l_3$ meets $l_2, l_6$ at $P_{12}, P_{16}$; $l’_3$ is the line through $B$ and parallel to $l_5$ meets $l_2, l_4$ at $P_{32}, P_{34}$, $l’_5$ is the line through $C$ and parallel to $l_1$ meets $l_4, l_6$ at $P_{54}, P_{56}$. Then show that six points $P_{12}$, $P_{16}$, $P_{32}$, $P_{34}$, $P_{54}$, $P_{56}$ lie on a new circle. - -REPLY [6 votes]: I here is a proof. -Note first, that the six points lie on a conic. Indeed, the opposite sides of the "hexagon" $P_{12}$, $P_{32}$, $P_{34}$, $P_{52}$, $P_{56}$, $P_{16}$ intersect again in points $A, B, C$, which lie on one line. Hence they lie on a conic by the converse to Pascal's theorem. So, we only need to prove that this conic is a circle. -Next, let's do the same thing as what is done when one deduces Pascal's theorem from Bezout. Consider cubic polynomials $F_{red}=L_1\cdot L_3\cdot L_5$, $F_{blue}=L_2\cdot L_4\cdot L_6$, $ F_{red}'=L_1'\cdot L_3'\cdot L_5'$ that are product of linear polynomials $L_i$ and $L_i'$ such that $L_i=0$ defines $l_i$ and $L_i'=0$ defines $l_i'$. Finally, let $L$ be the linear polynomial that vanishes at $A,B,C$. -From Bezout theorem it follows that for a unique value of $c$, $F_{red}+cF_{blue}$ is divisible by $L$ and $\frac{F_{red}+cF_{blue}}{L}=0$ is the equation of the original circle. And also for some $c'$, $F_{red}'+cF_{blue}$ is divisible by $L$. I claim that $c=c'$. If this is proven, then the desired statement is proven, because by the assumptions, the cubic term of $F_{red}'$ coincides with the cubic term on $F_{red}$ (indeed, their zeros a two triples of parallel lines). So it would follow that $\frac{F_{red}+cF_{blue}}{L}$ and $\frac{F_{red}'+cF_{blue}}{L}$ have the same quadratic term, and so both define circles. -So, it remains to show that $c=c'$. Let us denote by $\bar F_{red}$, $\bar F_{red}'$, $\bar F_{blue}$ the cubic terms of $ F_{red}$, $\ F_{red}'$, $ F_{blue}$. Let $\bar L$ be the linear term of $L$. Note that both $\bar F_{red}+ c\bar F_{blue}$ and $\bar F_{red}'+ c'\bar F_{blue}$ are divisible by $\bar L$. However, as we saw above $\bar F_{red}=\bar F_{red}'$. It follows that $c=c'$. QED.<|endoftext|> -TITLE: 3-primary part of $\pi_\bullet S^3$ -QUESTION [8 upvotes]: I am looking for the most updated calculation of $\pi_\bullet S^3$ at the prime 3 (and also at any other odd primes). -The motivation of this question is just curiousity: since $\pi_\bullet S^3$ has exponent $p$ at any odd prime, these homotopy groups are completely determined by the number of generators and I was wondering how this sequence of numbers looks like. -On Toda book the computation goes to the 19+3 group. I have also found computation up to 64, but only at the prime 2. - -REPLY [3 votes]: Gregory Arone told me that there is an unstable Adams spectral sequence, which I was not aware of. I found the $E_2$ page for the 3-primary part of $S^3$. - -I think it doesn't contain the differentials. Without differentials, one can be sure only up to the 19-stem, as in Toda's Composition Methods in Homotopy Groups of Spheres, but at least it gives a partial answer up to the 81-stem, until one will find a copy of Toda's Unstable 3-Primary Homotopy Groups of Spheres. -The number of copies of $\mathbb{Z}_3$ in $\pi_{3+*}S^3_{(3)}$ up to the 19-stem is 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1.<|endoftext|> -TITLE: Exchanges between independent sets of a matroid -QUESTION [5 upvotes]: Let $I, J$ be two bases of a matroid. For every $x$ in $I$, there is some $y$ in $J$ such that, if we exchange $x$ with $y$, then both resulting sets ($I \setminus x \cup y$ and $J \setminus y \cup x$) are bases (this is the strong basis exchange property). -Can we extend this property as follows: there exists a bijection $f$ between $I\setminus J$ and $J\setminus I$, such that for every $x$ in $I$, if we exchange $x$ with $f(x)$, then both resulting sets are bases? -The nearest result I found was in lecture notes by Goemans. In Lemma 5, he proves that there is a perfect matching between $I\setminus J$ and $J\setminus I$ in a bipartite graph that he denotes by $D_M(I)$. This means that for every $x$ in $I$, if we exchange $x$ with $f(x)$, then $I \setminus x \cup f(x)$ is a base. But, it does not imply that $J \setminus f(x) \cup x$ is a base too. - -REPLY [5 votes]: No, not every matroid satisfies this property. For example, it is known to fail for the cycle matroid of $K_4$. The matroids that satisfy your property are called base orderable matroids. There are important classes of matroids that are base orderable, such as transversal matroids. Moreover, base orderability is a minor-closed property, but Ingleton proved that there are actually an infinite number of excluded minors. See these slides -of Joseph Bonin for more information. For example, the slides include a proof that $M(K_4) $ is not base orderable.<|endoftext|> -TITLE: Are polynomials bounded on the primes possible? -QUESTION [9 upvotes]: If $\{p_i\}$ is the sequence of all primes, is it possible that there exist a non constant $P\in \mathbb{Z}[x_1,\dots x_n]$ such that $P(p_i,p_{i+1},\dots p_{i+n-1})$ is bounded in $i$? -More precisely, can widely believed conjectures, or even heuristic arguments, help make such a claim (even more) unlikely. - -REPLY [12 votes]: Here is a proof that such a polynomial does not exist assuming that every admissible $n$-tuple occurs infinitely often in the sequence of primes. -To see this let $a:=(0, a_1, \dots, a_{n-1})$ be an admissible $n$-tuple. Suppose $P \in \mathbb{Z}[x_1, \dots, x_n]$ is such that the function $f(i):=P(p_i,p_{i+1},\dots p_{i+n-1})$ is bounded. Replacing $x_i$ by $x_1+a_{i-1}$ for all $i \in \{2, \dots, n\}$ we obtain a polynomial $Q_a \in \mathbb{Z}[x_1]$. Assuming that the $n$-tuple $a$ occurs infinitely often, we have $Q_a(p_i)=f(i)$ for infinitely many $i$. Since $f(i)$ is bounded, $Q_a$ is a constant. Thus $Q_a=p(a)$ where $p$ is a non-constant polynomial only depending on $P$. Since $f$ takes only finitely many values, $p(a)$ only takes on finitely many values over all admissible $n$-tuples $a$. However, this is impossible. -As Terry Tao notes in the comments below: - -One can make this argument unconditional by noting that $O(\log^{n−1−o(1)}X)$ of the $O(\log^{n−1}X)$ admissible tuples $a=(0,a_1, \dots ,a_{n−1})$ with $a_1, \dots, a_{n−1}=O(\log X)$ will be associated to consecutive primes $p,p+a_1, \dots,p+a_{n−1}$ for some $p∼X$ (because $∼X/\log X$ primes will generate a tuple by Markov's inequality and each tuple is associated to $O(X/\log^{n−o(1)}X)$ primes by e.g. Selberg sieve). On the other hand, a polynomial constraint on these tuples would instead force at most $O(\log^{n−2}X)$ of these tuples to be admissible (Schwartz-Zippel lemma).<|endoftext|> -TITLE: Physical motivation for tight/overtwisted dichotomy -QUESTION [9 upvotes]: I'm learning about tight vs. overtwisted contact structures in contact geometry. I understand that we care about the existence/nonexistence of overtwisted disks in a contact structure in part because the distinction has proved useful (e.g., in classification). -But since contact geometry has a lot of applications to physics, I'm curious: Are there any big physical reasons for why we care about this distinction? -(reposted from math.SE) - -REPLY [4 votes]: In the physics of fluids, a reason for caring about tightness of the contact structure is the idea/conjecture that overtwisted discs raise the energy of the fluid. -The velocity field of an inviscid, incompressible fluid flow on a Riemannian manifold corresponds to a contact 1-form in dimension three.$^\ast$ In this context one can investigate how the topological tight/overtwisted dichotomy for contact structures relates to the physical properties of the fluid like its energy. Is the energy minimizing flow necessarily a tight contact structure? Ghrist and Komendarczyk, in Overtwisted energy-minimizing curl eigenfields, show that the answer is "no" in general, but "yes" under additional symmetry conditions. -$^\ast$ For an introduction and overview of this correspondence, see On the contact topology and geometry of ideal fluids.<|endoftext|> -TITLE: Topological complexity of ultrafilters in $2^\kappa$ for uncountable $\kappa$ -QUESTION [6 upvotes]: It is a well known fact that if $\mathcal{F}$ is a non-principal ultrafilter on $\omega$, then the set $\{ \alpha \in 2^\omega : \alpha \in \mathcal{F}\}$ (conflating binary strings with subsets of $\omega$) is not a Borel subset of $2^\omega$ with its standard product topology. -The proof of this that I am familiar with goes through showing that $\mathcal{F} \subseteq 2^\omega$ is not a measurable subset of $2^\omega$ by noting that if it were it would have density $\frac{1}{2}$ everywhere, contradicting the Lebesgue density theorem. -I am curious about the analogous statement with regards to ultrafilters on $\kappa$, considered as subsets of $2^\kappa$ with its compact product topology. I have difficulty imagining that a non-principal ultrafilter on $2^\kappa$ could be Borel (where by Borel I mean specifically an element of the $\sigma$-algebra generated by open sets, not just the $\sigma$-algebra generated by clopen sets), but I can't find a proof of this and the Lebesgue density theorem argument seems difficult to generalize to $2^\kappa$, even though there is a natural regular Borel measure on $2^\kappa$. - -REPLY [4 votes]: A Borel ultrafilter would have the property of Baire. Therefore either $\mathcal F$ or its complement $2^\kappa\setminus\mathcal F$ is comeager relative to some basic open set. Since $\mathcal F$ is invariant under finite changes, this would mean $\mathcal F$ or its complement is comeager. Since $\mathcal F$ is the image of $2^\kappa\setminus\mathcal F$ complement under a homeomorphism of $2^\kappa,$ this means both sets are comeager and must therefore intersect, which is impossible. -Fun exercise: show that a Borel map $2^\kappa\to[0,1]$ that vanishes on finite subsets of $\kappa$ cannot be a finitely additive probability measure.<|endoftext|> -TITLE: Graph embedding that locally minimizes total edge lengths -QUESTION [7 upvotes]: I consider a graph $G$ (possibly infinite, but locally finite) embedded in the Euclidean plane $\mathbb{E}^2 \cup \{\infty\}$ such that each local perturbation of the embedding "increases the total length". That is, for any sufficiently small neighborhood $U \subset \mathbb{E}^2$, the total length of $G \cap U$, if not empty, is minimized subject to fixed boundary points $G \cap \partial U$. -So all edges must be straight. If a vertex has degree 3, then it must be the Fermat point of its neighbors. If a vertex has degree 4, the adjacent edges must form two collinear pairs. In general, the outgoing unit vectors along the edges around a vertex must sum up to 0. I consider such embeddings as an analogy to minimal surfaces. Examples include, of course, infinite Steiner trees --- or finite Steiner trees if I use the point at infinity wisely. -I failed to find reasonable literature about such embeddings, hence would like to ask for references. In particular, I wonder if such an embedding may appear as (a nice supgraph of) the graph of some quasi-crystallographic tiling. - -REPLY [2 votes]: see -https://arxiv.org/pdf/1803.03728.pdf , -https://arxiv.org/pdf/1904.00483.pdf and -https://arxiv.org/pdf/1902.07872.pdf . -Here it is called geodesic nets.<|endoftext|> -TITLE: Do vector spaces without choice satisfy Cantor-Schroeder-Bernstein? -QUESTION [41 upvotes]: If $V \hookrightarrow W$ and $W \hookrightarrow V$ are injective linear maps, then is there an isomorphism $V \cong W$? -If we assume the axiom of choice, the answer is yes: use the fact that every linearly independent set can be extended to a basis and apply the usual Schroeder-Bernstein theorem. -If we don't assume the axiom of choice, and we work in ZF, say (or some other formalism with excluded middle), then vector spaces don't necessarily have bases (in fact, Blass showed that there must be a vector space without a basis over some field), so we can't use the same proof strategy. Nevertheless, there's room for optimism, since Schroeder-Bernstein still holds for sets in ZF. So one might hope that it also holds for vector spaces in ZF. -Question: Work in ZF (or some other formalism with excluded middle but without choice). If $V \hookrightarrow W$ and $W \hookrightarrow V$ are injective linear maps of vector spaces over a field $k$, then is there an isomorphism $V \cong W$? -Variation 1: What if we assume that $k$ is finite, or even that $k = \mathbb F_p$ for a prime $p$? -Variation 2: What if we assume that $V$ is a direct summand of $W$ and vice versa? -The following consequence of Bumby's theorem appears to be constructive: If $k$ is a ring and every $k$-module is injective, then $k$-modules satisfy Schroeder-Bernstein. But the condition "every module over a field is injective" sounds pretty choice-ey to me. I suppose it's worth noting, though: -Variation 3: Does "Every vector space over any field is injective" imply choice? How about "Every vector space over $\mathbb F_p$ is injective"? - -REPLY [29 votes]: There are models of ZF+DC in which every subset of every Polish space has the property of Baire (I can try to add references later, I think to Solovay and Shelah, but these are pretty well known). This implies that every linear map between Banach spaces is continuous. -So we can then take $\ell^\infty$ and $\ell^1$. It is very easy to construct (continuous) linear injections either way: the identity map from $\ell^1$ into $\ell^\infty$, and to go the other way, map $x_n$ to $2^{-n} x_n$. -But if there were a linear isomorphism between them, it would be a homeomorphism, and this is impossible because $\ell^1$ is separable and $\ell^\infty$ isn't. -(As a tie-in to Jeremy's answer, in this model $\ell^1$ is reflexive, and $\ell^\infty / c_0$ is a Banach space with no nonzero linear functionals.)<|endoftext|> -TITLE: Deformations of Hopf manifolds -QUESTION [8 upvotes]: Recall that a Hopf manifold is a quotient $\mathbb C^n\setminus 0$ by a free action of $\mathbb Z$ where the generator is acting by a holomorphic contraction. -Question 1. Is it true that any deformation of a Hopf manifold (as a complex manifold) is again a Hopf manifold for $n\ge 3$? -Question 2. Is there some kind of classification of Hopf manifolds and their deformations in dimension $\ge 3$ (for example $n=3$). -Note that for $n=2$ the answer to both questions is positive https://en.wikipedia.org/wiki/Hopf_surface - -REPLY [7 votes]: Both questions are answered affirmatively for sufficiently small deformations in this paper by Haefliger.<|endoftext|> -TITLE: When do infinitary compactness numbers exist? -QUESTION [10 upvotes]: For a logic $\mathcal{L}$, let the compactness number of $\mathcal{L}$ (if it exists) be the least $\kappa$ such that every $(<\kappa)$-satisfiable $\mathcal{L}$-theory is satisfiable. Note that there is no restriction here on the cardinality of the language of the theory in question. -For example, an uncountable cardinal $\kappa$ is strongly compact iff it is the compactness number of its own infinitary logic $\mathcal{L}_{\kappa,\kappa}$; more interestingly, Magidor showed that $\mathsf{SOL}$ has a compactness number iff there is an extendible cardinal, in which case its compactness number is the least extendible cardinal. -My question is: - -What is the strength of "For every $\kappa$, the compactness number of $\mathcal{L}_{\kappa,\kappa}$ exists?" - -EDIT: Originally I said that I didn't know anything relevant, but I just noticed that one of the suggested related questions is very relevant, namely this one: there it is shown for example that the existence of a compactness number for $\mathcal{L}_{\omega_1,\omega_1}$ already implies the existence of a measurable cardinal, or more technically that the existence of a compactness number for $\mathcal{L}_{\omega_1,\omega_1}$ is equivalent to the existence of an $\omega_1$-strongly compact cardinal. A natural guess based on that is that the principle in question is equivalent to "For every $\kappa$ there is a $\kappa$-strongly compact cardinal," but I haven't had a chance to read through the argument in detail so I'm not too confident here. - -REPLY [11 votes]: The compactness number for $\mathcal L_{\kappa,\kappa}$ is equal to the least $(\kappa,\infty)$-strongly compact cardinal. A cardinal is $(\kappa,\infty)$-strongly compact if for every set $X$, there is a $j : V\to M$ such that $\text{crit}(j)\geq \kappa$, and $j[X]$ can be covered by and element of $M$ of $M$-cardinality less than $j(\delta)$. I sketch a proof at the end because I don't know the reference. -But first: it follows easily that your hypothesis is equivalent to the existence of a proper class of almost strongly compact cardinals, which are (resp. should be) defined to be cardinals $\kappa$ such that for all $\gamma < \kappa$ every $\kappa$-complete filter can be extended to a $\gamma$-complete (resp. $\gamma^+$-complete) ultrafilter. Whether this is equivalent to the existence of a proper class of strongly compact cardinals is an open question. -The true consistency strength is probably a proper class of supercompacts: all three of these hypotheses are equivalent under the Ultrapower Axiom. There is some evidence that the equivalence between a proper class of almost strong compacts and a proper class of strong compacts is a theorem of ZFC: the first almost strongly compact cardinal above an ordinal $\gamma$ is either strongly compact or else has countable cofinality (although the truth is I needed a little SCH to handle the case $\gamma = 0$). This is in Some combinatorial properties of Ultimate $L$ and $V$. -Now the proof. In one direction, you show that $\mathcal L_{\kappa,\kappa}$ is $\delta$-compact for any $\kappa$-strongly compact $\delta$. Fix a $\delta$-consistent theory $T$ in the signature $\tau$. Cover $j[T]$ by a theory $S\subseteq j(T)$ in $M$ of $M$-cardinality less than $j(\delta)$. You get a model $\mathfrak A$ of $S$ in $M$ by $j(\delta)$-consistency of $j(T)$. Take the reduct of $\mathfrak A$ to $j[\tau]$. This is essentially a model of $T$: more precisely, $j : T \to j[T]$ is an isomorphism of $\mathcal L_{\kappa,\kappa}$-theories because $\text{crit}(j)\geq \kappa$. -Conversely, if $\delta$ is the compactness number of $\mathcal L_{\kappa,\kappa}$, then for any set $X$ and any $\delta$-complete filter base $\mathcal B$ on $X$, you can build a $\delta$-consistent theory whose models are $\kappa$-complete ultrafilters on $X$ extending $\mathcal B$. (A $\delta$-complete filter base is a family of sets such that the intersection of any ${<}\delta$-sized subfamily is nonempty.) -The signature has constants for all subsets of $X$ along with a predicate $W$. The theory contains the axiom "$W(A)$" for each $A\in \mathcal B$ and the axiom "If $W(\bigcup \mathcal P)$, then $\bigvee_{A\in \mathcal P}W(A)$" for every partition $\mathcal P$ of $X$ with $|\mathcal P| < \kappa$. The theory is $\delta$-consistent since if one takes a set $\mathcal A\subseteq P(X)$ of cardinality less than $\delta$, one obtains a model of the axioms in the signature restricted to constants from $\mathcal A$ by letting $W$ be the principal ultrafilter concentrated at $x\in \bigcap(\mathcal A\cap \mathcal B)$. -It follows that for any set $X$, there is a $\kappa$-complete ultrafilter on $P_{\delta}(X)$ extending the filter base $\langle A_x \rangle_{x\in X}$ where $A_x = \{\sigma \in P_\delta(X): x\in \sigma\}$. Such an ultrafilter is, by definition, fine. The associated ultrapower embedding $j : V\to M$ has critical point at least $\kappa$ and closure under $\kappa$-sequences by $\kappa$-completeness. Finally $\text{id}_\mathcal U$ is a cover of $j[X]$ by fineness, and $\text{id}_\mathcal U$ has $M$-cardinality less than $j(\delta)$ since it is an element of $j(P_{\delta}(X))$ by the definition of $M$-membership. So $\delta'$ is $(\kappa,\infty)$-strongly compact. But it is not too hard to show that the least $(\kappa,\infty)$-strongly compact cardinal is a limit cardinal, so $\delta$ must be $(\kappa,\infty)$-strongly compact.<|endoftext|> -TITLE: Convexity and Lipschitz continuity -QUESTION [9 upvotes]: It is probably an easy question, but somehow I am stuck. -Question Is the following statement true? If yes, how to prove it? - -Suppose that $f\in C^1(\mathbb{R}^n)$ is convex and -$$ -\langle\nabla f(x)-\nabla f(y),x-y\rangle \leq L|x-y|^2 -$$ -for some $L>0$ and all $x,y\in\mathbb{R}^n$. Does it follow that -$$ -|\nabla f(x)-\nabla f(y)|\leq L|x-y| -$$ -for all $x,y\in\mathbb{R}^n$? - -REPLY [4 votes]: This answer is a small modification of the answer of Denis Serre. I added for reader's convenience: (1) the result is slightly more general; (2) the answer contains much more details; (3) I am using a convolution by mollification approximation instead of inf-convolution. -Since convex functions satisfy -$$ -\langle \nabla f(x)-\nabla f(y),x-y\rangle\geq 0, -$$ -it suffices to prove the following more general result. - -Theorem. -Let $f\in C^1(\mathbb{R}^n)$ and let $L>0$.Then the following conditions are equivalent: -\begin{equation} -(1)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\langle\nabla f(x)-\nabla f(y),x-y\rangle|\leq L|x-y|^2 -\quad -\text{for all $x,y\in\mathbb{R}^n$.} -\end{equation} -\begin{equation} -(2)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\nabla f(x)-\nabla f(y)|\leq L|x-y| -\quad -\text{for all $x,y\in\mathbb{R}^n$.} -\end{equation} - -Proof. -While the implication (2) to (1) is obvious the other is not so we will prove the implication from (2) to (1) now. -Assume first that $f\in C^\infty(\mathbb{R}^n)$. For $|u|=1$, (1) yields, -$$ -\left|\left\langle\frac{\nabla f(x+tu)-\nabla f(x)}{t},u\right\rangle\right|\leq L, -$$ -so passing to the limit as $t\to 0$ gives -$$ -|\langle D^2f(x)u,u\rangle|\leq L. -$$ -Since $D^2 f(x)$ is a symmetric matrix, the spectral theorem implies that the operator norm of the matrix $D^2f(x)$ satisfies -$$ -\Vert D^2f(x)\Vert = \sup_{|u|=1}|\langle D^2f(x)u,u\rangle|\leq L. -$$ -This estimate however, easily implies the result -\begin{equation} -\begin{split} -& |\nabla f(x)-\nabla f(y)|= -\left|\int_0^1\frac{d}{dt}\nabla f(y+t(x-y))\, dt\right|\\ -&\leq |x-y|\int_0^1\Vert D^2f(y+t(x-y))\Vert\, dt\leq L|x-y|. -\end{split} -\end{equation} -This completes the proof when$f\in C^\infty$. Assume now that $f\in C^1$ and let $f_\epsilon=f*\varphi_\epsilon$ be a standard approximation by convolution. Recall that $f_\epsilon\in C^\infty$ and $\nabla f_\epsilon=(\nabla f)*\varphi_\epsilon$. We have -\begin{equation} -\begin{split} -& -|\langle \nabla f_\epsilon(x)-\nabla f_\epsilon(y),x-y\rangle|= -\Big|\Big\langle\int_{\mathbb{R}^n} (\nabla f(x-z)-\nabla f(y-z))\varphi_\epsilon(z)\, dz,x-y\Big\rangle\Big|\\ -&\leq -\int_{\mathbb{R}^n} -\big|\big\langle \nabla f(x-z)-\nabla f(y-z)),(x-z)-(y-z)\big\rangle\big|\, \varphi_\epsilon(z)\, dz -\leq L|x-y|^2, -\end{split} -\end{equation} -where the last inequality is a consequence of (1) and $\int_{\mathbb{R}^n}\varphi_\epsilon=1$. -Since $f_\epsilon\in C^\infty$, the first part of the proof yields -$$ -|\nabla f_\epsilon(x)-\nabla f_\epsilon(y)|\leq L|x-y| -$$ -and the result follows upon passing to the limit as $\epsilon\to 0$.<|endoftext|> -TITLE: Possible new theorem in plane geometry encompassing 5 famous geometry theorems -QUESTION [8 upvotes]: I am looking for a proof of a generalization Napoleon theorem, Bottema theorem and Brahmagupta theorem and van Aubel theorem, and Finsler–Hadwiger theorem in one configuration, as follows: -Let four points $A, B, C, D$ in the plain, the perpendicular bisector of $AB$ meets the perpendicular bisector of $CD$ at $P$. then always exist only one point $S$ such that: - -$(\overrightarrow{\rm PD}, \overrightarrow{\rm PC})\equiv 2(AD, AS) \equiv 2(BS, BC)$ and $(\overrightarrow{\rm PB}, \overrightarrow{\rm PA})\equiv 2(CB, CS) \equiv 2(DS, DA)$ (in the figure). - -The line through $S$ meets the perpendicular bisector of $CD$ at $E$ then $ES \perp AB$ if only if $(\overrightarrow{\rm EC}, \overrightarrow{\rm ED})\equiv 2(SC, SB) \equiv 2(SA, SD)$ - - - -Application: -If the problem was proved. Then we can apply the theorem to proof two famous theorem: - -A Proof of the Napoleon theorem: Apply the theorem part one with $\beta=120^\circ$ and $\alpha=30^\circ$ (See Figure 1). - -A Proof of the Bottema theorem: Apply the theorem part one with $\beta=45^\circ$ and $\alpha=90^\circ$ (See Figure 1). - -A proof of Van Aubel theorem. Apply the theorem part one, with $\alpha=\gamma=45^\circ$ - -A roof of Finsler–Hadwiger theorem. Apply the theorem part one, with $\alpha=\gamma=45^\circ$ - -A Proof of the Brahmagupta theorem: Apply the theorem part two with $\gamma=90^\circ$ (See Figure 1). - - -See also: - -Relative a generalization Bottema theorem - -REPLY [2 votes]: Let $O$ be a circumcentre of $\triangle ADS$. Then $\triangle OAS \sim \triangle PAB$, thus $\triangle POA\sim \triangle BSA$ and $BS:PO=SA:OA$, $\angle (BS,PO)=\angle (SA,OA)$. Analogously $CS:PO=SD:OD$ and $\angle (CS,PO)=\angle (SD,OD)$. This gives that $BS:CS=SA:SD$ and by angle chasing $\angle BSC=\gamma$, so $\triangle BSC\sim \triangle ASD$ (with different orientation), and we get the first part of your question. -Now let $K$ by symmetric to $C$ with respect to line $BS$. Let $E$ be a point such that $\triangle CED\sim \triangle CSK$, so that $E$ lies on the perpendicular bisector to $CD$ and $\angle CED=2\gamma$. We need to prove that $SE\perp AB$. We get $\triangle CES\sim \triangle CDK$, thus $\angle (ES,DK)=\angle (CE,CD)=\pi/2-\gamma$. But $\triangle DSK\sim \triangle ASB$, thus $\angle (DK,AB)=\angle (DS,AS)=\gamma$, and $\angle (ES,AB)=\angle (ES,DK)+\angle (DK,AB)=\pi/2$.<|endoftext|> -TITLE: Slices for certain $C_p$-spectrum -QUESTION [5 upvotes]: By the work of Hill-Yarnall, for the group $G=C_p,$ all the slices for any spectrum, in particular, for $S^V \wedge H\underline{\mathbb{Z}}$, are classified. Here $V$ is a representation of $C_p.$ -Again, following Yarnall's work, we know the spectrum $S^n \wedge H\underline{\mathbb{Z}}$ has the $n$-slice of the form $S^{W(n)}\wedge H\underline{\mathbb{Z}}$. Here $W(n)$ is a certain representation defined in Definition 3.5 of Yarnall's work "The slices of $S^n \wedge H\underline{\mathbb{Z}}$ for cyclic $p$-groups. -$\mathbf{Question:}$ If we take any $C_p$-representation $V=m+n\xi$. Is it true the $\dim(V)$-slice of the spectrum $S^V \wedge H\underline{\mathbb{Z}}$ is of the form $S^{U(m,n)}\wedge H\underline{\mathbb{Z}}$? If so, can we write explicitly what this representation $U(m,n)$ is (may be in terms of $W(m)$ and $n$)? -Thank you so much in advance. Any help will be appreciated. - -REPLY [3 votes]: This follows from the Hill-Yarnall formula for slices (I guess they do regular slices, so you have to deal with a shift if you want the classical ones). The reason is as follows: the slice of $X$ in dimension $n$ is given by first applying some algebraic procedure to $\pi_WX$ where $W$ is a certain representation of dimension $n$, and then suspending that Mackey functor by $W$. It turns out that, in our case, this algebraic procedure will always split out a Mackey functor which is equivalent to a suspension of $\underline{\mathbb{Z}}$. -Specifically, we have $S^V \wedge \underline{\mathbb{Z}}$ and you want the $\mathrm{dim}(V)$-slice. So we'll need to compute some $\pi_{W-V}\underline{\mathbb{Z}}$ where $W-V$ has dimension zero. You either get $\underline{\mathbb{Z}}$ or you get the transferred version $\underline{\mathbb{Z}}_{\mathrm{tr}}$ (when $p=2$ there is one further possibility, which is the Mackey functor that has $\mathbb{Z}$ on underlying with the sign representation and $0$ on fixed points). Now you do one of three possible algebraic procedures: (i) nothing, (ii) mod out the kernel of the restriction, or (iii) take the submackey functor generated by the transfer. The possible results of these algebraic procedures are again either $\underline{\mathbb{Z}}$ or $\underline{\mathbb{Z}}_{\mathrm{tr}}$ (or that extra possibility at $p=2$). So the $\mathrm{dim}(V)$-slice is given by the $W$-suspension of either $\underline{\mathbb{Z}}$ or $\underline{\mathbb{Z}}_{\mathrm{tr}}$. But now $\underline{\mathbb{Z}}_{\mathrm{tr}} \simeq \Sigma^{2-\lambda}\underline{\mathbb{Z}}$, so that's secretly also a suspension of $\underline{\mathbb{Z}}$. (Similarly, when $p=2$ you have the additional equivalence that $\Sigma^{1-\sigma}\underline{\mathbb{Z}}$ is the Mackey functor with $0$ fixed point part and $\mathbb{Z}$ with the sign representation on underlying). -You can work out what $W$ is explicitly (there's just gonna be several cases depending on writing the dimension of $V$ in the form $rp+2k+\varepsilon$ and splitting up into subcases depending on whether some quantity is positive or negative and so on...)<|endoftext|> -TITLE: Weakly homogenously Souslin sets and the measurability of $\omega_1$ -QUESTION [7 upvotes]: I found this intriguing remark at the end of Woodin's Supercompact cardinals, sets of reals, and weakly homogeneous trees (1988): - -The assertion that every set of reals, in $L(\mathbb{R})$, is the projection of a weakly homogeneous tree has consequences beyond the usual regularity properties such as Lebesgue measurability. For example by results of Kechris it follows that $L(\mathbb{R})\vDash$ ''$\omega_1$ is measurable''. - -Today, we would prove the last sentence by noting that if every set of reals in $L(\mathbb{R})$ is the projection of a weakly homogeneous tree (i.e., is weakly homogeneously Souslin), then $L(\mathbb{R})\vDash$ AD (and therefore $L(\mathbb{R})\vDash$ ''$\omega_1$ is measurable''). But in 1988 this was not known, so there should be a different proof, that uses just the structure of weakly homogeneously Souslin sets. Is there a source somewhere for such a proof? Or maybe it is an easy consequence of a previous result by Kechris? - -REPLY [8 votes]: This result is implied by a result in Kechris's paper "Subsets of $\aleph_1$ constructible from a real". Kechris proves the following: If there is a measurable cardinal, then every subset of $\omega_1$ is constructible from a real if and only if every subset of $\omega_1$ is weakly homogeneously Suslin in the codes. This implies that the closed unbounded filter is an ultrafilter, since if $x^\#$ exists, then the closed unbounded filter restricts to an $L[x]$-ultrafilter. The measurable cardinal is unnecessary for the direction you're interested in, because if there is a weakly homogeneously Suslin set that is not $\mathbf{\Sigma}^1_1$, then there is a measurable cardinal.<|endoftext|> -TITLE: From Topoi to Grothendieck categories -QUESTION [9 upvotes]: This question is mostly about a reference request. Let $\mathcal{E}$ be a Grothendieck topos. I am looking for a reference of the following two facts. I am aware that $(2) \Rightarrow (1)$ by Gabriel-Popescu. I was sure I had seen $(1)$ in Bourceux's encyclopedia, but I can't find it anymore. - -The category of internal abelian group objects $\mathsf{Ab}(\mathcal{E})$ is a Grothendieck category. -Call $\mathsf{Set}[\mathsf{Ab}]$ the classifying topos of abelian groups, and let $\mathcal{E} \simeq \mathsf{Sh}(C,J)$. Then $$\mathsf{Ab}(\mathcal{E}) \simeq \mathsf{Cocontlex(\mathsf{Set}[\mathsf{Ab}], \mathcal{E})} \simeq \mathsf{lex}(\mathsf{Ab}_\omega,\mathcal{E}) \simeq \mathsf{lex}(\mathsf{Ab}_\omega,\mathsf{Sh}(C,J)) \simeq \mathsf{Sh}(C,\mathsf{Ab}).$$ - -REPLY [9 votes]: Johnstone's "Topos theory" has a chapter on cohomology; your (1) is the first result the author proves. I don't remember about (2), but maybe it's also there?<|endoftext|> -TITLE: Definition of an n-category -QUESTION [24 upvotes]: What's the standard definition, if any, of an $n$-category as of 2020? The literature I can tap into is quite limited, but I'll try my best to list what I had so far. -In [Lei2001], Leinster demonstrated 10 different definitions for an $n$-category, and made no comment on whether they are equivalent or not. In [BSP2011], the authors set up axioms and claimed that all (many?) definitions of an $(\infty,n)$-category so far satisfy their axioms, and therefore are equivalent (up to some action). I include those definitions here for completeness: - -(a) Charles Rezk’s complete Segal Θn-spaces, -(b) the n-fold complete Segal spaces, -(c) André Hirschowitz and Simpson’s Segal n-categories, -(d) the n-relative categories of Clark Barwick and Dan Kan, -(e) categories enriched in any internal model category whose -underlying homotopy theory is a homotopy theory of (∞, -n)-categories, -(f) when n = 1, Boardman and Vogt’s quasicategories, -(g) when n = 1, Lurie’s marked simplicial sets, and -(h) when n = 2, Lurie’s scaled simplicial sets, - -However, all cases in [Lei2001] do not seem to be covered, and there are even more here. What's the crucial difference between defining an $n$-category and an $(\infty,n)$-category? -Question -In short, there are many definitions for higher categories.. so which one should we use? Is there a list of all definitions made, and a discussion on which is equivalent to which under which sense? Are there also discussions on which definition satisfies the three hypotheses - -stabilization hypothesis -tangle hypothesis -cobordism hypothesis - -postulated in [BD1995]? -Reference - -[Lei2001]: A Survey of Definitions of n-Category-[Tom Leinster]-[Theory and Applications of Categories, Vol. 10, 2002, No. 1, pp 1-70] -[BSP2011]: On the Unicity of the Homotopy Theory of Higher Categories-[Clark Barwick and Christopher Schommer-Pries]-[arXiv:1112.0040] -[BD1995]: Higher-dimensional Algebra and Topological Quantum Field Theory-[John C. Baez and James Dolan]-[arXiv:q-alg--9503002] - -Related - -Translating developments over different foundations - -REPLY [19 votes]: First of all, there are important differences between the notions of strict $n$-category, weak $n$-category, and $(\infty,n)$-category. The easiest notion is that of a strict $n$-category, and there's no doubt about the definition there: a strict $0$-category is a set, and by induction a strict $n$-category is a category enriched in the category of $(n-1)$-categories. -It's good that you cited Baez and Dolan's paper, which introduced an early model for the notion of a weak $n$-category. Between 1995 and 2001 there was a huge proliferation of other models. Morally, they should be categories weakly enriched in the category of weak $(n-1)$-categories, but there are many ways to define a weak enrichment, because there are many ways of keeping track of higher cells and how they combine. In 2004 there was a conference to try to get everyone together and figure out the commonalities between the models, and which were equivalent to which others. It did not result in one emerging as the "standard" model, and I don't think you should expect that to happen any time soon. However, we now know that models for weak $n$-categories broadly fall into two camps. Wikipedia says it nicely: - -There are basically two classes of theories: those in which the higher cells and higher compositions are realized algebraically (most remarkably Michael Batanin's theory of weak higher categories) and those in which more topological models are used (e.g. a higher category as a simplicial set satisfying some universality properties). - -Wikipedia also says "Several definitions have been given, and telling when they are equivalent, and in what sense, has become a new object of study in category theory." This matches my understanding of the field as it currently stands. I think of higher category theory as being interested in questions about the many models for weak $n$-categories. That's different from the study of $(\infty,n)$-categories, which is situated more in homotopy theory. -Now, others might come along and say "$(\infty,n)$-categories are the right thing" because MathOverflow has a larger representation of homotopy theorists than higher category theorists. You might get the same feel from reading the nLab, again based on who writes there. But if you go hang out in Sydney, Australia, where higher category theory is alive and well, you will not hear people saying $(\infty,n)$-categories are the "right" model or that the unicity theorem for $(\infty,n)$-categories solves the problem from 2004 of figuring out which models of weak $n$-categories are equivalent. -There is also plenty of ongoing work related to the stabilization hypothesis, tangle hypothesis, and cobordism hypothesis in various models of weak $n$-categories. For example, Batanin recently proved the stabilization hypothesis for Rezk's model based on $\Theta_n$-spaces. Then Batanin and I gave another proof that holds for a whole class of definitions of weak $n$-categories, including Rezk's model. Way back in 1998, Carlos Simpson proved the stabilization hypothesis for Tamsamani's definition of weak n-categories. Gepner and Haugseng proved the stabilization hypothesis for $(\infty,n)$-categories and the type of weak enrichment you'd get using Haugseng's PhD thesis (on enriched $\infty$-categories). Of course, famously, Lurie wrote thousands of pages towards proving the cobordism hypothesis for $(\infty,n)$-categories, and Ayala and Francis gave a shorter proof using factorization homology. -I'm sure there's lots of literature I missed, and I'm sure some will disagree with me in saying "yes, it is still valuable to study models of weak $n$-categories instead of only studying $(\infty,n)$-categories." But you asked for references so here are a bunch to get you started.<|endoftext|> -TITLE: Tree-width of graphs in which any two cycles touch -QUESTION [6 upvotes]: Let $G$ be a graph s.t. any two cycles $C_1, C_2 \subseteq G$ either have a common vertex or $G$ has an edge joining a vertex in $C_1$ to a vertex of $C_2$. Equivalently: for every cycle $C$ the graph obtained from $G$ by deleting $C$ and all neighbors of $C$ is acyclic. Let's denote the class of all such graphs by $\mathcal{G}$. -The cycle $C_n$, the complete graph $K_n$ and the complete bipartite graph $K_{s,t}$ are rather trivial examples of such graphs. -Are there constants $g, t$ such that all $G \in \mathcal{G}$ of girth at least $g$ (that is, all cycles of $G$ have length $> g$) have tree-width at most $t$? -UPDATE: As pointed out in the comments, the desired conclusion that $G$ has tree-width at most $t$ may as well be replaced by "contains at most $t$ disjoint cycles". - -REPLY [3 votes]: I tried to prove the statement for a while and I think I managed to narrow it down to one particularly difficult case. In the end, it led me to a counter example, showing there are no such values $g$ and $t$. This came as a bit of a surprise for me. The construction goes as follows. -(1) For every $n \geq 1$ there is a cycle $C$ and a labeling $\varphi: V(C) \to [n+1]$ such that $|\varphi^{-1}(n+1)| = 1$ and for every non-trivial path $P = xPy \subseteq C$ and all $i < \min\{\varphi(x), \varphi(y)\}$, $P$ contains a vertex labeled $i$. -proof: By induction on $n$, the case $n =1$ being trivial. In the inductive step, start from $(C, \varphi)$ for $n$, and obtain $C'$ from $C$ by subdividing every edge. Let $\varphi'(x) = \varphi(x)+1$ for $x \in C$ and $\varphi'(x) = 1$ for $x \in C' \setminus C$. -(2) Let now $n$ be given. Start with the disjoin union of $n$ copies $C_1, \ldots, C_n$ of the labled cycle from (1). Subdivide every edge of each cycle $n$ times, leaving the new vertices unlabeled. For every $i$, let $x_i \in C_i$ be the unique vertex labeled $n+1$. Join $x_i$ to all vertices on $\bigcup_{i < j \leq n} C_j$ labeled $i$. -It's easy to see that every cycle $D$ must contain at least one of $x_1, \ldots, x_n$. Let the minimum $1 \leq i \leq n$ with $x_i \in D$ be the index $\mathcal{idx}(D)$ of $D$. Moreover, we can see that $D$ contains a neighbor of $x_i$ for all $i < \mathcal{idx}(D)$. -Let $D_1, D_2$ be two cycles of $G$, wlog $\mathcal{idx}(D_1) \leq \mathcal{idx}(D_2)$. If equality holds, then $D_1 \cap D_2$ is non-empty. If $\mathcal{idx}(D_1) <\mathcal{idx}(D_2)$, then there is an edge from $D_1$ to $D_2$. Either way, any two cycles touch. -Moreover, since $G$ has disjoint pairwise touching cycles $C_1, \ldots , C_n$, the tree-width of $G$ is at least $n-1$. Since every cycle must contain an edge of at least one cycle $C_i$, the girth of $G$ is at least $n$.<|endoftext|> -TITLE: Functions on Stone spaces as "enveloping algebra" of Boolean algebra -QUESTION [7 upvotes]: I'm looking for references for the following closely related facts: -Given a Boolean algebra $B$, I denote by $\mathbb{Z}[B]$ the free ring generated by symbols $e_b$ such that $e_b e_{b'} = e_{b \cap b'}$ and $e_b + e_{b'} = e_{b \cup b'}+ e_{b \cap b'}$. -Then: - -The $e_b$ are the only idempotent of $\mathbb{Z}[B]$. - -$\mathbb{Z}[B]$ identifies with the algebras of continuous function from the Stone spectrum of $B$ to $\mathbb{Z}$. - - -Ideally, I would like a proof that is constructive (using the localic Stone spectrum) and applies to ``non-unital'' Boolean algebras, but the closest approximation would already be good. - -REPLY [3 votes]: Given a Boolean algebra, unital or non-unital, and a commutative ring $K$, the $K$-algebra $K[B]$ given by the generators and relations you give is isomorphic to the ring $C_c(\widehat B,K)$ of locally constant $K$-valued functions on the Stone space $\widehat B$ with compact support. -The intuitive reason is that if $\mathfrak p$ is a prime ideal of $K[B]$, then the only idempotents of $K[B]/\mathfrak p$ are $0$ and $1$ and so the elements of $B$ going to $1$ form an ultrafilter in your Boolean algebra. So basically a prime ideal is coordinatized by an ultrafilter on $B$ and a prime ideal of $K$. Now if you apply the Pierce sheaf representation (or if you work with the usual structure sheaf I suppose but this looks messier when $K$ is not a field) you will get to the result you want. -There is also the down and dirty proof. I’ll write $b$ instead of $e_b$, for $b\in B$. Note your relations makes the $0$ of the Boolean algebra the zero of the ring. Namely, you have a map from $K[B]$ to $C_c(\widehat B, K)$ taking $b$ in $B$ to the characteristic function $\chi_b$ of the compact open set of all ultrafilters containing $b$. The characteristic functions clearly satisfy your relations and span $C_c(\widehat B,K)$. So you just need that the natural map is injective. This is pretty easy. -If $\sum_{i=1}^nc_i\chi_{b_i}=0$, then we can use the standard disjointification trick and find disjoint non zero elements $b_1',\ldots,b_m'$ of the boolean algebra generated by $b_1,\ldots b_n$ such that each of $b_1,\ldots, b_n$ can be expressed as joins of $b_1’,\ldots, b_m’$ (take atoms from this finitely generated and hence finite boolean algebra for example). Using the defining relations of $K[B]$ you can then write $\sum_{i=1}^nc_ib_i=\sum_{i=1}^md_ib_i’$ for appropriate $d_i$. -Now we get $0=\sum_{i=1}^md_i\chi_{b_i’}$ but since this is a sum of characteristic functions of disjoint sets and only $0\in B$ maps to the empty set because every each nonzero element of $B$ is contained in an ultrafilter, we deduce each $d_i=0$. Therefore, $\sum_{i=1}^nc_ib_i=\sum_{i=1}^md_ib_i’=0$ in $K[B]$ and the map is injective. -If $K$ is a connected ring (with no idempotents except $0$ and $1$), then clearly the only idempotent elements of $C_c(\widehat B,K)$ are the characteristic functions of compact open sets, which by Stone duality are just the $\chi_b$ with $b\in B$ and hence the $b\in B$ are the only idempotents of $K[B]$ when $K$ is connected like $\mathbb Z$. -I think this result is just folklore, and as I said in the comments it is at least implicit in a number of locations such as Pierce and Keimel. In Theorem 3.11 of my paper, B. Steinberg, A groupoid approach to discrete inverse semigroup algebras I give a variation of the above proof to give a presentation for algebras of Hausdorff etale groupoids with Stone unit space. Unfortunately, that was omitted from the journal version of the paper. The same presentation was generalized to arbitrary etale groupoids with Stone unit space in an unpublished note of Buss-Meyer, but a different proof in the setting of tight algebras of inverse semigroups is given in Corollary 2.12 of B. Steinberg, N. Szakacs, Simplicity of inverse semigroup and étale groupoid algebras but this is further a field. The point is $K[B]$ is the tight algebra of the meet semilattice reduct of $B$.<|endoftext|> -TITLE: A question on the stability of $\operatorname{Cat}$ in $\operatorname{Cat}_\infty$ -QUESTION [12 upvotes]: $\DeclareMathOperator\Cat{Cat}$Suppose we have a span in $\Cat$ -$$ \require{AMScd} -\begin{CD} -A @> G>> X -\\ @VVFV -\\ B -\end{CD} -$$ -We can view this as a span in $\Cat_\infty$. What useful conditions can we impose to ensure the pushout is still a 1-category? -As a specific example, is either of the following conditions sufficient? - -$F$ is injective on objects and arrows -$F$ is injective on objects and arrows, and every isomorphism of the form $F(X) \cong F(Y)$ is in the image of $F : A(X,Y) \to B(F(X), F(Y))$ -Both $F$ and $G$ satisfy the property above - -Remark: This second proposition is the property $F$ is a monomorphism in $Cat_\infty$ together with the proposition that $F$ is an isocofibration in $Cat$ so that if the pushout in $Cat_\infty$ is a 1-category, it's given by taking the the pushout in $Cat$. For the question as asked we can drop the isocofibration condition. -Being injective on objects is not sufficient, since we have a pushout square -in $\Cat_\infty$ -$$ \require{AMScd} -\begin{CD} -S^1 @>>> 1 -\\ @VVV @VVV -\\ 1 @>>> S^2 -\end{CD} $$ -and $S^1 \to 1$ can be given by a functor between 1-object categories. -As @AchimKrause points out in the comments, injective on objects and arrows is not sufficient either. - -An example of a sufficient condition that does work (but is too restrictive for me) is if $A$, $B$, $X$ are all free categories and $F$ is obtained from an inclusion of the generating graphs. -In this case, we can compute this in the Bergner model structure on simplicially enriched categories. The map $A \to B$, when viewed in simplicial categories, is a cofibration between cofibrant objects (it is $\mathfrak{C}[-]$ applied to the inclusion of the generating graphs viewed as simplicial sets), and $X$ is cofibrant as well, and thus the pushout (which is obviously a 1-category) is a homotopy pushout, and thus computes the pushout in $\Cat_\infty$. -Another case that works, as described in the comments, is when $A$ and $B$ are groupoids and $F$ is a monomorphism in $Cat_\infty$; in this case, $B \cong A \amalg A'$, and thus the pushout in $Cat_\infty$ is $X \amalg A'$. - -REPLY [10 votes]: Martina Rovelli and I have indeed thought about the case of Dwyer morphisms before. Originally, we were also trying to employ Barwick-Kan, but I think there is the following subtle point there. -You have to specify how to look at a category as a particular relative category, and the natural way is to assign to a category $\mathcal{A}$ the pair $(\mathcal{A}, \mathrm{iso}(\mathcal{A}))$. However, I think this functor does not take Dwyer morphisms in categories to Dwyer morphisms in relative categories. Indeed, I think that already the inclusion of the object $a$ into the category ${a -TITLE: Is the forgetful functor $\mathrm{Mod}_R \mathrm{Sp} \rightarrow \mathrm{Sp}$ faithful? -QUESTION [9 upvotes]: $\DeclareMathOperator{\Sp}{\mathrm{Sp}}$I am taking a special case $\Sp$ here, mainly because it has nice categorical properties. -Let $R$ be an $E_\infty$-ring spectrum. In Higher Algebra, Lurie proves we have a forgetful functor (part of monadic adjunction) -$$ U_R:\operatorname{Mod}_R(\Sp) \rightarrow \Sp$$ -where $\Sp$ is in the $\infty$-category of spectra. -$U_R$ reflects equivalences. But is $U_R$ faithful in the sense that that the induced map of -$$Map(x,y)\rightarrow Map(U_Rx,U_Ry)$$ -mapping spaces is $-1$-truncated in the $\infty$-category of spaces. i.e. the homotopy fibers are $-1$-truncated. - -One categorically, $U$ is faithful in many cases, i.e. if we replace $\Sp$ with $\mathrm{Ab}$. -Perhaps the answer is false in $\infty$-categories. -I'd like to understand what goes wrong. Some comments on the following would be helpful: - -A counter example where $U_R$ is not faithful. (i.e. is it faithful when $R=H\Bbb Z$? ) -A brief/reference explanation for what accounts of this. - -REPLY [13 votes]: $U_R$ obviously preserves delooping, so if that were the case, because $\pi_0 map(X,Y) = \pi_1 map(X, \Sigma Y)$, you would also get an isomorphism on $\pi_0$, so an equivalence of mapping spaces. -In other words, $U_R$ is faithful if and only if it is fully faithful. But now for a map of ring spectra $R\to S$, the forgetful $Mod_S \to Mod_R$ is fully faithful if and only if $R\to S$ is an epimorphism of ring spectra (good examples are localizations - be careful that classical examples such as $R\to R/I$ for a usual ring $R$ tend to fail). -This is to say that "being an $S$-module" becomes a property of an $R$-module, rather than additional structure - so of course you can expect that to be very rare. -In your example of $H\mathbb Z$, it doesn't hold at all - you can for instance detect it on the level of the ring of stable cohomology operations of singular cohomology, which is bigger than just $\mathbb Z$ (look at the (co)homology of Eilenberg-MacLane spaces) - -REPLY [9 votes]: In general, the functor $U_R$ does not induce isomorphisms on higher homotopy groups of mapping spaces. Let $R=H(\mathbf{Z}/2)$. -Then $\pi_*(map(R,R))$ is the Steenrod algebra $\mathcal{A}^*$ where $map$ denotes the mapping spectrum. -The mapping spectrum $map(R,R)$ therefore has non-zero homotopy groups in negative degrees and differs from the mapping spectrum of $R$-module maps from $R$ to itself, which is just $R$ again, whose homotopy groups consist of $\mathbf{Z}/2$ concentrated in degree zero. -To see this difference directly in terms of mapping spaces as opposed to mapping spectra, we consider maps from $R$ to deloopings of $R$. For example, -$$\pi_1(Map_{R-Mod}(R, R[2])) \cong \pi_0(Map_{R-Mod}(R, \Omega R[2])) \cong \pi_0(Map_{R-Mod}(R, R[1])) \cong \mathrm{Ext}^1_R(R,R) = 0$$ -but -$$\pi_1(Map_{Sp}(R,R[2])) \cong \pi_0(Map_{Sp}(R, \Omega R[2])) \cong \pi_0(Map_{Sp}(R,R[1])) = \mathcal{A}^1 \cong \mathbf{Z}/2$$ -so the induced map on $\pi_1$ is not surjective.<|endoftext|> -TITLE: A “paradox” about the inner model problem -QUESTION [16 upvotes]: As stated in Woodin, Davis, and Rodriguez - The HOD dichotomy, a longstanding open problem in set theory is to construct a canonical inner model for supercompactness. In general there are various ways of spelling out what a “canonical inner model of large cardinal axiom A” means. The linked paper, as well as other work of Woodin, suggests the following weak thesis: -Thesis 1: If there are canonical inner models of supercompactness, at least one of them should be a weak extender model (as defined in 1). -As it is said in the last paragraph on page 16 in the linked paper, supported by Example 27, the exhibition of a weak extender model of supercompactness is not sufficient to solve the inner model problem for supercompacts. Of course, $V$ is trivially such a model. This is why Thesis 1 is a weak thesis. -Now, due to difficulties involved in spelling out in a general context what a canonical inner model is, the consensus view seems to be that a “canonical inner model of large cardinal axiom A” should be a proper-class iterate of a mouse. A mouse is a small iterable model of a sufficient fragment of ZFC + A, which is canonical because of the Comparison Lemma. In particular, it makes sense to speak of “the smallest mouse for ZFC + A”. For example $M_1^\sharp$ is the smallest mouse above a Woodin cardinal, and $M_1$ is the canonical inner model of one Woodin cardinal obtained by iterating the top measure of $M_1^\sharp$ out of the ordinals. So based these general considerations, we have: -Thesis 2: All canonical inner models of sufficiently large cardinals are iterates of mice. -The low-level versions of these theses are in harmony. For example, assuming the existence of a measurable and a sharp for measurability, a weak extender model for measurability is also an iterate of a mouse. But they seem to come into conflict around supercompactness. It is shown in 1 that if $\delta$ is extendible and $M$ is a weak extender model for the supercompactness of $\delta$, then there is no $j : M \to M$ with critical point $\geq \delta$. On the other hand, iterates of mice (through all ordinals) are, by their construction, self-embeddable with arbitrarily high critical point. -Question: How do we resolve the “paradox” and clarify the inner model problem? - -REPLY [12 votes]: Inner model theorists use the word "canonical" to explain the problem in intuitive terms, it is indeed a vague problem, though it is as precise as anything in the region of superstrong cardinals. - -Inner model theorists understand the importance of the problem, the way it sits within our understanding of models of set theory, which is why they always aim to explain what the problem is about instead of saying what solving it would mean. - -One could very easily make the inner model problem to be any of the test questions Gabe had, and indeed answering those questions is the main motivation. - -While reading the comment above, I didn't feel offended, I did find it amusing. I am a bit puzzled why forcing people are not registering that the inner model problem is as much a problem of inner model theory as forcing. - -Continuing 4, how about "do large cardinals imply that $\omega_1$ carries a precipitous ideal"?, how about is there a poset that kills all precipitous ideals? this is an open problem from FMS. - -Is $\Sigma^2_2$ absoluteness conditioned to generic diamond true? Is the $\Omega$ conjecture true? - -Does $\sf PFA$ imply there is an inner model with a supercompact? - -Does $\sf MM^{++}$ imply that there are no divergent models of $\sf AD$? - -Is $\mathsf{AD}^{L(\Bbb R)}+\Theta^{L(\Bbb R)}>\omega_3$ consistent? - -Can you have 4 consecutive measurable cardinals (under ZF)? - -Can you force $\sf GCH+\neg\square_{\omega_3}+\neg\square(\omega_3)$ from a large cardinal weaker than a Woodin cardinal that is a limit of Woodin cardinals? - -The only known consistency proof of one of Hamkins' maximlaity principles is forcing over $\rm HOD$ of a model of $\sf AD_\Bbb R+\Theta$ is reg. - -I do share the view that IMT is different, the reason being that it feels more like doing physics than doing math. Within the subject there are two groups (not necessarily disjoint). There are those who think about the models themselves and those who think about how to build these models. Doing the second is much more like doing physics than doing math. The first is as mathematical as it can be. Most talks one hears these days are on constructing the models rather than studying them. - -While I understand that the set theoretic community respects the area, I do feel that people often are being unfair to it (I do not mean anyone here). The fact that inner model theorists put the time to explain their area seems to be not fully appreciated, while often one sits through talks that give incredibly technical constructions and the only motivation for it is that "it is interesting to have such a thing". It would have been nice if all mathematicians tried to explain their area, the main problems and the importance. By this I don't mean history of the problems, but why one actually wants to solve it. Then many areas also would look quite vulnerable. I do agree with the point that "canonical inner model" is bad terminology, but if you have the courage to explain why you want to do something you inevitably run into the issue of sounding imprecise and vague.<|endoftext|> -TITLE: Lower bound for probability of getting exactly one head with pairwise independence -QUESTION [6 upvotes]: Say we toss $d$ pairwise independent coins, each with probability $1/d$ of getting a head. What is the highest lower bound one can give for the probability of getting exactly one head? -If they had been fully independent then I believe the answer is $1/e$. - -REPLY [10 votes]: The highest lower bound is $1/d$. -Indeed, for each $j\in[d]:=\{1,\dots,d\}$, let $A_j$ denote the event of the head on the $j$th coin and let $X_j:=1_{A_j}$. Let $S:=X_1+\dots+X_d$. Then the event of getting exactly one head is $\{S=1\}$. -Note that $EX_j=p$ and (by the pairwise independence) $EX_jX_k=p^2+pq\,1(j=k)$ for all $j,k$ in $[d]$, where $p:=1/d$ and $q:=1-p$. So, by Chebyshev's inequality, -$$P(S\ne1)=P(|S-1|\ge1) \\ -\le E(S-1)^2 \\ -=ES^2-2ES+1\\ -=\sum_{j,k\in[d]}EX_jX_k-2\sum_{j\in[d]}EX_j+1 \\ -=d^2p^2+dpq-2dp+1\\ -=1-1/d,\tag{1}$$ -whence -$$P(S=1)\ge1/d,\tag{2}$$ -so that $1/d$ is indeed a lower bound on the probability of getting exactly one head. -It is also easy to see that this lower bound is attained. Indeed, consider the events -$$B_j:=\{X_j=1,S-X_j=0\}=\{X_j=1,S=1\},\\ -C_{j,k}:=\{X_j=X_k=1,S-X_j-X_k=0\}=\{X_j=X_k=1,S=2\}$$ -for $j,k$ in $[d]$ such that $jdp(1-dp)=c(1-c)$, so that $P$ remains bounded away from $0$, just as in the case when the $A_i$'s are assumed to be independent. Thus, there is a "phase transition" at $c=1-$. (The case $d=1$ is, of course, trivial.)<|endoftext|> -TITLE: The set of isomorphism classes of Z/nZ-equivariant line bundles over a 2 dimensional Z/nZ-CW complex -QUESTION [7 upvotes]: Suppose I wish to find the set of isomorphism classes of $\mathbb{Z}/n\mathbb{Z}$-equivariant line bundles over a 2-dimensional, compact $\mathbb{Z}/n\mathbb{Z}$-CW-complex $X$, i.e. $\mathrm{Vect}^{1}_{\mathbb{Z}/n\mathbb{Z}}(X)$, I can write it using equivariant homotopy theory as -$\mathrm{Vect}^{1}_{\mathbb{Z}/n\mathbb{Z}}(X)\approx[X,B_{\mathbb{Z}/n\mathbb{Z}}GL_{1}(\mathbb{C})]_{\mathbb{Z}/n\mathbb{Z}}$. -Nonequivariantly, $B_{\mathbb{Z}/n\mathbb{Z}}GL_{1}(\mathbb{C})$ is $K(\mathbb{Z},2)$ so the above would seem like a Bredon cohomology group $H^2_{\mathbb{Z}/n\mathbb{Z}}(X;M)$ with $M$ some coefficient system and I could try to use the universal coefficient spectral sequence to attempt at computing it, however it seems like an awful lot of machinery for one of the simpler cases. -If the action is relatively nice, though it has fixed points, is there a simpler way of finding -$\mathrm{Vect}^{1}_{\mathbb{Z}/n\mathbb{Z}}(X)$? -If not, which coefficient system $M$ is the right one? -Thanks for your time - -REPLY [4 votes]: You can replace $GL_1(\mathbb C)$ with its maximal compact subgroup, which is $S^1$. Since $S^1$ is an abelian compact Lie group, there is a natural $\mathbb Z/n$-equivariant equivalence -$$B_{\mathbb Z/n}S^1\xrightarrow{\simeq} \mbox{Map}(E\mathbb Z/n, BS^1).$$ -See, for example, this MO question for a discussion of this equivalence, with references and some generalizations. -It follows that for a compact $\mathbb Z/n$-complex $X$, $\mbox{Vect}^1_{\mathbb Z/n}(X)$ is $\pi_0$ of the space -$$\mbox{Map}(X\times E\mathbb Z/n, K(\mathbb Z, 2))^{\mathbb Z/n}\simeq \mbox{Map}(X\times_{\mathbb Z/n} E\mathbb Z/n, K(\mathbb Z, 2))$$ -So the answer is the second cohomology group of the homotopy orbit space $X\times_{\mathbb Z/n} E\mathbb Z/n$, rather than a Bredon cohomology group of $X$ with coefficients in a Mackey functor. -The Leray-Serre spectral sequence for computing the (co)homology of the homotopy orbit space is often tractable.<|endoftext|> -TITLE: A strong form of Mostow rigidity without geometrization? -QUESTION [9 upvotes]: Mostow rigidity theorem says that two closed hyperbolic manifolds with isomorphic fundamental groups are isometric. -Here is my question: suppose that $M$ and $N$ are two closed 3-manifolds such that $M$ and $N$ are homotopy equivalent and such that $N$ is hyperbolic. Is it possible to prove that $M$ and $N$ are homeomorphic (diffeomorphic) without using geometrization theorem? - -REPLY [10 votes]: Gabai proved that homotopy hyperbolic 3-manifolds are virtually hyperbolic, in the paper of that name: - -Gabai, David, Homotopy hyperbolic 3-manifolds are virtually -hyperbolic. -J. Amer. Math. Soc. 7 (1994), no. 1, 193–198. - -I suspect this is the best you can do without geometrisation.<|endoftext|> -TITLE: An elementary inequality for three complex numbers -QUESTION [7 upvotes]: The following problem arose in asymptotic analysis of difference equations. - -Numerical maximization suggests that for all nonzero complex numbers $a,b,c$ we have -$$h\big(r(a,b,c),r(b,c,a),r(c,a,b)\big)\le2,\tag{1}$$ -where -$$r(a,b,c):=\Big|\frac{a b + a c - b c}{a^2}\Big|$$ -and $h(\cdot,\cdot,\cdot)$ is the harmonic mean, -with the equality in (1) iff $a,b,c$ are the vertices of an equilateral triangle centered at $0$. -Is this true? - -Remark 1: If the above conjecture is true, then obviously it will also hold if the harmonic mean $h\big(r(a,b,c),r(b,c,a),r(c,a,b)\big)$ is replaced by $\min\big(r(a,b,c),r(b,c,a),r(c,a,b)\big)$. -Remark 2: For nonzero real $a,b,c$, it appears that the best upper bound $2$ on $h\big(r(a,b,c),r(b,c,a),r(c,a,b)\big)$ should be replaced by $3/2$, "attained in the limit" iff two of the three real numbers $a,b,c$ are equal to each other while the remaining one goes to $0$. -Remark 3: For nonzero real $a,b,c$, Mathematica says that the best upper bound on $\min\big(r(a,b,c),r(b,c,a),r(c,a,b)\big)$ is $1$. - -A correct proof of the main, "harmonic-complex" conjecture or of the conjecture stated in Remark 2 or a "human" proof of the fact stated in Remark 3 would be enough for acceptance. - -REPLY [12 votes]: I will prove the original inequality. -First, performing the change of variables $x=1/a$, etc., and inverting the harmonic mean, we need -$$ - \sum \left|\frac{yz}{x(y+z-x)}\right|\geq \frac32. -$$ -Next, denoting $p=y+z-x$, etc., we transform the inequality to -$$ - \sum\left|\frac{(p+q)(p+r)}{p(q+r)}\right|\geq 3, -$$ -or -$$ - \sum\left|2+\frac{(p-q)(p-r)}{p(q+r)}\right|\geq 3. -$$ -Now we perform the last change -$$ - u= \frac{(p-q)(p-r)}{p(q+r)}, \quad\text{etc.,} -$$ -and notice that -$$ - \sum\frac 1u=\sum \frac{p(q+r)}{(p-q)(p-r)} =-1 -$$ -(which is just a three-variable identity). Thus, it suffices to show that -$$ - \sum |2+u|\geq 3 \quad \text{whenever}\quad \sum\frac1u=-1. -$$ -Denote $|2+u|,|2+v|,|2+w|$ by $r_{1,2,3}$, respectively. -Assume first that $r_i\leq 2$ for all $i$. The inverse image of $|2+z|=r$ (for $r\leq 2$) is a circle whose rightmost point is $-1/(r+2)$, hence, say, -$$ - \Re \frac1u\leq - \frac1{r_1+2}. - \qquad(*) -$$ -Adding up three such inequalities, we obtain -$$ - -1\leq-\sum\frac1{r_i+2} - \leq-\frac 9{\sum(r_i+2)}, -$$ -whence $\sum(r_i+2)\geq 9$, as desired. -Assume now that, say, $r_1>2$ (but $r_1<\sum r_i<3$, arguing indirectly). Then $\frac1u$ lies outside the disk having diameter $[-1/5,1]$ (as $|2+u|<3$). On the other hand, we have $r_2+r_3<1$, so $\frac1v, \frac1w$ lie in the open disks having diameters -$$ - \left[-\frac 1{2-r_i},-\frac1{2+r_i}\right]. -$$ -Since -$$ - \frac1{2-r_2}+\frac1{2-r_3}\leq \frac32 - \quad\text{and}\quad - \frac1{2+r_2}+\frac1{2+r_3}\geq \frac45, -$$ -point $\frac1u =-1-\frac1v-\frac1w$ lies in the open disk having diameter -$$ - \left[-1+\frac45,-1+\frac32\right] - = \left[-\frac15, \frac12\right]. -$$ -which contradicts the previous region indicated for $\frac1u $. -Equality arises only for $u=v=w=-3$, which seems to be rolled back easily.<|endoftext|> -TITLE: Has any open/difficult problem in ordinary mathematics been solved only/mostly by appeal to set theory? -QUESTION [26 upvotes]: We know that many (if not all) mathematical notions can be reduced to the talk of sets and set-membership. But it nevertheless sounds like a grueling task (if at all possible) to actually get advanced results in many branches of ordinary mathematics if we only work with sets and set-membership relation in our language, or otherwise only rely on set theory. To put it differently: it seems that in order to get results in many branches of mathematics one might not need to be very familiar with set theory at all, let alone being able to translate everything to the language of sets or to heavily rely on set theory. -I'm wondering if there are cases where an open/a difficult problem in other branches of mathematics (e.g., number theory or real analysis) has been solved mostly/only because of the insight that set theory has offered, directly or indirectly (say, through branches that heavily appeal to set theory, such as model theory). Even a historical incident will be helpful: a problem of the sort that was first solved thanks to set theory, but later on more accessible solutions have been found that don't deal much with sets. -Thank you very much! - -REPLY [11 votes]: Dehornoy order in braid theory was first discovered using large cardinal axioms. For a while it was an open problem if large cardinal axioms were actually required, although since then elementary approaches have been discovered.<|endoftext|> -TITLE: Do filtered colimits commute with finite limits in the category of pointed sets? -QUESTION [5 upvotes]: It seems to be the case that filtered colimits commute with finite limits in the category Set (for instance, this is shown in Why do filtered colimits commute with finite limits?), but does the same hold for the category of pointed sets? -For the case of when the finite limit is a binary product, the OP in the linked post explains why the claim holds in Set and generally any Cartesian closed category with filtered colimits. But the category of pointed sets is not Cartesian closed (namely because the category of pointed sets has a zero object and the only Cartesian closed category with a zero object is the trivial one). -So, first, is it even true that filtered colimits commute with binary products in the category of pointed sets (if not, is there an easy counterexample?), and if so, can we replace "binary products" with any "finite limit"? - -REPLY [11 votes]: I find it sometimes useful to remember that the stated commutation property holds in any ind-category. Pointed sets is ind-(finite pointed sets). -Of course, to verify your particular case by hand, working with elements, literally takes 5 minutes. But who wants to waste five minutes on elements?<|endoftext|> -TITLE: What is the meaning of the $L$-group? -QUESTION [5 upvotes]: Langlands' functoriality conjecture predicts that to a suitable homomorphism of $L$-groups -$$ -\psi : ^LG \to ^LH -$$ -there should be a transfer of automorphic representations from $G$ to $H$. For the purposes of discussion, let's take $^LG$ to be the Weil form -$$ -^LG = \hat{G}(\mathbb C) \rtimes W_{\mathbb Q} -$$ -where $W_{\mathbb Q}$ is the Weil group of $\mathbb Q$. This conjecture, as we know, has revealed many connections between disparate objects in representation theory, geometry, and number theory, and also works to explain various phenomena that we observe. My question is more on a philosophical level: setting aside the reasoning along the lines of "we believe it because it works," why should functoriality be true? -To narrow the question a little, what is the meaning of the $L$-group? How should we think of the semidirect product? What category does it live in? It blends a complex reductive group with the arithmetic of $\mathbb Q$, which is crucial to the entire framework of the Langlands program. As Casselman pointed out here, Langlands' letter to Weil already established that Langlands understood the centrality of the $L$-group, but this fact seems to have revealed itself through Langlands' deep experimental knowledge of Eisenstein series. Later work in geometric and $p$-adic Langlands reveal that the geometry of the $L$-group certainly realizes functoriality in certain senses, but I don't think it quite explains (for me, at least) the question of why. -The picture gets even muddier if we replace $W_\mathbb Q$ by the conjectural automorphic Langlands group $L_\mathbb Q$ as Langlands' reciprocity conjecture (perhaps) demands. -EDIT: To clarify a little further based on David Loeffler's answer. I realize that on some level it is a little bit of a fool's errand to ask such a meta question, but I will try to justify it. Certainly after over 50 years after Langlands' conjectures there is little doubt that they should be true, and as was pointed out, the $L$-group seems to arise in some natural sense especially in light of the Satake isomorphism (and its geometric variant too). This is along the line of what I mean by "we know it because it works." I think what I am trying to ask is in what sense might Langlands' Functoriality principle (as Arthur calls it) be more like an actual functor than just a principle? Here I am thinking of the usual local/global Langlands correspondences (which Langlands calls "reciprocity") as the special case of functoriality where $G$ is trivial. So at the base level, we have a functor from $$ -\{\text{admissible $L$-homomorphisms of $L$-groups}\} \to \{\text{packets of automorphic representations of reductive groups}\} -$$ -up to necessary equivalences, in a way that captures reciprocity as a special case, as Langlands originally formulated. (I understand that $p$-adic Langlands, among others, has discovered much more intricate data and Arthur's conjectures too, so I'd be happy to receive input on how to update this picture. Inded, people working on questions related to modularity have thought a lot about category theoretic, and nowadays derived, approaches, but not at the level of Functoriality as far as I kow.) -But the basic question is to what extent can we understand this in a more category theoretic way, so that this map might be an actual functor? With this in mind, this seems to lead quickly to the question of how should I think about the $L$-group arises from trying to make sense of the left-hand side in some meaningful way. Of course people have sought to study things like the stack of Langlands parameters, or quasicoherent sheaves on $\text{Rep}(^LG)$, but this all still seems to take the $L$-group for granted (with good reason of course), but if I think of the LHS as a homs of a category, what kind of category am I looking at? Is there some topological or geometric way in which it arises "naturally"? - -REPLY [5 votes]: I find this question somewhat strange; you ask "what is the meaning of the L-group?", but the survey article of Casselman which you link to is largely devoted to explaining the historical and conceptual motivation of the L-group -- in particular, how it naturally arises even in the relatively simple situation of [EDIT: unramified representations of] unramified reductive groups over nonarchimedean local fields that has been well-understood since the 1970's (no need for Eisenstein series, global Langlands groups, p-adic or geometric Langlands, or any other any other fancy machinery here). Are you saying that you don't understand Casselman's explanations, or that you don't find them convincing? -In fact, I had typed into this very box a rough sketch of how the L-group arises naturally when you consider the Satake isomorphism: first for split reductive groups over nonarch local fields, where $\hat{G}$ naturally arises; and more generally for unramified quasi-split ones, where ${}^L G$ naturally pops out. Then I checked the link and realised that this is exactly the content of sections 3-5 of Casselman's article, who explains it with far more clarity and authority than I ever could. If that doesn't answer your question, then perhaps you could clarify a little what the question actually is?<|endoftext|> -TITLE: Zariski's main theorem for non-representable morphisms? -QUESTION [6 upvotes]: Let $f:\mathcal{X}\to \mathcal{Y}$ be a separated quasi-finite map of qcqs Deligne-Mumford stacks. Is there a version of Zariski's main theorem that makes sense in this context? Rydh proved a version of this in the case where the map $f$ is also assumed to be representable, in which case we recover a stacky version of the classical factorization of $f$ as the inclusion of an open substack into a finite stack over $\mathcal{Y}$. -Obviously we can't hope for exactly such a factorization, since this would make $f$ automatically quasi-affine (and therefore representable). But I was wondering if maybe there is a factorization of $f$ into something like a (locally constant?) gerbe over an open substack of a finite stack over $\mathcal{Y}$. - -REPLY [5 votes]: You can take the relative coarse map to get a factorization of $f$ into $\mathcal{X} \to X \to \mathcal{Y}$ where $g : X \to \mathcal{Y}$ is representable and $\pi : \mathcal{X} \to X$ is proper + quasi-finite with $\mathcal{O}_X \to \pi_*\mathcal{O}_{\mathcal{X}}$ an isomorphism. Then you can apply the representable case of ZMT to $g$ to obtain a factorization -$$ -X \hookrightarrow \overline{X} \to \mathcal{Y} -$$ -where $X \hookrightarrow \overline{X}$ is an open immersion and $\overline{g} : \overline{X} \to \mathcal{Y}$ is finite. -Putting this together, we get that any such $f$ factors into -$$ -\mathcal{X} \xrightarrow{\rho} \overline{X} \xrightarrow{\overline{g}} \mathcal{Y} -$$ -where - -$\overline{g}$ is finite and in particular representable, -$\mathcal{X}$ is proper + quasi-finite over an open substack $i : X \subset \overline{X}$, and -$\rho_*\mathcal{O}_\mathcal{X} = i_*\mathcal{O}_X$. - -I think conditions $2 + 3$ can be replaced by something like $\mathcal{O}_\overline{X} \to \rho_*\mathcal{O}_\mathcal{X}$ is injective and integrally closed. -The existence of the relative coarse space is guaranteed under your assumptions by Theorem 3.1 here. Indeed the relative inertia stack is proper over $\mathcal{X}$ by the separated assumption and quasi-finite by the DM assumption. -I think by universality of the relative coarse map this is essentially the best you can do. In general the kernel of the map on inertia can jump so I don't think you can expect the first map to be a gerbe over an open substack, e.g., if $f$ itself the coarse space of a separated DM stack that is not a gerbe.<|endoftext|> -TITLE: Which singular homology classes can be represented by embedded manifolds? -QUESTION [6 upvotes]: Given a connected CW-complex $X$ I'm interested in if a given homology class $\sigma \in H_n(X)$ can be represented by a manifold meaning if there is a map $f : M^n \to X$ from a oriented manifold $M$ for which $f_*([M^n]) = \sigma$. Obviously this is always true for $n = 1$ and I could prove it for $n = 2$, but it seems this doesn't hold for any $n$. -For example I found this answer which talks about the case where $X$ is itself a manifold. It says there are cases where $\sigma$ is not represented by a manifold for $n = 7$. Are there similar results for $X$ that are not necessary manifolds? -I'm especially interested in the simpler case where $H_i(X) = 0$ for $1 < i < n$ - -REPLY [16 votes]: The question in the title differs from the question spelled out in the post: In the title, you ask for embedded manifolds, in the post you ask for just maps from manifolds. I think the version of the question asking for embedded manifolds but $X$ an arbitrary CW-complex is not very well-behaved, so let me answer the question in the post. -One way to think about this is that there is also a homology theory based on oriented manifolds mapping to $X$, called oriented bordism, $\operatorname{MSO}_*(X)$. The construction which assigns to a class represented by an oriented manifold with map to $X$ the image of its fundamental class in $H_*(X)$ comes as a natural transformation -$$ -\operatorname{MSO}_*(X) \to H_*(X) -$$ -of homology theories. In fact, it lifts to a map of spectra, $\operatorname{MSO}\to H\mathbb{Z}$, and this is the bottom map in the Postnikov tower for $\operatorname{MSO}$. This way, the question of when homology classes of $X$ are in the image of this natural transformation relates this to differentials in the Atiyah-Hirzebruch spectral sequence for $\operatorname{MSO}_*(X)$. The existence of homology classes which are not in the image corresponds to the fact that the map $\operatorname{MSO}\to H\mathbb{Z}$ does not split, but one can in fact work out explicit obstructions which lead to the examples you referred to. All this only depends on the homotopy type of $X$ (as opposed to the embedding question).<|endoftext|> -TITLE: Are generators defined in Tohoku paper equivalent to that defined in Wikipedia (Which I believe is a more widely used definition) -QUESTION [8 upvotes]: As I was reading Grothendieck's Tohoku paper(translated by M.L.Barr and M.Barr), I found that the definition of a generator in the category differs from that defined in wikipedia. -Let $\mathbf{C}$ be a category(It may be necessary that $\mathbf{C}$ is a locally small category), a family of generators {$U_i$}$_{i\in I}$ with $I$ being an index set, according to Tohoku paper, are a collection of objects such that for any object $A$ and any subobject $B \neq A$, there is $i\in I$ and a morphism $u\colon U_i \rightarrow A$ which does not come from $U_i \rightarrow B$. While in wikipedia, it is defined in a way such that for any $f,g\colon A\rightarrow B$ with $f\neq g$, there is an $i\in I$ and $u\colon U_i\rightarrow A$, such that $f\circ u \neq g\circ u$. -What I would like to know is that are these 2 definitions equivalent, or is the definition in Wikipedia stronger than that in Tohoku paper? - -REPLY [18 votes]: In this answer, let me use the terms generator and extremal generator for the Wikipedia and Tohoku definitions respectively. -In general, these two definitions are not equivalent, and neither implies the other. -Example 1. For an example of an extremal generator which is not a generator, consider a non-trivial group seen as a category with one object. The empty family is an extremal generator in this category (because there are no proper subobjects) but not a generator. -Example 2. An example of a generator which is not an extremal generator is the singleton family consisting of the terminal object (i.e. the singleton poset) in the category of posets. To see that it is not an extremal generator, observe that the discrete poset $\{0,1\}$ is a proper subobject of the poset $\{0 < 1\}$. -Nevertheless, there are many categories in which these implications do hold. -Proposition 1. In any category with equalisers, every extremal generator is a generator. -Proposition 2. In any balanced category (i.e. a category in which "monomorphism" + "epimorphism" => "isomorphism"), every generator is an extremal generator. -Hence in any balanced category with equalisers, the two notions of generator and extremal generator coincide. This includes all abelian categories (as in Grothendieck's paper) and all (pre)toposes.<|endoftext|> -TITLE: Atiyah-Singer for Riemannian and Kaehler manifolds -QUESTION [6 upvotes]: I am trying to understand the proof of the Atiyah--Singer index theorem, and would like to see how it works for two "simple" examples. Could somebody direct me to a proof for the special case of - -A Riemannian manifold and its associated Dirac operator -$$ -d+d^*: \Omega^\bullet \to \Omega^\bullet, -$$ -a Kaehler manifold $(M,g)$ and its Dirac operator -$$ -(\overline{\partial} + \overline{\partial}^*): \Omega^{0,\bullet} \to \Omega^{0,\bullet}. -$$ - -REPLY [10 votes]: I highly recommend the discussion in Shanahan's book, The Atiyah-Singer Index Theorem (An introduction), Lecture Notes in Math 638. In addition to a sketch of the proof, he gives a nice discussion of how the formidable general statement of the theorem gives the answers for your two examples, plus the (spin) Dirac operator and the signature operator. There are other treatments if you want to learn all of the details of the proof, but that book is excellent for the purpose of your question. -(Added later) A second reading of the question suggests that you are asking for a complete proof for these two cases, rather than instructions on how to deduce these cases from the full A-S theorem. -For (1), the ingredients are the Hodge theorem to identify the kernel and cokernel as the de Rham cohomology in even and odd dimensions. Then you need the de Rham theorem to identify these cohomology groups as (say) singular cohomology. This shows that the index is the Euler characteristic. Finally, you need to identify the Euler characteristic as the evaluation of the Euler class on the fundamental cycle of your manifold. You can find this latter in many places, eg Milnor-Stasheff. -I don't know that there is as direct a proof of (2), which is essentially the Hirzebruch-Riemann-Roch theorem.<|endoftext|> -TITLE: What is first-order logic with Dedekind-finite sets of variables? -QUESTION [10 upvotes]: The usual set up of first-order logic is with an infinite reservoir of variables which we can use in formulas. This is one of the annoying reasons why we need to put $\aleph_0$ into the cardinal equations, but it also provides us with the expressive freedom that we need. -It is not hard to see that there is no reason to consider any other cardinal, except $\aleph_0$, in this case: since formulas are finite, and proofs are finite, in any kind of proof there will only be finitely many variables. So anything larger than $\aleph_0$ is kind of irrelevant. But this assumes that all cardinals are comparable, and the Axiom of Choice makes things kinda nice. -Assuming that the Axiom of Choice fails, and badly, what happens when we substitute the reservoir of variables with some Dedekind-finite set? In particular, a set $A$ whose finite subsets (and finite injective sequences) form a Dedekind-finite set, or even an amorphous set? -Can we prove some interesting results (read: not entirely equal to standard first-order logic) in $\sf ZF$ (+ whatever set we needed exists), or at least some consistency results? For example, we don't need choice to prove that every theory in a finite language has a complete theory extending it, or that it has a model. What happens when we switch to this abominable version of first-order logic? - -REPLY [7 votes]: For sentences, nothing will change. The reason is that every sentence in the abominable language is canonically logically equivalent to a formula in the usual language, since every formula provides an ordering of the variables appearing in it, and they can then be replaced with ordinary variables from an $\omega$-indexed set of variables. -Because every sentence is logically equivalent to any of its notational variants (where we use different formulas), we can transfer any theory to an equivalent theory in the well-behaved language. -For example, a theory in the abominable language will be consistent if and only if its translation is consistent, since the proof system can validate any instance of notational variance, which involves only finitely many variables at a time. -For this reason, any consistent theory in the abominable language will have a completion: just translate it to the regular language, complete it there, and then take all notational variants of assertions of that completion. This will be complete in the abominable language. -A similar idea allows you to define the satisfaction relation for models and formulas in the abominable language. Just translate them to the regular language, use Tarski's definition, and then translate back. This will obey the Tarski recursion, even in the abominable language. -The only remaining issue seems to be when you have types involving infinitely many free variables. In this case, you are dealing with simultaneous assignments on those variables, where amorphicity might come into play.<|endoftext|> -TITLE: Famous cases of multiple papers by the same author published in same issue of same journal -QUESTION [6 upvotes]: I have been wondering if there are many cases of an author having published two (or more?) papers in the same issue of the same journal. I vaguely recall having seen one or two cases like this, maybe be old papers, but cannot vividly remember. I have the impression such a situation would make sense should the two papers be in the same topic, say, one is sort of a (substantial) continuation of the other, for instance (given the fact that in Mathematics there is some pressure against publishing too often in the same journal). I am of course asking this question for papers having a sole author (or maybe the same set of authors). - -REPLY [6 votes]: Edoardo Ballico published (until now) 404 articles in the International Journal of Pure and Applied Mathematics. He published many times two or more articles in the same issue. -For instance, in 2009 he published four articles in "Int. J. Pure Appl. Math. 57 (2009), no. 4,"<|endoftext|> -TITLE: How do the invariants of a group scheme action compare to the invariants of the group action by global sections -QUESTION [7 upvotes]: If $G$ is a group scheme over $S$ acting on an $S$-scheme $X$, I'd like to understand the algebra of invariants $(\mathcal{O}_X)^G$. Specifically, I'd like to understand its relation to invariants $(\mathcal{O}_X)^{G(S)}$. -To simplify notation, say everything is affine: $G = \operatorname{Spec}R$, $X = \operatorname{Spec}A$, and $S = \operatorname{Spec}k$, where $k$ is an arbitrary ring (not necessarily a field). If it helps we can assume $G$ is smooth. We work in the category of $k$-schemes. -The action is given by a map $\sigma : G\times X\rightarrow X$. Let $p : G\times X\rightarrow X$ be the projection map. Then there is a natural bijection $A = \operatorname{Hom}(X,\mathbb{A}^1)$, and by definition the subalgebra of invariants $A^G$ is the set of $f\in A$ whose corresponding map $F : X\rightarrow\mathbb{A}^1$ satisfies -$$F\circ\sigma = F\circ p$$ -Via $\sigma$, the group $G(k)$ acts on $X(k)$, and for any $k$-scheme $T$, $G(k)$ maps to $G(T)$ and hence also acts on $X(T)$, so $G(k)$ acts on $X$. Thus, we may also consider the ring of invariants $A^{G(k)}$. Certainly we have -$$A^G\subset A^{G(k)}$$ -My main question is: What is the clearest way to express this relationship? I'm looking for a statement of the form $f\in A$ is $G$-invariant if and only if it is fixed by $G(k)$ and some other conditions. -I think one can say that -$$A^G = \{f\in A| f\otimes_k 1\in A\otimes_k B \text{ is fixed by $G(B)$ for every $k$-algebra $B$}\}$$ -Is this correct? Is it possible to further restrict the class of $B$'s that you have to consider? Are there other ways of thinking about this? - -REPLY [5 votes]: Combining LSpice's (1 2) and my (1 2) comments into an answer: If $G(k)$ is Zariski dense in $G$, then $A^{G(k)} = A^G$. It is very common that $G(k)$ is Zariski dense in $G$: This happens whenever (1) $k$ is infinite and (2) $G$ is connected and either (3a) $G$ is reductive or else (3b) $k$ is perfect. See Density question in algebraic group.<|endoftext|> -TITLE: Probability of getting exactly one head and $k$-wise independence -QUESTION [5 upvotes]: Say we toss $d$ $k$-wise independent coins, each with probability $1/d$ of getting a head. What is the highest lower bound one can give for the probability of getting exactly one head? In the first instance, let us consider $k=3$ or $4$ but a result for general $k\geq2$ would be wonderful. If $k=d$ then $1/e$ is a lower bound independent of $d$. -Background -It turns out that for $k=2$ the answer is $1/d$ which is tight. If the probability of getting a head is instead $c/d$ for $0 k$. This will certainly be true $S=X_1+\ldots +X_d$ takes only values in $\{0,\ldots,k\}$. -We therefore try to find a distribution of the form -$$\mathbb{P}(X_1=x_1,\ldots,X_d=x_d)=a(x_1+\ldots+x_d)$$ -(the easiest case, the joint distribution depends only on the ``weight'' ) -with $a(i)=0$ for $i > k$. -We want $X_1,\ldots,X_d$ to be $k$-wise independent and identically distributed with $\mathbb{P}(X_1=1)=p$. -This will be true if for $j=0,...,k$ -$$\sum_{i=j}^d {d-j \choose i-j}a(i)=p^j\;\;.$$ -Solving this (triangular) system of linear equations we find that for $j=0,\ldots,k$ we have to set -$$a(j)=p^j\sum_{i=0}^{k-j}(-1)^i {d-j \choose i} p^i\;\;.$$ -These values will be a solution of the probabilistic problem if they are nonnegative. For $p=\frac{1}{d}$ this is easily shown to be the case. Thus in this case $\mathbb{P}(S=1)$ attains for even $k$ the lower bound given above. -Remarks: -(1) similarly, the construction achieves the corresponding ``Bonferroni upper bounds'' for odd $k$. -(2) for the case $k=d$ the probability -$$\mathbb{P}(S=1)=(1-\frac{1}{d})^{d-1}=\sum_{j=0}^{d-1}(-1)^j { d-1\choose j}\frac{1}{d^j}\;\;,$$ and the lower and upper bounds for the $k$-wise independent cases are just -the even resp. odd partial sums of the right hand side. -NOTE: for a given $k$ all $p\leq \frac{1}{d-k+1}$ give a probabilistic solution. The extreme cases $p=\frac{1}{d-k+1}$ of this construction -appear in problems 6 and 7 of the "Mathematical Preliminaries Redux" section in fascicle 5 of volume 4 of "The Art of Computer Programming".<|endoftext|> -TITLE: Elementary inhomogeneous inequality for three non-negative reals -QUESTION [5 upvotes]: I need the following estimate for something I am working on, but I don't immediately see how to establish it. -For $x, y, z \in \mathbb{R}_{\ge 0}$, show that -$$2xyz + x^2 + y^2 + z^2 + 1 \ge 2(xy + yz + zx),$$ -and I suspect the only point of equality is (1,1,1). -It feels like the sort of thing that ought to have a simple Olympiad-style proof using standard inequalities, but I haven't had any luck thus far; of course, I'll take any proof that I can get. -Also, if this inequality is reminiscent of any others, I would be grateful for references! - -REPLY [3 votes]: Another way. -Since $$\prod_{cyc}((x-1)(y-1))=\prod_{cyc}(x-y)^2\geq0,$$ we can assume that -$$(x-1)(y-1)\geq0,$$ which gives $$z(x-1)(y-1)\geq0$$ or -$$xyz\geq xz+yz-z.$$ -Id est, it's enough to prove that: -$$2xz+2yz-2z+x^2+y^2+z^2+1\geq2(xy+xz+yz)$$ or -$$(x-y)^2+(z-1)^2\geq0$$ and we are done!<|endoftext|> -TITLE: First order estimates of geodesic normal coordinates -QUESTION [6 upvotes]: Let $(M^n,g)$ be a complete Riemannian manifold with $|Rm| \le 1$. Can we find two positive constants $C$ and $\epsilon$, depending only on $n$, such that under the normal coordinates $(g_{ij})$ with respect to any point $p \in M$, we have -$$ -|\partial_k g_{ij}(x)| \le C -$$ -for any $|x| \le \epsilon$? -As pointed out in the comment, if the injectivity radius at $p$ is small, then the estimates should be understood for the pull-back of $g$ to the tangent space, which is always well-defined by the curvature bound. - -REPLY [9 votes]: The answer is 'no' for $n=2$ (and hence for all higher $n$). Here is how one can see this. -First, when $n=2$, recall that, by the Gauss Lemma, a metric $g$ in geodesic normal coordinates $(x,y)$ centered on $p$ takes the form -$$ -g = \mathrm{d}x^2 + \mathrm{d}y^2 - + h(x,y)\bigl(x\,\mathrm{d}y-y\,\mathrm{d}x)^2, -$$ -where the function $h$ is arbitrary, subject to the condition that $(x^2{+}y^2)h(x,y)+1>0$. -Letting $r^2 = x^2 + y^2$ and letting $R$ be the radial vector field $x\,\partial_x + y\,\partial_y$, one computes the formula for the Gauss curvature of $g$ to be -$$ -K = -\frac{2(1+r^2h)(RRh) - r^2(Rh)^2+2(5+3r^2h)(Rh) + 8r^2h^2+12h}{4(1+r^2h)^2}. -$$ -Thus, in the geodesic disk of radius $\epsilon>0$ about $p$, i.e., where $r^2=x^2 + y^2 \le\epsilon^2$, we can keep $|K|$ as small as we like merely by imposing sufficiently small bounds on $h$, $Rh$ and $RRh$, i.e., $h$ and its first two radial derivatives. More precisely, for any $M>0$, there exists a $\delta>0$ such that, if $|h|$, $|Rh|$ and $|RRh|$ are bounded by $\delta$ when $r\le\epsilon$, then $|K|\le M$ when $r\le \epsilon$. -Let $\rho(r)$ be a smooth function that is identically zero near $r=0$ and $r=\epsilon$ and, say, positive, at $r=\epsilon/2$, but satisfies the condition that, for any constant $\lambda$ with $|\lambda|\le 1$, the function $h(x,y) = \lambda\rho(r)$ yields a $K$ that satisfies the bound $|K|\le 1$. -Let $f(\theta)$ be any $2\pi$-periodic smooth function bounded by $1$ and consider the smooth function -$$ -h(r\,\cos\theta,r\,\sin\theta) = \rho(r)f(\theta). -$$ -Then $h$ and its radial derivatives are bounded in such a way that the Gauss curvature $K$ for the corresponding metric $g$ will be bounded in absolute value by $1$, but the 'angular derivative' of $h$, i.e., $xh_y-yh_x = \rho(r)f'(\theta)$, need not be bounded. In particular, by choosing $f$ appropriately (bounded by $1$ but with very large first derivatives), we can be sure that the coefficients of $g$ in this coordinate system, i.e., -$$ -g_{11} = 1 + y^2\,h(x,y),\qquad g_{12} = -xy\,h(x,y),\qquad g_{22} = 1+x^2\,h(x,y), -$$ -while bounded themselves, will have some very large first derivatives when $r = \epsilon/2$. In particular, there is no constant $C>0$ that would bound the first derivatives of these quantities independent of the choice of $f$.<|endoftext|> -TITLE: Assigning a "canonical geometry" to a Seifert surface -QUESTION [7 upvotes]: I originally posted this on stackexchange, but it hasn't gotten an answer. I hope it's not inappropriate for this forum. -Suppose I have a knot $K: S^1 \hookrightarrow S^3$ with minimal genus Seifert surface $S$. I would like to know whether we can endow $S$ with a geometry that is independent somehow of our embedding, assuming said embedding satisfies whatever conditions are appropriate. -I understand this is a naive/vague question, but I don't have any real background in geometric topology and so I'm not sure what conditions we would want the embedding to satisfy to even begin looking at geometric properties as knot invariants. I imagine there are some elementary theorems for 2-manifolds with boundary that would be useful here, but I'm not sure where to look for them. -I know that hyperbolic knots are characterized by the fact that their complements can be endowed with a geometry having constant curvature $-1$. Since we can embed $S$ in the complement as a smooth submanifold, does this also mean that all smooth Seifert surfaces for hyperbolic knots can likewise be given a geometry with constant curvature $-1$? Are we able to say anything at all about the surfaces for torus and satellite knots? - -REPLY [4 votes]: If $K$ is a non-trivial knot, then $\chi(S)<0$, so $S$ admits a hyperbolic structure as a surface. But in general, that metric does not arise from the emdedding into $S^3\setminus K$. -If $S^3\setminus K$ is hyperbolic, and $S$ is a properly embedded $\pi_1$-essential surface in $S^3\setminus K$, then $S$ is either virtually fibered, accidental, or quasi-fuchsian. See Bonahon, or Canary, Epstein, Green - Notes on notes of Thurston, or Thurston's notes. -In the case that $S$ is a minimal genus Seifert surface, then $S$ is either a fiber, or $S$ is quasi-fuchsian by work of Fenley. -Within the class of quasi-fuchsian surfaces are the surfaces which are totally geodesic. Totally geodesic Seifert surfaces will inherit a metric of constant curvature $-1$. This is quite rare for a Seifert surface, but examples are known due to Adams and Schoenfeld. -Perhaps the best bet for a canonical geometry on a Seifert surface $S$ in a hyperbolic knot complement, is to give $S$ the structure of a pleated surface. Again see Thurston's notes.<|endoftext|> -TITLE: What is the minimum size of a partial order containing all partial orders of size 5? -QUESTION [18 upvotes]: This earlier MO question asks to find the minimum size of a partial order that is universal for all partial orders of size $n$, i.e. any partial order of size $n$ embeds into it, preserving the order. In particular, the question asks if the minimum size $f(n)$ has a polynomial upper bound, to which the answer is no. -In this question, I am interested in some concrete values of $f(n)$ for small $n$. -So far, I know that: - -$f(0) = 0$ - -$f(1) = 1$ - -$f(2) = 3$ - -$f(3) = 5$ - -$f(4) = 8$ - -$f(n) \ge 2n - 1$ - -$f(n) \in \Omega(n^k)$ for all $k$ - - -Can we compute some additional values in this sequence? In particular, can we compute $f(5)$? -Notes - -I was able to verify $f(4) = 8$ using a computer-assisted proof using a SAT solver. I also tried naive enumeration of posets and checking for universality, but this fails at around $f(4)$. Computing $f(5)$ may require smarter enumeration, in particular better symmetry breaking. - -The sequence does not appear to be in OEIS yet (it does not appear to be any of the sequences beginning with 1, 3, 5, 8). I submitted this draft, and it was suggested that the sequence should be posted to MathOverflow to find more terms. - - -EDIT: New OEIS entry with f(5) = 11 here. - -REPLY [4 votes]: I will try to revive Sagemath's ticket #14110 and provide a Sagemath package for this enumeration (in fact, the C code, corresponding to the paper B. D. McKay and G. Brinkmann, Posets on up to 16 points, Order, 19 (2002) 147-179 - (mostly) due to Gunnar Brinkmann, which is using Brendan's McKay's nauty, is posted there.<|endoftext|> -TITLE: Generating functions of Collatz iterates? -QUESTION [11 upvotes]: Let $C(n) = n/2$ if $n$ is even and $3n+1$ otherwise be the Collatz function. -We look at the generating function $f_n(x) = \sum_{k=0}^\infty C^{(k)}(n)x^k$ of the iterates of the Collatz function. -The Collatz conjecture is then equivalent to: For all $n$: -$$f_n(x) = p_n(x) + x^{d+1} \frac{1+4x+2x^2}{1-x^3}$$ -where $d$ is the degree of the polynomial $p_n(x)$ with natural numbers as coefficients. -I have computed some of these generating functions. -Let -$$F_n(x) = (f_n(x), f_{C^{(1)}(n)}(x),\cdots,f_{C^{(l)}(n)}(x))$$ -where $l$ is the length of the Collatz sequence of $n$ ending at $1$. -The vector $F_n(x)$ when plugging in for $x$ a rational number, seems to parametrize an algebraic variety. Assuming that the Collatz conjecture is true. Can it be explained if or why this vector parametrizes an algebraic variety? -Here is an example for $n=3$: -The variety is given by the equations: -G^3 - H^3 - 4*G^2 + 4*G*H + H^2 + 4*G - 8*H = 0 -A*C - B^2 + 10*B - 3*C = 0 -F^2 - G*H - 4*F + G = 0 -F*G - H^2 - 2*F + H = 0 -F*H - G^2 + 2*G - 4*H = 0 -E - H - 7 = 0 - -and it is parametrized by: -A = (7*x^7 + 14*x^6 + x^5 + 2*x^4 - 13*x^3 - 5*x^2 - 10*x - 3)/(x^3 - 1) -B = (7*x^6 + 14*x^5 + x^4 + 2*x^3 - 16*x^2 - 5*x - 10)/(x^3 - 1) -C = (7*x^5 + 14*x^4 + x^3 - 8*x^2 - 16*x - 5)/(x^3 - 1) -D = (7*x^4 + 14*x^3 - 4*x^2 - 8*x - 16)/(x^3 - 1) -E = (7*x^3 - 2*x^2 - 4*x - 8)/(x^3 - 1) -F = (-x^2 - 2*x - 4)/(x^3 - 1) -G = (-4*x^2 - x - 2)/(x^3 - 1) -H = (-2*x^2 - 4*x - 1)/(x^3 - 1) - -where $F_3(x) = (A,B,C,D,E,F,G,H)$ -Here is some Sagemath script which does the computations. You can change the number $N=3$ in the script, but for $N=7$ it already takes a long time to compute the Groebner basis. -Edit: -Furthermore, the point $(n,C^{(1)}(n),\cdots,C^{(l)}(n))$ seem to always be a rational point of this variety. Example: -G^3 - H^3 - 4*G^2 + 4*G*H + H^2 + 4*G - 8*H = 0 -A*C - B^2 + 10*B - 3*C = 0 -F^2 - G*H - 4*F + G = 0 -F*G - H^2 - 2*F + H = 0 -F*H - G^2 + 2*G - 4*H = 0 -E - H - 7 = 0 -..... -A = 3 -B = 10 -C = 5 -D = 16 -E = 8 -F = 4 -G = 2 -H = 1 - -This last observation can be explained if the previous is true, because we can substitute $x=0$: -$$\forall m \text{ we have : } f_m(0) = m$$ -and hence: -$$F_n(0) = (n,C^{(1)}(n),\cdots,C^{(l)}(n))$$ -is a rational point on the variety. - -REPLY [5 votes]: Without assuming the Collatz conjecture, it can be shown that the generating functions satsify certain polynomial equations: -Observe that for all $n$: -$$f_{C(n)}(x) = \frac{f_n(x)-n}{x}$$ -hence: -$$f_{C^{(2)}(n)}(x) = \frac{f_{C(n)}(x)-C(n)}{x}$$ -Solving for $x$ and equating the two identities and letting $x_k:=f_{C^{(k)}(n)}(x)$, we find the polynomial equation: -$$\forall k=0,1,2,\cdots \text{ we have }: x_k x_{k+2}-C^{(k)}(n) x_{k+2}-x_{k+1}^2+C^{(k+1)}(n)x_{k+1} = 0$$ -which according to Wolfram Alpha each of these equations represent a "one-sheeted hyperboloid": -Wikipedia -Mathworld<|endoftext|> -TITLE: Why there is no 3-category or tricategory of bicategories? -QUESTION [12 upvotes]: I recently asked this question on MSE. -So I want to move it here in hope to gain a more wordy answer. -I have read around about bicategories, lax functor, lax natural transformation and modifications. I know that we have a 1category of Bicategories and lax functors. I know why we do not have a bicategory or 2category of bicategories, lax functor, and lax natural transformations, and I know that using ICONS instead of lax natural trasformations solves this problem. (or oplax, it doesn't matter) -What I cannot see is why we do not get a 3categories or tricategories of bicategories, Lax functors, ICONS and modifications? What fails? Where can I find a reference about it? - -REPLY [12 votes]: Edit: In this answer I missed that the question was about lax functors rather than pseudofunctors. See comments below. -Such a tricategory does exist, and in fact it is part of a richer structure. In Garner-Gurski The low-dimensional structures formed by tricategories (arxiv), Corollary 12 constructs a locally cubical bicategory of bicategories. This is a bicategory enriched over the monoidal 2-category of pseudo double categories, containing the following structure: - -its 0-cells are bicategories -its 1-cells are pseudofunctors -its "vertical 2-cells" are icons -its "horizontal 2-cells" are pseudonatural transformations -its 3-cells are "cubical modifications". - -Thus, if we discard the vertical 2-cells we obtain the usual tricategory of bicategories (although one has to do a bit of work to "lift" the coherences, which start life as icons, to pseudonatural transformations). If we instead discard the horizontal 2-cells, I believe we obtain the tricategory you're after. -It is a particularly strict sort of tricategory. This construction exhibits it as a bicategory enriched over the monoidal 2-category of strict 2-categories, but it might in fact be a strict 3-category; I have not checked carefully. -I suspect that your next question might be why no one has pointed this out before. Probably the answer is that no one had a use for it. One of the purposes of introducing icons was to reduce the categorical dimension of the structure containing bicategories, so putting the modifications back in would defeat that purpose. Also, modifications between icons may seem a priori to be of limited interest, since their components are endomorphisms of identity 1-cells — although of course such a judgment always evaporates when someone finds an application of them! Finally, the locally cubical bicategory seems more useful for most purposes: its categorical coherence dimension is equally limited (being an enriched bicategory, rather than a tricategory), while it contains strictly more information.<|endoftext|> -TITLE: The modularity theorem as a special case of the Bloch-Kato conjecture -QUESTION [35 upvotes]: In the homepage for the CRM's special semester this year, I found the interesting statement that the modularity theorem (formerly the Taniyama-Shimura-Weil conjecture) is a special case of the Bloch-Kato conjecture for the symmetric square motive of an elliptic curve. -The Bloch-Kato conjecture, as I know about it, is the following statement: -$$\text{ord}_{s=0}L(s,V)=\text{dim }H_{f}^{1}(K,V^{*}(1))-\text{dim }H^{0}(K,V^{*}(1))$$ -where $H_{f}^{1}$ is the Bloch-Kato Selmer group. I do not know much about motives and I do not know what the symmetric square motive of an elliptic curve is. In the formulation of the Bloch-Kato conjecture above I am taking $V$ to be a geometric Galois representation. My question is, how do we view the modularity theorem as this special case of the Bloch-Kato conjecture? - -REPLY [22 votes]: That is not what the link says. To quote (emphasis mine): - - ... in which this conjecture was reduced to a special instance of the Bloch-Kato conjecture for the symmetric square motive of an elliptic curve. - -That means something quite different. You could equally say that Wiles "reduced" the proof to the fact that $X(3)$ and $X(5)$ have genus zero, or that he "reduced" the proof to the Langlands-Tunnell theorem that (projective) $A_4$ and $S_4$ representations are automorphic. Shimura-Taniyama is no more a "special case" of these claims than it is of the Bloch-Kato conjecture. -The more relevant thing to say is that one (inductive) step in Wiles' argument required comparing the size of a certain congruence module (measuring congruences between one cuspform and other forms of a fixed level) and a relative tangent space (measuring congruences between one Galois representation and a certain prescribed family of deformations of that representation). Hida had already shown that the former quantity could be interpreted in terms of the special value of the adjoint L-function. So if one knew that the value of this adjoint L-function (divided by the correct period to obtain an integer) was divisible by the same power of $p$ as the order of the relative tangent space (which could be interpreted in terms of a Bloch-Kato Selmer group, then the inductive step would hold. This desired equality can indeed be interpreted as a special case of the Bloch-Kato conjecture, although not formulated for $\mathbf{Q}_p$ representations as you have done but in the more precise form by Bloch and Kato for $p$-adic lattices in Galois representations coming from motives $M$. Namely, in the context of your equation, the L-value on the LHS trivially doesn't vanish as it lies on the edge of the critical strip. So one can hope (given the motive in question) to define a suitable period $\Omega$ such that $L(s,M)/\Omega \in \mathbf{Q}^{\times}$, and then (after taking into account local fudge-factors) interpret the resulting integer (or at least the $p$-power part) as the order of a Bloch-Selmer group $H^1_f(\mathbf{Q}, T \otimes \mathbf{Q}_p/\mathbf{Z}_p)$, where $T$ is a $\mathbf{Z}_p$-lattice inside a Galois representation associated to $M$. -In this particular case, you can take the variety $E \times E$. Then the motive $M$ is a suitable piece of this. Then one is interested in the special value $L(M,2)$. The corresponding lattice $T$ can then be found inside -$$H^2(E \times E,\mathbf{Z}_p(1))$$ -Precisely, there is a lattice corresponding to the Tate module of $E$ at $p$. Denote the dual of this lattice by $\rho$. The lattice $\rho$ is isomorphic to $H^1(E,\mathbf{Z}_p)$. By the Kunneth formula, one then finds a copy of $\rho^{\otimes 2}$ inside $H^2(E \times E,\mathbf{Z}_p)$, and hence a copy of $\rho^{\otimes 2}(1)$ inside the group above. Since $\rho$ has dimension $2$, there is a decomposition (let $p$ be odd) $\rho^{\otimes 2} \simeq \mathrm{Sym}^2(\rho) \oplus \mathbf{Z}_p(-1)$, and then $T$ is identified with $\mathrm{Sym}^2(\rho)(1)$. (Apologies if I have got the twisting wrong, it is irritating to keep straight). -It might also be worth mentioning that Wiles famously didn't use any results towards the Bloch-Kato conjecture, but rather proved what he needed by himself and thus deduced some special cases of the Bloch-Kato conjecture, in particular an automorphic formula for the order of the group -$$H^1_f(\mathbf{Q},\mathrm{Sym}^2(\rho)(1)).$$ -(The Galois representation can also be identified the trace zero matrices $\mathrm{ad}^0(\rho)$ in the [lattice of the] adjoint representation.) Note that Wiles' formula manifestly implies that this group is finite. But even the finiteness of this formula was unknown for a general elliptic curve before the work of Wiles. (Although it was known in some cases by Flach for modular elliptic curves; not so useful for proving modularity.) Many modern modularity proofs (which use the same underlying mechanism that Wiles did) thus also end up resulting in some results towards Bloch-Kato conjecture for adjoint representations.<|endoftext|> -TITLE: Computing $(AA\otimes BB + AB \otimes BA)^{-1}$ -QUESTION [6 upvotes]: Can anyone suggest a way to numerically compute the following matrix vector product? -$$u=A^{-1}b=(AA\otimes BB + AB \otimes BA)^{-1}\operatorname{vec}(C)$$ -Here $AA,BB,AB,BA$ and $C$ are $d\times d$ matrices with $d\approx 1000$. A naive method is to expand Kronecker products and use linear solver for an answer in $O(d^6)$ time, but I need to do it in $O(d^3)$ time. -Not sure if helps, but $AA,AB,BA,BB$ are blocks of a covariance matrix, ie for partitioned random variable $(x,y)$ -$$ -\text{cov}\left( -\begin{array}{ll} -x\\y -\end{array}\right) -=\left(\begin{array}{cc}AA& AB\\BA& BB\end{array}\right)$$ -I tried Jacobi-like iterative method, writing $A=M-N$ with $M=AA\otimes BB$ and $N=-AB\otimes BA$ and iterating $u=M^{-1}Nu+M^{-1}b$ which is cheap because of Kronecker products, but this sometimes diverges. Empirically, divergence seems to happen when $x$ and $y$ are highly correlated with each other. Any suggestions/literature pointers are appreciated! - -REPLY [5 votes]: Your equation is equivalent to -$$ -(B^{-1}A)X + X(B^{-1}A)^T = B^{-1}A^{-1}CB^{-T}B^{-T}, -$$ -where $X$ is the matricization of $u$. This is a Sylvester equation, which can be solved for $X$ in $O(d^3)$ time. -You'll get some numerical ugliness if $A$ or $B$ are close to non-invertible, but that might be unavoidable in some sense: if either of $A$ or $B$ are ill-conditioned then so is the original coefficient matrix $AA \otimes BB + AB \otimes BA$.<|endoftext|> -TITLE: Interpolation theory and $C^k$-spaces -QUESTION [8 upvotes]: Consider the Banach spaces $C^k(M)$ ($k=0,1,2,\dots$), consisting of $k$times continuously differentiable functions $f:M\rightarrow \mathbb{C}$ on a closed manifold $M$ (or just the torus if that makes it easier). I have a few questions regarding their interpolation theory: - -Is $C^1(M)$ an interpolation space for the pair $(C^0(M),C^2(M))$? According to Bergh-Lofström, this means that any linear map $T:C^0(M)\rightarrow C^0(M)$ which leaves $C^2(M)$ invariant, also leaves $C^1(M)$ invariant. I don't see how one would prove this. The reason I am wondering is that this would be a sufficient (but not a necessary) condition for the association $(C^0,C^2)\mapsto C^1$ to extend to an interpolation functor on Banach spaces (Aronszajn-Gagliardo Theorem). -Can we identify the interpolation spaces $[C^k, C^l]_\theta$ or $[C^k,C^l]_{\theta,p}$ (where the brackets stand for complex and real interpolation respectively)? I only find results of this kind for Hölder-Zygmund spaces $C_*^k$, which differ from $C^k$ for integer values of $k$. Maybe one can even identify $C^k$ as member of some larger scale of spaces (Besov, Triebel, etc.)? - -REPLY [7 votes]: $C^1$ is not an interpolation space between $C$ and $C^2$. There is an example due to Mitjagin and Semenov of a sequence $T_n$ of uniformly bounded operators both in $C,C^2$ whose norm blow up in $C^1$. You find this example in the booklet by A. Lunardi "Interpolation Theory" (SNS Pisa publisher). In the edition in my hands, it is in Chapther I, after Example 1.3.3, pag 29.<|endoftext|> -TITLE: Is $G\mapsto \operatorname{Hol}(G)$ the object component of any functor on the category of groups? -QUESTION [5 upvotes]: On the objects of the category of groups we define the mapping $G\mapsto \operatorname{Hol}(G)$, the holomorph $G\rtimes \operatorname{Aut}(G)$ of $G$. Can we extend this mapping to a functor on this category? (Via extension to morphisms) - -REPLY [9 votes]: There is no such functor. Recall that a split epimorphism is a morphism $f : x \to y$ with a section (right inverse) $g : y \to x$, satisfying $fg = \text{id}_y$. Split epimorphisms, as their name suggests, are epimorphisms, and moreover they are absolute epimorpisms in that they are preserved by any functor whatsoever. -In $\text{Grp}$ every split epimorphism arises as a projection $N \rtimes G \to G$ where $N \rtimes G$ is a semidirect product, so any functor from groups to groups must preserve these. In particular if $\text{Hol}(-)$ were such a functor it would follow that $\text{Hol}(N \rtimes G)$ admits a split epimorphism to $\text{Hol}(G)$. -No such split epimorphism exists in general. Explicitly, take $N = C_2, G = C_2^3$ with the trivial action. Then -$$\text{Hol}(N \rtimes G) \cong C_2^4 \rtimes GL_4(\mathbb{F}_2)$$ -while -$$\text{Hol}(G) \cong C_2^3 \rtimes GL_3(\mathbb{F}_2).$$ -$GL_n(\mathbb{F}_2) \cong SL_n(\mathbb{F}_2) \cong PSL_n(\mathbb{F}_2)$ is simple for $n \ge 3$, so these two groups each have a unique nonabelian simple group in their Jordan-Holder decompositions, namely $GL_4(\mathbb{F}_2)$ and $GL_3(\mathbb{F}_2)$ respectively. Any non-solvable quotient of $C_2^4 \rtimes GL_4(\mathbb{F}_2)$ must also contain $GL_4(\mathbb{F}_2)$ in its Jordan-Holder decomposition, and since $GL_3(\mathbb{F}_2)$ and $GL_4(\mathbb{F}_2)$ have different orders they are non-isomorphic, so $\text{Hol}(G)$ is not a quotient (or even a subquotient) of $\text{Hol}(N \rtimes G)$. -A similar but simpler argument shows that there is no functor sending a group $G$ to its center $Z(G)$, since for example $\text{sgn} : S_3 \to C_2$ is a split epimorphism but there is no epimorphism $Z(S_3) \to Z(C_2)$ since the former is trivial and the latter is not. We can also show that there is no functor sending a group $G$ to its automorphism group $\text{Aut}(G)$ using the same counterexample $N = C_2, G = C_2^3$ as above, although $N = C_2, G = C_2^2$ also works and requires a slightly different argument. I used a similar but more complicated argument on math.SE recently to show that there is no functor sending a finite-dimensional vector space $V$ to $GL(V)$.<|endoftext|> -TITLE: Lower bound on growth for intersection of two subgroups of free group -QUESTION [6 upvotes]: Let $\Gamma_1$ and $\Gamma_2$ be two subgroups of the rank-$2$ free group $F_2$. Can then one find a nontrivial lower bound on the growth exponent of their intersection $\Gamma_1 \cap \Gamma_2$, in terms of the growth exponents of the two subgroups? -Here by the growth exponent of a subgroup $\Gamma \subset F_2$ I mean the real number -$$\lim_{n \to \infty} \frac{1}{n} \log \# \{w \in \Gamma \;\mid\; \ell(w) \leq n\},$$ -where $\ell$ denotes the word length inside $F_2$; so that this exponent lies between $0$ and $\log 3$. -For semigroups, clearly there is no hope for such a lower bound. Indeed, in the rank-$2$ free monoid $A^*$ (on the alphabet $A = \{a, b\}$), the two subsemigroups $S_1 := (aA^*)^+$ and $S_2 := (bA^*)^+$ (generated by all the words starting with $a$, resp. with $b$) are disjoint, but have maximal growth exponent (namely $\log 2$). -However this construction does not work at all for groups. I have tried to find pairs of subgroups in $F_2$ that have large growth exponents but trivial (or at least small) intersection, but I am not even sure where to start. Any help would be appreciated! - -REPLY [4 votes]: OK, in fact I figured this out as soon as I properly understood how Stallings graphs work. Here is the analogous construction for groups. -Let $\Gamma_0$ be the subgroup of the free group $F_2 = \langle a, b \rangle$ generated by the set $\{a^{k+1} b^{-1} a^{-k} \;\mid\; k \geq 0\}$. I claim that elements of $\Gamma_0$ can be characterized as follows: a reduced word $w(a,b,a^{-1},b^{-1})$ lies in $\Gamma_0$ if and only if the word obtained from $w$ by substituting an opening parenthesis to each $a$ and $b$, and a closing parenthesis to each $a^{-1}$ and $b^{-1}$, is well-parenthesized. -This is probably easy to show by hand, but the right way to see things is probably to use Stallings graphs (see e.g. these slides for a very quick visual introduction: http://www.lix.polytechnique.fr/combi/archivesSeminaire/transparents/150513_Frederique_Bassino.pdf, or this paper for a more detailed one: https://arxiv.org/abs/math/0202285 ). Their key property is that a word $w \in F_2$ belongs to some subgroup $\Gamma$ if and only if there exists a loop in the Stallings graph of $\Gamma$, starting and ending at its basepoint, and labeled by $w$. Now the Stallings graph of $\Gamma_0$ has vertices indexed by $\mathbb{Z}_{\geq 0}$ (with basepoint $0$) and, for each vertex $n$, an edge labeled $a$ and an edge labeled $b$ both going from $n$ to $n+1$; and the characterization follows immediately. -It is then not very hard to see that $\Gamma_0$ has maximal possible growth exponent, namely $\log 3$. In fact, the number of words of length $2n$ in $\Gamma$ is twice the sequence https://oeis.org/A059231 , which grows like a polynomial times $3^{2n}$ (OEIS gives the asymptotic estimate "a(n) ~ sqrt(2)3^(2n+1)/(8*sqrt(Pi)*n^(3/2))"). -Now we set $\Gamma_1 := a \Gamma_0 a^{-1}$ and $\Gamma_2 := b \Gamma_0 b^{-1}$. Of course these two groups still have growth exponent $\log 3$. However, their intersection is trivial: indeed, all nontrivial elements of $\Gamma_1$ (resp. of $\Gamma_2$) actually start with $a$ (resp. $b$) and end with $a^{-1}$ (resp. $b^{-1}$), even when written in reduced form. -This last statement is obvious from the characterization of $\Gamma_0$, but once again it is helpful to think in terms of Stallings graphs, in particular so as to bring into light the parallel with the semigroup construction. The Stallings graph of $\Gamma_1$ (resp. $\Gamma_2$) is obtained from the Stallings graph of $\Gamma_0$ by adding a new basepoint, with a single edge from that vertex to the old basepoint labeled by $a$ (resp. by $b$).<|endoftext|> -TITLE: Converse of Schreier theorem -QUESTION [5 upvotes]: I know that every subgroup of a free group is free (Schreier theorem). -I'm wondering that a (non-trivial) converse is true, that is, if every proper subgroup of an infinite group $G$ is free, then $G$ is free. -I think it is false but I cannot find counterexamples. -(I expect that some proper semi-direct product of the free group rank $n$ ($n \geq 2$) and $\mathbb{Z}/2\mathbb{Z}$ is counterexample but I cannot find yet.) -Any comments would be greatly appreciated. - -REPLY [2 votes]: As @YCor says in comments, there is a finitely generated example due to Ol'shanskii, which is essentially a kind of Tarski monster. However, Ol'shanskii's construction is very complicated. For "nicer" classes of groups, your question remains both important and open. As with most other kinds of Tarski monsters, I believe the answer to the following question remains unknown (though the answer is surely "yes"). - -Is there a non-free finitely presented group for which every proper subgroup is free? - -Indeed, the question remains open even in very nice classes of groups. - -Is there a non-free word-hyperbolic group in which every proper subgroup is free? - -An example would be huge news, since it would resolve in the negative TWO famous open questions, viz: - -Is every word-hyperbolic group residually finite? - -and - -Does every non-virtually-free hyperbolic group have a surface subgroup? - -Since two big questions in one go seems like too much to hope for, I prefer to specialise to the case of subgroups of infinite index. - -Is there a non-free, non-surface, infinite, word-hyperbolic group in which every proper, finitely generated subgroup of infinite index is free? - -For this last question, there are examples outside of the world of hyperbolic groups. The solvable Baumslag–Solitar groups -$BS(1,n)=\mathbb{Z}[1/n]\rtimes_n\mathbb{Z}$ -have the property that every nontrivial finitely generated subgroup of infinite index is infinite cyclic.<|endoftext|> -TITLE: Vector bundles on $\mathbf{P}^1$ and the Iwasawa decomposition -QUESTION [13 upvotes]: As everyone knows, every vector bundle on $\mathbf{P}^1$ splits as a direct sum of line bundles $\mathcal{O}(a_1)\oplus\cdots\oplus\mathcal{O}(a_n)$. This means that in the Weil-uniformisation description of vector bundles -$$\text{Bun}_{\text{GL}_n}(\mathbf{P}^1)\ =\ \text{GL}_n(k(\mathbf{P}^1))\backslash \text{GL}_n(\mathbf{A}_{\mathbf{P}^1})/\text{GL}_n(\mathbf{O}_{\mathbf{P}^1})$$ -the diagonal matrices $\text{diag}(t^{a_1},...,t^{a_n})$ form a complete set of double coset representatives, where $t$ is a uniformiser of some point in $\mathbf{P}^1$. -Question: Pretending that we knew nothing about vector bundles on $\mathbf{P}^1$, is it possible to get this result on double quotient representatives directly using the Iwasawa decomposition of $\text{GL}_n(\mathbf{A}_{\mathbf{P}^1})$, or something similar? - -Clarification: The question is “the right side is confusing, how can we massage it algebraically to make it obvious that it's just $\mathbf{Z}$"? I really love David Speyer's answer, a (different, non-adelic) way to compute $\text{Bun}_{\text{GL}_n}(\mathbf{P}^1)$, but it doesn't address that confusion. In my mind the question is still open. - -REPLY [4 votes]: Disclaimer: I am not fully confident in my understanding of the terminology here. Corrections are welcome. -$\def\GL{\mathrm{GL}}\def\PP{\mathbb{P}}\def\CC{\mathbb{C}}\def\ZZ{\mathbb{Z}}\def\Spec{\mathrm{Spec}\ }$My understanding is that this is Bruhat decomposition for the loop group of $\GL_n$. -Rather than your adelic formalism, I will cover $\PP^1$ with two open sets $\Spec \CC[t]$ and $\Spec \CC[t^{-1}]$, overlapping in $\Spec \CC[t, t^{-1}]$. Since every projective module over $\CC[t]$ is free, any vector bundle trivializes one each of our open sets (and hence on their overlap), and so vector bundles are classified by double cosets -$$\GL_n(\CC[t]) \backslash \! \GL_n(\CC[t, t^{-1}]) / \GL_n(\CC[t^{-1}]).$$ -Now (I might be using terminology wrongly here) the ind-group $\GL_n(\CC[t, t^{-1}]$ is a Kac-Moody group, called the loop group of $\GL_n$. A pair of opposite Borels can be taken to be -$$B_- := \{ g \in \GL_n(\CC[t]) : g \bmod t \ \mbox{is upper triangular} \}$$ -$$B_+ := \{ g \in \GL_n(\CC[t^{-1}]) : g \bmod t^{-1} \ \mbox{is lower triangular} \}.$$ -The Weyl group is $W:=S_n \ltimes \mathbb{Z}^n$, and can be represented by permutation matrices whose nonzero entries are monomials in $t$. We have Bruhat decomposition, $\GL_n(\CC[t, t^{-1}] = \bigsqcup_{w \in W} B_- w B_+$. So the elements of $W$ are double coset representatives for $B_- \backslash \GL_n(\CC[t,t^{-1}])/B_+$. -We want to quotient not by the Borels but by the parabolics, $P_- := \GL_n(\CC[t])$ and $P_+ = \GL_n(\CC[t^{-1}])$. So the elements of $W$ still represent all cosets, but redundantly. More specifically, the parabolic subgroup $W_P$ of $W$ corresponding to $P_{\pm}$ is $S_n$. Standard theory tells us that we can get a family of representatives for $B_- \backslash \GL_n(\CC[t,t^{-1}])/B_+$ by taking representatives for $W_P \backslash W / W_P$. -We compute $$W_P \backslash W / W_P = S_n \backslash (S_n \ltimes \ZZ^n) / S_n = S_n \backslash \ZZ^n$$ -$$ = \{ (a_1, a_2, \ldots, a_n) \in \ZZ^n : a_1 \geq a_2 \geq \cdots \geq a_n \}.$$ -Thus, isomorphism classes of vector bundles are indexed by $n$-tuples of decreasing integers, which is correct.<|endoftext|> -TITLE: Which groups are doubling? -QUESTION [6 upvotes]: A metric space $(M,d)$ is doubling if there exists $n$ such that every ball of radius $r$ can be covered by $n$ balls of radius $r/2$, for all $r$. For which f.g. groups $G$ and finite symmetric generating sets $S$, is $\mathrm{Cay}(G, S)$ doubling under the path metric? Groups like this have polynomial growth, so they are virtually nilpotent by Gromov's theorem. -So which virtually nilpotent groups are doubling, and for which generating sets? All, I suppose, but I got cold feet trying to do it, it seemed quite difficult straight from the definitions and I don't really know the Lie group stuff well enough. - -If $S$ is a finite symmetric generating set for a group $G$, is $\mathrm{Cay}(G, S)$ doubling precisely when $G$ is virtually nilpotent? - -I'll note that in general (undirected) graphs, doubling implies polynomial growth, but not the other way around, consider for example the comb graph with vertices $\mathbb{Z} \times \mathbb{N}$ and edges $\{\{(m,n), (m,n+1)\}, \{(m,0), (m+1,0)\} \;|\; m \in \mathbb{Z}, n \in \mathbb{N}\}$. But could be true for vertex-transitive graphs. - -REPLY [4 votes]: I think this follows from a standard ball-packing argument. -Suppose that $G$ with the metric $\rho$ induced from the Cayley graph has growth $V(R)=|B_R(1)| \sim R^d$, i.e. $\exists\ 0 -TITLE: Discriminant of characteristic polynomial as sum of squares -QUESTION [22 upvotes]: The characteristic polynomial of a real symmetric $n\times n$ matrix $H$ has $n$ real roots, counted with multiplicity. -Therefore the discriminant $D(H)$ of this polynomial is zero or positive. -It is zero if and only if there is a degenerate eigenvalue. -Thus $D(H)$ is a non-negative (homogeneous) polynomial in the $\frac12n(n+1)$ entries of $H$. -Some non-negative polynomials can be written as a sum of squares and I am interested in whether $D(H)$ can. -There is a concrete question at the end, but any insights into the general case are also welcome. -The size of the problem grows very quickly with dimension, so I will only look at $n=2$ (which I do understand) and $n=3$ (which I am yet to understand). -2D -In two dimensions it is pretty easy to write down the polynomial and its discriminant and see by eye that -$$ -D(H) -= -(h_{11}-h_{22})^2 -+ -4h_{12}^2, -$$ -which is indeed a sum of two squares. -Having a degenerate eigenvalue is a polynomial condition: it happens if and only if $D(H)=0$. -The discriminant is a second order polynomial, but writing it as a sum of squares leads to far simpler algebraic condition: $h_{11}-h_{22}=0$ and $h_{12}=0$. -Simple algebraic conditions for degeneracy are the goal here, but I thought the question would be of some interest in itself. -3D -In three dimensions the discriminant is pretty big: -$$ -D(H) -= -h_{22}^2h_{33}^4-2h_{11}h_{22}h_{33}^4+4h_{12}^2h_{33}^4+h_{11}^2h_{33}^4-2h_{22}h_{23}^2h_{33}^3+2h_{11}h_{23}^2h_{33}^3-8h_{12}h_{13}h_{23}h_{33}^3-2h_{22}^3h_{33}^3+2h_{11}h_{22}^2h_{33}^3+2h_{13}^2h_{22}h_{33}^3-8h_{12}^2h_{22}h_{33}^3+2h_{11}^2h_{22}h_{33}^3-2h_{11}h_{13}^2h_{33}^3-8h_{11}h_{12}^2h_{33}^3-2h_{11}^3h_{33}^3+h_{23}^4h_{33}^2+8h_{22}^2h_{23}^2h_{33}^2-10h_{11}h_{22}h_{23}^2h_{33}^2+2h_{13}^2h_{23}^2h_{33}^2+20h_{12}^2h_{23}^2h_{33}^2+2h_{11}^2h_{23}^2h_{33}^2+12h_{12}h_{13}h_{22}h_{23}h_{33}^2+12h_{11}h_{12}h_{13}h_{23}h_{33}^2+h_{22}^4h_{33}^2+2h_{11}h_{22}^3h_{33}^2+2h_{13}^2h_{22}^2h_{33}^2+2h_{12}^2h_{22}^2h_{33}^2-6h_{11}^2h_{22}^2h_{33}^2-10h_{11}h_{13}^2h_{22}h_{33}^2+20h_{11}h_{12}^2h_{22}h_{33}^2+2h_{11}^3h_{22}h_{33}^2+h_{13}^4h_{33}^2+20h_{12}^2h_{13}^2h_{33}^2+8h_{11}^2h_{13}^2h_{33}^2-8h_{12}^4h_{33}^2+2h_{11}^2h_{12}^2h_{33}^2+h_{11}^4h_{33}^2-10h_{22}h_{23}^4h_{33}+8h_{11}h_{23}^4h_{33}-36h_{12}h_{13}h_{23}^3h_{33}-2h_{22}^3h_{23}^2h_{33}-10h_{11}h_{22}^2h_{23}^2h_{33}-2h_{13}^2h_{22}h_{23}^2h_{33}-2h_{12}^2h_{22}h_{23}^2h_{33}+20h_{11}^2h_{22}h_{23}^2h_{33}-2h_{11}h_{13}^2h_{23}^2h_{33}-38h_{11}h_{12}^2h_{23}^2h_{33}-8h_{11}^3h_{23}^2h_{33}+12h_{12}h_{13}h_{22}^2h_{23}h_{33}-48h_{11}h_{12}h_{13}h_{22}h_{23}h_{33}-36h_{12}h_{13}^3h_{23}h_{33}+72h_{12}^3h_{13}h_{23}h_{33}+12h_{11}^2h_{12}h_{13}h_{23}h_{33}-2h_{11}h_{22}^4h_{33}-8h_{13}^2h_{22}^3h_{33}+2h_{12}^2h_{22}^3h_{33}+2h_{11}^2h_{22}^3h_{33}+20h_{11}h_{13}^2h_{22}^2h_{33}-10h_{11}h_{12}^2h_{22}^2h_{33}+2h_{11}^3h_{22}^2h_{33}+8h_{13}^4h_{22}h_{33}-38h_{12}^2h_{13}^2h_{22}h_{33}-10h_{11}^2h_{13}^2h_{22}h_{33}+8h_{12}^4h_{22}h_{33}-10h_{11}^2h_{12}^2h_{22}h_{33}-2h_{11}^4h_{22}h_{33}-10h_{11}h_{13}^4h_{33}-2h_{11}h_{12}^2h_{13}^2h_{33}-2h_{11}^3h_{13}^2h_{33}+8h_{11}h_{12}^4h_{33}+2h_{11}^3h_{12}^2h_{33}+4h_{23}^6+h_{22}^2h_{23}^4+8h_{11}h_{22}h_{23}^4+12h_{13}^2h_{23}^4+12h_{12}^2h_{23}^4-8h_{11}^2h_{23}^4-36h_{12}h_{13}h_{22}h_{23}^3+72h_{11}h_{12}h_{13}h_{23}^3+2h_{11}h_{22}^3h_{23}^2+20h_{13}^2h_{22}^2h_{23}^2+2h_{12}^2h_{22}^2h_{23}^2+2h_{11}^2h_{22}^2h_{23}^2-38h_{11}h_{13}^2h_{22}h_{23}^2-2h_{11}h_{12}^2h_{22}h_{23}^2-8h_{11}^3h_{22}h_{23}^2+12h_{13}^4h_{23}^2-84h_{12}^2h_{13}^2h_{23}^2+20h_{11}^2h_{13}^2h_{23}^2+12h_{12}^4h_{23}^2+20h_{11}^2h_{12}^2h_{23}^2+4h_{11}^4h_{23}^2-8h_{12}h_{13}h_{22}^3h_{23}+12h_{11}h_{12}h_{13}h_{22}^2h_{23}+72h_{12}h_{13}^3h_{22}h_{23}-36h_{12}^3h_{13}h_{22}h_{23}+12h_{11}^2h_{12}h_{13}h_{22}h_{23}-36h_{11}h_{12}h_{13}^3h_{23}-36h_{11}h_{12}^3h_{13}h_{23}-8h_{11}^3h_{12}h_{13}h_{23}+4h_{13}^2h_{22}^4+h_{11}^2h_{22}^4-8h_{11}h_{13}^2h_{22}^3-2h_{11}h_{12}^2h_{22}^3-2h_{11}^3h_{22}^3-8h_{13}^4h_{22}^2+20h_{12}^2h_{13}^2h_{22}^2+2h_{11}^2h_{13}^2h_{22}^2+h_{12}^4h_{22}^2+8h_{11}^2h_{12}^2h_{22}^2+h_{11}^4h_{22}^2+8h_{11}h_{13}^4h_{22}-2h_{11}h_{12}^2h_{13}^2h_{22}+2h_{11}^3h_{13}^2h_{22}-10h_{11}h_{12}^4h_{22}-2h_{11}^3h_{12}^2h_{22}+4h_{13}^6+12h_{12}^2h_{13}^4+h_{11}^2h_{13}^4+12h_{12}^4h_{13}^2+2h_{11}^2h_{12}^2h_{13}^2+4h_{12}^6+h_{11}^2h_{12}^4 -. -$$ -(I got this by Maxima.) -This is indeed a non-negative homogeneous polynomial of degree six in six variables, but it is too big for me to see any structure by eye and I cannot tell whether it is a sum of squares. -In the diagonal case $h_{12}=h_{13}=h_{23}=0$ the discriminant has a simpler expression: -$$ -D(H) -= -(h_{11}-h_{22})^2 -(h_{22}-h_{33})^2 -(h_{11}-h_{33})^2. -$$ -This form is not at all surprising, as it should be a sixth degree polynomial vanishing if and only if two diagonal entries coincide. -My concrete question is: -Is this $D(H)$ of the case $n=3$ a sum of squares (without assuming it is diagonal)? -If yes, what are the squared polynomials and how unique are they? -I have understood that there are computational tools for finding a sum of squares decomposition, but I have yet to find one that I could run with the software I have. -And I assume this particular polynomial has structure which simplifies matters: for example, the polynomial is invariant under orthogonal changes of basis and the non-negativity has a geometric meaning. -One can indeed diagonalize the matrix, but I cannot see a way to use this to understand what the polynomial is in terms of the original basis. -The 2D case and the diagonal 3D case suggest that being a sum of squares is a reasonable guess. - -REPLY [4 votes]: We know that $H$ is symmetric, and therefore, diagonalizable, as $H = Q^TDQ$ for some orthogonal matrix $Q$. Moreover, $D$ and $Q$ have the same eigenvalues, and thus the same characteristic polynomials. Perhaps this can be used? -In any case, this reference by Domokos mentions the other answers and references as well. It gives some explicit expressions in the 3x3 case, both in five squares (theorem 7.3) and in seven squares (theorem 7.4), showing that the decomposition is not unique.<|endoftext|> -TITLE: History of Sylvester's resultant? -QUESTION [11 upvotes]: Suppose that we have two polynomials that split: -$$\begin{align*} -f(x)=\sum_{k=0}^d a_{d-k}x^k&=\prod_{i=1}^d (x-\lambda_i),\\ -g(x)=\sum_{k=0}^e b_{e-k}x^k&=\prod_{j=1}^e (x-\mu_j).\\ -\end{align*}$$ -Then the following result is often attributed to James Joseph Sylvester: -$$ -\det\begin{pmatrix} a_0 & a_1 & \cdots & a_d && \\ & \ddots &\ddots &&\ddots & \\ && a_0 & a_1 & \cdots & a_d \\ b_0 & b_1 & \cdots & b_e && \\ & \ddots & \ddots && \ddots & \\ && b_0 & b_1 & \cdots & b_e -\end{pmatrix} = \prod_{i,j} (\lambda_i-\mu_j). -$$ -Can someone help me track down the original reference? - -REPLY [19 votes]: The resultant of the Sylvester matrix first appeared in J. J. Sylvester, Philos. Magazine 16, 132–135 (1840): A method of determining by mere inspection the derivatives from two equations of any degree. - -The last line is the resultant of the two equations at the top, which vanishes when two roots coincide. - I like the "ought to, and is" phrasing... - -The Sylvester matrix also appeared in an 1840 publication by Cauchy, who acknowledged the priority of Sylvester: - -1876 reprint of Cauchy's 1840 article "Mémoire sur l'élimination d'une variable entre deux équations algébriques". -The name "resultant" originates from -E. Bézout (1764): Recherches sur le degré des équations résultantes de l'évanouissement des inconnues et sur les moyens qu'on doit employer pour trouver ces équations. -Research on the degree of equations resulting from the vanishing of unknowns and on the methods one should use to obtain these equations.<|endoftext|> -TITLE: The number of ways to merge a permutation with itself -QUESTION [9 upvotes]: Let $\sigma$ be a permutation of $[k]=\{1,2, \dots , k\}$. Consider all the ordered triples $(\pi, s_{1},s_{2})$, such that $\pi$ is a permutation of length $2k-1$ that is a union of its two subsequences $s_{1}$ and $s_{2}$, each of which is of length $k$ and is order-isomorphic to $\sigma$. -Example: -$\sigma = 312$, -If $\pi = 54213$, then there are $4$ such triples: - -$(\pi, 523,413)$ - -$(\pi, 513,423)$ - -$(\pi, 413,523)$ - -$(\pi, 423,513)$ - - -Indeed, each of the listed sequences $s_{1}$ and $s_{2}$, namely $523$, $413$, $513$ and $423$ are order isomorphic to $\sigma=312$, i.e., if the triple is $xyz$, then $x>z>y$. -Denote the number of these triples by $N_{2k-1}^{\sigma}$. Prove that $N_{2k-1}^{\sigma}>\binom{2k-1}{k}^{2}$ for every $\sigma$. -Example: $k=2$. It suffices to show that $N_{3}^{21}>\binom{3}{2}^{2}=9$ since $N_{3}^{21}=N_{3}^{12}.$ In fact, we have 10 triples that are listed below: -$\sigma = 321$: $(321,32,31)$, $(321,31,32)$, $(321,32,21)$, $(321,21,32)$, $(321,31,21)$, $(321,21,31)$. -$\sigma = 312$: $(312,31,32)$, $(312,32,31)$. -$\sigma = 231$: $(231,21,31)$, $(231,31,21)$. -Conjectured generalisation [showed to be false in the answer of @Ilya Bogdanov]: For $1\leq v \leq k$, denote by $N_{2k-v}^{\sigma}$ the number of the triples $(\pi, s_{1},s_{2})$ for which $\pi$ is of length $2k-v$ and $s_{1}$ and $s_{2}$ have $v$ common elements. Is it true that $N_{2k-v}^{\sigma}>\binom{2k-v}{k}^{2}$ for every $\sigma$. Note that for $v=k$, we always have $1$ triple and the conditions holds trivially. When $v=0$, we obviously have $N_{2k}^{\sigma} = \binom{2k}{k}^{2}$ for every $\sigma$ of length $k$. - -LAST EDIT: 2020-04-13. Below is an interpretation of the right-hand side that may lead to a new, intuitive proof: -Denote by $N_{2k-1}^{\sigma , \sigma'}$ the number of merges of length $2k-1$ for the two patterns $\sigma = \sigma_{1}\cdots\sigma_{k}$ and $\sigma'=\sigma'_{1}\cdots\sigma'_{k}$ of length $k$. Furthermore, let $f(i,j,k) = \binom{i+j-2}{i-1}\binom{2k-i-j}{k-i}$. Note that there exist exactly $f(i,j,k)$ merges of $\sigma$ and $\sigma'$, which have a common element corresponding to $\sigma_{i}$ and $\sigma'_{j}$. Consider a fixed $\sigma$ and $\sigma'$ chosen uniformly at random from $S_{k}$. By linearity of expectation: -$$ -\mathbb{E}(N_{2k-1}^{\sigma , \sigma'}) = \sum\limits_{i=1}^{k}\sum\limits_{j=1}^{k}[\mathbb{E}(f(\sigma_{i},\sigma'_{j},k))\cdot f(i,j,k)]. -$$ -Since $\sigma'_{j}$ has a uniform distribution over $[k]$, for every $j\in [k]$, we have: -$$ -\mathbb{E}(f(\sigma_{i},\sigma'_{j},k)) = \frac{1}{k}\sum\limits_{u=1}^{k}f(\sigma_{i},u,k) = \binom{2k-1}{k}, -$$ -since for every fixed $\sigma_{i} = x\in [k]$, -$$ -\sum\limits_{u=1}^{k}f(x,u,k) = \sum\limits_{u=1}^{k}\binom{x+u-2}{x-1}\binom{2k-x-u}{k-x} = \binom{2k-1}{k}. -$$ -Then, -$$ - \mathbb{E}(N_{2k-1}^{\sigma , \sigma'}) = \frac{1}{k}\binom{2k-1}{k}\sum\limits_{i=1}^{k}\sum\limits_{j=1}^{k}f(i,j,k) = \frac{1}{k}\binom{2k-1}{k}\sum\limits_{i=1}^{k}\binom{2k-1}{k} = \\ -\frac{1}{k}\binom{2k-1}{k}k\binom{2k-1}{k} = \binom{2k-1}{k}^{2}. -$$ -Therefore, we have to prove that -$$ -N_{2k-1}^{\sigma} > \mathbb{E}(N_{2k-1}^{\sigma , \sigma'}), -$$ -when $\sigma'$ is chosen uniformly at random. Is there a way to use this new form of the statement? -Note: The same interpretation of the RHS, as the given expectation, can be obtained combinatorially, as well. - -REPLY [7 votes]: By @Max Alexeyev's solution above $N_{2k-1}^{\sigma}=tr(M_{k}(P_{\sigma}M_{k}P_{\sigma}^{-1}))$. -The eigenvalues and eigenvectors of $M_k$ are given here: Result attribution for eigenvalues of a matrix of Pascal-type. -In particular $\mathbf{e}:=(1,\ldots,1)$ (the all-ones vector) is an eigenvector to the eigenvalue ${2k-1 \choose k}$ for $M_k$ (and then also for $P_\sigma M_k P_\sigma^{-1}$). Hence $\mathbf{e}$ is an eigenvector of $M_{k}(P_{\sigma}M_{k}P_{\sigma}^{-1})$ to the eigenvalue ${2k-1 \choose k}^2$. -The desired inequality follows (since the product of symmetric positive definite matrices has only positive eigenvalues). -UPDATE: the inequality was already proved in the same way here (Lemma 4.3) -https://doi.org/10.1016/j.ejc.2009.02.004<|endoftext|> -TITLE: Why are coroots needed for the classification of reductive groups? -QUESTION [12 upvotes]: As we know reductive groups up to isomorphism corresponds to root data up to isomorphism. My question is why in the definition of root data do we need the coroots? -Let's break it down to two questions: - -Can you give an example of two non-isomorphic reductive groups $G_1$ and $G_2$ for which one gets the same roots? (I.e., if $\Phi_1$ are the roots of the first root datum whose character group is $X_1$, and $\Phi_2$ is the roots of the second root datum whose character group is $\Phi_2$, then there exists and isomorphism $X_1\rightarrow X_2$ which reduces to a bijection of $\Phi_1$ with $\Phi_2$.) If I understand correctly, I don't think that's ever possible if $G_1$ and $G_2$ are centrally isogenous... And of course a minimal requirement for such an example is that $G_1$ and $G_2$ have the same rank. -Heuristically, what information do the coroots provide? - -REPLY [13 votes]: (1) As anon says, an example is $G_1 = \mathrm{GL}_2$ and $G_2 = \mathbb{G}_m \times \mathrm{PGL}_2$. We can identify the root lattice and co-root lattice with $\mathbb{Z}^2$ (with the pairing being the standard dot product) so that the root and coroot systems are -$$ \Phi_1 = \{ \pm (1,-1) \},\ \Phi_1^{\vee} = \{ \pm (1,-1) \} \qquad \Phi_2 = \{ \pm (1,0) \},\ \Phi_2^{\vee} = \{ \pm (2,0) \}.$$ -The automorphism $(x,y) \mapsto (x,x+y)$ of $\mathbb{Z}^2$ takes $\Phi_1$ to $\Phi_2$. However, no such automorphism can take $\Phi_1^{\vee}$ to $\Phi_2^{\vee}$, since the vectors in $\Phi_2^{\vee}$ are divisible by $2$ and those in $\Phi_1^{\vee}$ are not. -(2) Maybe this is too basic but: Fix a maximal torus $T$ in $G$. There are, up to conjugacy in the source, finitely many maps $\mathrm{SL}_2 \to G$ for which the maximal torus of $\mathrm{SL}_2$ lands in $T$. The coroots, thought of as one parameter subgroups of $T$, are the images of the torus of $\mathrm{SL}_2$ under those maps. - -REPLY [13 votes]: $\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Zent{Z}\newcommand\Q{\mathbb Q}\newcommand\Z{\mathbb Z}$The collections of roots and the coroots, as abstract root systems, provide the same information (each being recoverable as the dual of the other), which may be why it seems non-obvious that the co-roots matter. The point is that we are given not just $(R, R^\vee)$ but $(X, R, X^\vee, R^\vee)$, which is to say the way that the root and co-root systems sit inside dual integral lattices; or, to say it differently, we have $R^\vee$ not just as an abstract root system, but as a collection of elements of $\Hom(X, \Z)$. -For semisimple groups, this, too, is unnecessary: since $X \otimes_\Z \Q$ is the $\Q$-span of $R$, one can recover $R^\vee$ inside $X^\vee \otimes_\Z \Q = \Hom_\Z(X, \Q)$ in the sense of abstract root systems. (To say it less fancily, we know the pairing of $R^\vee$ with $R$, hence with $\Q R = X \otimes_\Z \Q$.) It is part of the structure theory that the resulting subset of $X^\vee \otimes_\Z \Q$ actually lies in $X^\vee = \Hom_\Z(X, \Z)$. -I had a hunch that $(X, R)$ was really telling us the ingredients of $(G/\Zent(G), \Zent(G))$, and that the rôle of additionally specifying $R^\vee$ as a subset of $X^\vee$ was to specify the particular extension $1 \to \Zent(G) \to G \to G/\Zent(G) \to 1$; but, thanks to comments of @JayTaylor and @DavidESpeyer, I realise that I was not quite right (although only finitely wrong, which I guess is a good amount to be wrong). In fact we recover $(G/\Zent(G)^\circ, \Zent(G)^\circ)$, and need to specify the extension there. -To be specific, there is a (maybe weakly?) terminal group with a given root system $(X, R)$, namely, the group $G(X \cap \Q R, R) \times D(X/X \cap \Q R)$, where $G(X \cap \Q R, R)$ is the semisimple group with the indicated root system, and $D(X/X \cap \Q R)$ is the torus with the indicated character lattice. Given any other group $G$ with the same root system, we simply map $G \to G/\Zent(G)^\circ \times G/[G, G]$. Now the natural map $\Zent(G)^\circ \to G/[G, G]$ is not an isomorphism, but it is an isogeny, and isogenous tori over an algebraically closed field are isomorphic—they have the same rank, and that's all that there is to say about a torus over an algebraically closed field. This is the precise sense in which I say that $(X, R)$ ‘exactly knows’ $(G/\Zent(G)^\circ, \Zent(G)^\circ)$. -Now to the additional information carried in the co-roots, viewed as elements of $X^\vee$. Since there is an almost-direct product decomposition, in the form of a canonical isogeny $[G, G] \times \Zent(G)^\circ \to G$, we have that there is also an isogeny $[G, G] \to G/\Zent(G)^\circ$. This isogeny is determined by the corresponding map on root data (not systems); and, whereas the root datum of $G/\Zent(G)^\circ$ is $(X \cap \Q R, R, X^\vee/R^\perp, R^\vee)$, that of $[G, G]$ is $(X/R^{\vee\,\perp}, R, X^\vee \cap \Q R^\vee, R^\vee)$, where $R^\perp = \{\lambda \in X^\vee \mathrel: \text{$\langle\alpha, \lambda\rangle = 0$ for all $\alpha \in R$}\}$ and $R^{\vee\,\perp} = \{\chi \in X \mathrel: \text{$\langle\chi, \alpha^\vee\rangle = 0$ for all $\alpha^\vee \in R^\vee$}\}$. The morphism is the canonical one. We see how what's important is exactly the way that $R^\vee$ acts on $X$. -To put it more heuristically—at least for my value of ‘heuristic’—what we're really gaining is the ability to see the different ways that the connected centre intersects the derived subgroup. Notice the manifestation of this in @anon's example: for $\operatorname{GL}_1 \times \operatorname{PGL}_2$, the connected centre is the first factor and the derived subgroup is the second factor, and they intersect trivially; whereas, for $\operatorname{GL}_2$, the connected centre is the subgroup of scalar matrices, which intersects the derived subgroup $\operatorname{SL}_2$ in a subgroup of order 2. Specifically, the reason that this is interesting is that the kernel of $[G, G] \to G/\Zent(G)^\circ$ is $\Zent(G)^\circ \cap [G, G]$; and the character lattice of $\Zent(G)^\circ \cap [G, G]$ is $X/((X \cap \Q R) + R^{\vee\,\perp})$.<|endoftext|> -TITLE: On the definition of A-theory -QUESTION [18 upvotes]: Waldhausen's A-theory is a version of algebraic K-theory of spaces. Concretely, for a (pointed) space $X$, he considers the 'Waldhausen category' $\mathcal R_f(X)$ of finite retractive CW-complexes over $X$, applies his $S_{\bullet}$ construction to it, and obtains an infinite loop space, $A(X)$. The functor $A$ is extremely important in high-dimensional geometric topology, for instance it prominently features in the definition of $Wh^{\text{Diff}}(X)$, and in the parametrized $h$-cobordism theorem by Waldhausen, Jahren, and Rognes. -For some time, I believed that $A(X)$ could equivalently be described (using modern machinery that wasn't available when Waldhausen developed his theory) as the $K$-theory of the ring spectrum $\sum^{\infty}_+ \Omega X$, suitable interpreted. -This week I learned that while this might work to understand the connected components, it does not give the right description on $\pi_0$: for any connected space $X$, $\pi_0A(X)$ is simply $\mathbb Z$, given by the relative Euler characteristic of the relative cell complex. Moreover, the canonical map $A(X) \to K(\mathbb Z\pi_1(X))$ induces the canonical map $\mathbb Z \to K_0(\mathbb Z\pi_1(X))$ on connected components, and the cokernel of this map is $\tilde{K}_0(\mathbb Z \pi_1)$, which is often non-trivial. (See Wall's finiteness obstruction). -So my question is the following: could $A$-theory have been defined in terms of finitely dominated as opposed to finite relative CW-complexes, and then what I believed actually holds? And we just have to keep in mind that there is this difference on $\pi_0$, but besides that all is fine? Or is there something more substantial going on? -Sorry if this question is maybe rather vague, but it is my feeling that I am not the only one who might be puzzled about this... - -REPLY [7 votes]: Since the question remains unanswered, let me copy Tom Goodwillie's comment: - -If you allow finitely dominated instead of finite, it changes only π0. Analogously, in defining K(R) if you use finitely generated projective modules instead of free, it changes only π0. I believe that this is discussed somewhere in Waldhausen's big foundational paper. And in the EKMM book the corresponding issue for connective ring spectra is discussed. - -In more detail, for $X$ a connected space the (∞-)category of perfect $\mathbb{S}[\Omega X]$-modules is the Spanier-Whitehead category of the category of finitely dominated retractive CW-complexes over $X$, and therefore it has the same algebraic K-theory. Restricting to finite retractive CW-complexes over $X$ corresponds to taking the stable subcategory of perfect $\mathbb{S}[\Omega X]$-modules generated by the free ones under colimit, and so by Waldhausen's cofinality theorem it just replaces the $\pi_0$ with $\mathbb{Z}$. -This is worked out in detail in Lecture 21 of Jacob Lurie's course Algebraic K-theory and manifold topology. Note in particular Warning~9 there, where Lurie remarks that his definition of A-theory differs from the "traditional" one only on $\pi_0$.<|endoftext|> -TITLE: Orbit counting polynomials over finite fields -QUESTION [5 upvotes]: Let $X$ be an affine variety defined over $\mathbb{Z}$ and let $G$ be an algebraic group defined over $\mathbb{Z}$. Let $q$ be a power of a prime number. We write $\mathbb{F}_q$ for the field with $q$ elements and we define $$n_q:= \text{number of } G(\mathbb{F}_q) \text{ orbits in } X(\mathbb{F}_q).$$ -Question 1: Under what condition $n_q$ has a closed formula which is a polynomial in $q$? -Question 2: In case of a positive answer for Question 1, do the coefficients of this polynomial relate to any known invariants of $X$ and $G$? -Question 3: given a polynomial with rational coefficients, is there any way to detect if it arises in the above situation? - -REPLY [3 votes]: If the group $G$ is connected then there is a nice cohomological expression for the quantity $\#X(\mathbf F_q)/\#G(\mathbf F_q)$. Indeed in this case Lang's theorem implies that this equals the number of $\mathbf F_q$-points of the quotient stack $[X/G]$ (counted, as always, weighted by their stabilizer group) which can be interpreted cohomologically by the Grothendieck--Lefschetz trace formula, which was generalized to stacks by Behrend, as alluded to by Qiaochu. There is a very large literature about schemes and stacks with the property that their number of $\mathbf F_q$-points is given by a polynomial in $q$. Unless there is some "unexpected cancellation" in the cohomology, this corresponds to all cohomology groups being of Tate type. -If $G$ is disconnected then the quantity $\#[X/G](\mathbf F_q)$ is still well-behaved, but it generally not equal to $\#X(\mathbf F_q)/\#G(\mathbf F_q)$, since an $\mathbf F_q$-point of $[X/G]$ no longer has any reason to lift to an $\mathbf F_q$-point of $X$. -I doubt that you will find any nice expression or general result concerning the actual number of orbits, beyond what you get by combining the Grothendieck--Lefschetz trace formula and Burnside's lemma.<|endoftext|> -TITLE: Is it possible to jump ship to mathematics? -QUESTION [9 upvotes]: A few years back, I did an undergrad double major in math and physics, and afterwards went to a top-tier university to work on a physics Ph.D. Around about my third year of the Ph.D., I started to feel like my research was not really going anywhere, and I had serious thoughts of dropping out. I sought out advice though, and the word I got was always "Everybody feels that way in their third year. Stick it out and it will get better." -Unfortunately, it did not get better. I'm now in my sixth year of the Ph.D., and in all the time I've been here, my entire research group has only published one paper. It may take another year or two before I get to be lead author on a paper, and even then, I don't think the paper will be very good. Moreover, I have little desire to do this work for the rest of my life. And so I am again thinking seriously about what alternative options I have. -While I've published no physics papers in my time at grad school, I did publish two small notes of original research in Amer. Math. Monthly on my own, and even though they're not terribly consequential papers, I'm happier about them than about my "official" research. I enjoy math research, but I don't think I know enough to do it professionally (my physics research is not something like gauge theory that is of direct interest to mathematicians). I've been considering dropping my current program and starting over in a math Ph.D. program, but it's not clear to me whether this is a viable option. -So my questions are whether good math Ph.D. programs would be willing to take a person like myself? How damaging to my credibility as a researcher would it be to drop out of a Ph.D. program? Should I even mention that I was once a physics Ph.D. student, or just leave a big blank spot in my resume for those years? And finally, if this course of action is realistic, how should I go about getting recommendations (I would not expect my advisor to be very happy if I dropped out, and I have no real connections in the math community at this point)? -Obviously, I'm not going to do this without careful consideration of many factors, but I wanted to solicit the view of people in math academia on whether I'm being at all reasonable, or if I'm just a crank. - -(I can provide some more situational details if it is necessary, but I would like to maintain anonymity and genericity to the extent possible.) - -REPLY [10 votes]: Certainly it is possible to "jump ship" and, as the comments have already pointed out, a few years in a physics PhD program (where you published math papers) will not damage your credibility. In fact, I'd encourage you to market it as a strength. When you apply to math grad school, the fact is that you already have math publications (most other applicants don't). You already know how to do research. You already know how to pass qualifying exams and give talks. You are a lot stronger than the other applicants. In terms of letters of recommendation, I'd pick your favorite professor and ask that person. I'm sure they care about your well-being in life and would not want you to do a career you hate. This professor can speak to your strengths as a student, your strengths as a researcher, your passion for math, and how you got along with people in the department. That's plenty enough to get into a PhD program in math. The letter doesn't have to be from your physics PhD advisor. -I'd also market your time in physics as a strength when you do eventually apply for jobs after finishing the math PhD. I've been on about 10 hiring committees and can remember several cases where people had jumped ship, including a few where the person had actually already finished their first PhD then jumped ship and gotten a second PhD. We were really impressed with these candidates, and I remember one case where the person got a job offer from another university before we could even schedule an on-campus interview. -So, instead of saying "I hated physics and quit" you want to say something to the effect of "I have experience doing research in both mathematics and physics. This enables me to do interdisciplinary research, and to create collaborations with professors both in and outside of your department. It also informs my teaching, as I can better connect with math students who favor a physical intuition. And, it informs my advising of students: how to make decisions based on their passion, how to trust their instincts, and how to be resilient." -Now, I want to be clear that I'm at a liberal arts college, where teaching and advising are super important, and where we like candidates with a breadth of knowledge and the possibility to do interdisciplinary research. Even if your math research has nothing to do with physics, it's a benefit that you have such a deep knowledge of physics. It could help, for example, in teaching math courses aimed at students trying to do engineering or physics grad school (like, div, grad, curl for physicists, heat equation, etc). These days even big research universities care about your teaching, so I don't think the physics time would hurt you. -One last point. You've already sunk six years into the physics PhD. All else being equal, it would be better to have finished it before starting the math PhD (which, by the way, you can probably do a lot faster than six years with all you've learned about research) rather than to leave with no degree. If you think you can finish it within a year or so, I'd encourage you to do that, even if you plan to jump ship.<|endoftext|> -TITLE: Is the derivative of $x^n + x^{n-1} + \dots + x + 1$ irreducible? -QUESTION [26 upvotes]: I am working on some combinatorics problems. One of my problems leads to the following question: -Is it true that the derivative of $x^n + x^{n-1} + \dots + x + 1,$ namely $nx^{n-1} + (n-1)x^{n-2} + \dots + 2x + 1$, is irreducible in $\mathbb{Z}[X]$? -I believe it is true, and I have test by computers that it is true for $n \leq 100.$ - -REPLY [10 votes]: (Turning the comments into a community wiki answer.) -This problem is discussed in Classes of polynomials having only one non-cyclotomic irreducible factor, by Borisov, Filaseta, Lam, and Trifonov (Acta Arithmetica 90 (1999), 121–153). They prove irreducibility in many special cases but the problem remains open (or at least, none of the papers that Google Scholar lists as citing this paper solves the problem).<|endoftext|> -TITLE: Intersection theory in analytic geometry -QUESTION [7 upvotes]: This might be a weird/stupid question, but it came to me a couple of times, and I would like to get an answer for that. -In some papers I read, constantly the authors define some analytic subspaces, say $X$ and $Y$, and then the authors take the intersection product of their cycles $[X]\cdot [Y]$ in the homology group, without requiring $x$ intersect $Y$ transversely. -My question is, it seems that there is an intersection theory applied here, but I don't know how it works. Does Fulton-MacPherson's intersection theory directly apply to analytic setting ? - -REPLY [10 votes]: You don't say what kind of space $X$ and $Y$ are subspaces of. But if they sit in an oriented manifold there's an easy way to define an intersection product in homology. Namely if $M$ is an oriented $d$-manifold then there is a Poincaré duality isomorphism -$$ H_i(M,\mathbf Z) \cong H^{d-i}_c(M,\mathbf Z)$$ -between homology and compactly supported cohomology. The cohomology with compact support is a ring (though typically a non-unital ring) for the same reason that the usual cohomology is a ring: use contravariant functoriality for the diagonal morphism. In this way we obtain an intersection product on homology. -One can also work with Borel-Moore homology which is sometimes even nicer: under Poincaré duality the Borel-Moore homology corresponds to the usual cohomology, and in particular we get a unital ring, the unit being given by the fundamental class. And fundamental classes exist very generally in Borel-Moore homology, for example any irreducible complex analytic space (not necessarily smooth, not necessarily compact) has a fundamental class. -This is all easier than what Fulton does in his book. But Fulton obtains an intersection product on Chow groups, which carries more refined information.<|endoftext|> -TITLE: Reference for quaternions and complexification -QUESTION [7 upvotes]: I am writing something on the complexification of a real associative algebra. There are two well-known isomorphisms about quaternions: $$\mathbb{H}\otimes_\mathbb{R}\mathbb{C}\simeq M_2(\mathbb{C})$$ and $$\mathbb{H}\otimes_\mathbb{R}\mathbb{H}\simeq M_4(\mathbb{R}).$$ I need a reference which includes both of these two isomorphisms. I also wonder if there is a reference dealing with the bimodules over the real division rings and their complexification. Anything about complexification and related Galois descent are also welcome! Thanks a lot! - -REPLY [6 votes]: What you are looking for is contained in -T. Y. Lam: Introduction to quadratic forms over fields, Graduate Studies in Mathematics 67. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-1095-2/hbk). xxi, 550 p. (2005). ZBL1068.11023. -In particular: - -the isomorphism $$\mathbb{H}\otimes_\mathbb{R}\mathbb{C}\simeq M_2(\mathbb{C})$$ is a consequence of the proof of parts (3), (4) of Proposition 1.1 (taking, in the Author's notation, $F=\mathbb{R}$, $E=\mathbb{C}$, $a=b=-1$), see the beginning of p. 53; -the isomorphism $$\mathbb{H}\otimes_\mathbb{R}\mathbb{H}\simeq M_4(\mathbb{R})$$ is the content of Exercise 9 p. 76, with $F=\mathbb{R}$.<|endoftext|> -TITLE: Endomorphisms of the Cuntz algebra -QUESTION [5 upvotes]: Consider the Cuntz algebra $\mathcal{O}_n$ with $n \geq 2$ and let $\text{End}(\mathcal{O}_n)$ be the set of all (unital) $\ast$-endomorphisms of $\mathcal{O}_n$. I was wondering if there exists an element $x \in \mathcal{O}_n$ such that the evaluation map $\text{End}(\mathcal{O}_n) \rightarrow \mathcal{O_n},$ $\phi \mapsto \phi (x)$ is injective. -If no, what is the smallest $k \in \mathbb{N}$ for which $x\in \mathcal{O}_n \otimes \mathbb{C}^k$ exists such that the map $\text{End}(\mathcal{O}_n) \rightarrow \mathcal{O}_n \otimes \mathbb{C}^k$ given by $\phi \mapsto (\phi \otimes \mathrm{id}) (x)$ is injective? Is it $k=n-1$? - -REPLY [5 votes]: Although not exactly what the OP has in mind, -there is another interesting characterization of endomorphisms of $\mathcal O_n$ in terms of single elements. Namely -there is a one-to-one correspondence between endomorphisms of $\mathcal O_n$ and unitary elements of $\mathcal O_n$ given -as follows: - -If $u$ is a unitary element, one defines an endomorphisms $\varphi _u$ by sending each generator $S_i$ to $uS_i$. - -Conversely, given an endomorphism $\varphi $, one defines the unitary element -$$ - u_\varphi = \sum_{i=1}^n\varphi (S_i)S_i^*. - $$ - - -It is in fact very easy to show that these correspondences are each other's inverse.<|endoftext|> -TITLE: Books on the relationship between the Socratic method and mathematics? -QUESTION [8 upvotes]: Apart from books on heuristics by George Polya. -When trying to engage with and understand mathematical concepts and when applying abstract mathematical concepts to model "continuum" or real world problems. -I find it fruitful to engage in Socratic questioning based on the Socratic method and principles of critical thinking such as: - -Analyzing thought (questioning the components of my thinking, for example questioning goals and purpose, questioning assumptions) - -Assessing thought (questioning standards of my thinking, for example questioning clarity, accuracy, precision, breadth and depth) - - -I understand the only way to "learn mathematics is to do mathematics", but if this is approached unthinkingly, the learning in my experience tends to be rather superficial. -Since the quality of our thinking is driven by questions, "doing mathematics" should be approached in the spirit of Socrates who acknowledged “The only true wisdom is in knowing you know nothing”. -I suppose if meaning is context dependent, most of Socrates questions were rigorously focused on definitions. Not sure if Godel's incompleteness theorem regarding logical inconsistency is somehow related to this? -The importance of asking essential questions was fundamentally important to Georg Cantor given that his 1867 Doctoral thesis was entitled "In mathematics the art of proposing a question must be held of higher value than solving it." -Does anyone know of any textbooks which focus on the relationship between the Socratic method and deep mathematical thinking? - -REPLY [2 votes]: The book: -Alfréd Rényi, Dialogues on Mathematics, Holden Day, San Francisco, California 1975 -is a very moving reading about the topic of Mathematics and the Socratic method.<|endoftext|> -TITLE: Are all monotonically normal manifolds of dimension at least two metrizable? -QUESTION [8 upvotes]: Alan Dow and Frank Tall recently proved the consistency of the statement Every hereditarily normal manifold of dimension at least two is metrizable. -See: Dow, Alan; Tall, Franklin D., Hereditarily normal manifolds of dimension greater than one may all be metrizable, Trans. Am. Math. Soc. 372, No. 10, 6805-6851 (2019). ZBL1427.54006. -This suggests a natural question: - -Is it true in ZFC that every monotonically normal manifold of dimension at least two is metrizable? - -The Long Line is an example of a one-dimensional non-metrizable monotonically normal manifold. -NOTE: By "manifold" I mean a Hausdorff space which is locally euclidean. - -REPLY [6 votes]: It would appear that that it is true. The result is due to Z. Balogh and E. Rudin and appears in their paper Monotone Normality, Top. App. 47, (1992), 115-127. The statement to quote is the following. - -Corollary 2.3.(e). A manifold of dimension $\geq2$ is metrizable if and only if it is monotonically normal. - -This follows easily from the first of the two main results in the paper. - -Theorem I. A monotonically normal space is paracompact if and only if it does not have a closed subspace homeomorphic to a stationary subset of a regular uncountable cardinal. - -It's worth mentioning that with a little thought it's not hard to repackage the dimension hypothesis so as to include all manifolds. - -A manifold is metrisable if and only if it is separable and monotonically normal.<|endoftext|> -TITLE: Is there a model of ZFC that can define a "longer" model of ZFC to which it is isomorphic? -QUESTION [14 upvotes]: Suppose $ZFC$ is consistent. Is there a model $\mathcal{M}$ of $ZFC$ and formulae $\varphi_D(x)$ and $\varphi_\in(x,y)$ that define (in $\mathcal{M}$) the domain and membership relation of a model $\mathcal{N}$ of $ZFC$ such that we have $\mathcal{N}\cong\mathcal{M}$ externally (i.e., in $V$) and yet $\mathcal{M}$ thinks $\mathcal{N}$ is well-founded but not set-like? -I think it's clear that if this were true $\mathcal{M}$ would need to be ill-founded. -I suspect the Ehrenfeucht-Mostowski theorem, which gives automorphisms for such models, can be adapted to obtain an embedding from some $\mathcal{N}$ onto an initial segment of itself, say $\mathcal{M}$, but where the respective spines of indiscernibles used to generate these models are (externally) isomorphic. If that works, I think we'd have $\mathcal{N}\cong\mathcal{M}$ but - among other things - I can't see that (an isomorphic copy of) $\mathcal{N}$ would be definable in $\mathcal{M}$. -If the issue were that assuming $ZFC$ was consistent was not sufficient, I'd still be interested in the assumption that did suffice. - -REPLY [13 votes]: The answer is yes. -Let $M$ be a countable computably saturated model of ZFC with a measurable cardinal $\kappa$. Let $N$ be the Ord-length iterated ultrapower of a measure on $\kappa$. The model $M$ thinks $N$ is a definable well-founded class structure, which is strictly taller than $M$. But $M$ and $N$ are two countable models of ZFC with the same theory, same standard system, and computably saturated, and this ensures that they are isomorphic (externally). The model $N$ is computably saturated, since any definable class in a computably saturated structure is also computably saturated. - -Here is my previous answer to your question. Here is an example of the dual situation, where a model of set theory can define a model that it thinks is strictly shorter, but to which it is actually isomorphic. I am posting this because I find it interesting and related, even though it doesn't answer your question. -Theorem. Every countable computably saturated model of ZFC is isomorphic to a rank-initial segment of itself. -Proof. Suppose that $M$ is a countable computably saturated model of ZFC. It follows that the theory of $M$ is in the standard system of $M$. That is, there is a natural number $t$ in $M$ that codes a sequence of natural numbers, such that the standard part of that sequence is exactly the Gödel codes of the sentences true in $M$. By reflection, increasing large standard fragments of this theory are true in the rank-initial segments $V_\alpha^M$, and so by overspill, there must be some $N=(V_\alpha)^M$ that satisfies a nonstandard fragment of the theory coded by $t$. Thus, $N$ and $M$ have the same theory, and the same standard system. From this, it follows that they are isomorphic. $\Box$ -The model $N$ is definable in $M$ from parameters, since $M$ thinks $N$ is just $V_\alpha$. -You can see various versions of this argument in my paper: - -Gitman, Victoria; Hamkins, Joel David, A natural model of the multiverse axioms, Notre Dame J. Formal Logic 51, No. 4, 475-484 (2010). DOI:10.1215/00294527-2010-030, ZBL1214.03035. - -Here is another variation on the theme: -Theorem. There is a countable model $M$ of ZFC that is isomorphic to a forcing extension $M[c]$ of itself. -This is a little closer to what you asked for, since $M[c]$ can define $M$, and it is isomorphic to $M$. But this model is still set-like, so ultimately it doesn't fulfill your requirement. -Proof. Let $M_0$ be a countable computably saturated model of ZFC, and let $M=M_0[d]$ be the model obtained by forcing to add a Cohen real $d$. Let $c$ be $M$-generic, and consider $M[c]$. Since both are obtained by forcing over $M_0$ to add a Cohen real, they have the same theory, and they are both computably saturated and have the same standard system, hence isomorphic. $\Box$<|endoftext|> -TITLE: $M = AA^t$ where $A$ has unit norm columns -QUESTION [9 upvotes]: Let $M \in \mathbb{R}^{k\times k}$ positive definite with $\operatorname{tr} M = m$, where $m$ is an integer such that $m \geq k$. I have found a way (using this answer) to decompose $M = AA^t$ with $A \in \mathbb{R}^{k \times m}$ such that $A = (a_1, \dots, a_m), a_i \in \mathbb{R}^k$ and $\|a_i\|_2 = 1, i=1,\dots,m$. - -Is there a name for such a decomposition? This is not Cholesky, although it looks similar. -Is this decomposition unique? We can always take $\hat{A} := AD$ where $D$ is a permutation matrix with $\pm 1$ entries. Then $\hat{A}\hat{A}^t = M$ and $\hat{A}$'s columns have unit norm. I am not sure if there is any other obstruction to uniqueness. -In my numerical experiments, I find that the some columns of $A$ are identical (up to a sign). Any reason for that? - -Example -Let $M =diag(1.5,1.5)$. One can verify that $M=AA^t$ for -$$ -A = -\begin{pmatrix} -\sqrt{3/4}& \sqrt{3/4}& 0 \\ --1/2& 1/2 & 1\\ -\end{pmatrix} -$$ -P.S. The assumption on the trace above is necessary because $\text{tr} M = \text{tr} AA^t = \text{tr}A^tA$ and $A^tA\in \mathbb{R}^{m \times m}$ has unit diagonal. -Reference -Using Raphael's answer below I was able to find the reference: -Peter A. Fillmore, On sums of projections, Journal of functional analysis 4, 146-152 (1969). - -REPLY [2 votes]: This decomposition is equivalent to write $M$ as a sum of rank one orthogonal projection $$ M = \sum_{i=1}^m a_i a_i^* $$ -with $\|a_i\|=1$. Indeed for any $x$ we have $$(Mx)_{s} = \sum_{i\leq m,t\leq k} A_{si}A^T_{it}x_t = \sum_{i\leq m} (a_i)_s \langle a_i,x\rangle -$$ -Remark that in form it is easy to see the invariance by permutation with $\pm 1$ entries and that $\text{Tr}(M)=m$. -We can consider the application $\phi:(\mathbb{S}^{k-1})^m\rightarrow \mathbb{R}^{k\times k}$, $\phi(a_1,\cdots,a_m)=AA^T=M$. Because $(\mathbb{S}^{k-1})^m$ is a manifold of dimension $m(k-1)$ and the subset of symetric matrices of trace $m$ is a manifold of dimension $\frac{k(k+1)}{2}-1$. It is clear that we don't have unicity in the general case if $m> \frac{k^2+k-2}{2(k-1)}=\frac{k+2}{2}$.<|endoftext|> -TITLE: Replacing maximum degree with degeneracy in Reed's conjecture -QUESTION [7 upvotes]: Reed's conjecture says that $\chi(G)\leq \lceil\frac{\omega(G)+\Delta(G)+1}{2}\rceil$. One can think of $\lceil\frac{\omega(G)+\Delta(G)+1}{2}\rceil$ as the (rounded-up) average of the trivial lower bound and the trivial upper-bound on $\chi(G)$. An equally trivial upper-bound on $\chi(G)$ is $\mathrm{degen}(G)+1$ where $\mathrm{degen}(G)=\max\{\delta(H): H\subseteq G\}$ and clearly $\mathrm{degen}(G)\leq \Delta(G)$. -So I was just wondering if there are any simple examples which would disprove the stronger statement $\chi(G)\leq \lceil\frac{\omega(G)+\mathrm{degen}(G)+1}{2}\rceil$? - -REPLY [6 votes]: Here's a reasonably simple counterexample. -Take $C_9$, and label its vertices $v_0, \ldots, v_8$ along the cycle. Let $\mathcal{I}$ be the family of all independent sets of $C_9$ of size $3$. $\chi(C_9) = 3$, further: -Lemma. For any 3-coloring of $C_9$ there exists $I \in \mathcal{I}$ with vertices of all three colors. -Proof. Let $f$ be the 3-coloring. Following the sequence of colors $f(v_0), f(v_2), \ldots, f(v_8), f(v_1), \ldots, f(v_7), f(v_0)$, we can find a pair of vertices at distance $2$ with different colors, WLOG assume $f(v_0) = 0$, $f(v_2) = 1$. If any of $v_4, \ldots, v_7$ has color $2$, then we are done. Otherwise, $f(v_4), \ldots, f(v_7) \in \{0, 1\}$, and $f(v_4) = f(v_6) \neq f(v_5) = f(v_7)$. Since $f(v_1) = 2$, we can take $v_1, v_4, v_7$. -Now, create a graph $G$ as follows: take $C_9$, and for each $I \in \mathcal{I}$ create a new vertex $u_I$ connected to all elements of $I$. - -$w(G) = 2$, since there are no triangles (ensured by not connecting new vertices to vertices adjacent in $C_9$); - -$degen(G) = 3$. Indeed, for any subgraph $H \subseteq G$, $\delta(H) \leq 3$ if any $u_I \in H$, and $\delta(H) \leq 2$ if $H \subseteq C_9$. - -$\chi(G) = 4$. The upper bound is obvious. The lower bound follows from the lemma above: assume that $G$ is 3-colorable, then for $I = \{a, b, c\}$ produced by the lemma (for the 3-coloring restricted to $C_9$), the color of $u_I$ has to be distinct from (distinct) colors of $a, b, c$, a contradiction. - - -This violates the strong conjecture: $4 > \lceil \frac{2 + 3 + 1}{2}\rceil$.<|endoftext|> -TITLE: Is it possible to define an internal model of ZFC which is not set-like and which is not elementary equivalent to any definable set-like model? -QUESTION [14 upvotes]: (1) Are there formulae $\varphi_D(x)$ and $\varphi_\in(x,y)$ defining an internal model $\mathcal{N}$ of $ZFC$ where $\mathcal{N}$ is not set-like and no definable, set-like, internal model $\mathcal{M}$ is elementary equivalent to $\mathcal{N}$? -(2) Are there formulae $\varphi_D(x)$ and $\varphi_\in(x,y)$ defining an internal model $\mathcal{N}$ of $ZFC$ where $\mathcal{N}$ is not set-like and no definable, set-like, $\it{well-founded}$, internal model $\mathcal{M}$ is elementary equivalent to $\mathcal{N}$? -A simple way to obtain non set-like models of $ZFC$ is to take a normal ultrafilter $U$ and take the iterated ultrapower of $V$ through length $Ord$. This gives us a non set-like model of $ZFC$, but it is elementary equivalent to $V$. (This approach uses parameters, but they are easily removed.) -In a vague nutshell, is there another way to define "long" models without some form of iteration? - -REPLY [7 votes]: To complement Joel's answer that a positive outcome is consistently possible, let me show that the negative outcome is also possible, at least in the case of boldface definability. -For our background universe $V$, suppose that we have some second-order resources available to us, namely a global well-order and truth predicates for all class-sized structures. (I'll comment below on what this assumption entails.) This allows us to then carry out the standard argument for downward Löwenheim–Skolem. If $\mathcal N$ is a class-sized structure then we have Skolem functions for it and so can get a set-sized elementary submodel $\mathcal M$ of $\mathcal N$. In particular, $\mathcal M$ will be set-like and definable (with parameters), entailing a negative answer to your question. -Note, however, that this approach doesn't give you a negative answer for lightface definability, i.e. without parameters. Consider the case of $\mathcal N = (V,\in)$ and suppose we could find $\mathcal M$ a set-sized elementary submodel of $\mathcal N$ which is definable without parameters. But then we could define without parameters the set of true sentences in $(V,\in)$, contradicting Tarski's theorem on the undefinability of truth. -On the other hand, we can get lightface definability if we allow class quantifiers. If your global well-order is definable then, because the truth predicate is definable with class quantifiers, you get definable Skolem functions and so $\mathcal M$ is definable with class quantifiers. It's $\Delta^1_1$-definable, to be precise. Also, $\mathcal M$ is implicitly definable (without parameters), in the sense of Hamkins and Leahy - Algebraicity and implicit definability in set theory. -Let me now address what this assumption on $V$ entails, and why it doesn't apply to Joel's positive case. Having a global well-order is cheap—you can always add a generic one by class forcing, without adding any new sets. The easiest way to do this: just add a Cohen-generic subclass to $\mathrm{Ord}$. But having truth predicates comes with a cost. There's the consistency strength cost, of course, since having a truth predicate for $(V,\in)$ let's you see that ZFC is consistent. But we can say more. -Proposition (Essentially Krajewski): If the structure $(V,\in,\mathrm{Tr})$ satisfies ZF in the expanded language plus the assertion that $\mathrm{Tr}$ satisfies the Tarskian recursion for truth of $(V,\in)$, then $V$ contains a club of ordinals so that $V_\alpha$ is elementary in $V$. -Proof sketch: The point is, you can do the usual reflection argument but using $\mathrm{Tr}$ as a parameter to get the desired $V_\alpha$s, since you can express "$\mathrm{Tr}$ satisfies Tarski" as a first-order formula in the expanded language. There's a small bit more to be said about the $\omega$-nonstandard case, but it's not hard. -In particular, having truth predicates implies you have lots and lots of undefinable ordinals. So any Paris model—one whose ordinals are all definable without parameters—cannot admit a truth predicate. (Or rather—since of course every structure externally admits a truth predicate—if you add a truth predicate you destroy Replacement in the expanded language.) This includes Joel's pointwise definable $L_\lambda$. -Finally, let me note that since the $\mathcal M$ produced is a set, all we need for the counterexample is that $V$ can be extended to have the necessary second-order resources. So, for example, if your $V$ is $V_\kappa$ for some inaccessible $\kappa$, where you only look at definable classes, then you still get the counterexample. For you could expand your classes to the full powerset of $V$, apply the argument there to your definable class-sized structure $\mathcal N$, and thereby get $\mathcal M$ in $V$. But if it's in $V$, then we didn't need the extra classes to define it.<|endoftext|> -TITLE: Permutation function based on subsets -QUESTION [6 upvotes]: We have some subsets $A_1,\dots,A_k$ of $A=\{1,2,\dots,n\}$. For each permutation $\sigma$ of $A$, define $f(\sigma) = \sum_{i=1}^k g(\sigma,A_i)$, where if the earliest element of $A_i$ -in $\sigma$ appears in position $j$, then $g(\sigma,A_i)= 1/j$. Let $\sigma_1$ be the permutation maximizing $f(\sigma)$, breaking ties lexicographically. -Now, we add an element $r\not\in A_1$ to $A_1$, and let $\sigma_2$ be the permutation maximizing $f(\sigma)$. Does it always hold that $r$ appears no later in $\sigma_2$ than in $\sigma_1$? -A natural approach is to show that if $r$ appears later in $\sigma_2$ than in $\sigma_1$, then upon adding $r$ to $A_1$, $f(\sigma_1)$ increases by at least as much as $f(\sigma_2)$. But this may not be true, because there may already be an element in $A_1$ that appears in $\sigma_1$ before $r$. Still, it does not clearly lead to a counterexample either. - -REPLY [3 votes]: I have a counterexample, showing that $r$ can appear later in $\sigma_2$ than in $\sigma_1$. -I'm going to write this counterexample with weights on the sets, but because the weights are rational it can easily be converted to an unweighted counterexample by duplicating sets. -Counterexample: 3x{AC}, 2x{AB}, 2x{BD}, 2x{CD}, 11/8x{C}, 1x{A}, 1x{D}. -Calculations -The initial maximum value of $f$ is $215/24$, achieved by the permutations ADCB, CBAD, CBDA. By the tiebreaker, $\sigma_1$ is ADCB. -Now, we will add the element D to the set {A}, changing it to the set {AD}. -The new maximum value of $f$ is $217/24$, achieved by the permutations CBDA, CDAB, CDBA. By the tiebreaker, $\sigma_2$ is CBDA. -We added the element D to a set, but $\sigma_1$ has D in position 2, while $\sigma_2$ has D in position 3. - -Note that while this counterexample relies on the tiebreaker, it can be lightly modified to not rely on the tiebreaker. -To do so, add the subsets {A} and {B} each with an equal weight $\epsilon$ to the set system, for some $\epsilon$ very close to 0. This increases the value of ADCB by $(5/4)\epsilon$, while CBAD and CBDA increase by $(5/6)\epsilon$ and $(3/4)\epsilon$, respectively. As a result, ADCB is $\sigma_1$ without a tiebreaker needed. -The values of CDAB and CDBA increase by $(7/12)\epsilon$, which is smaller than CBDA's $(3/4)\epsilon$, so CBDA is $\sigma_2$ with no tiebreaker needed. - -The way I came up with the counterexample is by starting with the 2 element sets, which I designed to force the optimal permutation to start with either AD or CB. At this point, adding a D to an A subset would benefit a CBD permutation but not an AD permutation, which would make the CBD permutation the new optimum, and form a counterexample. I then futzed around with the weights on the single subsets to get the optima to work out right.<|endoftext|> -TITLE: Conjecture vs open question on a paper -QUESTION [23 upvotes]: Suppose that you are concluding a paper and writing the open questions (or further research directions) section, and suppose that you have several claims related to the work in the paper that your intuition is telling you they are true. Maybe you are also able to give heuristic reasons why you think the claims are likely true, but you are not yet able to prove them; or you may have some ideas on how to attack them, but that would bring you off-topic from the main content of the paper (or maybe it would make the paper unnecessarily long). -I was curious to hear what sort of criteria you use in deciding whether to give the claims as open question or rather as conjectures (or maybe not give them at all) in your paper? -PS. I have never asked a soft question on MO, but I have read many, so I hope it's okay if I post one. Please let me know if this is too much off-topic. - -REPLY [30 votes]: I doubt you'll get a definitive answer, so I'll frame my answer as appeal to expertise (that I happen to agree with), knowing that others may disagree. Most essays about mathematical writing encourage you to find your own voice and to think about the reader as a guiding principle. I'd add that it's wise to think about the state of your research area. Including conjectures and open questions in your papers gives other researchers something to work on. Phrasing them as conjectures also helps your field, because when some young person proves a Conjecture, it can help their career. They'll be more likely to get a job, win a grant, get tenure, etc. This is part of the argument Clark Barwick made in his essay on The Future of Homotopy Theory. In item 3 he writes (with reference to the field of homotopy theory): - -We do not have a good culture of problems and conjectures. The people at the top of our field do not, as a rule, issue problems or programs of conjectures that shape our subject for years to come. In fact, in many cases, they simply announce results with only an outline of proof - and never generate a complete proof. Then, when others work to develop proofs, they are not said to have solved a problem of So-and-So; rather, they have completed the write-up of So-and-So's proof or given a new proof of So-and-So's theorem. The ossification of a caste system - in which one group has the general ideas and vision while another toils to realize that vision - is no way for the subject to flourish. - -For me, the take-away is to include named Conjectures when I'm pretty sure the result is true, include Open Questions when I'm not that sure, and at all costs avoid Remarks where I claim things are true that I have not actually carefully proven. I think a Remark where one sketches a proof idea is fine, but in the interests of young people in the field, it's important to be clear that the Remark is not a complete proof and (ideally) to include the statement to be proven as a Conjecture. In my early papers, I sometimes used Remarks to advertise future papers. I'm not going to do that any more, because there are plenty of examples where someone did this and then never wrote the future paper, leading to the kind of issue Clark raises above. I'm grateful that Clark gave us something to aspire to, so we can make our field better for young people. -As for when to make something a Conjecture vs an Open Problem, Clark Barwick answers that, too, in his Notes on Mathematical Writing. On page 3, he defines - -A conjecture is an assertion that meets all the following criteria. - -It is precise and unambiguous. -The author strongly suspects that it is the case. -The author considers the assertion interesting or difficult. -The author has seriously attempted to prove it. -Nevertheless, the author does not know how to prove it. - -Anything that satisfies the final condition but not all of the others is a Question or a Problem. Do not fear formulating plenty of Questions and Problems. - -REPLY [12 votes]: As David White says, you're unlikely to get a definitive answer, since to some extent, it is a matter of taste and personal preference. There are many people who use conjecture to mean the five criteria in David's answer. There are others who use conjecture as a way to get people's attention, treating it almost as a provocation: "See if you can prove me wrong". The former have been known to write things such as "the evidence for conjecture XXX is more voluminous than it is convincing" for conjectures made primarily on numerical experiments, or even to refer to the "so-called conjecture YYY" when there is actually quite a bit of theoretical evidence. Some assertions should clearly be posed as questions, e.g., when you are unsure yourself whether you believe it is true or false; while others are very reasonably posed as conjectures because you've hit all five bullet points. For the ones in between, especially where you're basing the assertion partly on intuition gleaned from experience in your field, I'd propose this as a guiding principle: Pose it as a conjecture if you wouldn't mind seeing a paper entitled "A counter-example to a conjecture of Moreschi." But if you'd find that embarrassing, then pose it as a question!<|endoftext|> -TITLE: Almost free circle actions on spheres -QUESTION [9 upvotes]: $\DeclareMathOperator{\Fix}{\operatorname{Fix}}$I am looking for any reference regarding the following problem: -Problem: Consider a smooth almost-free action of $S^1$ on a smooth sphere $S^n$. Then for all finite subgroups $\mathbb{Z}_m\subseteq S^1$, the fixed point set $\Fix(\mathbb{Z}_m)\subseteq S^n$ is either empty or a rational homology sphere. -This is well-known to be true when the action is linear, or by Smith theory when $m=p^k$ is the power of a prime $p$. -A stronger statement would be that for every action of $\mathbb{Z}_m$ on a sphere $S^n$, the fixed point set $\Fix(\mathbb{Z}_m)$ is a rational homology sphere. What I know is that when $m$ is not the power of a prime, then $\Fix(\mathbb{Z}_m)$ is not necessarily a smooth sphere, but I have not found any indication about its possible (rational) homotopy types. -What boggles me the most is that, I had imagined this problem to either well known to be true, or well known to be false, or extremely interesting – however, I have not found any evidence for any of these. - -REPLY [2 votes]: Back in the early 1950's Ed Floyd and Pierre Conner found that composite order cyclic groups could act rather differently that cyclic p-groups. In their tradition, the publication [J. M. Kister, Examples of periodic maps on Euclidean spaces without fixed points. Bull. Amer. Math. Soc. 67 (1961), 471–474] shows that the cyclic group of order 6 can act on a big enough Euclidean space with no fixed points. One point compactifying, one gets an action on a sphere with one fixed point. -There are quite a few survey papers on this subject. Here is one that I found quickly, that has lots of references: Reinhard Schultz, Bull. A.M.S., volume 11, October 1984. See also lots of other papers by him, and also Bob Oliver.<|endoftext|> -TITLE: Is there a definition of reduced $E_\infty$ ring? -QUESTION [5 upvotes]: [Edit: I have completely changed the question in response to the replies given] -I am curious if there is well defined notion of reduced $E_\infty$-ring. -Let $CAlg$ denote the $\infty$-category of $E_\infty$-ring, $CAlg_1$, the one category of communicative rings. I would like to define the analog for reduced ring. -One categorically, $ CAlg_1^{red} \hookrightarrow CAlg_1$ admits a left adjoint $A \mapsto A^{red}:=A /nil(A)$. - -We can define $$CAlg^{red} \hookrightarrow CAlg$$ as the $\infty$-cat. of $E_\infty$-rings whose underling ring is reduced. -Does there exist a left adjoint? As mentioned in comments by Marc, this is false. - -Question[Edit]: What should be the notion of $E_\infty$-ring? Harry in the comment says that this should be an ordinary reduced ring. I would appreciate if some explanation could make this precise. - -REPLY [9 votes]: As Harry points out, the usual approach (For example in Lurie's SAG) is to think of the map $R\to \pi_0(R)$ for a connective $E_\infty$ ring as some kind of quotient by a nil-ideal. This is justified by the fact that $R = \operatorname{lim}_n \tau_{\leq n} R$ is a limit of square-zero extensions $\tau_{\leq n} R\to \tau_{\leq n-1} R$. So if you work with connective $E_\infty$ rings, I think the only reasonable thing is to say that reduced ones are concentrated in degree $0$, and the reduction of $R$ is given by the (ordinary) reduction of $\pi_0(R)$. -This of course breaks down when allowing nonconnective rings. I think in that case there is no reasonable concept of reducedness, but of course it depends a lot on what you want to achieve with such a notion.<|endoftext|> -TITLE: Equations defining hyperbolic geodesics in $\mathbb C \setminus\{0,1\}$ -QUESTION [26 upvotes]: Let $X=\mathbb C\setminus\{0,1\}$, equipped with the hyperbolic structure it inherits from Klein's modular $\lambda$ function $\lambda:\mathbb H \to X$. In each (non-peripheral and nontrivial) free-homotopy class of loop $S^1\to X$, there is a unique hyperbolic representative. -For example, the figure-eight curve $\alpha$ with winding number $+1$ around $1$ and $-1$ around $0$, pictured below, has a unique hyperbolic geodesic representative. - -In fact, the curve drawn in the above linked picture is precisely the path traced by the hyperbolic geodesic representative of $\alpha$, which is given by the lemniscate with equation $$16(x^2+y^2)((x-1)^2+y^2)=1.$$ -Proof sketch: The conformal transformation $w(z)=(2z-1)^2$ is an orbifold covering map from $X$ to a hyperbolic orbifold $Y$, where $Y$ is $\mathbb C \setminus\{1\}$ with a single orbifold point at $0$ with angle $\pi$. The geodesic representative of $\alpha$ projects to a geodesic $\eta$ that begins on the orbifold point, travels once counterclockwise around $1$ and hits the orbifold point, then travels once clockwise around $1$ before finishing again at the orbifold point. The anticonformal automorphism $w\mapsto \frac {\bar{w}}{\bar{w}-1}$ is an orientation-reversing isometry of $Y$ that preserves the homotopy class of $\eta$, so it's not hard to see that $\eta$ must have image given by the fixed-point set of this isometry, namely $|w-1|^2=1$. Putting the pieces together, we find that the points in the image of the geodesic representative of $\alpha$ satisfy $|(2z-1)^2-1|^2=1$, which simplifies to the equation above. -Question: Are all closed hyperbolic geodesics on $X$ algebraic curves? -I admit that it is a bit rash to ask the question in this way — one could more modestly ask which hyperbolic geodesics have polynomial defining equations — but I do want to emphasize that I am not aware of any negative results about the defining equations (the other two figure-eight curves are easy to work out as above). Of course, it's possible that the lemniscate equation arises as some low-dimensional accident due to the special symmetry present in the figure-eight curve, but it also seems entirely plausible that all hyperbolic geodesics on $X$ are somehow very convenient from the viewpoint of the canonical complex coordinate of $X$. -I've tried just looking at other examples using Mathematica to see if I might guess some equation for them (Mathematica's built-in ModularLambda function makes it easy to draw hyperbolic geodesics on $X$), but perhaps I am not fluent enough in solution sets of rational equations in the plane to make anything work. -Two other comments: - -I've tagged this question with "modular forms", because one could think of the above polynomial equation as a polynomial identity relating the real and imaginary parts of the modular function $\lambda$ along the hyperbolic geodesics in $\mathbb H$ that project to closed curves. - -There is an obvious analogy with this very recent MathOverflow question, which is a knot theory version of the same idea. - -REPLY [12 votes]: $\def\CC{\mathbb{C}}\def\HH{\mathbb{H}}\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}\def\Id{\mathrm{Id}}$These geodesics are always algebraic. We can understand their equations using the classical modular curve. -Let $PGL_2(\RR)$ act on $\CC$ in the usual way by Mobius transformations. Note that matrices of positive determinant take the upper half plane to itself, where as matrices of negative determinant switch the upper and lower half planes. We write $\HH$ for the upper half plane. Note also that the action of $PGL_2(\RR)$ commutes with complex conjugation. -Let $A \in GL_2(\RR)$ be a hyperbolic element, there is a unique geodesic through its fixed points. Let $R = A - \tfrac{Tr\ A}{2} \Id$. The geodesic through the fixed points of $A$ can be described as $\{ z : Rz = \bar{z} \}$. For example, let $A = \left[ \begin{smallmatrix} 1&2 \\ 0&1 \end{smallmatrix} \right]\left[ \begin{smallmatrix} 1&0 \\ 2&1 \end{smallmatrix} \right] = \left[ \begin{smallmatrix} 5&2 \\ 2&1 \end{smallmatrix} \right]$, this should give the lemniscate. Then $R = \left[ \begin{smallmatrix} 2&2 \\ 2&-2 \end{smallmatrix} \right]$. It is convenient to replace $R$ by $R/2$, since we are working with $PGL_2$ anyway; let $S = R/2 = \left[ \begin{smallmatrix} 1&1 \\ 1&-1 \end{smallmatrix} \right]$. The set $\{ S z = \bar{z} \}$ can be described explicitly as the circle of radius $\sqrt{2}$ around $1$. -Let $\Gamma(2)$ be the group of integer matrices with determinant $1$ which are $\Id \bmod 2$, and let $P\Gamma(2)$ be the quotient of $\Gamma(2)$ by the central subgroup $\pm \Id$. We can identify the quotient $P\Gamma(2) \backslash \HH$ with $\CC \setminus \{0, 1 \}$ by sending $z \in \HH$ to $\lambda(z)$ in $\CC \setminus \{0, 1 \}$, where $\lambda$ is the modular lambda function. -Note that, if $A \in \Gamma(2)$ then $Tr\ A \equiv 2 \bmod 4$, which shows that all the entries in $A - \tfrac{Tr\ A}{2} \Id$ are even integers, so we will always be able to factor out a $2$ as we did above. More generally, let's write $ A - \tfrac{Tr\ A}{2} \Id = k S$ where $k \in \ZZ$ and $S$ is an integer matrix whose entries have no common factor. Note that we will always have $S^2 = D \Id$ for some $D > 0$, and $\det S = -D$. The cokernel of $S$ will be a cyclic group of order $D$. (Cyclic because we took out common factors of the entries of $R$.) So, our general problem is to find an equation for $\{ \lambda(z) : S z = \bar{z} \}$. -Now, consider the elliptic curves $E_1 = \CC/\langle 1, z \rangle$ and $E_2 = \CC / \langle 1, Sz \rangle$. The matrix $S$ gives a degree $D$ cyclic isogeny from $E_1$ to $E_2$. This means that the $j$-invariants of $E_1$ and $E_2$ must obey the classical modular equation $\Phi_D(j_1, j_2)=0$. But, since $Sz = \bar{z}$, we have $j_2 = \bar{j_1}$. So $\Phi_D(j_1, \bar{j_1})=0$. -Take the equation $\Phi_D(j, \bar{j})=0$ and substitute $j = \tfrac{256 (1-\lambda+\lambda^2)^3}{\lambda^2 (1-\lambda)^2}$. -This gives a polynomial equation $\Psi_D(\lambda, \bar{\lambda})=0$. This will probably be reducible, but at least it will contain the geodesic. - -I carried this out for the lemniscate. The curve $\Phi_2(j_1, j_2)$ is given by the equation: -$$-157464000000000 + 8748000000 j_1 - 162000 j_1^2 + j_1^3 + 8748000000 j_2 + 40773375 j_1 j_2 + 1488 j_1^2 j_2 - 162000 j_2^2 + 1488 j_1 j_2^2 - j_1^2 j_2^2 + j_2^3=0$$ -I made the substitution $j_r = \tfrac{256 (1-\lambda_r+\lambda_r^2)^3}{\lambda_r^2 (1-\lambda_r)^2}$, cleared out denominators, and put $\lambda_1 = x+iy$ and $\lambda_2 = x - i y$. The resulting polynomial factored in $\CC[x,y]$ into $9$ irreducible polynomials: $3$ defined over $\RR$ and $3$ complex conjugate pairs. The three real factors were -$$-16 + 32 x - 16 x^2 + x^4 - 16 y^2 + 2 x^2 y^2 + y^4=0$$ -$$1 - 4 x - 10 x^2 - 4 x^3 + x^4 - 14 y^2 - 4 x y^2 + 2 x^2 y^2 + y^4=0$$ -$$-1 + 16 x^2 - 32 x^3 + 16 x^4 + 16 y^2 - 32 x y^2 + 32 x^2 y^2 + 16 y^4=0.$$ -Here they are plotted, with black dots at $0$ and $1$: - -The third one is the lemniscate. I am guessing the others correspond to $(A,S) = \left( \left[ \begin{smallmatrix} 3&4 \\ 2&3 \end{smallmatrix} \right],\ \left[ \begin{smallmatrix} 0&2\\1&0\end{smallmatrix} \right] \right)$ and $(A,S) = \left( \left[ \begin{smallmatrix} 3&2 \\ 4&3 \end{smallmatrix} \right],\ \left[ \begin{smallmatrix} 0&1\\2&0\end{smallmatrix} \right] \right)$. - -I have some ideas about when two matrices $S_1$ and $S_2$ would lie on same factor of $\Psi_D$, but they aren't finished. However, let me point out an example which I expect to be interesting. Consider -$$ A_1 = \begin{bmatrix} 11&60\\2&11\\ \end{bmatrix} \qquad A_2 = \begin{bmatrix} 11&-4\\-30&11\\ \end{bmatrix}$$ -giving -$$S_1 = \begin{bmatrix} 0&30 \\ 1&0 \end{bmatrix} \qquad S_2 =\begin{bmatrix} 0&-2 \\ -15 & 0 \\ \end{bmatrix}. $$ -Using the class group of $\ZZ[\sqrt{30}]$, I have checked that $A_1$ and $A_2$ are not conjugate, so these should be distinct geodesics. However, $S_1$ and $S_2$ are $2$-adically very close. I predict that both of these geodesics occur as components of the same irreducible factor of $\Psi_{30}(x+iy, x-iy)$. -Another similar choice, with larger trace but smaller $D$, would be $(A_1, S_1) = \left( \left[ \begin{smallmatrix} 19 & 60 \\ 6 & 19 \\ \end{smallmatrix} \right],\ \left[ \begin{smallmatrix} 0 & 10 \\ 1 & 0 \\ \end{smallmatrix} \right] \right)$ and $(A_2, S_2) = \left( \left[ \begin{smallmatrix} 19 & 12 \\ 30 & 19 \\ \end{smallmatrix} \right],\ \left[ \begin{smallmatrix} 0 & 2 \\ 5 & 0 \\ \end{smallmatrix} \right] \right)$.<|endoftext|> -TITLE: Are there mistakes in Kovalev's "Twisted connected sums and special Riemannian holonomy"? -QUESTION [8 upvotes]: This is kind of a strange and vague question... sorry about that. -I am really interested in $G_2$ Twisted Connected sums as described in this paper: https://arxiv.org/abs/math/0012189 "Twisted connected sums and special Riemannian holonomy" by Alexei Kovalev. I would like to use those constructions to come up with examples to test new ideas. -Some people have told me that this paper is "not taken seriously" because there are mistakes in it. These people were not sufficiently familiar with the paper to tell me what the mistakes were and whether or not they had been corrected. -I have found maybe one or two mistakes but they are very small. More like typos. -Does anyone here know what the "mistakes" are (if there are any) or what the story is behind these rumors? - -REPLY [12 votes]: The error in Kovalev's paper is described in arXiv:1206.227 (see the discussion following theorem 2.6). An alternative proof is in arXiv:1212.6929. - -Building on the previous work of Tian–Yau, Kovalev claimed to prove -the existence of exponentially asymptotically cylindrical Calabi–Yau -manifolds, improving substantially the asymptotics previously -established by Tian–Yau. Unfortunately Kovalev’s proof of the improved -asymptotics contains an error (it crucially relies on a Euclidean type -Sobolev inequality that fails for any volume growth rate less than the -maximal one). Other errors in Kovalev occur in the construction of -hyper-Kähler rotations (especially Lemma 6.47 which is used in the -proof of the main Theorem 6.44).<|endoftext|> -TITLE: Union closed conjecture induction -QUESTION [10 upvotes]: I would like an example showing that one of the most basic induction approaches to the union-closed conjecture fails. If, for any union-closed family $\mathcal{A}$ of subsets of a finite set $X$, there is some $x \in X$ such that each $y \in X$ has $|\{A \in \mathcal{A} : A \ni y \text{ and } A \ni x\}| \ge \frac{1}{2}|\{A \in \mathcal{A} : A \ni x\}|$, then we can merely use induction applied to the union-closed family $\{A \in \mathcal{A} : A \not \ni x\}$ to get some $y \in X$ in at least half of the sets of $\{A \in \mathcal{A} : A \not \ni x\}$, and by our choice of $x$, we then see that $y$ is in at least half the sets of $\mathcal{A}$. -I have to think that there is a known example showing this approach doesn't work, i.e., there is an $\mathcal{A}$ with no such $x$. But I couldn't think of an example. So,: - -Give an example of a finite set $X$ and a union-closed family $\mathcal{A} \subseteq \mathcal{P}(X)$ such that, for each $x \in X$, there is some $y \in X$ with $$|\{A \in \mathcal{A} : A \ni y \text{ and } A \ni x\}| < \frac{1}{2}|\{A \in \mathcal{A} : A \ni x\}|.$$ (Or prove the union-closed conjecture!) - -I avoid degenerate cases, like $X = \emptyset$, $\mathcal{A} = \emptyset$, or $\mathcal{A} = \{\emptyset\}$. - -REPLY [10 votes]: Let $X = 123456$, and $\mathcal{A} = \{134, 1345, 1346, 13456, 256, 2356, 2456, 23456, 123456\}$. Let $f_x$ and $f_{x, y}$ be the number of sets in $\mathcal{A}$ containing $x$, or both $x$ and $y$ respectively. Say that $y$ is rare for $x$ if $f_{x, y} < f_x / 2$. Then: - -$f_1 = f_2 = 5$, but $f_{1, 2} = 1$, thus $1$ and $2$ are rare for each other; - -$f_3 = f_4 = 7$, but $f_{3, 2} = f_{4, 2} = 3$, thus $2$ is rare for $3$ and $4$; - -similarly, $1$ is rare for $5$ and $6$.<|endoftext|> -TITLE: Do mathematicians use notebooks to keep their results these days? -QUESTION [7 upvotes]: Mathematicians work a lot and are usually inspired by many things. In their lifetimes they get to publish only portions of their results. There have been stories of how Gauss, Euler, Ramanujan, Einstein, Hilbert etc kept notebooks and in it we can find glimpses to their thought processes and glimpses to results they did not publish and nevertheless would have turned to be important if in fact they had done. These are known classical examples. - -Are there notable example of recent mathematicians who kept such notebooks? - -I find lot of mathematicians have blogs but these are on published results. Do mathematicians still have such notebooks and if so when do they transfer their results usually to their notebooks and when do they transfer such results to publication (in other words I am asking for organizing principles for organizing thought processes and progress)? - -REPLY [5 votes]: Yes, mathematicians keep notebooks. Sometimes, after their death, notebooks are published. -Here is an example: http://www.claymath.org/publications/quillen-notebooks - -Here is another example, though not so recent: https://www.math.uu.se/collaboration/beurling/unpublished-manuscripts/ -I believe many mathematicians have a lot of unpublished stuff. Sometimes their friends and students publish this after their death, exactly as it happened in the past. - -Mathematicians discuss on their blogs and web pages not only published results. There are many examples. Much material exists in the form of correspondence, lecture notes, and preprints which are not officially made public (not available to everyone, like blogs) and which mathematicians share among their colleagues.<|endoftext|> -TITLE: "Simple" condition that would prove a function transcendental -QUESTION [5 upvotes]: I've already asked the question on MSE but there are still no answers, so I'm going to ask it here. -I conjectured that for every algebraic function $f(x)$ that is differentiable on $\mathbb{R}$, its $\lim_{x\to\infty}$ is either $\infty$, $-\infty$, or a finite value, so: - -If $f(x)$ is differentiable everywhere on $\mathbb{R}$ and its $\lim_{x\to\infty}$ is not $\infty$, $-\infty$, nor a finite value, then $f(x)$ is transcendental. - -If this is true, how could it be proved? - -REPLY [4 votes]: $\newcommand\R{\mathbb R}$Your conjecture is true. Indeed, suppose that a function $f\colon\R\to\R$ is continuous and algebraic, so that -$$\sum_{j\in[n]_0}p_j(x)f(x)^j=0\tag{1}$$ -for all $x\in\R$, where $[n]_0:=\{0,\dots,n\}$, $n$ is a natural number, and, for each $j\in[n]_0$, $p_j$ is a polynomial function of some degree $m_j$, so that for some real $a_j\ne0$ -$$p_j(x)=(a_j+o(1))x^{m_j}\tag{2}$$ -as $x\to\infty$. -Suppose now that for some real $c$ and some sequence $(x_m)$ in $\R$ converging to $\infty$ we have $f(x_m)\to c$. Then, by (1) and (2), -$$p(c):=\sum_{j\in J}a_j c^j=0,\tag{3}$$ -where $J$ is the (nonempty) set of all $j\in[n]_0$ such that for all $i\in[n]_0$ we have $m_j\ge m_i$. So, $c$ must be in the finite set of the roots of the polynomial $p$. -On the other hand, if your conjecture were false, then, by the intermediate value theorem, there would be infinitely (even uncountably) many real $c$ such that for some sequence $(x_m)$ in $\R$ converging to $\infty$ we have $f(x_m)\to c$. -So, your conjecture is true.<|endoftext|> -TITLE: Reference for permanent integral identity -QUESTION [7 upvotes]: $\DeclareMathOperator\perm{perm}\DeclareMathOperator\diag{diag}$Using MacMahon's master theorem, the properties of complex gaussian integrals, and Cauchy's integral theorem one can show that the permanent of a matrix $A$ satisfies -\begin{equation} -\perm(A) = \frac{1}{\pi^N} \int_{\mathbb{R}^{2N}} d^N\mathbf{u} \, \exp\Bigl( -\sum_{i=1}^N \mathbf{u}_i^2 \Bigr) \prod_{i=1}^{N} \Bigl(\mathbf{u}_i\cdot\sum_{j=1}^NA_{ij} \mathbf{u}_j\Bigr), -\end{equation} -where $\mathbf{u}_i \equiv (u_{i1}, u_{i2})$ and $d^{N}\mathbf{u} \equiv \prod^{N}_{i =1}du_{i1} \wedge du_{i2}$. -My guess is that this result is not new, but I'm not familiar with the literature on such identities. Is this identity known? Alternatively, where would be a good place to check for similar such identities? - -Proof of Identity -The truth of the identity could be inferred from noting that we need our gaussian to integrate a function where $\mathbf{u}_i$ for each $i$ appears exactly twice. -To prove it directly, we note that MacMahon's master theorem states that the permanent of a matrix $A$ is the $x_1x_2\dotsm x_N$ coefficient of the quantity -\begin{equation} -\frac{1}{\det(I - XA)}, -\end{equation} -where $I$ is the $N\times N$ identity matrix and $X \equiv \diag(x_1, x_2, \dotsc, x_N)$. Thus, we have -\begin{equation} -\perm(A) = \frac{1}{(2\pi i)^N} \oint \left[\prod_{i=1}^{N} \frac{dq_i}{q_i^2} \right] \frac{1}{\det(I- QA)}, -\label{eq:perm_def} -\end{equation} -where $Q = \diag(q_1, q_2, \ldots, q_N)$, $q_i$ is a complex variable, and we are performing $N$ contour integrations in sequence. Defining a set of $N$ two-dimensional vectors $\{\mathbf{u}_i\}$ as $\textbf{u}_i \equiv (u_{i1}, u_{i2})$, we have -\begin{equation} -\frac{1}{\det(I-QA)} = \frac{1}{\pi^N} \int_{\mathbb{R}^{2N}} d^{N}\mathbf{u}\,\exp\Bigl(- \sum_{i, j=1}^N \mathbf{u}_i \cdot \left(\delta_{ij} - q_{i} A_{ij}\right) \mathbf{u}_{j}\Bigr). -\label{eq:gauss_result} -\end{equation} -Using this determinant expression in MacMahon's theorem and performing the contour integrations yields the stated result. - -REPLY [3 votes]: In case anyone from the future comes to this post: -I wasn't able to find relevant literature that answered my question so I wrote a preprint on it (https://arxiv.org/abs/2106.11861). It turns out that the gaussian above is more complicated than necessary. The main result of the paper is a generalization of the identity above: - -Let $p_X: \Omega_X \to \mathbb{R}$ be a probability distribution defined over the domain $\Omega_X$ with zero mean and unit variance. Let $A$ be an $n\times n$ matrix with elements $a_{i, j}$. Then the permanent of $A$ is -\begin{equation} -\text{perm}(A) = \int_{\Omega^n_{X}} d^n\textbf{x}\, \prod_{i=1}^n p_{X}(x_i) \, x_{i} \sum_{j=1}^n a_{i, j} x_j, -\label{eq:fund_thm} -\end{equation} -where $\Omega^n_X = \Omega_X \otimes \cdots \otimes \Omega_X$ is the $n$-factor product over the single-variable domain of integration. In condensed notation, we can write this result as the expectation value -\begin{equation} -\text{perm}(A) = \left\langle \prod_{i=1}^nx_{i} \sum_{j=1}^n a_{i, j} x_j \right\rangle_{x_i \sim p_X}, -\end{equation} -where the average is over $\{x_i\}$, a set of independent identically distributed random variables each of which is drawn from $p_X$. - -When $p_X = N(0, 1)$, we get something like the equation in the question.<|endoftext|> -TITLE: Is there a strong evidence, that the Zeeman conjecture is false? -QUESTION [6 upvotes]: In a lot of articles I have read, that the Zeeman Conjecture implies the Andrews-Curtis-Conjecture and if we disprove the AC-Conjecture then Zeeman is obviously also false (as the sources state). But what makes mathematicians think, that the Zeeman Conjecture could possibly be false? -Are there any (functioning) links to articles arguing about this question? - -REPLY [5 votes]: Here is one counter argument by Sergei Matveev:<|endoftext|> -TITLE: Does ODE uniqueness unconditionally implies the flow continuity? -QUESTION [9 upvotes]: Suppose we have a (say compactly supported) $C^0$-vector field $X:\mathbb R^n\to\mathbb R^n$ such that for every $x\in\mathbb R^n$ there is a unique $C^1$-curve $\gamma:\mathbb R\to\mathbb R^n$ solving $\dot\gamma_x(t)=X(\gamma_x(t))$ with $\gamma_x(0)=x$. -Then the ode flow $\mathcal F_X$ is pointwisely defined in the way that $\mathcal F_X(t,x)=\gamma_x(t)$. -My question is: Must $\mathcal F_X$ be a continuous map with respect to $x$? -Certainly there is no problem if $X$ is Lipschitz or just satisfies the Osgood condition. Since we have the regularity estimate of ODE flows with respect to their modulus of continuity. -When $X$ is H"older and somehow its ODE is uniquely solvable at every point, could their be continuous dependence? And if not how does the blow up occur. - -REPLY [5 votes]: Yes, the uniqueness of solutions implies continuous dependence on initial conditions and parameters. See Theorem 3.2 in Hartman's "Ordinary differential equations".<|endoftext|> -TITLE: Is there a stable structure on $[0,1]$ that approximates every continuous function? -QUESTION [8 upvotes]: The $n$-dimensional form of the Weierstrass approximation theorem is the statement that polynomial functions are dense under the $\ell_\infty$-norm in the space of continuous functions on $[0,1]^n$ for any $n<\omega$. -A trivial restatement of this fact is this: If we let $M= ([0,1],\dots)$ be the induced structure on the definable set $[0,1]$ as a subset of $\mathbb{R}$ as an ordered field, then for any continuous function $f:[0,1]^n\to [0,1]$ and any $\varepsilon > 0$, there is a definable function $g:M^n \to M$ such that $\left\lVert f-g\right\rVert < \varepsilon$. (Incidentally we don't actually need multiplication for this. The ordered group structure is enough.) -Since RCF is NIP, the induced structure on $[0,1]$ is NIP as well (and in fact o-minimal). I'm curious if this approximating property can be accomplished in a stable theory. I can think of more variations of this question than I should I put in an MO question, but I think the following two are reasonable to consider first. - -Question 1: Does there exist a structure $M$ whose underlying set is $[0,1]$, whose theory is stable, and which has the property that for any continuous function $f:[0,1]^n \to [0,1]$ and any $\varepsilon > 0$, there is a definable function $g : M^n \to M$ such that $\left\lVert f -g \right\rVert_\infty < \varepsilon$? - - -Question 2: Assuming the first question has a positive answer, is there such a structure in which the witnessing $g$'s are continuous? - -Note that the question doesn't depend on whether we interpret 'definable' as $\varnothing$-definable or definable with parameters. - -REPLY [8 votes]: The answer is positive if you don’t require $g$ to be continuous. Indeed, continuous functions $[0,1]^n\to[0,1]$ can be approximated by piecewise constant functions whose pieces are boxes with rational endpoints. Any such function is definable in the structure -$$M=([0,1],\{I_q:0 -TITLE: Does every open orientable even-dimensional smooth manifold admit an almost complex structure? -QUESTION [9 upvotes]: Does every open orientable even-dimensional smooth manifold admit an almost complex structure? - -REPLY [24 votes]: If $M$ admits an almost complex structure, then the odd Stiefel-Whitney classes vanish and the even Stiefel-Whitney classes admit integral lifts, namely $c_i(M) \equiv w_{2i}(M) \bmod 2$. These two conditions give restrictions on the smooth manifolds which can admit almost complex structures. -The first restriction, namely that $w_1(M) = 0$, is equivalent to orientability. If $M$ is orientable, then the second restriction, namely that $w_2(M)$ admits an integral lift, is equivalent to the manifold being spin$^c$. -An example of an orientable non-spin$^c$ manifold is the Wu manifold $SU(3)/SO(3)$ which has dimension five. Therefore $M = (SU(3)/SO(3))\times\mathbb{R}^{2k+1}$ is an open orientable even-dimensional manifold which does not admit an almost complex structure. -Note that $\dim M = 2k + 6$, so this gives examples in all positive even dimensions other than two and four. It turns out that in dimensions two and four, there are no examples. - -In dimension two, a manifold is almost complex if and only if it is orientable. -In dimension four, an open manifold admits an almost complex structure if and only if it is spin$^c$, and every orientable four-manifold is spin$^c$, see this note by Teichner and Vogt.<|endoftext|> -TITLE: When is a diffeomorphism a bundle map? -QUESTION [5 upvotes]: Let $F\rightarrow E_0 \rightarrow B$ and $F\rightarrow E_1\rightarrow B$ be two smooth fiber bundles. Suppose $E_0$ and $E_1$ are diffeomorphic. -What are the obstructions for $E_0$ and $E_1$ to be diffeomorphic via a bundle map? -To be more specific, I am interestid in the following cases: - -Vector bundles ($F=\mathbb{R}^n$) -Vector bundles over spheres ($B=S^n$) -Sphere bundles over spheres ($F=S^m$, $B=S^n$) - -REPLY [3 votes]: I actually found a partial answer to the case of vector bundles over spheres (that is, under certain dimensional restrictions): https://www.sciencedirect.com/science/article/pii/S0040938399000397<|endoftext|> -TITLE: Normal invariants -QUESTION [7 upvotes]: I am having a hard time finding examples of computations of normal invariants of surgery theory (or more generally the set of homotopy classes of maps $[X,G/O]$). Does anybody have good references? - -REPLY [4 votes]: If you want specifically low-dimensional calculations, the Kirby-Taylor article A survey of 4-manifolds through the eyes of surgery does this for 4-manifolds. The discussion highlights the difference between the topological and smooth (= PL in this dimension) cases.<|endoftext|> -TITLE: Discrete logarithm for polynomials -QUESTION [10 upvotes]: Let $p$ be a fixed small prime (I'm particularly interested in $p = 2$), and let $Q, R \in \mathbb{F}_p[X]$ be polynomials. -Consider the problem of determining the set of $n \in \mathbb{N}$ such that $X^n \equiv R$ in the ring $\mathbb{F}_p[X] / Q$. It is easy to see that this is either the empty set $\emptyset$, or a singleton set $\{ a \}$, or an arithmetic progression $\{ a + cm : m \in \mathbb{N} \}$. -In the special case where $Q$ is a primitive polynomial, $\mathbb{F}_p[X] / Q$ is a finite field and this is just the ordinary discrete logarithm problem, and there's an efficient quasi-polynomial-time algorithm for solving this: https://eprint.iacr.org/2013/400.pdf -What about when $Q$ is not a primitive polynomial? Can this more general problem be reduced to solving instances of the ordinary discrete logarithm, and therefore also be solved in quasi-polynomial time? -By the structure theorem for finitely-generated modules over a PID, we can factor $\mathbb{F}_p[X]/Q$ as the direct sum of rings of the form $\mathbb{F}_p[X]/S^k$, where $S$ is an irreducible polynomial. If we could solve the discrete logarithm problem in each of these direct summands, then the Chinese Remainder Theorem would determine the solution in the original ring. Hence, it suffices to consider the case where $Q$ is the power of an irreducible polynomial. But I can't see how to proceed any further, because 'power of an irreducible polynomial' is not the same thing as 'primitive polynomial'. -Motivation: when $p = 2$, this is equivalent to efficiently searching the output of a linear feedback shift register PRNG to determine where (if at all) a specific bit-string arises. - -REPLY [3 votes]: Since you bring up binary linear feedback shift registers at the end, I feel that it may be of interest to describe a relevant simple special case I once worked out when a slightly perpelexed colleague (a telcomm engineer) asked me about it. Just to get the ball rolling. If you have seen this much, then I apologize for wasting your time. -Alas, this answer does not cover an awful lot of ground. -This is about the very special case $p=2$, $Q=S^k$, $k=2^\ell$. In other words, the multiplicity of the irreducible polynomial $S$ as a factor is a power of $p$. -Let $L$ be the order of $X$ in the ring $\Bbb{F}_2[X]/S$. Then $L$ is also the period of the LFSR-sequence gotten with feedback polynomial $S$. The starting point is the simple observation that as -$$ -X^L\equiv1\pmod S, -$$ -freshman's dream then implies that $X^{2L}+1=(X^L+1)^2$ is divisible by $S^2$. Repeating the dose, we see that $X^{2^jL}+1$ is divisible by $S^k$ for all $k\le 2^j$. In other words, if $j$ is the smallest natural number such that $2^j\ge k$ it easily follows that the smallest period of the LFSR sequence generated by $S^k$ is $2^jL$. We know that the multiplicity of zeros of $X^{2^jL}+1$ in $\overline{\Bbb{F}_2}$ is exactly $2^j$, so a proper factor won't do. - -This means that among the remainders of $X^i$ modulo $S^k$, $i\in\Bbb{N}$, each coset modulo $S$ appears only $2^j$ times. The possibilities are thus severely limited. - -Next I restrict myself to the case $k=2^j$. Let's take a look at a small case $k=4$, -$S=X^2+X+1$, $L=3$, to see what's going on. We have $S^4=X^8+X^4+1$, so the cyclic subgroup generated by $X$ modulo $S^4$ contains the following $12$ cosets. -$$\{1,X,X^2,X^3,X^4,X^5,X^6,X^7,X^4+1,X^5+X,X^6+X^2,X^7+X^3\}.$$ -You see that all the elements only contain terms of degrees in a single residue class modulo $4$. As $S(X)^4=S(X^4)$ a moments reflection shows that this must always be the case. -In terms of a linear feedback shift register this means that we really should view the register of 8 bits consisting of four parts (or blocks) of bits at positions $B_i:=\{i,i+4\}, i=0,1,2,3$. A single clock tick moves the contents of $B_i, i=0,1,2,$ to the block $B_{i+1}$, but the contents of the block $B_3$ become the new content of the block $B_0$ only after being exposed to the LFSR with feedback polynomial $S$. -Viewed differently, if we run the clock four ticks, multiplying the coset by $X^4$, then each block is mapped back to its own position, but all four of them took turns getting exposed to the feedback polynomial $S$. -I'm sure this point of view let's you solve the discrete logarithm problem modulo $S^{2^j}$ if you can solve it modulo $S$. I'm afraid I don't remember what can be said about the cases when the exponent is not a power of $p$. I will add more, if I can think of something. -Notice that similar partitioning of the shift register may take place with certain other feedback polynomials also. For example the ninth cyclotomic polynomial $\Phi_9(X)=X^6+X^3+1$ remains irreducible modulo $2$. It is highly non-primitive as $L=9$ instead of the primitive case $2^6-1=63$. Anyway, the shift register of six bits is partitioned into three parts of the form $B_i=\{i,i+3\}, i=0,1,2,$ with similar behavior.<|endoftext|> -TITLE: Why is the regulator of a number field normalized the way it is? -QUESTION [7 upvotes]: The regulator of a number field is essentially the covolume of the unit group embedded into the vector space $\{(x_1, \ldots, x_{r+s}): \sum_i x_i=0\}$ under the log embedding: $$x \mapsto (\log |\sigma_1(x)|, \ldots \log |\sigma_r(x)|, \log |\sigma_{r+1}(x)|^2, \ldots , \log |\sigma_{r+s}(x)|^2).$$ -But you need to take some care in terms of what the right Euclidean structure is for defining volumes in $\{(x_1, \ldots, x_n): \sum_i x_i = 0\}$. What you're supposed to do is to use the Euclidean measure induced by any coordinate projection $\{(x_1, \ldots, x_{r+s}): \sum_i x_i = 0\} \rightarrow \mathbb{R}^{r+s-1}$. Equivalently, you could use the subspace Euclidean measure, but then normalize by dividing by $\sqrt{r+s}$. -My question is why is this the best normalization? -So far the only answer I have is that for the quadratic real case this makes the regulator exactly the log of the fundamental unit, which seems a very sensible convention. I guess it also makes the analytic class number formula slightly cleaner, but it's not obvious to me why this normalization exactly comes into the class number calculation. -My apologies if this question is too elementary. - -REPLY [4 votes]: Let $V$ be a vector space and $W$ a subspace. -Given a volume on $V$ and a volume on $V/W$, we can define a volume on $W$. (Here "volume" = "translation-invariant measure" but no real measure theory is being used. We could also define a volume as an element of the top wedge power of the dual vector space.) -Geometrically, this says that given a basis $v_1,\dotsc, v_m$ of $W$ which extends to a basis $v_1,\dotsc, v_n$ of $V$, the volume of the parallelepiped spanned by $v_1,\dotsc, v_m$ in $W$ is equal to the volume of the parallelepiped spanned by $v_1,\dotsc, v_n$ in $V$ times the volume of the parallelepiped spanned by $v_{m+1},\dotsc,v_n$ in $V/W$. (The same thing works with any shape in $V$ whose fibers under the projection map to $V/W$ are all translates of a fixed shape in $W$.) -We can express this algebraically with top forms as well, as François Brunault explains in the comments. -For $V$ the vector space of tuples $(x_1,\dotsc, x_{r+s})$, and $W$ the subspace $\sum_i x_i=0$, it is natural to identify $V/W$ with $\mathbb R$ under the map $\sum_i x_i$ and take the standard volume on $\mathbb R$. When $x_i$ are the logs of the individual coordinates, this identification corresponds to the norm map. -The reason this is the most commonly used normalization in number theory problem has to do with the fact that the norm map is used often in number theory. In particular, the norm is used in defining the zeta function, explaining why this approach gives nice formulas for the residue of the zeta function.<|endoftext|> -TITLE: How to use the Prime Number Theorem in order to prove Selberg's Formula? -QUESTION [7 upvotes]: I`m reading Melvin B. Nathanson's "Elementary Methods in Number Theory" -and I can't think of a way of deducing Selberg's formula (9.3) from the prime number theorem. -This is one of the tasks left for the reader at the end of chapter 9.3 (The Elementary Proof). -$$(9.3) \qquad \qquad \vartheta(x) \log x + \sum_{p \leq x}(\log p) \vartheta(x/p) = 2x\log x +O(x)$$ -Simply replacing $\vartheta(x)$ by $x + o(x)$ results in error terms of order $o(x\log x)$ which are -strictly worse than error terms of order $O(x)$. -I feel like I'm missing something rather simple, but I've been wasting just a little too much time on -that exercise. -Thanks in advance! - -REPLY [6 votes]: The $O(x)$ term can be refined by completely elementary means that are not at all complicated. Filip Saidak (On the prime number lemma of Selberg, Math. Scand. 103 (2008), no. 1, 5–10) proved a much clearer connection between Selberg's result and the prime number theorem. He stated this for $\psi(x) = \sum_{n\leq x}\Lambda(n)$, but this holds with trivial modifications for $\vartheta(x)=\sum_{p\leq x}\log p$. -If $\vartheta(x):=\sum_{n\leq x}\Lambda(n) = x + E(x)$, then -$\vartheta(x)\log x + \sum_{p\leq x} (\log p)\vartheta(x/p)=2x\log x -(2\gamma+1)x+O(E(x)(\log x)^2)$, -where $\gamma$ is the Euler-Mascheroni constant. So if all that you know is that $E(x) = o(x)$, then you cannot recover Selberg's result. Selberg's breakthrough that $-(2\gamma+1)x+O(E(x)(\log x)^2) = O(x)$ appears to be equivalent to the stronger estimate $E(x) \ll x/(\log x)^2$. - -REPLY [2 votes]: Edit: Answer to GH from MO by author. -Edit2: Changed a sign. -Let -$$l(n)=\left\{\begin{array}{11} \log p, & if\ n = p\ prime \\ 0, & else. \end{array}\right.$$ -Then -$$\sum_{pq\leq x}(\log p)(\log q)=\sum_{n\leq x}(l\ast l)(n)$$ -So far, so good. However, using Dirichlet's hyperbola identity, I only obtain -$$\sum_{n\leq x}(l\ast l)(n)=2\sum_{p\leq x^\frac{1}{2}}(\log p)\vartheta(\frac{x}{p})-\vartheta(x^\frac{1}{2})^2 $$, where obviously $\vartheta(x^\frac{1}{2})^2=O(x)$. -Thus I am back at my original problem, as I would have to show -$$\sum_{p\leq x^\frac{1}{2}}(\log p)\vartheta(\frac{x}{p})=\frac{1}{2}x\log x+O(x)$$ -Where do I fail to see the shortpath? Thanks for answering anyways, appreciate it!<|endoftext|> -TITLE: The group of isometries of a manifold is a Lie group, isn't it? -QUESTION [18 upvotes]: Let $M$ be a connected finite dimensional topological manifold and $g$ be any metric on it that induces the topology of $M$ ($g$ is not a Riemannian metric). How to prove that the group of isometries of $(M,g)$ is a finite dimensional Lie group? -PS. Note that in case $M$ is a Riemannian manifold, this statement is classical. - -REPLY [15 votes]: The assertion (for connected topological manifolds) is equivalent to the Hilbert–Smith conjecture. The latter (in one of its equivalent versions) asserts that every continuous action of every profinite group on every topological metrizable manifold has an open kernel. -Indeed, let $G$ be a profinite group acting continuously on a connected topological metrizable manifold $X$. Let $d$ be a proper distance on $X$ (inducing the topology). Then $d'(x,y)=\int_G d(gx,gy)$, integrating with respect to the Haar measure, defines another proper distance, continuous on $X^2$, and which in addition is $G$-invariant. Hence it defines the topology and is complete. If OP's assertion holds, then the isometry group being a Lie group, the map has an open kernel. -The Hilbert–Smith conjecture is known in dimension $\le 3$ ($\le 2$ is classical, and $3$ is more recent work of John Pardon). -The assertion is also known to hold when the isometry group acts transitively. - -Added remarks on the Hilbert–Smith conjecture: -If "green" qualifies topological groups, say that a topological group $G$ has "no small green subgroups" if there is a neighborhood of the identity $V$ in $G$ such that the only green subgroup of $G$ contained in $V$ is $\{1\}$. -For instance, the solution to Hilbert's 5th problem yields that a locally compact group has no small subgroups iff it is isomorphic (as topological group) to some Lie group. -One reformulation of the Hilbert–Smith conjecture is: - -For every connected topological (metrizable) manifold $M$, the topological group $\mathrm{Homeo}(M)$ (with the compact-open topology) has no small compact subgroups. - -Newman's theorems (see T. Tao's blog) implies that it has no small finite subgroups. The too optimistic hope would be that $\mathrm{Homeo}(M)$ has no small subgroups, but this is false: as soon as $M$ has positive dimension, it is easy to check that $\mathrm{Homeo}(M)$ has small discrete infinite cyclic subgroups.<|endoftext|> -TITLE: Averaging weakly almost periodic Schur multipliers -QUESTION [6 upvotes]: Let $\Gamma$ be a countable (discrete) group and let $\varphi:\Gamma\times\Gamma\to\mathbb{C}$ be a (non-equivariant) Schur multiplier. See Chapter 5 of [2] for details. Assume that, for all $t\in\Gamma$, the function -\begin{align*} -s\longmapsto\varphi(st,s) -\end{align*} -is weakly almost periodic. Let $m$ be the unique invariant mean on WAP$(\Gamma)$; see Section 3 of [1]. -Question: Is it true that the function $\psi:\Gamma\to\mathbb{C}$ given by -\begin{align*} -\psi(t)=m(s\mapsto\varphi(st,s)) -\end{align*} -is a Herz-Schur multiplier on $\Gamma$? Can we estimate the norm of $\psi$ in terms of the norm of $\varphi$? -First naive attempt: By Theorem 5.1 in [2], there is a Hilbert space $H$ and bounded functions $\xi, \eta:\Gamma\to H$ such that -\begin{align*} -\varphi(t,s)=\langle\xi(s),\eta(t)\rangle,\quad\forall s,t\in\Gamma. -\end{align*} -This allows us to write -\begin{align*} -\psi(s^{-1}t)=m\left(r\mapsto\langle\xi(rs),\eta(rt)\rangle\right),\quad\forall s,t\in\Gamma, -\end{align*} -but I don't know if this can be expressed as a scalar product on a suitable Hilbert space. - -[1] Uffe Haagerup, Søren Knudby, and Tim de Laat. A complete characterization of connected Lie groups with the approximation property. Ann. Sci. Éc. Norm. Supér. (4), 49(4):927-946, 2016. -[2] Gilles Pisier. Similarity problems and completely bounded maps, volume 1618 of Lecture Notes in Mathematics. Springer-Verlag, Berlin, expanded edition, 2001. Includes the solution to “The Halmos problem”. - -REPLY [6 votes]: Ignacio, the answer is yes. Indeed, there is a net $m_i$ of probability measures on $G$ such that $m(f) = \lim_i \int f dm_i$ for every $f \in \textrm{WAP}(\Gamma)$. One justification of this is as follows: extend $m$ to a (perhaps non-invariant) mean on $\ell_\infty(\Gamma)$, and use the standard weak-* density of $\ell_1(\Gamma)$ in its bidual. See page 1 of Greenleaf's Invariant Means on Topological Groups and their Applications. -The answer to your question follows directly: we have $\psi(s^{-1}t) = \lim_i \int \varphi(rs,rt) dm_i(r)$, so $(s,t) \mapsto \psi(s^{-1}t)$ is a pointwise limit of a net of Schur multipliers with norm at most the norm of $\varphi$, so is a Schur multiplier with norm at most the norm of $\varphi$.<|endoftext|> -TITLE: When does a topos satisfy the axiom of regularity? -QUESTION [10 upvotes]: In categorical set theory, we observe that certain topoi satisfy (suitable versions of) certain axioms from set theory. For example, Lawvere's $\mathsf{ETCS}$ asserts that $\mathbf{Set}$ is a well-pointed topos with a natural numbers object, satisfying the (internal) axiom of choice. $\mathsf{ETCS}$ is known to be equivalent to $\mathsf{BZC}$, a fragment of $\mathsf{ZFC}$ which doesn't include regularity. -My question is: what does it take for a topos to satisfy (a suitably phrased version of) the axiom of regularity? Or perhaps some statement which is equivalent (in the presence of the other $\mathsf{ZFC}$ axioms), as I understand regularity is not intuitionistically acceptable. - -REPLY [21 votes]: The relationship between toposes and set theories was studied comprehensively in - -Steve Awodey, Carsten Butz, Alex Simpson, Thomas Streicher: Relating first-order set theories, toposes and categories of classes. -Annals of Pure and Applied Logic, Volume 165, Issue 2, February 2014, Pages 428-502 - -Regularity is discussed under the name "well-foundedness". You can find a lot of details in the paper, I am just going to quickly review the setup. -We work in an elementary topos $\mathcal{E}$. -The first step is to cook up a notion of $\in$-membership. -Define a membership graph to be a triple $G = (|G|, A_G, r_G)$ where $|G|$ an $A_G$ are objects and $r_G : |G| \to A_G + P|G|$ a morphism. We think of $|G|$ as a set of vertices with each vertex $x \in |G|$ being either an atom $a$ (in case $r(x) = \mathrm{inl}(a)$ for $a : A_G$) or a branching vertex with adjacency set $d \subseteq |G|$ (in case $r(x) = \mathrm{inr}(d)$ for $d : P|G|$). We may define a bisimilarity relation $\sim_{G,H}$ between two membership graphs expressing the fact that, up to reordering and repetition, $G$ and $H$ represent the same $\in$-membership structure. In other words, $\sim_{G,H}$ expresses extensional equality of sets represented by $G$ and $H$. In particular, $\sim_{G,G}$ is an equivalence relation which quotients the membership graph to give a membership relation that is extensional. -Next, we define a new topos $\mathcal{E}_\mathrm{nwf}$ whose objects are triples $(D, m, G)$ where $G$ is a membership graph in $\mathcal{E}$, $D$ an object of $\mathcal{E}$, and $m : D \to |G|$ a mono in $\mathcal{E}$. A suitable notion of morphism is devised that takes into account the bisimilarity relations $\sim_{G,H}$. We then have: -Theorem 11.7: $\mathcal{E}_\mathrm{nwf}$ is equivalent to $\mathcal{E}$. -We may understand the theorem as saying that we enriched the topos $\mathcal{E}$ with membership relations to get $\mathcal{E}_\mathrm{nwf}$, which changed the topos only up to equivalence (so not in any essential way from the point of view of topos theory). -So far we allow both atoms and non-well-founded membership relation. The next step is to define what it means for a membership graph $G = (|G|, A_G, r_G)$ to be well-founded. Here there are no surprises, as we can use the internal language of the topos to state when $X : P|G|$ satisfies the property "$X$ contains all the atoms $A_G$ and is hereditarily closed under the membership relation $r_G$" (see the formula after Corollary 11.2). -We define a third topos $\mathcal{E}_\mathrm{wf}$ as the full subcategory of $\mathcal{E}_\mathrm{nwf}$ of those objects whose membership graphs are well-founded. -Proposition 11.3: The equivalence between $\mathcal{E}$ and $\mathcal{E}_\mathrm{nwf}$ cuts down to an equivalence between $\mathcal{E}$ and $\mathcal{E}_\mathrm{wf}$. -The moral of the story is that, firstly, we may define a notion of extensional membership relation on objects in a topos, and secondly, that restricting to the well-founded part of the topos does not change the topos in a way that is relevant to topos theory. -In a sense the answer to the question "when does a topos satisfy the axiom of regularity" is "it does not matter". This is further discussed in the paper in Section 11.4, where the authors conclude with - -... it follows that any topos can be construed both as a model of BIZFA− and as a model of BINWFA−. - -Here BIZFA- is their formulation of ZF-like set theory with foundation (regularity) and BIZWFA- a formulation of ZF-like set theory with anti-foundation. -Disclaimer: please do not take the above summary as a satisfactory description. The paper contains many more details and explanations, and it should be consulted for thorough understanding of the topic.<|endoftext|> -TITLE: Definition of subcoalgebra over a commutative ring -QUESTION [9 upvotes]: Let $k$ be commutative ring and $(C, \Delta)$ be a coalgebra over $k$. Let $D$ be a $k$-submodule of $C$. -Notes I'm reading give the following definition: - -$D$ is called subcoalgebra of $C$ if the comultiplication $\Delta: C - \to C \otimes C$ restricts to a mapping $$\Delta\vert _D: D \to D - \otimes D$$ - -However, this seems a bit odd to me. Indeed, in general the inclusion $i: D \otimes D \to C \otimes C$ need not be injective so it might not be possible to view $D\otimes D \subseteq C \otimes C$. -What would be the appropriate definition in this context? I see several options: -(1) Assume that we work with flat modules (for example, vector spaces) so that the above makes sense. This seems too restrictive. -(2) Ask that $D$ itself is a coalgebra such that the inclusion $j: D \hookrightarrow C$ is a morphism of coalgebras. -Which definition (if any of these) is the one I should use? - -REPLY [4 votes]: You are right. -In the case of an $R$-submodule $D$ of an $R$-coalgebra $C$, the correct definition for $D$ being a subcoalgebra of $C$ is your definition (2) and not the one posted in your notes. This is standard in the contemporary literature: see for example p.11, sect. 2.7 of 1. -The definition mentioned in your notes is valid (as you have already mentioned in the OP) for vector spaces and special cases for $R$-coalgebras (when $D$ is flat or pure as an $R$-submodule for example). -The problem is indeed the one you mention: In general, the restriction of the comultiplication of $C$ to $D$, that is $\Delta\vert _D: D \to C \otimes C$, does not necessarily "lift" to a map $\Delta\vert _D: D \to D \otimes D$. This can happen, even in cases in which $\Delta(D)$ is contained inside the image of the canonical map $D\otimes_R D\to C\otimes_R C$. -A relevant example (attributed to Warren Nichols) is the following: Consider the $\mathbb{Z}$-module $C=\mathbb{Z}/8\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$ and denote $x=(1,0)$, and $z=(0,1)$. A comultiplication $\Delta:C\to C\otimes_\mathbb{Z} C$ is defined by $\Delta(x)=0$ and $\Delta(z)=4x\otimes x$. So $\Delta(z)$ has order $2$ in $C\otimes_\mathbb{Z} C$. Then, let $y=2x$ and take the $\mathbb{Z}$-submodule $D=\mathbb{Z}y+\mathbb{Z}z\subset C$. Now, $\Delta(z)=y\otimes y$ and $\Delta(D)$ is contained in the image of $D\otimes_\mathbb{Z} D\to C\otimes_\mathbb{Z} C$ but the restriction $\Delta|_{D}$ does not lift (or: does not factor) to a comultiplication $\Delta:D\to D\otimes_\mathbb{Z} D$, since we can see that any preimage of $y\otimes y$ in $D\otimes_\mathbb{Z} D$ has order $4$. -(This example was first suggested by Nichols in 2, p.56 and is also cited as an exercise in 1). -Edit: Another interesting phenomenon, related to the definition, of an $R$-subcoalgebra (definition (2) of the OP), is the fact that following this, the $R$-subcoalgebra $D$ of the $R$-coalgebra $C$ is not uniquely determined: In other words, the same $R$-submodule may correspond to non-isomorphic subcoalgebras. This has already been mentioned and sufficiently explained in Adrien's answer. I thought it might be interesting to include an example of this "phenomenon" (which is again suggested in 2 and cited as an exercise in 1): -Consider the $\mathbb{Z}$-module $C=\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}$ and make it a coalgebra by setting $g=(1,0)$ to be a grouplike and $x=(0,1)$ to be $g$-primitive. Then, take the submodule $D\subset C$ which is spanned by $g$ and $2x$. So, $D\cong\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}\cong\mathbb{Z}\oplus 2\mathbb{Z}/4\mathbb{Z}$. There are (at least) two different ways to make $D$ into a coalgebra: Either consider $g$ to be a grouplike and $2x$ to be $g$-primitive (this corresponds to the restriction of the original map of $C$ in $D$), or take again $g$ to be grouplike and define -$$ -\Delta(2x)=g\otimes(2x)+(2x)\otimes(2x)+(2x)\otimes g \ \ \ and \ \ \ \epsilon(2x)=0 -$$ -Both options make $D$ into a coalgebra and in both cases one can easily show that the inclusion $D\hookrightarrow C$ is a coalgebra morphism. So, $D$ becomes a $\mathbb{Z}$-subcoalgebra of $C$ in (at least) two different ways. Furthermore, these are not isomorphic to each other, since the second structure has an additional grouplike element; this is $g+(2x)$. -References: - -Corings and comodules, T. Brzezinski, R. Wisbauer, -Nichols, W.D., Sweedler, M., Hopf algebras and combinatorics, AMS, in: "Umbral Calculus and Hopf algebras", Contemp. Math. 6, 49–84 (1982)<|endoftext|> -TITLE: Does Pinelis' inequality (1994) exist? -QUESTION [18 upvotes]: I am reading a paper on stochastic optimization. And in this paper, the proofs are based on the Pinelis' 1994 inequality. I read the paper by Pinelis for more information and it is with great frustration that I was not able to find the inequality corresponding to that mentioned in the paper I am reading. -Here's the inequality in the article I'm reading: -$$ \begin{array}l -\text{(The Pinelis' 1994 inequality). Let } X_1,\dots, X_T \in \mathbb{R}^d \\ \text{be a random process satisfying } \mathbb{E}[X_t\mid X_1,\dots,X_{t−1} ] = 0 \text{ and} \\ \|X_t\| ≤ M. \text{ Then } \mathbb{P}[\| X_1 + \cdots + X_T\|^2 > 2 \log(2 /\delta)M^2T] \leq \delta. -\end{array} -$$ -Nowhere in Pinelis' 1994 paper is this inequality mentioned. -So I ask myself, does this inequality really exist? Do you know it? - -REPLY [78 votes]: As noted in Carlo Beenakker's comment, your inequality is a direct application of Theorem 3.5 in the linked paper: in that theorem, take $d_j=X_j$, $r=\sqrt{2M^2 T\ln(2/\delta)}$, $b_*^2=M^2T$, and $D=1$.<|endoftext|> -TITLE: 4-cliques of pythagorean triples graph and its connectivity -QUESTION [11 upvotes]: Let natural numbers $a, b > 2$ be adjacent if $|a^2 - b^2|$ is a square number. One can find a 3-clique. -For example 153, 185, 697. The questions are: does there exist a 4-clique? Is this graph connected? - -REPLY [9 votes]: Seems that here is a proof that the graph is connected -Let $C$ be a component. - -$C$ contains an even number, as an odd number $2k+1$ is adjacent to $2k(k+1)$. - -$C$ contains a number divisible by $6$: if $2k\in C$ with $3\nmid k$, then $k^2-1\in C$ as well. - -If $3k\in C$ and $p$ is a prime, then $3kp\in C$. Induction on $p$. The base cases $p=2,3$ follow from adjacencies $3k-5k-13k-12k-15k-17k-8k-10k-6k$ and $\dots-15k-9k$. For the step, use the hypothesis to see that $C$ contains $3k(p^2-1)$, then $6kp$, and then $3kp$ by the base case. - -The graph is connected. Indeed, any two components contain some numbers $3k$ and $3n$, and then both contain $3kn$.<|endoftext|> -TITLE: When quotient stacks (for nonsmooth group) are algebraic and related questions -QUESTION [5 upvotes]: Let $k$ be a field. Consider a group $k$-scheme $G$ and let $X$ be a $k$-scheme equipped with an action of $G$. Then one can define the quotient stack $[X/G]$. Objects of $[X/G]$ over $k$-scheme $T$ are pairs $(\pi, \alpha)$ such that $\pi:P\rightarrow T$ is a locally (with respect to fpqc topology) trivial $G$-bundle and $\alpha:P\rightarrow X$ is a $G$-equivariant morphism. -Now in Olson's book Algebraic spaces and stacks in Example 8.1.12 the author assumes that $G$ is smooth in order to derive that the canonical map $X\rightarrow [X/G]$ is smooth and in result to infer that $[X/G]$ is an algebraic stack. It seems that the other part of his argument, which shows that the diagonal $\Delta_{[X/G]}:[X/G]\rightarrow [X/G]\times_k[X/G]$ is representable holds for any group $k$-scheme. -Here are some immediate questions. - -Under what conditions on $G$ the stack $[X/G]$ is algebraic? Is smoothness essential? -If $G$ is affine over $k$, then is $\Delta_{[X/G]}$ representable by quasi-affine morphism of algebraic spaces? If not, then what one should impose on $X$ to know that this is the case? - -REPLY [6 votes]: About 1. : no, smoothness isn't essential. -"Flat is enough" : De Jong's slogan to express this result due to M.Artin. -https://www.math.columbia.edu/~dejong/wordpress/?p=1584 -I quote : -"Given a flat, finitely presented group scheme $G$ over $S$ acting on a scheme $X$ over $S$, then the quotient stack $[X/G]$ is algebraic."<|endoftext|> -TITLE: Condition for a matrix to be a perfect power of an integer matrix -QUESTION [26 upvotes]: I have a question that seems to be rather simple but for I got no clue so far. -Let's say I have a matrix $A$ of size $2\times 2$ and integer entries. I want to know if there is a kind of test or characterization that can tell me if there exists an integer matrix $B$ such that $B^k = A$. -So far the only thing I got was the obvious restriction on the determinant of $A$ (it has to be an integer that is a perfect $k$-th power), but I was wondering if maybe some other (stronger) restriction on $A$ helps. -In my particular case, I had $k=8$. It seems to be that it could be possible that some characterization could exist for arbitrary $k$ (a characterization depending on $k$, certainly). - -REPLY [10 votes]: Here is the necessary and sufficient condition, in terms of $\det A$ and $\text{tr}(A)$, in order that a $2\times2$ matrix $A$ be the $k$-th power of some matrix with integer coefficients. -Edit.Edit, 11.11.2020 The proof.The proof, is essentially routine; yet I post everything for convenience, before I forget all details. -It is convenient to introduce the polynomials -$$P_k(x,y):=\sum_{j\ge0}(-1)^j{ k-j\choose j }x^{k-2j}y^j;$$ -their relevance in this context being that the polynomial $z^2-xz+y$ divides the polynomial $$z^{k+2}-xP_{k+1}(x,y)z+yP_{k}(x,y)$$ (even as elements of $\mathbb{Z}[x,y,z]$; see below for other properties we need). -Characterization of the $k$-th powers in $ M_2(\mathbb Z)$. Let $k\ge0$. A matrix $A\in M_2 (\mathbb Z)$ is a $k$-th power of an element of $ M_2 (\mathbb Z)$ if and only if there are $t,d$ in $\mathbb Z$ such that -1. $\det(A)=d^k$ -2. $\text{tr}(A)=P_k(t,d)-P_{k-2}(t,d)d$ -3. $P_{k-1}(t,d)$ divides $A+ P_{k-2}(t,d)d I $ -Precisely, if $B\in M_2 (\mathbb Z)$ verifies $B^k=A$ then $d:=\det(B)$ and $t:=\text{tr}(B)$ satisfy (1,2,3). -Conversely, if $(d,t)$ satisfy (1,2,3), there exists a $B\in M_2 (\mathbb Z)$ with characteristic polynomial $z^2-tz+d$. -Precisely, if $P_{k-1}(t,d)\ne0$, it is unique, namely -$$B:=\frac1{P_{k-1}(t,d)}\Big(A+ P_{k-2}(t,d)d I \Big).$$ -If $P_{k-1}(t,d)=0$, then $A=mI$, is an integer multiple of the identity, and all the infinitely many $B\in M_2 (\mathbb Z)$ with -with characteristic polynomial $z^2-tz+d$ satisfy $B^k=A$. -Proof. Assume $A=B^k$ and $B\in M_2 (\mathbb Z)$ and set $t:=\text{tr}(B)$ and $d:=\det(B)$. Then (1) is $\det(A)=\det(B^k)=\det(B)^k=p^k$. As seen above, the characteristic polynomial of $B$, $p_B(z):=z^2-tz+d $ divides the polynomial $z^k-P_{k-1}(t,d)z+P_{k-2}(t,d)d$, and since by Cayley-Hamilton $B^2-tB+d=0$, we also have -$$B^k-P_{k-1}(t,d)B+P_{k-2}(t,d)d I=0,$$ -so, taking the trace, we have $\text{tr}(A)= P_{k-1}(t,d)t -2P_{k-2}(t,d)d$, which is (2), while $P_{k-1}(t,d)B= A+P_{k-2}(t,d)d I$, is (3). -Conversely, assume the above conditions (1,2,3) hold for integers $t,d$. Consider first the case $P_{k-1}(t,d)\ne0$. So one can define -$$B:=\frac1{P_{k-1}(t,d)}\Big(A+ P_{k-2}(t,d)d I \Big),$$ -an element of $M_2 (\mathbb Z)$ thanks to (3). -The trace and determinant of $B$ are then by (1,2), hidding the variables $(t,d)$ in the $P_j$ -$$\text{tr}(B)=\frac{\text{tr}(A)+ 2P_{k-2}d}{P_{k-1} } = \frac{P_k +P_{k-2} d }{P_{k-1} } =\frac{ t P_{k-1} }{P_{k-1} } =t $$ -$$\det(B)= \frac{\det\Big(A+ P_{k-2} d I \Big)}{P_{k-1} ^2} -= \frac{P_{k-2}^2d^2 + \text{tr}(A)P_{k-2} d +\det(A) }{P_{k-1}^2}=$$ $$=\frac{P_{k-2}^2d^2 + \big(P_k-P_{k-2}d\big)P_{k-2}d +d^k }{P_{k-1}^2}= $$ -$$=\frac{ P_kP_{k-2}d +d^k }{P_{k-1}^2}= d,$$ -because $P_{k-1}^2 -P_kP_{k-2} =d^{k-1}$. Thus the characteristic polynomial of $B$ is $z^2-tz+d$, which implies $B^k=P_{k-1}B-P_{k-2}d\, I=A$. -Finally, consider the case $P_{k-1}(t,d)=0$. By (3) $A$ is then a multiple of the identity, $A=m I$, for $m:=-P_{k-2}(t,d)d$. If $m=0$, any nilpotent $B$ has the wanted properties. If $m\ne 0$, let $\lambda$ and $\mu$ be the roots of $z^2-tz+d$, so $t=\lambda+\mu$ and $d=\lambda\mu$. Then we have $\lambda\neq\mu$, otherwise $0=P_{k-1}(\lambda+\mu,\lambda\mu)=k\lambda^{k-1}$ and $\lambda=\mu=0=t=d$ and $A=0$. Also (see below) -$$0= P_{k-1}(\lambda+\mu,\lambda\mu)= \frac{\lambda^k-\mu^k}{\lambda-\mu} $$ -whence $\lambda^k=\mu^k$, and -$$m =-\lambda\mu P_{k-2}(\lambda+\mu,\lambda\mu)=-\frac{\lambda^k\mu-\mu^k\lambda}{\lambda-\mu} = \lambda^k=\mu^k.$$ -Let $B$ one of the infinitely many matrices in $M_2 (\mathbb Z)$ with characteristic polynomial $z^2-tz+d$. Since $\lambda\ne \mu$, $B$ is diagonalizable, $B=Q^{-1}\text{diag}(\lambda,\mu)Q$, so $$B^k=Q^{-1}\text{diag}(\lambda^k,\mu^k)Q=Q^{-1}mIQ=mI=A,$$ -ending the proof. - -More details. The sequence of polynomials $P_k(x,y)\in\mathbb{Z}[x,y]$ is defined by the two-term recurrence -$$\cases{P_{k+2}=xP_{k+1}-yP_k\\ -P_0=1 \\ P_{-1}=0.}$$ -One easily verifies by induction the expansion -$$P_k(x,y):=\sum_{j\ge0}(-1)^j{ k-j\choose j }x^{k-2j}y^j;$$ -in fact $P_k$ may also be presented in terms of the Chebyshev polynomials of the first kind as $P_k(x,y^2)=y^kxT_k\big(\frac{x}{2y}\big)\in\mathbb{Z}[x,y^2]$. -They verify -$$P_k(u+v,uv)=\frac{u^{k+1}-v^{k+1}}{u-v}=\sum_{j=0}^{k} u^jv^{k-j},$$ -and, related to that, for all $k\ge0$ one has: -$$z^{k+2}-P_{k+1}(x,y)z+yP_{k}(x,y)=\big(z^2-xz+y\big) \sum_{j=0}^kP_{k-j}(x,y)z^j, $$ both easily verified by induction. Finally, since they solve a two-term linear recursion, the Hankel determinant of order $2$ must be a $1$-term linear recurrence, and one finds -$$P_{k}(x,y)^2-P_{k+1}(x,y)P_{k-1}(x,y)=y^{k}.$$<|endoftext|> -TITLE: Outer automorphism group of posets -QUESTION [5 upvotes]: Let $X$ be a finite poset (we can assume it is connected) and $A_K(X)$ the incidence algebra of $X$ over a field $K$. -The following result is well known, see for example corollary 7.3.7 in the book "Incidence algebras" by Spiegel and O'Donnell. - -Theorem: If there exists an element $x \in X$ such that every element is comparable to $x$, then every automorphism of $A_K(X)$ is the composition of an inner automorphism of $A_K(X)$ and an automorphism of $X$. - -Note that that theorem includes many posets, for example any poset with a global maximum or minimum. -In particular, this gives an explicit description of the outer automorphism group of $A_K(X)$, which is always finite for such posets $X$. - -Question: Is a classification of finite posets $X$ such that the outer automorphism group of $A_K(X)$ is always finite for any field $K$ known? In particular, what is any easy example such that the outer automorphism group can be infinite? - -There is a more general result in the book (Theorem 7.3.6.) but it uses multiplicative automorphisms and I do not have a feeling how many there are up to inner automorphisms. - -REPLY [2 votes]: Theorem 2 of the paper http://www-math.mit.edu/~rstan/pubs/pubfiles/5.pdf states the following. Let $H$ denote the Hasse diagram of $X$, considered as a graph. Let $r$ denote the dimension of the mod 2 cycle space $V$ of $H$, and let $t$ denote the dimension of the subspace of $V$ generated by cycles consisting of two unrefinable (or saturated) chains of $X$ with the same endpoints. Then $\mathrm{Aut}(A_K(X))$ is isomorphic to a semidirect product of $(K^*)^{r-t}$ by $\mathrm{Aut}(X)$, where $K^*$ is the multiplicative group of $K$. -The smallest example where $r-t>0$ is the four-element "butterfly" poset, with elements $x,y,z,w$ and cover relations $x -TITLE: Profound but not popular mathematical topics and notions -QUESTION [7 upvotes]: The algebraic Theory of Invariants used to be a hot topic until David Hilbert proved his two theorems about invariants. Then for tens of years, the popularity of the topic went down a long time before it picked up again. -Question What are today's mathematical known topics and/or notions that are profound but not popular? Together with an example, could you add an explanation of such a situation? - -Example from Computer Science -- geometric SIMD (fine grain parallel processing) was a popular and hot topic till the middle of 1980'. Then you hardly hear about it while the idea is as fundamental as always. -The explanation is two-folded but very simple. On the one hand, there is some learning and new understanding involved in geometric SIMD processing; one needs to acquire new habits, new reflexes. On the other hand, the technology progress was such that people were satisfied with the results obtained without bothering with the SIMD ways. (Underneath, the new traditional computer architecture is not that traditional -- these days, it incorporates quite a bit of parallelity). We see that the geometric SIMD is not popular for the wrong reasons, and a lot of potentials is wasted. - -REPLY [10 votes]: This hasn't yet been revitalized, but I think John von Neumann's work on Continuous Geometry is rather deep, but there really doesn't seem to be much major work on this topic besides what you see in the references in the link above. -Oddly, even though von Neumann was explicitly aiming to de-emphasize the notion of point in geometry via this work, and the axioms for a continuous geometry are quite similar to those of the notion of frame in the theory of locales, when I have read about locales I have never seen von Neumann's work referred to as a precursor to the theory. (Of course frames only require finite meets.) I'm surprised about this since undoubtedly Marshall Stone was involved with the prehistory of locale theory as is reflected in Elements of the History of Locale Theory by Peter Johnstone -I've seen it mentioned on MO that during the East Coast Operator Algebra Symposium a while back concentrated on the $\mathbb{F}_{1}$ approach to RH, Alain Connes outlined how von Neumann's continuous geometry may have something to say about this approach. In the subsequent years, of course, Connes and Consani have found the Arithmetic Topos... -This might be opinionated, as feared, but I'd be interested in knowing what happened to this idea of von Neumann over the years, and how one can trace it through the literature (I'd like a lead...beyond von Neumann's text...)<|endoftext|> -TITLE: K-theory on finite-dimensional (possibly not finite) CW complexes -QUESTION [8 upvotes]: I am trying to understand why (at least my most elementary understanding of) topological K-theory breaks down for non-compact things (which I have seen asserted in various places). In particular, as someone who mainly thinks about manifolds, I would like to understand what fails on non-compact manifolds. But let's be a tad more general and consider any CW complexes of finite dimension. For simplicitly, I will only say things over $\mathbb C$. -On page 118 of Husemöller's Fiber Bundles, he asserts that $\tilde K^0(X)=[X,G_n(\mathbb C^{2n})]$ when $X$ is a CW complex of dimension $\leq 2n$. He then goes on to discuss $BU$ and proves that $\tilde K^0(X)=[X,BU]$ when $X$ is a finite CW complex. (I realize that I am being sloppy here in writing these as equalities of sets, rather than isomorphisms of functors.) What I would like to know is: why not upgrade this latter result to any finite-dimensional CW complex? It seems to me that all we need is for any map $X\rightarrow BU$ to factor though some finite-dimensional Grassmannian $G_n(\mathbb C^k)\subset BU.$ But $G_n(\mathbb C^{2n})$ contains the $2n$-skeleton of $BU$ (with the usual cell structure), so cellular approximation should suffice in this regard, right? -Where do things break down? So far, I have only been able to find mention of infinite-dimensional counterexamples. In this question, Dmitri Pavlov gives an answer which seems to indicate that there is some fundamental issue with non-compactness, even in finite dimension. However, my knowledge of sheaves is limited and I am having trouble resolving this with the cellular map discussion above. I was thinking that maybe the issue is with some Eilenberg-Steenrod axiom failing, but maps into the $KU$-spectrum should still define a cohomology theory. Is there some issue with how the correspondence $\tilde K^0(X)=[X,BU]$ relates to this cohomology theory? - -REPLY [3 votes]: I am posting an answer to my own question, so that it will be marked as answered and may possibly resolve a similar confusion for someone else, in the future. This question was of the form "People keep saying that something goes wrong and I can't figure out what!" So the answer of "Nothing goes wrong" from Tom Goodwillie was necessarily quite brief, but still very much appreciated. -First, the main point: $K$-theory is a well-defined cohomology theory on the category of finite-dimensional CW complex, which can be described by virtual vector bundles in the same way as it is for compact Hausdorff spaces (again, I am just doing this for complex $K$-theory, but it should be similar in other cases). In the compact Hausdorff setting, we use compactness to conclude that any map $X\rightarrow BU$ factors through some $G_n(\mathbb C^k)\subset BU.$ In the setting of finite-dimensional CW complexes, we instead use cellular approximation (since $G_n(\mathbb C^{2n})$ contains the $n$-skeleton of $BU$). The fact that we can always homotope to a map $X\rightarrow G_n(\mathbb C^{2n})$ (for a suitable $n$) also means that any bundle over $X$ is stably equivalent to one of dimension $\leq n$. This all can be phrased in terms of the "bundle dimension" discussed in Exercise 3.3 of Husemöller's Fiber Bundles. -The natural isomorphism of functors $\tilde K^0(–)=[–,BU]$ (where we can also talk about ring structure) is the only needed connection between classifying maps and virtual bundles. After that, any $\tilde K^i(X)=[X,\Omega^{|i|}BU]=[\Sigma^{|i|}X,BU]$ is described in the same way in terms of virtual bundles over $\Sigma^{|i|}X$, which is still a finite-dimensional CW complex. -So what didn't go wrong? As I mentioned above, I could only find one explicit mention of things possibly going wrong in finite dimensions, which was given by Dmitri Pavlov here. While my understanding is still limited (e.g. I don't know what a Grothendieck topology is), I think I understand the basic idea of his answer (which I'm quite grateful for, as I've never explicitly thought about $K$-theory from a sheafy perspective before). -The idea seems to be that virtual bundles form a presheaf and the question becomes "Is this a sheaf?" As usual, it seems that locality is straightforward, while gluability requires some more thought. We can piece together two virtual bundles that are compatible on their overlap, but given an infinite collection of open sets, the dimensions of the bundles may be unbounded. This is a big problem for gluing. But if we have a compact base, then we can restrict to a finite subcover, after which gluing goes through fine. If our base is a finite-dimensional CW complex, then we may not have a finite subcover, but our virtual bundles are stably equivalent to bundles of bounded dimension. Thus gluing still works without any issue of unbounded dimension. -Lastly, what can go wrong? In this answer, I don't want to give the impression that $K$-theory doesn't face issues in being generalized outside of the compact setting. Certainly, it does! My point is that, over CW complexes, the issue is infinite-dimensionality rather than non-compactness. Over infinite-dimensional CW complexes, the functorial correspondence $\tilde K^0(–)=[–,BU]$ (where the left-hand side is defined in terms of virtual bundles) breaks down, because maps into $BU$ no longer need to factor through a finite stage. Perhaps the simplest example is the inclusion $\mathbb C\mathbb P^\infty\subset BU$. We can still try to define $K$-theory in terms of virtual vector bundles, but it is no longer described by this spectrum. In fact, over infinite-dimensional CW complexes, this $K$-functor does not satisfy the Eilenberg-Steeenrod axioms, so it is not a cohomology theory and is not represented by any spectrum. -For more information on issues with infinite-dimensionality than I am able to provide, you can check out "The $K$-theory of Eilenberg-Maclane spaces" (Anderson and Hodgkin), which I got from Mike Doherty's answered to the afore-linked question. Alternatively, in the comments to this question, there is a link provided by user43326 to a postscript file of "Vector Bundles over Classifying Spaces" by Bob Oliver. A longer paper on the same topic, by Oliver and Jackowski, can be found here.<|endoftext|> -TITLE: What are the applications of modular forms in number theory? -QUESTION [5 upvotes]: I am new to the topic, so I'm trying to get an overview. I am aware of the relation between modular forms and $L$-series (but don't know what that does) and FLT. -Are there other applications of modular forms other than counting problems (by obtaining the coefficients of a series) in number theory? -A short list would be sufficient but a little more detail with that would be helpful. -EDIT: -I am aware of this post but my question is specifically on number theory. - -REPLY [4 votes]: Modular forms are used to solve Fermat's last theorem, Mock modular forms are used in black holes theory. I read also in Quanta magazine that Eisenstein series are used to compute what we call Monster group. -https://d2r55xnwy6nx47.cloudfront.net/uploads/2017/08/symmetry-algebra-and-the-monster-20170817.pdf -Here is an interesting article to see the beautiful application of modular forms in astronomy https://m.facebook.com/story.php?story_fbid=795545447309716&id=247304225467177<|endoftext|> -TITLE: Who is the "young student" André Weil is referring to in his letter from the prison? -QUESTION [60 upvotes]: I am reading a nice booklet (in Italian) containing the exchange of letters that André and Simone Weil had in 1940, when André was in Rouen prison for having refused to accomplish his military duties. -Of course, among these letters, there is the famous one where André describes his mathematical work to his sister, whose English translation was published in 2005 in the Notices AMS. Referring to this translation, at page 340 we can read - -[...] this bridge exists; it is the theory of the field of algebraic functions over a finite field of constants (that is to say, -a finite number of elements: also said to be a Galois field, or earlier “Galois imaginaries” because -Galois first defined them and studied them; they -are the algebraic extensions of a field with p elements formed by the numbers 0, 1, 2,...,p − 1 -where one calculates with them modulo p, p = -prime number). They appear already in Dedekind. -A young student in Göttingen, killed in 1914 or -1915, studied, in his dissertation that appeared in -1919 (work done entirely on his own, says his -teacher Landau), zeta functions for certain of these -fields, and showed that the ordinary methods of -the theory of algebraic numbers applied to them. - -My Italian book contains a note at this point, saying - -Di questo "giovane studente" non abbiamo altre notizie - -that can be translated as - -We have no further information about this "young student". - -This seems a bit strange to me: if an important result on zeta functions is really due to this student, his name should be known, at least among the experts in the field. So let me ask the following: -Who is the "young student in Göttingen, killed in 1914 or 1915" André Weil is talking about? - -REPLY [73 votes]: This must have been Heinrich Kornblum (1890-1914). - -[note by E. Landau in German, my translation] -$^1$ The author, born in Wohlau on August 23, 1890, had before the war -independently made the discovery that Dirichlet's classic proof of the theorem -of prime numbers in an arithmetic progression (along with the later elementary -reasons for the non-vanishing of the known series) had an analogue in -the theory of prime functions in residue classes with a double module ($p,M$). -His doctoral dissertation on this self-chosen topic was -already essentially finished when, as a war volunteer, he fell in October 1914 at Роёl-Сареllе. -Only recently I received from his estate the manuscript (known to me since 1914). I hereby publish the most beautiful and interesting parts. The Kornblum approach is characterized by high elegance -and shows that science has lost in him a very promising researcher. - -REPLY [51 votes]: That was most certainly Heinrich Kornblum. His paper titled 'Über die Primfunktionen in einer arithmetischen Progression' was published in Math. Z. 5 (1919) pp 100–111 (EuDML), see the zbMath revew. In the paper he establishes the analogue of Dirichlet's theorem on primes in arithmetic progressions, in the polynomial setting (with natural density). -His history and contribution are mentioned several times in Roquette's historical account of the Riemann Hypothesis in positive characteristic.<|endoftext|> -TITLE: Another characterization of matroids -QUESTION [14 upvotes]: Has anyone seen the following characterization of matroids? -Let $\Delta$ be a simplicial complex on finite ground set $E$. Then $\Delta$ is a matroid complex if and only if, for every $X\subseteq E$ and any two facets (maximal faces) $\sigma,\tau$ of the induced subcomplex $\Delta|_X$, the links of $\sigma$ and $\tau$ in $\Delta$ are equal. -This seems to be nontrivial, but not too hard to prove (perhaps a [2] or [2+] on Stanley's scale). The $\implies$ direction is conceptually clearer: the two links are both the matroid contractions of $A$, and the proof is by basis exchange and induction. For $\impliedby$, one can prove the contrapositive, using the criterion that $\Delta$ is a matroid complex iff every induced subcomplex is pure. -I have not seen this in the literature. Has anyone come across it before? - -REPLY [7 votes]: Yes though this is commonly stated using clutters and their associated minors, (identify your ASC with its clutter of facets) basically you're describing a forbidden minor characterisation of a matroid basis clutter (the forbidden clutter minor is $M=(\{1,2,3\},\{\{1\},\{2,3\}\})$ explaining why you can rephrase the absence of said minor in terms of the links of facets in all restrictions coinciding). Seymour for example gave a forbidden clutter minor characterisation of not just the basis clutters of matroids but of binary matroids as well as many other interesting clutters: -https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/jlms/s2-12.3.356 -Its a cool identity, though yeah its at least 40-50 years old, for example here are some course notes I found online that give your result under the header "proposition 2.1" on page 3 here: -https://sites.lafayette.edu/traldil/files/2010/06/matroids.pdf -I doubt you'll be able to find when it was originally noticed and definitely not using that notation since I think Tutte was still calling undirected cycles in graphs "polygons" during the 1960s which suggests to me that the terminology for simplicial complexes was not even standardized in usage. Though I am surprised no one has answered your question in these past two months, I am very new to matroid theory myself, though I've been trying to read an article on MFMC matroids and many of the lemmas were from an old paper on clutters which is why I recognised your identity.<|endoftext|> -TITLE: Graph that minimizes the number of b/w colorings where white vertices have an odd number of black -QUESTION [17 upvotes]: motivated from a physical context, we are currently interested in the following graph coloring problem: -Given a connected graph $G_n$ with $n$ vertices, how many colorings exist such that all white vertices have an odd number of black neighbors? We call this number $\omega(G_n)$. -(To the best of our knowledge, this particular coloring problem has not yet been studied, but if you have seen something like this, please let us know!) -As an example, consider the cycle graph with four vertices. There exist five such colorings: - -In constrast, the fully connected graph of four vertices has nine such colorings: - -Indeed, using completely different tools from physics, one can show that $\omega(G_n)$ is maximized by the fully connected graph, yielding the upper bound $\omega(G_n) \leq \begin{cases}2^{n-1} & n\text{ odd} \\ 2^{n-1}+1 & n\text{ even} \end{cases}$. Note that there is a total of $2^n$ colorings, thus, only about half of them can have the desired property. -Now we are interested in finding graphs that minimize $\omega(G_n)$ for a fixed number of vertices $n$. Exhaustively checking all connected graphs up to size 10 [Edit: this was a mistake, we only checked up to $n=8$. See Gordon's reply for a counter example with $n=9$ vertices] suggests that the graph minimizing $\omega(G_n)$ is given by the cycle graph of size $n$, but we have no clue on how to prove it. -Our questions are as follows: -i) Is there an equivalent, similar or related graph coloring problem known in the literature? -ii) Is there a graph theoretical argument that the cycle graph (among others) minimizes $\omega(G_n)$ for fixed $n$? -Lots of thanks in advance! - -REPLY [4 votes]: There is no graph-theoretical argument that the local equivalence class containing the cycle $C_n$ always minimises the value of $\omega$, because it is not true. -My smallest example is on 9 vertices with the following graph, which is just two pentagons merged at a single vertex. - -This $9$-vertex graph has just 28 valid colourings, whereas $\omega(C_9) = 31$. -I verified this with some naive code in SageMath that simply tests each subset of the vertices of graph to see whether it can be the black vertices of a valid colouring - i.e., whether each vertex outside the subset is adjacent to an odd number of vertices in the subset. -def isValid(g,black): - for v in g.vertices(): - if v not in black: - val = len([w for w in g.neighbors(v) if w in black]) - if val % 2 == 0: - return false - return true - -def validSetList(g): - return [s for s in Subsets(g.vertices()) if isValid(g,s)] - -(I can't figure out how to enable syntax highlighting.) -Then the next couple of lines create the graph, call the function to return the list of valid sets, and determine how many there are: -h=Graph('HPXP?E@') -len(validSetList(h)) - -So I think the actual minimiser is going to be something like a bunch of 5-cycles merged at a single vertex, but I need to experiment some more.<|endoftext|> -TITLE: Finitely presented modules admitting projective covers -QUESTION [6 upvotes]: A ring $R$ is called semi-perfect if every finitely generated $R$-module has a projective cover, and it can be proved that this is equivalent to say that the category consisting of the finitely generated projective $R$-modules is Krull-Schmidt. I was wondering, and what about the rings $R$ such that every finitely presented $R$-module has a projective cover? Do these rings have a special name, and are there characterizations of these rings, just like there are for semi-perfect rings? - -REPLY [4 votes]: Such rings were called "$F$-semiperfect", and more recently (thanks to rschweib for the information) "semiregular". -One characterization is that these are the rings $R$ such that $\bar{R}=R/J(R)$, the quotient by the Jacobson radical, is von Neumann regular and idempotents lift from $\bar{R}$ to $R$. This is analogous to the characterization of semiperfect rings as those for which $\bar{R}$ is semisimple and idempotents lift from $\bar{R}$ to $R$. -Some old references: -Oberst, Ulrich; Schneider, Hans-Jürgen, Die Struktur von projektiven Moduln. (The structure of projective modules.), Invent. Math. 13, 295-304 (1971). ZBL0232.16020. -Azumaya, Goro, F-semi-perfect modules, J. Algebra 136, No. 1, 73-85 (1991). ZBL0717.16005.<|endoftext|> -TITLE: Does the purported proof of Rota's conjecture provide an algorithm for calculating the forbidden minors of matroids over arbitrary finite fields? -QUESTION [6 upvotes]: About six years ago there was a proof announced and later outlined in a notice from AMS. However right now I can only seem to find forbidden minor characterizations for matroids linearly representable over $\mathbb{F}_2,\mathbb{F}_3,\mathbb{F}_4$ and some for $\mathbb{F}_5$. Now understanding the outline given by Geelen, Gerards, and Whittle is hard enough for me as I am not well versed in matroid theory, also a full proof hasn"t even been written yet so to go further I'd have to scour through the 20 something papers they wrote and used results from (most of which I don't even partially understand). However I'm curious as to how constructive their proof was and if it was in such a way that an algorithm can be derived from it as a collaroy allowing one to just run it over all finite fields up to some very large prime power on a super computer so we can get insight at least empirically as to what they look like. -I think this would be interesting because unlike other minor theorems for graphs like say the most famous Robertson–Seymour theorem, these give us insight to the class of graphs closed under the graph minor operation, yet this class is so large that it lacks any real 'neat structure' - its just graphs closed under minors. In contrast the class of matroids linearly representable over finite fields is much smaller then say the class of matroids closed under the matroid minor operation (also we know that an analogue of the Robertson–Seymour theorem for minors is false e.g. there exists matroids closed under minors without any finite set of forbidden minors) so id guess these adhere to some kind of general structure. Also knowing the minors of the first say 100 finite fields explicitly might give better insight into them and allow for interesting theorems to be derived from those particular matroids. For example the matroids representable over the first finite field $\mathbb{F}_2$ are called binary matroids and there are all sorts of special theorems for them e.g. an Euler theorem and factor critical theorem graph theory analogue which doesn't necessarily hold for matroids over other finite fields. - -REPLY [6 votes]: As far as I understand, the purported proof does not give an algorithm that given a finite field $\mathbb{F}$, computes the excluded minors for $\mathbb{F}$-representability. This is because it relies on well-quasi-ordering arguments, and therefore does not yield explicit upperbounds on the size of the excluded minors. Note that if one could prove that there exists a computable function $c: \mathbb{N} \to \mathbb{N}$ such that every excluded-minor for $\mathbb{F}$-representability has size at most $c(|\mathbb{F}|)$, then this would give a naive brute force algorithm, but it is unknown if such a computable function exists. Indeed, even for minor-closed classes of graphs, it is known that the problem of computing excluded minors is undecidable. So it may be that such a computable function $c$ does not exist. -See my other answer for more information about the undecidability results for computing the excluded minors of a minor-closed class of graphs. Finally, you might be interested in this recent post by Rutger Campbell on the Matroid Union Blog about a strategy to compute the excluded minors for the five element field.<|endoftext|> -TITLE: Picard group vs class group -QUESTION [7 upvotes]: The question. -Let $R$ be a commutative ring. Let $M$ be an $R$-module with the property that there exists an $R$-module $N$ such that $M\otimes_R N\cong R$. Does there always exist an ideal $I$ of $R$ such that $M$ is isomorphic to $I$ as an $R$-module? (I suspect not in this generality) -The background. -Let $R$ be a commutative ring. Here are two groups that one could associate to $R$. - -The "class group". - -The first group is "inspired by number theory". One takes the ideals of $R$ and observes that they have a natural multiplication defined on them. One defines two ideals $I$ and $J$ to be equivalent if there exist nonzerodivisors $s$ and $t$ such that $sI=tJ$. This relation plays well with the multiplication, giving us a multiplication on the equivalence classes (unless I screwed up; my reference is "back of an envelope calculation"). This makes the equivalence classes into a commutative monoid, and one could define the class group of $R$ to be units of this monoid, i.e. elements with an inverse. -Note: one could instead use fractional ideals. The theory of fractional ideals is often set up only for integral domains, and if I did screw up above then maybe I should have restricted to integral domains. A fractional ideal is defined to be an integral ideal with a denominator so I don't think this changes the group defined here. - -The Picard group. - -The second group is "inspired by geometry" -- it's the Picard group of $\operatorname{Spec}(R)$. More concretely, take the collection (it's not a set) of isomorphism classes of $R$-modules $M$. This has a multiplication coming from tensor product, and satisfies the axioms of a monoid except that it's not a set. The units of this monoid however are a set, because another back of an envelope calculation seems to indicate that if $M\otimes_R N\cong R$ and we write $1=\sum_i m_i\otimes n_i$, a finite sum, then the $m_i$ generate $M$ as an $R$-module, giving us some control over the size of the units of the monoid -- they're all isomorphic to a quotient of $R^n$ so we have regained control in a set-theoretic sense. The units of the monoid are the second group. -The question comes from me trying to convince myself that these groups are not equal in general (for I don't really expect them to be equal in general). If $R$ is a Dedekind domain (so $\operatorname{Spec}(R)$ is a smooth affine curve) then we have here the classical definition and the fancy definition of the class group of $R$, and the answer to the question is "yes". This is because every rank 1 projective $R$-module is isomorphic to an ideal of $R$; if I recall correctly then more generally every rank $n+1$ projective $R$-module is isomorphic to $I\oplus R^n$ for some ideal $I$ (this is true at least for the integers of a number field) which enables you to compute the zero'th algbraic $K$-group (Grothendieck group) of $R$. But more generally than this I am not sure what is going on. -On the divisors Wikipedia page I read "Every line bundle $L$ on $X$ on an integral Noetherian scheme is the class of some Cartier divisor" which makes me think that the result might be true for Noetherian integral domains, but I don't see the proof even there (perhaps it's standard). The way it's phrased makes me wonder then whether there are non-Noetherian counterexamples. - -REPLY [2 votes]: Here is an attempt to generalize somewhat the above-mentioned proof from Hartshorne. -Claim: Let $R$ be a ring whose total ring of fractions $R_{\mathrm{tot}}=S_{\mathrm{nzd}}^{-1}R$ is Artinian. Then any invertible $R$-module is isomorphic to an invertible ideal. -(The hypothesis holds at least in the following two "natural" cases: - -$R$ is a domain, which corresponds to the case of integral schemes, -$R$ is a Noetherian ring without embedded components, i.e. $\mathrm{Ass}\,R$ is precisely the set of minimal primes, in which cases the spectrum of $R_{\mathrm{tot}}$ consists precisely of these minimal primes, hence is $0$-dimensional.) - -Proof: One proceeds as in the proof from Hartshorne. Given an invertible module $M$, this is a locally free module of constant rank $1$, and so is the $R_{\mathrm{tot}}$-module $M \otimes_R R_{\mathrm{tot}}$. As $R_{\mathrm{tot}}$ is Artinian, any locally free module of rank $1$ is actually a free module of rank $1$, and thus we have -$$M=M\otimes_R R\hookrightarrow M\otimes_R R_{\mathrm{tot}}\simeq R_{\mathrm{tot}}.$$ -This realizes $M$ as an $R$-submodule $M'$ of $R_{\mathrm{tot}}$. It is finitely generated (because $M$ is), let us call these generators $a_1/s_1, \dots, a_n/s_n \in R_{\mathrm{tot}}.$ But then $s=s_1s_2 \dots s_n$ is a non-zero divisor of $R$, and we have $sM' \subseteq R$. Thus, $M$ is isomorphic to the invertible ideal $I:=sM'$. $\square$ -(I guess that the assumption can be relaxed a bit more, by assuming that $R_{\mathrm{tot}}$ is just a finite direct product of local rings (edit: Actually, even more by simply assuming that $\mathrm{Pic}(R_{\mathrm{tot}})=1$). But I don't know any new "natural" cases that this would provide.)<|endoftext|> -TITLE: Cayley-Hamilton over super rings -QUESTION [6 upvotes]: If $R$ is a commutative ring, then the Cayley-Hamilton theorem states that any endomorphism $\phi: R^{n} \rightarrow R^{n}$ of a rank $n$ free module satisfies its own characteristic polynomial, in particular satisfies a monic degree $n$ polynomial. -Now suppose that instead $R = R_{0} \oplus R_{1}$ is a commutative super ring (that is, it is $\mathbb{Z}/2$-graded, with even degree elements in the center, and odd degree elements anticommuting with each other). -Does every endomorphism $\phi: R^{p, q} \rightarrow R^{p, q}$ of a free module on $p$ even and $q$ odd variables satisfy a monic polynomial over $R$ of degree $p+q$? Does it satisfy some monic polynomial? - -REPLY [6 votes]: The case when $R$ is purely even but the module has a $Z/2$-grading was studied before, see for example - -"On the Cayley-Hamilton equation in the supercase" by Issai Kantor and Ivan Trishin, Comm. in Algebra 27:1 (1999), 233-259 -"Berezinians, Exterior Powers and Recurrent Sequences", by -H. M. Khudaverdian and TH. TH. Voronov, -Letters in Mathematical Physics, 74 (2005), 201–228 -Already in this case it seems that you should not expect the polynomial to be monic.<|endoftext|> -TITLE: Can we take a supremum over all Hilbert spaces? -QUESTION [19 upvotes]: In my paper On the optimal error bound for the first step in the method of cyclic alternating projections, I defined functions $f_n:[0,1]\to\mathbb{R}$, -$n\geqslant 2$, by -$$ -f_n(c)=\sup\{\|P_n\dotsm P_2 P_1-P_0\|\,|\,c_F(H_1,\dotsc,H_n)\leqslant c\}, -$$ -where -(1) the supremum is taken over all complex Hilbert spaces $H$ and systems of closed subspaces $H_1,\dotsc,H_n$ of $H$ such that the Friedrichs number $c_F(H_1,\dotsc,H_n)$ is less than or equal to $c$. The Friedrichs number is a quantitative characteristics of a system of closed subspaces, -$c$ is a given number from $[0,1]$. -(2) $P_i$ is the orthogonal projection onto $H_i$, $i=1,2,\dotsc,n$ and $P_0$ is the orthogonal projection onto the subspace $H_1\cap H_2\cap\dotsb\cap H_n$. -Now I have some doubts in the validity of this definition. -Namely, I know that all sets do not constitute a set and if I understand things right, all Hilbert spaces do not constitute a set (and even all one dimensional Hilbert spaces do not constitute a set). So, can I take the supremum? -Please help me; I will be very grateful for any comments, remarks, and answers. -In response to Mike Miller's comment: -Unfortunately, I also know almost nothing about set theory. -Yes, I understand that in fact I take the supremum of a "set" of real numbers, namely, of the "set" $A=A_n(c)$ -which consists of all $a\in\mathbb{R}$ -for which there exist a complex Hilbert space $H$ and a system of closed -subspaces $H_1,\dotsc,H_n$ of $H$ such that $c_F(H_1,\dotsc,H_n)\leqslant c$ and -$\|P_n\dotsm P_2 P_1-P_0\|=a$. -But I do not understand why the "set" $A$ is a set. - -REPLY [6 votes]: The purpose of this answer is to recall that the classical proof of the existence of suprema of bounded families of reals (as found in any analysis textbook) continues to work without any modifications whatsoever for proper classes. (In particular, we do not assume that the existence of suprema of bounded nonempty subsets of reals has been proved yet.) -In fact, the proof works in the Zermelo set theory, -without the axiom of replacement and the axiom of choice. -Theorem. -Suppose $C$ is a nonempty class (possibly proper) and $g\colon C\to \mathbf R$ is a map bounded from above. -Then $\sup_{c\in C} g(c)$ exists. -Proof. -Denote by $U$ the set of upper bounds for $g$, -i.e., $U=\{u\in\mathbf R\mid \forall c\in C\colon g(c)\le u\}$, -which exists by the axiom of separation. -Since $\sup_{c\in C} g(c)$ is by definition the smallest upper bound for $g$, -we have to show that $U$ has the smallest element. -By assumption, $U$ is nonempty. -Since $C$ is nonempty, the set $U$ is bounded from below, -namely, it is bounded from below by $g(c)$ for any $c\in C$. -Thus, $U$ is a nonempty upward-closed subset of reals bounded from below, -i.e., an upper Dedekind cut, so $\inf U$ exists. -It remains to show that $\inf U \in U$. -Indeed, $\inf U \in U$ is equivalent to -$$\forall c\in C\colon g(c) \le\inf U,$$ -which in its turn (by definition of $\inf$) is equivalent to -$$\forall c\in C \forall u\in U\colon g(c) \le u,$$ -equivalently, -$$\forall u\in U \forall c\in C\colon g(c) \le u,$$ -i.e., -$$\forall u\in U\colon u\in U,$$ -which is tautologically true. -Thus, $\inf U \in U$, so $\inf U$ is the smallest upper bound for $g$, which, therefore, exists.<|endoftext|> -TITLE: Converse to Erdős' conjecture on arithmetic progressions -QUESTION [10 upvotes]: I apologise in advance if this has been asked here before. I did a search and did not find anything obvious. Erdős' conjecture states that if $A\subseteq {\bf N}$ is such that $\sum_{n\in A} n^{-1}$ diverges, then $A$ contains arbitrarily long arithmetic sequences. -I was wondering if anything is known about the converse statement; i.e., if $\sum_{n\in A} n^{-1}$ is finite, is it true that $A$ will not have $k$-term arithmetic progressions for $k$ large enough? - -REPLY [17 votes]: Unfortunately such a simple converse cannot be possible because -one can "plant" long arithmetic progressions in $A$ while keeping it -sparse overall. For example, let $A$ consist of all integers of the form -$10^{n!}+m$ with $1 \leq m \leq n$ (which even makes $\sum_{n\in A} 1 / \log n$ -converge). -[I see that GH from MO posted a very similar answer while I was editing mine.] - -REPLY [16 votes]: It is not true. Take, for example, $A=\bigcup_{n\in\mathbb{N}}\{n^3,n^3+1,\dots,n^3+n\}$.<|endoftext|> -TITLE: Reference request: Origins of differential homological algebra -QUESTION [8 upvotes]: Differential homological algebra in its initial formulation is due to Eilenberg and Moore, who published the homological version of the Eilenberg–Moore spectral sequence in 1965 (and the cohomological version never), sufficiently long after its date of discovery that other accounts had been published in the meantime. -Paul Baum's 1962 thesis gives an account, but Moore had spoken on this theory at least as early as the 1959–60 Seminaire Henri Cartan, and in that appearance, says he and Eilenberg had worked it out jointly and it had already appeared in notes from the 1957–8 Princeton seminar on algebraic topology, which thus may be the earliest written source. -Are these still extant? -If so, where can they be found? -(This question has also been posted on the History of Mathematics StackExchange without resounding success.) - -REPLY [6 votes]: A copy of the mimeographed notes by J.C. Moore from the 1957-1958 Princeton Seminar on algebraic topology is in the Princeton University library: https://catalog.princeton.edu/catalog/1693293 -It says "available upon request", although that seems limited to faculty and students. Perhaps some kind soul with access would be willing to scan the notes and place them online?<|endoftext|> -TITLE: Prove or disprove that $\sum_{n=1}^{\infty}\frac{\lambda(n)\mathbb{E}_{n\in\mathbb{N}}[a_n]}{n}=0$ for any choice of $(a_n)$ -QUESTION [21 upvotes]: I've been obsessed with this one problem for many months now, and today is the sad day that I admit to myself I won't be able to solve it and I need your help. The problem is simple. We let -$$\mathbb{E}_{n\in\mathbb{N}}[f(n)]:=\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}f(n)$$ -denote the expected value of a function $f(n)$ (if it exists). This means that for some fixed choice of sequence $(a_n)_{n=1}^{\infty}$ the quantities $\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]$ will give the "average value of every $k$" elements. For example, $\mathbb{E}_{n\in\mathbb{N}}[a_{3n}]$ will be the average of $a_{3},a_{6},a_{9}$ etc... . If all of these quantities exist, we will call a sequence $a_n$ to be "sequentially summable". The problem is as follows: - -Show that for any bounded sequentially summable $a_n$ -\begin{equation}\sum_{k=1}^{\infty}\frac{\lambda(k)\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]}{k}=0\tag{1}\end{equation} -where $\lambda(n)$ denotes the Liouville function - -Swapping which side the term $\mathbb{E}_{n\in\mathbb{N}}[a_n]$ is on, this gives an absolutely lovely formula for $\mathbb{E}_{n\in\mathbb{N}}[a_n]$ from $\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]$ ($k\geq 2$) which can be appreciated by even people who have done no mathematics. Similarly, I would expect that the conjecture holds for $\lambda(n)$ replaced by its "twin" $\mu(n)$, the Mobius function. Throughout the rest of this question, I will give all of the partial results and a general outline of how they are obtained. -This first partial results answers the question "why on earth would these values be 0???": - -Partial Result $\#1$: For any bounded sequentially summable sequence $a_n$, it holds that -\begin{equation}\lim_{s\to1^+}\sum_{k=1}^{\infty}\frac{\lambda(k)\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]}{k^s}=\lim_{s\to1^+}\sum_{k=1}^{\infty}\frac{\mu(k)\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]}{k^{s}}=0\tag{2}\end{equation} -and thus if the sum in (1) converges it must converge to $0$ - -This result is obtained by inverting the summation order and exploiting the fact that -$$\sum_{d|n}\frac{\mu(n)}{n^s}=\prod_{p}\left(1-\frac{1}{p^s}\right)>0$$ -which means that the triangle inequality does not "lose" any sign cancelation. The following partial result is much stronger: - -Partial Result $\#2$: There exists an absolute constant $c_0$ such that for any sequentially summable sequence $a_n$ and $N>0$ -\begin{equation}\left|\sum_{k=1}^N\frac{\lambda(k)\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]}{k}\right|1$ we have that $\mathbb{E}_{n\in\mathbb{N}}[a_{kn}]=0$ and thus our sum converges, but not to $0$. -I add as well that -$$\sum_{k=1}^{N}\frac{\mu(k)}{k}\frac{1}{[N/k]}\sum_{n=1}^{[N/k]}a_{kn}=\frac{a_1}{N}$$ -due to simple inversion of summation order, and so since -$$\frac{1}{[N/k]}\sum_{n=1}^{[N/k]}a_{kn}\approx \mathbb{E}_{n\in\mathbb{N}}[a_{kn}]$$ -we get further intuition for the result. - -REPLY [10 votes]: Here is a more classical (more streamlined) version of Terry's nice argument. It avoids any reference to generalized limits (hence the axiom of choice). -We shall only use that $a_n$ is sequentially summable and bounded in square mean: -$$\sum_{n\leq N}|a_n|^2\ll N.$$ -By assumption, the limits -$$f(k):=\lim_{N\to\infty}\frac{1}{N}\sum_{n\leq N}a_{kn}$$ -exist, and we want to show that -$$\sum_{k=1}^\infty\frac{\lambda(k)f(k)}{k}=0.\tag{$\ast$}$$ -Lemma. The multiplicative convolution $g:=\mu\ast f$ satisfies -$$\sum_{q=1}^\infty\frac{|g(q)|^2}{\varphi(q)}<\infty.$$ -Proof. For the exponential sums -$$S_N(x):=\sum_{n\leq N} a_n e(nx),$$ -we observe that -$$\sum_{q\mid k}\sideset{}{^*}\sum_{b\bmod q}S_N(b/q)=\sum_{c\bmod k}S_N(c/k)=k\sum_{\substack{n\leq N\\k\mid n}}a_n=k\sum_{m\leq N/k}a_{km}.$$ -Dividing by $N$ and letting $N\to\infty$, the right hand side tends to $f(k)$. Therefore, -$$f(k)=\lim_{N\to\infty}\frac{1}{N}\sum_{q\mid k}\sideset{}{^*}\sum_{b\bmod q}S_N(b/q).$$ -It follows that -$$g(q)=\lim_{N\to\infty}\frac{1}{N}\sideset{}{^*}\sum_{b\bmod q}S_N(b/q).$$ -In particular, the limit on the right hand side exists. Using the Cauchy-Schwarz inequality and the large sieve inequality, we infer -$$\sum_{q\leq Q}\frac{|g(q)|^2}{\varphi(q)}\leq\limsup_{N\to\infty}\frac{1}{N^2}\sum_{q\leq Q}\sideset{}{^*}\sum_{b\bmod q}|S_N(b/q)|^2\leq\limsup_{N\to\infty}\frac{1}{N}\sum_{n\leq N}|a_n|^2\ll 1.$$ -Letting $Q$ tend to infinity, the inequality in the lemma follows. $\Box$ -For $g$ as in the Lemma, we have $f=1\ast g$, hence also -$$\sum_{k\leq K}\frac{\lambda(k)f(k)}{k}=\sum_{k\leq K}\frac{\lambda(k)}{k}\sum_{q\mid k}g(q)=\sum_{q\leq K}g(q)\sum_{\substack{k\leq K\\q\mid k}}\frac{\lambda(k)}{k} -=\sum_{q\leq K}\frac{g(q)\lambda(q)}{q}\sum_{\ell\leq K/q}\frac{\lambda(\ell)}{\ell}.$$ -In particular, -$$\left|\sum_{k\leq K}\frac{\lambda(k)f(k)}{k}\right|\leq -\sum_{q\leq K}\frac{|g(q)|}{q}\left|\sum_{\ell\leq K/q}\frac{\lambda(\ell)}{\ell}\right|.$$ -Let us fix a large $L>0$. We cut the $q$-sum into two parts $q\leq K/L$ and $q>K/L$, and we apply Cauchy-Schwarz to both parts: -$$\left|\sum_{k\leq K}\frac{\lambda(k)f(k)}{k}\right|^2\ll -\left(\sum_{q\leq K/L}\frac{|g(q)|^2}{\varphi(q)}\right)\left(\sum_{q\leq K/L}\frac{\varphi(q)}{q^2}\left|\sum_{\ell\leq K/q}\frac{\lambda(\ell)}{\ell}\right|^2\right) -+\left(\sum_{K/L -TITLE: Is there a way to describe the image of the $n$-fold residue map from $H^0(Y,\Omega_Y^n(\log E))$? -QUESTION [5 upvotes]: Let $X$ be an $n$-dimensional smooth algebraic variety, and let $Y$ be a compactification with $E=Y\setminus X$ simple normal crossings. There is the natural quotient map -$$\Omega_Y^n(\log E)\to \mathcal{H}^n(\Omega_Y^\bullet(\log E))\to 0.$$ -I want to understand the image of this map on global sections. If I am not mistaken this map can be identified with the $n$-fold residue map -$$\Omega_Y^n(\log E)\to \oplus_{p_i}\mathbb{C}\to 0,$$ -where the $p_i$ are all $n$-fold intersections of the components of $E$. The kernel of the residue map is $W_{n-1}\Omega_Y^{n}(\log E)$. So the image of the residue map on global sections is equal to the kernel of -$$\oplus_{p_i}\mathbb{C}\to H^1(Y,W_{n-1}\Omega_Y^n(\log E)).$$ -Is there a way to understand this map and its kernel more concretely? -Edit: is it maybe true, as in the case of Riemann surfaces, that the image is exactly equal to $\{\sum r_i=0\}\subset \oplus_{p_i}\mathbb{C}$? - -REPLY [4 votes]: Your guess in the last sentence is not quite right, but there is a natural higher-dimensional generalization of the statement. What happens is that you have a "boundary map" -$$ \bigoplus_{p_i} \mathbb C \to \bigoplus_{C_j} \mathbb C$$ -where the points $p_i$ are all the $n$-fold intersections of boundary components and the curves $C_j$ are all the $(n-1)$-fold intersections; the summand corresponding to $p_i$ is mapped to those summands $C_j$ such that $p_i \in C_j$. The image is precisely the kernel of this map. -All of these things come from Deligne's Hodge II. You have two filtrations on $\Omega^\bullet(\log E)$ giving you two different spectral sequences, one of which degenerates immediately (giving you the Hodge filtration) and one of which degenerates after the first differential (giving you the weight filtration), the maps you care about are seen in these spectral sequences and you have to think about what things mean. The upshot is first of all that $\Gamma(Y,\Omega^n(\log E)) = F^n H^n(X,\mathbb C)$. This then maps surjectively onto $\mathrm{Gr}^W_{2n} H^n(X,\mathbb C)$ which in turn injects via an edge map in the Leray spectral sequence into $\bigoplus_{p_i} \mathbb C$. Since the Leray spectral sequence degenerates after the first differential, the image of this map will just be the kernel of the differential from this spot of the spectral sequence, and that differential is the thing I wrote above.<|endoftext|> -TITLE: Does there exist a process to build a list of numbers whose standard deviation is an integer? -QUESTION [7 upvotes]: Or rephrased, is there a way to make a list of numbers whose sample variance is a square number? I'm interested in sequences of arbitrary length with integer elements. -(I come from a computer science background so my apologies if this question is a poor one). - -REPLY [7 votes]: Essentially, this problem is to generate integers $n,s,a_1,\dots,a_n$ such that $n\ge1$ and -$$(n-1)s^2=a_1^2+\dots+a_n^2.\tag{1}$$ -To do this, let $k:=\lfloor n/4\rfloor$ and take any nonnegative integers $t_1,\dots,t_k$ such that -$$(n-1)s^2=t_1+\dots+t_k.\tag{2}$$ -For each $j\in[k]:=\{1,\dots,k\}$, by Lagrange's four-square theorem, there exist integers $a_{j1},\dots,a_{j4}$ such that -$$t_j=\sum_{i=1}^4 a_{ji}^2,$$ -so that -$$(n-1)s^2=\sum_{j=1}^k\sum_{i=1}^4 a_{ji}^2.\tag{3}$$ -The latter expression is the sum of $4k\le n$ squares of nonnegative integers. Complementing this sum by the sum of $n-4k$ squares of $0$, we get representation (1) -- for any given natural $n$ and integer $s$. - -This is illustrated by the the following image of a Mathematica notebook, finding a representation (1) for $n=15$ and $s=5$; -$$(15-1)5^2=4^2 + 0^2 + 8^2 + 6^2 + 10^2 + 0^2 + 4^2 + 2^2 + 5^2 + 0^2 + 8^2 + - 5^2 + 0^2 + 0^2 + 0^2.$$ - - -Of course, the $a_i$'s in (1) are the deviations of the sample values from the sample mean. Previously, I forgot to take into account that the sum of the $a_i$'s must of course be $0$. -$\newcommand\ep\epsilon$One way to get that is to try to attach "signs" $\ep_i\in\{-1,1\}$ to the $a_i$'s so that $\sum_{i=1}^m\ep_i a_i=0$. I think usually this will be possible to do when $\sum_{i=1}^m a_i$ is even, as is the case with the above example (continued below). Otherwise, repeat with other $t_1,\dots,t_k$ in (2) and/or with other instances of the $a_{j1},\dots,a_{j4}$'s. - -The problem of the existence of "balancing" signs $\ep_i\in\{-1,1\}$ is obviously equivalent to the so-called partition problem, which is easily solved for not too big values $n$ and $\sum_{i=1}^m a_i$ in view of the simple recurrence relation, whose application to the particular $a_i$'s considered in the mentioned Mathematica notebook is illustrated here: - -This again confirms that for our particular $a_i$'s "balancing" signs $\ep_i\in\{-1,1\}$ exist. -The problem of actually finding an appropriate partition (which is equivalent to finding "balancing" signs $\ep_i\in\{-1,1\}$) was solved by Korf. -For our particular $a_i$'s there actually are $18$ different "balancing" sign assignments $(\ep_i)$, not counting sign assignments to the $0$'s. - -Another way to make the sum of the deviations $0$ is as follows. If $(n-1)s^2$ is even, find integers $b_1,\dots,b_{4l}$ such that -$$(n-1)s^2/2=b_1^2+\dots+b_{4l}^2$$ -(cf. (3)), where $l:=\lfloor n/8\rfloor$. Then, letting -$$(a_1,\dots,a_n):=(b_1,\dots,b_{4l},-b_1,\dots,-b_{4l},0,\dots,0),$$ -we have both (1) and the balance condition $\sum_{i\in[n]}a_i=0$ satisfied. Of course, one can also replace $(a_1,\dots,a_n)$ by any permutation of this $n$-tuple. - -REPLY [2 votes]: Alternatively, given rational numbers $u_1,\dotsc,u_n$ you can define $v=\sum_iu_i^2$ and $w_i=2u_i/(1+v)$ for $i\leq n$ and $w_{n+1}=(1-v)/(1+v)$. Then the numbers $w_j$ are rational with $\sum_jw_j^2=1$. By clearing denominators you get an equation $\sum_{j=1}^{n+1}a_j^2=b^2$ in integers.<|endoftext|> -TITLE: Does $0 -TITLE: Looking for a reference on the Euler characteristic of the manifold of fixed rank matrices -QUESTION [9 upvotes]: Let $\mathcal{M}_r$ be the set of $n \times m$ matrices over $\mathbb{R}$ or $\mathbb{C}$ of rank $r$. What is the Euler characteristic of $\mathcal{M}_r$? Can someone point me towards a reference for this calculation? - -REPLY [14 votes]: $\mathcal M_r$ can be described as a fiber bundle over the product $Gr(r,n) \times Gr(r,m)$ of the Grassmanian of $r$-dimensional subspaces in an $n$-dimensional vector space with the Grassmanian of $r$-dimensional subspaces in an $m$-dimensional vector space, where the fibers are all isomorphic to $GL_r$. (These are vector spaces over the field $\mathbb C$ or $\mathbb R$ as appropriate.) So its Euler characteristic is $$\chi( Gr(r,n) ) \chi( Gr(r,m)) \chi(GL_r).$$ -Over the complex numbers $\chi ( GL_r(\mathbb C))=0$ for $r>0$, making the product zero, and for $r=0$ the space $\mathcal M_r$ is a point, with Euler characteristic $1$. -Over the real numbers, $\chi(GL_r(\mathbb R))$ may be calculated by observing that $GL_r (\mathbb R)$ maps to $\mathbb R^r - \{0\}$ by a fibration whose fibers are all $\mathbb R^{r-1}$-bundles over $GL_{r-1}(\mathbb R)$, so $$ \chi(GL_r(\mathbb R))=\chi(GL_{r-1}(\mathbb R)) \chi ( \mathbb R^r - \{0 \} ) $$ which is $0$ for $r \geq 2$ since $\mathbb R^2 - \{0\}$ has Euler characteristic zero. -So the product vanishes for $r \geq 2$ and is $1$ for $r=0$. -Finally, for $r=1$, $\chi(GL_1(\mathbb R))=2$, and $Gr(1,n) = \mathbb R \mathbb P^{n-1}$ which has Euler characteristic $\frac{ 1 + (-1)^{n-1} }{2}$, so $$\chi (\mathcal M_1) = \frac{ (1+ (-1)^{n-1} ) (1+ (-1)^{m-1} ) }{ 2} $$ and is always even, as Arun Debray noted.<|endoftext|> -TITLE: Biased random Fibonacci sequences -QUESTION [11 upvotes]: I have recently been toying (very superficially) with the random Fibonacci sequence, i.e., defined by $F_0=1=F_1=1$ and -$$ -F_{n} = F_{n-1} + \varepsilon_n F_{n-2} -$$ -where $(\varepsilon_n)_{n\geq 2}$ is a sequence of i.i.d. Rademacher random variables. -It is known that this sequence is such that -$$ -|F_n|^{1/n} \xrightarrow[n\to\infty]{\rm a.s.} \nu -$$ -where $\nu \approx 1.13$ is the Viswanath constant (D. Viswanath, 1999). -I am interested in the case where the $(\varepsilon_n)_{n\geq 2}$ are no longer uniform in $\{-1,1\}$, but instead $1$ with probability $p\in[0,1]$ (so the standard Fibonacci sequence corresponds to $p=1$, and the standard random Fibonacci sequence to $p=1/2$). Then my understanding is that the convergence proof goes through, i.e., -$$ -|F_n|^{1/n} \xrightarrow[n\to\infty]{\rm a.s.} \nu(p) -$$ -I am interested in the behavior of $\nu(p)$ as a function of $p$. Is there anything known about it? -Through some basic simulations, the behavior seems very regular, but I am not sure what else to infer from it.$\nu(p)$ as a function of p" /> -Note that the paper by D. Viswanath does (briefly) discuss a biased version (see, e.g., Figure 5); however, the recurrence considered there is of the form $F_{n} = \varepsilon_n F_{n-1} + \varepsilon_n' F_{n-2}$ where $\varepsilon'_n,\varepsilon_n$ are i.i.d. The result is the same (because of the absolute values) when $p=1/2$, but it is not the case for $p\neq 1/2$. - -REPLY [6 votes]: This example is discussed in my 1990 PhD thesis, where is it is shown that $\nu(p)$ is a real analytic function of $p$. The thesis was written in Hebrew, but this result was generalized in two papers [1], [2]. Note that the existence of the limit you call the Viswanath constant (and more generally, $\nu(p)$) is an immediate consequence of the Furstenberg-Kesten Theorem [3] on the existence of the top Lyapunov exponent for random matrix products. This is because the vector $(F_n,F_{n+1})$ is obtained by multiplying $(F_{n-1},F_n)$ by a random matrix that takes two possible values. -[1] Peres, Yuval Analytic dependence of Lyapunov exponents on transition probabilities. Lyapunov exponents (Oberwolfach, 1990), 64–80, Lecture Notes in Math., 1486, Springer, Berlin, 1991. -[2] Peres, Yuval Domains of analytic continuation for the top Lyapunov exponent. Ann. Inst. H. Poincaré Probab. Statist. 28 (1992), no. 1, 131–148. (Reviewer: Philippe Bougerol) http://www.numdam.org/article/AIHPB_1992__28_1_131_0.pdf -[3] Furstenberg, Harry, and Harry Kesten. "Products of random matrices." The Annals of Mathematical Statistics 31, no. 2 (1960): 457-469.<|endoftext|> -TITLE: Complete Boolean algebras of subsets of $\mathbb N$ -QUESTION [5 upvotes]: Let $\mathfrak A$ be a subset of $\mathrm{Pow}(\mathbb N)$, the powerset of $\mathbb N$. Assume that $\mathfrak A$ is a complete Boolean algebra in the induced order, i.e., the inclusion order. Does it follow that $\mathfrak A$ is atomic? -A complete Boolean algebra $\mathfrak A$ is said to be atomic in case every nonzero element $A \in \mathfrak A$ is above a minimal nonzero element. We do not assume that the suprema and infima in $\mathfrak A$ are also those in $\mathrm{Pow}(\mathbb N)$. However, in the case of interest, $A_1 \wedge A_2 = 0$ does imply that $A_1 \cap A_2 = \emptyset$, for all $A_1, A_2 \in \mathfrak A$. - -REPLY [14 votes]: The answer is negative. Let $A$ be the completion of the denumerable atomless BA $B$. Then $A$ is complete and atomless. $A$ can be isomorphically embedded in $\mathrm{Pow}(\omega)$. In fact, $B$ can be isomorphically embedded in $\mathrm{Pow}(\omega)$, and by Sikorski's extension theorem, this embedding can be extended to an embedding of $A$ into $\mathrm{Pow}(\omega)$. -$B$ can be embedded in $\mathrm{Pow}(\omega)$ because $\mathrm{Pow}(\omega)$ has an independent subset of size $\omega$. Even of size $2^\omega$.<|endoftext|> -TITLE: Can every element of a homotopy group of a smooth manifold be represented by an immersion? -QUESTION [13 upvotes]: I originally posted this on MSE but didn't get much of a response, so I'll attempt to post it here. Let me know if this is not appropriate. -Let $M$ be a smooth manifold of dimension $n$. Let $\alpha \in \pi_i (M)$, for some $i \leq n$, and $f:S^i \to M$ be a map of the sphere representing the homotopy class $\alpha$. Can $f$ always be chosen to be an immersion? I.e. does every homotopy class of the maps from the sphere $S^i$ to $M$ contain an immersion? -In the comments of my original post, it is claimed that it doesn't always work when $i=n$, and that it should work for small enough $i$. What about for other values in between? - -REPLY [8 votes]: There exists a simply-connected closed $6$-manifold $M$ with a homotopy class $\alpha\in \pi_4(M)$ which does not contain an immersion. The following argument is due to Diarmuid Crowley, after we realized that my argument with Stiefel-Whitney classes could not produce examples. -According to Wall, -Wall, C. T. C., Classification problems in differential topology. V: On certain 6- manifolds, Invent. Math. 1, 355-374 (1966); Corrigendum. Ibid. 2, 306 (1967). ZBL0149.20601 -there exists a $1$-connected closed $6$-manifold $M$ such that $H^*(M)\cong H^*(\mathbb{C}P^3)$, the cup product $H^2(M)\times H^2(M)\to H^4(M)$ is trivial, and the first Pontryagin class $p_1(M)\in H^4(M)$ is non-zero. The relevant classification result is Theorem 5 in the linked paper. -Claim: The Hurewicz map $\pi_4(M)\to H_4(M)$ is onto. -Proof: $M$ has the homotopy type of a CW-complex with one cell in each even dimension betoween $0$ and $6$. Since the cup square of the generator in $H^2(M)$ is trivial, it follows that the attaching map $S^3\to S^2$ of the $4$-cell has trivial Hopf invariant, therefore is trivial. So $M^{(4)}\simeq S^2\vee S^4$. The claim follows. -Now let $\alpha\in \pi_4(M)$ be a class whose Hurewicz image is a generator $x\in H_4(M)$. Suppose $\alpha$ is represented by an immersion $f:S^4\looparrowright M$. The normal bundle of $f$ is a rank $2$ bundle $\nu(f)$ over $S^4$, hence is trivial. Also, $TS^4$ is stably trivial. The isomrphism of bundles $\nu(f)\oplus TS^4 \cong f^*(TM)$ and naturality of Potryagin classes now implies that $f^*(p_1(M))=0\in H^4(S^4)$. Thus -$$ -0=\langle f^*(p_1(M)),[S^4]\rangle = \langle p_1(M),f_*[S^4]\rangle = \langle p_1(M),x\rangle -$$ which implies $p_1(M)=0$, a contradiction.<|endoftext|> -TITLE: A diffeomorphism of the torus with constant singular values -QUESTION [7 upvotes]: Let $\mathbb{T}^2=\mathbb{S}^1 \times \mathbb{S}^1$ be the flat $2$-dimensional torus, and let $0<\sigma_1 < \sigma_2$ satisfy $\sigma_1 \sigma_2=1$. -Does there exist an area-preserving diffeomorphism $f:\mathbb{T}^2 \to \mathbb{T}^2$ whose singular values are constant $\sigma_1 , \sigma_2$? - -An immediate family of such diffeomorphisms which comes to mind are the affine (geodesic-preserving) maps which are induced by elements of $SL_2(\mathbb{Z})$. However, this family does not cover the entire range of pairs $\{ (\sigma_1,\frac{1}{\sigma_1}) \, | \, \sigma_1 \in (0,1) \}$, since it's countable. Furthermore the set of $\sigma_1$ which are admissible in this affine family is discrete away from zero, which is its only accumulation point. -  -Are there any non-affine examples? - -Edit: -Robert Bryant has given an answer which shows that there is no non-affine $C^3$ example. I wonder what happens if we allow reduced regularity, say Lipschitz maps whose differential has a.e. the singular values $\sigma_1 , \sigma_2$. - -REPLY [7 votes]: There are no non-affine examples with $f$ smooth (or even $C^3$). This follows from the fact that such an $f:\mathbb{T}^2\to\mathbb{T}^2$ would lift to a smooth (local) diffeomorphism $F:\mathbb{R}^2\to\mathbb{R}^2$ with constant singular values, and the argument I gave in my answer to this question implies that such a map $F$ would have to be affine.<|endoftext|> -TITLE: Representation theory of $\operatorname{SO}(n)$ for large $n$ -QUESTION [9 upvotes]: $\DeclareMathOperator{\SO}{\operatorname{SO}}\DeclareMathOperator{\SU}{\operatorname{SU}}$Background: In the quantum field theory literature people commonly consider "expansions" of $\SO(n)$ or $\SU(n)$ field theories in $1/n$ (see this wikipedia article). While this essentially restricts to an expansion on the level of perturbative QFT, more quantitative approaches exist for Spin-$\mathrm{O}(n)$ models, by comparison with spherical models. The general idea is that in the limit of large $n$, the field theory becomes Gaussian. -I was hoping to gain additional understanding into this "large spin-dimension limit" by studying the representation theory for $\SO(n)$ in the limit of large $n$. Morally, I would like to know if/how the group and representation theory of $\SO(n)$ "simplifies" for large $n$. More precisely, I am hoping for (1) bounds on Clebsch-Gordan coefficients in terms of $1/n$ and (2) some form of "approximate commutativity". In general, however, I have the following broad question: -Question: What is known on the asymptotic behavior of $\SO(n)$ as $n\to \infty$ in terms of its representation and group theory? - -REPLY [16 votes]: In this answer I'll focus on the representation theory of $SO(n)$ as $n \to \infty$ (rather than the group theory). Strictly speaking, I want to discuss $O(n)$ rather than $SO(n)$, but I hope that it's good enough for your purposes. There are two approaches that I would like to discuss. Both of them involve a category that "interpolates" representations of $O(n)$ as $n$ varies, and in both cases, the objects that play the role of irreducible representations are indexed by "irreducible representations of $O(n)$" as $n \to \infty$ (depending on your preferred conventions, these can be thought of as partitions of any size). -Also, I don't know what is meant by "approximate commutativity", but if you could tell me, I might be able to comment. -Approach number 1: Deligne Categories -Reference: Representation theory in complex rank, II - Etingof, see also Deligne Categories in Lattice Models and Quantum Field Theory, or Making Sense of O(N) Symmetry with Non-integer N - Binder and Rychkov -In this approach, we construct a category using a symbol $V$ which is supposed to be the usual representation of $O(n)$ on $\mathbb{C}^n$, but we don't specify what $n$ is. In more detail, consider a category whose objects are indexed by natural numbers $r$, and denoted $V^{\otimes r}$. The morphisms $\hom(V^{\otimes r}, V^{\otimes s})$ are supposed to imitate representations of orthogonal groups, where these hom-spaces are spanned by Brauer diagrams with a composition rule that involves $n$ as a parameter. For example, the endomorphism algebras are Brauer algebras, whose composition depends on a parameter $n$. The true hom-spaces in $Rep(O(n))$ are quotients (of the space with basis consisting of all Brauer diagrams) by the tensor ideal of negligible morphisms. In particular if you fix $r$ and $s$, then $\hom((\mathbb{C}^n)^{\otimes r}, (\mathbb{C}^n)^{\otimes s})$ is spanned by these Brauer diagrams, but if $n$ is too small (something like $n < r+s$), there may be linear dependence relations between the maps corresponding to those diagrams. For example, the endomorphism ring of $\mathbb{C}^n \otimes \mathbb{C}^n$ has three linearly independent basis vectors, due to the decomposition -$$ -\mathbb{C}^n \otimes \mathbb{C}^n = S^2(\mathbb{C}^n) \oplus \bigwedge \nolimits^2(\mathbb{C}^n) = (S^2(\mathbb{C}^n)/\mathbb{C}) \oplus \mathbb{C} \oplus \bigwedge \nolimits^2(\mathbb{C}^n), -$$ -where in the last term $\mathbb{C}$ denotes a trivial representation (embedded in $S^2(\mathbb{C}^n)$ via the invariant bilinear form), which makes $\mathbb{C}^n \otimes \mathbb{C}^n$ the sum of three non-isomorphic irreducibles. However, if $n=1$, then the first and last factors are zero, and the true endomorphism space is only one-dimensional. -The Deligne category $\underline{Rep}(O(n))$ is the Karoubian envelope of the category we've described (whose objects are $V^{\otimes n}$, morphisms are linear combinations of Brauer diagrams, and whose composition depends polynomially on the parameter $n$). I want to emphasise that $n$ can be taken to be any element of the ground ring/field, and does not have to be an integer (in contrast to the definition of orthogonal groups). This Karoubian envelope is a formal completion that allows us to take direct sums and direct summands. -This should be thought of by analogy to compact Lie groups where every irreducible can be found as a summand of $V^{\otimes n} \otimes (V^*)^{\otimes m}$ for some $m,n \in \mathbb{N}$, where $V$ is a faithful representation of the Lie group in question. For us, $V$ (mimicking $\mathbb{C}^n$) is self-dual because we're working with orthogonal groups, so including tensor powers of $V^*$ is redundant. -The indecomposable objects of the Deligne category are indexed by primitive idempotents in the Brauer algebras, which means they can be thought of as being indexed by "irreducible representations of $O(n)$ as $n \to \infty$". These objects can have a dimension associated to them using the formalism of tensor categories, which turns out to be a polynomial interpolating the usual dimensions of irreducible objects. For example, $\bigwedge \nolimits^2 (V)$ is a summand of $V \otimes V$ whose dimension is ${n \choose 2}$. -When $n$ is a positive integer, there's a functor from $\underline{Rep}(O_n)$ to $Rep(O(n))$ which sends $V^{\otimes r}$ to $(\mathbb{C}^n)^{\otimes r}$ and each Brauer diagram to the map of $O(n)$ representations that it ordinarily defines. The main way that $n$ enters the story is via composition of morphisms, and idempotents in the Brauer algebras (which have some kind of normalising factor that is a rational function of $n$). -Since you've specifically asked about asymptotics as $n \to \infty$, I think this might be relevant to you. Calculation of particular Clebsch-Gordan coefficients might be difficult because they ultimately depend on a choice of basis of the representations in question, while we are working with hom-spaces that are insensitive to choices of bases in representations. However, if you are able to phrase your exact question in terms of $\underline{Rep}(O(n))$, it seems likely that you could easily show that the coefficients that come our are rational functions of $n$. -Approach number 2: Stable Representation Categories. -Reference: Stability patterns in representation theory - Sam and Snowden -In this approach, one works explicitly with the group $O(\infty)$. If $GL(\infty)$ is the group of $\infty \times \infty$ matrices differing from the identity matrix in finitely many entries, then $GL(\infty)$ acts on $\mathbb{C}^\infty$ in the usual way. Then $O(\infty)$ is the subgroup of $GL(\infty)$ preserving a non-degenerate symmetric bilinear form on $\mathbb{C}^\infty$. This group is "too big" to understand all its representations, so we restrict to the category $Rep(O(\infty))$ of subquotients of direct sums of tensor powers of $\mathbb{C}^\infty$. -This is analogous to the statement about all irreducibles of a compact Lie group appearing in $V^{\otimes n} \otimes (V^*)^{\otimes m}$ for some $m,n \in \mathbb{N}$. In the case of $O(n)$, $\mathbb{C}^n$ is self-dual so there's no need to consider $(\mathbb{C}^n)^*$. While this reasoning doesn't apply to $O(\infty)$ (e.g. it's infinite dimensional so it's not a Lie group, $\mathbb{C}^\infty$ isn't self-dual because it's infinite dimensional), hopefully it's sufficient to convince you that $Rep(O(\infty))$ is a sensible category to consider. -The key difference between representations of $O(n)$ and representations of $O(\infty)$ is that there are fewer maps of representations than you might expect. For example, for $O(n)$, you have an embedding of the trivial representation in $\mathbb{C}^n \otimes \mathbb{C}^n$ given essentially by scalar multiples of the bilinear form preserved by $O(n)$. So for the usual bilinear form $\langle (x_i), (y_i) \rangle = \sum_i x_i y_i$, the invariant vector would be $\sum_i e_i \otimes e_i$ (where $e_i \in \mathbb{C}^n$ is the $i$-th basis vector). This fails for $O(\infty)$ because $\sum_i e_i \otimes e_i$ is now an infinite sum and so doesn't define an element of $\mathbb{C}^\infty \otimes \mathbb{C}^\infty$. Nevertheless there is a quotient map from $\mathbb{C}^\infty \otimes \mathbb{C}^\infty$ to the trivial representation given by applying the invariant bilinear form. So the trivial representation is a quotient, but not a submodule of $\mathbb{C}^\infty \otimes \mathbb{C}^\infty$, which illustrates that $Rep(S(\infty))$ is not semisimple (in constrast to the representation categories of orthogonal groups which it is imitating). -What I suspect is of particular interest to you is the existence of "specialisation functors" $\Gamma_n: Rep(O(\infty)) \to Rep(O(n))$. This functor admits a fairly concrete realisation given roughly as follows. Fix an $n$-dimensional subspace $\mathbb{C}^n$ of $\mathbb{C}^\infty$ on which the bilinear form restricts to another nondegenerate symmetric bilinear form. Take the orthogonal complement, which is abstractly $\mathbb{C}^\infty$, but retains an action of $H_n = O(\infty)$ (which you might write as $O(\infty-n)$ to emphasise that it acts on coordinates whose indices are shifted by $n$). Then $O(n)$ and $H_n$ may be viewed as commuting subgroups of $O(\infty)$, and so the subspace of $H_n$-invariants retains an action of $O(n)$. In fact this is precisely the functor in question; $\Gamma_n(V) = V^{H_n}$. Note in particular that $\Gamma_n(\mathbb{C}^\infty) = \mathbb{C}^n$. -So this could be relevant to your situation by considering a construction in $Rep(O(\infty))$, and then applying a specialisation functor $\Gamma_n$ to land in $Rep(O(n))$. This, in a certain sense "becomes faithful as $n \to \infty$". So the construction on the level of $Rep(O(\infty))$ would be the "$n \to \infty$ limit". One reason why this might not be what you are looking for is that $n$ is not a parameter in $Rep(O(\infty))$, and maps in $Rep(O(n))$ that depend on $n$ in a nontrivial way typically don't interpolate to $Rep(O(\infty))$, so I'm not sure how you would obtain any asymptotics as $n \to \infty$.<|endoftext|> -TITLE: A canonical bijection from linear independent vectors to parking functions -QUESTION [14 upvotes]: Call an $n$-vector $v$ in $\mathbb{Z}^n$ cool when it has only entries 0 or 1 and the ones appear in only one block. Thus there are $n(n+1)/2$ such vectors. For $n=3$ they are: -[ <[ 1, 0, 0 ]>, <[ 1, 1, 0 ]>, <[ 0, 1, 0 ]>, <[ 1, 1, 1 ]>, <[ 0, 1, 1 ]>, <[ 0, 0, 1 ]> ]. -Let $X_n$ be the set of cool $n$-vectors. Call a subset $U \subset X_n$ cool when $U$ has $n$ elements that are linearly independent. -There should be $(n+1)^{n-1}$ cool subsets of $X_n$. For $n=3$ they are: -[ [ <[ 1, 0, 0 ]>, <[ 1, 1, 0 ]>, <[ 1, 1, 1 ]> ], -[ <[ 1, 0, 0 ]>, <[ 1, 1, 0 ]>, <[ 0, 1, 1 ]> ], -[ <[ 1, 0, 0 ]>, <[ 1, 1, 0 ]>, <[ 0, 0, 1 ]> ], -[ <[ 1, 0, 0 ]>, <[ 0, 1, 0 ]>, <[ 1, 1, 1 ]> ], -[ <[ 1, 0, 0 ]>, <[ 0, 1, 0 ]>, <[ 0, 1, 1 ]> ], -[ <[ 1, 0, 0 ]>, <[ 0, 1, 0 ]>, <[ 0, 0, 1 ]> ], -[ <[ 1, 0, 0 ]>, <[ 1, 1, 1 ]>, <[ 0, 0, 1 ]> ], -[ <[ 1, 0, 0 ]>, <[ 0, 1, 1 ]>, <[ 0, 0, 1 ]> ], -[ <[ 1, 1, 0 ]>, <[ 0, 1, 0 ]>, <[ 1, 1, 1 ]> ], -[ <[ 1, 1, 0 ]>, <[ 0, 1, 0 ]>, <[ 0, 1, 1 ]> ], -[ <[ 1, 1, 0 ]>, <[ 0, 1, 0 ]>, <[ 0, 0, 1 ]> ], -[ <[ 1, 1, 0 ]>, <[ 1, 1, 1 ]>, <[ 0, 1, 1 ]> ], -[ <[ 1, 1, 0 ]>, <[ 0, 1, 1 ]>, <[ 0, 0, 1 ]> ], -[ <[ 0, 1, 0 ]>, <[ 1, 1, 1 ]>, <[ 0, 1, 1 ]> ], -[ <[ 0, 1, 0 ]>, <[ 1, 1, 1 ]>, <[ 0, 0, 1 ]> ], -[ <[ 1, 1, 1 ]>, <[ 0, 1, 1 ]>, <[ 0, 0, 1 ]> ] ] - -Question: Is there a canonical bijection from cool subsets of $X_n$ to parking functions (that are counted by the same number $(n+1)^{n-1}$)? - -Background: The cool vectors correspond to the indecomposable representations of the $A_n$-quiver algebra $A$ and the cool subsets to the bases of the Grothendieck group $K_0(A)$ of $A$. I'm interested in a "canonical" bijection to parking functions to enter some statistics from homological algebra into findstat : findstat.org that has several statistics and maps for parking functions. -I can not really say what canonical means but it should behave nice under some standard statistics from homological algebra. -For example for such a canonical bijection, the number of simple vectors (those having only one non-zero entry) or the number of projective vectors (those having the last entry nonzero) in U should probably correspond to something nice for parking functions. - -REPLY [19 votes]: They are in canonical bijection with the spanning trees of the complete graph $K_{n+1}$ (for which the bijections with parking functions are well known). -Indeed, let $K_{n+1}$ be the complete graph on the ground set $\{0,1,\ldots,n\}$. Denote $f_0=0$ and consider $n$ linearly independent vectors $f_1,\ldots,f_n$. Denote further $e_j=f_j-f_{j-1}$ for $j=1,\ldots,n$. They form another basis of the same $n$-dimensional space $W$ as $f_j$'s. For an edge $\epsilon=ij$, $i -TITLE: Closedness of linear image of positive L1 functions -QUESTION [9 upvotes]: Let $\mathcal X$ be the Banach space of $L^1$ functions on some probability space, $\mathcal Y$ be some other Banach space, $T:\mathcal X\to \mathcal Y$ be some surjective continuous linear map, $\mathcal X_+$ be the set of all elements of $\mathcal X$ with a nonnegative version (a closed convex cone), and $\mathcal Y_+:=T(\mathcal X_+)$ (a convex cone). -Question: - -Is $\mathcal Y_+$ necessarily closed in $\mathcal Y$? -If not, are there nice, easily verifiable conditions on $\mathcal Y_+$ that are sufficient for it to be closed? -Do either of these answers change if I'm willing to assume that $\mathcal Y$ is itself a closed subspace of some $L^1$ space, and $T$ is positive? - -REPLY [9 votes]: Take $\mathcal X = L^1(\Omega,\mu)$ where $\Omega = \{1,2,3,\dots\}$ and measure $\mu(\{k\}) = p_k$ with $p_k > 0$, -$\sum p_k = 1$. The norm in $\mathcal X$ is -$$ -\|f\|_{\mathcal X} = \sum_k |f(k)|p_k . -$$ -Let $\mathcal Y = \mathbb R^2$ with norm -$$ -\|(x,y)\|_{\mathcal Y} = \frac{1}{2}|x|+\frac{1}{2}|y|. -$$ -Thus $\mathcal Y$ is also $L^1$ of a probability space. -Let $(t_k)_{k=1}^\infty$ be a sequence or reals in $(0,1)$, so that $t_k \to 0$. We may assume $t_1 = 2/3, t_2 = 1/3$. Define $T : \mathcal X \to \mathcal Y$ as follows. -Suppose $f \in \mathcal X$; that is $\sum_k |f(k)| p_k < \infty$. Then define -$$ -T(f) = \left(\sum_k f(k)p_k t_k , \sum_k f(k)p_k (1-t_k)\right) \in \mathcal Y. -$$ -Then: -$\bullet\quad$ $T$ is linear -$\bullet\quad$ $T$ is bounded -$\bullet\quad$ $T$ is positive -$\bullet\quad$ for each $k \in \mathbb N$, we have $(t_k,1-t_k) \in T(\mathcal X_+)$ -$\bullet\quad$ $(2/3,1/3),(1/3,2/3) \in T(\mathcal X)$, so $T(\mathcal X) = \mathcal Y$ -$\bullet\quad$ $(0,1) \notin T(\mathcal X_+)$ -Thus the convex cone $T(\mathcal X_+)$ is not closed in $\mathcal Y$.<|endoftext|> -TITLE: Can the strongest Hensel lemma over integer rings imply smoothness over $\mathbb Z_p$? -QUESTION [5 upvotes]: Let $X \rightarrow \mathbb Z_p$ be a flat finite type morphism, with reduced special fiber and smooth generic fiber. -Assume $X(O_K) \rightarrow X(O_K/m_K)$ is surjective for all fintie extension $K$ of $\mathbb Q_p$. Then $X \rightarrow \mathbb Z_p$ can still be non-smooth, an example is $X=\mathbb Z_p[x,y,z]/(xy-p^2z)$. -Now assume $X(O_K/m_K^{n+1}) \rightarrow X(O_K/m_K^n)$ is surjective with same size fibers for all $n$ and all $K$ (the strongest Hensel lemma), then is there a counterexample that $X \rightarrow \mathbb Z_p$ is non-smooth? - -REPLY [4 votes]: This forces $f \colon X \to \operatorname{Spec} \mathbf Z_p$ to be smooth. In fact, we will only use the criterion for $n = 1$ and $K = \mathbf Q_q(\sqrt{p})$ the ramified extension of ramification index $2$ with residue field $\mathbf F_q$. Note that -$$\mathcal O_K/\mathfrak m_K^2 = \mathcal O_K/(p) \cong \mathbf F_q[\varepsilon]/(\varepsilon^2).$$ -We first show that the special fibre $X_0$ is smooth over $\mathbf F_p$. Indeed, if $x \in X_0(\mathbf F_q)$ is an $\mathbf F_q$-point, then the fibre of $X(\mathbf F_q[\varepsilon]/(\varepsilon^2)) \to X(\mathbf F_q)$ above $x$ is $\mathfrak m_x/\mathfrak m_x^2 \otimes_{\kappa(x)} \mathbf F_q$. Since these have the same size for any $q$ and any $x \in X_0(\mathbf F_q)$, we conclude that $\dim_{\kappa(x)} \mathfrak m_x/\mathfrak m_x^2$ is the same for all closed points $x \in X_0$, hence $X_0$ is smooth by Tags 00TR (which also holds over any perfect field by Tag 00TU) and 00NX. -Since $f$ is flat and $f_0 \colon X_0 \to \operatorname{Spec} \mathbf F_p$ is smooth, we conclude that $f$ is smooth, at least if $X$ has no irreducible components lying over the generic point of $\operatorname{Spec} \mathbf Z_p$. (For example, the smooth locus is open and contains the special fibre by Tag 02V4, so it must be everything.) -Remark. The only fact we needed about $\mathbf Z_p$ is that it is a DVR with perfect residue field. We used flatness of $f$, but we don't need to assume a priori that $X_0$ is reduced. -Of course if $X$ has components over $\mathbf Q_p$, we cannot say anything about those as they are not detected by $X(\mathcal O_K/\mathfrak m_K^n)$.<|endoftext|> -TITLE: Is there a real nonintegral number $x >1$ such that $\lfloor x^n \rfloor$ is a square integer for all $n \in \mathbb{N}$? -QUESTION [15 upvotes]: This question was inspired by the following: -https://math.stackexchange.com/questions/3882691/lfloor-xn-rfloor-lfloor-yn-rfloor-is-a-perfect-square -Is there a real nonintegral $x>1$ s.t. $\lfloor x^n \rfloor$ is square integer for all positive integers $n$? I am asking because the question is interesting in and of itself, but also because the proof techniques should also be interesting. - -REPLY [20 votes]: At the request of the OP, I am turning my comment into an answer. It is possible to have $\lfloor x^{n} \rfloor$ close to a square for all positive integers $n$. For example, if $x = \frac{7 + 3 \sqrt{5}}{2} = \phi^{4}$, where $\phi = \frac{1 + \sqrt{5}}{2}$, then $\lfloor x^{n} \rfloor + 3$ is a square for every positive integer $n$. For all integers $k$, $\phi^{k} + \left(-\frac{1}{\phi}\right)^{k} = L_{k}$, the $k$th Lucas number. Squaring this identity shows that $\phi^{2k} + 2 \cdot (-1)^{k} + \left(-\frac{1}{\phi}\right)^{2k} = L_{k}^{2}$ and so $L_{k}^{2} = L_{2k} + 2 \cdot (-1)^{k}$. -Thus -$$ - \lfloor x^{n} \rfloor + 3 = x^{n} + x^{-n} + 2 = L_{4n} + 2 = L_{2n}^{2} -$$ -is the square of an integer.<|endoftext|> -TITLE: Is this notion of finiteness closed under unions? -QUESTION [13 upvotes]: This was asked and bountied at MSE without success. -Throughout, we work in $\mathsf{ZF}$. -Say that a set $X$ is $\Pi^1_1$-pseudofinite if for every first-order sentence $\varphi$, if $\varphi$ has a model with underlying set $X$ then $\varphi$ has a finite model. (See here, and the answer and comments, for background.) Every $\Pi^1_1$-pseudofinite set is Dedekind-finite basically trivially, and with some model theory we can show that every amorphous set is $\Pi^1_1$-pseudofinite. Beyond that, however, things are less clear. -In particular, I noticed that I can't seem to prove a very basic property of this notion: - -Is the union of two $\Pi^1_1$-pseudofinite sets always $\Pi^1_1$-pseudofinite? - -I'm probably missing something simple, but I don't see a good way to get a handle on this. A structure on $X=A\sqcup B$ might not "see" that partition at all, and so none of the simple tricks I can think of work. - -REPLY [3 votes]: No, that class doesn't need to be closed under unions. I’ll describe a permutation model with two $\Pi_1^1$-pseudofinite sets whose disjoint union is not $\Pi_1^1$-pseudofinite. You can use Jech-Sochor to get a ZF model. -Fix a finite field $K.$ Consider the class of tuples $M=(X^M,Y^M,e^M)$ such that $X^M$ and $Y^M$ are finite $K$-vector spaces, and $e^M$ is a bilinear map $X^M\times (K\oplus Y^M)\to K.$ This data can be encoded in a language $\mathcal L.$ I think any encoding would be slightly unwieldy so I’ll just call these 3-tuples $\mathcal L$-structures. -I claim this class satisfies the conditions of Fraïssé's theorem. There is an initial object “$0$” defined by $X^0=Y^0=\{0\}$ and $e^0(0,(\lambda,0))=0.$ So the joint embedding property will follow from amalgamation. For amalgamating $A\to B$ and $A\to C,$ by choosing a splitting $X^B\cong X^A\oplus X_1$ etc we can assume $B$ is $(X^A\oplus X_1,Y^A\oplus Y_1,e^B)$ and $C$ is $(X^A\oplus X_2,Y^A\oplus Y_2,e^C),$ with the embedding maps being the direct sum inclusions. An amalgamation $D$ is defined by $X^D=X^A\oplus X_1\oplus X_2$ and $Y^D=Y^A\oplus Y_1\oplus Y_2,$ with the direct sum inclusions as embeddings from $B$ and $C,$ and -$$e^D((x_0,x_1,x_2),(\lambda,(y_0,y_1,y_2)))=e^B((x_0,x_1),(\lambda,(y_0,y_1)))+e^C((x_0,x_2),(\lambda,(y_0,y_2)))-e^A(x_0,(\lambda,y_0))$$ -The Fraïssé limit of this class gives us a structure $L.$ I’ll drop the superscripts so $(X,Y,e)=(X^L,Y^L,e^L).$ The theory $T_L$ of $L$ is $\omega$-categorical and, since Fraïssé limits are ultrahomogenous, $T_L$ has quantifier elimination. -Let $N$ be the permutation model with atoms $X\cup Y,$ automorphism group the $\mathcal L$-automorphisms, with open subgroups $G_{\bar{s}}$ for each $\bar{s}\in (X\cup Y)^{<\omega},$ consisting of the automorphisms fixing $\bar{s}.$ I’ll always argue externally, using ZFC. -A relation $R\subseteq X^n$ in $N$ is fixed by some $G_\bar{s}.$ I claim that $R$ is definable in $L$ with parameters $\bar{s}.$ Because $T_L$ is $\omega$-categorical, there is a partition of $X^n$ into sets $X_1,\dots,X_r,$ each defined by a formula with parameters $\bar{s},$ such that any two elements within the same part $X_j$ have the same complete type over $\bar{s}.$ For any $x,y\in X_j$ there is an automorphism $\pi\in G_{\bar{s}}$ with $\pi x=y,$ and hence $x\in R \iff y\in R.$ By taking a conjunction, $R$ is definable with parameters $\bar{s}.$ -By quantifier elimination, $R$ is definable by a quantifier free formula. This formula will be in the language $\mathcal L’$ of a $K$-vector space with constants for each $x\in X_{\bar s}:=X\cap \operatorname{rng}(\bar s),$ and (suitably encoded) unary functions $e_y(x)=e(x,y),$ for $y\in Y_{\bar s}:=Y\cap \operatorname{rng}(\bar s).$ The true theory of $X$ in this language is the theory of an infinite vector space with a finite number of constants and certain linear functionals. -I will argue that this is a pseudofinite theory. For any $n,$ pick a finite set of vectors $x\in X$ attaining each realizable combination of values for $(e_y(x))_{y\in Y_{\bar s}},$ and $n$ vectors in $X$ linearly independent from these choices and from $X_{\bar s}.$ Call the span of these vectors $X’.$ Duplicator can win the $n$-round Ehrenfeucht–Fraïssé game played on $X$ and $X’$ in the language $\mathcal L’$; at each round the choice is either forced by a linear dependency, or we can pick a vector linearly independent from previous choices with the right combinations of $(e_y(x))_{y\in Y_{\bar s}}.$ -The above arguments show that $X$ is $\Pi_1^1$-pseudofinite. A similar argument shows that $Y$ is $\Pi_1^1$-pseudofinite. -In $N,$ the set $X\cup Y$ is not $\Pi_1^1$-pseudofinite because $T_L$ satisfies the non-degeneracy conditions -$$(\forall x\in X)(\exists y\in Y) f(x,(0,y))\neq 0\vee x=0$$ -$$(\forall \lambda\in K)(\forall y\in Y)(\exists x\in X)f(x,(\lambda,y))\neq 0\vee (\lambda,y)=(0,0).$$ -These force any $\mathcal L$-structure $M$ to satisfy $\dim Y^M\geq \dim X^M\geq 1+\dim Y^M.$<|endoftext|> -TITLE: Bound for $GL(3)$ symmetric square -QUESTION [14 upvotes]: Let $\pi$ be an automorphic representation of $GL(3)$ over a number field. Let $a_n$ be the coefficients of $L(s, \pi, \mathrm{sym}^2)$. Do we know if -$$\sum_{n>0} \frac{|a_n|}{n^s}$$ -and -$$\sum_{n>0} \frac{\left( \sum_{k \mid n} |a_k|\right)^2}{n^s}$$ -converge for $\Re(s)>1$? or even for $\Re(s)>1+\delta$ for a quite small $\delta$? -This essentially amounts to say that the coefficients are constant on average. It seems to be known for Gelbart-Jacquet lifts by Rankin-Selberg properties, but is it known or at least expected for non-Gelbart-Jacquet lifts? - -REPLY [4 votes]: Let $\pi$ be a cuspidal automorphic representation of $\mathrm{GL}_m$ with unitary central character. In work of Takeda, it is shown that the (unramified part of the) $L$-function $L(s,\pi,\mathrm{Sym}^2)$ is holomorphic in the region $\mathrm{Re}(s) > 1-\frac{1}{2m}$. Taking $m=3$, we find that your first series converges for $\mathrm{Re}(s)>1$. -The second series is obviously a little trickier, but we can get there in the special case where $\pi$ is self-dual. In this case, $\pi$ is a Hecke character twist of the symmetric square lift of a cuspidal automorphic representation $\pi'$ on $\mathrm{GL}_2$. Suppose (for now) that $\pi=\mathrm{Sym}^2\pi'$ and $\pi$ has level 1 (so $\pi'$ is nondihedral). Then $\mathrm{Sym}^2\pi = 1\boxplus \mathrm{Sym}^4 \pi'$, in which case -$L(s,\pi,\mathrm{Sym}^2) = \zeta_F(s) L(s,\mathrm{Sym}^4\pi')$. -Note that $\mathrm{Sym}^4\pi'$ is a cuspidal automorphic representation of $\mathrm{GL}_5$. Thus $\Pi = 1\boxplus \mathrm{Sym}^4 \pi'$ is an automorphic representation of $\mathrm{GL}_6$, and the $L$-function -$L(s,\Pi\times\tilde{\Pi}) = \zeta_F(s)L(s,\mathrm{Sym}^4\pi')^2 L(s,\mathrm{Sym}^4\pi'\times \mathrm{Sym}^4\pi')$ -converges absolutely for $\mathrm{Re}(s)>1$. The $n$-th Dirichlet coefficient $\lambda_{\Pi\times\tilde{\Pi}}(n)$ is bounded below by $|a_n|^2$. A Dirichlet convolution calculation shows that if $s>1$, then the series is bounded above by -$L(s,\Pi\times\tilde{\Pi})\zeta_F(s)^2$ -(or maybe 2 needs to be replaced with a higher power?), which converges absolutely for $\mathrm{Re}(s)>1$. -Takeda, Shuichiro, The twisted symmetric square (L)-function of (\mathrm{GL}(r)), Duke Math. J. 163, No. 1, 175-266 (2014). ZBL1316.11037.<|endoftext|> -TITLE: Geometric intuition for $R[x,y]/ (x^2,y^2)$, kinematic second tangent bundle, and Wraith axiom -QUESTION [6 upvotes]: This is a sort of continuation of this question. -In synthetic differential geometry (SDG), we have $D\subset R$ comprised of the second order nilpotents. The Kock-Lawvere axiom (KL axiom) implies that a function $D\times D\to R^n$ is of the form $a_0+a_1d_1+a_2d_2+a_3d_1d_2$. This is like a 2-jet without square terms $f(a)+\partial_xf |_ad_1+\partial_yf|_ad_2+\partial_{xy}f|_ad_1d_2$. -In SDG, the infinitesimal rectangle $D\times D$ represents the second tangent bundle. In light of the KL-axioms I expect the classical second tangent bundle $\mathrm T^2X=\mathrm {TT}X$ of a $C^\infty$ manifold admits the following kinematic description: elements are equivalence classes of germs of $C^\infty$ maps $I^2\to X$ where $I$ is an interval about zero, and we identify such germs if upon composing with any germ in $C_{X,x}^\infty$ the partials and mixed partials coincide. Let us call such things "microsquares". They formalize the "2-jets without square terms" above. -If correct, this kinematic description is very geometric. For instance, it allows to define the flip on $\mathrm T^2X$ by flipping the $x,y$ coordinates of $I^2$. The two maps $\mathrm T^2X\rightrightarrows \mathrm TX$ given by $\mathrm T\pi_X,\pi_{\mathrm TX}$ are respectively given by restricting a microsquare to the $x$-axis and the $y$-axis. These fiber $\mathrm T^2X$ in two different ways: the fiber of $\mathrm T\pi_X$ over a kinematic tangent $\dot \gamma$ consists microsquare which restricts to $\gamma$ on the $x$-axis, and analogously for $\pi_{\mathrm TX}$. -The vertical lift applied to the tangent bundle gives a bundle isomorphism $\mathrm T(\mathrm TX/X)\cong \mathrm TX\times_X\mathrm TX$ over $\mathrm TX$, where the LHS is the vertical bundle of the tangent bundle, i.e the kernel of $\mathrm T\pi_X$. For all vector bundles this acts by taking a kinematic tangent (to a fiber of the bundle) to its derivative (which is a vector in the fiber). -Question 1. How to geometrically interpret the vertical lift for a "vertical microsquare"? A microsquare lies in the vertical bundle if its restriction to the $x$-axis is "constant", i.e the derivative of the restriction is zero. This is like saying the associated "2-jet without square terms" has $\partial_xf|_a=0$. What is the vertical lift doing with a microsquare that only makes sense if its restriction to the $x$-axis is zero? -My question is motivated by another one about a seeming discrepancy between SDG and the classical $C^\infty$ world: - -In the $C^\infty$ world, the vertical lift $ -\mathrm T(\mathrm TX/X)\cong \mathrm TX\times_X\mathrm TX$ is defined on any vertical microsquare. There is no further requirement for also being in the kernel of $\pi _{\mathrm TX}$ (restriction of a microsquare to its $y$-axis), and I see no reason for these kernels to coincide. - -In SDG, the Wraith axiom says that a function $D\times D\to R^n$ which is constant on the axes uniquely factors through the multiplication map $D\times D\to D$. This factorization takes such a function to a tangent vector, and this is the analog of the vertical lift. The $C^\infty$ version of being constant on the axes is having the $\partial_x,\partial_y$ coefficients of the '2-jet without square terms' vanish $\partial_xf|_a=0=\partial_yf|_a$. The remaining mixed partial term indeed factors through the multiplication map because that's how Taylor series are. The point is that the Wraith axiom asks for both partials to vanish, as opposed to the vertical bundle which involves only vanishing $\partial_x$. - - -Question 2. What is going on here, geometrically? Why does SDG want both partials to vanish while the $C^\infty$ world only cares about one of the partials? -Lastly and perhaps most fundamentally: I don't understand the geometric meaning of a microsquare. I understand 2-jets since we retain the information of the Hessian, but retaining only the mixed partials - I don't get it. -Question 3. What is the geometric content of a microsquare / an element in the second tangent bundle? - -REPLY [2 votes]: I will try to address your questions, and then point to some general cartegorical phenomena that are at play here. -Answer 1/2: In the category of smooth manifolds, or a proper model of synthetic differential geometry where your base number line has negatives, the two axioms are equivalent for vector bundles. The Wraith axiom may seem stronger, but suppose you have $x: TE$ that it is over the kernel of $T\pi$ and so it splits as $(x_1,x_2):E \times_M E$ - then if it were over the kernel of both $T\pi, p_E$ you would have $x_2 = 0$, giving the Wraith axiom. Showing the other axiom holds using the Wraith axiom requires subtraction (this is why the Cockett and Cruttwell used this version of the universality of the vertical lift in their definition of tangent categories). As far as I can tell, one of the reasons the Wraith axiom was introduced in synthetic differential geometry was so that the Lie bracket on vector fields can be constructed, and given a connected the Dombrowski splitting theorem would hold $T^2M \cong T(M) \times_M T(M) \times T(M)$. -Answer 3: -I think a good way to think about these things is based on the Weil functors approach, which can be found in Natural Operations in Differential Geometry. The first tangent bundle corresponds to an action by $R[x]/x^2$, the second by $R[x,y]/(x^2,y^2)$, and the bundle of 2-jets is represented by $R[x]/x^3$ which can be seen as the equalizer of endomorphisms on $R[x,y]/(x^2,y^2)$ (the identity $id$ and the flip $c$ that flips the variables $(x,y) \mapsto (y,x)$). -Generalities on the vertical lift: -The universality of the vertical lift is interesting - I'm not sure I would call it a geometric condition, in my experience it seems more algebraic. Kirill MacKenzie showed that a lot of the properties of the vertical lift on the tangent bundle, or a vector bundle, are satisfied by the core of a double vector bundle (the core is the subbundle of the apex $E$ the projects down to $0$ on each of the side-bundles $E^H, E^V$). In fact, there's a general universal property for triple vector bundles that gives you exactly the Jacobi identities. -Strictly speaking, you don't need the local triviality properties of vector bundles. For any commutative semiring $R$, you an define the limit sketch $RBun$ (an $R$-module bundle), and a double $R$-bundle is a model of the sketch $RBun \otimes RBun$. The first thing you can do is observe that for double $R$-bundles in any complete category, you can take the core of the double $R$-bundle; if you chose a commutative ring, then you can prove the core satisfies the stronger vertical lift axiom (it satisfies the Wraith axiom by definition). -You can also see that vertical connections pop up here in a somewhat surprising way. The vertical lift can be seen as a coreflection of $R$-bundles into the category of double $R$-bundles; vertical connections are sections of this coreflection. Once again, if you chose a commutative ring $R$, this is (probably) sufficient to use an analogue of the Dombrowski splitting theorem.<|endoftext|> -TITLE: Reference request: Examples of research on a set with interesting properties which turned out to be the empty set -QUESTION [44 upvotes]: I've seen internet jokes (at least more than 1) between mathematicians like this one here about someone studying a set with interesting properties. And then, after a lot of research (presumably after some years of work), find out such set couldn't be other than the empty set, making the work of years useless (or at least disappointing), I guess. -Is this something that happens commonly? Do you know any real examples of this? -EDIT: I like how someone interpreted this question in the comments as "are there verifiably true examples of this well-known 'urban legend template'?" - -REPLY [7 votes]: The odd-order theorem states that every finite group of odd order is solvable, and the proof involves developing a very large theory explaining what the smallest counterexample looks like, and to ultimately deduce that it cannot exist. -The odd-order theorem has been formalised (pdf) in Coq, a computer theorem prover, and the formalisation is to date one of the largest bodies of formalised mathematics. This makes it appealing to AI researchers, who go and train their deep learning networks using the collection of theorems proved in the formalisation, hoping that one day computers will start to be able to compete with humans in the realm of theorem-proving. -I find it amusing that, as a consequence, these networks are being trained to recognise a whole bunch of facts about an object which doesn't exist.<|endoftext|> -TITLE: Non-homeomorphic connected $T_2$-spaces with isomorphic topology poset -QUESTION [5 upvotes]: What are examples of non-homeomorphic connected $T_2$-spaces $(X_i,\tau_i)$ for $i=1,2$ such that the posets $(\tau_1, \subseteq)$ and $(\tau_2,\subseteq)$ are order-isomorphic? - -REPLY [9 votes]: Here's an answer without familiarity with locales, which I started after reading Todd Trimble's answer (so his answer is the right one to accept). -Let $X$ be a $\mathsf{T}_1$ topological space, $\tau_X$ the lattice of open subsets, and $\Phi_X$ the opposite lattice, which identifies to the lattice of closed subsets. Let's reconstruct $X$ from $\Phi_X$. -Denote by $0$ "zero" the unique minimal element in $\Phi_X$. Let $\Phi_X^\min$ be the set of minimal elements in $\Phi_X\smallsetminus\{0\}$. Let $i$ be the map $x\mapsto\{x\}$. Since $X$ is $\mathsf{T}_1$, $i$ is a well-defined injective map $X\to\Phi_X$, and its image is exactly $\Phi_X^\min$. (This already retrieves the cardinal of $X$.) -Now we wish to retrieve the topology. Namely, I claim that for $K\subset\Phi_X^\min$, $i^{-1}(K)$ is closed if and only if there exists $F\in\Phi_X$ such that $K=\{Z\in\Phi_X^\min\,:\,Z\le F\}$. -Indeed, suppose that $i^{-1}(K)$ is closed: define $F_K=i^{-1}(K)$ (so $K=i(F_K)$): then $\{Z\in\Phi_X:Z\le F_K\}=\{\{z\}:z\in F_K\}=i(F_K)=K$. Conversely, suppose $K=\{Z\in\Phi_X:Z\le F\}$ for some $F\in\Phi_X$. So $K=\{\{z\}:z\in F\}=i(F)$, so $i^{-1}(K)=F$ is closed. -Hence, for any $\mathsf{T}_1$ topological spaces $X,Y$, every isomorphism $\tau_X\to\tau_Y$ is induced by a unique homeomorphism $X\to Y$. -This also shows that the automorphism group of $\Phi_X$ is canonically isomorphic to the self-homeomorphism group of $X$. -This works without assuming $X$ to be sober. For example, it applies for the cofinite topology, $\Phi_X$ consisting of $X$ and its finite subsets, which is non-sober as soon as $X$ is infinite. (Todd's answer also encompasses non-bijective maps, which I didn't address; soberness is then probably important. Also, there are non-$\mathsf{T}_1$ sober spaces.)<|endoftext|> -TITLE: Goodwillie derivatives of $X \mapsto \Sigma^\infty X^{\wedge n}/\Sigma_n$ -QUESTION [13 upvotes]: I'm following Arone's lectures on Goodwillie calculus from Munster 2015. There he left an exercise: - -Find $\partial_kF$ for $F: \text{Top}_* \to \text{Sp}$ given by $F(X) = \Sigma^\infty X^{\wedge n}/\Sigma_n$ where $X^{\wedge n} = X \wedge \cdots \wedge X$ and $\Sigma_n$ is the symmetric group, which acts on $X^{\wedge n}$ by permuting factors. - -In the lectures he solves the case $n = 2$, that gives $\partial_1 F \simeq \Sigma^\infty\Sigma\mathbb{RP}^\infty$, $\partial_2F \simeq \Sigma^\infty$ and $\partial_nF \simeq *$. -I do not know how to work out the general case. I know that $\partial_kF \simeq \text{hocolim }\Omega^{km}\text{cr}_kF(S^m, \dots, S^m)$. -For $k = 1$ then $\partial_1F \simeq \text{hocolim }\Omega^mF(S^m)$. My idea is to write $F(S^m) = S^m \wedge G(S^m)$ for some $G$ in order to cancel $\Omega^m$. This can be done by choosing a $\phi: (S^m)^{\wedge n} \to (S^m)^{\wedge (n-1)}$ such that $(x_1, \dots, x_n) \mapsto (x_1 + \cdots + x_m, \phi(x_1, \dots, x_m))$ is a homemorphism. In this case $(S^m)^{\wedge n}/\Sigma_n \cong S^m \wedge (S^m)^{\wedge (n-1)}/\Sigma_n$ where the $\Sigma_n$-action on $(S^m)^{\wedge (n-1)}$ depends on $\phi$. Unfortunately, I was not able to find a $\phi$ such that this action looks nice or familiar as in the case $n = 2$. -Alternatively, I tought $(S^m)\text{^}^n/\Sigma_n \cong \text{subset}_{\leq n}(\mathbb{R^m})^+ \cong S^m \wedge \text{subset}_{\leq n, 0}(\mathbb{R^m})^+$ -where $(\cdots)^+$ means compactification, $\text{subset}_{\leq n}(\cdots)$ is the space of subsets of cardinality $\leq n$ and $\text{subset}_{\leq n, 0}(\mathbb{R^m})$ is the space of those with barycenter 0. Then $\partial_1F \simeq \Sigma^\infty\text{subset}_{\leq n, 0}(\mathbb{R^\infty})$. This agrees with the previous result for $n = 2$ since any two unordered points can be identified by the line joining them ($\mathbb{RP}^\infty$) and the distance from 0 ($\Sigma$, after compactification). But - -It does not seem to be the "right solution". -What about $\partial_nF$ for $n > 1$? I do not know how to proceed. - -Thanks in advance! -[original post] - -REPLY [13 votes]: Your solution is correct as far as it goes. Let $\rho^{n-1}\cong {\mathbb R}^{n-1}$ denote the reduced standard representation of $\Sigma_n$. One way to define $\rho^{n-1}$ explicitly is to say that it is the orthogonal complement of the diagonal in $\mathbb R^n$. Let's write it out: -$$ -\rho^{n-1}=\{(x_1, \ldots, x_n)\in\mathbb R^n\mid x_1+\cdots+x_n=0\} -$$ -Let $\hat S^{n-1}$ denote the one-point compactification of $\rho^{n-1}$. This is a sphere of dimension $n-1$ with a specified action of $\Sigma_n$. More generally, for any $m\ge 0$, let us write $\hat S^{(n-1)m}$ for $(\hat S^{n-1})^{\wedge m}$. -Furthermore, $\hat S^{(n-1)m}$ is the one-point compactification of the representation $(\rho^{n-1})^m$. It follows that $\hat S^{(n-1)m}$ is homeomorphic to the unreduced suspension of the unit sphere in $(\rho^{n-1})^m$. Let me denote the unit sphere by $\tilde S^{(n-1)m-1}$. We have a $\Sigma_n$-equivariant homeomorphism $$\hat S^{(n-1)m}\cong \Sigma \tilde S^{(n-1)m-1},$$ where $\Sigma$ indicates unreduced suspension. -Now let us take colimit as $m\to \infty$. We obtain the following equivalence, where both sides are just notation for the appropriate colimits -$$\hat S^{(n-1)\infty}\cong \Sigma \tilde S^{(n-1)\infty-1}.$$ -Let $F_n(X)=\Sigma^\infty X^{\wedge n}_{\Sigma_n}$. By your calculation, $$\partial_1(F_n)\simeq \Sigma^\infty \Sigma {\tilde S^{(n-1)\infty-1}}/_{\Sigma_n}.$$ -To analyze this further, let us calculate the fixed point set ${\tilde S^{(n-1)\infty-1}}^H$, where $H\subset \Sigma_n$. The following lemma is a quite easy exercise -Lemma: Let $H$ be a subgroup of $\Sigma_n$. Define $o(H)$ to be the number of orbits of the action of $H$ on $\{1, \ldots, n\}$. There is a homeomorphism $${\tilde S^{(n-1)-1}}^H\cong S^{o(H)-2}.$$ -Note that if $H$ is a transitive subgroup of $\Sigma_n$, then $o(H)=1$ and the right hand side of this equation is $S^{-1}$, which is taken to be the empty set. -More generally for $m\ge 1$, there is a homeomorphism -$${\tilde S^{(n-1)m-1}}^H\cong S^{(o(H)-1)m-1}.$$ -Taking limit as $m\to \infty$ we obtain the following -Corollary $(\tilde S^{(n-1)\infty-1})^H$ is contractible if $H$ is non-transitive and is empty if $H$ is transitive. -This means that $\tilde S^{(n-1)\infty-1}$ us a universal space for the family of non-transitive subgroups of $\Sigma_n$. Let us therefore introduce the notation $E_{\mathrm{nt}}\Sigma_n=\tilde S^{(n-1)\infty-1}$ and $B_{\mathrm{nt}}\Sigma_n={E_{\mathrm{nt}}\Sigma_n}/_{\Sigma_n}$. $B_{\mathrm{nt}}\Sigma_n$ is sometimes called the classifying space for the familty of non-transitive subgroups of $\Sigma_n$. We have proved the following -$$ -\partial_1 F_n \simeq \Sigma^\infty \Sigma B_{\mathrm{nt}}\Sigma_n -$$ -Note that when $n=2$, the family of non-transitive subgroups of $\Sigma_2$ consists of just the trivial group. Because of this, $B_{\mathrm{nt}}\Sigma_2=B\Sigma_2=\mathbb R P^\infty$. So we get the expected answer for $n=2$. -But there is another interesting way to think of the spectra $\partial_1 F_n$. Let $Sp^n(X)=X^n/_{\Sigma_n}$, where $X^n$ denotes $n$-fold cartesian product of $X$ with itself. It follows that for a pointed space $X$, $X^{\wedge n}/_{\Sigma_n}\cong Sp^n(X)/Sp^{n-1}(X)$. The construction $Sp$ extends from spaces to spectra. Let $\mathtt S$ denote the sphere spectrum. There is a sequence of spectra -$$ -\mathtt S=Sp^1(\mathtt S)\to\cdots \to Sp^n(\mathtt S)\to \cdots Sp^\infty(\mathtt S)=H\mathbb Z -$$ -Note that $Sp^\infty(\mathtt S)=H\mathbb Z$ by the Dold-Thom theorem. Let us call this sequence the symmetric powers filtration of the Eilenberg Mac Lane spectrum. -The point of this is that it is easy to see that the spectra $\partial_1 F_n$ are equivalent to the subquotients of the symmetric powers filtration. -$$ -\partial_1 F_n\simeq Sp^n(\mathtt S)/Sp^{n-1}(\mathtt S) -$$ -Remark The equivalence $Sp^n(\mathtt S)/Sp^{n-1}(\mathtt S)\simeq \Sigma^\infty \Sigma B_{\mathrm{nt}} \Sigma_n$ was initially discovered by Kathryn Lesh. -The spectra $Sp^n(\mathtt S)/Sp^{n-1}(\mathtt S)$ have been studied quite a lot, from various perspectives. The following result summarizes some facts about them that were known for a long time. -Theorem The spectrum $Sp^n(\mathtt S)/Sp^{n-1}(\mathtt S)$ and therefore also $\partial_1 F_n$ has the following properties - -It is rationally trivial for $n>1$ -It is integrally trivial unless $n$ is a power of a prime. -If $n$ is a power of a prime $p$ then it is a $p$-local spectrum. - -When $n=p^k$, the mod $p$-homology of $Sp^{p^k}(\mathtt S)/Sp^{p^k-1}(\mathtt S)$ is well-understood. It was initially calculated by Nakaoka. You can learn more cool details about these spectra for example in these notes of a lecture by Mike Hopkins. - -Now let's consider $\partial_k F_n$ for $k\ge 1$. You can express $\partial_k F_n$ in terms of $\partial_1 F_m$ (for varying $m$) using Goodwillie's model for $\partial_k$ as the multi-linearized cross-effect. The $k$-th cross effect of $F_n$ is the following functor of $k$ variables -$$ -\bigvee_{n_1+\cdots + n_k=n, n_i>0} F_{n_1}(X_1)\wedge \ldots \wedge F_{n_k}(X_k) -$$ -After multi-linearization, we obtain the following formula -$$ -\partial_k F_n\simeq \bigvee_{n_1+\cdots + n_k=n, n_i>0} \partial_1 F_{n_1}\wedge \ldots \wedge \partial_1 F_{n_k} -$$ -where the action of $\Sigma_k$ is self-evident. For example -$$ -\partial_2 F_4 \simeq \partial_1 F_3 \wedge \partial_1 F_1 \vee \partial_1 F_1 \wedge \partial_1 F_3 \vee \partial_1 F_2\wedge \partial_1 F_2 -$$ -Where $\Sigma_2$ acts by permuting the first two wedge factors, and also permuting the two smash factors of the last wedge factor.<|endoftext|> -TITLE: Degeneration twisted Hodge to de Rham spectral sequence -QUESTION [8 upvotes]: Let $X$ be a proper and smooth scheme over $\mathbf{C}$ and let $\mathbb{L}$ be a local system of finite dimensional $\mathbf{C}$-vector spaces. By the Riemann Hilbert correspondence, to $\mathbb{L}$ one can associate a locally free sheaf $\mathcal{F}$ of $\mathcal{O}_X$-modules with an integral connection $\nabla \colon \mathcal{F} \to \mathcal{F} \otimes \Omega^1_X$ such that $\mathbb{L} = \mathcal{F}^{\nabla = 0}$. In particular the de rham complex of $\mathcal{F}$: -$$ -0 \to \mathcal{F} \to \mathcal{F} \otimes \Omega^1_X \to \cdots -$$ -gives a resolution of $\mathbb{L}$. This resolution is not injective, but still we can use hypercohomology of the complex to compute the cohomology of $\mathbb{L}$, and we get a spectral sequence of hypercohomology. -If $\mathbb{L} = \mathbf{C}$ is constant, this is the usual Hodge to de Rham spectral sequence, that degenerates, immediately. What can be said in general? I guess that it not true that the sequence always degenerates. (But for example it should be true if $\mathbb{L} = f_\ast \mathbf{C}$ for a proper and smooth morphism $f \colon Y \to X$.) - -REPLY [6 votes]: Let me supplement Chris' answer with a few additional remarks. When $\mathcal{F}$ underlies a polarizable complex variation of Hodge structure, then it carries a filtration $F^\bullet\mathcal{F}$ which induces one on the de Rham complex -$$F^p\mathcal{F}\to F^{p-1}\mathcal{F}\otimes \Omega_X^1 \to \ldots$$ -The argument of Deligne sketched in Zucker's paper, implies that the spectral sequence -$$E_1= H^{p+q}(Gr^p_F(\mathcal{F}\otimes \Omega_X^\bullet))$$ -associated to the above filtration degenerates at $E_1$. If $\mathcal{F}$ is a flat unitary bundle, then this is the same as spectral sequence implicit in your question, but not in general. In fact, I would expect that the more naive spectral sequence would fail to degenerate for a reasonable complicated example.<|endoftext|> -TITLE: Singularities of PL embedding of surface in a contractible 4-manifold -QUESTION [6 upvotes]: I am trying to understand the article "A solution to a conjecture of Zeeman" by Akbulut, but I am not an expert in PL-geometry. -As far as I understand, two statements should be true, but I cannot find a reference: -Notation: -$B^n$ is the $n$-dimensional disk (with boundary). -Given PL manifolds $M^m,N^n $ with $m -TITLE: Are there "natural" sequences with "exotic" growth rates? What metatheorems are there guaranteeing "elementary" growth rates? -QUESTION [38 upvotes]: A thing that consistently surprises me is that many "natural" sequences $f(n)$, even apparently very complicated ones, have growth rates which can be described by elementary functions $g(n)$ (say, to be precise, functions expressible using a bounded number of arithmetic operations, $\exp$, and $\log$), in the sense that $\lim_{n \to \infty} \frac{f(n)}{g(n)} = 1$. I'll write $f \sim g$ for this equivalence relation. Here are some examples roughly in increasing order of how surprising I think it is that the growth rate is elementary (very subjective, of course). - -The Fibonacci sequence $F_n$ satisfies $F_n \sim \frac{\phi^n}{\sqrt{5}}$. -The factorial $n!$ satisfies $n! \sim \sqrt{2\pi n} \left( \frac{n}{e} \right)^n$. -The prime counting function $\pi(n)$ satisfies $\pi(n) \sim \frac{n}{\log n}$. -The partition function $p(n)$ satisfies $p(n) \sim \frac{1}{4n \sqrt{3}} \exp \left( \pi \sqrt{ \frac{2n}{3} } \right)$. -The Landau function $g(n)$ satisfies $\log g(n) \sim \sqrt{n \log n}$. I don't know whether or not it's expected that $g(n) \sim \exp(\sqrt{n \log n})$. -For $p$ a prime, the number $G(p^n)$ of groups of order $p^n$ satisfies $\log_p G(p^n) \sim \frac{2}{27} n^3$. - -I know some metatheorems guaranteeing elementary asymptotics for some classes of sequences. The simplest one involves sequences with meromorphic generating functions; this gives the Fibonacci example above as well as more complicated examples like the ordered Bell numbers. I have the impression that there are analogous theorems for Dirichlet series involving tauberian theorems that produce the PNT example and other similar number-theoretic examples. There's a metatheorem involving saddle point bounds which gives the factorial example and at least heuristically gives the partition function example. And I don't know any metatheorems relevant to the Landau function or $p$-group examples. So, questions: - -Q1: What are some "natural" sequences $f(n)$ which (possibly conjecturally) don't have elementary asymptotics, in the sense that there are no elementary functions $g(n)$ such that $f(n) \sim g(n)$? - -Right off the bat I want to rule out two classes of counterexamples that don't get at what I'm interested in: $f(n)$ may oscillate too wildly to have an elementary growth rate (for example $f(n)$ could be the number of abelian groups of order $n$), or it may grow too fast to have an elementary growth rate (for example $f(n)$ could be the busy beaver function). Unfortunately I'm not sure how to rigorously pose a condition that rules these and other similar-flavored counterexamples out. At the very least I want $f(n)$ to be a monotonically increasing unbounded sequence of positive integers, and I also want it to be bounded from above by an elementary function. -The kinds of sequences I'm interested in as potential counterexamples are sequences like the Landau function above, as well as combinatorial sequences like the Bell numbers $B_n$. The Bell numbers themselves might be a potential counterexample. Wikipedia gives some elementary bounds but expresses the growth rate in terms of the Lambert W function; it seems that the Lambert W function has elementary growth but I'm not sure if that implies that $B_n$ itself does. - -Q2: What are some more metatheorems guaranteeing elementary growth rates? Are there good organizing principles here? - - -Q3: What are some "natural" sequences known to have elementary growth rates but by specific arguments that don't fall under cases covered by any metatheorems? - -Apologies for the somewhat open-ended questions, I'd ask a tighter question if I knew how to state it. -Edit: Wow, apparently I asked almost exactly this question almost exactly $10$ years ago, and I even gave three of the same examples... - -REPLY [2 votes]: nombre mentioned the work by van den Dries, Macintyre and Marker on transseries, but it seems no one has mentioned the paper Logarithmic-exponential power series by the same authors, in which they proved: - -The compositional inverse to $x\mapsto(\log x)(\log\log x)$ is not asymptotic to a composition of semialgebraic functions, log and exp. - -confirming a conjecture of Hardy. Denote the inverse by $f(x)$. They mentioned in the paper that $e^{f(x)}$ had been shown to have non-elementary growth rate by Shackell. Also $f(x)=e^{e^{W(x)}}$ if my algebra is right. -This is a real function instead of an integer sequence, but $\lfloor f(n)\rfloor$ should work since $\lim_{n\rightarrow\infty}\frac{f(n+1)}{f(n)}=1$, so if $\lfloor f(n)\rfloor$ were asymptotic to $g(n)$ for some elementary function $g$, then $f(x)$ would be asymptotic to $g(x)$ (using the fact that $g(x)$ must be eventually increasing, mentioned in an answer to your old question). -I realized that I completely ignored trignometric functions; fortunately you also excluded them in your question. It seems even just adding $\cos x$ makes the situation a lot more complicated. In A Simple Example for a Theorem of Vijayaraghavan, the authors showed that for any $\phi:[0,\infty)\rightarrow\mathbb{R}$, there exists a suitable irrational number $\alpha$ such that $\frac{1}{2-\cos x-\cos\alpha x}$ is greater than $\phi$ on a sequence that tends to infinity.<|endoftext|> -TITLE: Generalized Birman exact sequence for surfaces with boundaries -QUESTION [6 upvotes]: Let $S_g^n$ be a surface of genus g with n boundaries and let $Mod(S_g^n)$ be its mapping class group. -We will also denote by $S_{g,m}^n$ a surface of genus g with n boundaries and m punctures. -The classical Birman exact sequence is -$$ 1 \to \pi_1(UT(S_g^{n-1}) \to Mod(S_g^n) \to Mod(S_g^{n-1}) \to 1$$ -where $UT(S_g^{n-1})$ is the unitary tangent bundle over $S_g^{n-1}$ and the last map corresponds to capping one of the boundaries by a disk. -I am wondering what is the kernel of the map $f \colon Mod(S_g^n) \to Mod(S_g^{n-k})$ obtained by capping $k$ boundaries by $k$ disks. -It is a known fact, proven by Birman, that the kernel of the map $Mod(S_{g,m}) \to Mod(S_{g,m-k})$ obtained by forgetting $k$ punctures can be identified with $B_k(S_{g,m})$, the braid group on $k$ strands over $S_{g,m}$. -Therefore, if $PB_k(S_{g,m})$ is the pure braid group on $k$ strands over $S_{g,m}$, my best guess is that the kernel of $f$ is isomorphic to $PB_k(S_{g,m}) \oplus \mathbb{Z}^k$, where $\mathbb{Z}^k$ corresponds to the free abelian group with generators the Dehn twists around the boundaries we capped. Indeed, we can factor $f$ into $f = f_1f_2$ where $f_2$ caps the boundaries by punctures disks and $f_1$ forgets the punctures. -Is this correct? If so, is there another expression for this kernel? (Similarly to how $\pi_1(S_g^{n-1}) \oplus \mathbb{Z}$ can be identified with $\pi_1(UT(S_g^{n-1})$). - -REPLY [6 votes]: It's almost right, your guess. -The easiest description is geometric, generalizing Birman's description $\pi_1(UT(S_g^{n-1}))$. You look instead at the configuration space parametrizing $k$ distinct points in $S_g^{n-k}$, together with unit tangent vectors at each of these $k$ points. So this is an open subspace of the cartesian product $UT(S_g^{n-k})^k$. The fundamental group of this space is the group you care about. It is sometimes called the framed braid group. -The framed braid group is an extension of the pure braid group $PB_k(S^m_g)$ by $\mathbf Z^k$, as one sees geometrically by fibering the framed configuration space over the usual configuration space with fiber $(S^1)^k$. For an open surface the tangent bundle is trivial, which implies that this $(S^1)^k$-bundle is trivial, and thus the extension is trivial, too. In this case your guess is correct. But for closed surfaces you get a nontrivial central extension.<|endoftext|> -TITLE: mixed Hodge structure of general linear group -QUESTION [5 upvotes]: Is there any literature about mixed Hodge structure on $H^*(GL(n,\mathbb{C}))$ ? I think it is a vary basic problem in mathematics, but there are no literature about it? - -REPLY [14 votes]: Yes, there is literature about it, it is Théorème 9.1.5 of Deligne's Hodge III. He gives a precise description more generally for a connected algebraic group $G$. In brief: from general Hopf algebra theory one knows that $H^\ast(G,\mathbf Q)$ is an exterior algebra on a finite set of odd degree generators. Deligne shows that this is also true for the mixed Hodge structure: as a mixed Hodge structure the cohomology is an exterior algebra, and the generators are purely of Tate type, with the generators in degree $2i-1$ of Hodge type $(i,i)$.<|endoftext|> -TITLE: Category theory and arithmetical identities -QUESTION [31 upvotes]: My question concerns categorical "proofs" or "guessing" of arithmetical identities using category theory. I’m not at all a specialist of CT but I try to put my hands on and to understand something in this fascinating subject. Trying to prove something concrete, the exercise was to prove the identity -$$ \gcd(n,m) \cdot \operatorname{lcm}(n,m) = n\cdot m$$ -Using the concepts of product and coproduct in a specific category (objects are integers and arrows is "divisibility") I've been able to prove it with very minimalistic assumptions, only the notion of divisibility of integers. No use of prime numbers or unique factorization with primes or other specific theorems like Bezout theorem, etc... -I'm wondering if someone has other examples like that one. Someone gave me the reference - -Marcelo Fiore, Tom Leinster, Objects of Categories as Complex Numbers, Advances in Mathematics 190 (2005), 264-277, doi:10.1016/j.aim.2004.01.002, arXiv:math/0212377, - -which is interesting but it doesn't really answer my question. In that paper, if I've correctly understood, they show that if something is true with complex numbers it is true in more general categories. -I would be interested if someone has some "simple" examples of concrete applications of CT proving or "guessing" arithmetical identities -Thanks a lot for any clue - -REPLY [26 votes]: A classic paper on this topic is Andreas Blass’s Seven trees in one (1994). -It starts with the observation that there is an explicit, constructive bijection between (finite) binary trees and seven-tuples of such trees: $T \cong T^7$. It then shows how this bijection can be obtained from the algebraic proof in $\mathbb{N}[t]$ that $t = 1 + t^2$ implies $t = t^7$. (Here $t = 1 + t^2$ expresses the defining property of finite binary trees: “each tree is either just a leaf, or a joined pair of two trees”.) Finally, it generalises this result, showing that polynomial identities in $\mathbb{N}[t]$ correspond to constructively-definable bijections between suitable polynomial functors on sets, or their initial algebras.<|endoftext|> -TITLE: Extension of a holomorphic curve in $B^4$ to one in $\mathbb{C}P^2$ -QUESTION [5 upvotes]: Let $B^4$ be the closed unit ball in $\mathbb{C}^2$ and $J$ an almost complex structure sufficiently closed to the standard complex structure on $\mathbb{C}^2$ in the $C^0$-topology. Let $u \colon S \rightarrow B^4$ be a smooth $J$-holomorphic curve transverse to $\partial B^4$. Can $u$ extend to a $\tilde{J}$-holomorphic curve in $\mathbb{C}P^2$? Here $\tilde{J}$ is an extension of $J$ on $\mathbb{C}P^2$. -Any hint and comment are really appreciated. Thank you in advance. - -REPLY [3 votes]: Yes, you can. -The "boundary" $\partial B \cap {\rm Im}(u)$ of $u$ is a link $K$ that is transverse to the contact structure $JTS^3\cap TS^3$, which is isotopic to the standard contact structure on $S^3$. John Etnyre and I proved that every transverse link in $(S^3,\xi_{st})$ can be capped off by a smooth symplectic surface $\hat C$ in $\mathbb{CP}^2\setminus B^4$ (here $B^4$ is a Darboux ball). We call any such cap a hat (because "relative symplectic cap" was too long). -As Chris Gerig mentions in his comment above, you can extend $J$ to an almost complex structure $\tilde J$ on $\mathbb{CP}^2$ for which $C = \hat C \cup u(S)$ is pseudo-holomorphic. -The argument of the proof of the existence of $\hat C$ isn't very hard: one constructs a cobordism to a torus link using local moves on braids, and then caps it off with a piece of an algebraic curve.<|endoftext|> -TITLE: Converse of the Lee-Yang circle theorem for polynomials with unitary roots -QUESTION [12 upvotes]: The Lee-Yang circle theorem states that if $\left( a_{ij} \right)$ is a Hermitian square $n \times n$ matrix whose entries are in the closed unit disc, then the polynomial $$ P\left(Z \right) = \sum_{S\subseteq\left[n\right]}\left(\prod_{i\in S,j\notin S}a_{ij}\right)Z^{\left|S\right|} $$ has all of its roots on the unit circle. Here $\left[n\right] = \{1,2,\dots,n\}$. -Let $P\left( Z \right)$ be the polynomial above, and denote the coefficient of $Z^k$ by $\alpha_k$. Note that $\alpha_0 = \alpha_n = 1$. Then it follows from the Lee-Yang circle theorem that for every real $\lambda$ with $\left|\lambda \right| \le 1$, the polynomial $$P_\lambda \left( Z \right)= \sum_{k=0}^n{\alpha_k \lambda^{k\left(n - k\right)} Z^k}$$ has all of its roots on the unit circle (this follows from replacing the matrix $\left( a_{ij} \right)$ with the matrix $\left( \lambda a_{ij} \right)$). -My question: suppose $$Q\left( Z \right) = \sum_{k=0}^n{\beta_k Z^k}$$ is a polynomial with complex coefficients, such that $\beta_0 = \beta_n = 1$, and such that for every $\lambda$ with $\left| \lambda \right| \le 1$ the polynomial $$ Q_\lambda\left(Z\right) = \sum_{k=0}^n{\beta_k \lambda^{k \left(n - k\right)} Z^k}$$ -has all of its roots on the unit circle. Then is it possible to write $Q\left(Z\right)$ as $P\left(Z\right)$ above for some Hermitian matrix $\left(a_{ij}\right)$ with entries in the closed unit disc? Is there some algorithm to find such matrix? - -REPLY [7 votes]: It looks to be enough to require this for $\lambda=1$. -Indeed, assume that the polynomial $Q(Z)=\sum_{k=0}^n \beta_k Z^k$, $\beta_0=\beta_n=1$, has all roots on the unit circle. Denote the roots by $-\theta_i^{-1}$, $i=1,\ldots,n$. Then $Q(Z)=\prod (1+\theta_i Z)$ and $\prod_i \theta_i=1$. -There exists an Hermitian matrix $(a_{ij})$ with $|a_{ij}|=1$ such that $\theta_i=\prod_{j\ne i} a_{ij}$. Indeed, we may put $a_{ij}=1$ if $\max(i,j) -TITLE: Is there a quantum analog of Kolmogorov Complexity? -QUESTION [5 upvotes]: Kolmogorov Complexity (interpreted in terms of shortest program computing a string) and Shannon Entropy are quite similar. -Since there is a quantum entropy is it reasonable to ask if there is quantum analog of Kolmogorov complexity that gives rise to some sort of shortest 'quantum' program interpretation or another suitable interpretation that corresponds to quantum entropy? -What is the right analogy and the right equivalence to quantum entropy? - -REPLY [2 votes]: Markus Mueller's Phd thesis is about quantum Kolmogorov complexity. -Quantum Kolmogorov Complexity and the Quantum Turing Machine -Here is his definition: -Given a Quantum Turing Machine $M$ and a finite error $\delta>0$, the finite-error quantum Kolmogorov complexity of a qubit string $|x\rangle$ is -$K_\delta^Q(x)=\min\limits_p\Big\{\ell(p): \|x-M(p)\|_{\mathrm{tr}}<\delta\Big\}$ -and the approximate-scheme quantum Kolmogorov complexity of $|x\rangle$ is -$K^Q(x)=\min\limits_p\left\{\ell(p): \forall k\in\mathbb{N}: \|x-M(p,k)\|_{\mathrm{tr}}<\frac{1}{k}\right\}$ -where $\|\cdot\|_{\mathrm{tr}}$ is the trace norm, i.e. $\|\rho-\sigma\|_{\mathrm{tr}}:=\frac{1}{2}\operatorname{Tr}\left(\sqrt{(\rho-\sigma)^\dagger(\rho-\sigma)}\right)$. -Another version of quantum Kolmogorov complexity of $|x\rangle$ with respect to quantum Turing machine $M$ is -$K^Q(x)=\min\limits_p\left\{\ell(p)+\lceil -\log\|\langle z|x\rangle\|^2\rceil: M(p)=|z\rangle\right\}$ -You can find this version in Ming Li and Paul Vitanyi's book: An Introduction to Kolmogorov Complexity and Its Applications<|endoftext|> -TITLE: Is there a finite dimensional algebra with left finitistic dimension different from its right finitistic dimension? -QUESTION [5 upvotes]: Let $\Lambda$ be finite dimensional algebra over a field $k$. The (left) finitistic dimension of a finite dimensional algebra is defined as -$$\operatorname{findim}(\Lambda)=\sup\{\operatorname{pd}M | M \in \operatorname{mod}\Lambda,\operatorname{pd}M < \infty\}$$ -where $\operatorname{mod}\Lambda$ is the category of finitely generated left $\Lambda$-modules, and $\operatorname{pd}M$ is the projective dimension of $M$. The right finitistic dimension is simply $\operatorname{findim}(\Lambda^{op})$, which by applying the duality $\operatorname{Hom}_k(-, k)$ is equal to -$$\operatorname{findim}(\Lambda^{op})=\sup\{\operatorname{id}M | M \in \operatorname{mod}\Lambda,\operatorname{id}M < \infty\}.$$ -I have seen people mention that $\operatorname{findim}(\Lambda)$ doesn't necessarily equal $\operatorname{findim}(\Lambda^{op})$, but I could not find an example by searching online. -I believe I have managed to find an example myself, so I'm answering my own question here in case anyone else is looking for an example in the future. If anyone has more examples or more information about the difference between the two dimensions, I will appreciate your answers. - -REPLY [7 votes]: The left and right finitistic dimension can take any pair of values. This is shown in Example 2.3 of -Green, Edward L.; Kirkman, Ellen; Kuzmanovich, James, Finitistic dimensions of finite dimensional monomial algebras, J. Algebra 136, No. 1, 37-50 (1991). ZBL0727.16003. -The simplest example where they differ is slightly simpler than the example given in Jacob FG's answer. Let $Q$ be the quiver with two vertices, a loop at the first vertex, and one arrow from the first to the second vertex. Then $\Lambda=kQ/\operatorname{rad}^2(kQ)$ has $\operatorname{findim}(\Lambda)=1$ and $\operatorname{findim}(\Lambda^{\text{op}})=0$. The examples in the cited paper are generalizations of this.<|endoftext|> -TITLE: Hyperplane arrangements whose regions all have the same shape -QUESTION [9 upvotes]: Suppose I have a (finite, real, central, essential) hyperplane arrangement $\mathcal{H}$ such that all regions "have the same shape": for any two regions $R,R'$, there is an orthogonal transformation taking $R$ to $R'$ (these transformations are not required to do anything nice to the rest of the arrangement). Is $\mathcal{H}$ necessarily a reflection arrangement? - -REPLY [11 votes]: This is a known open problem (for isometric regions), which, as far as I know, is still not settled. -The dimension 3 case was proved affirmatively in https://arxiv.org/abs/1501.05991, where also some history of the question is outlined. I am not aware of any progress since.<|endoftext|> -TITLE: Restricting representations to a principal $\mathfrak{sl}(2)$ -QUESTION [5 upvotes]: Let $\mathfrak{g}$ be a semi-simple Lie algebra over $\mathbb{C}$ with simply connected group $G$ and suppose that -$$\mathfrak{g} = \bigoplus_i\mathfrak{g}_i$$ is a $\mathbb{Z}$- or $\mathbb{Z}/n\mathbb{Z}$-grading on $\mathfrak{g}$, and consider the subalgebra $\mathfrak{g}_0$ and its representation on $\mathfrak{g}_i$ for $i\neq 0$. -Assume that $\mathfrak{g}_0$ is a non-abelian reductive Lie algebra. Suppose that $X$ is a regular nilpotent element of $\mathfrak{g}_0$ and fix a principal $\mathfrak{sl}(2)$-triple $\mathfrak{s}=\operatorname{span}_{\mathbb{C}}\{F,H,E\}\subset \mathfrak{g}$ associated to $X$. I am interested in understanding the restriction of the representation $\mathfrak{g}_n$ to $\mathfrak{s}$. -I am specifically interested in the following case: -Suppose $\mathfrak{g}_0=\mathfrak{m}$ is the Levi subalgebra of a maximal parabolic Lie algebra $\mathfrak{p}= \bigoplus_{i\geq 0}\mathfrak{g}_i$, where I am considering the grading induced by characters of the center of $\mathfrak{m}$. In this case, the representations $\mathfrak{g}_i$ are -irreducible (using maximality) prehomogeneous representations of the associated Levi subgroup $M\subset G$. Working out several cases, it seems that the restriction of $\mathfrak{g}_i$ is always multiplicity free as an $\mathfrak{s}$-module. - -Is the restriction of $\mathfrak{g}_i$ always multiplicity free as an $\mathfrak{s}$-module? [EDIT: No, see my answer below.] If not, is there an interpretation of when such a restriction is multiplicity-free? - -I recently ran into this problem studying certain distributions in the context of $p$-adic groups. Certain domain properties of these distributions turned out to be equivalent to the multiplicity-free property above on the Langlands-dual Lie algebra. This was surprising to me, and raises the question of whether there exists a broader interpretation of this type of decomposition. -I've seen this crop up a few different places for special representations, such as Kostant's work on the adjoint representation and Gross's work for minuscule representations (refs below), showing that these restrictions often encode deep information about $\mathfrak{g}$ and the representation. However, the general problem is mysterious to me. A vague question is - -What does this decomposition (say in terms of those weights arising and their multiplicities) encode about the representation $\mathfrak{g}_i$? - -Really, this is just probing to see if there is a nice interpretation out there. The question makes sense for an arbitrary representation, but the additional structure here makes me hope for a meaningful question. Any references would be greatly appreciated. -Kostant, Bertram, The principal three-dimensional subgroup and the Betti numbers of a complex simple Lie group, Am. J. Math. 81, 973-1032 (1959). ZBL0099.25603. -Gross, Benedict H., On minuscule representations and the principal SL(2), Represent. Theory 4, 225-244 (2000). ZBL0986.22011. - -REPLY [3 votes]: $\DeclareMathOperator\Sym{Sym}$After further calculation, I realized that the answer to the first question is NO, the restriction of $\mathfrak g$ to $\mathfrak s$ need not be multiplicity free: suppose $\mathfrak{g}=\mathfrak{e}_8$, and suppose that $\mathfrak{p}$ is the maximal parabolic subalgebra associated to the simple root indicated by the $\times$ below: -$\qquad\begin{matrix} -0 - 0 - \stackrel{\stackrel{\displaystyle 0}|}{\times} - 0 - 0 - 0 - 0 . -\end{matrix}$ -Then the Levi subalgebra $\mathfrak{m}=\mathbb{C}\oplus \mathfrak{sl}(3)\oplus \mathfrak{sl}(2)\oplus \mathfrak{sl}(5)$ acts on the $1^\text{st}$ graded piece $\mathfrak{g}_1$ via the tensor product of the standard representations. As a representation of a principal $\mathfrak{sl}(2)$-subalgebra of $\mathfrak{m}$, this decomposes as -$$ -\Sym^{7}(\mathbb{C}^2)\oplus \Sym^5(\mathbb{C}^2)^{\oplus2}\oplus\Sym^3(\mathbb{C}^2)^{\oplus2}\oplus\Sym^1(\mathbb{C}^2). -$$ -In fact, this generalizes the simplest counter example of $\mathfrak{g}=\mathfrak{so}(8)$ with the parabolic subalgebra $\mathbb{C}\oplus \mathfrak{sl}(2)\oplus \mathfrak{sl}(2)\oplus \mathfrak{sl}(2)$. In that case, $\mathfrak{g}_1$ decomposes as $\Sym^3(\mathbb{C}^2)\oplus \mathbb{C}^2\oplus \mathbb{C}^2$. -I still would be interested in an understanding of what such decompositions tell us.<|endoftext|> -TITLE: Explicit large finite fields in characteristic $2$ -QUESTION [11 upvotes]: Every finite field of characteristic $2$ ist given by $\mathbb{F}_2[x]/P(x)$ for some irreducible polynomial $P\in \mathbb{F}_2[x]$. -For small degree, a simple algorithm gives a way to find $P$. Is there a way to give explicitely the polynomial $P$ for large degrees? I would be interested in a formula like -$$x^n+a_{n-1} x^{n-1}+\cdots +a_1 x+a_0$$ -where $a_0,\ldots,a_{n-1}\in \mathbb{F}_2[n]$. This might be too much to ask for all degrees, so an infinite sequence of irreducible polynomials would be good. -EDIT: Some sequences have already be given. If you have more to share (with the proof of the irreducibility, either by giving the reference or explaining it), please go ahead. - -REPLY [5 votes]: For $n=3^k$, the polynomial $p=x^{2n}+x^n+1$ is irreducible over $\mathbb{F}_2$. -Proof: By Rabin's irreducibilty test, it suffices to check that $p|x^{2^{2n}}-x$ and $\gcd(p,x^{2^{2n/3}}-x)=1$. -Note that the order of $x$ mod $p$ is $3n=3^{k+1}$. Hence, since $3^{k+1}|4^{n}-1$ by lifting-the-exponent lemma, we have $p|x^{2^{2n}-1}-1$. -Again by lifting-the-exponent lemma, we have $4^{3^{k-1}}-1=3^km$ for $m$ not divisible by $3$. Hence $x^{4^{3^{k-1}}-1}=x^{nm}\equiv x^n\ \text{or}\ x^{2n} \pmod{p}$ since $x^n$ has order $3$. As $(x^n-1)x^n=(x^{2n}-1)x^{2n}=1$, this mean that $x^{2^{2n/3}-1}-1$ is invertible in $\mathbb{F}_2[x]/(p)$. As $x$ is also invertible, we have shown that $\gcd(p,x^{2^{2n/3}}-x)=1$, as wanted.<|endoftext|> -TITLE: A formula for this generating function that is similar to the $qt$-Catalan numbers -QUESTION [15 upvotes]: I came up with the following conjecture: -$$ -\sum_{n \ge 0} z^n \sum_{\mu \vdash n} \frac{ t^{\sum l}q^{\sum a}}{\prod (q^a - t^{l+1})(t^l - q^{a+1})} = \exp\left(\sum_{n \ge 1} \frac{z^n}{n(q^n-1)(t^n-1)}\right) -$$ -where each unlabeled sum and product is over the cells of the diagram of the partition, and $l$ and $a$ are the arm and leg lengths. -The coefficients on the left hand side are similar to the $qt$-Catalan numbers but are missing some factors. -I verified it up to order 10 computationally (see here). -With a little work, this conjecture would prove a special case of the conjecture in this question. -I hoping to get some ideas or tools about how to deal with things that look like this. - -REPLY [7 votes]: This identity can be proved using the results in Garsia and Haiman's paper "A Remarkable q,t-Catalan Sequence and q-Lagrange Inversion". In particular theorem 3.10(e) gives -$$\sum_{\mu \vdash n} \frac{ t^{\sum l}q^{\sum a}}{\prod (q^a - t^{l+1})(t^l - q^{a+1})}=\sum_{\mu \vdash n} \frac{ t^{\sum a}q^{\sum a}}{\prod (1 - q^h)(1 - t^h)}$$ -where $h=a+l+1$ denotes the usual hook length. Writing this expression in terms of principal specializations of Schur functions together with an application of the Cauchy identity gives -$$\sum_{n \ge 0} z^n \sum_{\mu \vdash n} \frac{ t^{\sum l}q^{\sum a}}{\prod (q^a - t^{l+1})(t^l - q^{a+1})}=\sum_{\mu}s_{\mu}(1,q,q^2,\dots)s_{\mu}(z,zt,zt^2,\dots)=\prod_{i,j\geq 0}\frac{1}{1-q^it^jz}$$ -and it remains to notice that -$$\prod_{i,j\geq 0}\frac{1}{1-q^it^jz}=\exp\left(\sum_{n\geq 1}\frac{1}{n}\sum_{i,j\geq 0} q^{ni}t^{nj}z^n\right)=\exp\left(\sum_{n \ge 1} \frac{z^n}{n(q^n-1)(t^n-1)}\right)$$ -as desired.<|endoftext|> -TITLE: Number of conjugacy classes of finite reductive groups -QUESTION [7 upvotes]: Let $G$ be a connected reductive group over $\mathbb{Z}$. Let $c_{G(\mathbb{F}_q)}$ be the number of conjugacy classes of $G(\mathbb{F}_q)$. -Question: Is it true that $c_{G(\mathbb{F}_q)}$ is a quasi-polynomial in $q$? I.e. is it true that there exists an integer $N$ (depending only on $G$) such that for every $i\in \{0,1,\dotsc,N-1\}$ there exists a polynomial $P_i$ such that $c_{G(\mathbb{F}_q)}=P_i(q)$ for all $q\equiv i \bmod N$? -If the answer is yes, then a type-independent explanation would be desirable. I'm mainly interested in the case $i=1$ but the general question seems worthwhile. -Frank Lübeck's tables at Character Degrees and their Multiplicities for some Groups of Lie Type of Rank < 9 illustrate that the question has a positive answer for (simply connected form of) exceptional groups. For instance, for $G_2$, $E_6$ and $E_7$, the site states that $N=6$ and gives the explicit polynomials. We also have a lot of positive evidence for classical groups coming from explicit computations in Macdonald - Numbers of conjugacy classes in some finite classical groups and Wall - On the conjugacy classes in the unitary, symplectic, and orthogonal groups among others. - -REPLY [15 votes]: Assume that $G$ is adjoint split over $\mathbb F_q$. Let $G^*$ be the (Langlands) dual group; it is also split over $\mathbb F_q$. For each semisimple $s \in G^*( \mathbb F_q)$ let $N_s$ be the number of unipotent representations -of the centralizer $Z_{G^*}(s)(\mathbb F_q)$. The number of conjugacy classes in $G(\mathbb F_q)$ is equal to the number of irreducible characters of $G(\mathbb F_q)$ and this equals (by my 1984 classification) the sum $\sum_s N_s$ where $s$ runs over the $G^*(\mathbb F_q)$-conjugacy classes of semisimple elements in $G^*(\mathbb F_q)$. Since the $N_s$ are very well behaved (much better behaved that number of unipotent classes) this reduces the problem to a problem of counting semisimple classes in $G^*(\mathbb F_q)$ with centralizer of fixed type. From this the desired result follows. -The main point is that the number of unipotent classes in $G(\mathbb F_q)$ behaves differently in small characteristic while the number of uniptent representations of a reductive group over $\mathbb F_q$ is independent of characteristic. This makes it easier to count irreducible representations than conjugacy classes.<|endoftext|> -TITLE: Can the methods of algebra characterize nonlinear PDE blow-ups? -QUESTION [6 upvotes]: Consider 2 simple differential equations: $x'(t)=x(t)^2, x(0)=1$ and $x'(t)=-x(t)^2, x(0)=1$. -As $t$ goes from $0$ to $\infty$, the first equation ($x(t)=1/(1-t)$) will lead to a finite-time blow-up, while the second ($x(t)=1/(1+t)$) doesn't. Even though, from looking at the algebra, there is only one difference between a plus and a minus. (Luckily, these admit exact solutions. Real-world problems don't.) -Is it therefore correct to think that the methods of algebra (loosely defined, as opposed to traditional analysis of PDEs) will never be able to help us with this kind of question? So far, the algebraic characterizations of PDEs I have seen include some kind of jet bundles (Gromov's partial differential relations, Vinogradov's theory of diffiety which generalizes algebraic geometry etc.). Vinogradov even managed to define spectral sequences and de Rham cohomology, and his school has used them to compute conservation laws / integrable structures of PDEs. Gromov's geometric h-principle is related to developments in fluid dynamics such as Onsager's conjecture. But that is the extent of my knowledge. All the blow-up proofs for PDEs I have seen seem to be mainly analytical in nature. -If you are familiar with such algebraic theories of nonlinear PDEs, do you know a lead that can help characterize PDE blow-ups? - -REPLY [4 votes]: There are classes of PDE's where blow-ups of solutions can be characterised. The examples I am aware of are those of integrable PDE's. To connect to classical algebraic theory, let us focus on so-called finite gap solutions. A typical example of integrable PDE's is given by the $\sinh$-Gordon equations -$$\Delta u=\mp \sinh 2u,$$ -where $u$ is a real-valued function defined on the complex plane. Doubly periodic finite gap solutions are given by a linear flow: there exists a (real) linear map from the torus determined by the period lattice into the Jacobian of the so-called spectral curve. This linear map has to take values inside a real (or quaternionic) component of the Jacobian in order for $u$ to be real-valued. The solution becomes singular (it has a blow up) where the linear map intersects the $\theta$-divisor inside the Jacobian. For some classes (e.g., the above sinh-Gordon equation with the - sign) one can prove that you will never intersects the $\theta$-divisor (see Hitchin, Harmonic 2-tori in the 3-sphere, Journal Diff Geo, 1990, Proposition 7.15), while for other classes (e.g. a certain $\cosh$-Gordon) equation you necessarily intersectthe $\theta$-divisor (see Babich-Bobenko, Willmore tori with umbilic lines, Duke Journal, 1993).<|endoftext|> -TITLE: A property forcing the Frobenius-Schur indicator to be positive -QUESTION [6 upvotes]: Let $G$ be a finite group. Two irreducible complex representations $V,V'$ of $G$ are called dual to each other if $V \otimes V'$ admits a trivial component, i.e. $\hom_G(V \otimes V',V_0)$ is positive dimensional (thus one-dimensional) with $V_0$ the trivial representation. Then the representation $V'$ is denoted $V^*$ (and $V \simeq V^{**}$). -Let $V, W$ be irreducible complex representations of $G$ such that: - -$W^* \simeq W$, -$\hom_G(V \otimes V^*,W)$ is odd-dimensional. - -Question: Is it true that the Frobenius-Schur indicator of $W$ is equal to $1$? -The odd-dimensional assumption cannot be extended to even (see counterexamples in Appendix). -It is conjectured to be true for every pivotal fusion category, see [a, Conjecture 4.26]. See also [b] and [c]. - -References -[a] Z. Wang, Topological quantum computation, CBMS Reg. Conf. Ser. Math. (112) xiii + 115pp, (2010). -[b] G. Mason, A brief history of the positivity conjecture in tensor category theory. Bull. Inst. Math. Acad. Sin. (N.S.) 14 (2019), no. 2, 149--153 -[c] J. Fuchs; I. Runkel; C. Schweigert. A reason for fusion rules to be even. J. Phys. A 35 (2002), no. 19, L255--L259. - -Appendix -Smallest counterexample among the finite groups: -gap> G:=PSU(3,2);; Order(G); -72 -gap> Indicator(CharacterTable(G),2); -[ 1, 1, 1, 1, -1, 1 ] -gap> M:=RepGroupFusionRing(G);; -gap> M[6][6]; # FS(V)=1 -[ 1, 1, 1, 1, 2, 7 ] - -Smallest counterexample among the finite simple groups: -gap> G:=PSU(3,5);; Order(G); -126000 -gap> Indicator(CharacterTable(G),2); -[ 1, -1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0 ] -gap> M:=RepGroupFusionRing(G);; -gap> M[8][8]; # FS(V)=1 -[ 1, 2, 2, 3, 3, 3, 9, 9, 11, 11, 11, 11, 12, 12 ] -gap> M[11][12]; # FS(V)=0 -[ 1, 2, 3, 4, 4, 4, 10, 13, 17, 17, 15, 15, 18, 18 ] - -with the following code: -RepGroupFusionRing:=function(g) - local irr,n,M; - irr:=Irr(g); - n:=Size(irr); - M:=List([1..n],i->List([1..n],j->List([1..n],k->ScalarProduct(irr[i]*irr[j],irr[k])))); - return M; -end;; - -REPLY [9 votes]: Here is a more general statement, see also Lemma 1.2 in [1]. - -Lemma: Let $Z$ be a self-dual $kG$-module which admits a non-degenerate $G$-invariant symmetric (alternating) bilinear form $b$. Suppose that $W$ is a self-dual irreducible $kG$-module. If $W$ occurs in $Z$ as a composition factor of odd multiplicity, then $W$ admits a non-degenerate $G$-invariant symmetric (alternating) bilinear form. - - -Proof: Induction on $\dim Z$. Let $Q \subset Z$ be an irreducible $G$-submodule of $Z$, so either $Q$ is non-degenerate or totally singular. -If $Q$ is non-degenerate, then $Z = Q \oplus Q^\perp$. If $Q \cong W$ the lemma follows, and if $Q \not\cong W$ the lemma follows by applying induction on $Q^\perp$. -If $Q$ is totally singular, then $b$ induces a non-degenerate bilinear form on $Q^\perp/Q$, which is of the same type as $b$. So we can apply induction on $Q^\perp / Q$ since $Z/Q^\perp \cong Q^*$. - -We can apply the lemma for $Z = V \otimes V^*$, since it always admits a non-degenerate $G$-invariant symmetric bilinear form. Identifying $V \otimes V^* \cong \operatorname{End}(V)$, one such form is given by $b(x,y) = \operatorname{Tr}(xy)$ for all $x, y \in \operatorname{End}(V)$. -So in your setting: if $W$ is irreducible, $W \cong W^*$, and $\operatorname{Hom}_G(V \otimes V^*, W)$ is one-dimensional, it follows from the lemma that $W$ admits a nondegenerate $G$-invariant symmetric bilinear form. In other words, the Frobenius-Schur indicator of $W$ is $1$. - -[1] R. Gow, W. Willems, Methods to decide if simple self-dual modules over fields of characteristic 2 are of quadratic type, J. Algebra 175 (1995) 1067–1081.<|endoftext|> -TITLE: Is $\Bigl\{ n \sum_{k=2}^{n-1} \frac{1}{k}\Bigr\}$ unique $\forall n \in \Bbb{N}, n>1$ -QUESTION [12 upvotes]: If we define $$f(n) = \Bigl\{ n \sum_{k=2}^{n-1} \frac{1}{k}\Bigr\}$$ -is it true that $f(n) \ne f(m)$ whenever $n \ne m, \forall m,n \in \Bbb{N}$ (where the curly braces denote the fractional part)? -I wanted to explore coming up with a kind of "global residue" concept for a number, based on all the numbers less than it. - -REPLY [13 votes]: It follows from Bertrand's postulate that $mH_m - nH_n$ is an integer only in the case $n=m$. Suppose $nm/2$. The only term with $p$ in the denominator in $mH_m$ is $m/p$ which is not an integer (since $m/2

n$. -If $p\le n$, then $nH_n$ has the unique term $n/p$ with $p$ in the denominator. But the only way $m/p - n/p$ can be an integer is if $n=(m-kp)$ (for some integer $k\ge 1$) and we also have $p\le n$, but this is impossible since $p >m/2$.<|endoftext|> -TITLE: Question about an example in symplectic geometry -QUESTION [6 upvotes]: Let M be a coadjoint orbit of dimension 6 of $SU(3)$, and let T be the maximal torus in $SU(3)$. If we denote $\mu : M \longrightarrow \mathbb{R}^2$ the moment map associated to the action of T on M, then the image of the moment map is a hexagon with vertices are image of $M^T$ by $\mu $. -My questions are: -$1.$ What is $M^T$? (My attempt was to choose a regular element $ X \in \mathfrak{t} \simeq \mathfrak{t}^*$, and consider M to be the orbit of X, and then I get $M^T=\lbrace y=gxg^{-1} \in M, ty=yt ,\forall t\in T \rbrace= \mathfrak{t} ?).$ -$2.$ why is the image of the moment map a hexagon? Well, I know from convexity theorem that the image of the moment map, $\mu(M)$, is the convex hull of $\lbrace \mu(F)$, F connected component of $M^T\rbrace$, and that $\mu$ is constant on each connected component of $M^T$ and this implies that the set $\mu(M^T)$ is finite, but how can we find the components of $M^T$ and the cardinal number of the set $\mu (M^T)$ without having an explicit formula of $\mu$?. -Any feedback would be greatly appreciated! - -REPLY [4 votes]: I'll preserve your notation: $M$ is the coadjoint orbit of a regular semisimple element $X \in \mathfrak t^*$ (which you seem to also call $x$). I also assume we're working in characteristic $0$, or at least not $3$. -The orbit $M$ is neither contained in, nor contains, $\mathfrak t^*$. Rather, a conjugate of $X$ lies in $\mathfrak t^*$ if and only if it is a conjugate by the Weyl group $W = \operatorname N_{\operatorname{SU}(3)}(T)/T$. Thus, since regular elements in this case are strongly regular, $M^T = M \cap \mathfrak t^*$ has order $6$. These are the vertices of your hexagon.<|endoftext|> -TITLE: Can a product of Cohn matrices over the Eisenstein integers with non-zero, non-unit coefficients be a Cohn matrix? -QUESTION [5 upvotes]: For $k > 1$, is it possible that $\begin{pmatrix} a_1 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} a_2 & 1 \\ -1 & 0 \end{pmatrix}\ldots \begin{pmatrix} a_k & 1 \\ -1 & 0 \end{pmatrix} = \pm \begin{pmatrix} b & 1 \\ -1 & 0 \end{pmatrix}$ if $a_1,a_2,\ldots a_k,b$ are Eisenstein integers and $|a_i| > 1$ for $i=1,2,\ldots k$? -If the Eisenstein integers are replaced with the Gaussian integers, this is possible. -$\begin{pmatrix} 3 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 - i & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 + i & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 - i & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} -2 & 1 \\ -1 & 0 \end{pmatrix} = -\begin{pmatrix} i & 1 \\ -1 & 0 \end{pmatrix}$ -This problem came up in my research; I am primarily interested in the case where $b = (\pm 1 \pm \sqrt{-3})/2$, but I suspect that there might not be a solution for any choice of $b$. -I originally asked this question on math.stackexchange (https://math.stackexchange.com/q/3903567/202799), but it seems to be more difficult than I first thought. - -REPLY [3 votes]: To my surprise, not only is there a solution for some $b$, there is actually a very simple infinite family of solutions for every $b$. Let $\omega = \frac{1 + \sqrt{-3}}{2}$. Then -$\begin{pmatrix} a_0 + a_1\omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2 -\omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 + \omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2 - \omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 + \omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2 - \omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} b_0 + b_1\omega & 1 \\ -1 & 0 \end{pmatrix} = -\begin{pmatrix} a_0 + b_0 - 1 + (a_1 + b_1 - 1)\omega & 1 \\ -1 & 0 \end{pmatrix}$ -I wish I could say that there was some clever trick to finding this solution, but I cannot: it was found by a brute force search through $\approx 5$ million possibilities using Python after I realized that the bottom right coordinate of the product of $n$ Cohn matrices only depends on the inner $n - 2$ matrices. -I do not know if there are shorter solutions. One can prove that there are no solutions with $k < 5$, but there could be solutions with $k = 5$ or $k = 6$. If there are, however, they have to involve some elements with squared norm at least $4$, as I checked everything smaller.<|endoftext|> -TITLE: How can one recover/obtain information from the renormalization group procedure? -QUESTION [6 upvotes]: I know the basic idea behind the renormalization group approach as it is used in mathematical physics to study both QFT and statistical mechanics. However, I have trouble understanding how can one recover the information one was trying to obtain using this technique. Let me elaborate. -Although there is no such thing as a 'general approach' to RG, I want to try to sketch the ideas from a generic model. As stated in Brydges & Kennedy's article, one starts with integrals of the form: -\begin{eqnarray} -Z(\varphi) = \int d\mu_{C}(\psi) e^{-V_{0}(\psi+\varphi)} = (\mu_{C}*e^{-V_{0}})(\varphi) \tag{1}\label{1} -\end{eqnarray} -where $\varphi = (\varphi_{x})_{x\in \Lambda}$ is a Gaussian process with joint distribution $\mu_{C}$, mean zero and covariance $C$. Suppose we can write $C$ as a sum $C=C_{1}+C_{2}$. Then: -\begin{eqnarray} -Z(\varphi) = \int d\mu_{C_{1}+C_{2}}(\psi)e^{-V_{0}(\psi+\varphi)} = \int d\mu_{C_{2}}(\zeta)\int d\mu_{C_{1}}(\psi) e^{-V_{0}(\psi+\varphi+\zeta)} = \int d\mu_{C_{2}}(\zeta) (\mu_{C_{1}}*e^{-V_{0}})(\varphi+\zeta) = \int d\mu_{C_{2}}(\zeta)e^{-V_{1}(\varphi+\zeta)} \tag{2}\label{2} -\end{eqnarray} -where: -\begin{eqnarray} -V_{1} = -\ln \mu_{C_{1}}*e^{-V_{0}} \tag{3}\label{3} -\end{eqnarray} -Thus, we can define a map on the (informal) space of actions, called renormalization group map and denoted by $RG$, such that $RG: V_{0} \to V_{1}$. Analogously, if $C=C_{1}+\cdots +C_{n}$, $n \ge 2$, then sucessive applications of (\ref{2}) lead to: -\begin{eqnarray} -Z(\varphi) = \int d\mu_{C_{n}}(\zeta_{n})(\mu_{C_{n-1}}*e^{-V_{n-1}})(\varphi+\zeta_{n}) = \int d\mu_{C_{n}}(\zeta_{n})e^{-V_{n}(\varphi+\zeta_{n})} \tag{4}\label{4} -\end{eqnarray} -where $V_{n}= RG(V_{n-1})=\cdots = RG^{n-1}(V_{0})$. We thus defined a 'trajectory' $V_{0}\to V_{1}\to V_{2}\to \cdots \to V_{n}$. -All these being said, I believe the main idea of the process is to (luckily) prove that the above trajectory ends up in a fixed point. In other words, luckily we have $RG^{n}(V_{0}) = V^{*}$ for every $n$ suficiently large. -The above scenario, although very generic, appears in some discussions on the topic. As an example, see Salmhofer's book. -Now comes my questions. -(1) How can one recover the information about $Z(\varphi)$ once we was lucky and obtained $V^{*}$? See, $Z(\varphi)$ was our object of study in the first place, right? But I don't see how to get back and obtain it. -(2) One situation in which the covariance splits into a sum of covariances is when one is trying to approach the continuum limit from a scaled lattice. This can be done in either QFT and statistical mechanics, but I believe it is more common for QFT models. But when we think about statistical mechanics, can one obtain critical temperatures, critical exponents and other thermodynamics entities from the above process? Is it possible to ilustrate how it could be done considering this very generic model? - -REPLY [3 votes]: The limiting function $V^\ast$ is such that any further convolutions of $e^{-V^\ast}$ with $\mu$ return $e^{-V^\ast}$, so $Z^\ast(\phi)=e^{-V^\ast(\phi)}$. - -To obtain critical properties, you need the correlator $K(x,x')=\langle\phi(x)\phi(x')\rangle$. The decay length of the correlator diverges at the critical temperature $T_c$ as a power law $(T-T_c)^{-\alpha}$ and the power $\alpha$ is the critical exponent. The correlator is obtained by adding a source term $\lambda\psi(x)\psi(x')$ to the exponent in the definition of $Z(\phi)$ and then evaluating $dZ/d\lambda$ at $\phi=0$. - - -In reference to the title of the post: "How can one recover/obtain information from the renormalization group procedure?" Information that depends on features that appear at small distances cannot be recovered, it is lost in the renormalization flow (which is not reversible). The information that remains refers to features that persist at large distances, such as a diverging correlation length and the critical exponents associated with it.<|endoftext|> -TITLE: Concentration inequalities for very rare events on a multiplicative scale -QUESTION [9 upvotes]: Let $E_1, \dots, E_N$ be independent events, each of probability $p$, where $p$ is very close to $0$. Let $A_N = \frac{1}{N} ( 1_{E_1} + \dots + 1_{E_N} )$ be the proportion of the events $E_i$ that occur. We expect $A_N$ to be tightly concentrated around its mean $p$. -Suppose we want to estimate something like $\mathbb{P}(A_N > p^{1/2})$. On the one hand, the multiplicative difference $p^{1/2}/p$ is huge, but on the other hand the additive difference $p^{1/2} - p$ is very small. All the standard concentration inequalities (Azuma-Hoeffding, Chernov, etc.) give an upper bound for the above probability in terms of the additive difference, which gives only a very slow exponential decay rate as $N \to \infty$ for the above probability if $p$ is very small. -My question is: which phenomenon is closer to the truth? Should the event $\{A_N > p^{1/2}\}$ be very rare because $p^{1/2}/p$ is huge, or should it be not so rare because $p^{1/2} - p$ is tiny? If the former, are there any references out there for concentration inequalities that capture that? - -REPLY [7 votes]: Let $n:=N$. Let us show that for all natural $n$ and all $p\in(0,1)$ -$$P(A_n>\sqrt p)\le\frac{\sqrt p+p}{1+p},\tag{1}$$ -so that $P(A_n>\sqrt p)\to0$ whenever $p\downarrow0$. -Consider first the case when $n\ge1/\sqrt p$, so that $1/n\le\sqrt p$. -In view of Cantelli's inequality, -$$\begin{aligned} -P(A_n>\sqrt p)&\le\frac{p(1-p)/n}{p(1-p)/n+(\sqrt p-p)^2} \\ -&\le\frac{p(1-p)\sqrt p}{p(1-p)\sqrt p+(\sqrt p-p)^2} \\ -&=\frac{\sqrt p+p}{1+p}, -\end{aligned} -$$ -so that (1) holds if $n\ge1/\sqrt p$. -In the remaining case, when $n<1/\sqrt p$, we have $\sqrt p<1/n$ and hence -$$P(A_n>\sqrt p)=P(A_n>0)=1-(1-p)^n\le np<\sqrt p<\frac{\sqrt p+p}{1+p},$$ -so that (1) again holds. -Thus indeed, (1) holds for all natural $n$ and all $p\in(0,1)$. (It actually holds for all $p\in[0,1]$.)<|endoftext|> -TITLE: Decomposition of real algebraic varieties into manifolds -QUESTION [6 upvotes]: I apologize in advance if this question is too elementary for MO. I am new to the field of algebraic geometry. -I am dealing with a (real) algebraic variety $V$ of (Krull) dimension $n$. I keep reading that $V$ can be decomposed into differentiable manifolds of dimensions $0, 1, ..., n$. -While this seems plausible I could not find this statement in the literature. Is it true? I would also be glad about a reference. - -REPLY [7 votes]: I believe this follows from results in Chapter 9 of Real Algebraic Geometry by Bochnak, Coste, and Roy. In particular, Proposition 9.1.8 implies the strata are Nash manifolds, satisfying certain "niceness" conditions.<|endoftext|> -TITLE: Künneth formula for de Rham cohomology with respect to an integrable connection -QUESTION [6 upvotes]: I am reading through https://stacks.math.columbia.edu/tag/0FM9 which proves that for $X,Y$ schemes over some base $S$ and $X \times _S Y \overset{p}{\rightarrow} X$ resp. $X \times _S Y \overset{q}{\rightarrow} Y$ the projection morphisms, there is an isomorphism of complexes $$ \mathrm{Tot}(p^{-1} \Omega^{\bullet}_{X/S} \otimes_{f^{-1}\mathcal{O}_S} q^{-1} \Omega^{\bullet}_{Y/S}) \rightarrow \Omega_{X\times_S Y} ^{\bullet},$$ where $f:X \times_S Y \rightarrow S$ is the structure map. Now, I am wondering whether we can show something like this for de Rham complexes with respect to integrable connections, i.e. let $(E, \nabla_E)$ and $(F,\nabla_F)$ be vector bundles over $\mathcal{O}_X$ and $\mathcal{O}_Y$ with integrable connections. We then get an integrable connection $p^{\ast}\nabla_E \otimes q^{\ast}\nabla_F$ on $p^{\ast}E \otimes q^{\ast}F$ do we have an isomorphism of complexes $$\mathrm{Tot}(p^{-1}(E \otimes \Omega^{\bullet}_{X/S}) \otimes_{f^{-1}\mathcal{O}_S} q^{-1}(F \otimes \Omega^{\bullet}_{Y/S})) \rightarrow p^{\ast}E \otimes q^{\ast}F \otimes \Omega_{X\times_S Y} ^{\bullet},$$ with the differentials induced by the connections? -EDIT: There seem to be some problems with this StacksPorject section since 0FLS is wrong. Nevertheless, the isomorphism of complexes should still exist if we replace all inverse image functors by pull-backs and tensor over $\mathcal{O}_{X\times_S Y}$ instead of $f^{-1}\mathcal{O}_S$. - -REPLY [3 votes]: The corrected formulation of the K"unneth formula in the Stacks project can also be used to given an answer this question. Let me explain. -Observe that the differentials in the de Rham complex $E \otimes \Omega^\bullet_{X/S}$ of $(E, \nabla_E)$ are differential operators of finite order and similarly for $(F, \nabla_F)$. The construction in Section 0G4A therefore produces a complex $T^\bullet$ on $X \times_S Y$ whose terms are -$$ -T^n = \left(E \boxtimes F\right) \otimes_{\mathcal{O}_{X \times_S Y}} \left( \bigoplus_{n = a + b} \Omega^a_{X/S} \boxtimes \Omega^b_{Y/S}\right) = (E \boxtimes F) \otimes_{\mathcal{O}_{X \times_S Y}} \Omega^n_{X \times_S Y} -$$ -A local calculation shows that the differentials of this complex are coming from the integrable connection $\nabla_{E \boxtimes F}$ we obtain on $E \boxtimes F$ using $\nabla_E$ and $\nabla_F$. -Asssume $S = \text{Spec}(A)$ is affine for simplicity. The discussion in Section 0G4A also constructs a cup product map in the derived category D(A) -$$ -R\Gamma(X, E \otimes \Omega^\bullet_{X/S}) \otimes_A^\mathbf{L} R\Gamma(Y, F \otimes \Omega^\bullet_{Y/S}) \longrightarrow R\Gamma(X \times_S Y, (E \boxtimes F) \otimes \Omega_{X \times_S Y/S}) -$$ -on the (total) de Rham cohomology of the given modules with integrable connections. Of course this won't always be an isomorphism! But Lemma 0FLT shows that is an isomorphism when $X \to S$ and $Y \to S$ are smooth, quasi-compact, and have affine diagonal. If in addition $A$ is a field or more generally if the de Rham cohomology modules $H^i_{dR}(E) = H^i(R\Gamma(X, E \otimes \Omega^\bullet_{X/S}))$ are flat $A$-modules for all $i$, then we obtain the more familiar -$$ -H^m_{dR}(E \boxtimes F) = \bigoplus_{m = i + j} H^i_{dR}(E) \otimes_A H^j_{dR}(F) -$$ -Hope this helps.<|endoftext|> -TITLE: Throwing a fair die until most recent roll is smaller than previous one -QUESTION [12 upvotes]: I roll a fair die with $n>1$ sides until the most recent roll is smaller than the previous one. Let $E_n$ be the expected number of rolls. Do we have $\lim_{n\to\infty} E_n < \infty$? If not, what about $\lim_{n\to\infty} E_n/n$ and $\lim_{n\to\infty} E_n/\log(n)$? - -REPLY [9 votes]: Yet another derivation of the formula $E_n=\left(\frac{n}{n-1}\right)^n$ given by Kasper Andersen. -Write $[m]=\{1,2,\dots, m\}$. -There are ${n+k-1 \choose k}$ non-decreasing functions from $[k]$ to $[n]$. -( Why? The number of such functions $f$ is the same as the number of strictly increasing functions $g$ from $[k]$ to $[n+k-1]$ -- there is a bijection given by $g(i)=f(i)+i-1$. The number of those is ${n+k-1 \choose k}$, since for every subset of $[n+k-1]$ of size $k$, there is precisely one strictly increasing function from $[k]$ to $[n+k-1]$ with that image set. ) -Let $X$ be the number of rolls before the first descent is observed. Then $X\geq k$ if and only if the first $k$ rolls give a non-decreasing function from $[k]$ to $[n]$. The total number of possibilities for the first $k$ rolls is $n^k$, and they are all equally likely, so the -probability that $X\geq k$ is -$\frac{1}{n^k}{n+k-1 \choose k}$. -Then -\begin{align*} -E_n &= \mathbb{E}(1+X)\\ -&=1+\mathbb{E}(X)\\ -&=1+\sum_{k=1}^\infty \mathbb{P}(X\geq k)\\ -&=1+\sum_{k=1}^\infty \frac{1}{n^k}{n+k-1 \choose k}\\ -&=1+\frac{1}{n}{n\choose 1} + \frac{1}{n^2}{n+1\choose 2} -+\frac{1}{n^3}{n+2\choose 3} +\dots\\ -&=\left(1-\frac{1}{n}\right)^{-n} -\end{align*} -by the binomial theorem.<|endoftext|> -TITLE: Autobiographies and correspondences of mathematicians -QUESTION [11 upvotes]: Lately I have enjoyed reading several autobiographies and correspondences of mathematicians. I'd like to find more, so I thought I'd ask here which others you have come across and enjoyed. -P.S. I have collected suggestions from answers to an earlier similar question Autobiography of mathematicians and this one into the following list: -Autobiographies -(1) Souvenirs d'apprentissage, André Weil (In English: Apprenticeship of a Mathematician) -(2) Récoltes et Semailles, Alexander Grothendieck -(3) Un mathématicien aux prises avec le siècle, Laurent Schwartz (In English: A Mathematician Grappling with His Century) -(4) The Map of My Life, Goro Shimura -(5) A Mathematician's Apology, G. H. Hardy -(6) Adventures of a Mathematician, Stanislaw Ulam -(7) I Am a Mathematician, Norbert Wiener -(8) Love and Math, Edward Frenkel -(9) I Want to Be a Mathematician: An Automathography in Three Parts, Paul Halmos -(10) Un mathématicien juif, Häim Brezis -(11) La forme d'une vie: Mémoires (1924-2010), Benoit Maldelbrot (In English: The Fractalist: Memoir of a Scientific Maverick) -(12) Théorème Vivant, Cédric Villani (In English: Birth of a Theorem: A Mathematical Adventure) -(13) To Talk of Many Things, An Autobiography, Dame Kathleen Ollerenshaw -(14) Will to Freedom, Egon Balas -(15) So hab ich's erlebt, Walter Rudin (In English: The Way I Remember It) -(16) Eine Frau und die Mathematik 1933--1940: der Beginn einer wissenschafltichen Laufbahn, Helene Braun -(17) Wspomnienia I zapiski, Hugo Steinhaus (In German: Erinnerungen und Aufzeichnungen) -(18) A Mathematical Autobiography, Saunders Mac Lane -(19) Random Curves: Journeys of a Mathematican, Neal Koblitz -(20) Enigmas of Chance: An Autobiography, Marc Kac -(21) Vospominanifa detstva, Sofya Kovalevskaya (In English: A Russian Childhood) -(22) De Vita propria, Girolamo Cardano (In English: The Book of My Life) -(23) Ex-Prodigy: My Childhood and Youth, Norbert Wiener -(24) Eye of the Hurricane, Richard Bellman -(25) Passages from the life of a philosopher, Charles Babbage -(26) Autobiography of Sir George Biddell Airy -(27) Schrijf dat op, Hans. Knipsels uit een leven, Hans Freudenthal (In English: Write that down, Hans. Excerpts from a life) -(28) A Half Century of Polish Mathematics: Remembrances and Reflections, Kuratowski, Kazimierz -(29) La mia vita di matematico attraverso la cronistoria dei miei lavori, Francesco Giacomo Tricomi -(30) Pushing limits : From West Point to Berkeley & Beyond, Ted Hill -(31) My Search for Ramanujan: How I Learned to Count, Ken Ono and Amir D. Aczel -Correspondence -(1) Grothendieck-Serre Correspondence (AMS/SMF) -(2) Correspondance Serre-Tate (SMF) -(3) Ramanujan: Letters and Commentary (AMS) - -REPLY [3 votes]: I think you will enjoy Norbert Wiener's autobiographic "I Am a Mathematician". - -REPLY [3 votes]: Edward Frenkel's "Love and Math" is a mix of popular maths book, autobiography, and general declaration of love towards mathematics. - -REPLY [3 votes]: Paul R. Halmos: I Want to Be a Mathematician: An Automathography. -Benoit B. Maldelbrot: La forme d'une vie: Mémoires (1924-2010). -Cédric Villani: Théorème vivant. - -REPLY [3 votes]: Haïm Brezis: Un mathématicien juif. -I don't know whether it has been translated in English.<|endoftext|> -TITLE: Gödel on pure mathematics and medieval theology -QUESTION [10 upvotes]: I was watching this youtube video recently where Gregory Chaitin paraphrases something from one of Gödel's unpublished essays (apparently published now). It is at the 4:48 mark of the video Gregory Chaitin - Is Mathematics Invented or Discovered? -Does anyone know which particular essay Chaitin is referring to? As a pure mathematician I found the whole interview to be fascinating and would like to read more about this from Gödel himself. I did some quick searches from some volumes containing Gödel's essays but did not find these passages. - -REPLY [12 votes]: I am quite certain a paraphrase of the quote "the only place where medieval theology survives is pure math", attributed to one of Gödel's essays by Chaitin, does not appear in his collected works (see Vol III: Unpublished essays and lectures). (At least a search for "medieval" and "theology" only returns references to Gödel's ontological proof for the existence of God.) -However, if I understand the question in the OP more generally as an interest in this line of thought of Gödel, then the 1961 essay The modern development of the foundations of mathematics in the light of philosophy develops it as follows: - -I would like to attempt here to describe, in terms of philosophical -concepts, the development of foundational research in mathematics -since around the turn of the century, and to fit it into a general -schema of possible philosophical world-views. For this, it is -necessary first of all to become clear about the schema itself. I -believe that the most fruitful principle for gaining an overall view -of the possible world-views will be to divide them up according to the -degree and the manner of their affinity to or, respectively, turning -away from metaphysics (or religion). In this way we immediately obtain -a division into two groups: skepticism, materialism and positivism -stand on one side, spiritualism, idealism and theology on the other. -[...] -Now it is a familiar fact, even a platitude, that the development of -philosophy since the Renaissance has by and large gone from right to -left - not in a straight line, but with reverses, yet still, on the -whole. Particularly in physics, this development has reached a peak in -our own time. [...] It would truly be a miracle if this (I would like -to say rabid) development had not also begun to make itself felt in -the conception of mathematics. Actually, mathematics, by its nature as -an a priori science, always has, in and of itself, an inclination -toward the right, and, for this reason, has long withstood the spirit -of the time that has ruled since the Renaissance. - -In connection with the topic of Gregory Chaitin's interview, the question whether mathematics is invented or discovered: Gödel was inclined to the latter, and in a 1951 essay Some basic theorems on the foundations of mathematics and their implications cited Hermite: There exists, unless I am mistaken, an entire world consisting of the totality of mathematical truths, which is accessible to us only through our intelligence, just as there exists the world of physical realities; each one is independent of us, both of them divinely created.<|endoftext|> -TITLE: Applications of Robinson's consistency theorem in algebra? -QUESTION [11 upvotes]: This is crossposted from MSE. It's also my first time asking on MO, so please let me know if there's anything you need from me! - -There are a family of results which, in many model theory books, are proven around the same time, often as corollaries of each other. These are - -The Craig interpolation theorem -The Robinson consistency theorem -The Beth definability theorem - -For many "early" topics in model theory, there are some obvious results in algebra which admit easy proofs using this machinery. Compactness, completeness, and Lowenheim-Skolem all come to mind. Marker even finds low-hanging applications of o-minimality! However, I can't seem to find any algebraic applications of these theorems, even though they seem just as applicable as the other concepts I've mentioned. -I'm sure there are special cases where one might like to know that a relation (say, the ordering of a group, etc.?) is not definable from the rest of the structure (here Beth's Theorem and Padoa's Method might be useful). Similarly, knowing that we can find a model of $T_1 \cup T_2$ provided there's no obvious obstruction seems eminently useful (here I have in mind the version of Robinson's theorem that doesn't rely on completeness of $T_1 \cap T_2$). I'm not so surprised that I can't come up with any applications myself (sometimes being creative is hard), but I am surprised that searching my usual references, as well as spending some time on google, hasn't turned up any results. -Does anyone have any fun applications of these theorems in algebra? -Thanks in advance ^_^ - -REPLY [5 votes]: I don't know of a nice purely algebraic application off the top of my head, I don't even know of many applications to model theory. I was actually pretty happy with I found some application of Craig interpolation to model theory a few years back with Minh Chieu Tran and Alex Kruckman. -I can mention an application to model-theoretic algebra. Koeniggsmann and Jahnke show that if K is a field which admits a non-trivial Henselian valuation and is neither separably nor real closed then K defines a valuation which induces the Henselian topology. (Any two non-trivial Henselian valuations on a field induce the same topology, and they do not show that the definable valuation is Henselian.) Their proof makes crucial use of Beth definibility. -This is in the paper "Uniformly defining p-Henselian valuations", which you can find here: https://arxiv.org/pdf/1407.8156.pdf. It may also be worth mentioning that if K is either separably closed or real closed then K does not admit a non-trivial definable valuation (in either case this follows from the relevant quantifier elimination.) -It is my understanding that Craig interpolation has applications to computer science, but I'm not very familiar with this.<|endoftext|> -TITLE: Is it true that $\{x^4+y^2+z^2:\ x,y,z\in\mathbb Z[i]\}=\{a+2bi:\ a,b\in\mathbb Z\}$? -QUESTION [18 upvotes]: Recall that the ring of Gaussian integers is -$$\mathbb Z[i]=\{a+bi:\ a,b\in\mathbb Z\}.$$ -Clearly -$$(a+bi)^2=a^2-b^2+2abi\ \ \mbox{and}\ \ (a+bi)^4=(a^2-b^2)^2-4a^2b^2+4ab(a^2-b^2)i.$$ -Question. Is it true that $\{x^4+y^2+z^2:\ x,y,z\in\mathbb Z[i]\}=\{a+2bi:\ a,b\in\mathbb Z\}$? -Evidence. Via Mathematica I have found that -\begin{align} &\{x^4+y^2+z^2:\ x,y,z\in\{r+si:\ r,s\in\{-14,\ldots,14\}\}\} -\\&\quad \supseteq\{a+2bi:\ a,b\in\mathbb Z\ \mbox{and}\ |a+2bi|\le 50\}. -\end{align} -For example, -$$43+22i=2^4+(14-11i)^2+(11-13i)^2$$ -and -$$-34+26i=(2+i)^4+(13-i)^2+(1+14i)^2=(4+i)^4+(11-11i)^2+(1+14i)^2.$$ -Motivation. The question is motivated by my following result (cf. my 2017 JNT paper) -$$\{x^4+y^2+z^2+w^2:\ x,y,z,w=0,1,2,\ldots\}=\{0,1,2,\ldots\}$$ -which refines Lagrange's four-square theorem. -I conjecture that the question has a positive answer, but I'm unable to prove this. Your comments are welcome! - -REPLY [27 votes]: Yes, it is true that $\{ x^{4} + y^{2} + z^{2} : x, y, z \in \mathbb{Z}[i] \} = \{ a + 2bi : a, b \in \mathbb{Z} \}$. Indeed, one can even take $x$ to be either $0$ or $1$ in all cases. Because $y^{2}+z^{2} = (y+iz)(y-iz)$ is reducible, this is analogous to the statement that every integer can be written in the form $x^{2}-y^{2}$ or $x^{2}-y^{2} + 1$ with $x, y \in \mathbb{Z}$. -Suppose that $a, b \in \mathbb{Z}$ and $a$ is odd. Then -$$ - a + 2bi = 0^{4} + \left(\frac{a+1}{2} + bi\right)^{2} + \left(b - \left(\frac{a-1}{2}\right)i\right)^{2}. -$$ -If $a, b \in \mathbb{Z}$ and $a$ is even, then $a-1$ is odd, and the identity above allows one to write $a-1 + 2bi = y^{2} + z^{2}$ with $y, z \in \mathbb{Z}[i]$. Hence, $a + 2bi = x^{4} + y^{2} + z^{2}$ with $y, z \in \mathbb{Z}[i]$ and $x = 1$.<|endoftext|> -TITLE: Arithmetic zeta functions of products and fibrations -QUESTION [5 upvotes]: Suppose $X$ and $Y$ are schemes of finite type over $\mathbb{Z}$. -How is the arithmetic zeta function of their product, $\zeta_{X \times Y} (s)$, related to their individual zeta functions, $\zeta_X(s)$ and $\zeta_Y(s)$? More generally if $Z$ is a fiber bundle over $X$ with fiber $Y$, does a similar relation hold? -(I have been wondering if $\ln \zeta(s)$ is a ring homomorphism to an appropriate target ring from the Grothendieck ring of schemes of finite type over $\mathbb{Z}$.) - -REPLY [3 votes]: We have the relation $\zeta_{X\times Y}(s) = \zeta_X(s)* \zeta_Y(s)$ where $*$ is the Witt product in the Witt ring of $\mathbb Z[[t]]$. For any commutative ring A, the (big) Witt ring $W(A)$ is defined by: - -Under addition, $W(A), +$ is isomorphic to the group $(1 + tA[[t]],\times)$ -The multiplication $*$ is uniquely determined by $$(1-at)^{-1}(1-bt)^{-1} = (1-abt)^{-1}.$$ - -That this is indeed true can be verified on each Euler product: the $k$-points of $X\times Y$ are in bijection with pairs of $k$-points on $X,Y$. In fact, the same relation holds if we only assume that $Z$ is a fiber bundle over $X$ with fiber $Y$ by the same fact about points: to give a point $P$ on $Z$ is to give a point on $X$ (the image of $P$ under projection) and a point on the fiber. -A reference for this is https://arxiv.org/abs/1407.1813 and can also be found on the Wikipedia page: https://en.wikipedia.org/wiki/Arithmetic_zeta_function -[The reference only talks about this formula over finite fields but I think it holds for the zeta function of any scheme $X$ finite type over $\mathbb Z$ because such a zeta function is the product of a zeta function $X/\mathbb F_p$ over every finite field. As an example, the Riemann zeta function is the product of the zeta function of a point over every $\mathbb F_p$. Since the formula is proven Euler product by euler product, this is not an issue.<|endoftext|> -TITLE: Do you know which is the minimal local ring that is not isomorphic to its opposite? -QUESTION [5 upvotes]: The most popular examples are non-local rings and minimal has 16 elements. I am interested in knowing examples of local rings not isomorphic to their opposite. - -REPLY [6 votes]: I learned this example from MO-user Johannes Hahn: -The algebra is $A=K/(x^3,y*x,y^2,x^2*y)$ over a field $K$ with 2 elements. -Then $A$ as an $A$-module as 20 submodules, but $A^{op}$ as an $A^{op}$-module has 16 submodules. -Thus $A$ and $A^{op}$ are not isomorphic. -This also gives an example where $A$ and $A^{op}$ are not derived equivalent (since local algebras are derived equivalent iff they are isomorphic). Another argument (that works for any field $K$) is that $\Omega_A^{1}(I)$ has dimension 5 but $\Omega_{A^{op}}^{1}(I)$ has dimension 10 when $I$ is the indecomposable injective module. -One might wonder whether a finite local algebra over a finite field is isomorphic to its opposite algebra if and only if the number of submodules of the regular module coincide. -It might also be interesting to see a selfinjective local algebra not isomorphic to its opposite algebra.<|endoftext|> -TITLE: Constructing M-curves à la Hilbert -QUESTION [6 upvotes]: I have been reading some text about Harnack's theorem. The theorem basically says that for degree $d$, the maximal number of connected components in the real (projective) plane of a plane curve with degree $d$ is $(d-1)(d-2)/2 + 1$. A curve with this maximal number of component is called an M-curve. I am trying to construct such a curve with maximal number of components. There are several algorithms known and the oldest ones are from Harnack himself and from Hilbert. The algorithm of Hilbert for even degree curves is as follows: -Let us say $d=4$, so we expect an M-curve to have 4 components (it is proven that such components are oval, i.e. they are closed loop in the affine plane). To construct such a curve Hilbert proposes to first take two ellipses (say defined by equation $e_1(x,y)=0, e_2(x,y)=0$) intersecting in 4 real points $P_1,P_2,P_3,P_4$ (preferably with major axes orthogonal to each other) in an order such that the ellipses are partitioned into arcs $P_1P_2, P_2P_3, P_3P_4, P_4P_1$. Then we take lines defined by equations $l_1=0,\dots,l_4=0$ such that they intersect say the arc $P_1P_2$. We then take a real number $\epsilon$, with $|\epsilon|$ rather small and now construct the zero set of the equation $e_1e_2 + \epsilon l_1l_2l_3l_4 = 0$ (perturbation of the two ellipse with 4 lines). According to several literature, we should be able to construct 4 ovals shown in the figure below (the result should look like the dashed ovals in the figure). I got the image and the the algorithm from the survey of Gudkov : Topology of real projective varieties. -$P_1P_2$ is the $z_3z_4$ in the image" /> -My problem with this is that I can construct 4 oval components. But it seems that no matter how I try, the ovals will not be the same as the special oval in the figure with 4 "prongs" (I hope you know what I mean). My result often looks as follows: - -I want to be able to do this, because it seems relevant for the recursive procedure of the algorithm (to construct an M-curve with degree $d=6$ from our result with $d=4$), we need one of the original ellipse intersect this special oval at 8 points. I was wondering if anyone was able to construct a similar image using a computer algebra system. I share my Mathematica code (below) here with the result. I understand that I have to tweak the perturbation with the lines and where the lines exactly lie to get different shape and I even did this interactively but I don't seem to get the image of Gudkov (probably originally of Hilbert). Am I missing something here? Is there something wrong with my interpretation of this construction? Even if someone is able to get this 4 components as illustrated by Gudkov, I was wondering if there is any intuition on how we choose our perturbation and the slopes of these lines that could help me find the correct parameter much easier than how I currently do (which is mostly by trial and error). These lines do act as "attractors" so that their distances from each other and from the previous M-curve help shape these ovals in a way that I like. This is the only "intuition" I have of them. Anyone has a tip on how to efficiently construct these M-curves using Hilbert's algorithm? -e1 = x^2/4+y^2/2-1; -e2 = x^2+y^2/4-1; -f=e1*e2; - -(*horizontal lines that intersect the arc at 4 points *) -l=(y+1)(y+0.5)(y-0.5)(y-1); -eps = 0.01; -ContourPlot[{l==0,f==0},{x,-5,5},{y,-5,5}] - -(* 4 components! *) -ContourPlot[f+eps*l==0,{x,-5,5},{y,-5,5},MaxRecursion->5] - -(*vertical lines that intersect the arc at 4 points *) -l=(x+1)(x+4)(x+3)(x+3.5); -eps = 0.001; -ContourPlot[{l==0,f==0},{x,-5,5},{y,-5,5}] - -(* 4 components! *) -ContourPlot[f+eps*l==0,{x,-5,5},{y,-5,5},MaxRecursion->5] - -REPLY [2 votes]: As Zach Teitler hinted, I shouldn't have checked it visually. In fact, if you make the construction correctly, it is quite easy to show that you have 8 real intersection with one of the ellipse and one component of the M-curve: -For instance, if we use four horizontal lines all lying in the y-range of say the above arc. Then by the nature of the position of the ellipses, there are four real intersections of the lines and the (top arc of the) horizontal ellipse in the first quadrant (thus 8 intersection in total with this ellipse). These four intersections are also the zeros of the new equation $e_1e_2+\epsilon l_1l_2l_3l_4$. This more or less justifies the 4 prongs. Visually I tested it in Mathematica, if I still have the 4 components (I made sure they are visible by modifying the ellipse in my orgiinal post a bit): -e1 = x^2/4 + 2 y^2 - 1; -e2 = x^2/3 + y^2/4 - 1; -f = e1*e2; -(* make sure that the horizontal lines are intersecting the top arc of the horizontal ellipse *) -NSolve[{e1==0,e2==0}] -l=(y-9/14)(y-5/7)(y-7/11)(y-5/8); -eps = 1/100; -(* visually check if we indeed have 4 components *) -ContourPlot[f+eps*l==0,{x,-3,3},{y,-2,2},MaxRecursion->5] - - -I used rational numbers to make sure the CAS will not use any numerical methods to solve. Now testing in Maple if the 4 "prongs" really exists. I usually don't do this (i.e. I use Groebner basis) but since the system of equations is rather easy, Maple will not miss any solutions if you just use "solve": -e1 := x^2/4 + 2*y^2 - 1: -e2 := x^2/3 + y^2/4 - 1: -f=e1*e2: -l:=(y-9/14)*(y-5/7)*(y-7/11)*(y-5/8); -evalf(l); -eps := 1/100; -#check intersection with horizontal ellipse -solve({e1,e1*e2+eps*l}); -#yields: {x = 1/2*RootOf(2*_Z^2-7), y = 5/8}, {x = 2/7*RootOf(_Z^2+1), y = 5/7}, {x = 2/11*RootOf(_Z^2-23), y = 7/11}, {x = 1/7*RootOf(_Z^2-34), y = 9/14}<|endoftext|> -TITLE: What is the Levi-Civita connection trying to describe? -QUESTION [42 upvotes]: I have seen similar questions, but none of the answers relate to my difficulty, which I will now proceed to convey. -Let $(M,g)$ be a Riemannian manifolds. The Levi-Civita connection is the unique connection that satisfies two conditions: agreeing with the metric, and being torsion-free. -Agreeing with the metric is easy to understand. This is equivalent to the parallel transport associated with the connection to satisfy that the isomorphism between tangent spaces at different points along a path are isometries. Makes sense. -Let's imagine for a second what happens if we stop with this condition, and take the case of $M=\mathbb{R}^2$, with $g$ being the usual metric. Then it's easy to think of non-trivial ways to define parallel transport other than the one induced by the Levi-Civita connection. -For example, imagine the following way to do parallel transport: if $\gamma$ is a path in $\mathbb{R}^2$, then the associated map from $TM_{\gamma(s)}$ to $TM_{\gamma(t)}$ will be a rotation based with angle $p_2(\gamma(s))-p_2(\gamma(t))$, where $p_i$ is the projection of $\mathbb{R}^2$ onto the $i^\text{th}$ coordinate. -So I guess torsion-free-ness is supposed to rule this kind of example out. -Now I'm somewhat confused. One of the answers to a similar question that any two connections that satisfy that they agree with the metric satisfy that they have the same geodesics, and in that case choosing a torsion-free one is just a way of choosing a canonical one. That seems incorrect, as $\gamma(t)=(0,t)$ is a geodesic of $\mathbb{R}^2$ with the Levi-Civita connection but not the one I just described... -Let's think from a different direction. In the case of $\mathbb{R}^2$, if $\nabla$ is the usual (and therefore Levi-Civita) connection then $\nabla_XY$ is just $XY$, and $\nabla_YX$ is just $YX$. So of course we have torsion-free-ness. -So I guess one way to think of torsion-free-ness is saying that you want the parallel transport induced by the connection to be the one associated with $\mathbb{R}^n$ via the local trivializations. -Except that this seems over-simplistic: torsion-free-ness is weaker than the condition that $\nabla_XY=XY$ and $\nabla_YX=YX$. So why this crazy weaker condition that $\nabla_XY-\nabla_YX=[X, Y]$? What does that even mean geometrically? Why is this sensible? How would say that in words that are similar to "it means that the connection is the connection induced from the trivializations" except more correct than that? - -REPLY [15 votes]: Without loss of generality (Nash embedding theorem) we may assume the Riemannian manifold is an embedded submanifold of Euclidean space: its metric at any point is just the restriction of the Euclidean inner product to the tangent plane. Imagine we live on this submanifold (just like we live on a sphere called Earth) and we want to calculate things, such as our acceleration as we run around our planet. -Remember, the metric gives us a means of measuring distances and angles, but no direct way of computing rates-of-change of vector fields. A connection is what determines the rates-of-change of vector fields (such as acceleration, which is the rate-of-change of velocity vectors). And connections are just "infinitesimal limits" of parallel transport. So the question becomes, given a submanifold of Euclidean space, is there a canonical way of defining parallel transport which is useful in some way? -Often things are "useful" if they correspond to what happens in the real world. So how should parallel transport be defined on our planet? How is it defined on Earth? -The very first thing might be to agree on what path we would take if we are told to walk in a straight line. If we did this on Earth, we would walk along a great circle even though we think we are walking in a straight line. Why? Because after each level step we take, gravity pulls our foot back down to Earth. We think we are going straight, but gravity causes our path to curve in the ambient Euclidean space. (For what it is worth, we tend to interpret this "curve" that gravity induces in our path, as the least change required to keep us on the surface of our planet, so to speak.) -Requirement 1: When we are told to walk in a straight line, the curve we actually trace out (due to gravity, or mathematically, due to Euclidean projection back to the submanifold) should be a geodesic, i.e., have zero acceleration. -Now, imagine as we walk, we are holding a lance. Maybe the lance is pointing straight ahead, but maybe it is pointing to our left. Regardless, we are told not to move the lance as we walk in a straight line. Now, from the perspective of the ambient Euclidean space, where the lance points is going to change as we walk. But from our perspective, we are very comfortable being told to walk without moving the lance. We want the evolution of the lance's position to correspond to parallel transport. Indeed, parallel transport defines how a vector is moved along a curve, and it is quite natural/useful to define parallel transport to be what results if we are told to walk with the lance/vector in our hand without moving it at all. The curvature of the Earth causes it to move, but we believe we are not moving it. -Requirement 2: Parallel transport corresponds to carrying a "vector" with us as we walk along a path without consciously moving the vector. (This actually includes Requirement 1 as a special case when the vector is our own velocity vector.) -These requirements uniquely define the Levi-Civita connection and explain why it is natural/useful. It corresponds to the world we live in. -Now, a few words can be said about the usual axioms used to define the Levi-Civita connection: metric connection with zero torsion. The metric connection means when we parallel transport vectors, their norms and the angles between them do not change. Certainly, if we are carrying two lances and told not to move them, we expect the angle between them to stay the same, and we expect the length of each lance to stay the same too. This on its own is not enough for geodesics to be the "correct" curves, i.e., those curves that result when we are told to walk in a straight line. Torsion actually decomposes into two parts (see Millman's 1971 paper "Geodesics in Metrical Connections"). One part controls what geodesics look like, and the other part determines whether parallel transport will cause a vector to spin orthogonal to the direction of motion along a geodesic. If we start holding a lance straight up (it wouldn't be in the tangent plane but ignore this technicality or think in higher dimensions), but as we walk straight ahead, we rotate the lance so it goes from pointing up to pointing right, then down, then left, then up etc, then our parallel transport has torsion. Hence, taken together, a metric connection with zero torsion gives us the definition of parallel transport corresponding to "do not move the vector as you walk along the curve". This is the Levi-Civita connection. -ps. In Appendix 1.D of the second edition of "Mathematical Methods of Classical Mechanics" by Arnold, a geometric way of constructing parallel transport to have no torsion is explained. Given a tangent vector at a point on a geodesic, the aim is to transport it without altering it any more than necessary, as explained above. Without a Euclidean embedding, this can be done intrinsically by considering families of geodesic curves (see Appendix 1.D of Arnold's book). The infinitesimal requirement reduces to the no-torsion equation $\nabla_X Y - \nabla_Y X = [X,Y]$. Thus, the geometric meaning of $\nabla_X Y - \nabla_Y X = [X,Y]$ is that parallel transport will not induce any extraneous movement of the tangent vector. (The geometric picture in Appendix 1.D of Arnold takes a few paragraphs to explain even though the concept itself is straightforward enough.)<|endoftext|> -TITLE: What is the subgroup of $\mathrm{SL}(n,\mathbb{C})$ which preserves the discriminant? -QUESTION [8 upvotes]: $\DeclareMathOperator{\SL}{\operatorname{SL}}$Let $\mathcal{P}_{n-1}$ be the space of complex polynomials in one variable, say $z$, of degree at most $n-1$. As a complex vector space, it is clearly $n$-dimensional. Consider the basis $1$, $z,\ldots,z^{n-1}$ of $\mathcal{P}_{n-1}$. This allows us to identify $\mathcal{P}_{n-1}$ with $\mathbb{C}^n$. -The complex Lie group $\SL(n,\mathbb{C})$ acts on $\mathbb{C}^n$, and thus, via our identification, it acts on $\mathcal{P}_{n-1}$. The discriminant is a homogeneous polynomial on $\mathcal{P}_{n-1}$ of degree $2n-2$. -My question can now be formulated. What is the subgroup of $\SL(n,\mathbb{C})$ which preserves the discriminant (using our identification of $\mathcal{P}_{n-1}$ with $\mathbb{C}^n$)? I have a feeling it is the image of $\SL(2,\mathbb{C})$ under a principal homomorphism from $\SL(2,\mathbb{C})$ to $\SL(n,\mathbb{C})$. -More specifically, $\SL(2,\mathbb{C})$ acts simultaneously on all linear factors of a polynomial of degree $n-1$, and this action preserves the discriminant. I think these are all "projective" transformations which preserve the discriminant, but I am not sure how to show that. Edit: I should really write that I think these are all transformations in $\SL(n,\mathbb{C})$ which preserve the discriminant (see the remark by @NoamD.Elkies below). -Edit 2: here is a conceptual proof for polynomials of degree at most $2$ (i.e. for $n=3$ using my notation). Let $L$ be the map on $\mathbb{C}P^2$ induced by some element $g \in \SL(3,\mathbb{C})$ which preserves the discriminant of quadratic polynomials. Note that we can think of $\mathbb{C}P^2$ as the set of unordered pairs of points on $\mathbb{C}P^1$ (namely the roots of the corresponding polynomial, viewed up to scaling). Note also that polynomials of vanishing discriminant correspond to an unordered pair of coinciding points on $\mathbb{C}P^1$. -Let $f$ be the holomorphic map from $\mathbb{C}P^1$ to $\mathbb{C}P^1$ obtained essentially by restricting $L$ to coinciding pairs of points on $\mathbb{C}P^1$, in turn corresponding to the vanishing locus of the discriminant. Note that $f$ induces a holomorphic map $\tilde{f}$ on the set of unordered pairs of points on $\mathbb{C}P^1$, and thus on $\mathbb{C}P^2$. -Moreover, $\tilde{f}$ and $L$ agree on the conic in $\mathbb{C}P^1$ coinciding to the vanishing locus of the discriminant, and thus they must be equal, since they are two linear isomorphism maps from $\mathbb{C}P^2$ to itself which agree on a conic in $\mathbb{C}P^2$. -We have thus shown the claim for $n=3$. I am hoping this can be generalized to higher $n$'s. I think it is doable. -Edit 3: the previous argument can be generalized using an additional hypothesis, that the element $g \in \SL(n,\mathbb{C})$ not only preserves the discriminant, but also maps any polynomial with a single root of multiplicity $n-1$ to a polynomial with a single root of multiplicity $n-1$. Under this additional hypothesis, essentially the same argument proves that $L$, which is the map on $\mathbb{C}P^{n-1}$ induced by $g$, is induced by some holomorphic automorphism of $\mathbb{C}P^1$, and thus by some element of $\SL(2,\mathbb{C})$. But is it necessary to make this assumption? Is my conclusion false without it? Or is it perhaps the case that this hypothesis can be proved? - -REPLY [9 votes]: At the request of the OP, I post my comment as an answer. View $\mathcal{P}_n$ as the space of homogeneous polynomials of degree $n$ in 2 variables. Let $\Delta _p\subset \mathcal{P}_n$ be the locus of polynomials with one linear factor of multiplicity $\geq p$. One can show that -the singular locus of $\Delta _p$ is $\Delta _{p+1}$. Therefore the subgroup $G$ of $\operatorname{GL}(\mathcal{P}_n) $ preserving the discriminant hypersurface $\Delta _2$ preserves $\Delta _n$. Now the image of $\Delta _n$ in $\mathbb{P}(\mathcal{P}_n)\cong \mathbb{P}^n$ is a rational normal curve, that is, the image of the $n$-th Veronese embedding $V_n:\mathbb{P}^1\hookrightarrow \mathbb{P}^n$. Thus up to homotheties, $G$ is the group of automorphisms of $\mathbb{P}^n$ preserving -$V_n(\mathbb{P}^1)$, which maps isomorphically to $\operatorname{Aut}(\mathbb{P}^1)=\operatorname{PGL}(2,\mathbb{C}) $.<|endoftext|> -TITLE: Is there an example of a non measurable function with a measurable graph? -QUESTION [8 upvotes]: Let $(S, \Sigma)$ be a measurable space. Let $f: S \rightarrow \mathbb{R}$ be function and let $\mathcal{B}(\mathbb{R})$ be Borel $\sigma$-algebra on $\mathbb{R}$. Let $G(f)$ be the graph of $f$, that means -$ G(f)=\{(x,f(x)) \in S\times \mathbb{R} : x \in S\} $. We write $ \Sigma \otimes \mathcal{B}(\mathbb{R})$ to indicate the product $\sigma$-algebra of $\Sigma$ and $\mathcal{B}(\mathbb{R})$. -We can prove that: - -If $f$ is $\Sigma$-$\mathcal{B}(\mathbb{R})$ measurable then $G(f) \in \Sigma \otimes \mathcal{B}(\mathbb{R})$. -Assuming that $S$ is a Polish space and $\Sigma$ is its Borel $\sigma$-algebra, if $G(f) \in \Sigma \otimes \mathcal{B}(\mathbb{R})$ then $f$ is $\Sigma$-$\mathcal{B}(\mathbb{R})$ measurable. - -The proof of item 1 is rather easy. The proof of item 2 is more advanced and uses analytic sets. One such proof can found in here (section 3 proposition 6). -Question: Can we prove that if $G(f) \in \Sigma \otimes \mathcal{B}(\mathbb{R})$ then $f$ is $\Sigma$-$\mathcal{B}(\mathbb{R})$ measurable in the general case (that is, just having $(S, \Sigma)$ a measurable space)? If not, is there a counterexample (that is, an example where $G(f) \in \Sigma \otimes \mathcal{B}(\mathbb{R})$, but $f$ is not $\Sigma$-$\mathcal{B}(\mathbb{R})$ measurable)? -(Of course changing the $\sigma$-algebra in the counter-domain makes it trivial to have counter-examples). -Remark: Examples where $G(f) \notin \Sigma \otimes \mathcal{B}(\mathbb{R})$ and $f$ is not $\Sigma$-$\mathcal{B}(\mathbb{R})$ measurable are rather trivial to produce. They are not what I am asking for. - -Let $X=[0,1]$, $\Sigma$ be the Borel $\sigma$-algebra in $[0,1]$, let $A\subseteq [0,1]$ be a set not in $\Sigma$. Let $f$ be $\chi_A$ (the indicator function) of $A$. Its is easy to see that $G(f) \notin \Sigma \otimes \mathcal{B}(\mathbb{R})$, and $f$ is not $\Sigma$-$\mathcal{B}(\mathbb{R})$ measurable. - -REPLY [8 votes]: See Srivastava, A Course on Borel Sets, in the section "Solovay's Coding of Borel Sets", beginning at "We now proceed to give an example of a function with domain coanalytic whose graph is Borel and that is not Borel measurable." Google books preview (There is a typo: it should say $E\subseteq C\times \mathbb R\times 2^{\mathbb N}$ instead of $E\subseteq \mathbb R\times 2^{\mathbb N}.$) -The example is a function $f:C\times \mathbb R\to2^{\mathbb N}$ where $C$ is a set of codes for Borel sets. Of course $\mathbb R$ is Borel isomorphic to $2^{\mathbb N}.$<|endoftext|> -TITLE: Do you know an elegant proof for this expression for a Schur function? -QUESTION [14 upvotes]: I know that the identity -$$ -s_\mu = \sum_{\mu-\lambda \text{ is a horizontal strip}} \;\sum_{\alpha\vdash|\lambda|} \frac{\chi^\lambda_\alpha}{z_\alpha} \prod_i(p_i-1)^{a_i} -$$ -holds. -Here $\alpha=1^{a_1}2^{a_2}\dotsb$ in exponential notation; in other words, $a_i$ is the number of times that $i$ occurs in $\alpha$. -I know this because I can show that it is essentially equivalent to the formula I.5 of Garsia and Goupil EJC 2009. -Note that the equality of top-degree terms is the well-known expansion of a Schur function in terms of power sum symmetric functions, so it suffices to show that all lower degree terms vanish. -I feel that his simple-looking formula should have a direct and elegant proof. Is there one? - -REPLY [6 votes]: Here is another argument. Let $\chi^{\mu/k}$ denote the skew character -of the symmetric group $\mathfrak{S}_n$ corresponding to the skew -shape $\mu/k$. Then by Pieri's rule, -$$ \sum_{\mu-\lambda\ \mathrm{is\ a\ horizontal\ strip}} - \sum_{\alpha\vdash|\lambda|}\frac{\chi^\lambda_\alpha}{z_\alpha} - \prod_i p_i^{a_i} - = \sum_k \sum_{\alpha\vdash n-k} - \frac{\chi^{\mu/k}_\alpha}{z_\alpha}p_\alpha $$ -$$ \qquad = \sum_k s_{\mu/k}. $$ -Let $s_k^\perp$ denote the linear operator taking $s_\lambda$ to -$s_{\lambda/k}$. Let $\psi$ denote the linear operator taking $f$ to -$\sum_k s_k^\perp f$. Let $\theta$ denote the -algebra automorphism taking $p_i$ to $p_i-1$. -I claim that $\psi$ and $\theta$ are inverses (so in fact $\psi$ is an -algebra automorphism). By linearity it suffices to show that -$$ \psi p_\lambda = \prod_i (p_{\lambda_i}+1).\ \ \ (*) $$ -Now it is not hard to see that $s_k^\perp p_\lambda$ is equal to $\sum -p_\nu$, where $\nu$ is obtained from $\lambda$ by removing a set of -parts (regarding equal parts as distinguishable) summing to $k$. From -this (*) is immediate. -We therefore have -$$ \left.\sum_k s_{\mu/k}\right|_{p_i\to p_i-1} = \theta\psi s_\mu = - s_\mu, $$ -as desired.<|endoftext|> -TITLE: Graph in which no cycle has two crossing chords -QUESTION [13 upvotes]: Let $G$ be a graph which does not contain a simple cycle $v_1\ldots v_k$ and two "crossing" chords $v_iv_j$ and $v_pv_q$, $i -TITLE: Closed form for $\sum_{j=0}^{\infty}{\alpha \choose j} {\beta \choose j}x^j$ -QUESTION [5 upvotes]: As stated, I wonder if there is a closed form for the generating function $F_{\alpha,\beta}(x):=\sum_{j=0}^{\infty}{\alpha \choose j} {\beta \choose j}x^j$ where $\alpha,\beta \in\mathbb{N}$. Calling this a generating function is slightly misleading since ${\alpha \choose j}=0$ when $j>\alpha$ so this is really a finite sum. A few cases are known already: In the case $x=1$ we have that -$$F_{\alpha,\beta}(1)=\sum_{j=0}^{\infty}{\alpha \choose j} {\beta \choose j}={\alpha+\beta \choose \alpha}$$ -by Chu-Vandermore. The main approach I would take for this sort of problem would be to use the recurrence relations on binomial coefficients to get a differential question. In this case, this yields the equation -$$\left(x^{2}-x\right)F''\left(x\right)-\left(1+\left(\alpha+\beta-1\right)x\right)F'\left(x\right)+\alpha\beta F\left(x\right)=0$$ -This equation is really similar to an Euler differential equation, but the fact that the terms are polynomials with terms of varying degrees messes it up. I can't solve it, and Wolfram Alpha gives a useless answer in terms of a function that takes in $9$ arguments and a seperarte function that takes $4$. This feels like the sort of problem which would have a nice solution, but fixing $\alpha$ and $\beta$ and looking at the polynomials generated they do not seem to be very simple nor do they have roots at rational numbers. - -REPLY [4 votes]: Here is a "metaproof" that no simple closed form exists. -A conjecture by Carnevale and Voll states that: -For nonnegative integers $\alpha,\beta$ with $\alpha>\beta$, we have that -$$ -F_{\alpha,\beta}(-1)\neq 0. -$$ -As far as I know, the conjecture is still open! -For recent work in this direction see this article by Habsieger.<|endoftext|> -TITLE: Spin networks as functionals on the moduli space of connections modulo gauge transformations on a graph -QUESTION [8 upvotes]: I have just read a big part of John Baez's nice article Spin network states in Gauge theory. The definitions are quite clear in that article. However, there is a part which is not explained explicitly there, in my humble opinion. Let us say you have a spin network corresponding to a compact connected Lie group $G$ (the gauge group), which is a finite oriented graph with each edge labeling an irreducible representation of $G$ and each vertex $v$ labeling an intertwining operator (for the action of $G$) from the tensor product of the (irreducible) representations labeled by the set of all incoming edges at $v$ to the tensor product of the reps labeled by the set of all outgoing edges at $v$. -What I would like to understand though, is how to regard a spin network as an element of $\mathrm{L}^2(\mathcal{A}/\mathcal{G})$, where $\mathcal{A}$ denotes the space of connections on the graph (here regarded as parallel transport maps associated to each edge, please see the article for more detail) and $\mathcal{G}$ is the group of gauge transformations which acts on $\mathcal{A}$. -Given a spin network and a "connection" $A$ on the graph, how do we "evaluate" the spin network on the connection $A$, or rather on the equivalence class of $A$ under the group $\mathcal{G}$ of gauge transformations? The description in that article is via identifications and the Peter-Weyl theorem. Could someone perhaps spell it out for me in more concrete terms? - -REPLY [2 votes]: In Spin Networks in Nonperturbative Quantum Gravity, John Baez explains what I am trying to understand in section $2$. Essentially, you form a big tensor by tensoring out the intertwining operators at each vertex, and you form another big tensor by tensoring out the "parallel transport maps" associated to each edge when given a connection $A$ on the graph, and then you just contract each upper (resp. lower) index of the first big tensor with a lower (resp. upper) index of the second big tensor. The result is the "evaluation" of the spin network at the connection $A$. It is explained in more detail in the paper linked to in this answer. I thank the author for sending me a link to this paper.<|endoftext|> -TITLE: How does one rigorously lift a map $Sp \rightarrow Sp$ of spectra to equivariant spectra? -QUESTION [6 upvotes]: This is in part motivated from my attempt to understand tate diagonal in III.1 of Thomas Nikolaus, Peter Scholze, On topological cyclic homology, arXiv:1707.01799. I just want to make my understanding precise. - -Particular goal: -How the map (III.1) $T_p: Sp \rightarrow Sp$ -$$ X \mapsto (X \otimes \cdots \otimes X)^{tC_p}$$ -is defined rigorously. What I could define: I could define a map -$$ Sp \rightarrow Sp^{\times n} \rightarrow Sp$$ -$$ X \mapsto (X,\ldots, X) \mapsto X \otimes \cdots \otimes X $$ -using monoidal structure of $Sp^\otimes$ of spectra. -Question How do I rigorously lift this to map $Sp^{BC_p}$? ( allowing me to apply the Tate functor $(-)^{tC_p} : Sp^{BC_p} \rightarrow Sp$. ) - -EDIT: Most of my question of this goal have been resolved in the replies below (of which all are nice answers). I still have the following confusion - -how does one prove formulas fo the underlying(under the notation of Maxime) the adjunction: -$$ U:Sp^{BG} \rightarrow Sp:Ind, CoInd $$ -of "forgetful"/"inclusion"? where Ind and CoInd are left and right adjoint respectively. i.e. -It seems that -$$ \bigoplus_g X \simeq UInd X $$ -$$ U CoInd X \simeq ?? $$ - -In particular I am confused about the computation $CoInd(Sp) \simeq Sp^{\times n}$. - -REPLY [6 votes]: Let $C$ be a complete $\infty$-category. -Let $U:Fun(BC_n,C)\to C$ denote the forgetful functor, $\mathrm{CoInd}$ its right adjoint, and $(-)^{triv}$ the functor given by precomposition along $BC_n\to *$. -Then we have a canonical equivalence $U(X^{triv})\to X$ which yields, by adjunction, a map $X^{triv}\to \mathrm{CoInd}(X)$ which is $C_n$-equivariant. -Apply this to $C= Cat_\infty$ and $X=Sp$ yields a $C_n$-equivariant map $Sp \to \mathrm{CoInd}(Sp)$. Now $\mathrm{CoInd}(Sp) = Sp^{\times n}$ with the permutation action. -Now $Sp$ can be canonically seen as a commutative monoid in $Cat_\infty$, that is, a certain type of functor $Fin_*\to Cat_\infty$, which we can then obviously restrict to $Fin$ to get $Fin\to Cat_\infty$, informally given by $n\mapsto Sp^{\times n}$. -In particular, we get a $\Sigma_n$-equivariant map $Sp^{\times n}\to Sp$ corresponding to the smash product, and the action of $\Sigma_n$ on $Sp^{\times n}$ restricts to the permutation action of $C_n$ -You can prove this by dealing with the universal case. -Another way to do that, which certainly agrees, is to note that in $CAlg(Cat_\infty)$, products and coproducts agree, i.e. it is preadditive and so induction and co-induction agree. In particular you get for free a $C_n$-equivariant map $\mathrm{CoInd}(Sp)\to Sp$ (from $Sp\to U(Sp^{triv})$) which is also given by smash product. -Anyways, it follows that both $Sp\to Sp^{\times n}$ and $Sp^{\times n}\to Sp$ are $C_n$-equivariant -Your left Kan extension construction will not work. Left Kan extending along $*\to BG$ is left adjoint to the forgetful functor, i.e. it is induction - when composed with the forgetful map, this looks like $\bigoplus_{g\in G}$, so if you left Kan extend $X\mapsto X\otimes ... \otimes X$, you will get $\bigoplus_{g\in C_p}X\otimes... \otimes X$, and no permutation action. -As Harry already pointed out, the answer to your side question is "yes", the inclusion has both a left and a right adjoint, in particular it preserves limits and colimits. - -REPLY [4 votes]: There are several ways to do this, depending on how much technology one is interested in using. -One way to do it is to use the fact that the $\infty$-category of commutative monoid objects in a category with finite products, $\mathsf{CMon}(\mathcal{C})$, can be computed as $\mathsf{Fun}^{\times}(\mathsf{Span}(\mathrm{Fin})^{op}, \mathcal{C})$- i.e. the $\infty$-category of product-preserving presheaves on the $(2,1)$-category of spans of finite sets, with values in $\mathcal{C}$. (See, e.g., Theorem 6.5 in the paper of Nardin https://arxiv.org/pdf/1608.07704.pdf for a proof in a more general context; essentially the proof is by right Kan extending from the restriction to finite pointed sets.) Now, $\mathrm{Map}_{\mathsf{Span}(\mathrm{Fin})}(\bullet, \bullet)$ is then the groupoid of finite sets, and in particular receives a map $\mathrm{B}\Sigma_n \to \mathrm{Map}_{\mathsf{Span}(\mathrm{Fin})}(\bullet, \bullet)$. Now take $\mathcal{C}=\mathsf{Cat}_{\infty}$ and consider the functor $J \mapsto \mathsf{Sp}^{\times J}$ given by the symmetric monoidal structure on $\mathsf{Sp}$. Composing we get: $\mathrm{B}\Sigma_n \to \mathrm{Map}_{\mathsf{Span}(\mathrm{Fin})}(\bullet, \bullet) \to \mathrm{Map}_{\mathsf{Cat}_{\infty}}(\mathsf{Sp}, \mathsf{Sp})$. That's the same as a functor $\mathsf{Sp} \to \mathsf{Sp}^{\mathrm{B}\Sigma_n} = \mathsf{Sp}^{h\Sigma_n}$, which is what you're after (you can take $n=p$ and restrict to $C_p$ if you want). -But maybe you don't want to use that fact about spans. That's fine. You can follow the approach from the beginning of section 2.2 in DAGXIII (https://www.math.ias.edu/~lurie/papers/DAG-XIII.pdf). The point is this: if $\mathcal{D}$ is a symmetric monoidal $\infty$-category, i.e. a commutative monoid/algebra object in $\mathsf{Cat}_{\infty}$, then we automatically get a functor $\mathrm{Sym}(\mathcal{D}) \to \mathcal{D}$ from the free commutative algebra object. The free commutative algebra object is computed as $\coprod \mathcal{D}^{\times n}_{h\Sigma_n}$. Restricting to the $n$th summand gives a map $\mathcal{D}^{\times n}_{h\Sigma_n} \to \mathcal{D}$ which refines the functor $(X_1, ..., X_n) \mapsto X_1\otimes \cdots \otimes X_n$. We may view it just as well as a functor $\mathcal{D}^{\times n} \to \mathcal{D}$ in $\mathsf{Fun}(\mathrm{B}\Sigma_n, \mathsf{Cat}_{\infty})$ (since taking colimits is adjoint to the constant diagram functor). On the other hand, we also have a diagonal map $\mathcal{D} \to \mathcal{D}^{\times n}$ which is obtained by applying $\mathrm{Fun}(-, \mathcal{D})$ to the $\Sigma_n$-equivariant map of sets $\{1, ..., n\} \to \bullet$, and hence is $\Sigma_n$-equivariant. Composing gives $\mathcal{D} \to \mathcal{D}^{\times n} \to \mathcal{D}$ with a $\Sigma_n$-equivariant structure that does what you want. - -Also: - -This functor $X \mapsto (X^{\otimes p})^{tC_p}$ is not 'the Tate diagonal'. The Tate diagonal is a natural transformation (the unique lax symmetric monoidal natural transformation) $X \to (X^{\otimes p})^{tC_p}$.<|endoftext|> -TITLE: Is there a real valued function whose limit exists only on irrational numbers? -QUESTION [8 upvotes]: I have been trying to find a function $f : \mathbb R \to \mathbb R$ such that $\lim_{x \to c} f(x)$ exists when $c$ is irrational and the limit doesn't exist when $c$ is rational. -I tried variations of the Dirichlet function and Thomae's function, but I couldn't get anywhere. -I also tried proving that such a function cannot exist, using the fact that both the rationals and the irrationals are dense in real numbers. But I couldn't get a satisfying proof that way either. - -REPLY [9 votes]: Arrange rationals in a sequence $q_n$, and set -$$f(x) = \sum_{n = 1}^\infty 2^{-n} \mathbb{1}_{[q_n,\infty)}(x),$$ -where -$$\mathbb{1}_{[q_n,\infty)}(x) = \begin{cases} 1 & \text{if $x \geqslant q_n$,} \\ 0 & \text{if $x < q_n$.} \end{cases} $$ -In other words, -$$f(x) = \sum_{n : q_n \leqslant x} 2^{-n} .$$ -By the dominated convergence theorem, we have -$$\lim_{x \to a^-} f(x) = \sum_{n : q_n < a} 2^{-n}$$ -and -$$\lim_{x \to a^+} f(x) = \sum_{n : q_n \leqslant a} 2^{-n} = f(a) .$$ -It follows that -$$\lim_{x \to q_n^+} f(x) = 2^{-n} + \lim_{x \to q_n^-} f(x)$$ -and hence $f$ has no limit at each $q_n$, but $f$ is continuous at every irrational point.<|endoftext|> -TITLE: Sheaf of relative Kähler differentials intuitively -QUESTION [10 upvotes]: Let $f: X \to Y$ be a separated morphism between $k$-varieties or more general schemes -of finite type. The most common way in standard literature on algebraic -geometry to define the sheaf of relative Kähler differentials $\Omega_{X/Y}$ -is to observe that the diagonal map $\Delta: X \to X \times_Y X$ is a closed -embedding (we assume $f$ separated) and let $I \subset O_{X \times_Y X}$ -ideal sheaf define image $\Omega(X)$. The sheaf of relative Kähler differentials -is defined as -$$ \Omega_{X/Y}:= \Delta^* (I/I^2) $$ -and I'm interested in the geometric motivation behind this definition. As is so often -the case, the origins are in differential geometry. -Let $f: X \to Y$ a equidimensional surjective map between connected $k$-manifolds $Y, X$ -for $k= \mathbb{R}, \mathbb{C}$ and we moreover assume that every -fiber $F:= f^{-1}(y) \subset X$ for $y \in Y$ is also a connected -submanifold of same dimension. Most natural examples: a vector bundle $X$ over a manifold $Y$ or a fibration with 'nice' behavior on fibers. We obtain an exact sequence of -tangent spaces -$$ 0 \to T_{X/Y} \to T_X \to f^*T_Y \to 0 $$ -where $T_{X/Y}$ is the kernel of induced map of tangent bundles. intuitively, for every -$x \in f^{-1}(y)$, $(T_{X/Y})_x$ is the tangent space of the fiber at $x$. The relative space -of Kähler differentials $\Omega_{X/Y}$ is defined as the dual of $T_{X/Y}$ and sits in the sequence which we -obtain if we dualize the previous sequence above of tangent spaces: -$$ 0 \to f^*\Omega_Y \to \Omega_X \to \Omega_{X/Y} \to 0 $$ -Now let us translate the first definition from modern algebraic geometry also to framework of -differential geometry: -Let $f: X \to Y$ as above surjective map between connected manifolds with -equidim submanifold fibers and now let us embed $X$ via diagonal map -$\Delta: X \to X \times_Y X$ into the fiber product (a problem: -does the fiber product exist as manifold of $X,Y$ nice enough?). -Since $\Delta(X) \subset X \times_Y X$ is a smooth submanifold, we can pick local coordinates $( x_1 , ... , x_n )$ -around a $x \in \Delta(X)$ such that $ \Delta(X) $ is locally defined by $x_{k + 1} = ... = x_n = 0$; -then with this choice of coordinates -$$T_x X \times_Y X = k{\frac{\partial}{\partial x_1} \vert _x,... -\frac{\partial}{\partial x_n} \vert _x }$$ -$$T_x \Delta(X) = k{\frac{\partial}{\partial x_1} \vert _x,... -\frac{\partial}{\partial x_k} \vert _x }$$ -$$(N_{X \times_Y X/\Delta(X)})_x = k{\frac{\partial}{\partial x_{k+1}} \vert _x,... -\frac{\partial}{\partial x_n} \vert _x }$$ -and the ideal sheaf is locally generated by $x_{k+1}, ..., x_n $. The -bundle $N_{X \times_Y X/\Delta(X)}$ is the normal bundle of embedding -$\Delta$ and fits in following canonical sequence: -$$0 \to T_{\Delta(X)} \to T_{X \times_Y X} \to N_{X \times_Y X/X} \to 0$$ -Also, it is well known that the pairing -$$(I/I^2)_x \times (N_{X \times_Y X/\Delta(X)})_x \to k$$ -is perfect and therefore $I/I^2 \cong (N_{X \times_Y X/\Delta(X)})^{\vee}$. As -pointed out above in algebraic geometry we define -sheaf (or bundle in more old fashioned language) -of relative Kähler differentials $\Omega_{X/Y}$ as -$ \Omega_{X/Y}:= \Delta^* (I/I^2) $. -Therefore it is reasonable to conjecture although I nowhere found a proof that if there is any justice n this world then these two definitions -of relative differential bundles should coinside in the setting -of differential geometry. That is for -$f: X \to Y$ surjective map between connected manifolds with -equidim submanifold fibers, we should have -$$T_{X/Y} \cong \Delta^* N_{X \times_Y X/\Delta(X)}$$ -Can we write down an explicit isomorphism and -understand what is geometrically going on there? I would like -also remark that I asked almost identical question in MathStackExchange -and suppose that possibly MO might be a better place to ask about -it. - -REPLY [3 votes]: This question may have interest for many people. In general if $k \rightarrow A \rightarrow^f B$ is an arbitrary sequence of maps of commutative unital rings, there is a canonical right exact sequence of $B$-modules -S1. $B\otimes \Omega^1_{A/k} \rightarrow \Omega^1_{B/k} \rightarrow \Omega^1_{B/A} \rightarrow 0$. -In Matsumura, Commmutative ring theory Thm 25.1 you may find this construction. If $B$ is $0$-smooth over $A$ it follows when you add a $0$ on the left it follows S1 is split exact. When you dualize S1 you get the sequence -S2. $0 \rightarrow Der_A(B) \rightarrow Der_k(B) \rightarrow^{Tf} B\otimes_A Der_k(A)$ -where $Tf$ is the "tangent morphism" of the map $f$. If $X:=Spec(B)$ and $Y:=Spec(A)$ -it follows the relative tangent bundle $T_{X/Y}$ is the sheafification of the module $Der_A(B)$. When you sheafifiy S2 you get the sequence -S3. $0 \rightarrow T_{X/Y} \rightarrow T_{X/k} \rightarrow^{Tf} f^*T_{Y/k}$ -and S3 is an exact sequence of $\mathcal{O}_X$-modules in general. There is a long exact sequence continuation of the sequence S3. -If $k$ is the field of complex numbers and $X,Y$ are finitely generated and regular algebras over $k$ it follows $T_{X/k}$ and $f^*T_{Y/k}$ are finite rank locally free sheaves on $X$. The "relative tangent sheaf" $T_{X/Y}$ is related to properties of the morphism $f$. -Definition. Assume $X,Y$ are smooth irreducible schemes of finite type over $k$ the field of complex numbers (this means the local rings of $X,Y$ are regular at all points). The morphism $f$ is "smooth of relative dimension $n$" iff - -$f$ is flat -$dim(X)=dim(Y)+n$ -The cotangent sheaf $\Omega^1_{X/Y}$ is locally free of rank $n$. - -By Hartshorne, Prop III.10.2 this implies that the fibers $f^{-1}(y):=X_y$ of $X$ at any point $y\in Y$ are regular of dimension $n$. Hence the cotangent sheaf $\Omega^1_{X/Y}$ "measures" when the fibers of $f$ are regular/smooth of the same dimension $n$. -Example: Let $\pi:X:=\mathbb{V}(\mathcal{E}^*)\rightarrow Y$ be a finite rank geometric vector bundle on $Y$. Locally it follows the sheaf $\mathcal{E}^*$ is a free $A$-module $E:=A\{x_1,..,x_n\}$ on the elements $x_i$ hence locally the map $\pi$ is the following map: -S4. $A \rightarrow B:=A[x_1,..,x_n]$ -where the latter is a polynomial ring on $n$-variables. In this case the relative module of Kahler differentials $\Omega^1_{B/A}$ becomes -S5. $\Omega^1_{B/A}\cong B\{dx_1,..,dx_n\}$ -which is the free $B$-module of rank $n$ on the elements $dx_i$. Hence the relative -cotangent sheaf $\Omega^1_{X/Y}$ is in this case locally trivial of rank $n$ and the morphism $\pi$ is smooth of relative dimension $n$. The fibers are affine spaces and they are smooth. Hence the relative cotangent sheaf is related to properties of the morphism. The relative tangent sheaf is the dual of the relative cotangent sheaf, but when you dualize you loose information. Hence the cotangent sheaf is more fundamental. -Note: When you write down the relative tangent sequence, this is not a sequence of "tangent spaces", it is an exact sequence of coherent sheaves on $X$. The relative "tangent bundle" is not a vector bundle in general. In your situation it is a "coherent sheaf". If the relative cotangent sheaf is locally trivial it follows the fibers of your morphism are smooth manifolds (if you consider holomorphic maps of complex projective manifolds). -Example: Ramification. For a holomorphic map $f:X \rightarrow Y$ of complex projective manifolds the same constructions applies because of a classical result saying $f$,$X$ and $Y$ are algebraic. Moreover any coherent analytic sheaf $E$ on $X$ is algebraic. -Hence for $f$ we get a short exact of $\mathcal{O}_X$-modules -S6 $f^*\Omega^1_{Y/k} \rightarrow \Omega^1_{X/k} \rightarrow \Omega^1_{X/Y} \rightarrow 0.$ -Here $\mathcal{O}_X$ is the sheaf of regular functions on the complex algebraic manifold $X$. If the map $f$ is flat it follows $\Omega^1_{X/Y}$ is locally free of rank $n$ iff the fibers of $f$ are complex projective manifolds of dimension $n$. In general the coherent sheaf $\Omega^1_{X/Y}$ is generically locally free, hence it is related to the "ramification" of the map $f$. Hence the cotangent sheaf $\Omega^1_{X/Y}$ and more generally the $k$'th order sheaf of principal parts $\mathcal{P}^k_{X/Y}$ is used in the study of the ramification and discriminant of the morphism $f$ -Example: Etale morphisms. In Hartshorne Ex. III.10.3 they prove that a morphism $f:X\rightarrow Y$ of finite type over a field $k$ is etale iff -$f$ is flat and $\Omega^1_{X/Y}=0$. -Exampe: If $X$ is a complex projective manifold it follows the sheaves $\Omega^1_{X/\mathbb{C}}$ and $T_{X/\mathbb{C}}$ are locally free of finite rank and dual of each other: $\Omega^1_{X/\mathbb{C}} \cong T_{X/\mathbb{C}}^*$. In the non smooth situation there is a canonial morphism -$i:\Omega^1_{X/\mathbb{C}} \rightarrow T_{X/\mathbb{C}}^*$ but $i$ is not an isomorphism in general. The relative cotangent bundle $\Omega^1_{X/Y}$ is a coherent $\mathcal{O}_X$-module in general.<|endoftext|> -TITLE: Regularity of convex sets in $\mathbb{R}^n$ -QUESTION [11 upvotes]: The following result is Proposition 2.4.3 in [1]: - -Theorem. Let $K\subset\mathbb{R}^n$ be a bounded convex set with the non-empty interior. Then $\partial K\in C^{1,1}$ if and only if -there is $r>0$ such that $K$ is the unioin of balls of radius $r$. - -Question. Do you know who is the author of this result? -Hörmander does not provide any reference so it could be that his book is the first place where it was proved. However, the statement seems so classical that I tend to believe the result had been known before. -It is very likely that I will refer to this result in my research and I would like to give a proper reference to it. -Edit. It seems that there are earlier references than the one in the answer, see the comments below the answer. That means the questions might still have a better answer. -[1] L. Hörmander, -Notions of convexity. -Progress in Mathematics, 127. Birkhäuser Boston, Inc., Boston, MA, 1994. - -REPLY [5 votes]: To my knowledge, this is the first place where I saw the result being claimed: -C. O. Kiselman. Regularity classes for operations in convexity theory. Kodai Math. J. 15. 1992. -In particular on the first page the author states: - -To describe the simplest case of our results, let $A$ be a compact set in $R^n$. If the boundary of $A$ is of class $C^{1,1}$, then $A$ is a union of Euclidean balls with radii bounded from below; if $A$ is convex, the converse holds.<|endoftext|> -TITLE: Does $\text{AC}_{\text{WO}}$ prove $\Theta \neq \aleph_{\omega+1}?$ -QUESTION [11 upvotes]: The choice principle $\text{AC}_{\text{WO}}$ proves a large amount of cardinal arithmetic. It's well-known to imply DC, that successor cardinals are regular, and that for all $X$, there is $\lambda$ such that $\aleph(X)=\aleph^*(X)=\lambda^+.$ Furthermore, we have the following: - -($\text{ZF + AC}_{\text{WO}}$) If $\aleph(^{\text{cf}(\kappa)} \kappa) = \lambda^+,$ then $\text{cf}(\lambda)>\text{cf}(\kappa)$ or $\text{cf}(\lambda)=\omega.$ - -Pf: Fix $\lambda < \aleph(^{\text{cf}(\kappa)} \kappa)$ such that $\omega_1 \le \text{cf}(\lambda) \le \text{cf}(\kappa)$ and a cofinal sequence $\langle \alpha_{\xi}: \xi<\text{cf}(\lambda) \rangle \subset \lambda.$ Notice that the domination ordering $\le^*$ on $^{\text{cf}(\lambda)} \lambda$ is well-founded, since it has no descending sequences and DC holds. Any well-ordering $\prec$ of $\lambda$ embeds into $(^{\text{cf}(\lambda)} \lambda, \le^*)$ by $\gamma \mapsto (\xi \mapsto \text{ot}({\gamma \downarrow_{\prec}} \cap \alpha_{\xi}, \prec)),$ so $\text{rk}(^{\text{cf}(\lambda)} \lambda, \le^*) \ge \lambda^+.$ -We thus have $\lambda^+ < \aleph^*(^{\text{cf}(\lambda)} \lambda) = \aleph(^{\text{cf}(\lambda)} \lambda) \le \aleph(^{\text{cf}(\lambda) \cdot \text{cf}(\kappa)} \kappa) = \aleph(^{\text{cf}(\kappa)} \kappa).$ $\square$ -Of course, AC proves this result without the uncountable cofinality hypothesis, which raises the question, - -Does $\text{ZF + AC}_{\text{WO}}$ prove that $\aleph(^{\text{cf}(\kappa)} \kappa) = \lambda^+$ for some $\lambda$ with $\text{cf}(\lambda) > \text{cf}(\kappa)?$ - -Note that a failure of this implication has consistency strength at least a Woodin cardinal, because then $\lambda$ is a singular cardinal with countable cofinal sequence $A,$ and $\text{HOD}(A)$ computes $\lambda^+$ incorrectly. -Of particular interest is the simplest non-trivial case: - -Does $\text{ZF + AC}_{\text{WO}}$ prove $\aleph(\mathbb{R})=\Theta \neq \aleph_{\omega+1}?$ - -A proof from a stronger partition principle like PP or WPP would also be of interest. - -REPLY [5 votes]: ($\text{ZF + AC}_{\text{WO}}$) For any cardinals $\kappa_1, \kappa_2,$ there is $\lambda$ such that $\aleph(^{\kappa_2}\kappa_1)=\lambda^+$ and $\text{cf}(\lambda)>\kappa_2.$ -Pf: Let $\lambda$ be such that $\aleph(^{\kappa_2}\kappa_1)=\lambda^+,$ and fix a cofinal sequence $\langle\gamma_{\xi}: \xi<\text{cf}(\lambda) \rangle \subset \lambda.$ Choose injections $f_{\alpha}: \alpha \rightarrow \lambda$ for $\alpha<\lambda^+.$ For such $\alpha,$ we recursively define $g_{\alpha}: \text{cf}(\lambda) \rightarrow \lambda$ by setting $g_{\alpha}(\xi) = \min(\lambda \setminus \{g_{\beta}(\xi): \beta \in f_{\alpha}^{-1}(\gamma_{\xi})\}).$ -Notice that $\alpha \mapsto g_{\alpha}$ injects $\lambda^+$ into $^{\text{cf}(\lambda)} \lambda,$ so $\aleph(^{\kappa_2} \kappa_1)=\lambda^+ < \aleph(^{\text{cf}(\lambda)} \lambda) \le \aleph(^{\text{cf}(\lambda) \cdot \kappa_2} \kappa_1).$ Clearly $\text{cf}(\lambda)> \kappa_2.$ $\square$ -Corollary: $\text{AC}_{\text{WO}}$ does prove $\Theta \neq \aleph_{\omega+1}.$<|endoftext|> -TITLE: A generalization of partition function to the sums of squares -QUESTION [7 upvotes]: The well known partition function $p(n)$ is defined as the number of ways to represent $n$ as the sum of natural numbers. An asymptotic formula for $p(n)$ is -$$p(n)\sim\frac{1}{4n\sqrt{3}}\exp\left(\pi\sqrt{\frac{2n}{3}}\right)$$ -which was obtained by Ramanujan. Recently an interesting idea came to me: generalizing the partition function. The number of ways of representing $n$ as the sum of four squares is known, and many similar things like number of ways of representing a number as a sum of two squares, etc. are known. But that didn't satisfy me. I wanted to truly generalize it. -So first, I took $p_2(n)$, the number of ways of representing $n$ as the sum of squares. It is obvious that $p_2(n)\le p(n)$. It has been conjectured that -$$p_2(n)\sim c\cdot n^{\alpha}\exp(\beta\cdot n^{1/3})$$ -where - -$\alpha=-\frac{7}{6}$ -$\beta=\frac{3}{2}\frac{\pi}{2}^{1/3}\zeta\left(\frac{3}{2}\right)^{2/3}$ -$c=\frac{\zeta(3/2)^{2/3}}{\sqrt{3}(4\pi)^{7/6}}$ -and the generating function of $p_2(n)$ is -$$\prod_{m\ge1}\frac{1}{1-n^{m^2}}$$ -I found these in an article which was not at all about $p_2(n)$ but these two were given for some reason. So my main questions are: -What more is known about $p_2(n)$? -What is the generating function of $r_k(n)$? What is $r_k(n)$ is mentioned below. - -And the questions which are not necessary to answer but they would be useful for me are: - -How was the conjecture even formulated? I don't think it was formulated because of computational evidence because the formula is too much complicated. -How can we prove that generating function formula? - -Update: $p_2(n)$ is on OEIS as entry A001156. From that page, I found that -$$p_2(n)\sim3^{-1/2}(4\pi n)^{-7/6}\zeta\left(\frac{3}{2}\right)^{2/3}\exp(3\cdot2^{-4/3}\pi^{1/3}\zeta\left(\frac{3}{2}\right)^{2/3}n^{1/3})$$ -Which was proven by Hardy and Ramanujan. Can anyone link an article containing the proof of this asymptotic formula? - -See the paper - -G. H. Hardy, S. Ramanujan, Asymptotic formulæ in combinatory analysis, Proceedings of the London Mathematical Society (series 2) 17 (1918) pp75—115, doi:10.1112/plms/s2-17.1.75, (scanned pdf, retypeset pdf). - -$p_2(n)$ is not a standard notation; but a standard notation if $r_k(n)$ (many authors use it), which denotes the number of ways of representing $n$ as a sum of $k$ squares. - -REPLY [4 votes]: You also asked about the generating function. Write $r^k(n)$ for the number of partitions of $n$ with each part the $k$th power of a positive integer. That generating function is -$$\sum_{n=0}^\infty r^k(n)q^n = \prod_{m=1}^\infty \frac{1}{1-q^{m^k}}$$ -since the $m$th factor on the right is a geometric series $(1+q^{m^k}+q^{2m^k}+\cdots)$ accounting for $0,1,2,\ldots$ copies of $m^k$. (This is at the beginning of the Gafni paper referenced in Thomas Bloom's answer.) -You want to keep track of how many $k$th powers are used. Write $r_j^k(n)$ for the number of partitions of $n$ with exactly $j$ parts, each the $k$th power of a positive integer. To keep track of the number of $k\text{th}$ powers, use Euler's trick of including a tracking variable: -$$\sum_{n=0}^\infty r_j^k(n)q^nz^j = \prod_{m=1}^\infty \frac{1}{1-zq^{m^k}}$$ -where the geometric series is now $(1+zq^{m^k}+z^2q^{2m^k}+\cdots)$.<|endoftext|> -TITLE: Does there exist an ordering-functor? -QUESTION [14 upvotes]: This sounds like a very silly question which should have have a negative answer but I don't see an argument. The precise question is this: - -Does there exist a covariant functor $ord$ from the category of sets and functions to the category of totally ordered sets and increasing functions such that every set $X$ is the underlying set of $ord(X)$? - - -Of course, I assume the axiom of choice so that every set can be even well-ordered. Since the well-ordering is based on rather arbitrary choices there does not seem to be a reason why it should be in any sense "compatible" with all functions. My guess is that such a functor does not even exist in the subcategory of finite sets. - -REPLY [19 votes]: Conceptual answer. -There can be no such functor. Let $C$ be any concrete category of finite sets and mappings such that the only automorphisms in $C$ are trivial. I claim there is no underlying set preserving functor $F$ from the category $\mathbf{FSet}$ of finite sets to $C$. The category of finite totally ordered sets and order-preserving maps is of this form. -Let us call the rank of a mapping the size of its image. -First note in the category $\mathbf{FSet}$ of finite sets, all idempotents split. That is, if $e\colon X\to X$ is an idempotent then there are morphisms $f\colon X\to Y$ and $g\colon Y\to X$ with $gf=e$ and $fg=1_Y$ for some finite set $Y$. Here you can take $Y$ to be the image of $e$ and take $g$ to be the inclusion and $f$ to be the "corestriction" of $e$. Since $F(f)F(g)=F(fg)=F(1_Y)=1_Y$, we deduce that $F(f)$ is surjective and $F(g)$ is injective and so $F(e)=F(g)F(f)$ has image of size $|Y|$. Thus $F$ preserves the ranks of idempotents, i.e., rank $e$ = rank $F(e)$. -Next note that if $|X|\geq 3$, then there is an idempotent of rank $|X|-2$ which is a product of conjugate rank $|X|-1$ idempotents. To ease notation assume $|X|=n$. For $i\neq j$ let $e_{i,j}$ be the rank $n-1$ idempotent with $e_{i,j}(i)=i=e_{i,j}(j)$ and which fixes all other elements. Let $k\neq i,j$. Then $f=e_{i,j}e_{j, k}$ sends $i,j,k$ to $i$ and fixes all other elements and so is a rank $n-2$ idempotent. -Notice that $e_{i,j}$ and $e_{j,k}$ are conjugate by the permutation $(i,j,k)$: we have $e_{j,k} = (i,j,k)e_{i,j}(k,j,i)$. Since $F((i,j,k))=1_X$, we must have $F(e_{i,j})=F(e_{j,k})$. Since these are idempotents, $F(f)=f(e_{i,j})$, contradicting that $F$ preserves rank of idempotents. -Original answer. I think there is no functor from finite sets to finite totally ordered sets with order preserving maps preserving underlying sets. Here is a proof. Let $F$ be such a functor. It must send every permutation of a finite set $X$ to the identity as there are no non-trivial automorphisms of a finite totally ordered set. Next I claim that every non-invertible endomorphism $f\colon X\to X$ has $F(f)$ a constant mapping. For this, I use the well known fact that the monoid of self-maps of $X$ is generated by the symmetric group $S_X$ and any idempotent $e$ with image of size $|X|-1$. Hence, since $F(S_X)$ is the identity, every non-invertible mapping gets sent to $F(e)$. But if $f\colon X\to X$ is a constant mapping, we have that $f= gh$ with $h\colon X\to \{1\}$ and $g\colon \{1\}\to X$. Thus $F(f)=F(g)F(h)$ has image of size $1$. Thus $F(e)$ above must have image of size $1$ and so every non-invertible mapping of $X$ to $X$ is sent to a constant mapping. This is a problem. -Consider $f\colon \{1,2\}\to \{1,2,3\}$ given by the inclusion and $g\colon \{1,2,3\}\to \{1,2\}$ given by $g(1)=1$, $g(2)=2$ and $g(3)=2$. Then $gf$ is the identity but $fg$ is non-invertible. Thus $F(gf)$ is an identify mapping and $F(fg)$ is a constant mapping, which is impossible as $gf=gfgf$ and so $F(gf)=F(g)F(fg)F(f)$ is both a constant map and an identity for a two element set.<|endoftext|> -TITLE: Can non-split extension be isomorphic to the split one as objects -QUESTION [6 upvotes]: Is it possible to have a non-split short exact sequences of vector bundles (on some smooth variety) $0\rightarrow V_1 \rightarrow V_2 \rightarrow V_3 \rightarrow 0$. Such that $V_2\cong V_1\oplus V_3$ as vector bundles? - -REPLY [10 votes]: $\newcommand{\cO}{\mathcal{O}}$Consider exact sequence of trivial vector bundles $$0\to\cO\xrightarrow{\left(\begin{matrix}x \\ y\end{matrix}\right)}\cO\oplus\cO\xrightarrow{\left(\begin{matrix}y & -x\end{matrix}\right)}\cO\to 0$$ on $X=\mathbb{A}^2_{x,y}\setminus\{0\}$. One checks easily that it is exact on stalks (but the same sequence on $\mathbb{A}^2$ is not exact at $(x,y)=(0,0)$). It is the pullback of $0\to\cO(-1)\to \cO\oplus\cO\to \cO(1)\to 0$ from $\mathbb{P}^1$. A splitting of this sequence would be a pair of functions $f_1,f_2\in H^0(\mathbb{A}^2\setminus\{0\},\cO)=k[x,y]$ such that $yf_1-xf_2=1$ but there is no such pair. - -On the positive side, any such sequence has to be split if $X$ is proper. In this case the spaces $Hom(V_3, V_1)$ and $Hom(V_3, V_3)$ are finite-dimensional over $k$. Applying $Hom(V_3,-)$ to this exact sequence we get a left exact sequence of finite-dimensional vector spaces $$0\to Hom(V_3, V_1)\to Hom(V_3,V_2)\to Hom(V_3,V_3)$$ The dimension of the vector space in the middle $Hom(V_3,V_2)\simeq Hom(V_3, V_1\oplus V_3)$ is equal to the sum of the dimensions of first and third terms. Therefore, the sequence has to be exact on the right and, in particular, the identity $Id_{V_3}\in Hom(V_3, V_3)$ lifts to a morphism $Hom(V_3, V_2)$ that gives a splitting.<|endoftext|> -TITLE: How strong is "all up-classes are infinitarily definable"? -QUESTION [6 upvotes]: Working in MK (or some other not-too-strong class theory if you prefer), say that an up-class is a class of structures $\mathfrak{X}$ which is definable in $V$ (allowing parameters from $V$) and such that whenever $\mathcal{A}\in\mathfrak{X}$ and $i:\mathcal{A}\rightarrow\mathcal{B}$ is an embedding then $\mathcal{B}\in\mathfrak{X}$. -Consider the following principle of infinitary axiomatization of up-classes: - -(IAU) Suppose $\mathfrak{X}$ is an up-class. Then there is some $\mathcal{L}_{\infty,\infty}$-sentence $\varphi$ such that $\mathfrak{X}=\{\mathcal{A}:\mathcal{A}\models\varphi\}$. - -By a result of Makowsky, we get that Vopenka's principle implies IAU. Specifically, suppose $\mathfrak{X}$ is an up-class. Per Makowsky there is some $\kappa$ such that whenever $T$ is a finitary first-order all of whose models are in $\mathfrak{X}$ then there is some $T_0\subseteq T$ of size $<\kappa$ all of whose models are in $\mathfrak{X}$. Meanwhile, note that whenever $\mathcal{A}\in\mathfrak{X}$ we have - since $\mathfrak{X}$ is an up-class - that the every model of the atomic diagram of $\mathcal{A}$ is also in $\mathfrak{X}$. Putting these together, letting $\sigma$ be the disjunction of all $\mathcal{L}_{\kappa,\kappa}$-sentences which only have models in $\mathfrak{X}$ we get $Mod(\sigma)=\mathfrak{X}$. -However, the actual strength of IAU itself is unclear to me: - -How strong is IAU? - -In particular, I'm curious whether IAU is compatible with V=L. I strongly suspect the answer is negative, but I don't see how to prove it at the moment. -Note that one point of flexibility here is that IAU says nothing about the syntactic complexity of the defining sentence, whereas the argument via Vopenka gives a very simple form: the $\sigma$ it builds is a disjunction of size $\le 2^\kappa$ of sentences, each of which consists of an existential quantifier over a $<\kappa$-length tuple followed by a $<\kappa$-length conjunction of literals - basically "$\Sigma_2$ in the sense of $\mathcal{L}_{\infty,\infty}$." - -REPLY [7 votes]: I think IAU is equivalent to Vopěnka's principle. For the other direction, assume -Vopěnka's principle fails. Then there is a proper class of structures (WLOG graphs), none of which embeds into any other. Because this is a proper class, there is an injection of $\mathcal{L}_{\infty,\infty}$ into it. In other words, for each sentence $\sigma \in \mathcal{L}_{\infty,\infty}$ we may choose a structure $\mathcal{M}_\sigma$ in such a way that for any two distinct sentences $\sigma,\tau \in \mathcal{L}_{\infty,\infty}$, neither $\mathcal{M}_\sigma$ nor $\mathcal{M}_\tau$ embeds into the other. -Now we define the "diagonal" class of structures $D = \{\mathcal{M}_\sigma \mid \sigma \in \mathcal{L}_{\infty,\infty} \wedge \mathcal{M}_\sigma \not\models \sigma\}$. I claim that its upward closure $D\mathord{\uparrow}$ does not have the form $\operatorname{Mod}(\tau)$ for any sentence $\tau \in \mathcal{L}_{\infty,\infty}$. This is because $\mathcal{M}_\tau \in D\mathord{\uparrow} \iff \mathcal{M}_\tau \in D \iff \mathcal{M}_\tau \not\models \tau \iff \mathcal{M}_\tau \notin \operatorname{Mod}(\tau)$. -The key point is that $\mathcal{M}_\tau \in D\mathord{\uparrow}$ implies $\mathcal{M}_\tau \in D$ because no other structure $\mathcal{M}_\sigma \in D$ can embed into $\mathcal{M}_\tau$. -Edit: The argument can be modified to avoid relying on global choice, as follows. -Assume that we have a counterexample to Vopěnka's principle, meaning a proper class $\mathcal{C}$ of structures (WLOG graphs), none of which embeds into any other. For every sentence $\sigma \in \mathcal{L}_{\infty,\infty}$, define the structure $\mathcal{M}_\sigma = \langle V_\lambda; \in, \{\sigma\}, \mathcal{C} \cap V_\lambda\rangle$ for the least limit ordinal $\lambda$ that is greater than the rank of $\sigma$ (meaning just its Von Neumann rank as a set) and has the additional property that $A \cap \lambda$ has order type $\lambda$ where $A$ is the proper class of ordinals $\{\text{rank}(\mathcal{M}) \mid \mathcal{M}\in C\}$. -Note that for any two distinct sentences $\sigma, \tau \in \mathcal{L}_{\infty,\infty}$ there is no elementary embedding $j : \mathcal{M}_\sigma\to \mathcal{M}_\tau$. If there were, then letting $\kappa$ be the critical point of $j$ (which exists because $j$ maps $\sigma$ to $\tau$) and letting $\mathcal{M} \in \mathcal{C}$ be a structure of rank equal to the $\kappa$th element of $A$, the restriction $j \restriction \mathcal{M}$ is an embedding of $\mathcal{M}$ into some other structure $j(\mathcal{M}) \in \mathcal{C}$ whose rank equals the $j(\kappa)$th element of $A$ and which is therefore different from $\mathcal{M}$. This contradicts the assumption that $\mathcal{C}$ was a counterexample to Vopěnka's principle. -To remove the word "elementary" above and ensure that for any two distinct sentences $\sigma, \tau \in \mathcal{L}_{\infty,\infty}$ there is no embedding $j : \mathcal{M}_\sigma\to \mathcal{M}_\tau$, we just need to modify the structures $\mathcal{M}_\sigma$ in the usual way by adding a relation for every first-order formula. (I didn't want to complicate the notation with this when defining the structures initially.) -Then we can define the diagonal class of structures $D = \{\mathcal{M}_\sigma \mid \sigma \in \mathcal{L}_{\infty,\infty} \wedge \mathcal{M}_\sigma \not\models \sigma\}$ and show that its upward closure $D\mathord{\uparrow}$ does not have the form $\operatorname{Mod}(\tau)$ for any sentence $\tau \in \mathcal{L}_{\infty,\infty}$ as before.<|endoftext|> -TITLE: Is $\mathsf{R}$ axiomatizable by finitely many schemes? -QUESTION [9 upvotes]: Recall that $\mathsf{R}$ is the theory of arithmetic consisting of the quantifier-free theory of $(\mathbb{N};+,\times,0,1,<)$ together with, for each $k\in\mathbb{N}$, the sentence $$\forall x[(\bigwedge_{i\le k}x\not=\underline{i})\rightarrow \underline{k} -TITLE: A limit of conjugate representations in $\mathrm{GL}(n,\mathbb C)$ -QUESTION [5 upvotes]: $\DeclareMathOperator{\GL}{GL}$Let $G$ be a finitely presented group with generators $g_1,\dotsc, g_k$. Suppose we have a family of representations $\rho_t:G\to \GL(n,\mathbb C)$ with $t\in [0,1]$ smoothly dependent on $t$. Suppose that for any $t\in [0,1]$ and $i\in\{1,\dotsc,k\}$, $\rho_t(g_i)$ is diagonalisable, and for $t>0$ all representations $\rho_t$ are conjugate. -Question. Is it true that $\rho_0$ is conjugate to $\rho_t$ for $t>0$? -(PS Yves gave a counter-example in a comment.) -Is the answer positive, if we additionally ask $\rho_t(G)\subset \GL(n,\mathbb C)$ to be conjugate to a subgroup of $\operatorname U(n)\subset \GL(n,\mathbb C)$ for all $t$? - -REPLY [7 votes]: It is proven in Lemma 1.25 of the book "Varieties of representations" by Lubotzky and Magid (Memoirs of AMS, vol. 336, 1985) that each $GL(n,{\mathbb C})$-conjugation orbit of a semisimple representation $\rho\in R_n={\mathrm Hom}(\Gamma, GL(n, {\mathbb C}))$ is Zariski-closed in $R_n$. If a representation is unitarizable, then it is completely reducible, i.e. semisimple. Thus, you get the property you are after. Note that you do not need finite presentability of $\Gamma$, only finite generation.<|endoftext|> -TITLE: Inverse Galois problem for non-Galois extensions -QUESTION [5 upvotes]: The inverse Galois problem asks whether every finite group appears as the Galois group of a Galois extension of the rational numbers. -Is anything known about the anologous problem, where the extensions are not required to be Galois? In other words, for a finite group $G$, does there exist a finite field extension $K$ of $\mathbb{Q}$ such that $\mathrm{Aut}(K/\mathbb{Q})=G$? -Is this suspected to be as difficult as the inverse Galois problem or easier? - -REPLY [7 votes]: The answer to this is positive. The first correct proof, it seems, was given in -Michael D. Fried. A note on automorphism groups of algebraic number -fields. Proc. Amer. Math. Soc., 80(3):386–388, 1980. -For a generalization to Hilbertian fields and some history see for example -F. Legrand and E. Paran. Automorphism groups over Hilbertian fields. Journal of Algebra -Volume 503, 2018. journal website<|endoftext|> -TITLE: Spectral decomposition of product of modular functions -QUESTION [9 upvotes]: The eigenfunctions of the Laplacian on $SL(2,\mathbb Z)\backslash \mathbb H$ are known to be given by three types: the constant function, the real analytic Eisenstein series (which come in a continuous family), and the Maass cusp forms (which come in a discrete family -- the first few given e.g. here). Let us denote $E_s(\tau, \bar\tau)$ as the real analytic Eisenstein series which scales as $y^s + \frac{\Lambda(1-s)}{\Lambda(s)}y^{1-s}$ as $y \rightarrow \infty$, where $y = \text{Im}(\tau)$, $\Lambda(s) = \pi^{-s} \Gamma(s)\zeta(2s)$; and $\nu_i$ denote the $i^{\text{th}}$ discrete cusp form.  -My question is: Is there a simple expression for $\int_{SL(2,\mathbb Z)\backslash \mathbb H} \frac{dxdy}{y^2} E_{s_1}(\tau, \bar\tau) E_{s_2}(\tau, \bar\tau) \overline{\nu_i(\tau,\bar\tau)}$? (We can take $s_i \in 1/2 + i \mathbb R$.) For example, if there were only one $E_s$ in the previous expression, this would vanish due to orthonormality. What about when there is a product of two of them? -The motivation is the following --- it is known that any modular invariant function in $L^2$ can be expanded in Eisensteins on the principal series and Maass cusp forms. If we take the product of two modular invariants in $L^2$ each without support on the discrete cusp forms, does the product necessarily also have no support on the discrete series (let us assume the product remains in $L^2$, or at least has good decomposition..)?  -Similarly are there simple or known expressions for: -$\int_{SL(2,\mathbb Z)\backslash \mathbb H} \frac{dxdy}{y^2} E_{s_1}(\tau, \bar\tau) E_{s_2}(\tau, \bar\tau) \overline{E_{s_3}(\tau,\bar\tau)}$ -$\int_{SL(2,\mathbb Z)\backslash \mathbb H} \frac{dxdy}{y^2} E_{s_1}(\tau, \bar\tau) \nu_{i_1}(\tau, \bar\tau) \overline{\nu_{i_2}(\tau,\bar\tau)}$ -$\int_{SL(2,\mathbb Z)\backslash \mathbb H} \frac{dxdy}{y^2} \nu_{i_1}(\tau, \bar\tau) \nu_{i_2}(\tau, \bar\tau) \overline{\nu_{i_3}(\tau,\bar\tau)}$ -known in the literature? -(The first of these three may not converge but hopefully can be written with delta functions or something.) Thank you! - -REPLY [10 votes]: I'm guessing you're a physicist by your unusual notation! The real-analytic Eisenstein series is usually denoted $E(z,s)$ (where $z = \tau$ in your notation), not $E_s(\tau,\bar{\tau})$ (why does $\bar{\tau}$ appear?). I'll also let $f(z)$ denote a cusp form rather than $\nu_i(\tau,\bar{\tau})$ (I've never seen this notation used). - -Anyway, the question asks (among other things) the following. Given two cusp forms $f_1,f_2$ and an Eisenstein series $E(z,s)$, determine a closed formula for -$$\int_{\Gamma \backslash \mathbb{H}} f_1(z) f_2(z) E(z,s) \, d\mu(z).$$ -Here $\Gamma = \mathrm{SL}_2(\mathbb{Z})$ and $d\mu(z) = y^{-2} \, dx \, dy$. -This question has a nice answer, as does the same question where one allows any triple product involving various cusp forms and Eisenstein series. When at least one form is an Eisenstein series, this is the Rankin-Selberg method; when all three are cusp forms, this is known as the Watson-Ichino triple product formula. In Section 2.2 of this paper of mine, I write out the exact formulae for these in several cases, and provide some references. - -Let me give a more detailed explanation of this in the case written above. It is known, going back to work of Hecke, that any Maass cusp form $f$ can be written as a finite linear combination of Hecke-Maass cusp forms; these are eigenfunctions of the Hecke operators and have Fourier expansions of the form -$$f(z) = \rho_f \sum_{n = 1}^{\infty} \frac{\lambda_f(n)}{\sqrt{n}} W_{0,it_f}(4\pi ny) (e^{2\pi i nx} + \epsilon_f e^{-2\pi i nx}).$$ -Here $\rho_f$ is a normalising constant, $\lambda_f(n)$ is the $n$-th Hecke eigenvalue (which is multiplicative as an arithmetic function and conjecturally satisfies $\lambda_f(p) \in [-2,2]$ for any prime $p$), $t_f$ is the spectral parameter of $f$ (so that $f$ has Laplacian eigenvalue $\lambda_f = 1/4 + t_f^2$), $W_{\alpha,\beta}$ denotes the Whittaker function (in this case, it is basically just a $K$-Bessel function), and $\epsilon_f \in \{1,-1\}$ is the parity of $f$ (so that $f(-\overline{z}) = \epsilon_f f(z)$). -So we may assume that $f_1$ and $f_2$ are of this form. One then unfolds the integral by inserting the identity -$$E(z,s) = \sum_{\gamma \in \Gamma_{\infty} \backslash \Gamma} \Im(\gamma z)^s,$$ -using the fact that $f_1(\gamma z) f_2(\gamma z) = f_1(z) f_2(z)$, and that $\Gamma_{\infty} \backslash \mathbb{H} \cong \{z : x \in [0,1], \ y > 0\}$. One then inserts the Fourier expansions of $f_1$ and $f_2$ and evaluates the $x$ integral, then the $y$ integral; this latter integral is essentially the Mellin transform of the product of two $K$-Bessel functions, which has an explicit expression in terms of products of gamma functions. -All in all, one finds that the triple product is equal to -$$\frac{\rho_{f_1} \rho_{f_2} \pi^{-s}}{\Gamma(s)} \prod_{\pm_1, \pm_2} \Gamma\left(\frac{s \pm_1 it_{f_1} \pm_2 it_{f_2}}{2}\right) \sum_{n = 1}^{\infty} \frac{\lambda_{f_1}(n) \lambda_{f_2}(n)}{n^s}.$$ -(At least if $\epsilon_{f_1} = \epsilon_{f_2}$; otherwise, the gamma functions are changed slightly.) Here we are initially assuming that $\Re(s) > 1$; however, by analytic continuation, this extends meromorphically to all of $\mathbb{C}$ (with a simple pole at $s = 1$ if $f_1 = f_2$ and no poles otherwise). Let $\zeta(s) = \sum_{n = 1}^{\infty} n^{-s}$ denote the Riemann zeta function. Then we define the Rankin-Selberg $L$-function $L(s,f_1 \times f_2)$ to be -$$L(s,f_1 \times f_2) = \zeta(2s) \sum_{n = 1}^{\infty} \frac{\lambda_{f_1}(n) \lambda_{f_2}(n)}{n^s}.$$ -Again, this is initially defined for $\Re(s) > 1$ in this way, but extends meromorphically to $\mathbb{C}$. Moreover, this has a nice functional equation, which can be proven using the functional equation for $E(z,s)$. - -Next, let me discuss the same problem for -$$\int_{\Gamma \backslash \mathbb{H}} f(z) E\left(z,\frac{1}{2} + it\right) E(z,s) \, d\mu(z).$$ -The key point is that $E(z,1/2 + it)$ has a nice Fourier expansion; $t$ replaces $t_f$, while $\lambda_f(n)$ is replaced by $\lambda(n,t) = \sum_{ab = n} a^{it} b^{-it}$. The exact same method works, except that the ensuing Rankin-Selberg $L$-function factorises as the product of two $L$-functions: -$$\zeta(2s) \sum_{n = 1}^{\infty} \frac{\lambda_f(n) \lambda(n,t)}{n^s} = \sum_{m = 1}^{\infty} \frac{\lambda_f(m)}{m^{s + it}} \sum_{n = 1}^{\infty} \frac{\lambda_f(n)}{n^{s - it}} = L(s + it,f) L(s - it,f).$$ - -Now we consider -$$\int_{\Gamma \backslash \mathbb{H}}^{\mathrm{reg}} E\left(z,\frac{1}{2} + it_1\right) E\left(z,\frac{1}{2} + it_2\right) E(z,s) \, d\mu(z).$$ -I have written $\int^{\mathrm{reg}}$ to denote the fact that we must regularise (or renormalise) this integral in order for it to converge. This goes back to work of Zagier. Here not only does the Rankin-Selberg $L$-function factorise into the product of two $L$-functions, but each of these $L$-functions factorises further into the product of two copies of the Riemann zeta function. - -Finally, let me consider -$$\int_{\Gamma \backslash \mathbb{H}} f_1(z) f_2(z) f_3(z) \, d\mu(z).$$ -Here the Rankin-Selberg method does not apply, since we cannot unfold as there is no Eisenstein series present. Nonetheless, this triple product can be shown (via methods coming from representation theory) to be associated to $L$-functions. This was done in Watson's thesis and was further expanded upon by Ichino. What they show is an identity relating the square of the absolute value of this integral to a certain $L$-function, namely $L(1/2, f_1 \times f_2 \times f_3)$.<|endoftext|> -TITLE: Map from Bruhat stratification to Catalan stratification for the space of totally nonnegative upper-triangular matrices -QUESTION [10 upvotes]: $\DeclareMathOperator\SL{SL}$This question came up in a class "Total Positivity and Cluster Algebras" being taught by Chris Fraser. -Let $N^+$ denote the space of uni-upper-triangular matrices in $\SL(n,\mathbb{R})$, and $N^+_{\geq 0} \subseteq \SL_{\geq 0}(n,\mathbb{R})$ the totally nonnegative parts of these spaces. -There are two stratifications of $N^+_{\geq 0}$ that I want to compare. -The first is what I'll call the Catalan stratification. It's the stratification based on the location of nonzero entries for a matrix $M \in N^+_{\geq 0}$, which can easily be seen to necessarily lie below/to the left of a Dyck path. E.g., one stratum in the case $n=5$ is: -$$ \begin{pmatrix} 1 & * & * & 0 & 0 \\ 0 & 1 & * & * & 0 \\ 0 & 0 & 1 & * & 0 \\ 0 & 0 & 0 & 1 & * \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} $$ -where the $*$'s denote the nonzero entries. So there are $C_n = \frac{1}{n+1}\binom{2n}{n}$ many strata here, and the closure relation among these strata is just containment of Dyck paths. -The second is more sophisticated; I'll call it the Bruhat stratification. For $1 \leq i \leq n-1$, let $x_i(t)$ denote the matrix with $1$'s along the main diagonal, $t$ in the $i$th spot right above the main diagonal, and $0$'s elsewhere. It's a theorem of Lusztig that for any permutation $w \in S_n$ and any reduced word $(i_1,i_2,\dotsc,i_\ell)$ of $w$, the map $(t_1,\dotsc,t_\ell)\to x_{i_1}(t_1)\dotsb x_{i_\ell}(t_\ell)$ is a homeomorphism of $\mathbb{R}_{>0}^{\ell}$ onto $Y^o_w := B^{-1}wB^{-1} \cap N^+_{\geq 0}$, where $B^-$ are the lower-triangular matrices in $\SL(n,\mathbb{R})$. (See Hersh - Regular cell complexes in total positivity for background on this.) So we have a stratification of $N^+_{\geq 0}$ indexed by permutations in $S_n$, and the closure relations among these strata is the (strong) Bruhat order on these permutations. -Question: Is it true that every Catalan stratum is a union of Bruhat strata? If so, what's the resulting map from permutations to Dyck paths? -I suspect the map from permutations to Dyck paths might be something like the "Cambrian congruences" of Nathan Reading. But I also suspect that this question may have already been studied somewhere. - -REPLY [7 votes]: The Catalan strata are unions of Bruhat strata, and the resulting map from permutations to Dyck paths is indeed given by taking the left-to-right maxima. -There is a way to parametrize totally nonnegative matrices by writing them as Lindström-Gessel-Viennot matrices for a certain weighted directed graph. This is explained nicely in the first section of "Total positivity: tests and parametrizations", by Fomin and Zelevinsky, where the directed graph is depicted in figure 2. They explain the parametrization for general matrices, but if you want to restrict to uni-upper-triangular ones then simply keep just the part of the graph with blue edges. -This combinatorial picture is then refined to also parametrize $Y^o_w$. They construct a directed graph for every reduced word $(i_1,\dots,i_{\ell})$. See figures 4 and 5, where for your case you only have blue edges. -Now, for every permutation $w$ you have a canonical reduced word of the form -$$(m_1,m_1+1,\dots,n_1)(m_2,m_2+1,\dots n_2)\cdots (m_k,m_k+1\dots n_k)$$ -where $n\geq n_1>n_2>\cdots >n_k\geq 1$ and $n_i\geq m_i\geq 1$, which is just the first reduced word for $w$ in the reverse lex order. By drawing the associated directed graph we can find out exactly which matrix entries will be nonzero! The $ij$ entry is nonzero if and only if there is a way to travel from source $i$ to sink $j$. This already means that each $Y^o_w$ sits inside a unique Catalan stratum, and therefore each Catalan stratum is a union of Bruhat strata. -In order to get the description of the map from permutations to Dyck paths there are probably many ways to see it. I reasoned as follows: a pair $(m_i,n_i)$ from above is "redundant" if there is some other pair $(m_j,n_j)$ with $n_j>n_i$ and $m_j\le m_i$. Removing all the redundant blocks from the reduced word can be seen not to affect which matrix entries are nonzero, therefore it stays within the same Catalan stratum. This produces a 321 avoiding permutation which has the same convex hull as $w$ from the upper right corner (See corollary 5.8 in "Some Combinatorial Aspects of Reduced Words -in Finite Coxeter Groups"). Incidentally, this 321 permutation also represents the lowest dimensional Bruhat cell in the Catalan stratum.<|endoftext|> -TITLE: What makes dependent type theory more suitable than set theory for proof assistants? -QUESTION [138 upvotes]: In his talk, The Future of Mathematics, Dr. Kevin Buzzard states that Lean is the only existing proof assistant suitable for formalizing all of math. In the Q&A part of the talk (at 1:00:00) he justifies this as follows: - -Automation is very difficult with set theory -Simple type theory is too simple -Univalent type theory hasn't been successful in proof assistants - -My question is about the first of these: Why is automation very difficult with set theory (compared to dependent type theory)? - -REPLY [7 votes]: I'll answer just the automation question since the other answers gave nice broad overview, but didn't seem focus on that narrow question. My own direct automation experience is wrt to ACL2, Lean and SMT-based solvers. -Strictly speaking, I don't know if there's any foundational argument for why set theory would be better or worse than the type theory-based approach in Lean. -The strengths that Lean have from my perspective are: an expressive explicit type system, a relatively simple core language for representing terms, and a attention to how terms are represented for efficient manipulation. -With regards to typed core logics, most automation in theorem provers is tailored to specific common theories that are widely used in mathematics. When writing such automation, it is important to know the types and operations involved. For example, in writing a decision procedure for linear arithmetic in an untyped language, one needs to carefully check that any transformations still make sense even if the expressions do not denote numbers. By having a typed and typechecked expression language, one gets from the theorem prover itself and does not have to pay the additional runtime and complexity costs. -A second strength of Lean is ensuring that the core language is simple, but expressive so that one can represent proofs compactly. When using automation such as SMT solvers, the "proof terms" generated as evidence can be very large and the core proof language needs to be designed to compactly represent proofs while still be amenably to efficient checking. I'm not sure if Lean has an advantage to Coq or other solvers here per se but it is a factor in Lean's design. -A third strength of Lean is that the language for writing tactics and creating definitions and theorems are one and the same. There is a bit of syntactic sugar for the tactic sequences and a tactic-specific library, but by having the same language one does not have to learn multiple languages just to get started writing tactics. Lean is also not unique here -- ACL2 is similar, but it is a strength of Lean still. It will also become even more relevant with Lean 4 thanks to the efficient compiler being developed.<|endoftext|> -TITLE: Colimits of short exact sequences of C*-algebras -QUESTION [5 upvotes]: Assume I have an inductive system of short exact sequences of $C^{\ast}$-algebras (i.e., short exact sequences $0 \to A_n \to B_n \to C_n \to 0$ together with transformations from the $n$-th to the $(n+1)$-st short exact sequence so that all squares commute). If I form now the colimit of the $C^{\ast}$-algebras, is the resulting sequence $$0 \to \varinjlim A_n \to \varinjlim B_n \to \varinjlim C_n \to 0$$ still exact? Note that I do not want to assume here that the connecting maps in the colimits I form are injective. - -REPLY [3 votes]: My notation -$$ - i_n:A_n\to B_n, - $$ -$$ - p_n:B_n\to C_n, - $$ -$$ - i:\displaystyle \lim_\to A_n\to \displaystyle \lim_\to B_n, - $$ -$$ - p:\displaystyle \lim_\to B_n\to \displaystyle \lim_\to C_n, - $$ -$$ - \beta _n:B_n\to\displaystyle \lim_\to B_n. - $$ -I suppose the only contentious point is to prove that $\text{Ker}(p) \subseteq \text{Im}(i)$, so suppose that this fails. For -each $\varepsilon >0$ we may then choose some $b\in \displaystyle \lim_\to B_n$, such that - -$\|p(b)\|<\varepsilon $, - -$\text{dist}(b,\text{Im}(i)) > 1-\varepsilon $. - - -Since the union of the images of the $B_n$ is dense in $\displaystyle \lim_\to B_n$, we may assume that $b=\beta_n(b_n)$, for some $b_n\in B_n$. -Increasing $n$, if necessary, we may assume that moreover $\|p_n(b_n)\|<\varepsilon $. But this is a contradiction since -$$ - \varepsilon >\|p_n(b_n)\| = \text{dist}(b_n,\text{Im}(i_n))\geq $$ $$ \geq\text{dist}(\beta _n(b_n),\text{Im}(i))= - \text{dist}(b,\text{Im}(i)) >1-\varepsilon . - $$<|endoftext|> -TITLE: Unitary representation is strictly continuous -QUESTION [7 upvotes]: Let $G$ be a compact group and $u: G \to B(H)$ be a strongly continuous unitary representation on the Hilbert space $H$. Then is $u: G \to B(H)$ strictly continuous? -That is, give $B(H)$ the topology induced by the $*$-isomorphism $M(B_0(H))\cong B(H)$. Explicitely, a net $(x_i)$ in $B(H)$ converges strictly to $x$ iff $\|x_i y -x y\|\to 0$ and $\|yx_i \to yx\| \to 0$ for all compact operators $y \in B_0(H)$. On bounded subsets this agrees with the $*$-strong topology. - -REPLY [5 votes]: As you note, on bounded sets, the strict topology and the strong-$\ast$ topology agree on bounded sets. As the set of unitary operators is bounded, we can just work with the strong-$\ast$ topology. If $(u_i)$ is a net of unitary operators converging strongly to $u$ a unitary, then for $\xi\in H$, -$$ \| u_i^\ast(\xi) - u^\ast(\xi)\|^2 = \|u_i^\ast(\xi)\|^2 - (u_i^\ast(\xi)|u^\ast(\xi)) - (u^\ast(\xi)|u_i^\ast(\xi)) + \|u^\ast(\xi)\|^2 $$ -Here I write $(\cdot|\cdot)$ for the inner product. As $u_i,u$ are unitary, this is equal to -$$ 2\|\xi\|^2 - (\xi|u_iu^\ast(\xi)) - (u_iu^\ast(\xi)|\xi). $$ -As $u_i(\eta)\rightarrow u(\eta)$ for any $\eta$, this converges to -$$ 2\|\xi\|^2 - (\xi|u u^\ast(\xi)) - (u u^\ast(\xi)|\xi) = 0. $$ -Thus $u_i\rightarrow u$ strong-$\ast$. This is just a proof that the strong and strong-$\ast$ topologies agree on the set of unitaries; this is surely in standard textbooks, if you look hard.<|endoftext|> -TITLE: A conjecture harmonic numbers -QUESTION [17 upvotes]: I will outlay a few observations applying to the harmonic numbers that may be interesting to prove (if it hasn't already been proven). -From the Online Encyclopedia of Positive Integers we have: -$a(n)$ is the number of permutations $p$ of $\{1,\ldots,n\}$ such that the minimum number of block interchanges required to sort the permutation $p$ to the identity permutation is maximized. -$1, 1, 5, 8, 84, 180, 3044, 8064, 193248, 604800,\ldots$ (https://oeis.org/A260695) -Consider the following harmonic numbers: -$$1 + 1/2 = (1 + 2)\cdot 1/2!$$ -$$1 + 1/2 + 1/3 + 1/4 = (1 + 2 + 3 + 4)\cdot 5/4!$$ -$$1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 = (1 + 2 + 3 + 4 + 5 + 6)\cdot 84/6!$$ -$$1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8)\cdot 3044/8!$$ -and so on. -Therefore, the following generalization suggests itself: -$$1 + 1/2 + 1/3 + 1/4 + \cdots + 1/2n = (2n^2 + n)\cdot a(2n - 1)/(2n)!$$ -If this hasn't been proven, then I will leave it as a conjecture. -Cheers, -Robert - -REPLY [4 votes]: The sequence OEIS A260695 can be seen as the interweaving of nonzero Hultman numbers $\mathcal{H}(n,k)$ for $k=1$ and $k=2$. Namely, -$$a(n) = \mathcal{H}(n,1+(n\bmod 2)).$$ -It is known that $\mathcal{H}(n,k)$ is nonzero only when $n-k$ is odd, in which case its value is given by -$$\mathcal{H}(n,k) = \frac{c(n+2,k)}{\binom{n+2}2},$$ -where $c(\cdot,\cdot)$ are unsigned Stirling numbers of first kind. -Noticing that $c(n+2,2)=(n+1)!H_{n+1}$, we obtain the same formulae as in Mikhail's answer.<|endoftext|> -TITLE: Characteristic polynomial of a matrix related to pairs of elements generating $\mathbb{Z}/n\mathbb{Z}$ -QUESTION [8 upvotes]: Fix $n\geq 2$. Let $A$ be the matrix whose rows and columns -are indexed by pairs $(a,b)\in \mathbb{Z}/n\mathbb{Z}$ such -that $a,b$ generate $\mathbb{Z}/n\mathbb{Z}$ (the number of -such pairs is $\phi(n)\psi(n)$, where $\phi(n)$ is the Euler -phi-function and $\psi(n)$ is the Dedekind $\psi$-function), -with the $((a,b),(c,d))$ entry defined by -$$ A_{(a,b),(c,d)}=\left\{ \begin{array}{rl} - 1, & \mathrm{if}\ c=a\ \mathrm{and}\ d=b-a\\ - 1, & \mathrm{if}\ c=a-b\ \mathrm{and}\ d=b\\ - 0, & \mathrm{otherwise}. \end{array} \right. $$ -The characteristic polynomial of $A$ factors quite a -bit. What is the explanation for this behavior? Is $A$ -diagonalizable (over $\mathbb{C}$)? What is the rank of $A$? This matrix -arises in the congruence properties mod $n$ of the entries -of Stern's triangle. -Here are these polynomials for $n\leq 11$. I write $[k]$ for -an irreducible polynomial (over $\mathbb{Q}$) of degree $k$. -$$ n=2:\ \ (x+1)(x-1)(x-2) $$ -$$ n=3:\ \ x^2(x-1)^3(x-2)(x^2+x+2) $$ -$$ n=4:\ \ x^2(x+1)^3(x-1)^4(x-2)(x^2-x+2) $$ -$$ n=5:\ \ x^4(x+1)^2(x-1)^7(x-2)(x^2+1)^2(x^2-x+2)(x^4-x^2+4) $$ -$$ n=6:\ \ - x^6(x+1)^4(x-1)^7(x-2)(x^2+2x+2)(x^2-2x+2)(x^2+x+2) $$ -$$ n=7:\ \ x^{10}(x+1)^4(x-1)^{11}(x-2)(x^2+2)(x^4+3)^2[4]^2[4] - $$ -$$ n=8:\ \ - x^8(x+1)^8(x-1)^{13}(x-2)(x^2+1)^2(x^2-x+2)(x^4-x^2+4)[4][4] - $$ -$$ n=9:\ \ - x^{14}(x+1)^6(x-1)^{17}(x-2)(x^2-2x+2)^2(x^2+x+2)^3[6]^2[6]^2 - $$ -$$ n=10:\ \ - x^{12}(x+1)^{10}(x-1)^{15}(x-2)(x^2-x+2)(x^2+1)^2 - [4]^2[4][4][4][4] $$ -$$ n=11:\ \ - x^{20}(x+1)^{10}(x-1)^{21}(x-2)(x^2+2x+2)(x^2-2x+2) - (x^2+x+2)[3]^2[4]^2[4]^2[8]^2[12]^2. $$ -The factor $x-2$ is clear since the all 1's vector is an -eigenvector with eigenvalue 2. - -REPLY [2 votes]: The corank is $\phi(n)\psi(n)/6$ for all $n>3$. In particular, your computations imply that the matrix is not diagonalizable for $n=6,7,9$. -This follows from the fact that all nonzero entries are accumulated in the $6\times 6$ blocks with rows indexed by -$$ - (a,b),\; (a-b,a),\; (-b,a-b),\; (-a,-b),\; (b-a,a), \; (b,b-a) -$$ -and columns indexed by -$$ - (a-b,b),\; (-b,a),\; (-a,a-b),\; (b-a,-b),\; (b,-a),\; (-a,b-a) -$$ -(and all indices are paritioned into such 6-tuples!). Each such block has the form -$$ - \begin{pmatrix} - 1& 1&\\ - & 1& 1\\ - & & 1& 1\\ - & & & 1& 1\\ - & & & & 1& 1\\ - 1& & & & & 1\\ - \end{pmatrix}, -$$ -hence it is of corank 1. -This can be seen from the following observation. Move along your matrix, passing alternately from a one to another one in the same row, and to another one in the same column. A pair of moves performs the row change $(a,b)\mapsto (a-b,a)$, and a column change $(c,d)\mapsto (-d,c+d)$. Both these maps have order $6$, and (for $n>3$) they have no orbit of smaller length. -For $n=2$ and $n=3$, there appear smaller cycles.<|endoftext|> -TITLE: Books that teach other subjects, written for a mathematician -QUESTION [124 upvotes]: Say I am a mathematician who doesn't know any chemistry but would like to learn it. What books should I read? -Or say I want to learn about Einstein's theory of relativity, but I don't even know much basic physics. What sources should I read? -I am looking for texts that teach subjects that are not mathematics, but I do not want to read through standard high school, undergraduate (and beyond) material. I am looking for recommendations of sources that teach a scientific theory from a basic level, but not from a basic mathematical level. Strong preference would be to concise, terse texts that are foundational but totally rigorous. -Not sure if these exist, but I often wish they did. - -REPLY [2 votes]: Robin Giles, Mathematical Foundations of Thermodynamics. From the preface: - -This monograph is an attempt to give an account of the foundations of thermodynamics which is more than usually rigorous, not only in its logical structure but also in the "rules of interpretation" in which physical meaning is assigned to the theoretical terms.<|endoftext|> -TITLE: Why does an invertible complex symmetric matrix always have a complex symmetric square root? -QUESTION [5 upvotes]: Chapter XI Theorem 3 from here implicitly states that an invertible complex symmetric matrix always has a complex symmetric square root. -It's clear that a square root exists, by appealing to the Jordan Normal Form and the fact that the matrix is invertible. But why should this also be symmetric? -Note also that not all square roots of invertible symmetric matrices are symmetric: For instance, one square root of $\begin{bmatrix}-1 & 0 \\ 0 & -1\end{bmatrix}$ is $\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$, which is not complex symmetric. -What I've tried: By adapting the proof of the Spectral Theorem to complex symmetric matrices, one gets that "most" symmetric matrices can be diagonalised using an orthogonal matrix. For those matrices, the square root claim is true. What do I mean by most (as opposed to all) such matrices can be diagonalised? Call a vector $v$ a null vector if $v^T v = 0$. The adapted Spectral Theorem states that as long as a complex symmetric matrix has no null eigenvectors, it must be diagonalisable by an orthogonal matrix. The adapted Spectral Theorem is in fact false for matrices which have null eigenvectors. I don't know how to prove the square root claim for those matrices. - -REPLY [10 votes]: Higham, in Functions of Matrices, Theorem 1.12, shows that the Jordan form definition is equivalent to a definition based on Hermite interpolation. That shows that the square root of a matrix $A$ (if based on a branch of square root analytic at the eigenvalues of $A$) is a polynomial in $A$. Therefore, if $A$ is symmetric so is its square root.<|endoftext|> -TITLE: Which abelian categories possess an exact faithful functor into abelian groups that respects products? -QUESTION [5 upvotes]: Let $A$ be an abelian category closed with respect to small coproducts (that is, and AB3 category). Which assumptions are sufficient to ensure the existence of an exact faithful functor from $A^{op}$ into abelian groups that respects products (i.e., it should send $A$-coproducts into the corresponding products)? -I suspect that in certain cases one can take a functor represented by an injective cogenerator in the category $\operatorname{Ind}-A$, but I don't understand when this works. Do exact $\alpha$-filtered colimits in $A$ (where $\alpha$ is a regular cardinal) help (say, if $A$ contains a generator)? -Upd. Sorry; I mistunderstood Qiaochu Yuan's answer below. His answer only treats the existence of functors that satisfy certain non-trivial additional conditions, whereas any functor from $A^{op}$ into abelian groups that respects products will be sufficient for my purposes. So, do there exist any re-formulations of this existence, or sufficient assumptions that imply it? - -REPLY [6 votes]: I find it less confusing to work directly with $A^{op}$ so let me do that; I'll rename it $C$. We have a complete abelian category $C$ (completeness is equivalent to being closed under small products) and we want to know when it admits an exact faithful functor $G : C \to \text{Ab}$ which respects products (equivalently, in the presence of exactness, continuous). If $G$ satisfies the solution set condition then by the (general) adjoint functor theorem $G$ has a left adjoint $F : \text{Ab} \to C$. This left adjoint is determined (by cocontinuity) by $F(\mathbb{Z})$, which I'll name $P$. The adjunction gives that $P$ represents $G$, and exactness and faithfulness gives that $P$ is a projective generator. Conversely, given a projective generator $P$, $\text{Hom}(P, -)$ is an exact faithful functor which respects products. Going back to $A$, you want an injective cogenerator (in $A$, not in $\text{Ind}(A)$). -So if you want more general examples than this then $G$ can't satisfy the solution set condition.<|endoftext|> -TITLE: An integral identity -QUESTION [9 upvotes]: $\newcommand\la\lambda\newcommand\w{\mathfrak w}\newcommand\R{\mathbb R}$Numerical calculations and other considerations (The min of the mean of iid exponential variables) suggest that -$$\int_\R \frac{1-e^{itu}}{e^{itu}-1-it}\,\frac{dt}t=\pi i\,\frac u{1-u}$$ -for $u\in(0,1)$, with the integral understood in the principal value sense. However, I have not been able to prove this, even with the help of Mathematica. -How can this be proved? - -REPLY [8 votes]: $\newcommand\eps\varepsilon$ We want to show that, under $R\to\infty$ and $\eps\to 0+$, we have -$$\int_{(-R,-\eps)\cup(\eps,R)} \frac{1-e^{itu}}{e^{itu}-1-it}\,\frac{dt}t=\pi i\,\frac u{1-u}+o(1).$$ -Equivalently, -$$\int_{(-R,-\eps)\cup(\eps,R)}\left(\frac{1-e^{itu}}{e^{itu}-1-it}+1\right)\,\frac{dt}t=\pi i\,\frac u{1-u}+o(1).$$ -In other words, -$$\int_{(-R,-\eps)\cup(\eps,R)}\frac{dt}{e^{itu}-1-it}=\pi\,\frac u{u-1}+o(1).$$ -The integrand is holomorphic in an open set containing $\{t\in\mathbb{C}:\text{$\Im(t)\geq 0$ and $t\neq 0$}\}$, hence by Cauchy's theorem it suffices to show that -$$\int_{\gamma(R)}\frac{dt}{e^{itu}-1-it}=-\pi+o(1)\qquad\text{and}\qquad -\int_{\gamma(\eps)}\frac{dt}{e^{itu}-1-it}=\frac{\pi}{u-1}+o(1),$$ -where $\gamma(r)$ is the semicircle in $\{t\in\mathbb{C}:\Im(t)\geq 0\}$ going from $r$ to $-r$. For large $r$, the integrand on $\gamma(r)$ is $i/t+O(1/t^2)$. For small $r$, the integrand on $\gamma(r)$ is $-i/(t(u-1))+O_u(1)$. The result follows.<|endoftext|> -TITLE: An identity for the Lambert $W$ function -QUESTION [5 upvotes]: Expressing the integral in An integral identity in terms of residues, we come to the following supposed identity: -$$\sum_{k=-\infty}^\infty\frac1{1 + W_k(x)}=\frac12$$ -for all $x\in(-1/e,0)$, where $W_k$ is the $k$th branch of the Lambert $W$ function. -How can this be proved? - -REPLY [2 votes]: This identity is now established, since -An integral identity cited in the above post is now proved.<|endoftext|> -TITLE: Statements related to Thurston's work on the surface -QUESTION [5 upvotes]: If we have simple closed curves $\alpha$ and $\beta$ on a surface $\Sigma_g$, the intersection number $i(\alpha ,\beta)$ is defined to be the minimal cardinality of $\alpha_1\cap\beta_1$ as $\alpha_1$ and $\beta_1$ range over all simple closed curves isotopic to $\alpha$ and $\beta$, respectively. We say $\alpha$ and $\beta$ intersect minimally if -$i(\alpha ,\beta) = |\alpha\cap\beta|\,$. - -How to see that $\alpha$ and $\beta$ intersect minimally if there are no pairs of $p,q\in\alpha\cap\beta$ such that the arc joining $p$ to $q$ along $\alpha$ followed by the arc from $q$ back to $p$ along $\beta$ bounds a disk in $\Sigma_g$? -Maybe a sketch of the proof idea? - -I think the converse is also true : "that $\alpha$ and $\beta$ intersect minimally only if there are no pairs of $p,q\in\alpha\cap\beta$ such that the arc joining $p$ to $q$ along $\alpha$ followed by the arc from $q$ back to $p$ along $\beta$ bounds a disk in $\Sigma_g$." - -REPLY [10 votes]: This is called the "bigon criterion". For a discussion, see Section 1.2.4 (and in particular Proposition 1.7) of the "Primer on mapping class groups" by Farb and Margalit. -The Google search "bigon criterion" also finds various references and lecture notes. For example, here is the top hit: -https://math.stackexchange.com/questions/1646340/proof-of-the-bigon-criterion<|endoftext|> -TITLE: Does the Fourier expansion of the j-function have any prime coefficients? -QUESTION [20 upvotes]: Title asks it: Does the Fourier expansion of the j-function have any prime coefficients? -A superabundance of congruences involving primes up to 13 rule out many candidates, but calculation suggests that primes $p>13$ occur as divisors at frequencies (about?) $1/p$. -But -$$c_{71}=278775024890624328476718493296348769305198947=(353) (5533876049689057963) (142708463580969897033673)$$ -so that might count as a near miss. -That said, though composite, none of $c_{1319},c_{1559},c_{1871},c_{2111},c_{2231},c_{3239},c_{3551}, -c_{4271}, c_{4799}, c_{5471}$,... has a factor less than 100. - -REPLY [21 votes]: There are seven prime values (passing a BPSW test) of $c_n$ with $n \le 2 \cdot 10^7$, at indices 457871, 685031, 1029071, 1101431, 9407831, 11769911, and 18437999. -For a writeup about the computations, source code, and the prime numbers themselves, see: -https://github.com/fredrik-johansson/jfunction -The first prime $c_{457871}$ is the following 3689-digit number: -30801636514011810817665318374658229108845878479020424564852145048550477094266397753615620563343543122809859646561082297428127434662591162587465688189470673559500946738365120052237063677089165938711865291826029106118944935275575747924692016273156986404534483801538929131770513483035012136870394657959401728135298020394188493171098233244180987908576898694823463564573986521378977434641833939268448907892425327423931702787985965542437137823321051154070271812702040044018249180573203850835848455588503282760349067467783417988439399728717693269653344946303351200847243267479008953480983893357261716684979093234554524907570654175153716092013381021033631140271230374127359907711968372522745844402375355449952404485703394467619855566837872874898909971592754384802113052943616244616228982238122927893868687205058321497136583843454449483568143398274104240118934930464297655670020151675237805373543868606164624359906981163728280265160187710553222515223577715624809615213830941718538338330718318212962253165154784990727034694055424785446779353587418545233599512401297307560287330719685869795522920270669609498639686158362603820371777436303198350821510562028940026062291473250204511985677264132923030166903832835499326969147438699601413777550018800825631136927531190121891919023537853918707520543367800297170322643557011053113196699442392677721251164071912545596204918917566579754931649583289900867072630307177448327512748573948720052532426026582228531694404885199310535255097152264726060203165225244982020696634699998363392968546865623955653923292400203238704016331998301098882719280237294694368706187727194309664172903851082095220551328392538250319846241916349485122880753951745944930504836900929002975743487891022913325212927877123283110371004511930228354316650588540693893785256904915901446318107624914828072059779823732136657855374843120155979700684426795681799949828354496446059153417627020927839115150227802126261918915561804876068184778844391781134858921611498880947743310539662060436623031296503497690935576519684398269951037522254928920040638913033916447480995451172728711833839757712418893132702856601768693707767408660072694546475350179623544013680656723320135856226549734491873157405584154845513711221581080660024268777736123563250061564832634510929167076111162819878757339574946474890631427413494266622011134541083389565676533915214975782362935488902596227893270340685189472412328774565836766141144309983845007818359559400419277334332676913295096744471767890050480297665094988593656013858404531156346968039028944208325870737014099138274367178622028282421548639862174878971583355345514824564041789417768026191489674685601130884770272042320774286085741557976163690129647685531361972350643878463985533720888119015787469601005630621659687749812528402893568495431951749454436584549713954945990865871059056586397539337257860037142247888078726810588665972975365062119719322236620399267595421045468328581519496916075539958247075033250213251987450634988759698630929366797978086037651833328039578740169593342409162117195011259429537049154618053581738781731898468529654264215451362063751162610494664217028387863630499363451607405876105244341222047086047630946033922722790116254686279821386138836326264599571591967081705529398847838399554521937548523334357827287180353919911626556542110462725508733594289737987833526697420630616509010365868192637613354631155273335447830149024265581051436057754643260886600218454986343035650276055047068090749733940363539512760186969985697586068479926077515927351409212093813617210155525035138051400826519032339949113881712130873854263099726139035786383205124051644483087596049043497124737908938233702928805190964560673396398573801431769834039050534841422792714076699<|endoftext|> -TITLE: In M-theory, what can hypothesis H tell us that quantization in ordinary cohomology cannot? -QUESTION [26 upvotes]: In classical field theory, many fields and related objects are described as differential -forms. For example, in electromagnetism, the field $F := B - \mathrm dt\wedge E$ is a 2-form, and Maxwell's -equations ask that it be closed. Quantization imposes integrality constraints on $F$, which may concisely be -pacakged as asking for $F$ to be a (cocycle representative of a) differential cohomology class $\check F\in \check -H^2(M; \mathbb Z)$ (here $M$ is spacetime). -In some theories, it turns out that it makes more sense to use differential generalized cohomology to perform -this quantization. For example, in superstring theory, the Ramond-Ramond field is quantized in differential -$K$-theory. Determining which generalized cohomology theory $X$ to use is a bit of an art: it's part of the data of -formulating the theory, making it harder to conclusively prove that one choice is right or wrong. Instead, what one -can do is prove that quantizing a given field in differential $X$-theory recovers as a theorem something whose -physics implication is something that physicists already believe about the theory, and that other choices of -generalized cohomology theories do not yield that conclusion. -My specific question is about M-theory (or more precisely, its low-energy limit given by supergravity). There is a -field called the $C$-field, which classically is a 3-form. There are two proposals that I know of for quantizing the -$C$-field in differential generalized cohomology: - -Traditionally, one uses $X = H\mathbb Z$, i.e. ordinary cohomology, twisted by the orientation local system. This recovers some facts about M-theory that physicists expect, including the appearance of $K$-theory in superstring theory. -Alternatively, Fiorenza-Sati-Schreiber posit "hypothesis H," that the -$C$-field is quantized in $J$-twisted stable cohomotopy. That is, $X = \mathbb S$, twisted on a spacetime -manifold $M$ by the composite of maps $M\to B\mathrm O$ given by the stable normal bundle, then $B\mathrm - O\to B\mathrm{GL}_1(\mathbb S)$ given by the $J$-homomorphism. In several papers, Fiorenza-Sati-Schreiber -prove that hypothesis H implies several facts about M-theory that physicists expected, providing evidence in -favor of hypothesis H. - -To what extent does the ordinary cohomology proposal fail to satisfy these hypotheses? That is, what is an example -of a claim which physicists believe to be true about M-theory which is true under hypothesis H, but not under the -ordinary cohomology hypothesis? - -REPLY [17 votes]: Traditional approach. Notice that what is considered in [DMW00; DFM03] and elsewhere to quantize the C-field flux $G_4$ is not just ordinary cohomology, but ordinary cohomology with bells and whistles added as need be: -Foremost there is the half-integral quantization of the $G_4$-flux, mentioned as (3.2) in [DMW00]. Ordinary cohomology may be modified ("shifted") to accommodate this, as maybe first formalized in [HS02] and used in [DFM03]. -Beyond this, [DMW00] argue that the "M-theory path integral" imposes a further condition on the shifted integral cohomology class of $G_4$, namely that it be in the kernel of the Steenrod operation $Sq^3$. Whether this "integral equation of motion" is enforced by flux quantization in some further modification of ordinary cohomology is not discussed there. -Instead, the observation made is that the constraint $Sq^3 [F_4] = 0$ is also the first condition that appears in the Atiyah-Hirzebruch spectral sequence for lifting an ordinary cohomology class $[F_4]$ to complex topological K-theory, as demanded by the widely accepted conjecture that RR-field fluxes $F_{n}$ in string theory are quantized in topological K-theory. -Since, moreover, $F_4$ is meant to come from $G_4$ as one lifts string theory to M-theory, the observation of [DMW00] is hence that the "integral equation of motion" in M-theory reproduces one of the constraints on one of the RR-field components given by flux quantization in K-theory. -Suggestive as this is, this consistency check is arguably not a "derivation of K-theory" in string theory. In fact, the conjecture that string theoretic RR-flux really is quantized in K-theory remains itself being debated (for instance, it's not clear how to make it compatible with S-duality). The problem here is that little is known with certainty about non-perturbative string theory beyond a web of interlocking conjectures. -Hypothesis H. In view of this situation it seems worthwhile to try a strict top-down approach where a unified generalized cohomology theory in M-theory is postulated and its consequences on flux quantization rigorously derived. -There are good hints what this M-theoretic cohomology theory ought to be: Its image in rational cohomology must see the trivialization of the cup square of the $G_4$-flux demanded by 11d supergravity -- this being the M-theoretic lift of the twisted Bianchi identities that motivate the twisted K-theory conjecture [MSa 03, Sec. 4.2; FSS 16, Sec. 3]. The condition happens to be exactly the relation that identifies the Sullivan model of the 4-sphere, thus suggesting that M-brane charge is quantized in 4-Cohomotopy theory; due to [Sa13, Sec. 2.5]. -Indeed, cohomotopical charge quantization in M-theory, on the rational level, follows from a first-principles analysis of the super $p$-brane scan, and is as such the direct M-theoretic analogue of a computation that derives the twisted K-theory classification of D-brane charge ([FSS15], reviewed in [FSS19a, Sec. 7]). -This means that any cohomology theory which quantizes M-brane charge should rationally coincide with Cohomotopy theory. Accordingly, it is quite natural to consider the hypothesis (dubbed "Hypothesis H") that M-brane charge is in Cohomotopy theory itself [Sa13] (suitably twisted by the tangent bundle [FSS19b; FSS19c]). -Implications. Indeed, one finds that the assumption of Hypothesis H, that M-brane charge is in J-twisted Cohomotopy theory, readily implies both the half-integral shifted flux quantization on $G_4$ as well as the "integral equation of motion" -- together with a list of further expected constraints [FSS19b, Table 1]. -This way Hypothesis H implies as much "derivation of K-theory"; though one should push further to an actual derivation of the implied flux quantization of RR-fields. Derivation from Hypothesis H of more of the fine-print in the K-theory conjecture is in [SS19a; BSS18; BMSS18]. -Moreover, Hypothesis H sees the Hořava-Witten Green-Schwarz mechanism in the presence of M5-branes [FSS19d; SS20b], reveals fine-print in the M5-brane anomaly cancellation argument [FSS19c; SS20a], and seems to see a zoo of subtle brane charge effects expected in Hanany-Witten systems [SS19b]. -Certainly none of these effects follows from flux quantization in just ordinary cohomology (nor in K-theory, for that matter). -By way of outlook, we think we see now that there is a natural chromatic character map on twisted Cohomotopy which exhibits the M5-brane partition function as charge-quantized in elliptic cohomology, matching traditional discussion of M5-brane ellitptic genera. This is work in progress. -Conclusion. In summary, rigorous derivation of the implications of Hypothesis H suggests that twisted Cohomotopy theory sees a fair number of subtle effects that have previously been argued informally to appear in the elusive non-perturbative completion of string theory. This may be indication that, going beyond the traditional approach of hard-coding M-theoretic folklore into a putative "C-field model", charge quantization in twisted Cohomotopy might get to the native heart of the elusive mathematical nature of "M-theory". But, of course, more analysis is necessary. -Exposition. More exposition of motivation and development of Hypothesis H may be found in talk notes of the recent meeting at NYU AD M-Theory and Mathematics: [Sa20; Sc20]. -References. - -[DMW00] E. Diaconescu, G. Moore, E. Witten, $E_8$ Gauge Theory, and a Derivation of K-Theory from M-Theory, Adv. Theor. Math. Phys. 6:1031-1134, 2003 (arXiv:hep-th/0005090), summarised in: A Derivation of K-Theory from M-Theory (arXiv:hep-th/0005091) - -[HS02] M Hopkins, I. Singer, Quadratic Functions in Geometry, Topology,and M-Theory, J. Differential Geom. Volume 70, Number 3 (2005), 329-452 (arXiv:math.AT/0211216, euclid:1143642908) - -[DFM03] E. Diaconescu, D. Freed, G. Moore, The $M$-theory 3-form and $E_8$-gauge theory, chapter in: Elliptic Cohomology Geometry, Applications, and Higher Chromatic Analogues, Cambridge University Press 2007 (arXiv:hep-th/0312069, doi:10.1017/CBO9780511721489) - -[MSa03] V. Mathai, H. Sati, Some Relations between Twisted K-theory and E8 Gauge Theory, JHEP 0403:016, 2004 (arXiv:hep-th/0312033, doi:10.1088/1126-6708/2004/03/016) - -[Sa13] H. Sati, Framed M-branes, corners, and topological invariants, J. Math. Phys. 59 (2018), 062304 (arXiv:1310.1060) - -[FSS15] D. Fiorenza, H. Sati, U. Schreiber, The WZW term of the M5-brane and differential cohomotopy, J. Math. Phys. 56, 102301 (2015) (arXiv:1506.07557, doi:10.1063/1.4932618) - -[FSS16] D. Fiorenza, H. Sati, U. Schreiber, Rational sphere valued supercocycles in M-theory and type IIA string theory, J. Geom. Phys., Vol 114 (2017) (arXiv:1606.03206, doi:10.1016/j.geomphys.2016.11.024) - -[BMSS18] V. Braunack-Mayer, H. Sati, U. Schreiber, Gauge enhancement of super M-branes via parametrized stable homotopy theory, Comm. Math. Phys. 371: 197 (2019) (arXiv:1806.01115, doi:10.1007/s00220-019-03441-4) - -[BSS18] S. Burton, H. Sati, U. Schreiber, Lift of fractional D-brane charge to equivariant Cohomotopy theory, J. Geom. Phys., 2020 (in print) (arXiv:1812.09679) - -[FSS19a] D. Fiorenza, H. Sati, U. Schreiber, The rational higher structure of M-theory, in: Proceedings of Higher Structures in M-Theory 2018, Fortsch. Phys. 2019 (arXiv:1903.02834, doi:10.1002/prop.201910017) - -[FSS19b] D. Fiorenza, H. Sati, U. Schreiber, Twisted Cohomotopy implies M-Theory anomaly cancellation on 8-manifolds, Comm. Math. Phys. 377(3), 1961-2025 (2020) (arXiv:1904.10207, doi:10.1007/s00220-020-03707-2) - -[FSS19c] D. Fiorenza, H. Sati, U. Schreiber, Twisted Cohomotopy implies level quantization of the full 6d Wess-Zumino-term of the M5-brane, Comm. Math. Phys. 2020 (in print) (arXiv:1906.07417) - -[FSS19d] D. Fiorenza, H. Sati, U. Schreiber, Twistorial Cohomotopy Implies Green-Schwarz anomaly cancellation (arXiv:2008.08544) - -[SS19a] H. Sati, U. Schreiber, Equivariant Cohomotopy implies orientifold tadpole cancellation J. Geom. Phys. Vol 156, 2020, 103775 (arXiv:1909.12277, doi:10.1016/j.geomphys.2020.103775) - -[SS19b] H. Sati, U. Schreiber, Differential Cohomotopy implies intersecting brane observables via configuration spaces and chord diagrams (arXiv:1912.10425) - -[SS20a] H. Sati, U. Schreiber, Twisted Cohomotopy implies M5-brane anomaly cancellation (arXiv:2002.07737) - -[SS20b] H. Sati, U. Schreiber, The character map in equivariant twistorial Cohomotopy implies the Green-Schwarz mechanism with heterotic M5-branes (arXiv:2011.06533) - -[Sa20] H. Sati, M-theory and cohomotopy, talk at M-Theory and Mathematics, NYUAD 2020 (pdf) - -[Sc20] U. Schreiber, Microscopic brane physics from Cohomotopy theory, talk at M-Theory and Mathematics, NYUAD 2020 (pdf)<|endoftext|> -TITLE: degree five genus one curves without rational points? -QUESTION [9 upvotes]: Let $X$ be a smooth genus one curve over $k$. I don't call it elliptic curve because it will have no rational points. -By index of $X$ we mean the smallest degree of a closed point on $X$; equivalently by Riemann-Roch that's the same as the smallest positive degree of a divisor, or the greatest common divisor of all degrees of closed points. -Of course, the index of a curve equals one if the field is algebraically closed, so we consider non-closed fields here. For example, a smooth cubic curve in $\mathbb{P^2}$ can have index one or three. -Here is the question: can one characterize fields $k$ which admit genus one curves of index $d$? -I am especially interested in the $d = 5$ case; then the model of such curve over algebraic closure is a linear section of the Grassmannian $\mathrm{Gr}(2,5)$. For which fields do we have genus one curves of index $5$? Is there a way to figure this out without writing explicit equations? -UPDATE: I actually don't fully understand the case of number fields. Are all number fields allowed? - -REPLY [9 votes]: I'll address the case $d = 5$ over any number field, without recourse to Gross-Zagier formulas and Tate-Shafarevich groups. If $X$ has index 5, then it has order 5 in $H^1(k,E)$, hence comes from $H^1(k,E[5])$, where $E$ is the Jacobian. I assume the converse is not true (presumably some 5-torsion classes come from torsors of index 25), but I'll give examples of cohomology classes which are visibly of index 5. -Over any field there exists at least one elliptic curve with a rational point of order 5, or by duality, a subgroup (scheme) isomorphic to $\mu_5$. Over $\mathbb{Q}$ there is even an elliptic curve (of conductor 50 I believe) that contains both. In the first case, consider the induced map -$$\mathrm{Hom}(G_k,\mathbb{Z}/5\mathbb{Z}) = H^1(k,\mathbb{Z}/5\mathbb{Z}) \stackrel{i}{\to} H^1(k, E[5])$$ -In the dual case, consider $H^1(k, \mu_5) \simeq k^{\times}/k^{\times 5}$. The kernel of $i$ comes from the $k$-rational points in the kernel of the dual isogeny, so it has $\mathbb{F}_5$-dimension at most 1, in either case. The image of $i$ gives you torsors which have index 1 or 5 (since they are twists of a 5-isogeny). The index is 1 precisely if the cohomology class "comes from rational points", i.e if its in the image of $E(k)/5E(k)$ under the coboundary map. If $k$ is a number field, then the latter is a finite group, whereas $H^1(k, \mathbb{Z}/5\mathbb{Z})$ or $H^1(k, \mu_5)$ are both infinite (see the explicit descriptions above). So there are infinitely many index 5 PHS $X$.<|endoftext|> -TITLE: A specific property of bi-adjunction -QUESTION [5 upvotes]: Let $$I: C \rightleftarrows D: F$$ be biadjoint [1] functors between categories $C, D$. That is, $I$ is the left and also the right adjoint of $F$ (thus vice versa). Put in notations, it's -$$ \cdots \vdash I \vdash F \vdash I \vdash F \cdots$$ -Then since $I$ is left adjoint to $F$, for each object $X$ in $D$ there exists a unit map -$$ \eta_X: X \to IFX.$$ -Similarly, since $F$ is left adjoint to $I$, there exists a unit map -$$ \eta'_{FX}: FX \to FIFX.$$ -Question: Given their type, one naturally asks if $F(\eta_X): FX \to FIFX$ equals to $\eta'_{FX}$, or is there any general relation? -(Or if you have some interesting examples of such functors, please let me know. I'm happy to check if they satisfy this condition or not.) -Attempt: This is a special case of an ambidextrous adjunction. The corresponding (co)monads are Frobenius monads. I've looked in some paper about them, and couldn't find this structure discussed. -I've also tried to prove it myself.. but these unit maps come from different adjunctions, so a priori there is no reason for them to relate. But I can't be sure. -Footnote -[1] "biadjunction" is not a standard term. see Mike Shulman's comment below - -REPLY [4 votes]: Qiaochu Yuan’s answer excellently explains the general phenomenon. Here is another slightly simpler concrete example: -Let $\newcommand{\Z}{\mathbb{Z}}\newcommand{\BZ}{{\mathbf{B}\Z}}\BZ$ be the group $\Z$ viewed as a one-object groupoid, with arrows $a^n : \ast \to \ast$ for $n \in \Z$. Then its identity functor $\newcommand{\id}{\mathrm{id}}\id_\BZ$ is self-adjoint in $\Z$-many ways: for any $n$, we can take $a^n$ as the unit and $a^{-n}$ as the counit. (Any identity functor is trivially self-adjoint, and then as Qiaochu explains, we can conjugate it by natural automorphisms of the identity functor, which in this case are precisely $\Z$.) -So now we can choose units+counits for the biadjunction $\id_\BZ \dashv \id_\BZ \dashv \id_\BZ$ in $\Z^2$-many ways. Writing $a^n$ and $a^m$ for the two choices of unit, your condition will hold just when $m=n$.<|endoftext|> -TITLE: Todd polynomials -QUESTION [10 upvotes]: Let $T_k(x_1,\ldots,x_n)$ be the Todd polynomials, $e_k(x_1,\ldots,x_n)$ the elementary symmetric polynomials and $p_k(x_1,\ldots, x_n)$ the power sums of degree $k$. -We have the following generating formulas -\begin{align*} -\sum_{k\geq 0}T_k(x_1,\ldots,x_n)t^k = \prod_{i=1}^n\frac{tx_i}{1-e^{-tx^i}}\,,\\ -\sum_{k\geq 0}e_k(x_1,\ldots,x_n)t^k = \prod_{i=1}^n(1+tx_i)\,,\\ -\sum_{k\geq 0}\frac{1}{k!}p_k(x_1,\ldots,x_n)t^k = \sum_{i=1}^ne^{tx_i}\,. -\end{align*} -There is an explicit relation between $(1/k!)p_k$ and $e_k$ in terms of Newton's identities which can be expressed using generating series (see e.g. wikipedia). Is there some similar expression for $T_k$ in terms of $e_k$ or $p_k$? -For example if $X$ is a hyperkähler complex manifold. Replacing $x_i$ with $\alpha_i$ the roots of $c(TX)$, one can show that (see (3.13)): -$$ -td(X) = \text{exp}\Big[-2\sum_{n\geq0} b_{2n}\text{ch}_{2n}(TX)\Big]\,, -$$ -where $b_{2n}$ are the modified Bernoulli numbers. Is there an explicit formula without any assumptions on the geometry? - -REPLY [14 votes]: We have -$$\log \sum_{k \ge 0} T_k t^k = \sum_{i=1}^n \log \frac{x_i t}{1 - e^{-x_i t}}$$ -so if we write -$$\log \frac{x_i t}{1 - e^{-x_i t}} = \log \sum_{k \ge 0} B_k^{+} x_i^k \frac{t^k}{k!} = \sum_{k \ge 1} b_k x_i^k \frac{t^k}{k!}$$ -(using the sign conventions explained on Wikipedia) then we straightforwardly have -$$\sum_{k \ge 0} T_k t^k = \exp \left( \sum_{k \ge 1} b_k p_k \frac{t^k}{k!} \right).$$ -WolframAlpha gives that the generating function of the series $b_k$ begins -$$b(t) = \log \frac{t}{1 - e^{-t}} = \frac{t}{2} - \frac{t^2}{24} + \frac{t^4}{2880} - \frac{t^6}{181440} \pm $$ -which gives $b_2 = - \frac{1}{12}, b_4 = \frac{1}{120}, b_6 = - \frac{1}{252}$. Plugging the denominators into the OEIS gives A006953, the sequence of denominators of $\frac{B_{2k}}{2k}$, which suggests the following. We have -$$b'(t) = \frac{d}{dt} \left( \log t - \log (1 - e^{-t}) \right) = \frac{1}{t} - \frac{e^{-t}}{1 - e^{-t}} = \frac{1}{t} \left( 1 - \frac{t}{e^t - 1} \right) = - \sum_{k \ge 1} B_k^{-} \frac{t^{k-1}}{k!}$$ -which gives $b_k = - \frac{B_k^{-}}{k}$ for $k \ge 1$. Altogether we have -$$\boxed{ \sum_{k \ge 0} T_k t^k = \exp \left( \sum_{k \ge 1} - \frac{B_k^{-}}{k} p_k \frac{t^k}{k!} \right) }.$$ -As a sanity check, expanding the terms in WolframAlpha up to $t^4$ in terms of elementary symmetric polynomials / Chern classes gives -$$T_1 = \frac{c_1}{2}$$ -$$T_2 = \frac{c_1^2 + c_2}{12}$$ -$$T_3 = \frac{c_1 c_2}{24}$$ -$$T_4 = \frac{-c_1^4 + 4c_2 c_1^2 + c_3 c_1 + 3c_2^2 - c_4}{720}$$ -which agrees with the formulas for the first few terms of the Todd class on Wikipedia. -Edit: I believe this is the same as the result you quote except that the linear term in your result is zero (equivalently, $c_1$ vanishes for a hyperkahler manifold).<|endoftext|> -TITLE: Subsets of a ball/sphere with the largest sum of distances -QUESTION [6 upvotes]: $\newcommand\R{\mathbb R}\newcommand\S{\mathbb S}$Let $B_d$ and $S_{d-1}$ denote, respectively, the closed unit ball and the unit sphere in $\R^d$. Let us say that a finite subset $F$ of $B_d$ is maximal if the sum of all pairwise Euclidean distances between the points in the set $F$ is the largest possible given $n:=\text{card}\,F$. -In a comment to a post now apparently deleted, Noam D. Elkies noted that, by convexity, (i) any maximal set must lie on $S_{d-1}$ and (ii) such a set must be the set of all vertices of a regular polygon if $d=2$. -This suggests the following questions: -Q1. For $d=3$ and $n$ such that there exists a regular polytope with $n$ vertices, is it true that every maximal set must be the set of all vertices of a regular polytope? -Q2. For $d=3$ and any given natural $n$, is a maximal set of $n$ distinct points unique up to an orthogonal transformation? -Q3. A weaker version of Q2: For $d=3$ and any given natural $n$, are there only finitely many, up to an orthogonal transformation, maximal sets of $n$ distinct points? -Q4. Same questions for any $d\ge3$. -Q5. Same questions with the Euclidean distances $|x-y|$ between points $x$ and $y$ replaced by increasing functions $f(|x-y|)$ of the distances; e.g., here one may take $f(r)\equiv r^2$ or $f(r)\equiv -1/r^{d-2}$. -The answer to the questions about regular polytopes seem to be positive at least for $n\le d+1$, with the regular simplexes as the regular polytopes. - -One of the correct and complete answers to any one of these questions will be accepted as an answer to this entire post. - -REPLY [6 votes]: Negative answers to some of those questions: -Q1 Not always; for $n=8$ a square antiprism is better than a cube. -For example, in radius $\sqrt 3$, the cube with vertices $(\pm 1, \pm1, \pm1)$ -gives $16(1 + 2\sqrt{2} + \sqrt{3}) = 88.9676+$ while changing -the $z=+1$ vertices to $(\pm\sqrt2, 0, 1)$ and $(0,\pm\sqrt2,1)$ gives -$16\bigl(\!\sqrt{8+\sqrt{8}} + \sqrt{8-\sqrt{8}}\bigr) = 89.0362+$. One can then -optimize the $z$-values further. [Similarly, in ions of the form -${\rm MF}_8^{n-}$, with eight fluorine atoms around a large central M atom -(such as xenon [sic] or rhenium), the F's form a square antiprism, not a cube.] -Q5 Again not always. Indeed for $\sum|x-y|^2$ we have -$$ -\mathop{\sum\!\sum}_{1 \leq i < j \leq n} - |x_i - x_j|^2 = n^2 - \Bigl| \sum_i x_i \Bigr|^2 -$$ -so the maximal configurations are precisely those whose center of mass -is the origin, and once $n$ is at all large there is a positive-dimensional -family of such configurations. There are also examples for other natural $f$ -where the 24-cell (the $(d,n)=(4,24)$ regular polytope) is suboptimal; see - -Henry Cohn, John H. Conway, Noam D. Elkies, and Abhinav Kumar: The $D_4$ root system is not universally optimal, Experimental Math. 16 (2007), 313--320, -arXiv:math/0607447.<|endoftext|> -TITLE: Defining arithmetic local systems -QUESTION [5 upvotes]: In his ICM 2018 lecture Venkatesh gives a "working" definition of arithmetic local systems of $K$-vector spaces for various fields $K$ as follows: [ICM 2018 Proceedings, Vol. 1, p. 278] - -While for $\mathbb{F}_p$ and $\mathbb{Q}_p$ there is no ambiguity in the definitions, for $\mathbb{Q}$ it is not easy to see over which field to construct representations of arithmetic étale fundamental groups because $GL_n(\mathbb{Q})$ has "too few" finite subgroups, so tentatively he proposes Chow motives, but adds (on p. 277) that with Ayoub and Cisinski-Déglise's theories of motivic sheaves now available it might be possible to give a better definition. For $\mathbb{C}$ he says he has no idea and deems it a fundamental problem of number theory. -Q. 1: For $\mathbb{Q}$ what would be the benefits of using motivic sheaves in place of Chow motives? Would they furnish more local systems as needed for Langlands correspondence, or just be easier to work with? Has his suggestion actually been worked out using any of several constructions of category of motivic sheaves? -Q. 2: For $\mathbb{C}$ is there no definition because there is as yet no motivic Grothendieck group whose representations these local systems would arise from? Or is it some other reason? Reading the referenced paper of Arthur hasn't clarified it for me even though the paper is as clear as it could be. - -REPLY [4 votes]: For Q1: the problem is that Chow motives are a little too restrictive, because they are all (conjecturally at least) semi-simple, whereas non-semisimple objects are hugely important for arithmetic. E.g. if you take an elliptic curve $E / \mathbf{Q}$, and you let $Y = E - P - \infty$ where $P$ is a (non-trivial) rational point of $E$, then the excision sequence in etale cohomology shows that the cohomology of $Y$ contains a Galois representation which is an extension of $V_p(E)$ by the trivial representation in the category of $\mathbf{Q}_p$-linear Galois reps; and this extension realises the cohomology class attached to $P$ by Kummer theory. However, this "punctured elliptic curve" isn't a proper variety so it doesn't give a Chow motive. The various approaches to "mixed motives", including Ayoub and Cisinski--Deglise's works, are an attempt to construct a bigger category which includes Chow motives but also has some interesting non-semisimple objects, and which can "see" non-proper or non-smooth varieties. -For Q2: perhaps what Venkatesh is getting at here is the fact that there are lots of automorphic objects arising in the Langlands program, e.g. non-algebraic Maass forms for $GL(2)$, which are clearly number-theoretically important but are not "definable over $\overline{\mathbf{Q}}$" in any natural way -- e.g. they give $L$-functions $\sum a_n n^{-s}$ which have analytic continuation and functional equations, but such that the $a_n$'s are complex numbers which are (as far as we can tell) transcendental. These should correspond to some kind of "local system" which we don't have any idea as yet how to define.<|endoftext|> -TITLE: Differential refinement of homology -QUESTION [9 upvotes]: Differential cohomology is a refinement of ordinary cohomology by differential data. It's construction comes down to the observation that $H^2(M, \mathbb{Z})$ is isomorphic to the space of isomorphism classes of line bundles via the Chern class, and a connection on such a line bundle thus provides a refinement of the Chern class. -I am looking for a reference to a similar refinement of ordinary homology, which is in some sense Poincare dual to differential cohomology. Is this discussed in the literature? -My motivation comes from the fact that the Chern class of a line bundle $L$ is Poincare dual to the zero locus of a generic section of $L$. Thus, on the level of ordinary cohomology/homology there exist a nice duality phrased in the language of line bundles. - -REPLY [4 votes]: Differential cohomology groups are computed as homotopy groups -$\hat{\rm H}^n(M)=π_0(F_n(M))$, where $F_n\colon{\sf Man}^{\rm op}\to{\sf Sp}$ -is a sheaf of spectra on the site of smooth manifolds. -(See Bunke–Nikolaus–Völkl's Differential cohomology theories as sheaves of spectra for more details.) -Thus, Verdier duality (§5.5.5 in Lurie's Higher Algebra) -suggests a very natural way to convert a differential cohomology theory -into a differential homology theory: -start with a sheaf of spectra $F_n$, -convert it into a cosheaf of spectra $G_n$, -then set $\hat{\rm H}_n(M):=π_0(G_n(M))$. -This yields a covariant functor that can be naturally called a differential homology theory.<|endoftext|> -TITLE: Isomorphisms of complete intersections -QUESTION [5 upvotes]: Let $X, Y\in \mathbb{P}^n$ be two singular Fano complete intersections of the same multidegree $(d_1,…,d_r)$. -If we assume there is an isomorphism $f\colon X\rightarrow Y$ are there any assumptions so that we can conclude that $f$ is induced by an action of the Automorphism group of $\mathbb{P}^n$, $\operatorname{PGL}(n+1)$? -I was reading the "Algebraic Hypersurfaces" note by J. Kollár (https://www.ams.org/journals/bull/2019-56-04/S0273-0979-2019-01663-2/S0273-0979-2019-01663-2.pdf) and I was wondering if there was a generalisation of Theorem $30$ for complete intersections. - -REPLY [7 votes]: This always holds for $\dim(X)\geq 3$. The point is that in this case the Picard group of $X$ is cyclic, generated by the line bundle $\mathscr{O}_X(1):= \mathscr{O}_{\mathbb{P}^n}(1)_{|X}$; this is SGA 2, Exp. 12, Cor. 3.7. Therefore any isomorphism $f:X\rightarrow Y$ induces an isomorphism $f^*\mathscr{O}_Y(1)\cong \mathscr{O}_X(1)$. -Since the embedding in $\mathbb{P}^n$ is given by the global sections of this line bundle, this implies that $f$ is the restriction of an automorphism of $\mathbb{P}^n$.<|endoftext|> -TITLE: almost complex $\mathbb{Z}^{6}$-action -QUESTION [5 upvotes]: Suppose we take an almost complex structure on $\mathbb{T}^{6}$ with $c_{1} \neq 0$ (there should be infinitely many homotopy classes satisfying this requirement). Now pull it back to the universal cover $\mathbb{R}^6$, giving an almost complex structure $J$, it should be invariant by the deck group action of $\mathbb{Z}^{6}$. -My question is there a case where we can explicitly describe $J$ and the group action of $\mathbb{Z}^6$ in coordinates? -(I asked this question in stack exchange in August https://math.stackexchange.com/questions/3789163/almost-complex-actions-on-mathbbr6 with no answer so I thought I would ask here.) - -REPLY [4 votes]: Here is a construction of a family of such examples that will work, but you will have to choose a particular map to get an explicit example. -Let's use coordinates $v_1,v_2,v_3,v_4,x,y$ (each periodic of period $2\pi$). Choose a smooth map $u=u(x,y):\mathbb{T}^2\to S^2$ that has nonzero degree. (This will ensure that it is not null-homotopic.) One possible example would be to make it a $2$-to-$1$ branched conformal double cover (i.e., it will have 4 branch points), the classical way to exhibit an elliptic curve as a branched double cover of $S^2=\mathbb{CP}^1$. -Write $u = (u_1,u_2,u_3)$ and consider the following $2$-form on $\mathbb{T}^6$ -$$ -\begin{aligned} -\omega &= u_1\,(\mathrm{d}v_1\wedge\mathrm{d}v_2+\mathrm{d}v_3\wedge\mathrm{d}v_4) - +u_2\,(\mathrm{d}v_2\wedge\mathrm{d}v_3+\mathrm{d}v_1\wedge\mathrm{d}v_4)\\ -&\qquad +u_3\,(\mathrm{d}v_3\wedge\mathrm{d}v_1+\mathrm{d}v_2\wedge\mathrm{d}v_4) - + \mathrm{d}x\wedge\mathrm{d}y -\end{aligned} -$$ -This $2$-form is compatible with the flat metric $g = {\mathrm{d}v_1}^2 + {\mathrm{d}v_2}^2 + {\mathrm{d}v_3}^2 + {\mathrm{d}v_4}^2 + {\mathrm{d}x}^2 + {\mathrm{d}y}^2$ and hence defines a $g$-orthogonal almost complex structure on $\mathbb{T}^6$ by the usual rule $\omega(w_1,w_2) = g(Jw_1,w_2)$ for all tangent vectors $w_1,w_2$. -I think this $J$ is probably not integrable for any nonconstant $u$ (and hence for $u:\mathbb{T}^2\to S^2$ of nonzero degree), but it will certainly be nonintegrable for `generic' $u$. The nonzero degree requirement on $u$ implies that $c_1\not=0$ for this almost complex structure.<|endoftext|> -TITLE: Inverting lower triangular matrix in time $n^2$ -QUESTION [32 upvotes]: I have a lower $n\times n$ triangular matrix called $A$ and I want to get $A^{-1}$ solved in $O(n^2)$. How can I do it? -I tried using a method called "forward substitution", but the inversion is solved in $O(n^3)$ for full $n\times n$ matrix. - -REPLY [89 votes]: No such method is known at present. -If one could invert lower triangular $n \times n$ matrices in time $O(n^2)$ -then one could multiply $N \times N$ matrices in time $O(N^2)$. -Indeed let $n=3N$ and apply the putative inversion algorithm to -the block matrix -$$ -\left( - \begin{array}{ccc} I & 0 & 0 \cr B & I & 0 \cr 0 & A & I \end{array} -\right) -$$ -for any $N\times N$ matrices $A,B$: the inverse is -$$ -\left( - \begin{array}{rrr} I & 0 & 0 \cr -B & I & 0 \cr AB & \!\!\! -A & I \end{array} -\right) \, , -$$ -so you could read $AB$ off the bottom left block. -It is still an open problem whether general matrix multiplication -can be done in time $O(N^2)$, or even $O(N^{2+o(1)})$. In particular -it follows that no method is known to do what you are asking. -In fact it is known that conversely an algorithm that takes -$O(N^2)$ or $O(N^{2+o(1)})$ time to multiply $N \times N$ matrices -would let us also invert $n \times n$ matrices in time -$O(n^2)$ or $O(n^{2+o(1)})$ respectively -(with a different $O$-constant, and not limited to triangular matrices). -So your question is in fact equivalent to the open question about -fast matrix multiplication. See for instance page 3 of -these -lecture notes by Garth Isaak, which also shows the block-diagonal trick -(in the upper- instead of lower-triangular setting). -POSTSCRIPT Strictly speaking, the reduction from -$O(N^c)$ matrix multiplication to $O(n^c)$ inversion of -triangular matrices means only that either we don't know how to attain -$c=2$ or $c=2+o(1)$ in the latter problem, or such an algorithm is known -but somehow nobody has noticed that this solves the former problem. -But the second possibility seem most unlikely, because fast matrix -multiplication is such a celebrated problem, and its reduction to -triangular-matrix inversion is quite well known.<|endoftext|> -TITLE: Is there a good algebraic model of random n-hypergraphs? -QUESTION [9 upvotes]: Suppose $F$ is a finite field and $-1$ is a square in $F$. Let $E$ be the binary relation on $F$ where $(a,b) \in E$ iff $a - b$ is a square. Then $(F,E)$ is called a Paley graph. Paley graphs are well-known algebraic models of random graphs (indeed Paley graphs are ``quasi-random" in a precise sense). -I am curious to know if there is also a "finite field model" for random $n$-hypergraphs. It would be natural to ask for a quasi-random algebraic family of $n$-hypergraphs, but apparently there are multiple notions of quasi-randomness for hypergraphs, and I don't know anything about them, so I don't want to be more precise. - -REPLY [2 votes]: For $G$ a quasirandom hypergraph and $k \geq 3$, let $H$ be a hypergraph whose vertices are that of $G$ and $(x_1, ..., x_k)$ is a hyperedge iff $(x_1, ..., x_k)$ is a clique in $G$. Then there is a one-to-one correspondence between the hyperedges of $H$ and the $k$-cliques of $G$ and between the copies of $M_k$ in $H$ and the copies of $L(Q_k)$ in $G$, where $M_k$, a $k$-hypergraph with $k{2^{k-1}}$ vertices and $2^k$ edges, is defined on p.4 of the paper "Weak quasi-randomness for uniform hypergraphs" and $L(Q_k)$ is the line graph of the hypercube graph. -The $\text{MIN}_d$ condition, a quasi-random condition defined on p.5 of "Weak quasi-randomness for uniform hypergraphs", states that a sequence of hypergraphs is quasirandom if the density of a hyperedge (i.e. probability of finding a hyperedge by choosing $k$ vertices) tends to a fixed $d$ and the density of $M_k$ tends to $d^{2^k}$. -The density of hyperedges on $H$ tends to $1/2^{k \choose 2}$ and the density of $M_k$ tends to $1/2^{{k \choose 2}2^k}$, by counting $k$-cliques and $L(Q_k)$ in the quasirandom graph $G$. Thus $H$ is quasidrandom if $G$ is. -So we can define such a class of $H$ by taking $G$ to be Paley graphs.<|endoftext|> -TITLE: Getting a model of $\mathsf{ZFC}$ that fails to nicely cover an inner model -QUESTION [12 upvotes]: Consider the following statement: - -$(\dagger)$ $\ $ There is an inner model $M$ such that $M \models \mathsf{GCH}+\square$ and for every countable $X \subseteq \mathrm{Ord}$, there is a countable $Y \in M$ such that $X \subseteq Y$. - -When I say $M$ is an ``inner model'' I mean that $M$ is a class (definable with parameters) such that $M \supseteq \mathrm{Ord}$ and $\langle M,\in \rangle \models \mathsf{ZFC}$. -Question: Does the negation of $(\dagger)$ have any large cardinal strength? That is, does the consistency of $\mathsf{ZFC}+\neg(\dagger)$ imply the consistency of large cardinals? And if so, what kind of large cardinals are required for $\neg(\dagger)$? -Here are a few observations: -$\bullet \ $ The statement of $(\dagger)$ seems close in spirit to the statement of Jensen's Covering Lemma. If we were to replace "countable" with "of size $\kappa$" for any uncountable $\kappa$, then this modified version of $(\dagger)$ would follow from the Covering Lemma by taking $M = \mathrm{L}$, and therefore its negation would imply the existence of $0^\sharp$. -$\bullet \ $ However, the Covering Lemma does not imply $(\dagger)$. Furthermore, if we modify $(\dagger)$ by insisting on $M = \mathrm{L}$, then we don't need the failure of the Covering Lemma, or any large cardinal strength at all, to get this modified version of $(\dagger)$ to fail. This is because of Namba forcing. If we start with $\mathrm{L}$ and add a Namba-generic filter $G$, then $\mathrm{L}[G]$ will fail to satisfy ``$(\dagger)$ with $M = \mathrm{L}$.'' However, (I'm fairly certain that) $\mathrm{L}[G]$ is itself still a model of $\mathsf{GCH}+\square$, which means that $\mathrm{L}[G]$ satisfies $(\dagger)$, simply by taking $M = \mathrm{L}[G]$. Therefore Namba forcing does not make $(\dagger)$ fail. -$\bullet \ $ The Singular Cardinals Hypothesis follows from $(\dagger)$. Hence one can force a failure of $(\dagger)$ by forcing $\neg\mathsf{SCH}$, which requires a measurable cardinal $\kappa$ of Mitchell order $\kappa^{++}$. So I know that $\neg(\dagger)$ is consistent relative to large cardinals -- I just don't know whether any large cardinals are actually necessary. -$\bullet \ $ If $(\dagger)$ holds in some ground model $V$, then it continues to hold in any $\omega$-distributive forcing extension of $V$. If $(\dagger)$ and Jensen's Covering Lemma both hold in $V$, then both continue to hold in any cardinal-preserving forcing extension of $V$. -$\bullet \ $ I suppose $(\dagger)$ isn't expressible as a first-order statement in the language of set theory, but it is expressible as a scheme in the metatheory. -My motivation for asking this question is that I've proved a topological theorem using $(\dagger)$ as a hypothesis. I'd like to know my hypothesis can be negated without assuming something with large cardinal strength, like the failure of $\mathsf{SCH}$. - -REPLY [10 votes]: The consistency strength of the failure of $(\dagger)$ is an inaccessible cardinal. -Building on the comment of Mohammad, if $\omega_2^V$ is a successor cardinal in $L$ then there is a set $X \subseteq \aleph_1^V$ such that $L[X]$ computes $\aleph_1, \aleph_2$ correctly, which (assuming $0^\#$ does not exist) is enough, since this model would satisfy $\mathrm{GCH}$. -On the other direction, in the paper "Inner Models from Extended Logics" by Kennedy, Magidor and Vaananen, Theorem 6.6 they show that starting with $V=L$ and an inaccessible cardinal $\kappa$, there is a (modification) of revised-countable-support iteration of a variant of the Namba forcing, such that $V[G] \models \kappa = \aleph_2 = 2^{\aleph_0}$ and $V[G] = (C^*)^{V[G]}$, where the model $C^*$ is $L[A]$, for $A$ the class of all ordinals of countable cofinality. -Since every model witnessing $(\dagger)$ would be able to compute the class $A$ and thus would contain $L[A]$, this model would witness the failure of $(\dagger)$.<|endoftext|> -TITLE: Are there numbers whose binary and ternary representations simultaneously have few digit transitions? How frequent are those numbers? -QUESTION [10 upvotes]: For a natural number $n$, let $c_b(n)$ denote the number of digit transitions in the representation of $n$ in base $b$. By a digit transition, we mean a pair of successive, unequal digits: for instance, the decimal number 114633366 has 4 transitions, given by 14, 46, 63 and 36, hence $c_{10}(114633366) = 4$. -Are there any known results surrounding the quantity $c_2(n) + c_3(n)$ (or, more generally, sums over different bases)? More specifically, I am trying to prove or disprove the following claim: -“For any $C > 0$, there exists a natural number $n$ such that, for any natural $k$ with $n \leq k \leq 2n$, $c_2(k) + c_3(k) > C$.” -In other words, I want to know whether it is possible to always find a number $n$ such that no number between $n$ and $2n$ has binary and ternary representations that are simultaneously nice (in the sense that they have, between them, fewer than $C$ digit transitions). -As a remark, I believe the numbers with few digit transitions in base $b$ can equivalently be described as the numbers expressible as $\sum e_ib^i$ where $i \geq 0$, $e_i \in \{-1, 0, 1\}$, and only few of the $e_i$ are non-zero. -Edit: As Steven Stadnicki pointed out below, this is not true for general $b$. A similar characterisation probably exists in general and is not too difficult, but I don't know if it helps. -As another remark, for a fixed number $k$ of transitions, when varying the number of digits $n$, the number of numbers with $k$ or fewer transitions is polynomial in $n$ (choose the places where the transitions happen, and the digits). The number of numbers, however, is exponential in $n$, so heuristically, one would expect the sets of numbers with few transitions in base 2 and base 3 respectively to have less and less opportunity to intersect as $n$ gets large; however, I am lacking intuition in the area, so I don't know if there is a way to make this rigorous. -My main field is not number theory, so I would be grateful for any advice of where to look for relevant results, no matter how obvious you think it is! -Update: I have checked the minimal value of $c_2(n) + c_3(n)$ between successive powers of 2. Between $2^i$ and $2^{i + 1}$, it is below 6 for $i \leq 22$. It then stabilises at 7 until $i = 30$ inclusive (except for $i = 25$ where it drops back to 6). It jumps to 9 for $i = 31$. - -REPLY [14 votes]: [Edited because I had misread $c_b(k)$ to be the number of non-zero digits in the base $b$ representation, rather than the number of digit transitions. The argument works for both variants. -T] -The claim is true, and one can in fact argue by purely Archimedean methods (with the only number theoretic input being the irrationality of $\log 2 / \log 3$). -If the claim were false, then there exists $C$ such that for every power of two $n = 2^j$, there exists $2^j \leq k \leq 2^{j+1}$ such that $c_2(k)+c_3(k) \leq C$. In particular, for every $j$ there exists a solution to the exponential Diophantine equation -$$ e_{j,1} (2^{j - a_{j,1}+1}-1) + \dots + e_{j,l} (2^{j - a_{j,l}+1}-1) $$ -$$= f_{j,1} (3^{j^* - b_{j,1}+1}-1)/2 + \dots + f_{j,m} (3^{j_* - b_{j,m}+1}-1)/2 \quad (1)$$ -with $j^* := \lfloor \frac{\log 2}{\log 3} j \rfloor$, $l+m \leq C$, -$$ e_{j,1},\dots,e_{j,l}, f_{j,1},\dots, f_{j,m} = \{-2,-1,0,1,2\},$$ -and -$$ O(1) = a_{j,1} < \dots < a_{j,l}$$ -and -$$ O(1) = b_{j,1} < \dots < b_{j,m}$$ -integers. (There are some additional constraints like $a_{j,l} \leq j, b_{j,m} \leq j_*$ that we shall simply discard. The $O(1)$ constraints are extremely explicit, for instance it would be safe to take $a_{j,1}, b_{j,1} \in \{-3,-2,-1,0,1,2,3\}$.) -We will show instead that the set of natural numbers $j$ for which there is a solution to (1) obeying the stated constraints in fact has natural density zero, giving the required contradiction (this is a version of the probabilistic method). -We perform the following induction. For any $l',m' \geq 1$, let $P(l',m')$ denote the claim that for any integers $a_1,\dots,a_{l'},b_1,\dots,b_{m'}$, the set of $j$ for which there is a solution to (1) obeying the stated constraints together with the additional constraints -$$ l \geq l'; m \geq m'$$ -$$ a_{j,i} = a_i \hbox{ for } i=1,\dots,l'$$ -$$ b_{j,i} = b_i \hbox{ for } i=1,\dots,m'$$ -has natural density zero. If we can prove $P(1,1)$ then we are done by the union bound (since there are only boundedly many choices for $a_{j,1}$ and $b_{j,1}$). On the other hand $P(l',m')$ is vacuously true when $l'+m' > C$. So it will suffice to show that $P(l',m')$ is true whenever $P(l'+1,m'), P(l',m'+1)$ are both true. -Fix $l',m',a_1,\dots,a_{l'},b_1,\dots,b_{m'}$, and let $K$ be a large number to be chosen later. By the induction hypothesis and the union bound, the set of $j$ for which one can find a solution to (1) with the indicated constraints will have density zero if we impose either the additional constraint -$$ a_{j,l'+1} \leq K$$ -or -$$ b_{j,m'+1} \leq K$$ -(adopting the convention that $a_{j,i}=\infty$ if $i>l$, and $b_{j,i}=\infty$ if $i>m$). Thus we may restrict attention to those solutions which instead obey the constraints -$$ a_{j,l'+1}, a_{j,m'+1} > K.$$ -The equation (1) and the given constraints (and the convergence of the geometric series $\sum_{n=0}^\infty 2^{-n}, \sum_{n=0}^\infty 3^{-n}$) then implies (for large $j$) that -$$ (1 + O(2^{-K})) (e_{j,1} 2^{j - a_{1}+1} + \dots + e_{j,l'} 2^{j - a_{l'}+1})$$ -$$= (1 + O(3^{-K})) (f_{j,1} 3^{j^* - b_{1}+1} + \dots + f_{j,m'} 3^{j_* - b_{m'}+1})/2$$ -which on taking logarithms and rearranging implies that -$$ \frac{\log 2}{\log 3} j = j_* + \frac{1}{\log 3} \log \frac{(f_{j,1} 3^{-b_{1}+1} + \dots + f_{j,m'} 3^{-b_{m'}+1})/2}{e_{j,1} 2^{-a_{1}+1}+\dots+e_{j,l'} 2^{-a_{l'}}} + O( 2^{-K} ),$$ -or on taking modulo $1$ to eliminate $j_*$ -$$ \frac{\log 2}{\log 3} j = \frac{1}{\log 3} \log \frac{(f_{j,1} 3^{-b_{1}+1} + \dots + f_{j,m'} 3^{-b_{m'}+1})/2}{e_{j,1} 2^{-a_{1}+1}+\dots+e_{j,l'} 2^{-a_{l'}+1}} + O( 2^{-K} ) \hbox{ mod } 1.$$ -Thus $\frac{\log 2}{\log 3} j \hbox{ mod } 1$ is constrained to the union of $O(O(1)^{m+m'})$ intervals of length $2^{-K}$. But $\frac{\log 2}{\log 3}$ is irrational, so $\frac{\log 2}{\log 3} j \hbox{ mod } 1$ is equidistributed. Thus the set of solutions to this system of constraints has natural upper density at most $O(O(1)^{m+m'} 2^{-K})$. Since $K$ can be taken arbitrary large, we obtain zero density for the original problem as required. -If one used some quantitative result on the irrationality of $\log 2 / \log 3$, such as that provided by Baker's theorem, one would presumably be able to get some quantitative control on how $C$ must necessarily grow with $n$, but I haven't tried to compute this exactly.<|endoftext|> -TITLE: What is the correct notion of morphism between statistical manifolds? -QUESTION [6 upvotes]: Given two statistical manifolds, is there a notion of "isomorphic"? What are morphisms? - -REPLY [5 votes]: Any smooth (resp. $C^1$) statistical manifold can be embedded into the space of probability measures on a finite set. -A probability measure can be viewed as a weakly averaging affine measurable functional taking values in the unit interval which preserves limits, so statistical manifolds can be viewed categorically as the same thing after their embedding into probability measures on finite sets. -From the abstract of the second linked paper: - -The probability measures on a space are the elements of a submonad of a double dualization monad on the category of measurable spaces into the unit interval, and this monad is naturally isomorphic to the Giry monad. We show this submonad is the codensity monad of a functor from the category of convex spaces to the category of measurable spaces. - -You could thusly take a morphism between statistical manifolds to be a morphism of monads between the appropriate associated monads in the bicategory of monads which commutes with the embeddings, and define an iso in the obvious manner. -Link 1: Lê, H.V. Statistical manifolds are statistical models. J. geom. 84, 83–93 (2006). -Link 2: Sturtz, Kirk, Categorical Probability Theory, arXiv:1406.6030 [math.CT] (2015)<|endoftext|> -TITLE: Motivations for the term "jet" in the context of viscosity solutions for fully nonlinear PDE -QUESTION [5 upvotes]: My question is very direct: - -What are the motivations for the name "jet"(subjet, superjet) in the context of viscosity solutions for second order fully nonlinear elliptic PDE? - -The definition of which can be seen in Crandall, Ishii, Lions: -Crandall, Michael G.; Ishii, Hitoshi; Lions, Pierre-Louis, User’s guide to viscosity solutions of second order partial differential equations, Bull. Am. Math. Soc., New Ser. 27, No. 1, 1-67 (1992). ZBL0755.35015. -see also https://arxiv.org/pdf/math/9207212.pdf. - -REPLY [3 votes]: The machinery of Jets were introduced by Ehresmann as an geometrically invariant way of describing PDEs and their solutions in analogy to the way that vector fields and integral curves invariantly describe ODEs and their solutions. -For example, there is a way of geometrically describing symmetries here and so Noether's theorem. -It's worth noting that sprays describe 2nd order ODEs and are a sort of midway house. It comes from iterating tangent bundles - but this doesn't appear to generalise well - at least I haven't seen any generalisations of such, except in passing. Whereas the notion of a jet generalises the equivalence class of curves of a certain tangent order.<|endoftext|> -TITLE: Is the hereditary version of this weak finiteness notion nontrivial? -QUESTION [7 upvotes]: Say that a set $X$ is $\Pi^1_1$-pseudofinite if every first-order sentence $\varphi$ with a model with underlying set $X$ has a finite model. The existence of infinite $\Pi^1_1$-pseudofinite sets is consistent with $\mathsf{ZF}$, since indeed every amorphous set is $\Pi^1_1$-pseudofinite. -Perhaps surprisingly, it is not immediately clear whether the class of $\Pi^1_1$-pseudofinite sets need be closed under finite unions. My question is what happens when we fix this weakness by brute force: - -Is it consistent with $\mathsf{ZF}$ that there is an infinite hereditarily $\Pi^1_1$-pseudofinite set - that is, an infinite set $X$ such that whenever $Y$ is $\Pi^1_1$-pseudofinite, $X\cup Y$ is also $\Pi^1_1$-pseudofinite? - -Note that the hereditarily $\Pi^1_1$-pseudofinite sets are closed under finite unions, so this actually does "fix" the situation above. One natural hope is that amorphous sets do the trick again, but I don't see how - certainly the argument linked above doesn't suffice. (Of course there are finiteness notions stricter than amorphousness - e.g. "in any partition into infinitely many pieces, all but finitely many of those pieces are singletons" - but to my knowledge they're all significantly more finicky to work with, so it would be very nice if we didn't have to go there.) - -REPLY [2 votes]: It is consistent that there are infinite hereditarily $\Pi_1^1$-pseudofinite sets. I'll just say "pseudofinite" instead of "$\Pi_1^1$-pseudofinite" for the rest of this post. - -Theorem 1. Let $N$ be a model of ZF-Foundation satisfying "pseudofinite violations of choice for pseudofinite sets": there is a pseudofinite set $A$ such that for all pseudofinite $X,$ there is an ordinal $\alpha$ and a surjection $A^{<\omega}\times\alpha\to X.$ -In $N,$ the class of pseudofinite sets is closed under finite unions. - -The reason I've stated it like this is because not only does the hypothesis hold in permutation models where the set of atoms is pseudofinite but, if I've stated it correctly now, it should also hold in typical ZF embeddings of these permutation models. In particular these hypotheses hold in the basic Fraenkel model, with $A$ being the set of atoms, and in the ZF model you get by applying Jech-Sochor to the basic Fraenkel model. -The theorem will follow from two easy observations. (Proof is at the bottom.) -Lemma 2. If $X$ is pseudofinite, $n\in\omega,$ and $f:X^n\to Y$ is a function, then $f[X]$ is pseudofinite. -This is because $f$ defines an interpretation of $f[X]$ in $X,$ and ZF lets us define the preimages of any relation on $Y.$ -Corollary 3. If $X$ is pseudofinite then there is no surjection from $X$ to an infinite linearly ordered set. -Corollary 3 is what François G. Dorais' answer mentioned; Tarski's "II-finite". It follows from Lemma 2 with $n=1$ because if a linear order has an $n$'th element for each $n,$ then it contains a copy of $\omega,$ which is not pseudofinite. And if it doesn't have an $n$'th element, it satisfies the statement "has $\geq n$ elements but no $n$'th element" which also contradicts pseudofiniteness. -In terms of order properties, I think every pseudofinite set $X$ is "P-finite" in the terminology of [2]: every partial order on $X$ has a maximal element. This is the strongest notion of finiteness considered there for non-amorphous sets, and you know amorphous is strictly stronger than pseudofinite. But order properties might be a red herring. If you take a permutation model with a homogeneous vector space $V$ over $\mathbb F_2$ with a generic symmetric bilinear form then $V$ won't be pseudofinite; finite models satisfy $(\exists z)(\forall y) B(y,y)=B(y,z),$ but $V$ won't. A concrete model is $V=\mathbb F_2^{\oplus\omega}$ with $B(x,y)=\sum_i x_iy_i.$ There doesn't seem to be any ordering involved as far as I can see. And if you instead take a symplectic bilinear form $B$ - so $B$ is non-degenerate and satisfies $B(v,v)=0$ - then I think $V$ will be pseudofinite. Specifically, Duplicator can win an $n$-step Ehrenfeucht–Fraïssé game for $V$ and a $2n$-dimensional symplectic space, by ensuring the chosen elements span isomorphic subspaces at each step. - -ZF models -Because of Corollary 3, the statement "if $x$ and $y$ are pseudofinite then so is $x\cup y$" is surjectively boundable and hence injectively boundable in the sense of [1]. So it transfers to ZF. -Ideally I'd want to rule out all models I can think of, not just Pincus'. Here is my sketchy understanding of how this should work for the Jech-Sodor embedding theorem. In these models, sets have a support of the form $(e,E)$ where $e$ is a set of indices for generic subsets of a regular cardinal $\kappa,$ and $E$ is a set of atoms of a permutation model. Not all permutation models have minimal supports, e.g. take an infinite homogenous vector space over $\mathbb F_3.$ But there is always a minimal set of reals $e.$ That part is where the work is; the section in Jech's "Axiom of Choice" on the embedding theorems has very similar statements though I don't see anything directly quotable. This gives a map $s:X\to \mathcal P^{<\omega}(\kappa)$ taking each $x\in X$ to the minimal set $e$ such that there is a pair $(e,E)$ that is a support for $x.$ The formal statement would be $p\Vdash \dot{r}\in \dot{s}(\dot{x})$ if there is a name $\dot{z}$ fixed by symmetries of the symmetric extension that fix $(e,E),$ satisfying $p\Vdash \dot{z}=\dot{x},$ and $\dot{r}$ is the canonical name for the set indexed by an element of $e,$ and we can't get a smaller set $e$ by refining $p$ and choosing a different $\dot{z}.$ Since the image of $s$ is linearly ordered, it has to be finite when $X$ is pseudofinite. Finite sets can be absorbed into "$\alpha$" so the only parameters that matter are the atoms. - -Proof of Theorem 1. -The following are equivalent for non-empty sets $X\in N$: - -$X$ is pseudofinite -There is a surjection $A^{p_1}\coprod \dots\coprod A^{p_k}\to X$ for some $k,p_1,\dots,p_k\in\omega.$ -There is a surjection $A^n\to X$ for some $n$. - -1⇒2: By our small violations of choice axiom, there is a surjection $f:A^{<\omega}\times \alpha\to X.$ Define $g:X\to \omega\times \alpha$ by $g(x)=\min\{(|w|,\beta):g(w,\beta)=x\},$ using the ordinal product ordering. By Corollary 3, the image of $g$ is a finite set $I\subset\omega\times\alpha.$ After some reindexing, this is of the required form: the surjection is defined on $\coprod_{i\in I} A^{i_1}$ by sending $w\in A^{i_1}$ (in the $i$'th component of the disjoint union) to $f(w,i_2).$ -2⇒3: set $n=2k+\max p_i$ and encode the component index $1\leq i\leq k$ using the equality relation on the first $2k$ variables -3⇒1: This is Lemma 2. -The class of sets defined by condition 2 is obviously closed under finite unions, as required. $\square$ - -[1]: David Pincus, Zermelo-Fraenkel Consistency Results by Fraenkel-Mostowski Methods, The Journal of Symbolic Logic, Vol. 37, No. 4 (Dec., 1972), pp. 721-743 -[2]: Omar De la Cruz, Damir D. Dzhafarov, and Eric J. Hall, Definitions of finiteness based on order properties, Fundamenta Mathematicae (2006), Volume: 189, Issue: 2, page 155-172. https://eudml.org/doc/282771<|endoftext|> -TITLE: Euler characteristic of the simplicial complex of sets of elements in a semilattice with non-zero meet -QUESTION [8 upvotes]: In a combinatorial computation, I came across the following quantity: -Consider a finite meet semilattice $L$, that is, a finite poset which is closed under $\min$. Denote the least element of $L$ by $0$. -Now, define $Z := \{ S \subset L : \min S = 0 \}$. I want to compute the quantity -$$ \chi := \sum_{S \in Z} (-1)^{|S|+1}. $$ -I noticed that the complement $Z^c$ of $Z$ in $2^{L}$ (the collection of subsets of $L$ whose meet is not $0$) is a simplicial complex on $L$. Moreover, since the Euler characteristic of $2^L$ is $0$, the above quantity $\chi$ is actually just equal to the Euler characteristic of $Z^c$: -$$ \chi = \sum_{S \in Z^c} (-1)^{|S|}. $$ -My question is: - -Is there any way to efficiently compute $\chi$ or bounds on it/approximations to it using some properties of $L$? (I already know that in many of the cases I'm interested in $\chi \ne 0$). -Since (1) seems likely quite difficult to answer, especially in this general context, are there any references that can tell me more about the relationship between $L$ and $Z$ or $Z^c$, $H^{\cdot}(Z^c), \chi$, etc? These seem like natural enough objects to study, but I wasn't able to find a reference on them. When I look up simplicial complexes associated to a semilattice, mostly I find references to the order complex, which seems very different. I could find something called "the ideal zero divisor complex" for rings which seems similar, but I am working with a semilattice, not a ring, and elements, not ideals. I could also find information about zero divisor graphs of semigroups, which is related, but still rather far from the idea I'm looking for. - -(And of course, I am actually only concerned about a very particular family of semilattices $L$, which happen to consist of certain partitions of {1,...,N}; the meet here is given by common refinement. I don't want to say too much about them here to avoid making the question too specific, but suffice to say I don't necessarily need a completely general answer to this question.) - -REPLY [4 votes]: This is a special case of the crosscut theorem. See e.g. Corollary 3.9.4 of Enumerative Combinatorics, vol. 1, second ed. Let $L'$ be $L$ with a top element $\hat{1}$ adjoined. In Corollary 3.9.4 take $X$ to be all elements of $L'$ not equal to $1$. We get $\chi=-\mu_{L'}(0,1)$, where $\mu_{L'}$ is the Möbius function of $L'$. There is lots of information about computing Möbius functions in the above reference.<|endoftext|> -TITLE: Example of an uncountable sequence of abelian groups with nonvanishing $\varprojlim^2$? -QUESTION [8 upvotes]: $\DeclareMathOperator{\op}{\mathrm{op}}\DeclareMathOperator{\Ab}{\mathsf{Ab}}\DeclareMathOperator{\Vect}{\mathsf{Vect}}$Question 1: What is an example of a sequence $(X_\alpha)_{\alpha<\kappa}$ of abelian groups such that $\varprojlim^2_{\alpha < \kappa} X_\alpha \neq 0$? -Here $\varprojlim^2_{\alpha<\kappa}$ is the second derived functor of the limit functor. Necessarily $\kappa$ will be uncountable, and of uncountable cofinality. I suspect it should be possible to give an example where the $X_\alpha$ are vector spaces and $\kappa = \omega_1$ is the first uncountable ordinal. - -This question likely sounds funny -- usually one only discusses $\varprojlim^n_{\alpha<\kappa}$ for $n\geq 2$ in more exotic abelian categories than $\Ab$ or $\Vect$. This is because one usually only deals with the case where $\kappa = \omega$ or at least has cofinality $\omega$, in which case the functor $\varprojlim_{\alpha < \kappa}^n : \Ab^{\kappa^{\op}} \to \Ab$, i.e. the $n$th derived functor of the limit functor, vanishes for $n \geq 2$. The usual proof uses a very natural 2-step resolution, which only works when $\kappa = \omega$, or by extension when $\kappa$ is of countable cofinality. -But when it comes to longer sequences, this resolution is not available. In fact, I've only seen $\varprojlim_{\alpha<\kappa}^n$ discussed for $\kappa$ of uncountable cofinality in Neeman's Triangulated Categories, appendix A, which contains methods of constructing resolutions in $\Ab^{\kappa^{\op}}$, but the resolutions are not of finite length. - -Another way of saying that $\varprojlim^2_{n<\omega} = 0$ in abelian groups is that $\varprojlim^1_{n<\omega}$ is right exact. So a closely related question is: -Question 2: What is an example of an epimorphism $(X_\alpha \to Y_\alpha)_{\alpha<\kappa}$ of inverse systems of abelian groups such that the induced map $\varprojlim^1_{\alpha<\kappa} X_\alpha \to \varprojlim^1_{\alpha<\kappa} Y_\alpha$ is not an epimorphism? -And by the way, -Question 3: What is the global dimension of the category $\Ab^{\kappa^{\op}}$ of $\kappa$-indexed inverse systems of abelian groups, for a given regular cardinal $\kappa$? How about $\Vect^{\kappa^{\op}}$, where $\Vect$ is the category of vector spaces over your favorite field? - -REPLY [10 votes]: A great survey on this and some related topics is Osofsky's "The subscript of $\aleph_n$, projective dimension, and the vanishing of $\varprojlim^{(n)}$." As far as I am aware, this 1974 paper still describes the state of the art on the matter. -Osofsky (exposing material from Mitchell's book on rings with many objects) gives the following example of a sequence such as you want. Fix any ring $R$ and let $\Delta R$ be the constant functor at $R$ indexed by $\aleph^{\mathrm{op}}_1$. Then $\Delta R$ can be shown to have homological dimension $2$ via a technical argument involving its nice ordered basis (see section 5 of Osofsky.) Now considering any projective resolution -$$\cdots\to P_2\to P_1\to P_0\to \Delta R\to 0,$$ -if $K$ is the kernel of $P_1\to P_0$ then the length-2 extension -$$0\to K \to P_1\to P_0\to \Delta R\to 0$$ -must be nontrivial, or else the kernel $K'$ of $P_0\to \Delta R$ would be a summand of $P_1$, making $\Delta R$ of projective dimension at most $1$. -That is, $\mathrm{Ext}^2(\Delta R, K)=\varprojlim^{(2)}_{\aleph_1^{\mathrm{op}}} K$ is nontrivial whenever $K$ is the kernel of the map between the first two projectives resolving $\Delta R$, for any ring $R$. The same arguments go through when $\aleph_1$ is replaced by $\aleph_n$ to get nontrivial $n$th derived functors of lim. So the answer to your question 3 is "$n+1$, if $\kappa=\aleph_n$, and $\infty$, if it's larger." These holds, passing to cofinalities, even when $\kappa$ is replaced with an arbitrary directed set, and I think even an arbitrary filtered category.<|endoftext|> -TITLE: What's the deal with Möbius pseudorandomness? -QUESTION [6 upvotes]: As I do more number theory, and in particular analytic number theory, I keep hearing more about the Möbius function $\mu(n)$ and how it is supposedly "pseudorandom". The values of the Möbius function at $n$ are determined by a formula, so what does this mean? Also, why is it so important that the Möbius function behaves pseudorandomly? -Just to be entirely clear, I am defining $\mu(n)$ by -\begin{equation} -\mu(n)= -\begin{cases} -(-1)^{\omega(n)}& n \,\,\mathrm{squarefree}\\ -0&\mathrm{otherwise} -\end{cases} -\end{equation} -CLARIFICATION: This question was made with the intention that I would answer it myself. Any criticism, commentary, or discussion is appreciated. - -REPLY [2 votes]: Here are a few (maybe helpful?) tidbits. In papers of 1931 and 1964 in the Comptes Rendus, Denjoy surmised that RH is true if the Mobius function was random for then your condition at (2) would be met. Also: true randomness would probably mean that the sequence needs to pass any statistical test that would be passed with probability one if the sequence consisted of statistically mutually independent random variables. In this latter case the order term in your condition (2) could actually be replaced by the square root of (x log log x) since the Law of the Iterated Logarithm would apply to it. There is also a time series character to this issue and for this recent work by Matomaki, Radziwill, Tao, Sarnak and others is relevant. Much remains unknown.<|endoftext|> -TITLE: Is there the longest geodesic? -QUESTION [9 upvotes]: Given a closed 2-surface $M$ together with a Riemannian metric $g$. -We pick a free homotopy class $\gamma \in \pi_1(M)$ and consider the set $C(\gamma)$ of all closed geodesics homotopic to $\gamma$. -Of course, the set $C(\gamma)$ may be infinite. The length of a geodesic loop gives a function on $C(\gamma)$. My question is whether this function is bounded (in terms of $g$)? - -REPLY [6 votes]: No such bound $C(\gamma)$ exists, even when $S$ is the two-sphere and even assuming that all geodesics considered are simple. Here is the example (which generalises to surfaces with genus). -Suppose that $S$ is the two-sphere. Pick four open disks $(D_i)_{i = 1}^4$ whose closures are closed disks and which are pairwise disjoint. (For example, use small round disks with respect to the usual round metric on $S$.) Let $P = S - \bigcup_i D_i$; so $P$ is a "four-holed sphere". We equip $P$ with a hyperbolic metric $g_P$ where all boundary components are geodesic. We now claim the following: - -$(P, g_P)$ has infinitely many closed simple geodesics (produced, for example, by a "braiding" construction). -All but four of these (namely, the curves of $\partial P$) are disjoint from $\partial P$. -In any infinite collection of these geodesics, their lengths are unbounded. - -We now choose any riemannian metric $g_S$ on $S$ that extends $g_P$. Note that all of the previous geodesics (except perhaps the components of $\partial P$) remain geodesic with respect to the metric $g_S$. Since all of those geodesics are null-homotopic in $S$, we are done. -Morally: The riemannian surface $(S, g_S)$ has four "mushrooms" (the disks $D_i$). The geodesics used above wander around the surface avoiding the tops of the mushrooms.<|endoftext|> -TITLE: Compact object and compact generator in a category -QUESTION [12 upvotes]: I found two definitions of compact object. - -(Lurie, Jacob (2009), Higher topos theory, p.392) Let $\mathcal{C}$ be a category which admits filtered colimits. An object $C \in \mathcal{C}$ is said to be compact if the corepresentable functor -$$ -\operatorname{Hom}_{e}(C, \bullet) -$$ -commutes with filtered colimits. - - -(Abelian Categories, Daniel Murfet, Definition 18) Let $\mathcal{C}$ be a category and $A$ an object of $\mathcal{C}$. We say that $A$ is compact (or sometimes small) if whenever we have a morphism $u: A \longrightarrow \bigoplus_{i \in I} A_{i}$ from $A$ into a nonempty coproduct, there is a nonempty finite subset $J \subseteq I$ and a factorisation of $u$ of the following form -$$ -A \longrightarrow \bigoplus_{j \in J} A_{j} \longrightarrow \bigoplus_{i \in I} A_{i}. -$$ - -I don't know how to show that they are equivalent, could you please help me? -In addition, we have the definition of the generator of an abelian category. - -(GENERATORS VERSUS PROJECTIVE GENERATORS INABELIAN CATEGORIES, CHARLES PAQUETTE, p.1) Let $\mathcal{A}$ be an abelian category. An object $M$ of $\mathcal{A}$ is a generator of $\mathcal{A}$ if for any object $X$ of $\mathcal{A}$, we have an epimorphism $\bigoplus_{i\in I} M\to X$ where $I$ is some index set. - -So what should the compact generator be? Is it a generator such that there is a factorisation of the following form? -$$ -\bigoplus_{i\in I} M \to \bigoplus_{i\in J} M \to X. -$$ -(all arrows are reversed??) -Thank you very much! - -REPLY [12 votes]: Part of the tricky thing about this circle of ideas is that several definitions are not equivalent in full generality but become equivalent with extra hypotheses. For example, a basic result about compact objects is the following characterization of module categories, which among other things provides a characterization of Morita equivalences. - -Theorem (Gabriel): A cocomplete abelian category $C$ is equivalent to the category $\text{Mod}(R)$ of modules over a ring $R$ iff it admits a compact projective generator $P$ such that $\text{End}(P) \cong R$. - -Both "compact" and "generator" in the statement of this theorem are individually ambiguous. "Compact" could mean either Lurie-compact or Murfet-compact, and "generator" can have something like ~7 different meanings, maybe ~3 of which are in common-ish use (?); see Mike Shulman's Generators and colimit closures (which discusses 5 possible definitions) and my blog post Generators (which discusses 6 possible definitions, 4 of which overlap with Mike's) for a discussion. -The happy fact is that nevertheless, the meaning of "compact projective" and of "compact projective generator" in the statement of Gabriel's theorem is unambiguous: - -in a cocomplete abelian category, "compact projective," using either Lurie-compactness or Murfet-compactness, is equivalent to the condition that $\text{Hom}(P, -) : C \to \text{Ab}$ commutes with all (small) colimits (this condition is also known as being tiny; see my blog post Tiny objects for a discussion), and -for compact projective objects in a cocomplete abelian category, nearly all of the definitions of "generator" that I'm aware of collapse and become equivalent. I'll limit myself to naming two: the weakest is that every nonzero object admits a nonzero map from $P$ (which I call "weak generator"; I forget if this name is standard), and the strongest is that every object can be written as the coequalizer of a pair of maps between coproducts of copies of $P$ (which I call "presenting generator"; this is not standard. In an abelian category coequalizers can be replaced with cokernels but this definition generalizes nicely to algebraic categories such as groups and rings). - -There is the additional nuance that in a stable $\infty$-categorical setting like the one Lurie works in it seems that one can drop projectivity but I'm not sure what the precise statements are. E.g. I believe there's a stable $\infty$-categorical analogue of Gabriel's theorem characterizing module categories over $E_1$ ring spectra and I believe that analogue involves compact generators. -Anyway, for what it's worth I would advocate for Lurie-compactness as the "default" meaning of compactness. Murfet-compactness is quite specific to the abelian setting, but Lurie-compactness is nice in many settings; for example, in the category of models of a Lawvere theory (groups, rings, etc.) an object is Lurie-compact iff it's finitely presented. Already this implies the not-entirely-obvoius fact that for modules being finitely presented is Morita invariant.<|endoftext|> -TITLE: To describe an invariant trivector in dimension 8 geometrically -QUESTION [11 upvotes]: $\newcommand\Alt{\bigwedge\nolimits}$Let $G=\operatorname{SL}(2,\Bbb C)$, and let $R$ denote the natural 2-dimensional representation of $G$ in ${\Bbb C}^2$. -For an integer $p\ge 0$, write $R_p=S^p R$; then $R_1=R$ and $\dim R_p=p+1$. -Using Table 5 in the book of Onishchik and Vinberg, I computed that the representation -$$ R_2\otimes\Alt^2 R_4 $$ -contains the trivial representation with multiplicity one. -I used the table as a black box. - -Question. Let $V\subset R_2\otimes\Alt^2 R_4$ denote the corresponding one-dimensional subspace. -How can one describe $V$ as a subspace geometrically? - -Motivation: I want to consider a $\operatorname{PGL}(2,k)$-fixed trivector -$$v\in V\subset R_2\otimes\Alt^2 R_4\subset \Alt^3(R_2\oplus R_4)$$ -of the 8-dimensional vector space $W=R_2\oplus R_4$ -over a field $k$ of characteristic 0, -and then to twist all this using a Galois-cocycle of $\operatorname{PGL}(2,k)$. For this end I need a geometric description of $V$. -Feel free to add/edit tags! - -REPLY [12 votes]: Here's another very nice (but still algebraic) interpretation that explains some of the geometry: Recall that $\operatorname{SL}(2,\mathbb{C})$ has a $2$-to-$1$ representation into $\operatorname{SL}(3,\mathbb{C})$ so that the Lie algebra splits as -$$ -{\frak{sl}}(3,\mathbb{C}) = {\frak{sl}}(2,\mathbb{C})\oplus {\frak{m}} -$$ -where ${\frak{m}}$ is the ($5$-dimensional) orthogonal complement of ${\frak{sl}}(2,\mathbb{C})$ using the Killing form of ${\frak{sl}}(3,\mathbb{C})$. Note that ${\frak{m}}$ is an irreducible ${\frak{sl}}(2,\mathbb{C})$-module, and that every element $x\in {\frak{sl}}(3,\mathbb{C})$ can be written uniquely as $x = x_0 + x_1$ with $x_0\in {\frak{sl}}(2,\mathbb{C})$ and $x_1\in{\frak{m}}$. Note also that $[{\frak{m}},{\frak{m}}]= {\frak{sl}}(2,\mathbb{C})$. -This defines the desired pairing ${\frak{sl}}(2,\mathbb{C})\times \bigwedge\nolimits^2({\frak{m}})\to\mathbb{C}$: Send $(x_0,y_1,z_1)$ to $\operatorname{tr}(x_0[y_1,z_1])$. Of course, this makes the $\operatorname{SL}(2,\mathbb{C})$-invariance of the pairing obvious.<|endoftext|> -TITLE: Is every surjective holomorphic self-map on a compact complex manifold finite-to-one? -QUESTION [10 upvotes]: I have already asked this question on stack exchange, but I didn’t get any answer. -Let $X$ be a compact connected complex manifold. - -Let $f:X \to X$ be a surjective holomorphic map. Is it true that $f$ is a finite map (i.e., every point has finitely many preimages)? - -If $X$ is one dimensional then the answer to the above question is yes. If $X$ is a complex projective space then the answer is again yes. -If we don’t assume surjectivity then we can easily construct non-constant, non-finite self-maps (just consider the product manifold and $f$ to be projection on one of the coordinates). - -REPLY [12 votes]: Let me give a sketch of proof for Gromov's claim in the case where $X$ is Kähler. More precisely, let me prove the following ${}$ - -Proposition [G03, p.223]. Let $X$, $Y$ be two complex manifolds (not necessarily compact) of the same dimension and having the same even Betti numbers. If $X$ is Kähler, then every proper surjective holomorphic map $f \colon X \to Y$ is finite-to-one. - -Proof. By a result of Wells [W74, Theorem 3.1], the map $f \colon X \to Y$ induces an injection $$f^* \colon H^{r}(Y, \, \mathbb{C}) \to H^{r}(X, \, \mathbb{C})$$ for all $r$. Using the duality $$H_r(X, \, \mathbb{C})=\mathrm{Hom}(H^r(X, \, \mathbb{C}), \, \mathbb{C})$$ we infer that the induced map in homology $$f_* \colon H_{r}(X, \, \mathbb{C}) \to H_{r}(Y, \, \mathbb{C})$$ is surjective for all $r$. -If $f \colon X \to Y$ contracts a subvariety $Z$ of positive complex dimension $i$ to a point, then its fundamental class would give an element $[Z]$ in the kernel of $$f_* \colon H_{2i}(X, \, \mathbb{C}) \to H_{2i}(Y, \, \mathbb{C})$$ -Now $Z$ is compact by the properness assumption and so, pairing with the Kähler -form on $X$, we can deduce that $[Z]$ is non-zero. Hence $b_{2i}(X) > b_{2i}(Y)$, contradiction. $\square$ -Remark. The first version of this answer did not assume that $X$ was Kähler (in fact, Gromov does not make this assumption). However, if $X$ is not Kähler it may happen that the fundamental class of a compact subvariety of $X$ is homologically trivial, see the example of Hopf's surface given in Michael Albanese answer to [MSE]: in this case, $X$ contains a torus $C$ whose fundamental class $[C]$ is zero in $H_2(X, \, \mathbb{C})$, simply because the last group is trivial. So, if $X$ is not Kähler, the last part of the proof breaks down. -In fact, I do not know how to make the argument work without the Kähler assumption for $X$, so I asked a new MO question about this problem. -References. -[MSE] https://math.stackexchange.com/q/1556561/456212 -[G03] Gromov, M: On the entropy of holomorphic maps, Enseign. Math. II. Sér. 49, No. 3-4, 217-235 (2003). ZBL1080.37051. -[W74] Wells, R. O. jun: Comparison of de Rham and Dolbeault cohomology for proper surjective mappings, Pac. J. Math. 53, 281-300 (1974). ZBL0261.32005.<|endoftext|> -TITLE: Random walk on $\mathbb{Z}^2$ going forward with probability $p$ -QUESTION [5 upvotes]: Consider a random walk on $\mathbb{Z}^2$ which goes forward (i.e. takes a step in the same direction as the last step) with probability $p$ and turns right and left with probability $\frac{1-p}{2}$ respectively. Is it recurrent for all $1 > p \geq 0$? - -REPLY [3 votes]: It looks like recurrence follows from Theorem 1 in Bender and Richmond, Correlated Random Walks, Ann. Probab. 12(1) (1984): 274–278 DOI:10.1214/aop/1176993392. (It gets late, though, so I may be getting something wrong.) -If $X_n$ is the "random walk" described in the question and $D_n$ is the "direction" in which the random walk is moving in step $n$, then $(D_n, X_{n+1} - X_n)$ is the "correlated random walk" according to the terminology of Bender and Richmond. Thus, we have to verify three conditions: - -The "correlated random walk" is drift-free, that is, $\mathbb E|X_n| = o(n)$. This seems to follow from exponential convergence of the distribution of $D_n$ to the uniform distribution over four admissible directions. - -Condition A holds: for some $d$ and $d'$, the support of the conditional distribution $X_{n+1} - X_n$ given $D_n = d$ and $D_{n+1} = d'$ is a linearly dense set in $\mathbb R^2$. This deepends on the exact definition of $X_n$, but even if this condition fails, the "correlated random walk" $(D_{2n}, X_{2n+2} - X_{2n})$ satisfies this condition. - -Condition B holds: the Markov chain $D_n$ (or $D_{2n}$) is "strongly irreducible": this is clearly true.<|endoftext|> -TITLE: Inducing a model structure using a cosimplicial object -QUESTION [7 upvotes]: In a recent paper, Hiroshi Kihara induced a model structure on the category of diffeological spaces. He generates the classes of fibrations, cofibrations, and weak equivalences by constructing a functor $d:\Delta \to \mathcal{D}$ (where $\mathcal{D}$ is the category of diffeological spaces, and $\Delta$ is the simplex category), horns $\Lambda^p_k$ are constructed as smooth deformation retracts of the cells $\Delta^p$. Weak equivalences are those maps $f:X \to Y$ so that the $d$-nerve $N_d(f) = \mathcal{D}(d-,f)$ is a weak equivalence of simplicial sets, and using the set of horn inclusions $\Lambda^p_k \to \Delta^p$ as the generating cofibrations. -Based on the results claimed in this arxiv preprint, it seems like he is using the nerve of the functor $d: \Delta \to \mathcal{D}$ to transfer the model structure from simplicial sets to differential objects. This makes me think that there is some notion of a "good enough" cosimplicial object that induces a model structure on a category (somewhat similar to how tangent category can be encoded as a nice functor from $\mathsf{Weil}^{op}$ into a cartesian closed category $\mathcal{E}$). - -REPLY [6 votes]: A somewhat general statement along these lines would be as follows. Suppose $\mathcal{D}$ is a cartesian closed locally presentable category and $d : \Delta \to \mathcal{D}$ is a cosimplicial object with associated geometric realization adjunction $|{\cdot}| : \mathrm{sSet} \to \mathcal{D}$, satisfying the following properties. - -The maps $|\Lambda^n_k| \to |\Delta^n|$ have retracts for each $n \ge 1$ and $0 \le n \le k$. -Then $\mathrm{Sing} X$ is a fibrant simplicial set for any $X \in \mathcal{D}$, so every object will be fibrant in the transferred model structure. - -The functor $|{\cdot}|$ preserves finite products. -Then $X^{|\Delta^1|}$ is a path object for $X = X^{|\Delta^0|}$, for any $X \in \mathcal{D}$. - - -This is enough to ensure that the transferred model structure exists, as described on the nLab. -In fact, this is essentially how Quillen originally constructed the classical model structure on Top--except that the category of all topological spaces does not quite satisfy most of these conditions; but it comes close enough for similar arguments to go through. -Of course condition 1 (and to a lesser extent, condition 2) is far from formal, and the kinds of model structures produced in this way are rather special. -As a trivial example, the Yoneda embedding $\Delta \to \mathrm{sSet}$ doesn't satisfy condition 1 (otherwise it would be automatic!) but obviously the model category structure on $\mathrm{sSet}$ can still be transferred across the identity functor. -We can be somewhat more precise about which model categories can be constructed by a variation of the above argument. Condition 1 is equivalent to every object of the transferred model structure being fibrant, because if $|\Lambda^n_k|$ is fibrant then the acyclic cofibration $|\Lambda^n_k| \to |\Delta^n|$ must have a retraction. Condition 2 is basically meant to ask for a simplicial structure on $\mathcal{D}$ to use for building path objects. One way to get this, not necessarily the only one, is for $\mathcal{D}$ to be cartesian closed and set $K \otimes X = |K| \times X$, and ask for $|{\cdot}|$ to preserve finite products. So, the model categories that can be built in this way are the simplicial ones transferred along a simplicial adjunction $\mathrm{sSet} \rightleftarrows \mathcal{D}$ with every object fibrant. -Having looked at the referenced paper more closely, it seems that the geometric realization functor considered there does not preserve finite products. Instead the acyclicity condition is checked in another way using deformation retracts, analogous to Lemma 2.4.8 in Mark Hovey's book. I'm not aware of a general framework for this kind of argument but it does seem relatively formal as well.<|endoftext|> -TITLE: Standard conjecture on u-invariants? -QUESTION [9 upvotes]: This is well beyond my expertise, but I just learned some of the history behind -$u$-invariants of fields $F$, -where ($u(F)+1$)-variable quadratic equations always have a non-trivial solution, -but $u(F)$-variable equations may not. -Here is the Wikipedia explanation. -In particular, although it seemed possible that the $u$-invariant of any field was -a power of $2$, it was shown by Merkurev in 1991 that there is a field -with $u$-invariant of any even number. But the hypothesis that it could never -be an odd number was contradicted by Izhboldin (in Fields of $u$-invariant $9$) in 2001 who constructed a field with $u$-invariant of $9$.1 - -Q. My question is: Where does this issue stand now, ~20 years later? -Is there some sense -among experts that there is a field with $u$-invariant for any odd number larger than $7$? -Or is it completely open, with no prevailing hypothesis? - -As FZaldivar pointed out in the comments, it was earlier known -that $u \neq 3,5,7$, so $u=9$ was the first realized odd number. - -1 -"Oleg Izhboldin died tragically ... at the age of 37 after submitting this article" -ZBL review. - -REPLY [8 votes]: For the classical $u$-invariant of fields of characteristic $\neq 2$, some known results are: - -The $u$-invariant of formally real fields is $\infty$. - -If $K$ is an algebraically closed field, its $u$-invariant is $1$. More generally, if $K$ does not have quadratic extensions its $u$-invariant is $1$. - -There are no fields of $u$-invariant $3, 5$ or $7$. This was proven by R. Elman, T. Y. Lam, ( Math. Z. 131 (1973), 283--304 and Invent. Math. 21 (1973), 125--137.) - -There are fields of any even $u$-invariant (A. S. Merkurev, Izv. Akad. Nauk. SSSR Ser. Mat. 55 (1991) No. 1, 218--224.) - -There is a field of $u$-invariant $9$ (O. T. Izhboldin, Ann. of Math. (2) 154 (2001), no. 3, 529--587) - -There is a field of $u$-invariant $2^r+1$ for every integer $r\geq 3$ (A. Vishnik, Algebra, arithmetic, and geometry: in honor of Yu. I. Manin. Vol. II, 661--685, -Progr. Math., 270, Birkhäuser Boston, Boston, MA, 2009.) - - -On the other hand, for certain families of fields, the $u$-invariant is known, and for other families, the computations are not complete. For example, if $F$ is a finite field of odd characteristic by the Chevalley-Waring theorem every quadratic form in three variables over $F$ represents $0$, that is, every $3$-dimensional form over $F$ is isotropic and thus $u(F)=2$.<|endoftext|> -TITLE: An inequality about unit vector orthogonal to $(1,1,...,1)$ -QUESTION [12 upvotes]: Does there exist a constant $\alpha>0$ such that the following holds? -$$\liminf_{n\to\infty}\inf_{x\in\mathbb{R}^n, \sum_{i=1}^nx_i^2=1, \sum_{i=1}^nx_i=0}\frac{\sum_{i0$ and all large enough $n$, where -\begin{equation*} - C:=\{(i,j)\in[n]^2\colon|i-j|\le n/4\}. -\end{equation*} -Suppose that $n\ge8$. Partition the interval $[n]$ into intervals $\de_1,\dots,\de_8$ in $\mathbb N$ of lengths as equal as possible, so that for any $k$ and $l$ in $[8]$ such that $k0$. Suppose otherwise: that $\tL=\tL_{\min}\le0$. -The second crucial point is that for any $m_{2,1},\dots,m_{2,8},m_{1,1},\dots,m_{1,8},p_1,\dots,p_8$ satisfying conditions (3)--(5), there exist r.v.'s $Y$ and $Z$ (defined, say, on the standard probability space over $\Om:=[0,1]^2$) and events $A_1,\dots,A_8,B_1,\dots,B_8$ such that, with the $I_k$'s and $J_k$'s still defined by (0.5), the nonuples $(Y,I_1,\dots,I_8)$ and $(Z,J_1,\dots,J_8)$ are iid, $(A_1,\dots,A_k)$ is a partition of $\Om$, $(B_1,\dots,B_k)$ is also a partition of $\Om$, $P(A_k)=P(B_k)=p_k$ for all $k\in[8]$ (cf. (-1)), and -$Y$ and $Z$ are iid zero-mean unit-variance r.v.'s such that -$EY^2I_k=EZ^2J_k=m_{2,k}$ and $EYI_k=EZJ_k=m_{1,k}$ (cf. (1.5)). -$\Big($Indeed, all these conditions can be satisfied e.g. as follows. Re-define here $\de_1,\dots,\de_8$ as non-overlapping subintervals $[0,1]$ of lengths $p_1,\dots,p_8$, and then re-define $A_1,\dots,A_8,B_1,\dots,B_8$ by the formula -\begin{equation*} - A_k:=\de_k\times[0,1],\quad B_k:=[0,1]\times\de_k; -\end{equation*} -cf. (-0.5). Finally, for each $k\in[8]$ partition the interval $\de_k$ (of length $p_k$) into two subintervals, say $\eta_k$ and $\zeta_k$, each of length $p_k/2$, and let -\begin{equation*} - Y:=\sum_{k=1}^8(u_k\,1_{\eta_k\times[0,1]}+v_k\,1_{\zeta_k\times[0,1]}), \quad - Z:=\sum_{k=1}^8(u_k\,1_{[0,1]\times\eta_k}+v_k\,1_{[0,1]\times\zeta_k}), -\end{equation*} -where -\begin{equation*} - u_k:=\frac{m_{1,k}-\sqrt{p_k m_{2,k}-m_{1,k}^2}}{p_k}, \quad - v_k:=\frac{m_{1,k}+\sqrt{p_k m_{2,k}-m_{1,k}^2}}{p_k}. -\end{equation*} -Then all the conditions on events $A_1,\dots,A_8,B_1,\dots,B_8$ and r.v.'s $Y,Z$ listed in the paragraph just above will be satisfied.$\Big)$ -So, the minimum value of $L$ coincides with $\tL_{\min}$, which was assumed to be $\le0$. Since $L\ge0$, we may now assume without loss of generality that $L=0$. -Looking now back at (1), we see that then $Y=Z$ almost surely (a.s.) on the event $D$, still defined as in (0). -Therefore and because $Y$ and $Z$ are iid, it follows (i) that $Y$ is a.s. constant on $A_{2j-1}\cup A_{2j}$ for each $j\in[4]$ and (ii) $Y$ is a.s. constant on $A_{2j}\cup A_{2j+1}$ for each $k\in[3]$. So, $Y$ is a.s. constant on $\Om$. Since $EY=0$, we have $Y=0$ a.s., which contradicts the condition that $Y$ is of unit variance. $\Box$ - -If one is curious, numerical minimization of $L$ given (3) and (4) with $p_1=\dots=p_8=1/8$ gives about the same result by three different methods: $L\ge0.038060$. So, it appears that, at least for large enough $n$, the best constant $c$ in (-2) is $>3/100$. -Here is an image of the corresponding Mathematica notebook:<|endoftext|> -TITLE: Finding the right map between cohomology with local coefficients and Čech cohomology -QUESTION [5 upvotes]: Let $X$ be a space which is paracompact, Hausdorff, and sufficiently nice that it has a universal covering space (and map) $p:\tilde{X}\to X$. Also, let $\pi:=\pi_1(X)$ and $A$ some $\mathbb{Z}[\pi]$-module, or if preferred, some abelian group on which $\pi$ acts. Under these conditions, there is a notion of cohomology with local coefficients which can be computed using cochain groups $C^n_\pi(X,A)=Hom_{\mathbb{Z}[\pi]}(C_n(\tilde X),A)$ where $C_n(\tilde{X})$ is the singular chain groups on $\tilde X$ and has an action of $\pi$ on it "inherited" from the action on $\tilde X$ itself by post-composition, that is, for $g\in \pi$, $\sigma \in C_n(\tilde X)$, $g\cdot \sigma$ is just $\sigma$ followed by the action of $g$. Let's call this cohomology $H^n_\pi(X,A)$. -On the other hand, there is another notion of cohomology with local coefficients whereby one defines a locally constant sheaf $\mathcal A$ on $X$ using the constant sheaf $A_\tilde{X}$ on $\tilde X$ and the action of $\pi$ on $A$, and takes the sheaf cohomology of that. It appears to be well known that (1) these two theories are essentially the same, and (2) for paracompact Hausdorff spaces, the Čech cohomology of a sheaf is the same as the sheaf cohomology of that same sheaf. That means that the Čech cohomology $\check{H}^n(X,\mathcal A)$ should be isomorphic to $H^n_\pi(X,A)$. -With all of this background out of the way, I can begin to describe my actual question. First of all, I cannot comprehend Sheaf cohomology, no matter how hard I try. Every reference I have found that describes it just seems impenetrable to me. On the other hand, I can more or less make sense of Čech cohomology, at least a lot better than "direct" sheaf cohomology. So: I would like to define some explicit map, $\check H^*(X ,\mathcal A)\to H^*_\pi (X, A)$ (or going the other way), which can exhibit this isomorphism more directly. -What appears natural for a map like this is to perhaps restrict to some convenient cofinal collection of covers of $X$ to get maps $C^*(\mathscr U,\mathcal A)\to C^*_\pi(X,A)$. A good candidate comes from the property of a universal cover, whereby every element $x\in X$ has a neighborhood $U_x$ so that $p^{-1}(U_x)$ composed of a disjoint union of open sets, each of which $p$ makes homeomorphic to $U_x$. It's not hard to show covers of the form $\mathscr U=\lbrace U_x : x\in X \rbrace$ with each $U_x$ a neighborhood like this is a cofinal collection, and then $p^{-1}(\mathscr U)$ is an open cover of $\tilde X$. There is a possibly useful property of Singular cohomology, too, which for any open cover $\mathscr V$ of $\tilde X$ makes the inclusion of the subcomplex $C_n(\mathscr V)=\langle \sigma: \text{im} (\sigma) \subset V\text{ for some }V \in \mathscr V \rangle$ a chain equivalence. That carries over to a cochain equivalence between $Hom_{\mathbb{Z}[\pi]}(C_n(\tilde X),A)$ and $Hom_{\mathbb{Z}[\pi]}(C_n(\mathscr V),A)$, and I'll call the latter group $C^n_\pi (\mathscr V , A)$. -My most promising candidate after several attempts, using these ideas, is to consider covers of $X$ as above, and for each one make a map $\psi_\mathscr{U}: \check{C}^n(\mathscr U , \mathcal A) \to C^n(p^{-1}(\mathscr U),A)$ by first applying a map $\iota_\mathscr{U}:\check{C}^n(\mathscr U , \mathcal A)\to \check{C}^n(p^{-1}(\mathscr U ), A_\tilde{X})$ defined by $(\iota_\mathscr{U} f)(p^{-1}(U_{x_0}),...,p^{-1}(U_{x_n}))=f(U_{x_0},...,U_{x_n})$, then picking out for each $\sigma \in C_n(p^{-1}(\mathscr U))$ the open set $U_\sigma:=U_{p(\sigma(v_0))}\cap ... \cap U_{p(\sigma(v_n))}$ and setting $(\psi_\mathscr{U} f)(\sigma):=[(\iota_\mathscr{U} f)(p^{-1}(U_\sigma))](\sigma(v_0))$. I say this is promising because evaluating at some point is needed in order to get the $\pi$-equivariance expected of an element of $C^n_\pi(p^{-1}(\mathscr U),A)$, and because it seems likely to give "injectivity" and "surjectivity" at least at the cohomology level after taking the direct limit defining Čech cohomology. -The only problem is that I'm not entirely sure this actually gives a cochain map. Running through the calculations, I've found that: - -$(\partial\psi_{\mathscr{U}}f)(\sigma)-(\psi_{\mathscr{U}}\partial f)(\sigma)=[(\iota_{\mathscr{U}}f)(p^{-1}(U_{p(\sigma(v_{1}))}),...,p^{-1}(U_{p(\sigma(v_{n+1}))}))](\sigma(v_{1}))-[(\iota_{\mathscr{U}}f)(p^{-1}(U_{p(\sigma(v_{1}))}),...,p^{-1}(U_{p(\sigma(v_{n+1}))}))](\sigma(v_{0}))$ - -This is more or less because the '0 boundary' of the simplex sends $v_0$ to the same place the original simplex sends $v_1$ to, while the rest of the boundaries send $v_0$ to $\sigma(v_0)$. Of course, this needs to be $0$ to get a cochain map, but I'm not sure these two terms are the same (or if they are, how to show it). -So: Is this the wrong map again? If it is, what's the right map? Or, is there some really simple way to see that those two terms are the same and the difference really is $0$? Are there any references that have attempted something similar? - -REPLY [4 votes]: Like I explained in the comments, there is no reason to expect the existence of a particularly explicit direct map between the Cech and singular complexes, since we naturally obtain a zig-zag -$$ -\check C^*(\{U_i\}_{i\in I},\mathcal A)\to \check C^*(\{U_i\}_{i\in I},\operatorname{Sing}^\bullet(\mathcal A))\leftarrow \operatorname{Sing}^\bullet(\mathcal A)(X) = \operatorname{Hom}_{\mathbb Z\pi}(\operatorname{Sing}^\bullet(\widetilde X),A) -$$ -where $\operatorname{Sing}^\bullet(\mathcal A)(-)$ is the presheaf which assigns to $U\in X$ the $\pi$-equivariant maps from $p^{-1}(X)$ to $A$ and $p:\widetilde X\to X$ is the universal cover. Note that the individual terms $\operatorname{Sing}^\bullet(\mathcal A)(-)$ are not sheaves: if you know the values on small simplices, i.e. those inside an open set of a cover, you can't recover the values on all simplices. However, the inclusion of small simplices is a quasiisomorphism since every simplicial chain is homologous to its barycentric subdivision, which will eventually be small. This shows that the left-pointing map is a quasiisomorphism; the right-pointing map is a quasiisomorphism if all intersections are empty or contractible, i.e. $\{U_i\}_{i\in I}$ is a good cover. For nice spaces (e.g. CW complexes) the Cech cohomology groups are computed by the Cech complex of a good cover. In that case, it is possible to get an explicit chain map from the singular cochain complex to the Cech complex as follows: -The good cover $\{U_i\}_{i\in I}$ determines an abstract simplicial complex $S_U$, i.e. a collection of finite subsets of $I$ (namely, those for which the intersection is nonempty) which is closed under taking subsets. We may inductively choose for $J\in S_U$ a simplex $\sigma_J:\Delta^{|J|-1}\to X$ whose image lands in $\bigcup_{j\in J} U_j$ and whose restriction to a face is given by the previous choice in the obvious sense (send the barycentre to a point in $\bigcap_{j\in J} U_j$ and iteratively "cone off" the maps on the lower-dimensional simplices, which is possible by contractibility). These singular simplices form a subcomplex of the singular chains, and it's straightforward to show that restricting singular cochains to this subcomplex gives a cochain map from these to the Cech complex. -In the other direction, one would have to pick a chain homotopy inverse to the inclusion of small simplices, which is not possible to do explicitly as far as I know.<|endoftext|> -TITLE: About a claim by Gromov on proper holomorphic maps -QUESTION [32 upvotes]: At p. 223 of his paper [G03], Mikhail Gromov makes the following claim: - -Let $X$, $Y$ be two complex manifolds (not necessarily compact or Kähler) of the same dimension and having the same even Betti numbers. Then every proper surjective holomorphic map $f \colon X \to Y$ is finite-to-one. - -In my answer to MO question 377353, I proposed a proof for this claim. However, checking the details I realized that such a proof requires that $X$ is Kähler. -In fact, the last step of my argument uses in an essential way the fact that the fundamental class of a compact subvariety of $X$ is non-trivial in homology, and this is in general false when $X$ is not -Kähler. For instance, in his answer to [MSE], Michael Albanese notices that the standard Hopf's surface $X$ contains a compact torus $C$ whose fundamental class $[C]$ is zero in $H_2(X, \, \mathbb{C})$, simply because the last group is trivial. -So, let me ask the - -Question. Is Gromov's claim true without the Kähler assumption for $X$? If so, what is a proof? - -References. -[G03] Gromov, M: On the entropy of holomorphic maps, Enseign. Math. II. Sér. 49, No. 3-4, 217-235 (2003). ZBL1080.37051. -[MSE] https://math.stackexchange.com/q/1556561/456212 - -REPLY [4 votes]: I think I've got it. This is only a sketch, and I am not an expert on singular complex analytic things, but it seems about right. -Let $f:X\to Y$ be a proper holomorphic map between complex manifolds of the same dimension $n$, and assume that some fiber has positive dimension. I want to find a rationally nontrivial element of the kernel of $f_\ast :H_{2k}(X)\to H_{2k}(Y)$ for some $k$. -Let $S\subset X$ be the union of all fibers having positive dimension. I presume that $S$ is a closed analytic subset of $X$. Let $Z$ be an irreducible component of $S$ of maximal dimension. Let $m$ be the dimension of $f(S)$. Let $k -TITLE: triviality of homology with local coefficients -QUESTION [5 upvotes]: Let $X$ be a manifold or a CW-complex. -Let -$\pi: \tilde X\longrightarrow X$ -be a covering map. -Let $\pi_1(X)$ be the fundamental group of $X$ and let $\rho: \pi_1(X)\longrightarrow O(n)$ be an orthogonal representation. -Define the $\rho$-twisted chain complex of $\tilde X$ by -$C_*(\tilde X,\rho)=C_*(\tilde X)\otimes_{\pi_1(X)} \mathbb{R}^n$ -where $\pi_1(X)$ acts on $C_*(\tilde X)$ from the right by deck transformations and acts on $\mathbb{R}^n$ from the left by orthogonal transformations. -In the book: Lecture Notes in Algebraic Topology by -James F. Davis and -Paul Kirk, Chapter 5, the homology with local coefficients is defined as the homology of the $\rho$-twisted chain complex -$H_*(\tilde X,\rho)=H_*(C_*(\tilde X,\rho))$. -Question. -Can we add some additional hypothesis on $X$, the covering space $\tilde X$, and the covering map $\pi:\tilde X\longrightarrow X$ such that for such $X$ and $\tilde X$, we can always find an $n\geq 2$ and a $\rho$ satisfying that $H_*(\tilde X,\rho)$ is trivial? -Thanks for guidance. - -REPLY [5 votes]: Maybe you are looking for something more interesting, but you can take $X=S^1$, universal cover $\tilde X$, and $\rho: {\mathbb Z}\to O(n)$ such that the image group has no fixed unit vectors in $R^n$. Then $H_*(\tilde X,\rho)=0$ (which is a nice exercise to work out if you are new to this material). A more challenging problem would be: -Construct a finite CW-complex $X$ such that for each $n\ge 2$ there exists a representation $\rho: \pi_1(X)\to SO(n)$ with vanishing homology. -If you are interested in 3-dimensional topology, here are two classes of examples you should be aware of: -a. Suppose that $X$ is a closed connected orientable 3-manifold with finite nontrivial fundamental group $\pi$ and $\tilde X\to X$ is its universal covering. Then for each $\rho: \pi\to O(4)$ such that $\rho(\pi)$ has no fixed unit vectors, $H_*(\tilde X,\rho)=0$. (Examples of such $\rho$ are given by the fact that $\pi$ embeds in $SO(4)$ so that the image group acts freely on $S^3$.) -b. Suppose that $X$ is a closed connected orientable arithmetic hyperbolic 3-manifold and $\tilde X\to X$ is its universal covering. Then there exists a representation $\rho: \pi_1(X)\to O(3)$ such that $H_*(\tilde X,\rho)=0$. -Edit. At the same time, there are spaces for which your property does not hold, for instance a space whose fundamental group admits only trivial orthogonal representations.<|endoftext|> -TITLE: Difficulty with "On fibering certain 3-manifolds" by Stallings -QUESTION [6 upvotes]: I am reading the paper "On fibering certain 3-manifolds" by John Stallings and I was hoping someone could help me through a certain detail. In particular, I am confused at the very end of the proof of Theorem 1 which is as follows: -Theorem (Stallings): -Let $M^3$ be a compact 3-manifold such that there is a finitely generated normal subgroup $G$ of $\pi_1(M)$ so that $\pi_1(M)/G \cong \mathbb{Z}$. Then $G$ is in fact the fundamental group of a 2-manifold embedded in $M$. -Stallings breaks the proof up into sections 2,3,4, and 5. I will briefly summarize how each of these sections goes, but my confusion is that I do not understand the alluded to "geometric construction" in section 5 at the end of the proof. -In section 2, Stallings tells us to look at a map $f: M \to S^1$ that topologically realizes the surjective homomorphism $\pi_1(M) \to \pi_1(M)/G \cong \mathbb{Z}$. We are then instructed to pick a point $p \in S^1$ and look at the surface $f^{-1}(p)$. Section 2 proceeds to give a method to ensure that $f^{-1}(p)$ is in fact connected. This uses the finitely generated property of $G$. -In section 3, Stallings look a the now connected surface $f^{-1}(p)$ and, if the inclusion map for this surface to the 3-manifold is not $\pi_1$-injective, he uses the loop theorem to find a disk to do surgery on $f^{-1}(p)$ and then he performs this surgery by homotoping $f$ and looking again at the preimage of $p$. -In section 4, Stallings repeats step 2 on the result of step 3 which has the effect of removing one of the two components of the surgered surface. Then by an Euler characteristic count, we say that this process of repeating steps 2 and 3 must terminate with a map $f : M \to S^1$ with $f^{-1}(p)$ connected and $\pi_1$-injective. (Here Stallings writes that $\pi_1(f^{-1}(p)) \to H_1(M)$ is injective - I imagine that is a typo?). -Now we have reached the section I do not understand. We know that the image of $\pi_1(f^{-1}(p)$ is in the kernel of the map $\pi_1(M) \to \mathbb{Z}$, and we want to argue that it is the entire kernel. Here, Stallings says, "If there is anything else in the kernel of $f_*$, then a geometric construction shows that the kernel of $f_*$ is the union of strictly increasing sequence of groups, and so could not be finitely generated." (Also he mentions in the previous sentence that this is due to Neuwirth and is in his thesis which I have not tracked down.) -What is this geometric construction? - -REPLY [8 votes]: I think about it like this. For convenience, I'll assume $M$ is closed. -Given a homomorphism $\phi:\pi_1M\to\mathbb{Z}$, Stallings explains how to find an essential surface $S\subset M$ with $\pi_1S\leq\ker\phi$, as you outline nicely in the question. -Now we construct the cyclic cover $M_\phi$ of $M$ corresponding to $\ker\phi$ as follows. Let $M_0$ be the result of cutting $M$ along $S$, so $M_0$ has two boundary components, $S_+$ and $S_-$, corresponding to $S$. Then $M_\phi$ can be thought of as an infinite gluing: -$M_\phi=\left(\bigcup_{n\in \mathbb{Z}} M_0\right)/\sim$ , -where $\sim$ glues $S_-$ in the $n$th copy of $M_0$ to $S_+$ in the $(n-1)$th copy of $M_0$. -Therefore, $\ker\phi$ decomposes as an infinite amalgamated product: -$\ker\phi= \ldots *_{\pi_1 S} \pi_1 M_0 *_{\pi_1 S} \pi_1 M_0 *_{\pi_1 S} \pi_1 M_0 *_{\pi_1S} \ldots $ . -In particular, it is infinitely generated unless the inclusion $S\to M_0$ induces an isomorphism of fundamental groups. But this only occurs if $M_0$ is a product $S\times [0,1]$, which is exactly the case in which $S$ is a fibre.<|endoftext|> -TITLE: Counting eigenvalues without diagonalizing a matrix -QUESTION [7 upvotes]: Today's arXiv has a paper by Pierpaolo Vivo, Index of a matrix, complex logarithms, and multidimensional Fresnel integrals, which asks the question whether it is possible to calculate the number $N(\lambda_1,\lambda_2)$ of eigenvalues of a real symmetric matrix $M$ that lie in the interval $(\lambda_1,\lambda_2)$ , without explicitly diagonalizing the matrix. The author points out that a certain formula in the literature based on the branch-cut structure of the complex logarithm generically fails to produce the correct result. -In a discussion with my colleagues (thank you, Fabian & Inanc) we could fix this branch-cut ambiguity by first differentiating the logarithm and then integrating, -$$N(\lambda_1,\lambda_2)=\frac{1}{2\pi i}\lim_{\epsilon\rightarrow 0^+}\int_{\lambda_1}^{\lambda_2}d\lambda\; \frac{d}{d\lambda}\left[\log\det(M-\lambda+i\epsilon)-\log\det(M-\lambda-i\epsilon)\right].$$ -As an example, applied to the matrix $M$ defined in equation 3 of the cited paper this gives the following plot for $N(-15,\lambda)$ as a function of $\lambda$, in agreement with the eigenvalues at $-11.03, -2.80, -0.63, 2.464$. - -Question: From a computational point of view, this method to count levels is unlikely to be efficient relative to a direct diagonalization. The cited paper develops a method based on Fresnel integrals. Is there an alternative more efficient approach? - -I note an earlier MO question along the same lines, but for the specific case of a random matrix with localized eigenstates; here the idea is to have a method that works generically. - -REPLY [16 votes]: Here is an efficient method. -First of all, I must quote that diagonalizing $M$ is not a method, because there is no explicit way to carry this out. It amounts to calculating the roots of a polynomial ! At best, one can do this in an approximate way. -Instead, I suggest to perform a preliminary step : put $M$ in Hessenberg form. This is done (method of Givens) in $O(n^3)$ operations, by using rotations in coordinate planes. This calculation is stable in the sense that the rounding errors, if any, are not amplified, so the spectrum is not polluted by more than the elementary rouding error of the machine. -Because $M$ was symmetric, its Hessenberg form $H$, still symmetric, is actually tridiagonal. Then you can form a Sturm sequence by taking the principal submatrices of $H$ and their characteristic polynomials $P_j$. Their calculation is extremely cheap in the tridiagonal case: say that -$$H=\begin{pmatrix} a_1 & b_1 & 0 & \cdots & 0 \\ b_1 & \ddots & \ddots & \ddots & \vdots \\ 0 & \ddots & \ddots & \ddots & 0 \\ \vdots & \ddots & \ddots & \ddots & b_{n-1} \\ 0 & \cdots & 0 & b_{n-1} & a_n \end{pmatrix}.$$ -Then -$$P_n(X)=(X-a_n)P_{n-1}(X)-b_{n-1}^2P_{n-2}(X).$$ -We may assume that the $b_j$'s are non-zero (otherwise, treat matrices of smaller sizes). Then $$(P_j(c)=0)\longrightarrow(P_{j+1}(c)P_{j-1}(c)<0),$$ -the Sturm assumption. -Now Sturm Theorem gives you the number of eigenvalues in $[a,b]$ as the difference $V(a)-V(b)$, where $V(x)$ is the number of sign changes in the sequence $(P_n(x),\ldots,P_0\equiv1)$.<|endoftext|> -TITLE: Matrix obtained by recursive multiplication and a cyclic permutation -QUESTION [9 upvotes]: Have you ever seen this matrix? Each row is obtained from the previous one by multiplying each element by the corresponding element of the next cyclic permutation of $(a_1,\dots, a_n)$: -$$\left( - \begin{array}{llllllll} - 1 & 1 & 1 & \dots & 1 & 1 \\ - a_1 & a_2 & a_3 & \dots & a_{n-1} & a_{n} \\ - a_1 a_2 & a_2a_3 & a_3 a_4 & \dots & a_{n-1} a_n & a_{n} a_1 \\ - a_1 a_2a_3 & a_2a_3 a_4 & a_3 a_4a_5 & \dots & a_{n-1} a_n a_1& a_{n} a_1 a_2 \\ - a_1 a_2a_3a_4 & a_2a_3 a_4 a_5& a_3 a_4a_5 a_6& \dots & a_{n-1} a_n a_1 a_2& a_{n} a_1 a_2a_3 \\ - \dots & \dots & \dots & \dots & \dots & \dots \\ - a_1 a_2a_3a_4 \dots a_{n-2}& a_2a_3 a_4 a_5\dots a_{n-1} & a_3 a_4a_5 a_6\dots a_n&\dots & a_{n-1} a_n a_1 a_2\dots a_{n-4}& a_{n} a_1 a_2a_3\dots a_{n-3} \\ - a_1 a_2a_3a_4 \dots a_{n-1}& a_2a_3 a_4 a_5\dots a_{n} & a_3 a_4a_5 a_6\dots a_1& \dots & a_{n-1} a_n a_1 a_2\dots a_{n-3}& a_{n} a_1 a_2a_3\dots a_{n-2} \\ - \end{array} -\right)$$ -I would like to know if there is a closed formula for the determinant; of course it is invariant (up to sign) under cyclic permutations of $(a_1,a_2,\dots,a_n)$. - -REPLY [8 votes]: Denote the matrix $A$, and index all $a_i$, and all rows and columns starting from $0$ for convenience. If, say, $a_0 = 0$, then $\det A = (-1)^{\lfloor (n - 1) / 2 \rfloor} \prod_{i = 1}^{n - 1} a_i^i$ by substituting and computing the remaining upper triangular determinant. Let's further assume that all $a_i$ are non-zero. -Let $p_k = \prod_{i = 0}^{k - 1} a_i$ ($p_0 = 1$ by convention). Put $a'_i = a_i p_n^{-1/n}$, and construct $A'$ similarly. Alternatively, $A'$ is $A$ with $i$-th row multiplied by $p_n^{-i/n}$. Defining $p'_k$ similarly, we now have $p'_k = p_k p_n^{-k / n}$, in particular $p'_n = 1$. -After multiplying $i$-th column of $A'$ by $p'_i$, we arrive at a Hankel matrix $$P' = \begin{pmatrix} 1 & p'_1 & \ldots & p'_{n - 1} \\ -p'_1 & p'_2 & \ldots & 1 \\ \ldots & \ldots & \ldots & \ldots \\ p'_{n - 1} & 1 & \ldots & p'_{n - 2}\end{pmatrix},$$ which is a row permutation of a circulant $(1, \ldots, p'_{n - 1})$. With all substitutions in mind, we have -$$\det A = (-1)^{\lfloor(n - 1) / 2\rfloor} \prod_{k = 0}^{n - 1} p_k \cdot \prod_{k = 0}^{n - 1}\left(\sum_{j = 0}^{n - 1} p'_j e^{i\frac{2\pi jk}{n}} \right).$$ -One can see that the problem is as general as arbitrary circulant determinant.<|endoftext|> -TITLE: Effective weight-monodromy conjecture -QUESTION [11 upvotes]: $\DeclareMathOperator\Gr{Gr}$Let $G$ be the absolute Galois group of a finite extension of $\mathbb{Q}_p$ with inertia subgroup $I$, and let $V$ be an $\ell$-adic representation of $G$. Grothendieck's Monodromy Theorem says that after passing to an open subgroup of $G$, the action of $I$ is unipotent, and moreover may be described by a nilpotent operator $N \colon V \to V$. -There is a monodromy filtration $M_i V$, such that $N M_i V \subseteq M_{i-2}$ and $N^i \colon \Gr^M_i V \to \Gr^M_{-i} V$ is an isomorphism. The Weight-Monodromy Conjecture states that if $V=H^n(X_{\overline{k}};\mathbb{Q}_{\ell})$ for a variety $X$ over a number field $k$, then the Frobenius action on $\Gr^M_i V$ is pure of weight $n+i$ (in particular, the case of good reduction is essentially the Weil conjecture). -Is there a version of this conjecture stating that $\Gr^M_i V = 0$ for $|i| > n$? I can't seem to find that in the literature (e.g., in Conjecture 1.13 of Scholze, or in Deligne's original article). -This is true for abelian varieties, and it seems implied in the literature, but I haven't found a statement like this. -The statement should more generally say that whenever $V$ is an effective motive of weight $n$, the weights are in the interval $[0,2n]$. - -REPLY [10 votes]: You can bound the filtration length (assuming the WM conj) using the weight spectral sequence of Rapoport--Zink. This is a sp seq converging to $H^*(X_{\overline{k}})$; and if the WM conj holds, then the monodromy filtration coincides with the filtration induced by this spectral seq. (There's a nice account of this in Scholl's paper https://www.dpmms.cam.ac.uk/~ajs1005/preprints/weil-preprint1.pdf.) The $E_1$ page of the spectral seq is explicitly given in terms of the components of the special fibre of a semistable model, so you can make explicit what region of the plane they're supported in and you get a bound on the length of the filtration. (I haven't checked whether that bound is the one you want though, the indexing is pretty barbarous.) -Edit. Prompted by Pol's comment, I did the computation: the $E_1^{ij}$ terms are supported in a parallelogram with vertices at $(0, 0), (0, d), (2d, 0), (2d, -d)$. So in particular the filtration on $H^n$ has at most $1 + \min(n, 2d-n)$ nonzero graded pieces for any $n$.<|endoftext|> -TITLE: Can this inequality be proved using weighted maximal function estimates? -QUESTION [8 upvotes]: I am trying to understand the following fact: - -Suppose $\{B_i\}_i$ are disjoint balls in $\mathbb R^n$, and $A_i \subset 100 B_i$ is a subset with $|A_i| \geq c |B_i|$. Then for any nonnegative $f$, we have $\sum_i |B_i| \inf_{A_i} f \lesssim \int_{\cup_i A_i} f$, where the implied constant depends only on $c$ and the dimension $n$. - -(Here, $|\cdot|$ denotes Lebesgue measure, and $100B$ denotes the ball with the same center as $B$ and $100$ times the radius.) -Is there a way to prove this with (some combination of) covering lemmas, maximal function estimates, or weighted inequalities? I couldn't see an easy way to prove this. - -Some background (which isn't needed for my question): The statement above is taken from Chapter 13 of David and Semmes's Singular integrals and rectifiable sets in $\mathbb R^n$. (It appears in the middle of a proof. They do not state this as a separate lemma.) -Here is a sketch of the proof in the book: -Let $p \in (1, \infty)$ and for each $i$, let $w_i$ be a function on $A_i$ (all TBD). By Holder, -\begin{align*} -\inf_{A_i} f -\leq -\left(\frac{1}{|A_i|}\int_{A_i} f^{1/p} \right)^{p} -\leq -\left(\frac{1}{|A_i|}\int_{A_i} f w_i \right) -\left(\frac{1}{|A_i|}\int_{A_i} w_i^{-p'/p} \right)^{p/p'} -\end{align*} -so -\begin{align} -\sum_i |B_i| \inf_{A_i} f -&\lesssim -\sum_i -\left(\int_{A_i} f w_i \right) -\left(\frac{1}{|A_i|}\int_{A_i} w_i^{-p'/p} \right)^{p/p'} -\\ -&\leq -\left(\int f \textstyle\sum_i 1_{A_i} w_i \right) -\left(\sup_i\frac{1}{|A_i|}\int_{A_i} w_i^{-p'/p} \right)^{p/p'} -\end{align} -To complete the proof, we just need to choose $p$ and $w_i$ so that (i) $\sum_i 1_{A_i} w_i \lesssim 1$ and (ii) $\sup_i\frac{1}{|A_i|}\int_{A_i} w_i^{-p'/p} \lesssim 1$. This can be accomplished as follows: -Let $p = 3$. Introduce an ordering on the indices so that $i \prec j$ if $|B_i| < |B_j|$ (and break ties arbitarily). Set $w_i(x)^{-1/2} = \sum_{j \preceq i} 1_{A_j}(x) = \# \{ j : x \in A_j \text{ and } j \preceq i\}$. -Note that if $j \preceq i$ and $A_j \cap A_i \neq \emptyset$, then $B_j \subset 300B_i$. This, with the disjointness of the $B_j$, implies $$\int_{A_i} w_i^{-1/2} \leq \sum_{j \preceq i, A_j \cap A_i \neq \emptyset} |A_j| \approx \sum_{j \preceq i, A_j \cap A_i \neq \emptyset} |B_j| \leq |300B_i| \approx |A_i|.$$ -This proves (ii). (Also, this implies $w_i(x) > 0$ for almost every $x \in A_i$.) -Finally, for any fixed $x$, if $w_i(x) = w_j(x) \neq 0$, then $i=j$. Since $w_i$ takes values in $\{m^{-2} : m \in \mathbb N\} \cup \{0\}$, we have the pointwise bound $\sum_i 1_{A_i} w_i \leq \frac{\pi^2}{6}$, which shows (i) holds and completes the proof. -I don't really have a good intuition for this proof, especially how to motivate the choice of $p$ and $w_i$ (other than "because it works"). In particular, I am mystified (and amazed) at how the authors use $\sum_{m=1}^\infty m^{-2} < \infty$ to control the overlap of the $\{A_i\}_i$. This is why I'd be interested to see if there was another proof. - -REPLY [11 votes]: It suffices to show that -$$ \sum_i |B_i| 1_{\inf_{A_i} f > t} \lesssim \int_{\bigcup A_i} 1_{f>t}$$ -for any $t>0$, since the claim follows by integrating in $t$ and using the Fubini-Tonelli theorem (i.e., use the layer cake decomposition). (Equivalently: to prove the claim, it suffices to do so in the special case when $f$ is an indicator function.) But one has -$$ M (1_{\bigcup A_i} 1_{f>t})(x) \gtrsim 1$$ -whenever $x \in B_i$ and $\inf_{A_i} f>t$, so the claim follows from the Hardy-Littlewood maximal inequality.<|endoftext|> -TITLE: For what modules is the endomorphism ring a division ring? -QUESTION [8 upvotes]: Let $k$ be a field, $A$ a $k$-algebra of finite length and $M$ an $A$-module of finite length. When does it happen, that $\text{End}(M)$ is a division ring? Notice if $M$ is simple, then it happens and if it happens, then $M$ but be indecomposable. So this property is something inbetween simple and indecomposable, and it came up in the context of torsion pairs. - -REPLY [13 votes]: Such modules are called bricks for finite dimensional algebras and there are in general very many of them. -Having a division ring as the endomorphism ring is equivalent to the condition that every non-zero endomorphism morphism is invertible. -For hereditary and tilted algebras they are quite interesting, see https://link.springer.com/article/10.1007/BF03323325 where the name brick might also appeared first. -A classification for given classes of such algebras is an interesting problem, see for example https://arxiv.org/pdf/1712.08311.pdf . -Theorem 1.4 in https://arxiv.org/pdf/1503.00285.pdf gives a characterisation when an algebra has only finitely many bricks.<|endoftext|> -TITLE: Yoneda lemma for monoidal categories -QUESTION [5 upvotes]: I am looking at the Yoneda lemma trying to see where the assumption of "locally small" really comes in. Obviously in order to define a functor to the category sets using $Hom$-spaces we need our $Hom$-spaces to be sets. However if we consider a enriched-category, enriched over some non-locally small monoidal category M, then for any element of the category, our $Hom$-sets give us a functor $Hom(A,-)$ to M. -In particlar, in the statement -$$ -Hom(Hom(-,A),F) \simeq F(A), -$$ -where $F$ is a set-valued functor, where does the assumption of "smallness" play a role. -In the answer to this question, it is stated that the category of sets can be replaced by any Grothendieck universe $U$. However, the definition of a Grothendieck universe assumes that $U$ is a set. Moreover, the enriched Yoneda lemma again assumes "smallness". In these answers, is smallness a necessary assumption? - -REPLY [10 votes]: The Yoneda lemma is a purely formal result that does not require any size assumptions. For any closed symmetric monoidal category $\mathbf{V}$, any $\mathbf{V}$-category $C$, any object $A\in C$, and any functor $F:C\to \mathbf{V}$, there is an isomorphism -$$ [C,\mathbf{V}](よ^A,F) \cong F(A). $$ -Here $よ^A$ denotes the hom-functor $C(A,-)$ and $[C,\mathbf{V}]$ denotes the $\mathbf{V}$-enriched hom-category. It is true that one needs $\mathbf{V}$ to have limits of the size of $C$ in order for $[C,\mathbf{V}]$ to exist as a $\mathbf{V}$-category, but even if this fails, the statement is true and provable in the following sense: if we write down the diagram whose limit would, if it existed, be the LHS, then the RHS is a limit of that diagram. -It is even possible to formulate and prove versions of the Yoneda lemma that do not require $\mathbf{V}$ to be closed or symmetric, and even that allow it to be a multicategory rather than a monoidal category. See, for instance, Lemma 5.29 of my paper enriched indexed categories, or proposition 8.2 of my paper with Richard Garner, enriched categories as a free cocompletion for the bicategorical case.<|endoftext|> -TITLE: Fermat's last theorem $\pm1$ -QUESTION [8 upvotes]: I'm planning a challenge over on Code Golf.SE about integers $a, b, c \ge 0$ such that -$$a^n + b^n = c^n \pm 1$$ -for a given integer $n > 2$. However, I'm interested in whether any non-trivial solutions to this exist for a given $n$. Here, I'm defining "non-trivial" solutions as triples $a, b, c$ such all three are unique and non-zero (i.e. to avoid $(a, 1, a)$ and $(a, 0, a)$, and related triples). -I've found this question which asks a related (and broader) question about the existence of such triples, and the accepted answer states - -I think that if $n\ge5$ (and assuming the ABCD conjecture), then for any $k$, the equation -$$ a^n + b^n - c^n = k $$ -has only finitely many solutions $a,b,c\in\mathbb{Z}$ with $|a|,|b|,|c|$ distinct and non-zero. - -However, this doesn't fully state whether there are a non-zero number of distinct, non-zero solutions. -This is a program which attempts to find such triples, with $0 \le a, b, c \le 100$, given an input $n$, but so far it hasn't found any for either $n = 4$ or $n = 5$, and it times out if you increase the upper limit by any significant amount. -Therefore, my question is: - -Can it be shown that, for all integers $n > 2$, the equation $a^n + b^n = c^n \pm 1$ has at least 1 non-trivial solution, for $a, b, c \ge 0$? -If not, does expanding the range for $a, b, c$ to $\mathbb{Z}$ affect or change this? - -REPLY [5 votes]: In a message "A conjecture related to Fermat's Last Theorem" sent to Number Theory List on Sep. 26, 2015, I wrote the following: -In 1936 K. Mahler discovered that -$$(9t^3+1)^3 + (9t^4)^3 - (9t^4+3t)^3 = 1.$$ -Clearly, -$$|1^n+1^n-2^n| = 2^n-2\ \mbox{for every}\ n = 4,5,6,\ldots$$ -and -$$13^5+16^5-17^5 = 371293+1048576-1419857 = 12 < 2^5-2.$$ -Here I report my following conjecture which can be viewed as a -further refinement of Fermat's Last Theorem. -CONJECTURE (Sept. 24-25, 2015). -(i) For any integers $n > 3$ and $x,y,z > 0$ with $\{x,y\}\not= -\{1,z\}$, we have -$$|x^n+y^n-z^n|\ge2^n-2,$$ -unless $n = 5$, $\{x,y\} = \{13,16\}$ and $z = 17$. -(ii) For any integers $n > 3$ and $x,y,z > 0$ with $z\not\in\{x,y\}$, -there is a prime $p$ with -$$x^n+y^n < p < z^n\ \ \mbox{or}\ \ z^n < p < x^n+y^n, $$ -unless $n = 5$, $\{x,y\} = \{13,16\}$ and $z = 17$. -(iii) For any integers $n > 3$, $x > y \ge0$ and $z > 0$ with $x\not=z$, there always exists a prime $p$ with -$$x^n-y^n < p < z^n\ \ \mbox{or}\ \ z^n < p < x^n-y^n. $$ -I have checked this new conjecture via Mathematica. For example, I -have verified part (i) of the conjecture for $n = 4,\ldots,10$ and $x,y,z=1,\ldots,1700$.<|endoftext|> -TITLE: Grothendieck group of the category of boundary conditions of topological field theory -QUESTION [8 upvotes]: In this paper also the journal front page, eq. 2.14, it introduces -the Grothendieck group $K^0$ of the category of boundary conditions of topological field theory. - - -My question is that - -what exactly is the Grothendieck group $K^0$ of the category of boundary conditions of topological field theory really mean? -Does this say that the boundary conditions of topological field theory can be related or classified by a Grothendieck group? Is this an abelian group or nonabelian group? ($K^0(BCondμ(M^{d−1}) )=?$) -Should the boundary conditions of topological field theory classified by some bimodule of certain modular tensor category? how is this related to a Grothendieck group? - -REPLY [12 votes]: To understand the possible spaces of boundary conditions for a TQFT, it is helpful to start in highest dimension. -Suppose you have a $(d+1)$-dimensional nonanomalous TQFT $\mathcal Q$. (The anomalous story is a bit more complicated, because it means that lots of things you might want to write down are only well-defined up to phase factors.) Given a $d$-manifold $M^d$, it makes sense to ask about boundary conditions placed on $M$. Physically, you imagine a cylinder $M \times \mathbb{R}_{\geq 0}$, and ask that the "bulk" $M \times \mathbb{R}_{>0}$ be flooded with your TQFT $\mathcal Q$, and then you try to write down some valid rules for the "boundary values of fields" (at least if your TQFT does come to you with a description in terms of fluctuating quantum fields). -There are a few things you could do: - -You could impose some local conditions on the values of the fields at the boundary. If you think in terms of classical field theory for a moment, then the "bulk" field theory is some $(d+1)$-dimensional PDE that the fields need to solve (this PDE is called the equations of motion EOM), and you could demand that the restrictions of the fields (and their normal derivatives) satisfy some additional $d$-dimensional PDE. You could add some new "boundary fields" that couple to the bulk. - -A typical example of this is in the Chern-Simons / WZW -correspondence. Classically, the bulk EOM say that you have a -$G$-bundle with flat connection. This is "topological": the flat connections are flat in any metric. The WZW boundary condition includes a field which is a section of this bundle, and the boundary EOM say that it is a holomorphic section. Holomorphicity is something that you can check locally: it is a PDE on the boundary. - -You could also impose some nonlocal conditions on the boundary. For example, you might ask that some field at some point in $M^d$ have value related in some way to the value of some field at some other far-away point in $M^d$. A more physical example occurs when you ask for the boundary values of a field to satisfy an integral equation, rather than a (partial) differential equation. For example, when $d+1=2$, a typical field theory might ask that the field be, say, a holomorphic function $f$. Given a circle $M^1 = S^1$, an interesting boundary value is to ask that the field extend holomorphically over the disk $D^2 \supset S^1$. Of course, this happens exactly when the negative Fourier modes of $f$ vanish. These Fourier modes are something like $\oint_{S^1} z^{-n} f(z) \mathrm{d}z$. - - -I bring up this distinction to emphasize that when people talk about the collection of "boundary conditions on $M$", they implicitly mean that nonlocal boundary conditions are allowed. -What type of thing is the collection of boundary conditions on $M^d$? Well, because the boundary conditions are allowed to be nonlocal, I can just as well work with the $(0+1)$-dimensional QFT produced by compactifying along $M$, i.e. $\int_{M}\mathcal{Q} := \mathcal{Q}(M \times -)$. But then we're just discussing boundary conditions for a (topological) quantum mechanics model, since that's what $(0+1)$-dimensional QFT is. A "boundary condition" is then just the same as a state in the model: it is a thing that you can prepare at time $0$ and then allow to propagate forward in time. -So the punchline is that the collection of boundary conditions on $M^d$ is a Hilbert space: specifically, it is the Hilbert space that $\mathcal{Q}$ assigns to $M^d$. -Ok, with that understood, now we can ask about a $(d-1)$-manifold $M^{d-1}$. If I tell you such a manifold, then you could look at the $d$-manifold $M^{d-1} \times \mathbb{R}$, if you want, and ask about boundary conditions that you can place on $M^{d-1} \times \mathbb{R}$. Well, that's not a very good question because of the noncompactness of $\mathbb{R}$ (you end up not defining elliptic operators). Rather, you could ask about boundary conditions placeable on $M^{d-1} \times \mathbb{R}$ which are local in the $\mathbb{R}$-direction. In other words, I will allow you to impose integral equations, but only if the integral is over something that only points in the $M^{d-1}$ direction. I also demand that the boundary conditions be invariant under translation in the $\mathbb{R}$-direction. I might forget to say "translation invariant" below, but I always want it. -Just as before, to understand the structure of this collection, you can compactify along $M$, and consider the now-$(1+1)$-dimensional (T)QFT $\int_{M^{d-1}}\mathcal{Q} := \mathcal{Q}(M^{d-1}\times-)$. -All I want to point out if that this collection is naturally a category. Specifically, suppose you have two local-in-the-$\mathbb{R}$-direction boundary conditions $X$ and $Y$ on $M^{d-1} \times \mathbb{R}$. Place one of them just on $M^{d-1} \times \mathbb{R}_{<0}$ and the other just on $M^{d-1} \times \mathbb{R}_{>0}$. Now you can ask to extend this configuration to something on all of $M^{d-1} \times \mathbb{R}$. Again I will allow you to write down expressions which integrate over $M^{d-1}$, if you want. The extension then looks like an "interface" between the boundary conditions. Anyway, this set of interfaces between $X$ and $Y$ will be $\hom(X,Y)$. If you place an interface at $M \times\{0\}$ and another one at $M \times \{1\}$, and then compactify out that interval $M \times [0,1] \to M$, then you get a (manifestly associative) composition of interfaces. -[Actually, the compactification is subtle, because the interface might secretly depend on a metric in the $\mathbb{R}$-direction. You can do it if the boundary conditions $X,Y$ are "topological in the $\mathbb{R}$-direction", but in general it is harder.] -Ok, fine, so this is why you have a "category of boundary conditions" assigned to $M^{d-1}$. And, just like we had Hilbert spaces in dimension $d$, we have actually linear categories in dimension $d-1$. This linearity is essentially the ability to perform "superpositions" in quantum theory. (In unitary quantum theory, it is in fact a Hilbert category.) -Now I can get to your actual question. First, of course, any linear category has an abelian group as its $K^0$. Let's understand it. Well, if you do have a local-in-the-$\mathbb{R}$-direction boundary condition on $M^{d-1} \times \mathbb{R}$, then, because it is local and translation invariant, it also defines a (partially local) boundary condition on $M^{d-1} \times S^1$. This assignment takes direct sums to sums. -In other words, there is a canonical map of abelian groups $K^0(\mathcal{Q}(M^{d-1})) \to \mathcal{Q}(M^{d-1} \times S^1)$. The codomain is the Grothendieck group of the category of local-and-translation-invariant-in-the-$\mathbb{R}$-direction boundary conditions on $M^{d-1} \times \mathbb{R}$, and the domain is the Hilbert space of boundary conditions on $M^{d-1} \times S^1$. -In many cases, this map induces an isomorphism $K^0(\mathcal{Q}(M^{d-1})) \otimes \mathbb{C} \cong \mathcal{Q}(M^{d-1} \times S^1)$. -This is true, for example, in any (extended) TQFT valued in any member of the Bestiary of 2-vector spaces. You can prove this simply by compactifying down to a $(1+1)$-dimensional TQFT, and then proving a universal statement for $(1+1)$d TQFTs. Cases where it fails have some pathology (nonunitarity or noncompactness or...) that make them less "physical".<|endoftext|> -TITLE: PDF readers for presenting Math online -QUESTION [14 upvotes]: In the current situation it seems especially important to be able to present your mathematical results online in a way that your audience does not fall asleep in front of their screens. But I am struggling to find suitable software one could use to enhance the usual beamer LaTeX presentations with some handwriting/drawing. -So I am searching for the following features: - -Allows to present PDFs, as generated by the beamer Latex package -Compatibility with usual video conference software (should be easy, just share your screen) -Easy to use and easy to access handwriting tools (pen, colors, laserpointer, etc.) -Drawing should stay on the slide you drew them on, so you can go back to them and save them after your talk -Program and Toolbar should not be visible to the audience (so maybe campatibility with 2 screens would be necessary here) -Additionally: Shows next slide, time, additional notes, ... (only for the speaker) - -What is the best tool you know for this purpose? -PS: I am not restricted to any operating system. - -REPLY [4 votes]: If your presentation is based on a Beamer file, I recommend PdfPc. It supports a hidden dashboard for the speaker, showing previous/next slides and some annotation and pointing tools. -Moreover, it allows you to jump among sections without having to run slide by slide - which is particularly cumbersome when a single frame corresponds to multiple pages on the pdf file. -In practice, when using PdfPc you then would only share the presentation window (but not the speaker dashboard) on your video conferencing software.<|endoftext|> -TITLE: Bounding the fourier coefficient field -QUESTION [14 upvotes]: Let $f = \sum_n a_n q^n \in S_2(\Gamma_0(N))$ be a normalized, non-CM, newform of weight $N \geq 1$ and level $2$. Let $K_f := {\mathbb Q}(\{a_n\}) \subset {\mathbb C}$ be the number field generated by its Fourier coefficients. -I was wondering if there is a bound known for the discriminant $\Delta_{K_f}$ of $K_f$, in terms of $N$ and (possibly) the dimension of the field $K_f$? -It is okay if this would be a terrible bound. I just need "any" bound. - -REPLY [7 votes]: Here's an approach for a really bad bound. -Updated below based on comments and further reflection, but still giving a very bad bound. - -First, $K_f$ is contained in the field generated by the eigenvalues of the Hecke operators $T_1$, ..., $T_m$, where $m$ is given by Sturm's bound. - -Using Deligne's bounds on the size of the Hecke eigenvalues of $T_n$, one can bound the discriminant of the number field $K_n$ obtained by adjoining all eigenvalues of $T_n$ to $\mathbb Q$. - -Now you can use those discriminant bounds to bound the discriminant of the compositum number field $K = K_2 K_3 \dots K_m$, which gives a bound on the discriminant of $K_f$. - - -This will give a bound that grows much faster than expontially in $N$.<|endoftext|> -TITLE: Diffeomorphisms of manifolds with boundary -QUESTION [5 upvotes]: I repeat this, which I posted in Math Stack, where it got some attention but no answer. -If two compact manifolds have diffeomorphic interiors and diffeomorphic boundaries, are they then diffeomorphic? Is it true for surfaces? Some context: there seems to exist an example by Barden and Mazur of a nontrivial cobordism between some oriented manifold $M$ and $-M$ whose interior is trivial (diffeo to $(0,1)\times M$). This would be a counterexample, but I cannot find the reference anywhere. - -REPLY [2 votes]: This is definitely true for surfaces (easy) and for 3-manifolds (harder). The argument for surfaces goes via classification: Two smooth surfaces with boundary are homeomorphic if and only if they are diffeomorphic. For a compact connected surface $S$ the full set of topological invariants is the triple: -$t(S)$= (orientability, Euler characteristic, number of boundary components) -It is an easy exercise to see that if $S, S'$ have homeomorphic interiors then $t(S)=t(S')$, hence, $S$ is homeomorphic to $S'$. -For 3-manifolds the proof is harder, see -C. Edwards, Concentricity in 3-manifolds. Trans. Amer. Math. Soc. 113 (1964) 406–423.<|endoftext|> -TITLE: How can the number of rational points depend on the choice of height function? -QUESTION [6 upvotes]: Let $V/\mathbb{Q}$ be a subvariety of $\mathbb{P}^n$. There are many plausible choices of height function, some differing only by constant factors: $\max |x_i|$ (for $(x_0,x_1,\dotsc,x_n)$, $\gcd(x_1,\dotsc,x_n)=1$, representing a given point $P$), or also $\sqrt{\sum_i |x_i|^2}$, etc. -Given a height function $H$, we can define $N_H(V;x)$ to be the number of points $P$ in $V(\mathbf{Q})$ with $H(P)\leq x$. For $H_1\ll H_2\ll H_1$, we clearly have $N_{H_1}(V;x)\ll O(1)^n N_{H_2}(V;x)$ and $N_{H_2}(V;x)\ll O(1)^n N_{H_1}(V;x)$. Could the behavior of $N_{H_1}(V;x)$ and $N_{H_2}(V;x)$ nevertheless be qualitatiely different? For example -- could it be that $N_{H_1}(V;x)$ has an asymptotic, whereas $N_{H_2}(V;x)$ does not (i.e. it oscillates)? - -REPLY [5 votes]: Surely this behaviour can never happen, but it will be near impossible to prove this. Conjectures of Manin and others predict that there is an asymptotic formula for these functions in many cases, and the shape of the asymptotic only depends on the choice of embedding and not the choice of norm. (Changing the norm just changes the leading constant in the asymptotic formula). -A nice trick however is that if you can prove an asymptotic for a "dense" set of norms, then you get an asymptotic for all norms via standard approximation arguments. In particular, e.g. it suffices to prove an asymptotic for all norms which are given by infinitely differentiable functions.<|endoftext|> -TITLE: Distance to a closed set. Is this result known? -QUESTION [5 upvotes]: Given a closed set $\varnothing\neq E\subset\mathbb{R}^n$, let $\operatorname{Unp}(E)$ be the set if points $x\in\mathbb{R}^n$ for which there is a unique point $y\in E$ nearest to $x$. -Clearly $E\subset\operatorname{Unp}(E)$. -On the other hand, it is known that the Lebesgue measure of the set $\mathbb{R}^n\setminus\operatorname{Unp}(E)$ equals zero, $\lvert\mathbb{R}^n\setminus\operatorname{Unp}(E)\rvert=0$ (cf. Set of points with a unique closest point in a compact set). -Erdős [1] proved a much stronger result: - -The set $\mathbb{R}^n\setminus\operatorname{Unp}(E)$ is contained in the sum of countably many surfaces of finite $(n-1)$-dimensional measure. - -This is a beautiful and I think, not very well known theorem. -However, from the results existing in the literature one can conclude: - -Theorem. -For any closed set $E\subset\mathbb{R}^n$ -there are countably many $C^2$-graphs $\{ G_j\}_{j=1}^\infty$ such that -$$ -\mathcal{H}^{n-1}\Bigl((\mathbb{R}^n\setminus\operatorname{Unp}(E))\setminus\bigcup_{j=1}^\infty G_j\Bigr)=0. -$$ - -Here $\mathcal{H}^{n-1}$ stands for the Hausdorff measure and -by a $C^2$-graph I mean the graph of a $C^2$ function $f:\mathbb{R}^{n-1}\to\mathbb{R}$. -$$ -\{x\in\mathbb{R}^n\mathrel: x_i=f(x_1,\dotsc,x_{i-1},x_{i+1},\dotsc,x_n)\}. -$$ -While the theorem can be concluded from what is in the literature, I was not able to find a straightforward reference to this statement. I think this result is of considerable interest, but most of people who would find it interesting would have difficulty to find it in the literature. - -Question. Do you know if this result (or something very similar) has been published anywhere? - -[1] P. Erdős, On the Hausdorff dimension of some sets in Euclidean space. -Bull. Amer. Math. Soc. 52 (1946), 107-109. - -REPLY [2 votes]: The following result can be found in -D. H. Fremlin, Skeletons and central sets, Proc. London Math. Soc. (3) 74 (1997), 701–720. -Let $\Omega$ be an open proper subset of $\mathbb{R}^n$, where $n > 1$. Its skeleton is -$$R:=\{ x \in \Omega: \text{ there are distinct } y, y'\in \mathbb{R}^n \setminus \Omega - \text{ such that } -\rho( x , y ) =\rho( x , y ) = \rho( x , \mathbb{R}^n \setminus \Omega)\},$$ -where we write $\rho$ for the Euclidean metric on $\mathbb{R}^n$. -Theorem 1G: Let $\Omega$ be an open proper subset of $\mathbb{R}^n$. Then its skeleton $R$ is the -union of a sequence of closed sets each of which is Lipschitz isomorphic to a subset -of $\mathbb{R}^{n-1}$. Consequently , the dimension of $R$ is at most $n-1$, whether we mean -inductive dimension , covering dimension , Hausdorff dimension or Minkowski -dimension.<|endoftext|> -TITLE: "Kronecker Product" for quasi-symmetric functions -QUESTION [5 upvotes]: Recall that the Kronecker product -$s_\lambda * s_\mu$ of two Schur functions $s_\lambda$ and $s_\mu$ is the symmetric function -whose expansion (in terms of Schur functions) is given by -\begin{equation} -\sum_{\nu \, \vdash \, n} g_{\lambda \mu}^\nu \, s_\nu -\end{equation} -where $\lambda$, $\mu$, and $\nu$ are partitions of $n$ and -$g_{\lambda \mu}^\nu$ is the Kronecker coefficient, which famously -counts the multiplicity of $V_\nu$ in the tensor product $V_\lambda \otimes V_\mu$ of the irreducible representations of the symmetric group $S_n$. -Switch now to the quasi-symmetric world: Given a composition $\alpha = (\alpha_1, \dots, \alpha_k)$ of $n$ let $L_\alpha$ be the fundamental quasi-symmetric functions defined by -\begin{equation} -L_\alpha = \sum x_{\ell_1} \cdots \,x_{\ell_k} -\end{equation} -where the sum is taken over all sequences -$1 \leq \ell_1 \leq \cdots \leq \ell_k$ such that $\ell_i < \ell_{i+1}$ whenever -$i = \alpha_1 + \cdots + \alpha_j$ for some $1 \leq j \leq k-1$. -The space of symmetric functions within the $\mathrm{QSym}_n :=\Bbb{Q}$-span of $\{ L_\alpha \, \big| \, \alpha \, \models \, n \}$ coincides with the -$\mathrm{Sym}_n:= \Bbb{Q}$-span of $\{ s_\lambda \, \big| \, \lambda \, \vdash \, n \}$ -the latter of which is endowed with the Kronecker $*$-product. -Question: -Can the Kronecker $*$-product on $\mathrm{Sym}_n$ be extended to all -of $\mathrm{QSym}_n$ so that there exist non-negative -integers $\tilde{g}_{\alpha,\beta}^{\, \gamma}$ for each -triple $\alpha$, $\beta$, $\gamma$ of compositions of $n$ -satisfying -\begin{equation} -L_\alpha * L_\beta = -\sum_{\gamma \, \models \, n} \tilde{g}_{\alpha,\beta}^{\, \gamma} -\, L_\gamma \quad \text{?} -\end{equation} -p.s. Covertly, I am asking whether or not there is some kind of tensor product structure (as in a symmetric tensor category) on the projective indecomposable representations of the 0-Hecke algebra $H_n(0)$. Any thoughts on that would also be appreciated. -thanks, ines. - -REPLY [2 votes]: One silly (or super wishful thinking approach) -is to use the formula -$$ -g_{\lambda \mu \nu} = \frac{1}{n!} \sum_{\sigma \in S_n} -\chi^{\lambda}(\sigma) -\chi^{\mu}(\sigma) -\chi^{\nu}(\sigma). -$$ -Perhaps some version of -$$ -g_{\alpha, \beta,\gamma} = \frac{1}{n!} \sum_{\tau \in S_n} -\chi^{\alpha}(\tau) -\chi^{\beta}(\tau) -\chi^{\gamma}(\tau) -$$ -where now $\chi^{\alpha}(\tau)$ are the coefficients which show up -when the quasisymmetric power sums are expanded in terms of -the Gessel quasisymmetric functions. -Here, I suppose that in the sum, one does not only consider -the cycle type, but decide to order the cycles by smallest element, -and then let $\alpha$ be the integer composition given by the cycle lengths in that order. -If the coefficients are non-negative by some miracle, then, well, you have something cool to start with. -Ballantine, Cristina; Daugherty, Zajj; Hicks, Angela; Mason, Sarah; Niese, Elizabeth, On quasisymmetric power sums, J. Comb. Theory, Ser. A 175, Article ID 105273, 36 p. (2020). ZBL1442.05241.<|endoftext|> -TITLE: Are $K(\pi_1,1)$ tangentially homotopy equivalent? -QUESTION [12 upvotes]: Is it known whether any two smooth, compact manifolds $X \simeq K(\pi_1,1) \simeq Y$ are tangentially homotopy equivalent, i.e. the pullback of the tangent bundle of $Y$ along some smooth homotopy equivalence $X \rightarrow Y$ is isomorphic to the tangent bundle of $X$? I suspect this may be difficult, it does not appear stronger or weaker than the Borel conjecture, because even if the Borel conjecture were true, we could have multiple smooth structures which are not tangentially homotopy equivalent. - -REPLY [14 votes]: I think the answer is no: there exists a pair of aspherical closed smooth manifolds which are homotopy equivalent but not tangentially homotopy equivalent. -Claim: Let $X$ be a smooth closed oriented 9-manifold such that $p_2(TX) = 0 \in H^8(X;\mathbb{Z}) = H_1(X;\mathbb{Z})$. For any $v \in H_1(X;\mathbb{Z})$ with $7 v = 0$, there exists a smooth manifold $Y$ and a PL homeomorphism $f: X \to Y$, such that $f^*(p_2(TY)) = v$. -If $v \neq 0$, there can then be no tangential homotopy equivalence $X \to Y$, since it would have to take $p_2(TY) \neq 0$ to $p_2(TX) = 0$. To get a concrete example we can take $X$ to be the product of $(S^1)^6$ and a closed aspherical 3-manifold with non-trivial 7-torsion in $H_1$. Even more concretely, the 3-manifold can be taken as the mapping torus of the diffeomorphism of $S^1 \times S^1$ corresponding to the matrix $\begin{bmatrix}1 & 7\\0 & 1\end{bmatrix}.$ -Proof of claim: The 7-torsion in $H^8(X;\mathbb{Z})$ agrees with the 7-torsion in $H^8(X;\mathbb{Z}_{(7)})$, and by smoothing theory it suffices to see that $(0,v)$ is in the image of the homomorphism $$[X,PL/O] \to [X,BO] \xrightarrow{(p_1,p_2)} H^4(X;\mathbb{Z}_{(7)}) \times H^8(X;\mathbb{Z}_{(7)}).$$ But the second map factors through an isomorphism from $[X,BO] \otimes \mathbb{Z}_{(7)}$, and in the domain we may therefore factor over $[X,PL/O] \otimes \mathbb{Z}_{(7)}$. But by the Kervaire-Milnor calculation of exotic spheres there is a map $PL/O \to K(\mathbb{Z}/7\mathbb{Z},7)$ inducing an isomorphism on homotopy groups in a large range (far beyond $9 = \dim(X)$) after tensoring with $\mathbb{Z}_{(7)}$. Furthermore, the connecting map $$H^7(X;\mathbb{Z}/7\mathbb{Z}) \xleftarrow{\cong} [X,PL/O] \otimes \mathbb{Z}_{(7)} \to [X,BO] \otimes \mathbb{Z}_{(7)} \xrightarrow{p_2} H^8(X;\mathbb{Z}_{(7)})$$ may be identified with the Bockstein homomorphism $\beta: H^7(X;\mathbb{Z}/7\mathbb{Z}) \to H^8(X;\mathbb{Z}_{(7)})$, which may in turn be identified with $\beta: H_2(X;\mathbb{Z}/7\mathbb{Z}) \to H_1(X;\mathbb{Z}_{(7)})$. But the image of that is precisely the kernel of multiplication by 7, i.e. the 7-torsion elements. $\Box$<|endoftext|> -TITLE: Examples of back of envelope calculations leading to good intuition? -QUESTION [57 upvotes]: Some time ago, I read about an "approximate approach" to the Stirling's formula in M.Sanjoy's Street Fighting Mathematics. In summary, the book used a integral estimation heuristic from spectroscopy -$$\int_{\mathbb{R_{\ge 0}}} f(x) dx \approx \max(f) * (\text{point where}\ \frac{1}{2} \max(f)\ \text{is achieved}) $$ -to estimate the Gamma function with $f(x) = f_t(x) = x^{t}e^{-x} $. This leads to the estimate -$$\Gamma(n) = \int_{\mathbb{R}_{\ge 0}} x^{n}e^{-x} dx \approx \sqrt{8 n} \left(\frac{n}{e}\right)^n$$ -which is an extremely good estimate (the "proportionality constant" $\sqrt{8}$ is correct to within 10% with correct order of growth.) This heuristic was very helpful in understanding the growth of the actual formula $\Gamma(n) \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n$. -I think approximations of this sort is useful because - -It gives a sense of what the answer "ought" to be. -When the approximation deviates from the actual answer, it's interesting to think about which part of the approximation failed. - -Another "back-of-envelope calculation" is the calculation for the Prime Number Theorem in Courant and Robbins, What is Mathematics? -My Question. I am looking for similar instances in mathematics where "back-of-envelope calculations" such as the above leading to good intuition in mathematics. -For the purpose of my question, let's require that the calculation addresses questions in pure mathematics (so, no physics, engineering, etc. since there seems to already be plenty of literature on this). -Edit: as per helpful feedback from Peter LeFanu Lumsdaine, I removed two requirements: "Does not require anything beyond, say undergraduate mathematics" and "Does not formalize into a rigorous proof." -Edit 2 (as per helpful discussion in the comments): part of what I am interested in is how people use various techniques to compute/approximate objects of interest. For instance, I think we can all agree that the use of integral approximation demonstrated above is quite creative (if not, a nonstandard way of approaching Stirling). In response to Meow, topological invariants for "similar" (homotopy equivalent, homeomomorphic, etc) mostly amounts to the "same sort" of argument, so I would count that as "one" approximation argument unless there is a particular example where the heuristic argument is highly nontrivial. - -REPLY [16 votes]: Minkowski theorem -The Poisson summation formula writes -$$\sum_{n \in \mathbb Z^n} \phi(n) = \sum_{n \in \mathbb Z^n} \widehat{\phi}(n)$$ -where $\hat{\phi}$ is the Fourier transform of $\phi$. Let's take $\phi = \mathbf 1_A$ the characteristic function of a set $A$. Roughtly, the outcome is expected to be -$$ -|A \cap \mathbb Z^n| = \sum_{x \in \mathbb Z^n} \mathbf{1}_A(x) = \sum_{x \in \mathbb Z^n} \widehat{\mathbf{1}_A}(x) \geqslant \widehat{\mathbf{1}_A}(0) = \mathrm{vol}(A), -$$ -and this would prove that there are two distinct points of the lattice $\mathbb Z^n$ in $A$ as soon as $\mathrm{vol}(A)>1$: this is the idea of Minkowski's theorem. Of course, $\mathbf 1_A$ is not an admissible function in the Poisson summation formula and this idea has to be massaged a bit. Doing so, we realize that we need to assume some nice properties on $A$ (viz. convex and symmetric) and that the volume has to be a bit larger (viz. $2^n$). -Trace formulas -More generally, trace formulas enjoy a lot from these heuristics. They are distributional equalities of the form -$$\sum_{\lambda \in \mathrm{spec}} \phi(\lambda) = \sum_{\lambda \in \mathrm{geom}} \widehat{\phi}(\lambda)$$ -where the left sum runs over "spectral" terms (e.g. automorphic forms, eigenvalues of the Laplacian), the right sum over "geometric terms" (e.g. geodesics, conjugacy classes) and $\hat{\phi}$ is an explicitly defined integral transform of $\phi$. They are in particular used to establish results on average, and the (illegal) use of characteristic functions on one side often gives you the right main term when estimating the trivial term on the other side (similarly to the $0 \in \mathbb Z^n$ above). Two examples on a compact surface $S$: - -if you take the characteristic function on the spectral side, you guess the Weyl law counting the eigenvalues of the laplacian $\Delta$: -$$|\{\lambda \in \mathrm{spec}(\Delta) \ : \ |\lambda| \leqslant X\}| \sim \frac{\mathrm{vol}(S)}{4\pi}X$$ -if you take the characteristic function on the geometric side, you guess the prime geodesic theorem counting the closed geodesics of bounded length $\ell$ on $S$: -$$|\{\gamma \text{ geodesic on } S \ : \ \exp(\ell(\gamma)) \leqslant X\}| \sim \mathrm{li}(X)$$ - -The extent of these ideas in geometry, number theory, automorphic forms, spectral theory, etc. is impressive, and these back of envelope calculations are a strong 'and reliable guide. (and, of course, turning these heuristics into proofs are another matter)<|endoftext|> -TITLE: A better version of Weyl's Law or uniform estimates of Laplacian higher eigenvalues -QUESTION [5 upvotes]: Let $(M^n,g)$ be a closed $n$ dimensional Riemannian manifold with $\mathrm{Ric}_g\ge -K$, $(K\ge 0)$. Weyl's law(along with Karamata Tauberian Theorem) asserts that the eigenvalue $\lambda_i$ of $-\Delta$ has the following asymptotic behavior -$$ -\lambda_i \sim c_n\left(\frac{i}{\mathrm{Vol}_g(M^n)}\right)^{2/n}\quad (i\to \infty) -$$ -Is there a finer version of Weyl's law so that for $i$ large enough, the following uniform bound is true? -$$ -\left|\lambda_i-c_n\left(\frac{i}{\mathrm{Vol}_g(M^n)}\right)^{2/n}\right|\le C(K,n,D,\mathrm{Vol}_g(M^n)) -$$ -Here $C(K,n,D, \mathrm{Vol}_g(M^n))$ is a constant depending on the Ricci lower bound $-K$, dimension $n$, diameter $D$ of $M^n$ and the volume $\mathrm{Vol}_g(M^n))$, the point is that it does not depend on $i$. If the conditions given at the beginning are not sufficient, what are the other conditions one can put on $M^n$ for the uniform estimates to hold (or any example to show it never holds)? -Thanks in advance. - -REPLY [9 votes]: It seems unlikely that such a bound holds except in very special cases. For instance, it fails for round spheres, which have very large multiplicity of eigenvalues. In fact, for spheres the eigenvalue counting function has jumps of order $\lambda^{n-1}$. -On a round 2-sphere, for a constant eigenvalue $\lambda $, the second term in your estimate will vary on order $\lambda$, which makes a uniform bound on the left hand side impossible. -For generic metrics where closed geodesics are "sparse," it's possible to strengthen the standard Weyl law, but the refinement isn't strong enough to give uniform bounds on the difference between eigenvalues and the asymptotic formula from Weyl's law. I believe the state of the art refinement of Weyl's law is due to Canzani and Galkowski, and their paper is a good reference. The relevant result is Theorem 7 on page 12.<|endoftext|> -TITLE: Picard group of connected linear algebraic group -QUESTION [5 upvotes]: Here's a statement: -Suppose $G$ is a connected linear algebraic group over a field $k$, then $Pic(G)$ is a finite group. -I know this is true when $k=\mathbb{C}$. My question is does this true for abitrary field $k$? If not, how about furthermore when $G$ is smooth or even reductive? Is there any reference? -Thanks for any help. - -REPLY [11 votes]: $\DeclareMathOperator\Pic{Pic}$The statement is false over most imperfect fields, even for smooth affine group schemes. -In particular, it is false over any separably closed imperfect field $k$. I will give an example over imperfect fields of characteristic at least $3$, but it is not difficult to adapt it to work in char. 2 as well. -The statement is, however, correct over fields of characteristic $0$ or if $G$ is reductive. This is proven in Prop. 4.5 of Knop, Kraft, Luna, Vust - Local properties of algebraic group actions (DOI). The proof is written under the assumption that $\operatorname{char} k=0$ but it can be adapted to char. $p$ in the case of reductive $G$. -Example: Let $k$ be a separably closed imperfect field of characteristic $p>2$, and $U=\operatorname{Spec} k[x,y]/(y^p-x-ax^p)$ for some $a\in k \setminus k^p$. -Remark: $U$ is a naturally a subgroup of $\mathbf{G}_a^2$. This is a so-called $k$-wound form of $\mathbf{G}_a$. This, in particular, means that $U$ is isomorphic to $\mathbf{G}_a$ over the algebraic closure of $k$, i.e. $U_{\bar{k}} \cong \mathbf{G}_{a, \bar{k}}$. -Claim: $\Pic(U)$ is infinite. -One easily checks that its Zariski closure $C\mathrel{:=}V(Y^p-XZ^{p-1}-aX^p)$ inside $\mathbf{P}^2_k$ is a regular curve of genus $\frac{(p-1)(p-2)}{2}>0$ such that $C\setminus U$ is a point $P$ with residue field $k(a^{1/p})$. So we have an exact sequence -$$ -\mathbf{Z}P \to \Pic(C) \to \Pic(U) \to 0 -$$ -that induces an inclusion $\Pic^0(C) \to \Pic(U)$. So it suffices to show that $\Pic^0(C)$ is infinite. Now the Picard functor $\Pic^0_{C/k}$ is representable by a $k$-smooth group scheme of dimension $\frac{(p-1)(p-2)}{2}$. Therefore, -$$ -\Pic^0(C)=\mathbf{Pic}^0_{C/k}(k) -$$ -is infinite as $\dim \mathbf{Pic}^0_{C/k}>0$ and $k$ is separably closed.<|endoftext|> -TITLE: The diversity of Riemannian metrics adapted to a given (1 dimensional) foliation, A Krein Millman view point -QUESTION [5 upvotes]: Let $X$ be a Kronecker vector field on the two dimensional torus $\mathbb{T}^2$. Let $K$ be the space of all 1- forms $\alpha$ of class $C^1$ on $\mathbb{T}^2$ which satisfy $d\alpha=0,\;\alpha(X)=1$. -Then $K$ is a convex closed subset of all $C^1$ 1-forms on $\mathbb{T}^2$. - -Is $K$ a compact subset of the space of 1-forms with respect to $C^1$ topology? If the answer is affirmative. according to the Krein Millman theorem, what is a presise description of its extreme points of $K$? - - -Does the topological structure of $K$ depends on chosing the vector field $X$ tangent to our initial Kronecker foliation of torus? Does the topological structure of $K$ depend on the slope of our Kronecker foliation? - -Motivation: -A motivation for this question is the following: -In this post and some other related linked posts we try to find a Riemannian metric compatible to orbits of a non vanishing vector fields. Choosing various metrics enable us to have different curvatuare functions. Possessing an appropriate curvature function is very essential for appllying the Gauss Bonnet theorem to the problem of limit cycles of vctor fields.(For counting them as closed geodesics). So this situation leads us to think about the diversity of closed differential 1-forms $\alpha$ with $\alpha(X)=1$. Under these conditions, in particular the propery of closed convexity of this set $K$. one is tempt to be curious about the presice description of possible extrem points of $K$. -Remark: For generalization of this question to $n$ dimensional space we should consider the spaceof all 1-form $\alpha$ with $i_X d\alpha=0,\;\alpha(X)=1$. - -REPLY [5 votes]: I don't think it is compact, but perhaps I miss a normalization condition? -Let $X=\partial_x+a\partial_y$, with $a$ irrational (doesn't actually matter for the following). Let $\alpha\in C$ (e.g. $\alpha=dx$) and let $\omega_\lambda=\lambda(a dx-dy)$, for $\lambda\in \mathbb R$. As $X$ lives in the kernel of $\omega_\lambda$, and $d\omega_\lambda=0$, then $\alpha+\omega_\lambda\in C$.<|endoftext|> -TITLE: An alternative to continued fraction and applications -QUESTION [8 upvotes]: This post is inspired by the Numberphile video 2.920050977316, advertising the paper A Prime-Representing Constant by Dylan Fridman, Juli Garbulsky, Bruno Glecer, James Grime and Massi Tron Florentin, involving an alternative to continued fractions. The goal of this post is to discuss the relevance of this alternative by asking whether it can prove the irrationality of numbers for which it was unknown before. -Let us first recall the notion of continued fraction. For a given number $\alpha>0$, consider the recurrence relation $u_0 = \alpha$ and $$ u_{n+1} = \begin{cases} -(u_n - \lfloor u_n \rfloor)^{-1} & \text{ if } u_n \neq \lfloor u_n \rfloor \\ - 0 & \text{ otherwise } -\end{cases}$$ -and let $a_n = \lfloor u_n \rfloor $. Then $$\alpha = a_0 + \frac{1}{a_1+\frac{1}{a_2+\frac{1}{\ddots}}}$$ denoted $[a_0; a_1, a_2, \dotsc]$. It is rational if and only if $a_n = 0$ for $n$ large enough. So it is a great tool to prove the irrationality of some numbers. For example, $\phi = [1;1,1, \dotsc]$ is the golden ratio, because $(\phi-1)^{-1}=\phi$. -Let $p_n$ be the $n$th prime, then we can consider the irrational number $[p_1;p_2,p_3, \dots] = 2.31303673643\ldots$ (A064442), which then compresses the data of all the prime numbers, in a more natural and efficient way than just taking $2.\mathbf{3}5\mathbf{7}11\mathbf{13}17\mathbf{19}\ldots$. -The paper mentioned above provides another interesting way to compress the primes numbers, which uses Bertrand’s postulate, i.e. $p_n < p_{n+1} < 2p_n$. This way is a kind of alternative to continued fractions. For a given number $\beta \ge 2$, consider the recurrence relation $u_1=\beta$ and $$u_{n+1} = \lfloor u_n \rfloor (u_n - \lfloor u_n \rfloor + 1).$$ Let $a_n= \lfloor u_n \rfloor $. Then $a_n \le a_{n+1} < 2a_n$ and the mentioned paper proves that $$\beta = \sum_{n=1}^{\infty}\frac{a_{n+1}-1}{\prod_{i=1}^{n-1}a_i}$$ denoted, let’s say, $(a_1,a_2,a_3, \dots )$. -By the mentioned paper: -Theorem 1: Let $(a_n)$ be a sequence of positive integers such that: - -$a_n < a_{n+1} < 2a_n$, -$\frac{a_{n+1}}{a_n} \to 1$ - -then $\beta := (a_1,a_2,a_3, \dots ) := \sum_{n=1}^{\infty}\frac{a_{n+1}-1}{\prod_{i=1}^{n-1}a_i}$ is irrational. -It follows that the number $(p_1,p_2,p_3,\dots) = 2.920050977316\ldots$ is irrational. -Question: Can Theorem 1 be proved by some previously known methods? -Remark: The first point of Theorem 1 can be relaxed to $a_n \le a_{n+1} < 2a_n$, when $(a_n)$ is not eventually-constant. -For a given non-constant polynomial $P \in \mathbb{Z}[X]$ with a positive leading term and $P(n) \neq 0$ for all $n \in \mathbb{N}_{\ge 1}$, consider $a_n=P(n)$. Then it is easy to deduce from Theorem 1 that the number $e_P\mathrel{:=}(a_1,a_2, \dotsc )$ is irrational. For example, take $P(X)=X^k$, with $k \in \mathbb{N}_{\ge 1}$, then $$e_k:= \sum_{n=1}^{\infty} \frac{(n+1)^k-1}{n!^k}$$ is irrational. Note that $e_1 = e$ is Euler’s number. -The following result applies for an alternative proof of the irrationality of $e_k$ for all $k$, and of $e_P$ for many $P$ (not all), but not for $(p_1,p_2,p_3, \dots)$ -Theorem 2: Let $(a_n)$ be a sequence of positive integers such that: - -$a_n \le a_{n+1} < 2a_n$, -$\forall k \in \mathbb{N}_{\ge 1}$, $\exists m$ such that $k$ divides $a_m$, - -then $\beta := (a_1,a_2,a_3, \dots ) := \sum_{n=1}^{\infty}\frac{a_{n+1}-1}{\prod_{i=1}^{n-1}a_i}$ is irrational. -proof: Assume that $\beta = \frac{p}{q}$. By assumption, there is $m$ -such that $q$ divides $a_m$. By the mentioned paper, if $u_1=\beta$ and $u_{n+1} = \lfloor u_n \rfloor (u_n - \lfloor u_n \rfloor + 1)$, then -$a_n= \lfloor u_n \rfloor $. It is easy to see that $u_n$ can always be written with a denominator equal to $q$ (possibly not simplified). It follows that $u_{m+1}=a_m(u_m-a_m+1)$ and that $a_m u_m$ is an integer. So $u_{m+1}$ is an integer. It follows that for all $n>m$ then $u_n=u_{m+1}$, and so $a_n=a_{m+1}$. But the second point of Theorem 2 implies that $a_n \to \infty$, contradiction. $\square$ -The following example will show that the condition $\frac{a_{n+1}}{a_n} \to 1$ is not necessary for the irrationality. -Consider $a_n=\lfloor \frac{3^n}{2^n} \rfloor + r_n$, with $0 \le r_n < n$ such that $n$ divides $a_n$. Adjust the sequence for $n$ small so that the first point of Theorem 2 holds. Then $\beta$ is irrational whereas $\frac{a_{n+1}}{a_n} \to \frac{3}{2} \neq 1$. -Bonus question: What is a necessary and sufficent condition for irrationality? -Joel Moreira suggested in this comment that it may be rational if and only if $(a_n)$ is eventually-constant. See the new post Do these rational sequences always reach an integer? dedicated to this question. -FYI, it is easy to compute that $$\pi = (3, 3, 4, 5, 5, 7, 10, 10, 13, 17, 31, 35, 67, 123, 223, 305, 414, 822, 1550, 2224, ...) $$ - -REPLY [3 votes]: I am sorry If the comment is misleading, and welcome to point out any mistakes in the following proof. This is a clarification of the previous comment. -And this is only a proof of the irrationality of $e_k$. -And the proof strategy is an imitation of Fourier's proof of the irrationality of Euler's number $e$. - -if $\forall n=\mathbb{N}^{*} \quad n$, $n$ suffice large, -$$ -\left(n!\right) \cdot a \notin \mathbb{Z} \quad \text { then } a \notin \mathbb{Q} \hspace{1cm}(1) -$$ -WLOG, in the following calculation we don't distinguish $x,y$ if $x-y\in \mathbb{Z}$. And we write $x=y+\mathbb{Z}$ iff $x-y\in \mathbb{Z}$. -$\begin{aligned} m ! e_{k} +\mathbb{Z}&=\sum_{n \geq m+1} \frac{(n+1)^{k}-1}{(m+1) \cdots(n-1) n)^{k}}+\mathbb{Z} \\ &=\sum_{n \geqslant m+2} \frac{(n+1)^{k}-1}{((n-1) \cdots(n-1) n)^{k}}+\frac{(m+2)^{k}-1}{(m+1)^{k}}+\mathbb{Z}\\ -&=\sum_{n \geq m+2} \frac{(n+1)^{k}-1}{((m+1) \cdots(n-1) n)^{k}}+\sum_{i=1}^{k-1} \frac{C_{k}^{i} \cdot(m+1)^{i}}{(m+1)^{k}}+1 +\mathbb{Z}\\ -&=\sum_{n \geqslant m+2} \frac{(n+1)^{k}-1}{( m+1) \cdots(n-1) n)^{k}}+\sum_{i=1}^{k-1} \frac{C_{k}^{i}}{(m+1)^{i}}+\mathbb{Z}\hspace{1cm}(*) -\end{aligned}$ -In fact in $(*)$ we have $\sum_{n \geqslant m+2} \frac{(n+1)^{k}-1}{((m+1) \cdots(n-1) n)^{k}}= O(\frac{1}{m^{k}})$, $\sum_{i=1}^{k-1} \frac{C_{k}^{i}}{(m+1)^{i}}=O(\frac{1}{m})$. -Now take $m$ suffice large, in fact $m=10000\cdot k^{100}$ is ok, -then -$$0< \sum_{n \geqslant m+2} \frac{(n+1)^{k}-1}{((m+1) \cdots(n-1) n)^{k}}+\sum_{i=1}^{k-1} \frac{C_{k}^{i}}{(m+1)^{i}}< 1$$ -So $(*)\neq \mathbb{Z}$, so $(1)$ is true, $ e_{k}$ is not rational.<|endoftext|> -TITLE: Primes in arithmetic progression $a \pmod q$ -QUESTION [5 upvotes]: Can we prove the "Bertrand postulate" for primes $a \pmod q$, namely: there is always a prime number $p\equiv a \pmod q$ betwen $nq$ and $nq^2$ for every $n>0$ and $(a,q)=1$. (This would mean that between $nq$ and $nq^2$ there are always at least $\varphi(q)$ primes, each belonging to a different residue class modulo $q$.) - -REPLY [5 votes]: This question appears to have been addressed by P. Moree in this paper. In the notation of that paper, he defines -$$\displaystyle B_m(z,d) = \liminf \{c : \forall x \geq c, (x,zx) \text{ contains at least } m \text{ primes} $$ -$$\displaystyle \text{ congruent to } a \pmod{d} \text{ for all } \gcd(a,d) = 1\}.$$ -In particular, he proves that provided the quantity $\sigma(d)$ defined to be -$$\displaystyle \sigma(d) = \sum_{\substack{p \leq d \\ p \nmid d}} \frac{1}{p}$$ -is strictly less than one, then one can obtain an explicit estimate for $B_m(z,d)$ whenever -$$\displaystyle z > \frac{d}{(d-1)(1 - \sigma(d))}.$$ -His methods generalize an elementary argument of Erdos (who gave an elementary proof of the original Bertrand's postulate), and so do not run into the problems that one encounters when attempting to use the known analytic theorems to attack the problem (i.e., possible existence of Siegel zeroes, etc.). Unfortunately, by Mertens' theorem the hypothesis that $\sigma(d) < 1$ is extremely strict: indeed, Moree shows that $\sigma(d) < 1$ implies that $d \leq 840$. -In principle, one can approach the problem by obtaining a reasonably explicit lower bound for the number of primes in an arithmetic progression, up to some bound $x$. For each pair $(a,q)$ with $\gcd(a,q) = 1$ define $N(a,q)$ to be a positive number with the property that whenever $x > N(a,q)$ we have -$$\pi(x; a,q) > \frac{2x}{3 \phi(q) \log x},$$ -say. Then for $q \geq 2$ and $nq > N(a,q)$ there will always be a prime congruent to $a \pmod{q}$ in the interval $(nq, nq^2)$, since by the Brun-Titchmarsh theorem we have -$$\displaystyle \pi(x; a,q) \leq \frac{2x}{\phi(q) \log (x/q)}$$ -for all $x \geq q \geq 2$. In particular, -$$\displaystyle \pi(nq^2; a, q) - \pi(nq; a, q) > \frac{2nq^2}{3\phi(q) \log(nq^2)} - \frac{2nq}{\phi(q) \log(n)}$$ -$$ = \frac{2nq}{3\phi(q)} \left(\frac{(2q -3) \log n - 6 \log q}{\log n(\log n + \log q^2)} \right)$$ -which is positive. Calculating $N(a,q)$ however seems to be a very hard problem.<|endoftext|> -TITLE: Maximum density of a set without a fixed pattern -QUESTION [8 upvotes]: Consider a finite set $S$ of nonnegative integers. -What is the maximum natural density of an infinite subset of $\mathbb{Z}$ which does not contain any translation of $S$? -Of course, this will depend on $S$, but maybe there is a simple algorithm or characterization. -I am also interested about the same question in $\mathbb{Z}^k$. -Have the above questions been researched in any form? I didn't come up with a search query which returns anything. - -REPLY [5 votes]: The question is equivalent to finding the minimum density of a covering of $\mathbb{Z}$ by translations of $-S$. This problem has been studied for the integers and also for other groups; see for example -Wolfgang M. Schmidt and David M. Tuller, Covering and packing in $\mathbb{Z}^n$ and $\mathbb{R}^n$, -http://dx.doi.org/10.1007%2Fs00605-009-0099-x -Béla Bollobás, Svante Janson and Oliver Riordan, On covering by translates of a set, -https://doi.org/10.1002/rsa.20346<|endoftext|> -TITLE: $\pi_{2p^kn - 1}(P^{2n+1}(p^r))$ contains a $\mathbb{Z}_{p^{r+1}}$ summand -QUESTION [10 upvotes]: I am reading Neisendorfer's paper Samelson products and exponents of homotopy groups and related papers. I am stuck on theorem 14.1 on page 21, which says that there exists a $\mathbb{Z}_{p^{r+1}}$ summand in $\pi_{2p^kn - 1}(P^{2n+1}(p^r))$ for any $k \geq 1$ where $p > 3$ is a prime number and $P^{2n+1}(p^r) = D^{2n+1} \cup_{p^r} S^{2n}$. -The overall idea of the proof is to show that there exists an element in position $2p^kn - 2$ in the mod $p$ homotopy Bockstein spectral sequence for $\Omega P^{2n+1}(p^r)$ such that survives-until and then dies-in the $(r + 1)^\text{th}$ page. The first step is to find such an element and to show that it survives. -The method consists in comparing the mod $p$ homotopy Bockstein ss $E_\pi^*$ with the mod $p$ homology Bockstein ss $E_H^*$ via the mod $p$ Hurewicz map $\phi$. Both the spectral sequences are differential graded Lie algebras via, respectively, the Samelson product and the commutator of the Pontryagin product; moreover the Hurewicz map is a morphism of differential graded Lie algebras. -The mod $p$ homology Bockstein ss is simple. To see this, first notice that $H^*(P^{2n}(p^r); \mathbb{Z}) \cong \mathbb{Z}_{p^r}\langle \alpha \rangle$ where $\deg \alpha = 2n$, hence $H_*(P^{2n}(p^r); \mathbb{Z}) \cong \mathbb{Z}_{p^r}\langle\beta\rangle$ where $\deg \beta = 2n - 1$ and finally $H_*(P^{2n}(p^r); \mathbb{Z}_p) \cong \mathbb{Z}_p\langle u, v \rangle$ where $\deg u = 2n - 1$ and $\deg v = 2n$. Then we have $E^1_H \cong H_*(\Omega P^{2n+1}(p^r); \mathbb{Z}_p) \cong H_*(\Omega\Sigma P^{2n}(p^r); \mathbb{Z}_p) \cong \mathbb{Z}_p^\text{Alg} \langle u, v\rangle$ by the Bott-Samelson theorem. Moreover $\beta^s_H = 0$ for $s < r$ and $\beta^r_Hv = u$, so $E^1_H = \cdots = E^r_H$ and $E^{r+1}_H = \cdots = E^\infty_H \cong \mathbb{Z}_p$. -The mod $p$ homotopy Bockstein ss is not that simple, but surely we have at least two elements $\mu, \nu$ respectively in degrees $2n - 1$ and $2n$ such that $\phi\mu = u$ and $\phi\nu = v$ for the mod $p$ Hurewicz theorem. Moreover $\beta^s_\pi = 0$ for $s < r$ and $\beta^r_\pi v = u$, so $E^1_\pi = \cdots = E^r_\pi$. -Now the Lie algebra structures come to play. In any graded differential Lie algebra $(L, d)$ over a field of characteristic $p > 2$ any element $x \in L$ of degree $2n$ gives rise to some cycles $\sigma_k(x), \tau_k(x)$ in degrees $2p^kn - 2$ and $2p^kn - 1$ for any $k \in \mathbb{N}$. Moreover, in the universal enveloping algebra of $L$ holds $dx^{p^k} = \tau_k(x)$. -Specializing to our Bockstein sss, this gives us the cycles $\sigma_k(\nu)$, $\tau_k(\nu)$ in $E_\pi^*$ and $\sigma_k(v), \tau_k(v)$ in $E_H^*$. Moreover, since they are constructed via the Lie algebra structures, we have that $\phi\sigma_k(\nu) = \sigma_k(v)$ and $\phi\tau_k(\nu) = \tau_k(v)$. The claim is that $\sigma_k(\nu)$ survives until $E^{r+1}_\pi$ for any $k$ and $\tau_k(\nu)$ survives until $E^{r+1}_\pi$ for any $k > 1$. -How can I see the survival of $\sigma_k(\nu)$? Probably I am missing something stupid, but at the moment I cannot see it. Thanks in advance to anyone. - -REPLY [7 votes]: Firstly, let me thank Gustavo Granja, who independently contacted Joe Neisendorfer in order to make me have an answer. Secondly, let me thank Joe Neisendorfer for his time and answers. -Briefly, the answer to my question is in proposition 9.6.2 (page 296) in Neisendorfer's Algebraic methods in unstable homotopy theory. First notice that the mod $p$ Hurewicz map $\phi^r: E^r_\pi \to E^r_H$ has image in $PE^r_H$, the module of primitive elements of $E^r_H$. Hence it factors as $\phi^r: E^r_\pi \to PE^r_H \hookrightarrow E^r_H$ and we have $\phi^{r+1}: E^{r+1}_\pi \to HPE^r_H \to E^{r+1}_H$. It is true that $\phi^{r+1}[\sigma_k(\nu)] = [\sigma_k(v)]$ is trivial when regarded in $E^{r+1}_H$ but it is no more trivial when regarded in $HPE^r_H$. Hence $[\sigma_k(\nu)]$ is nontrivial as well, so $\sigma_k(\nu)$ survives until $E^{r+1}_\pi$. -The nontriviality of $[\sigma_k(v)]$ in $HPE^r_H$ has nothing to do with topology, but it is a purely algebraic fact. In fact it always holds $0 \neq [\sigma_k(v)] \in HP(dv, v)$ and this applies here because $HPE^r_H = HPT(dv, v) = HPUL(dv, v) = HP(dv, v)$. -I post here Neisendorfer's complete answers, edited accordingly with the notations used here. Any error is due to my transcription. -First answer - -The secret to the question is to look at the fibration sequence $\Omega F^{2n+1}(p^r) \to \Omega P^{2n+1}(p^r) \to \Omega S^{2n+1}$. In mod p homology it looks like $T(\text{ad}^k(v)(u))_{k\geq 0} \to T(v,u) \to T(v)$, three tensor algebras in a row, the middle one on two generators is acyclic with the $r^\text{th}$ Bockstein differential, and the the left one is infinitely generated and decidedly not acyclic. -The important thing is that there are no powers $v^{p^k}$ on the left. They are in the middle and on the right. In the middle tensor algebra they make the Bockstein spectral sequence acyclic. Nothing can be seen after that. But the Bockstein spectral sequence on the left has all the information and it is visible. -The homology Bockstein spectral sequence does not become trivial on the left. In fact, $E^{r+1}$ there is a big polynomial tensor exterior algebra, that is, it is $\bigotimes E(\sigma_k(v)) \otimes P(\tau_k(v))$ for all $k \geq 1$. -Further analysis shows that the left has no torsion of order greater than $p^{r+1}$ and thus $\beta^{r+1} \tau_k(v) = \sigma_k(v)$ for all $k$. This is nontrivial. It needs a brilliant argument with chains. I can say "brilliant" since Moore thought of this and I don't think anyone else could have. Maybe Eilenberg could have. -The Hurewicz homomorphism shows that this is true in homotopy also. Of course, the elements $\text{ad}^k(\nu)(\mu)$ are relative Samelson products in homotopy. Full details are in the paper "Torsion in homotopy groups". -So the main idea is that the middle homology Bockstein spectral sequence vanishes since it is an acyclic tensor algebra (bad!) but the left one survives to the next stage since the $p^\text{th}$ powers are gone (good!). You can see what is happening there, but not in the middle! This was Moore's brilliant idea. It is the heart of the Cohen-Moore-Neisendorfer paper. - -Second answer - -Consider the mod $p$ Bockstein spectral sequences of $\Omega P^{2n+1}(p^r)$ and later those of $\Omega F^{2n+1}(p^r)$. -The Hurewicz map of both of these mod $p$ Bockstein spectral sequences $\phi: E^r_\pi \to E^r_H$ factors through the primitives $E^r_\pi \to PE^r_H \to E^r_H$. -Factoring through the primitives gives us a stronger representation than we would get by going all the way to homology. Taking Bockstein homology we get $E^{r+1}_\pi \to HPE^r_H \to E^{r+1}_H$ where $\sigma_k(\nu)$ has a nontrivial image in the middle for both spaces and $\tau_k(\nu)$ has a nontrivial image in the middle for the space $\Omega F^{2n+1}(p^r)$. Hence they are nontrivial on the left. The homology of the primitives gives us a nontrivial representation of mod $p$ homotopy at the $(r+1)^\text{th}$ stage. -Looking more closely at $\Omega F^{2n+1}(p^r)$, we can see that $\beta^{r+1} \tau_k(v) = \sigma_k(v)$ in mod $p$ homology. That is, this mod p homology Bockstein spectral sequence has two nontrivial differentials. In other words it has no torsion of order bigger than $p^{r+1}$ but it does have that. This is where we see the precise higher order torsion via the Hurewicz representation of the mod $p$ homotopy. Up to the kernel of the mod $p$ Hurewicz map $\beta^{r+1} \tau_k(\nu) = \sigma_k(\nu)$ in homotopy. -This is the way an integral class of order $p^{r+1}$ is represented in the Bockstein spectral sequence. Thus we see that $\sigma_k(\nu)$ represents integral torsion of order exactly $p^{r+1}$ and in that dimension. -For further details see section 9.6 of the book "Algebraic methods in unstable homotopy theory".<|endoftext|> -TITLE: Spherical harmonics – pointwise and L1 bounds -QUESTION [9 upvotes]: Let $\{ \phi _{d,m}\}_{m\geq 1}$ be multi-dimensional spherical harmonics, i.e., solutions of $\Delta \phi = E\phi$ on the sphere $S^d$ for $d>1$, arranged in an increasing order $E_1 \leq E_2 \leq \cdots \leq E_m \leq \cdots $, and normalized in $L^2(S^d)$, i.e., $\int_{S^d} |\phi_{d,m}(x)|^2 \, dx = 1 $. -Question: What are the best known upper and lower bounds on the $L^1$ and $L^{\infty}$ norms of $\phi_{d,m}$? What are good references for these bounds? -I am specifically interested in the dependence on $d$, so if the results are only known for the first few $m$ values, it would be of interest. - -REPLY [5 votes]: The sup-norm of a (normalized) spherical harmonic of degree $k$ is bounded by $C \sqrt d_k$, where $d_k$ is the dimension of all spherical harmonics of order $k$ and $C$ is also given. The estimate is precise. Everything is explicit and can be found, for example, in the notes by P. Garrett: Harmonic analysis on spheres or in Stein-Weiss: Introduction to Fourier Analysison Euclidean spaces, Chapter IV, Corollary 2(b).<|endoftext|> -TITLE: Mathematics of GANs (generative adversarial networks) -QUESTION [7 upvotes]: Generative Adversarial Networks were introduced in http://papers.nips.cc/paper/5423-generative-adversarial-nets and has more than 20000 citations. -The paper introduced key paradigm changes which require applications from modern areas of mathematics. I wanted to ask what are some the mathematics required to understand GANs as far as we know now and what are some key resources which provide accessible insight and a roadmap to learning? - -REPLY [6 votes]: I recommend some papers by Lars Mescheder: - -The paper The Numerics of GANs formalizes GANs as two-player games and analyzes their training dynamics, -the paper Which Training Methods for GANs do actually Converge? takes this analysis further and -the PhD thesis Stability and Expressiveness of Deep Generative Models cantains a quite thorough and mathematical introduction to GANs.<|endoftext|> -TITLE: Example of a $p$-divisible group that is not representable by a formal scheme -QUESTION [9 upvotes]: Let $R$ be a ring such that $p^nR=0$ for some integer $n$, and $G$ be a $p$-divisible group over $R$. -We think of a $p$-divisible groups as an fppf sheaf $G\colon \mathrm{Alg}^{op}_{R}\to \mathbf{Gps}$ such that -$1) \ G=\mathrm{colim} \ G[p^n]$, -$2) \ [p]\colon G \to G$ is surjective, -$3) \ G[p]$ is a finite, locally-free group scheme. -Question: Is $G$ represented by an $R$-formal scheme (with a finitely generated ideal of definition)? -The answer is positive if $G$ is 'etale or connected. More generally, the answer is positive if $G$ is isogenous to an extension of an 'etale $p$-divisible group by a connected $p$-divisible group (Lemma 3.3.1). In particular, this always holds if $R$ is a field. -However, Scholze and Weinstein write Section 3 of their paper as though there are examples of non-representable $p$-divisible groups but never provide an example of such a group. So, I guess the answer to the question above should be negative in general. But it will be nice to see a particular counter-example. -We can try to take $Y=\mathrm{Spec} \ R$ to be the affine modular curve over $\bar{\mathbf{F}}_p$ (with some full level structure) and $G$ the $p$-divisible group of the universal elliptic curve over $Y$. Then, presumably, $G$ should not be representable. But I don't know a rigorous way to prove it. - -REPLY [10 votes]: Your supposed example works indeed. More generally, I think whenever the étale part is not of locally constant height one will run into problems. -Here's a proof that the $p$-divisible group $G$ of the universal elliptic curve $E$ in characteristic $p$ (with auxiliary level structure) is not representable by a formal scheme. Assume it was, and look at an open affine subset $U\subset G$ containing a supersingular point (whose preimage in $G$ is still topologically just a point). Then $U$ necessarily contains the whole preimage of the generic point of the base $Y=\mathrm{Spec}\, R$, as this whole preimage specializes to the given supersingular point (by properness of all $G[p^n]$). But this preimage is not quasicompact, because the étale part of $G$ over the ordinary locus gives a decomposition into countably many connected components. On the other hand, $U$ being affine, it had to be quasicompact, giving a contradiction.<|endoftext|> -TITLE: Is every face exposed if all extreme points are exposed? -QUESTION [7 upvotes]: Let $C$ be a non-empty compact convex subset of ${\mathbb R}^d$ such that every extreme point of $C$ is an exposed point of $C$. Does it follow from this that every face of $C$ is an exposed face? - -REPLY [7 votes]: I don't think so. In dimension 3, first let $S$ be a "stadium" in the plane xy, namely: the segment joining $(1,1,0)$ to $(-1,1,0)$, the half-circle centered at $(1,0,0)$, radius $1$, with middle point $(2,0)$, and its opposite. Also consider the cube with vertices $(\pm 1,\pm 1,\pm 1)$. Consider the convex hull $K$ of this whole set of points. - -The extremal points then are: the 8 vertices of the cube, and the points in the open half circles. They are all exposed. (Notably, the boundary points $(\pm 1,\pm 1,0)$ of half circles, which are extremal and non-exposed within the stadium, are not extremal in $K$ as they're not vertices in the cube.) For instance, the vertex $(1,1,1)$ is exposed, using the halfspace $\{x+y+z\le 3\}$. For points in open half circles, vertical halfspaces do the job. -The vertical edges of the cube, say the one joining $(1,1,1)$ and $(1,1,-1)$, are faces. But are not exposed: indeed, looking around $(1,1,0)$, the only closed halfspace containing $K$ with $(1,1,0)$ as boundary point is the halfspace $\{y\le 1\}$. But its intersection with $K$ is a whole 2-dimensional cube face. - -REPLY [2 votes]: I believe I found a counterexample. -We first set $d=3$ and -$$ -C_1:=B_1(0)\cup [0,1]\times [-1,1]\subset \mathbb R^2 -\\ -C_2:=C_1\times [-1,1]\subset \mathbb R^3, -$$ -where $B_1(0)$ denotes the closed unit ball in $\mathbb R^2$. -Now, the set -$$ -F_1 := \{0\}\times \{1\}\times [-1,1] \subset C_2 -$$ -is a face of $C_2$ which is not an exposed face. -However, the points $(0,\pm 1,\pm 1)$ are extreme points of $C_2$ -which are not exposed points, -and thus $C_2$ is not a counterexample. -This can be solved by defining -$$ -C:= \{(x,y,z)\in C_2 \mid x+z \geq -1,\, x-z \geq -1\}. -$$ -With this modification, -the points $(0,\pm 1,\pm 1)$ are still extreme points of $C$ -are also exposed points. -It can also be checked that all other extreme points are exposed points. -However, the face $F_1$ is still a face of $C$ which is not an exposed face, -so $C$ constitutes a counterexample.<|endoftext|> -TITLE: A very elementary question on the definition of sheaf on a site -QUESTION [8 upvotes]: I'm now studying the etale cohomology with the book 'Introduction to Etale Cohomology' by Tamme. -In the page 26 of the book, 'a family of effective epimorphisms' is introduced. -'A family $\{ U_{i} \rightarrow V \}$ is a family of effective epimorphisms if the diagram -$Hom(V,Z) \rightarrow \prod_{i} Hom(U_{i}, Z) \rightrightarrows \prod_{i,j}(U_{i}\times_{U}U_{j},Z)$ is exact for all the objects $Z$. -The question is 'in the condition, do we restrict the pairs $i,j$ to be distinct?' -I gave a thought on this question, but I'm not sure whether the two versions give us the equivalent results or not. -Thank you very much in advance. - -REPLY [8 votes]: That exactness conditions can be rephrased more explicitely as: -$$ Hom(V,Z) = \left\lbrace (v_i) \in \prod_i Hom(U_i,Z) \ \middle| \ \forall i,j,v_i \circ \pi_1 = v_j \circ \pi_2 \right\rbrace $$ -where $\pi_1,\pi_2$ denotes the two projections $U_i \times_V U_j \rightrightarrows U_i,U_j$. -When you write it like this, the condition in the case $i=j$ is clearly vacuous when all the map $U_i \to V$ are monomorphisms. Indeed in this case $U_i \times_V U_j$ is justs the intersection of $U_i$ and $U_j$, so that $\pi_1=\pi_2$ when $i = j$. This case is very frequent, and you can very often restrict to it by considering the "image" of the $U_i$ in $V$. -But in some situation (for e.g. if you want to keep your objects $U_i$ be to in some specified site that do not admit image factorization like the étale site) it might not be the case. and in general you need the case $i=j$. Consider the case where the you only have a single map $U \to V$. Then the condition becomes -$$ Hom(V,Z) = \left\lbrace f \in Hom(U,Z) \ \middle| \ f\circ \pi_1 = f \circ \pi_2 \right\rbrace $$ -where $\pi_1$ and $\pi_2$ are the two projections $U \times_V U \rightrightarrows U$. -You can think of $U \times_V U \rightrightarrows U$ as a map $U \times_V U \to U \times U$ which is a monomorphisms and corresponds to the "equivalence relation such that $V$ should be the quotient of $U$ by this relations". Or rephrased this as $V$ being the coequalizer ('in the category of sheaves') $U \times_V U \rightrightarrows U \to V$, i.e. $V = U /R$ where $R$ is the equivalence relation $U \times_V U$. -And a function from $V \to Z$ can be described as a function $U \to Z$ which is compatible to the equivalence relation $R$ such that $U/R \simeq V$. -Also note that in the general case (with several map) you can think of the general condition as being in two part: you have the condition for $i=j$ that assert that each maps $U_i \to Z$ factors through "the image $V_i$ of $U_i$ in $V$" (if this make sense) , and the condition for $i \neq j$ that implement the usual compatibility condition. - -REPLY [7 votes]: Yes, you have to include the case $i=j$. Just look at what happens in the case of a single $U_1$, in order for this to boil down to the concept of a single effective epimorphism (introduced in the preceding paragraph in the book) you need to include $i=j=1$.<|endoftext|> -TITLE: Coordinate-free description of an alternating trilinear form on pure octonions -QUESTION [5 upvotes]: Let $O$ denote the division algebra of octonions over $\Bbb R$, and write $V$ for the 7-dimensional quotient space $O/{\Bbb R}$. -The compact group $G_2:={\rm Aut}(O)$ naturally acts on $V$, -and clearly the 7-dimensional representation of $G_2$ in $V$ is isomorphic to its representation in the space of pure octonions. -I know from a classification of alternating trilinear forms in dimension 7 that there exists a $G_2$-invariant alternating trilinear form -$\omega\in\Lambda^3 V^*$ - -Question. What is a coordinate-free description of a $G_2$-invariant alternating trilinear form on $V$? - -REPLY [8 votes]: The form $(x,y,z)\mapsto \mathrm{Re}(x(yz)+y(zx)+z(xy)−x(zy)−y(xz)−z(yx))$ is clearly invariant and alternating. It is nonzero, since its value at $(i,j,k)$ (which satisfy the quaternions relations) is $-6$. -Actually, it can be checked that the symmetrized form $\mathrm{Re}(x(yz)+y(zx)+z(xy)+x(zy)+y(xz)+z(yx))$ vanishes. So the invariant form $$(x,y,z)\mapsto\mathrm{Re}(x(yz)+y(zx)+z(xy)$$ is already alternating (and takes the value $-3$ at $(i,j,k)$: it's actually zero modulo $3$ on the basis).<|endoftext|> -TITLE: Two questions on the permutohedron -QUESTION [8 upvotes]: The $n$-dimensional permutohedron $P_n$ is the polytope given by the convex hull of all the possible permutations of the vector $(1,2,\dots,n+1)\in\mathbb{R}^{n+1}$. So it has $(n+1)!$ vertexes. -I would like to ask if there is a formula for the the number of integer points of $P_n$ and whether it is known that $P_n$ is a very ample polytope for any $n\geq 1$. -For instance, $P_2$ is an hexagon in $\mathbb{R}^3$ contained in the plane $x_1+x_2+x_3 = 6$. It has $7$ integer points and it is very ample. - -REPLY [15 votes]: The number of integer points in $P_n$ is the number of forests on $[n]$; see Section 3 of Stanley's Decompositions of rational convex polytopes. In fact you can see there a simple description of its entire Ehrhart polynomial in terms of forests. (See also Section 9.3 of Beck and Robins's book Computing the continuous discretely.) -The toric variety corresponding to the permutohedron is also very well-studied, and is often called the "permutohedral variety," but I don't know off the top of my head an answer to your question on ampleness. I would take a look at the book by Gelfand-Kapranov-Zelevinsky, which is a seminal source for a lot of this. -EDIT: Okay, I looked up the definition of very ample polytope. A lattice polytope $\mathcal{P}$ is very ample if for every $N \gg 0$, and $\mathbf{a} \in \mathbb{Z}^{n}\cap N\mathcal{P}$, there are $\mathbf{a}_1,\ldots,\mathbf{a}_N\in \mathbb{Z}\cap \mathcal{P}$ such that $\mathbf{a}=\mathbf{a}_1+\cdots+\mathbf{a}_N$. In fact, the (standard) permutohedron has a stronger property than this. It has the Integer Decomposition Property (IDP) which says that for every $N\geq 1$, and $\mathbf{a} \in \mathbb{Z}^{n}\cap N\mathcal{P}$, there are $\mathbf{a}_1,\ldots,\mathbf{a}_N\in \mathbb{Z}\cap \mathcal{P}$ such that $\mathbf{a}=\mathbf{a}_1+\cdots+\mathbf{a}_N$ (this property is also sometimes called normal, in the context of algebraic geometry, I think). The reason the permutohedron is IDP is because any lattice zonotope is IDP: every zonotope has a tiling by half-open parallelepipeds, and these parallelepipeds are easily seen to be IDP; and the permutohedron is a lattice zonotope- see Chapter 9 of the Beck-Robins book mentioned above. -As John Machacek linked to below, a recent paper of Higashitani and Ohsugi shows that in fact the class of Minkowski sums of unit simplices (which includes the regular permutohedra and many of the generalized permutohedra of Postnikov) have the IDP property.<|endoftext|> -TITLE: Prime numbers in a sparse set -QUESTION [7 upvotes]: Is there any set $X$ which is a density 0 subset of $N^*$ and we already know that there are infinitely many primes in it, beside examples which come from $x^2+y^4$(or its proof)? - -Problem1: In particular, is it already proved that there exist $c>1$, s.t. $A_c=\{n\in\mathbb{N}^*| \exists k\in\mathbb{N}^* , n=[k^c]\}$ contains infinitely many primes? - -If this problem can not be solved by existing methods, can the following one be solved by existing methods? - -Problem2: for all $c>1$, $\exists d(c), c>d(c)>0$, s.t. $A_c=\{n\in\mathbb{N}^*| \exists k\in\mathbb{N}^* , n=[k^{e}], e\in (c-d(c),c+d(c))\}$ contains infinitely many primes. - -This is only a naive question, thanks in advance. - -REPLY [6 votes]: I should mention that aside from primes represented by the polynomial $x^2 + y^4$, we may further thin the sequence by insisting that $y$ is also prime; this is a result of Heath-Brown and Li. Moreover, in 2001 Heath-Brown showed that the binary cubic form $x^3 + 2y^3$ represents infinitely many primes, thereby resolving an old question of Hardy which asked whether there are infinitely many primes which are sums of three cubes. Heath-Brown's result was subsequently generalized in joint work with B. Moroz in 2002.<|endoftext|> -TITLE: Why is this condition necessary for the existence of a transferred simplicial model structure? -QUESTION [5 upvotes]: In chapter II of Goerss and Jardine's text on simplicial homotopy theory, they give a general theorem, Theorem 6.8, for transfer of simplicial model structures across a simplicial adjunction. This theorem generalizes Theorem 4.1, which concerns transfer of the standard simplicial model structure on $\operatorname{sSet}$ to the category $sC$ for some bicomplete category $C$. Theorem 6.8 says that given an adjunction $F:C\leftrightarrows D:G$ where $C$ is a $\beta$-cofibrantly-generated simplicial model category (for some cardinal $\beta$), $D$ is an arbitrary simplicial category, and $F$ preserves the simplicial tensoring, the simplicial model structure transfers provided three conditions hold: - -$G$ preserves $\beta$-colimits; - -If $f\in Mor(D)$ is in the saturation of $F(I)$, where $I$ is the set of generating fibrations for $C$, then $Gf$ is a cofibration; - -A cofibration having the left lifting property with respect to all fibrations is a weak equivalence. - - -I'm trying to figure out why condition 2 is necessary. Conditions 1 and 3 are used in the standard fashion to do the small object argument, but the proof G&J outline makes no explicit use of 2. I went back to Theorem 4.1, and I couldn't find any use of a similar condition there, either. Intuitively, it doesn't really make sense to me: why should we care about the images of cofibrations under $GF$? -Is this condition truly necessary? If so, where and how is it used? - -REPLY [5 votes]: Since Goerss and Jardine do not give a (full) proof of this theorem, it is unclear how exactly this condition was intended to be used. -However, this type of construction (where weak equivalences and fibrations are created by a right adjoint functor) is known as a transferred model structure, -and the relevant results originate with Kan. -See, for example, Theorem 11.3.2 in Hirschhorn's book. -Assuming the small object argument can be performed (always -true for cofibrantly generated model structures on locally presentable categories), -the transferred model structure exists if and only if -the functor G (or the functors G_i if there is more than one) sends -transfinite compositions of cobase changes of elements of F(g) to weak equivalences in C, -where g runs over generating acyclic cofibrations of the original model category C. -And if G_i preserves transfinite compositions (as is the case here), then the criterion -can be further simplified to requiring that -G sends cobase changes of elements of F(g) to weak equivalences in C. -This condition is necessary and sufficient, so yields the most general -theorem of this type, and is considerably easier to verify than the conditions -of Theorem II.6.8. -Kan's theorem also easily implies the applications of Theorem II.6.8 in Goerss and Jardine's book. -(Goerss and Jardine's book was published in 1999, and a lot of simplification -happened in this area since then, e.g., the work of Smith on combinatorial model categories.)<|endoftext|> -TITLE: Is the Pierce spectrum useful elsewhere in Mathematics? -QUESTION [5 upvotes]: In Borceaux and Janelidze's Galois Theories, a construction of the Pierce spectrum is given. It is the poset of ideals in a Boolean ring. It's construction is reminiscent of the Zariski spectrum in commutative ring theory, however I've not seen it used elsewhere. -Q. Is the Pierce spectrum important elsewhere in mathematics? - -REPLY [6 votes]: For Boolean rings, the Pierce spectrum coincides with the Zariski spectrum -and is one of the functors implementing the -Stone duality between Boolean algebras and compact -totally disconnected Hausdorff spaces and Stonean duality -between complete Boolean algebras and compact extremally disconnected -Hausdorff spaces. -Restricting further to complete Boolean algebras of projections -of commutative von Neumann algebras produces -an equivalence of categories of commutative von Neumann algebras -and compact strictly localizable enhanced measurable spaces, -i.e., a version of the Gelfand duality in the setting of measure theory.<|endoftext|> -TITLE: Characterizing the elements of $(A-A)/(A-A)$, where $A$ is a Cantor-like subset of the integers -QUESTION [18 upvotes]: Short version of my question: I'm interested in the following fact. - -If $m,n$ are odd integers, then $m/n$ can be written as the ratio of two numbers of the form $\sum_{j=0}^\ell \epsilon_j 4^j$, where $\epsilon_j \in \{-1,0,1\}$. - - -More background: -Let $A = \{0,1,4,5,16,\ldots\}$ be the set of nonnegative integers whose base 4 expansion contains only the digits 0 and 1. -Given a number $r \in \mathbb Q \setminus \{0\}$, there is a unique way (up to sign) to write $r = 4^k \frac{m}{n}$, where (1) $k \in \mathbb Z$, (2) $m$ and $n$ are not divisible by 4, and (3) $\gcd(m, n) = 1$. -Then the following is true: - -Theorem: Let $r \in \mathbb Q \setminus \{0\}$. Then $r \in \frac{A-A}{A-A} = \{ \frac{a-b}{c-d} ~|~ a,b,c,d \in A\}$ if and only if both $m$ and $n$ are odd. - -The "only if" direction can be proved easily with elementary number theory. (Every nonzero element of $A-A$ contains an even number of factors of $2$.) -For the "if" direction, the only proof I know is in Section 10.3 of Mattila's Fourier Analysis and Hausdorff Dimension (and is based on Kenyon's 1997 paper Projecting the one-dimensional Sierpinski gasket). The proof looks at the Fourier transforms of measures defined on projections of the four-corner Cantor set. Actually, the textbook proves something else, and the theorem I stated above is a corollary. This leads me to wonder if there is an easier proof of the "if" direction of the theorem above (e.g., via elementary number theory). - -REPLY [13 votes]: You can find an elementary proof in this paper with a very honest title: - -J.H. Loxton, A.J. van der Poorten, "An awful problem about integers in base four" -Acta Arith., 49 (1987), pp. 193-203 - -In the paper the authors prove that any integer $m$ with an even number of $2$'s in its prime factorization can be written as a number of the form $\frac{A-A}{A-A}$. This is derived as a simple corollary of the main theorem which says that $A_k+nA_k$ eventually has fewer than $4^k$ terms, for large enough $k$. Here $A_k$ denotes the members of $A$ with at most $k$ digits. Their proof applies verbatim to the set $mA_k+nA_k$, which then implies the claim in your question about fractions $m/n$. -I personally find this elementary proof to still be quite similar in spirit to Kenyon's argument. Self similarity and Fourier transforms aren't mentioned, but they are respectively replaced by the recurrence relations -$$(mA_{k+1}+nA_{k+1})=(mA_k+nA_k)+4^{k}(mA_1+nA_1)$$ -$$(mA_{k+1}+nA_{k+1})=4(mA_k+nA_k)+(mA_1+nA_1)$$ -and the generating function $$\prod_{j=1}^k (1+\theta^{-m4^j})(1+\theta^{-n4^j})$$ -respectively.<|endoftext|> -TITLE: Decomposition of $\bigotimes^{m} \mathbb{C}^{n}$ under the action of $\operatorname{GL}_{n}\times \operatorname{S}_{m}$ -QUESTION [7 upvotes]: $\DeclareMathOperator\GL{GL}\DeclareMathOperator\S{S}$I want to know the proof of the following theorem. It is stated somewhere that, a proof can be found in: "Roger Howe, Perspectives on invariant theory: Schur duality, multiplicity-free actions and -beyond, The Schur lectures (1992) (Tel Aviv), Israel Math. Conf. Proc., vol. 8, Bar-Ilan Univ., -Ramat Gan, 1995, pp. 1–182, DOI 10.1007/BF02771542. MR1321638 (96e:13006)". -However, I am not able to access this. Could anyone help me out to find this paper, or may be, is there any other place where I can find the proof? -Theorem: Let $F^{\lambda}_{n}$ denote the irreducible rational representation -of $\operatorname{GL}_{n}$ with highest weight indexed by $\lambda$. Let $W^{\lambda}_{m}$ denote the irreducible complex representation of $\S_{m}$ indexed by $\lambda$. Under the joint action of $\GL_{n}\times \S_{m}$ on $\bigotimes^{m} \mathbb{C}^{n}$, we have the multiplicity free decomposition $$ \bigotimes^{m} \mathbb{C}^{n} \cong \bigoplus_{\lambda} F^{\lambda}_{n} \otimes W^{\lambda}_{m} $$ -where the sum is over all partitions $\lambda$ of $m$ with at most $n$ parts. Note that all -irreducible representations of $\S_{m}$ appear in the decomposition when $n \geq m$. - -REPLY [4 votes]: A good reference for the Schur-Weyl duality (and also for generalization of this to other classical groups) I can recommend Symmetry, Representations, and Invariants by Roe Goodman and Nolan Wallach.<|endoftext|> -TITLE: Is the filtered colimit topology on the space of signed Radon measures linear and locally convex? -QUESTION [5 upvotes]: Let $X$ be a compact Hausdorff space. In chapter 3 of Peter Scholze's Lectures on Analytic Geometry he considers the space of signed Radon measures on $X$ equipped with the filtered colimit (aka inductive limit) topology of the (in the weak$^*$-topology) compact absolutely convex subsets $\mathcal{M}(X)_{\leq c}$. Here $\mathcal{M}(X)_{\leq c}$ denotes the subset of measures with total variation norm less or equal than $c$ with $c>0$. -Then he states that the resulting topology is a locally convex vector topology. I was wondering if the subsets $\mathcal{M}(X)_{\leq c}$ form a neighborhood basis of the origin. If the answer is yes, then I do not see why the resulting topology is not the same as the one induced by the total variation norm. If the answer is no, then I do not see how to show that this topology is a locally convex vector topology. -Any clarification on this would be really appreciated. - -REPLY [7 votes]: This is a general, well-known fact about the dual of a Banach space. The finest topology which agrees with the weak$\ast$ topology on the bounded sets is locally convex. It is often called the bounded weak$\ast$ topology. It is complete and has the same convergent sequences as the weak$\ast$ topology. In non trivial situations, it is weaker than the norm topology—its dual is the original space. Yours is the case of the dual of a $C(K)$-space. For a general reference, look up the Banach-Dieudonné theorem in any standard text on Banach spaces.<|endoftext|> -TITLE: Probability of a deviation when Jensen’s inequality is almost tight -QUESTION [7 upvotes]: This is a cross-post to a yet unanswered question in Math StackExchange -https://math.stackexchange.com/questions/3906767/probability-of-a-deviation-when-jensen-s-inequality-is-almost-tight -Let $X>0$ be a random variable. Suppose that we knew that for some $\epsilon \geq 0$, -\begin{eqnarray} -\log(E[X]) \leq E[\log(X)] + \epsilon -\tag{1} \label{eq:primary} -\end{eqnarray} -The question is: if $\epsilon$ is small, can we find a good bound for -\begin{eqnarray*} -P\left( \log(X) > E[\log(X)] + \eta \right) -\end{eqnarray*} -for a given $\eta > 0$. One bound can be obtained this way: -\begin{eqnarray*} -P\left( \log(X) > E[\log(X)] + \eta \right) &=& P\left( X > \exp(E[\log(X)] + \eta) \right) \\ -& \leq & E[X] / \exp(E[\log(X)] + \eta) \\ -& = & \exp( \log E[X] - E[\log(X)] - \eta ) \\ -& \leq & \exp(\epsilon - \eta) -\end{eqnarray*} -where the first inequality follows from Markov’s inequality. This seems like a good bound due to the exponential decay with $\eta$, but upon closer examination it appears that it can be significantly improved. If we have $\epsilon = 0$, then this bounds gives -\begin{eqnarray} -P\left( \log(X) > E[\log(X)] + \eta \right) & \leq & \exp(-\eta) -\tag{2} \label{eq:good_but_not_best} -\end{eqnarray} -However, from Jensen's inequality applied to (\ref{eq:primary}) with $\epsilon = 0$ we obtain $\log(E[X]) = E[\log(X)]$ and therefore $X$ is a constant almost everywhere. As a consequence, for any $\eta>0$, -\begin{eqnarray*} -P\left( \log(X) > E[\log(X)] + \eta \right) = 0. -\end{eqnarray*} -which is (of course) infinitely better than (\ref{eq:good_but_not_best}). -It would appear that a better bound should decay to zero as $\epsilon$ decays, and ideally preserve the exponential decay with $\eta$. Any suggestions? -(I am aware a version of this question has been asked previously Quantitative Version of Jensen's Inequality?) - -REPLY [3 votes]: $\newcommand\ep\epsilon $Let $u:=\eta>0$, so that the probability in question is $P(\ln X>E\ln X+u)$. Note that this probability will not change if we replace there $X$ by $tX$ for any real $t>0$. So, without loss of generality -\begin{equation*} -E\ln X=0, \tag{-1} -\end{equation*} -and hence your condition (1) can be rewritten as -\begin{equation*} - EX\le e^\ep, \tag{0} -\end{equation*} -and then the probability in question simplifies to -\begin{equation*} - P(X>v), -\end{equation*} -where -\begin{equation*} - v:=e^u>1. -\end{equation*} -Take now any $z\in(0,v)$ and for all real $x>0$ let -\begin{equation*} - g(x):=ax-b\ln x+c, -\end{equation*} -where -\begin{equation*} - a:=a(z):=\frac{1/v}{h(r)},\quad b:=b(z):=az,\quad c:=c(z):=az\ln\frac ze, -\end{equation*} -\begin{equation*} - h(r):=1-r+r\ln r,\quad r:=z/v\in(0,1). -\end{equation*} -Note that the function $h$ is decreasing on $(0,1)$, with $h(1-)=0$. So, $h>0$ on $(0,1)$ and hence $a>0$ and $b>0$. So, the function $g$ is convex on $(0,\infty)$. Moreover, -\begin{equation*} - g(z)=g'(z)=0, \quad g(v)=1. -\end{equation*} -It follows that $g(x)\ge1(x>v)$ for all real $x>0$ and hence, in view of (-1) and (0), -\begin{equation*} - P(X>v)\le Eg(X)=a\,EX+c\le ae^\ep+c. \tag{1} -\end{equation*} -The latter expression, $ae^\ep+c$, in (1) can now be minimized in $z\in(0,v)$, with the minimizer expressed in terms of Lambert's $W$ function. -The suboptimal but simple choice $z=1$ in (1) yields -\begin{equation*} - P(\ln X>E\ln X+u)=P(X>v)\le\frac{e^\ep-1}{v-1-\ln v} -\end{equation*} -and hence -\begin{equation*} - P(\ln X>E\ln X+u)\le B_\ep(u):=\min\Big(1,\frac{e^\ep-1}{e^u-1-u}\Big). -\end{equation*} -The simple upper bound $B_\ep(u)$ has both of the desired properties: -(i) for each real $u>0$ -\begin{equation*} - B_\ep(u)\underset{\ep\downarrow0}\longrightarrow0; -\end{equation*} -(ii) uniformly over all $\ep\in(0,1)$ (say) -\begin{equation*} - B_\ep(u)=O(e^{-u}) -\end{equation*} -as $u\to\infty$.<|endoftext|> -TITLE: Building algebraic geometry without prime ideals -QUESTION [35 upvotes]: $\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\ev{ev}$Teaching algebraic geometry, in particular schemes, I am struggling to provide intuitive proofs. In particular, I find it counter-intuitive that points are prime ideals. I discovered a trick which I suspect is not new. Basically, you build the functor of points into the definition. I want to modify the definition of $\Spec(R)$ as follows: -As a set, $\Spec(R)$ is simply all pairs $x=(k_x, \ev_x)$ where $k_x$ is a field and $\ev_x:R\to k_x$ is a homomorphism. Then as usual, elements of $R$ are called functions and the value of a function $f\in R$ at a point $x$ is $f(x)\mathrel{:=}\ev_x(f)\in k_x$. Then it continues as usual: closed set is where some collection of functions vanishes. Basic open set is where some function is invertible. -Of course, there are some problems with this approach: - -The class of all fields is not a set. Technically, we can limit ourselves to some very large set of "test fields". So this can be swept under the rug. - -$\Spec(R)$ with this definition is not $T_0$. But after getting used to spaces being not Hausdorff it should be easy to take it to the next level with spaces being not $T_0$. Of course, to every non-$T_0$ space there is a canonically associated $T_0$ space where you identify topologically indistinguishable points, so you recover the usual construction of $\Spec(R)$ this way. - - -Nevertheless, I find this approach much more intuitive, because it seems like a natural question to solve some system of equations in some unknown field, rather then studying prime ideals (which is of course basically the same thing, language aside). -Is this not new? Are there any lecture notes following this approach? Of course, the full "functor of points" approach sort of contains this one, but notice that to do what I want I do not need Yoneda lemma, I do not ask for functoriality, so I do not need to sweep under the rug all the tedious checks of naturality. So I find it more basic than functor of points. -Here is an example. When we construct the localization of a ring $R$ with respect to a multiplicative set $S$ we prove that prime ideals of $S^{-1}R$ are in bijection with a subset of ideals of $R$. With this approach the corresponding statement is a simple consequence of the universal property of the localization, there is nothing more to prove. -Another example. Prove that the map $\mathbb{A}^1\to \mathbb{A}^3$ given by $t\to (t^3, t^4, t^5)$ has image $Z(xz-y^2, x^3-yz, x^2 y -z^2)$. This becomes simply high school algebra. - -REPLY [14 votes]: This nice approach to points on schemes in fact becomes crucial once one leaves the world of schemes and travels to the galaxy of stacks. -For an algebraic stack $X$, one defines a point of $X$ to be a morphism $\mathrm{Spec}(k) \to X$, modulo the natural equivalence relation discussed above. -See in particular: -https://stacks.math.columbia.edu/tag/01J5 -https://stacks.math.columbia.edu/tag/04XE<|endoftext|> -TITLE: Can one show corbordism invariance of the Crane-Yetter state-sum using simplicial methods / are there 'Pachner-like' moves for cobordisms? -QUESTION [12 upvotes]: Let $\mathcal{C}$ denote some Unitary Braided Modular Fusion Category. It is well known that the Crane-Yetter state-sum, $Z_{CY}(\bullet|\mathcal{C})$ is an oriented-cobordism invariant. In other words, let $N^5$ be a 5-manifold with $\partial N^5=M_1^4 \sqcup \bar{M}_2^4$, which is a cobordism between 4-manifolds $M_1,M_2$. Then $Z_{CY}(M_1|\mathcal{C}) = Z_{CY}(M_2|\mathcal{C})$. -Using the fact that the oriented cobordism group $\Omega_4^{SO}=\mathbb{Z}$ is generated by $CP^2$ and the bordism invariant of $M$ is the signature $\sigma(M)$, we can express $Z_{CY}(M)$ in terms of the signature of $M$ and the value of $Z_{CY}(CP^2|\mathcal{C})$, in particular as $Z_{CY}(CP^2|\mathcal{C})^{\sigma(M)}$. There are well-known expressions for $Z_{CY}(CP^2|\mathcal{C})$ in terms of the Braided Fusion Category data. -Is there a proof of the cobordism invariance using combinatorial/triangulation-based methods? The proofs I've seen all use fairly abstract skein theory arguments - it'd be illuminating if there were a more hands-on approach. -It would also be nice if there were versions of the Pachner move theorem for cobordisms. In particular, are there a set of additional moves to the Pachner moves that can generate cobordisms, as opposed to just PL homeomorphism? Even comments about some generating set of 4-manifold cobordisms (e.g. via connected sums with some set of manifolds) would be helpful, even if they're not explicity translated into the language of triangulations. Recall that the PL-invariance of $Z_{CY}(\bullet|\mathcal{C})$ can be proved via Pachner moves. - -REPLY [2 votes]: I know very little about this topic, but I think this may be answered in this paper that studies a cubical analogue of Pachner moves: -Karim Adiprasito and Gaku Liu, Normal crossing immersions, cobordisms and flips. - -Abstract: We study various analogues of theorems from PL topology for cubical -complexes. In particular, we characterize when two PL homeomorphic -cubulations are equivalent by Pachner moves by showing the question to -be equivalent to the existence of cobordisms between generic -immersions of hypersurfaces. This solves a question and conjecture of -Habegger, Funar and Thurston. - -Instead of working with manifolds broken up into $n$-simplices (triangulations), this result concerns manifolds which are broken up into $n$-cubes (a "cubulation") and defines combinatorial rules for going from one cubulation to another via making local changes (Definition 2.1).<|endoftext|> -TITLE: Origin of the symbol for the tensor product -QUESTION [9 upvotes]: I have recently realised that the Paleo-Hebrew (and Phoenician) graph for the Hebrew letter ט (Teth) is $\otimes$. This made me wonder if there is any relation between the choice of the symbol and the choice of the name (as far as I am aware, tensor product is meant to be a stretched product, coming from latin tendere although I confess I can't understand why the tensor product of two spaces, or modules, or elements, should be thought as being "stretched"), since ט is the letter used in modern and ancient Hebrew to denote the sound T in words borrowed from other languages. -I have checked the Archives Bourbaki, in particular the second redaction (n°034) of Algèbre. Chapitre II, algèbre linéaire and the corresponding discussion which seem to me the first occurrences of the name (still called "produit tensoriel (ou kroneckerien)", see ibid. p. 198) and of the symbol, but found no hints. The tensor product is presented as a special case of the bilinear product of two modules, which is denoted by $\odot$, itself not a graph that I am aware of in the Paleo-Hebrew alphabet: the circle in $\otimes$ could simply come from the one in $\odot$, but it might also be related to the first letter T of tensor. -Is there any reference I can look at for more information? - -REPLY [8 votes]: According to John Aldrich's list of "Earliest Uses of Symbols for Matrices and Vectors", the notation $\times$ for direct product (as well as the name itself) goes back to Wedderburn's 1934 Lectures on Matrices (page 74). - -Some further search gave a much earlier source, Hurwitz's 1894 Zur Invariantentheorie - -I still have to track down the step from $\times$ to $\otimes$. -Incidentally, On the history of the Kronecker product argues that it should more appropriately be called the Zehfuss product. -In any case, since tensor product is $\otimes$ and tensor sum is $\oplus$, it seems obvious that the $\times$ and the $+$ refer to the arithmetical operations of multiplication and addition, not to a letter.<|endoftext|> -TITLE: Kazhdan's property (T) for $\tilde{C}_2$-lattices -QUESTION [9 upvotes]: It is known that higher rank lattices have property (T) and also that lattices on 2-dimensional Euclidean buildings have property (T) provided the thickness $q+1$ of the building is large enough (which is a condition only in type $\tilde{C}_2$ and $\tilde{G}_2$). My question is about the best known bound that guarantees property (T). -Specifically Żuk states in his famous note that $q \ge 3$ is sufficient for type $\tilde{C}_2$ and that $q \ge 4$ is sufficient for type $\tilde{G}_2$. However, that statement is based on calculations performed by Garland and others (Garland's being the most readable) which only give sufficiency for $q \ge 7$ in type $\tilde{C}_2$ and for $q \ge 11$ in type $\tilde{G}_2$. Is there another justification for Żuk's claim? (Am I making a mistake?) -This is not purely hypothetical: Essert constructs lattices on $\tilde{C}_2$ buildings that lie in the region for which Żuk asserts property (T) but Garland's bounds do not guarantee it (as far as I can see). -Are there other references that give better bounds? -To turn all of this into a concrete question, let me ask this: -Do all of the groups described by Essert (Section 0.2, Theorem 0.4) -have property (T)? - -REPLY [8 votes]: I don't have access to Zuk's note, but I remember finding an error in it when I read it (so this could be the same problem you found). He did improve on Garland in terms of thickness by taking average of the eigenvalues of the Laplacian of the links of two connected vertices - see the paragraph after the proof of Theorem 2.5 in Ballmann and Światkowski. However, that improvement does not grantee the results you quoted. -I managed to do a little better - see Table 2 (after Remark 4.1) in link, but that still does not give you $q=3$ for $\widetilde{C}_2$ lattices (only $q=4$).<|endoftext|> -TITLE: On HTT's remark 2.4.1.9 -QUESTION [7 upvotes]: Background -I have some difficulty to understand remark 2.4.1.9 in Lurie's HTT, which says a $p$-Cartesian edge is determined up to equivalence. -To be more precise, let $p:X\to S$ be an inner fibration of simplicial sets, let $x$ be a vertex of $X$, and let $\bar{f}:\bar{y}\to p(x)$ be an edge of $S$ ending at $p(x)$. If there are two $p$-Cartesian edges $f:y\to x$ and $f':y'\to x$ in $X$ with $p(f)=p(f')=\bar{f}$, then we have that $f$ and $f'$ are strong final objects of the $\infty$-category $K\stackrel{\text{def}}{=}X_{/x} \times_{S{/p(x)}}\{\bar{f}\}$, and thus they are equivalent $K$. -That's true. Since for a $p$-Cartesian edge $f$, -$$q:X_{/f}\to X_{/x}\times_{S_{/p(x)}}S_{/p(f)}$$ -is a trivial fibration, and $K_{/\{f\}}\to K$ is a pullback of $q$. -My Question: -But since Lurie's original words are "($p$-Cartesian edge) is therefore determined up to equivalence by $\bar{f}$ and $x$", I think we should conversely have that an object who is equivalent with a $p$-Cartesian edge is also a $p$-Cartesian edge. And he indeed utilized such a claim in the proof of Proposition 2.4.2.4 in HTT. However I cannot figure out a detailed proof. I hope someone can help me. - -REPLY [6 votes]: Take a look at Proposition 55.6 in Charles Rezk's notes Stuff about quasi-categories, (which is the formal meaning of Lurie's words). -It says that fix an edge $\bar{f}:\bar{y}\to p(x)$ in $S$, the "space" of $p$-Cartesian lifts of $\bar{f}$ is contractible. -PS: I should have posted this as a comment, but I don't have enough reputation to do so...<|endoftext|> -TITLE: Finite group such that $K_{-1} (\mathbb Z G)$ has non-trivial torsion -QUESTION [8 upvotes]: According to Carters Lower K-theory of finite groups for a finite group $G$ we have -$$ K_{-1} (\mathbb Z G) = \mathbb Z^r \oplus \mathbb Z_2^s $$ -where $s$ is the sum over all irreducible representations over $\mathbb Q$ which have even Schur index, but odd Schur index over $\mathbb Q_p$ for every finite prime which divides the order of $G$. -The integer $s$ seems to be $0$ for a pretty wide class of groups and its hard for me to even find a single concrete example of a finite group with non-trivial $s$. Does anyone know of a result in that direction? - -REPLY [6 votes]: Many results in this direction can be found in the paper -B. A. Magurn: Negative (K)-theory of generalized quaternion groups and binary polyhedral groups, Commun. Algebra 41, No. 11, 4146-4160 (2013). ZBL1284.19004. -In particular, you can have a look at - -Corollary (p.4155) Given the dicyclic group $\mathsf{Q}_n$, the group $K_{-1}(\mathbb{Z} \mathsf{Q}_n)$ is torsion free if and only if $n$ is a power -of a prime in $3 + 4 \mathbb{Z}$ or $n=2$. - -All the other occurrences for $n$ provide an infinite series of examples with $s\neq 0$. The paper also contains explicit calculations for $s$. - -Theorem 5 (p.4155) The group $K_{-1}$ for binary polyhedral groups is as follows: $$K_{-1}(\mathbb{Z} \tilde{\mathsf{A}}_4)= \mathbb{Z}, - \quad K_{-1}(\mathbb{Z} \tilde{\mathsf{S}}_4)= \mathbb{Z}\oplus - \mathbb{Z}_2, \quad K_{-1}(\mathbb{Z} \tilde{\mathsf{A}}_5)= - \mathbb{Z}^2\oplus \mathbb{Z}_2$$ - -This yields two further examples with $s=1$.<|endoftext|> -TITLE: Convergence of the series involving Mobius functions $\sum_{k,d} \mu(d) x_{kd}$ -QUESTION [12 upvotes]: (I originally asked this question here, but the problem appears much more difficult than I think after a moment of thought, so I think it might be more suitable to post it here. Please tell me if this is not the right place to ask.) -Let $\sum x_n$ be an absolutely convergence sequence, then obviously, from the elementary properties of Mobius function $\mu$ that $\sum_{d|n} \mu (d)=0$ for $n\neq 1$, -$$ -\sum_{n=1}^\infty \sum_{d|n} \mu(d)x_n =x_1. -$$ -However, if we alter the order of summation, i.e. -$$ -\sum_{d=1}^\infty \sum_{k=1}^\infty \mu(d) x_{kd}, -$$ -where $kd$ is the product of $k$ and $d$, will we get the same sum, or does the sum diverge? -One way of proving that we can exchange the order of summation is that -$$ -\left|\sum_{kd\geq M, k\leq K,d\leq D} \mu(d) x_{kd}\right| <\epsilon -$$ -for all $K,D$ and sufficiently large $M$. But then $\mu$ does not cancel out as nicely, and the bound on the sum of $\mu$'s using Merten's function is too weak. -Since the proof seems difficult, I attempt to find a divergent counterexample. I try to use the fact proven in an answer below this MO question that the Dirichlet series of Mobius function $\mu$, -$$ -\sum_{n=1}^\infty \mu(n)n^{-s}, -$$ -diverges for $0<\Re s <1/2$. -Define the sequence $b_n$ by -$$ -b_n=n^{-1/4}-(n+1)^{-1/4}. -$$ -Clearly $\sum_{n=r}^\infty b_n = r^{-1/4}$. -Now define $a_{k}$ as follows. - -$a_1=b_1$ . -Suppose that $a_1,\ldots, a_{l}$ are already defined, such that $a_l=b_n$ for some $n$. Then let $l'>l$ be a multiple of $(n+1)!$, and $a_{l'}:=b_{n+1}$. Furthermore, let $a_{k}=0$ for $k$ strictly between $l$ and $l'$. - -This recursive definition gives a sequence $a_n$, such that the sequence of sums $S_n=\sum_{k=1}^{\infty} a_{nk}$ decreases slower than $n^{-1/4}$ (again $nk$ denotes the product of $n$ and $k$). However, this does not make $\sum_{d=1}^\infty \sum_{k=1}^\infty \mu(d) x_{kd}$ a Dirichlet series, so even if the decrease in magnitude is slow, we cannot conclude that it is divergent. Positive and negative terms might be cancelled out. -So, both proof and counterexample are difficult - the bounds and constructions are always a little bit worse then desired. -Are there any known results about the convergence of the sum $\sum_{d=1}^\infty \sum_{k=1}^\infty \mu(d) x_{kd}$? - -REPLY [10 votes]: This question was considered by Wintner (1945). On pages 16-18 of the linked document, he observes that $\sum_n 2^{\omega(n)}|x_n|<\infty$ implies the absolute convergence of the second display (here $\omega(n)=\sum_{p\mid n}1$), hence in this case the second display converges to $x_1$. Moreover, based on Toeplitz's norm principle, he constructs an example with $\sum_n |x_n|<\infty$ for which the OP's second display diverges.<|endoftext|> -TITLE: VC-dimension of disambiguated classes -QUESTION [8 upvotes]: Consider an "ambiguous" function class -$F^\star\subseteq\{0,1,\star\}^X$ (i.e., $F$ consists of Boolean functions acting on a set $X$ with some missing values, indicated by $\star$). -We say that -$F^\star$ shatters a set $S\subseteq X$ -if $F^\star(S)\supseteq\{0,1\}^S$. -Define $VC(F^\star)$ as the maximal size of any shattered set (possibly, $\infty$). -We say that $\bar f\in\{0,1\}^X$ is a disambiguation -of $f^\star\in F^\star$ if -the two functions agree on $x\in X$ -whenever $f^\star(x)\neq\star$. -We say that $\bar F\subseteq\{0,1\}^X$ is a disambiguation of -$F^\star$ if each $\bar f\in \bar F$ -is a disambiguation of -some $f^\star\in F^\star$ -and every $f^\star\in F^\star$ -has a disambiguated representative $\bar f\in \bar F$. -Conjecture: There is a universal constant $c$ such that for any ambiguous -$F^\star$ there is a disambiguation $\bar F$ such that -$$ -VC(\bar F) -\le c -VC(F^\star) -.$$ -Note: -This open problem was posed here: -https://arxiv.org/abs/1810.02180 . -It is known that $c$ must be $>1$, and Lemma 6.2 therein provides an analog of Sauer's lemma for $F^\star$. - -REPLY [5 votes]: This has been disproved in Theorem 11 of Alon, Hanneke, Holzman, and Moran. The proof is short and elegant (building on recent deep results of others').<|endoftext|> -TITLE: Topological groups in which all subgroups are closed -QUESTION [7 upvotes]: General question: does there exist a nondiscrete topological group $G$ such that all subgroups of $G$ are closed? Or, does there exist a nondiscrete topological vector space $V$ such that all vector subspaces of $V$ are closed? -I am interested specifically in topological abelian groups with linear topology, or in topological vector spaces with linear topology. A topological group $A$ is said to have linear topology if open subgroups form a base of neighborhoods of zero in $A$. A topological vector space $V$ is said to have linear topology if open vector subspaces form a base of neighborhoods of zero in $V$. -Hence my more specific question is: does there exist a nondiscrete topological abelian group $A$ with linear topology such that all subgroups of $A$ are closed? Or, does there exist a nondiscrete topological vector space $V$ with linear topology such that all vector subspaces of $V$ are closed? - -Motivation: I am trying to work out the very basics of the theory of topological abelian groups/vector spaces with linear topology. In particular, I am trying to understand closed maps. -For comparison, let us start with open maps. Let $B$, $C$ be topological abelian groups with linear topologies, and let $p\colon B\longrightarrow C$ be an abelian group homomorphism. Then $p$ is an open map (as a map of topological spaces) if and only if the image of every open subgroup in $B$ is an open subgroup in $C$. -Moreover, let $p\colon B\longrightarrow C$ be a surjective group homomorphism between topological groups. Then the topology of $C$ is the quotient topology of the topology of $B$ (via $p$) if and only if $p$ is an open continuous map. -The latter assertion is not at all true for topological spaces in general. It is easy to come up with an example of a topological space $Y$ and a surjective map of sets $p\colon Y\longrightarrow Z$ such that, when $Z$ is endowed with the quotient topology of the topology of $Y$ (via $p$), the map $p$ is not open. But topological groups are better behaved. - -Closed maps appear to be more complicated. I am interested specifically in injective closed maps. Let us start with an injective map between topological spaces $i\colon X\longrightarrow Y$. Then $i$ is a closed continuous map if and only if $i(X)$ is a closed subset in $Y$ and the the topology of $X$ is induced from the topology of $Y$ via the embedding $i$. -Now let $A$, $B$ be topological abelian groups with linear topologies and $i\colon A\longrightarrow B$ be a continuous group homomorphism. Assume that the image of any closed subgroup in $A$ is a closed subgroup in $B$. Does it follow that $i$ is a closed map (as a map of topological spaces), i.e., that the image of any closed subset in $A$ is a closed subset in $B$? -Suppose that we've managed to find a nondiscrete topological abelian group $B$ with linear topology such that all subgroups of $B$ are closed. Let $A$ denote the same abelian group as $B$, but endowed with the discrete topology; and let $i$ be the identity map. Then the image of any subgroup of $A$ is closed in $B$, but $i$ is not a closed map (and the topology on $A$ is not induced from $B$). So we obtain a counterexample to the previous question. - -REPLY [9 votes]: $\mathbf{Z}$ with the profinite topology has the property that every subgroup is closed. That's because every subgroup is an intersection of finite index subgroup. However it is not discrete (the profinite topology on an infinite group is never discrete). -If $K$ is any discrete field and $V$ an infinite-dimensional vector space over $K$, we can endow $V$ with the "pro-finite-dimensional" topology, for which a basis of neighborhoods of $0$ consists of finite-codimensional subspaces. Then every subspace is closed, but only finite-codimensional subspaces are open, so it's not discrete.<|endoftext|> -TITLE: Classification of (not necessarily connected) compact Lie groups -QUESTION [10 upvotes]: I am looking for a classification of compact (not necessarily connected) Lie groups. Clearly, all such groups are extensions of a finite "component group" $\pi_0(G)$ by a compact connected Lie group $G_0$: -$\require{AMScd}$ -\begin{CD} -0 @>>> G_0 @>>> G @>p>> \pi_0(G) @>>> 0 -\end{CD} -The classification of compact connected Lie groups is familiar to me, so my question is how to classify such extensions. - -UPDATE: I suspect the following is true (due to @LSpice, with my added requirement that $H$, $P$ are finite): -Hypothesis: $G$ can always be written as -$$ -G= \frac{G_0 \rtimes H}{P} -$$ -for finite groups $H,P$, where $P \subseteq Z(G_0 \rtimes H)$. -UPDATE 2: @LSpice has proven this below for the weaker requirement that $P$ intersects $G_0$ within $Z(G_0)$, and provided a counterexample where $P$ cannot be taken to be central. -UPDATE 3: See Improved classification of compact Lie groups for a follow-up question (which I won't write here to avoid excessive clutter.) - -A less useful claim from my original question: any such $G$ can be constructed from $G_0$ in three steps: - -Take the direct product of $G_0$ with a finite group. - -Quotient the result by a finite subgroup of its center. - -Extend a finite subgroup of $\mathrm{Out}(G_0)$ by the result. - - -(Step 3 may always be is not a semidirect product in general.) - -REPLY [7 votes]: $\DeclareMathOperator\U{U}$Consider the matrices $u = \begin{pmatrix} -0 & 1 \\ --1 & 0 \\ -&& 0 & 1 \\ -&& 1 & 0 -\end{pmatrix}$ and $v = \begin{pmatrix} -0 && 1 \\ -& 0 && 1 \\ --1 && 0 \\ -& -1 && 0 -\end{pmatrix}$. These belong to the finite group of signed permutation matrices, so the group that they generate is finite and we call it $H$. Put $G_0 = \left\{d(z, w) \mathrel{:=} \begin{pmatrix} z \\ & z^{-1} \\ && w \\ &&& w^{-1} \end{pmatrix} \mathrel: z, w \in \U(1)\right\}$. Since $u d(z, w)u^{-1} = d(z^{-1}, w^{-1})$ and $v d(z, w)v^{-1} = d(w, z)$, the group $G$ generated by $G_0$, $u$, and $v$ has $G_0$ as its identity component. Now let $G_0 \rtimes H \to G$ be any cover restricting to the inclusion $G_0 \to G$, and let $\tilde u$ be an element of $H$ whose image lies in $u G_0$; say the image is $u d(z, w)$. Then $\tilde u^2$ maps to $(u d(z, w))^2 = u^2 = d(-1, 1)$, so $d(-1, 1) \rtimes \tilde u^2$ lies in $\ker(G_0 \rtimes H \to G)$. If $\tilde v$ is an element of $H$ whose image lies in $v G_0$, then $\tilde v(d(-1, 1) \rtimes \tilde u^2)\tilde v^{-1}$ lies in $d(1, -1) \rtimes H$, hence does not equal $d(-1, 1) \rtimes H$. That is, $\ker(G_0 \rtimes H \to G)$ is not central in $G_0 \rtimes H$. - -What we can do is find (in general, not just for the specific example above) a finite subgroup $H$ of $G$ such that the multiplication map $G^\circ \times H \to G$ is surjective, and its kernel centralises $G^\circ$. (In the specific example above, we could take $H = \langle u, v\rangle$.) -$\DeclareMathOperator\Ad{Ad}\DeclareMathOperator\Gal{Gal}\DeclareMathOperator\Norm{Norm}\DeclareMathOperator\Weyl{W}\DeclareMathOperator\Zent{Z}\newcommand\C{{\mathbb C}}\newcommand\R{\mathbb R}\newcommand\adform{_\text{ad}}\newcommand\scform{_\text{sc}}\newcommand\X{\mathcal X}$To prove this, I'll use a few pieces of structure theory: - -All maximal tori in $G$ are $G^\circ$-conjugate. -All Borel subgroups of $G_\C$ are $G^\circ_\C$-conjugate. -For every maximal torus $T$ in $G$, the map $\Weyl(G^\circ, T) \to \Weyl(G^\circ_\C, T_\C)$ is an isomorphism. -If $G\scform$ and $(G_\C)\scform$ are the simply connected covers of the derived groups of $G^\circ$ and $G^\circ_\C$, then $(G\scform)_\C$ equals $(G_\C)\scform$. -Every compact Lie group has a finite subgroup that meets every component. - -I only need (4) to prove that, for every maximal torus $T$ in $G$, the map from $T$ to the set of conjugation-fixed elements of $T/\Zent(G^\circ)$ is surjective. This is probably a well known fact in its own right for real-group theorists. -Now consider triples $(T, B_\C, \X)$ as follows: $T$ is a maximal torus in $G$; $B_\C$ is a Borel subgroup of $G^\circ_\C$ containing $T_\C$, with a resulting set of simple roots $\Delta(B_\C, T_\C)$; and $\X$ is a set consisting of a real ray in each complex simple root space (i.e., the set of positive real multiples of some fixed non-$0$ vector). (Sorry about the pair of modifiers "complex simple".) I will call these 'pinnings', although it doesn't agree with the usual terminology (where we pick individual root vectors, not rays). I claim that $G^\circ/\Zent(G^\circ)$ acts simply transitively on the set of pinnings. -Once we have transitivity, freeness is clear: if $g \in G^\circ$ stabilises some pair $(T, B_\C)$, then it lies in $T$, and so stabilises every complex root space; but then, for it to stabilise some choice of rays $\X$, it has to have the property that $\alpha(g)$ is positive and real for each simple root $\alpha$; but also $\alpha(g)$ is a norm-$1$ complex number, hence trivial, for each simple root $\alpha$, hence for each root $\alpha$, so that $g$ is central. -For transitivity, since (1) all maximal tori in $G$ are $G^\circ$-conjugate, so (2) for every maximal torus $T$ in $G$, the Weyl group $\Weyl(G^\circ_\C, T_\C)$ acts transitively on the Borel subgroups of $G^\circ_\C$ containing $T_\C$, and (3) $\Weyl(G^\circ, T) \to \Weyl(G^\circ_\C, T_\C)$ is an isomorphism, it suffices to show that all possible sets $\X$ are conjugate. Here's the argument that I came up with to show that they are even $T$-conjugate; I think it can probably be made much less awkward. Fix a simple root $\alpha$, and two non-$0$ elements $X_\alpha$ and $X'_\alpha$ of the corresponding root space. Then there are a positive real number $r$ and a norm-$1$ complex number $z$ such that $X'_\alpha = r z X_\alpha$. Choose a norm-$1$ complex number $w$ such that $w^2 = z$. There is then a unique element $s\adform$ of $T_\C/\Zent(G^\circ_\C)$ such that $\alpha(s\adform) = w$, and $\beta(s\adform) = 1$ for all simple roots $\beta \ne \alpha$. By (4), we can choose a lift $s\scform$ of $s\adform$ to $(G\scform)_\C = (G_\C)\scform$, which necessarily lies in the preimage $(T_\C)\scform$ of (the intersection with the derived subgroup of) $T$, and put $t\scform = s\scform\cdot\overline{s\scform}$. Then -$$ -\alpha(t\scform) = \alpha(s\scform)\overline{\overline\alpha(s\scform)} = \alpha(s\scform)\overline{\alpha(s\scform)^{-1}} = w\cdot\overline{w^{-1}} = z, -$$ -and, similarly, $\beta(t\scform) = 1$ for all simple roots $\beta \ne \alpha$. Now the image $t$ of $t\scform$ in $G^\circ_\C$ lies in $T_\C$ and is fixed by conjugation, hence lies in $T$; and $\Ad(t)X_\alpha = z X_\alpha$ lies on the ray through $X'_\alpha$. -Since $G$ also acts on the set of pinnings, we have a well defined map $p : G \to G^\circ/\Zent(G^\circ)$ that restricts to the natural projection on $G^\circ$. Now $\ker(p)$ meets every component, but it contains $\Zent(G^\circ)$, so it need not be finite. Applying (5) to the Lie group $\ker(p)$ yields the desired subgroup $H$. Note that, as requested in your improved classification, conjugation by any element of $H$ fixes a pinning, hence, if inner, must be trivial.<|endoftext|> -TITLE: Two definitions of automorphic forms on Lie groups -QUESTION [6 upvotes]: My question is the about the equivalence of two definitions of automorphic forms on a semisimple Lie group. -The most common definition of automorphic forms on a semisimple Lie group $G$ with respect to a discrete subgroup $\Gamma$ is given by a smooth function $f:G\to \mathbb{C}$ with the following properties: - -left $\Gamma$-invariant, -$K$-finite, -$Z(\mathfrak{g})$-finite, -slowly increasing. - -Is this equivalent to the ones defined for $\operatorname{SL}(2,\mathbb{R})$ and $\operatorname{SL}(2,\mathbb{Z})$ where we have an automorphy factor $J(g,z)=cz+d$? -I also find a general definition with automorphy factors as follows. [Borel, Armand. "Introduction to automorphic forms." Proc. Symp. Pure Math. Vol. 9. No. 199210. 1966.] -I think given a classical automorphic form on $\operatorname {SL}(2,\mathbb{R})$, one could get a $\Gamma$-invariant form by multiplying an automorphy factor $j$. See page 283 of [Bump, Automorphic forms and representations]. -So I guess the two definitions differs by such a automorphic factor $j:G/K\times G\to K_\mathbb{C}$. - -REPLY [3 votes]: There are two notions: automorphic (or modular) form on a domain $G/K$, which would potentially have an "automorphy factor/cocycle" $\mu:\Gamma\times G/K\to \mathbb C^\times$ (or to a larger $GL_n$ for vector-valued automorphic forms), and automorphic forms on_a_group $G$, which have no cocycle, but instead are left $\Gamma$-invariant, and right $K$-equivariant. -When the cocycle extends from $\Gamma\times G/K$ to $G\times G/K$, the procedure in Borel's Boulder article gives an automorphic form on the group attached to an automorphic form on the domain. -A meaningful instance of a cocycle not extending from the discrete subgroup $\Gamma_o(4)$ to the natural ambient Lie group $SL_2(\mathbb R)$ is the example of half-integral-weight automorphic forms "on the upper half-plane", as in G. Shimura's papers around 1973.<|endoftext|> -TITLE: existence of birational morphism and divisors -QUESTION [5 upvotes]: The following result was metioned in a lecture: A nonsingular (or smooth) projective surface (variety of dimension 2) has a -birational morphism to the projective plane, if and only if there -exists an effective divisor D such that |D| is base point free -(i.e. for every point P there is a member D' of |D| with P not in D') -and D has self-intersection number 1. -However I didn't find the proof in any book. Dose some research paper contain the proof of the above result? - -REPLY [6 votes]: I don't know where to find a proof written, but it is not hard to give one here. -One direction is easy. If $S \rightarrow \mathbf P^2$ is a birational morphism, then let $D$ be the pullback of a hyperplane on $\mathbf P^2$. This is basepoint-free and has $D^2=1$, because both properties are preserved by pullback. -The converse is a little more work. Suppose $D$ is a divisor with the given properties. Since $|D|$ is basepoint-free, it defines a morphism $f: S \rightarrow \mathbf P^n$ for some $n$ such that $D$ is the preimage of a hyperplane. Note that $f(S)$ must have dimension 2, since if it had smaller dimension then we would get $D^2=0$. So $f$ is generically finite. Then $D^2$ equals $(\operatorname{deg}(f)) \cdot \operatorname{deg}(f(S))$. So $D^2=1$ implies that $\operatorname{deg}(f)=1$, which means $f$ is birational, and that $\operatorname{deg}(f(S))=1$, which means that $f(S)\cong \mathbf P^2$.<|endoftext|> -TITLE: Suprema of directed sets -QUESTION [16 upvotes]: Let $(X, \le)$ be a partially ordered set. We call a subset $S \subseteq X$... - -... a chain if each two elements in $S$ are comparable with respect to $\le$ (in other words, $S$ is linearly ordered with respect to $\le$). - -... directed if for all $x,y \in S$ there exists $z \in S$ that dominates $x$ and $y$. - - -Obviously, every chain is directed. -Question. Assume that every chain in $X$ has a supremum. Does it follow that every directed set in $X$ has a supremum? -Remarks. -(1) By Zorn's lemma, $X$ has a maximal element (in fact, every element of $X$ is dominated by a maximal element of $X$). -(2) Since the empty set is a chain and thus has a supremum, it follows that $X$ has a smallest element (though this doesn't seem to be particularly relevant to the question). -(3) Let $D \subseteq X$ be directed. We cannot apply Zorn's lemma directedly to $D$ since the supremum of a chain in $D$ might not be in $D$. What we can do is to add the set of all supremuma of subsets of $D$ (whenever they exist) to $D$, and thus obtain a new set $\tilde D$. Then $\tilde D$ is closed with respect to taking suprema, but I cannot see if (and why) $\tilde D$ is directed. -Actually, the answer to the question is yes if and only if this set $\tilde D$ is always directed: the implication "$\Rightarrow$" is trivial, and the implication "$\Leftarrow$" follows from applying Zorn's lemma to $\tilde D$ and from the fact that a maximal element in a directed set is always the supremum of this set. -(4) In general, a directed set does not necessarily contain a co-final chain. For instance, let $\mathcal{F}$ denote the set of all finite subsets of $\mathbb{R}$, ordered by set inclusion. Obviously, $\mathcal{F}$ is directed; but every union of a chain of finite sets if at most countable, so $\mathcal{F}$ does not contain a co-final chain. -(5) Let $D \subseteq X$ be directed. We can apply Zorn's lemma to the set $\mathcal{D}$ of all directed subsets of $D$ that have a supremum in $X$, or to the set $\mathcal{S}$ of all subsets of $D$ that have a supremum in $X$; so $\mathcal{D}$ and $\mathcal{S}$ both have a maximal element $D_{\max}$ and $S_{\max}$, respectively. But I see no way to show that $D_{\max}$ or $S_{\max}$ is co-final in $D$ (and thus equal to $D$). -(6) If $X$ is a lattice (i.e., every (non-empty) finite subset of $X$ has a supremum), then the answer to the question is yes: Indeed, let $D \subseteq X$ be directed, and let $\mathcal{S}$ denote the set of all subsets of $D$ that have a supremum in $X$. Then $\mathcal{S}$ contains all finite subsets of $D$, and $\mathcal{S}$ is stable with respect to monotone unions (i.e., unions of chains). This implies that $\mathcal{S}$ equals the power set of $D$, so in particular, $D \in \mathcal{S}$. -Motivation. In a preprint of mine I briefly considered a similar question in the context of ordered vector spaces, and I remarked that I do not know the answer in this specific vector space setting. Now, I'm about to submit a revision of this preprint, and I noted that I do not even know the answer for general partially ordered sets (without any vector space structure). - -REPLY [18 votes]: Yes, a poset that has suprema of all chains also has suprema of all directed sets. This is known, and I vaguely recall seeing it attributed to Solovay. The proof consists of showing, by induction on cardinals $\kappa$, that having suprema of all chains implies having suprema for all directed set of size $\leq\kappa$. -The case of finite $\kappa$ is trivial, since a finite directed set has a top element. The case of $\kappa=\aleph_0$ is almost trivial, since a countable directed set contains a cofinal chain. So assume from now on that $\kappa$ is uncountable and that $D$ is a directed subset of $X$ with $|D|=\kappa$. -Write $D$ as the union of an increasing (with respect to $\subseteq$) transfinite sequence of subsets $D_i$, each of cardinality $<\kappa$. (If $\kappa$ is a regular cardinal, then this sequence necessarily has length $\kappa$, but if $\kappa$ is singular then the sequence can be shorter.) We may assume that each $D_i$ is also directed; just choose for each pair in $D$ an upper bound in $D$, and close each $D_i$ under the "chosen upper bound" function. (The cardinality of the closure will be at most the maximum of $|D_i|^2$ and $\aleph_0$, so it is still $<\kappa$.) -So, by induction hypothesis, each $D_i$ has a supremum $s_i$ in $X$, and, since the $D_i$ form an $\subseteq$-increasing sequence, their suprema $s_i$ form a $\leq$-increasing sequence in $X$. By hypothesis, the sequence of $s_i$'s has a supremum $x$ in $X$, and it is easy to check that this $x$ also serves as the supremum of $D$.<|endoftext|> -TITLE: Automorphy of mixed Tate motives over $\mathbb{Z}$ -QUESTION [5 upvotes]: Deligne, Goncharov and Levine have constructed a Tannakian category of mixed Tate motives, MTM($\mathcal{O}_{K,S}$), over the ring of integers of a number field $K$ unramified outside a finite set of places $S$. -In particular there is a category MTM($\mathbb{Z}$) of mixed Tate motives unramified over $\mathbb{Z}$. Its objects are extensions of pure Tate motives $\mathbb{Q}(n)$ where $n$ is an integer. They arise from motivic sheaves on the moduli spaces of marked genus 0 curves. -The $\ell$-adic realization of the pure Tate motive $\mathbb{Q}(n)$ is the $Gal (\overline{\mathbb{Q}}/\mathbb{Q})$-module $\mathbb{Q}_\ell$, the Galois action being given by the $n$-th power of the $\ell$-adic cyclotomic character $\chi_\ell$. -In Langlands philosophy every motive should correspond to an automorphic form. For pure Tate motives $\mathbb{Q}(n)$ class field theory shows how: they correspond to powers of idèle class characters. - -Which automorphic form/representation does a truly mixed Tate motive correspond to? - -For example we know that $M^1_n := Ext^1(\mathbb{Q}(0), \mathbb{Q}(n)) = K_{2n-1}(\mathbb{Z})\otimes \mathbb{Q}$ for $n > 1$ where $K_.$ are the Milnor K-groups. What makes $M^1_n$ automorphic? - -REPLY [5 votes]: This answer is a slight addition to Joel's and David's. -In the theory of Galois representations, there is a general philosophy that $p$-adic phenomena (say in Hida families or eigenvarieties) reflect corresponding mod $p$ phenomena. So before asking if all extensions of $\mathbb Q_p(n)$ by $\mathbb Q_p$ can be constructed automorphically, one could ask the corresponding question mod $p$. Here is a slightly more general version, for the number field $\mathbb Q$: if $\psi$ and $\chi$ are two mod $p$ Dirichlet characters (of some conductor), thought of as Galois characters, whose product is odd, can any extension of $\chi$ by $\psi$ be realised in the reduction mod $p$ of a lattice in the (irreducible) $p$-adic Galois representation attached to some cusp form. -I think that the answer to this (and more general version of this) question is believed to be yes (folklore), and perhaps is even known to be yes in the particular case I just discussed. Questions of this type are certainly discussed in Skinner--Wiles: I think they will show that if you have a $p$-adic lattice in an irreducible representation whose reduction gives some extension class, then there is a cusp form giving rise to this extension. (They will need to assume ordinary at $p$, but actually recent work of L. Pan will let you replace ordinary by a more general Fontaine--Mazur-type condition.) Part of their technique is to move from one possible extension class to another. (Kisin used to call this ``Skinner--Wiles tunnelling''.) -I don't know if it's known that every extension of $\chi$ by $\psi$ actually can be lifted to a lattice that is ordinary (or, more generally, geometric) at $p$, though. If it were, one would get a positive answer to the mod $p$ version of the problem in some cases. -On a different note, I believe that Harder (and probably many others too) studied extension classes of automorphic motives arising from Eisenstein cohomology classes, and in this way constructed (or proposed construtions for) actual mixed motives, rather than just Galois representations. The first example of this would be looking at cohomology of open modular curves, or of complete modular curves relative to the cusps. One obtains mixed motives this way, but by Manin--Drinfeld, if you work over $\mathbb Q$ you get split extensions --- non-splitness only occurs with torsion coefficients. I think you can get more interesting things if you both delete some cusps, and then work relative to others (so that the weight filtration on cohomology has length three rather than just two), although it's been a long time since I thought about it (Mazur mentioned these ideas to me, in conjunction with Harder's name, when I was a grad student). I don't know what happens on higher dimensional Shimura varieties, but work of Harris and Zucker on mixed Hodge theory of Shimura varieties is probably relevant.<|endoftext|> -TITLE: Detecting nonorientability -QUESTION [11 upvotes]: Suppose that $M^m$ is an $m$-dimensional, compact, connected manifold without boundary, $m\geq 2$, not necessarily orientable. $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bZ}{\mathbb{Z}}$ -To a smooth map -$$ F: M\to\bR^m,\;\;F(p) =\big(\; f_1(p),\dotsc , f_m(p)\;\big) $$ -we associate the top degree form -$$ -\omega_F=df_1\wedge \cdots\wedge df_m= F^* (du^1\wedge \cdots \wedge du^m), -$$ -where $u^1,\dotsc, u^m$ are the canonical Euclidean coordinates on $\bR^m$. -For generic $F$, the zero locus of $\omega_F$ is a hypersurface $W_F$ in $M$ and the homology class $[W_F]\in H_{m-1}(M,\bZ/2)$ is the Poincare dual of the Stieffel-Whitney class $w_1(TM)\in H^1(M,\bZ/2)$. -The noncompact manifold $M_F=M\setminus W_F$ is oriented by the top degree form $\omega_F$. -If $M$ where orientable, then an orientation $or_M$ on $M$ induces an orientation on $M_F$ and -$$ -\int_{(M_F,or_M)}\omega_F=\int_{(M,or_M)} \omega_F =\int_{(M,or_M)}d(f_1df_2\wedge\cdots\wedge df_m)= 0. -$$ -Suppose that $M$ is nonorientable. Is it true that for any orientation $or$ on $M_F$ we have -$$ -\int_{(M_F,or)} \omega_F \neq 0? -$$ -Comment. In the initial form the question was ill posed as indicated by Robert Bryant and I modified it. Here is an alternate reformulation. -Let us point out that if we denote by $or_F$ the orientation on $M_F$ induced by $\omega_F$, then -$$ -\int_{(M_F,or_F)} \omega_F >0, -$$ -for any $M$, orientable or not. -If $(M_i)_{1\leq i\leq k}$ are components of $M_F$, then for any orientation $or$ on $M_F$ there exist $\epsilon_i=\pm$ such that the orientation $or_i$ on $M_i$ induced by $or$ satisfies $or_i=\epsilon_i or_F$. Thus the question can be rephrased as follows. -If $M$ is non orientable is it true that, for any $\epsilon_i=\pm 1$, we have -$$ -\sum_{i=1}^k \epsilon_i \int_{(M_i,or_F)} \omega_F\neq 0? -$$ - -REPLY [2 votes]: For certain $F$ and $\epsilon_i$ the answer is no. But it is probably yes for generic choices. -Here is an example with n=2 and $M$ the Klein bottle. We start with F being a standard projection of the Klein bottle into the plane. This is easy to visualize by googling for images of these glass Klein bottles that are everywhere. -With the standard projection we get that $W_F$ is an embedded circle and $M_F$ has just one component. So in that case the integral will be non-zero for each of the two possible orientations. -However now we are going to deform F slightly. We can imagine that our glass Klein bottle has a lump/wort/blister and that it is really quite large. This gives use a new projection from the Klein bottle to the plane. The image looks something like this: - -The new $W_F$ is the inverse image of the red curve. It is now two circles. $M_F$ now has two components, one of which is a disk. Topologically we have taken the old $M_F$ and removed a disk. Let me call these regions $M_0$ and $M_1$ (the disk) following the notation of the OP. -When we integrate $\omega_F$ over these two pieces we will get numbers which are essentially the areas of the images of these regions under $F$ (with signs depending on the $\epsilon_i$/$\textrm{or}_i$). Now we can imagine expanding the "wort" - making its image in the plane larger. This will change the magnitude of the integral over the two pieces $M_0$ and $M_1$. My original idea was that we could make it so that these numbers are exactly matched in magnitude. Then with the correct choice of $\epsilon_i$'s we can make these factors contribute oppositely so that for this particular $F$ and this particular choice of $\epsilon_i$'s the net integral vanishes. -However that does not quite work. The reason is the the "wort" is split into two halves. The, say, lower half contributes to $M_1$ (it essentially is the disk) but the upper half contributes to $M_0$. When we deform the "wort" the upper and lower half are deformed in matching ways. So we can't quite ensure that the magnitudes of the integrals become equal. -However, now what we can do is add a second "wort" onto the first. We make sure the second "wort" emerges from the lower half (the $M_1$ half). The corresponding image in the plane looks something like this: - -Now $W_F$ is three circles, and $M_F$ has a new component $M_2$ which is a disc which has been cut out of the old $M_1$. When we enlarge the new "wort" it again increases the magnitude of the integral on two regions, but now the two regions involved are $M_1$ and $M_2$. -So now we can adjust the size of the new wort so that the magnitude of the integral on $M_0$ exactly matches the sum of the magnitudes on $M_1$ and $M_2$. Now with appropriate choices of $\epsilon_i$'s, the total integral over $M_F$ will cancel. -I think it is always possible to do a similar trick/deformation.<|endoftext|> -TITLE: How aggressive is the fibrant replacement of $\mathrm{Bord}_n$? -QUESTION [18 upvotes]: Lurie (On the Classification of Topological Field Theories), with some corrections by Calaque and Scheimbauer (A note on the $(\infty,n)$-category of cobordisms), famously constructed a symmetric monoidal $(\infty,n)$-category $\mathrm{Bord}_n$ of $n$-dimensional smooth (co)bordisms. -Actually, the construction produces the following. Saying "such and such thing is an $(\infty,n)$-category" requires choosing a model of the $\infty$-category $\{\text{$(\infty,n)$-categories}\}$. By a "model" I mean a model category; "$(\infty,n)$-category" is then defined to mean fibrant-cofibrant object in the model. To foreshadow the question I'm going to get to, let me emphasize that a random object in the model category is not an $(\infty,n)$-category according to the model. Rather, it determines an $(\infty,n)$-category by (co)fibrant replacement. -Lurie and Calaque–Scheimbauer use the model of $(\infty,n)$-categories called "complete $n$-fold Segal spaces". The objects in the model are $n$-fold simplicial spaces, and the words "Segal" and "complete" refer to specific fibrancy conditions. Speaking approximately, "Segal" is a version of the requirement that composition be well-defined and single-valued: it rules out, for example, a situation where you have morphisms $f : X\to Y$ and $g : Y \to Z$, but no composition "$g\circ f$". "Completeness" has to do with invertibility: it says that a weak form of invertibility implies a strong form of invertibility. Fibrant replacement for the Segal condition goes through and adds in formal compositions of previously-incomposable pairs. Fibrant replacement for the completeness condition goes through and adds formal strong inverses to weakly-invertible things. -With this choice in mind, Lurie and Calaque–Scheimbauer build an $n$-fold simplicial space which deserves the name $\mathrm{Bord}_n$ in the following sense: its entries are moduli spaces of bordisms. They prove that $\mathrm{Bord}_n$ satisfies the Segal condition. However, they also emphasize that, for large enough $n$, the $n$-fold simplicial space $\mathrm{Bord}_n$ of Lurie and Calaque–Scheimbauer is not complete. The issue has to do with the h-cobordism theorem, which can be used to give morphisms that I am calling "weakly but not strongly invertible". -For the purposes of stating the Cobordism Hypothesis, this in some sense "doesn't matter". First, as I foreshadowed above, $\mathrm{Bord}_n$ has a completion $\widehat{\mathrm{Bord}}_n$: when people talk about "the $(\infty,n)$-category of cobordisms" and cite Lurie or Calaque–Scheimbauer, they really mean $\widehat{\mathrm{Bord}}_n$ and not $\mathrm{Bord}_n$. Second, if $\mathcal{C}$ is any true $(\infty,n)$-category, i.e. a complete $n$-fold Segal space, then $\hom(\mathrm{Bord}_n,\mathcal{C}) = \hom(\widehat{\mathrm{Bord}}_n,\mathcal{C})$. So a universal-property characterization like "$\hom(\mathrm{Bord}_n,\mathcal{C}) = \{\text{fully-dualizable objects in $\mathcal C$}\}$" will hold for both if it holds for either. -On the other hand, the replacement procedure to complete $\mathrm{Bord}_n \leadsto \widehat{\mathrm{Bord}}_n$ could be quite aggressive: in general, completion might require adding in lots of extra morphisms. For instance, one thing that I know that completion does is that it takes any h-cobordism $W$ between $M$ and $N$, slices it up $W$ parallel to the $M$- and $N$-boundaries, and then reads the slices as a new "smooth" family of manifolds connecting $M$ and $N$. In other words, this procedure goes through and potentially-aggressively modifies the moduli spaces of manifolds from what you originally thought they were. -The general version of my question is: - -Just how aggressive is the completion $\mathrm{Bord}_n \leadsto \widehat{\mathrm{Bord}}_n$? How much does it add to the moduli spaces of $k$-dimensional bordisms? - -Here is a specific version of my question. The existence of h-cobordisms is related to the existence of manifold factors, which are non-manifold spaces $X$, like the dogbone space, such that $X \times \mathbb R \cong \mathbb R^N$, or more generally $X \times M \cong N$ for some manifolds $M$ and $N$. - -Do manifold factors show up as $k$-morphisms in $\widehat{\mathrm{Bord}}_n$? Do fully-extended TFTs assign values to manifold factors? - -REPLY [17 votes]: The completeness condition is not really about making things invertible which weren't already. It is about where the information about invertible morphisms is stored. -We can already see this with $(\infty,1)$-categories of $n$-dimensional bordisms. -Since $\mathrm{Bord}_n$ satisfies the Segal conditions, it makes sense to talk about compositions and invertibility of morphisms (points in the space of morphisms). What completeness would require is that if you have an invertible morphism (i.e. an invertible bordism), then it is already represented by a path in the space of objects. -In the bordism category (assuming we avoid dimension 4) the invertible morphisms are exactly the h-cobordisms. -In the Lurie and Calaque–Scheimbauer models, the space of objects is a union of $\operatorname{BDiff}(M)$, and the paths in this space correspond to bordisms which are mapping cylinders. In general there are h-cobordisms which are not mapping cylinders. So the "completeness fibrant replacement" would change the space to of objects to something like the classifying space of the subcategory of h-cobordisms. -But all of this is just about manifolds and you are not adding anything really new to the underlying category. You are just rearranging the furniture in the house. ;) -In particular manifold factors do not show up. -As a side note, the Lurie and Calaque–Scheimbauer models are also not Reedy fibrant (which is a condition even before Segal and complete). This means that for example there can be a morphism from $X$ to $Y$, and a path in the space of objects from $X$ to $X'$, but no morphism from $X'$ to $Y$.<|endoftext|> -TITLE: Can every cancellative invertible-free monoid be embedded in a group? -QUESTION [5 upvotes]: A monoid is invertible-free if $xy=1$ implies $x=y=1$ for all $x,y$. -Question: Can every cancellative invertible-free monoid be embedded in a group? -I'm fairly sure that a quotient of the free product of such a monoid with its mirror (this is the monoid with the same elements and identity but reversed multiplication, i.e. $x\cdot y=yx$) is the "most general" group in which it can be embedded. -This is the non-commutative version of the construction of the integers from the natural numbers. -Does this appear anywhere in the literature as a problem / proposition / theorem? - -REPLY [5 votes]: No, it is not true even for finitely generated monoids. Take any semigroup $S$ which is cancellative and does not embed into a group (first examples were constructed by Malcev). Consider the monoid $S^1$ which is $S\sqcup\{1\}$ with $1$ a (new if $S$ is a monoid) neutral element. Then $S^1$ is an invertible-free monoid which does not embed into a group. It is cancellative iff $S$ does not have a neutral element.<|endoftext|> -TITLE: Inner model theory without choice -QUESTION [8 upvotes]: How much of the inner model project can be constructed without assuming the axiom of choice? I.e. which large cardinals provably have canonical inner models not assuming choice? - -REPLY [10 votes]: Addressing the first question: I should argue that Choice comes in almost at the beginning of the inner model project, if we regard proving Covering Lemmata as an integral part of that project: one defines a model under an anti-large cardinal assumption; proves that it is rigid; the Covering Lemma is then proven as holding over that model. One then has the contrapositive that if Covering fails, then so does your anti large cardinal hypothesis, and then we move on to define the ‘next’ inner model that accommodates that large cardinal. -With that in mind, then in ZF alone one can indeed prove that assuming $\neg O^\sharp$ then the (Jensen Strong) Covering Lemma holds over $L$. Dodd & Jensen then defined as the ‘next’ model $K^{DJ}$, the core model ‘below’ a measurable cardinal. The general Covering Lemma $CL$, $\Gamma$ say, then ran: -$\Gamma$: “ Assume $\neg O^{\dagger}$. Then either the Covering Lemma holds over $K^{DJ}$, or it holds over $L[\mu_0]$ where the latter is the least ‘$\rho$-model’, i.e. has a measure $\mu_0$ on some least possible ordinal $\kappa_0$. Or it holds over $L[C]$ where $C$ is a Prikry sequence over $L[\mu_0]$”. -If these three alternatives fail, then we have $O^{\dagger}$ and we look to build the next model, culminating in a model with two measurable cardinals {\em &c.} But $\Gamma$ can only be proven in ZFC. Reason: if one looks at $M = \bigcap_{i< \omega^2}L[\mu_i]$ (where $L[\mu_i]$ is the $i$t’h iterate of the least $\rho$-model, now on $\kappa_i$) this is a ZF model, but the set of Prikry sequences in the model cofinal in the ordinal $\kappa = \kappa_{\omega^2}$ is not wellorderable in $M$. One then ends up with $(\neg \Gamma)^M$.<|endoftext|> -TITLE: Two interpretations of implication in categorical logic? -QUESTION [13 upvotes]: I am a bit confused about the interpretation of "implication" in the standard treatment of categorical logic, for example in [Bart Jacobs 1999] "Categorical Logic and Type Theory". -(A). On the one hand, the book says that "terms give rise to morphisms between contexts, this is the canonical way to produce a category from types" (p.5) Under this interpretation, type contexts are the objects in the category, also called indices, such that fibers can be built over indices. Under the Curry-Howard isomorphism, this arrow would correspond to an implication $\Rightarrow$ in logic. -(B). On the other hand, "each fiber category $P(I)$ is a Boolean algebra... with propositional connectives such as $\wedge, \vee, \top, \bot, \neg$ in (Boolean) logic" (p.12, ibid). This seems to be the approach used by Lawvere when he formulated the notion of topos, with the sub-object classifier $\Omega$. Under this interpretation, one can also create the (Boolean) logic implication operator $A \Rightarrow B$ defined as $\neg A \vee B$ (in Boolean algebra, but one can also do similarly in Heyting algebra). Would this $\Rightarrow$ be in conflict with the $\Rightarrow$ in (A)? -Edit: More specifically, if there is an arrow $A \Rightarrow B$ in the fiber, as in (B), how does it correspond (via Curry-Howard) to some type or term in the base category? - -REPLY [9 votes]: The answers of varkor and Dmitri explain that in a given category, entailment corresponds to external homsets while implication corresponds to internal-homs. However, there's another thing going on here too that may be relevant to your question. When interpreting a logic over a type theory, the corresponding categorical structure is a fibration, in which there is both a base category (corresponding to types and terms) and a fiber category (corresponding to propositions and entailment). The Curry-Howard interpretation of "logic" happens only in the fiber category. Thus, the internal-hom in the base category corresponds to a function-type, while it is the internal-hom in the fiber category that corresponds to implication.<|endoftext|> -TITLE: Is this Laurent phenomenon explained by invariance/periodicity? -QUESTION [5 upvotes]: In Chapter 4 (subsection "Somos sequence update") of his Tracking the Automatic -Ant, David Gale -discusses three families of recursively defined sequences of numbers, all -due to Dana Scott and inspired by the Somos sequences: - -Sequence 1. Fix a positive integer $k\geq 2$. Define a -sequence $\left( a_{0},a_{1},a_{2},\ldots \right) $ of positive rational -numbers recursively by setting -\begin{align*} -a_{n}=1\qquad \text{for each }n -TITLE: A question about subspace in ${\bigwedge}^2({\mathbb R}^n)$ -QUESTION [33 upvotes]: Let $E$ be a linear subspace of ${\bigwedge}^2({\mathbb R}^n)$. What is the minimal dimension of $E$ that guarantees $E$ contains a nonzero element of the form $X\wedge Y$, with $X, Y\in{\mathbb R}^n$? -When $n=3$, dimension $1$ is enough. When $n=4$ we would need dimension $4$. For general $n$, it is easy to see $E$ having dimension $\frac{(n-1)(n-2)}{2}+1$ is sufficient, but I don't know if that is optimal. - -REPLY [13 votes]: One more exact answer: for $n=8$ the minimal dimension is $22$, -again attaining the "easy" bound ${n-1 \choose 2} + 1$ -and the same as the value for the next dimension $n=9$. -First to explain why ${n-1 \choose 2} + 1$ is enough, -not just for real vector spaces but for an $n$-dimensional vector space $V$ -over any field. -Fix nonzero $X_0 \in V$. Then $X_0 \wedge V = \{ X_0 \wedge Y : Y \in V \}$ -is a linear subspace of dimension $n-1$ in $\bigwedge^2 V$. -Thus if $E \subset \bigwedge^2 V$ is a linear subspace of codimension $n-2$ -then it must have nonzero intersection with $X_0 \wedge V$. -Now take $n=8$ and identify $V$ with the Cayley octonions. -Let $V_0 \subset V$ consist of the "purely imaginary" octonions, -so $V$ is the orthogonal direct sum of $\bf R$ with $V_0$. -Write $\bigwedge^2 V = ({\bf R} \wedge V_0) \oplus \bigwedge^2 V_0$, -and let $E$ be the kernel of the homomorphism $h: \bigwedge^2 V \to V_0$ -that takes $1 \wedge X$ to $X$ and $X \wedge Y$ to the imaginary part of -the octonion $XY$, for any $X,Y \in V_0$. -[This is well-defined because $XY + Y\!X \in \bf R$ for all $X,Y \in V_0$, -so $h(Y\wedge X) = - h(X \wedge Y)$.] Then $E$ has dimension $21$. -I claim that $E$ contains no nonzero pure tensors. Indeed a pure tensor in -$\bigwedge^2 V$ has the form $1 \wedge X$, $X \wedge Y$, or $(1+X) \wedge Y$ -for some $X,Y \in V_0$ which are linearly independent (so in particular nonzero). -Certainly $h(1\wedge X) = X$ is nonzero. So is $h(X \wedge Y)$, -because if $XY \in \bf R$ for $X,Y \in V_0$ then $X$ and $Y$ are proportional. -Finally $h\bigl((1+X) \wedge Y\bigr) = Y + h(X \wedge Y)$ cannot be zero because -the imaginary part of $XY$ is orthogonal to $Y$. -It also follows that for $n=6,7$ the minimal $\dim E$ is at least -${n \choose 2} - 6 = 9, 15$, and thus exceeds the lower bound -${n-2 \choose 2} + 1 = 7, 11$ coming from the dimension of the Grassmannian. -Replacing the Cayley octonions by the Hamilton quaternions -recovers the answer of $4$ for $n=4$.<|endoftext|> -TITLE: Do we have an algorithm for comparing $e^e$ with rationals? -QUESTION [6 upvotes]: Do we have an algorithm for comparing $e^e$ with rationals, with a known time to convergence? -In a non-constructive sense, there obviously is an algorithm. - -If $e^e$ is some rational $q_0$, then we can decide if a rational $q$ is bigger than or smaller than or equal to $e^e$ by comparing it with $q_0$. -If $e^e$ is irrational, then we calculate the sequences -$$z_n =\left(1+\frac{\small{1}}{n}\right)^n$$ -$$a_n = \left(1+\frac{z_n}{n}\right)^n$$ -$$\phantom{2+}b_n = \left(1+\frac{z_n}{n}\right)^{n+2}$$ -If $a_n>q$, then $e^e>q$; and if $b_n -TITLE: Improved classification of compact Lie groups -QUESTION [16 upvotes]: This question is a follow-up to Classification of (not necessarily connected) compact Lie groups. In the answer to that question, @LSpice proved that any compact, not necessarily connected Lie group $G$ takes the form -$$ -G = \frac{G_0 \rtimes R}{P} -$$ -where $G_0$ is the identity component of $G$, $R$ is a finite group, and $P$ is a finite, common subgroup of $G_0$ and $R$ that is central within $G_0$ (but need not be central within $R$). -Nonetheless, there are many possibilities for the semi-direct product. To narrow the list, it would be convenient to separate out those elements of $R$ that act by non-trivial outer automorphisms on $G_0$ and modify the rest so that they commute with $G_0$. -UPDATE: my original hypothesis (below) is false. A weaker, possibly correct version is: -Hypothesis: $R$ and $P$ can be chosen above such that every element of $R$ either (1) acts by a non-trivial outer automorphism on $G_0$ or (2) acts trivially on $G_0$. -UPDATE 2: @LSpice proved this in the updated answer to Classification of (not necessarily connected) compact Lie groups. A concise rephrasing of the proof is given in my answer below. - -By comparison, this is false: -Hypothesis: Any compact Lie group $G$ can be written in the form -$$ -G = \frac{(G_0 \times H) \rtimes R}{P} -$$ -where $H, R, P$ are finite groups and non-trivial elements of $R$ act by non-trivial outer automorphisms on $G_0$. -Counterexample: consider $G = U(1) \rtimes \mathbb{Z}_4$, where the generator $r$ of $\mathbb{Z}_4$ acts by the ``charge conjugation'' outer automorphism $r^{-1} e^{i \theta} r = e^{-i \theta}$ on $U(1)$. In any finite extension $G'$ of this group, elements of $\pi_0(G)$ that act by charge conjugation will never square to the identity in $G'$, so $G'$ never takes the required $(G\times H) \rtimes \mathbb{Z}_2$ form with $\mathbb{Z}_2$ acting on $U(1)$ by charge conjugation. - -REPLY [2 votes]: @LSpice has already proven my revised conjecture in the updated answer to Classification of (not necessarily connected) compact Lie groups, but let me give another, closely related proof. -Since $1\to \mathrm{Inn}(G_0) \to \mathrm{Aut}(G_0) \to \mathrm{Out}(G_0) \to 1$ always splits, see Does Aut(G) → Out(G) always split for a compact, connected Lie group G?, we can choose a subgroup $R_0 \subseteq \mathrm{Aut}(G_0)$ for which the restriction of $\mathrm{Aut}(G_0) \to \mathrm{Out}(G_0)$ is an isomorphism. The inverse image of $R_0$ under the map $f:G \to \mathrm{Aut}(G_0)$ induced by conjugation is a subgroup $K \subseteq G$ whose intersection with $G_0$ is $Z(G_0)$. -Multiplying any $g\in G$ by arbitrary $h \in G_0$ multiplies the associated $f(g) \in \mathrm{Aut}(G_0)$ by an arbitrary inner automorphism $f(h) \in \mathrm{Inn}(G_0)$, without changing $g$'s connected component. Thus, $K$ meets every connected component of $G$. -Using the result of In any Lie group with finitely many connected components, does there exist a finite subgroup which meets every component?, $K$ has a finite subgroup $R$ that meets every component of $K$, hence it meets every component of $G$ as well, and intersects $G_0$ within $Z(G_0)$. By design, the elements of $R$ either act by non-trivial outer automorphisms on $G_0$ or they act trivially on $G_0$. This proves my (revised) conjecture. - -COMMENT ADDED: An interesting, yet false, generalization is stated and disproven below. -It is well known that any compact, connected Lie group $G_0$ takes the form -$$G_0 = \frac{T^k \times G_1 \times \ldots \times G_\ell}{P}$$ -where $T^k$ denotes a $k$-torus, $G_1, \ldots, G_\ell$ are compact, simply connected, simple Lie groups, and $P$ is central. One might think that the quotients in the expressions for $G$ and $G_0$ could be combined, so that any compact Lie group $G$ would take the form: -$$ -G = \frac{(T^k \times G_1 \times \ldots \times G_\ell) \rtimes R}{P} -$$ -where as before each element of $R$ acts by a non-trivial outer or acts trivially on $T^k \times G_1 \times \ldots \times G_\ell$. However, this is false. -Counterexample: Consider $G=(\mathrm{SO}(2k) \rtimes \mathbb{Z}_4) / \mathbb{Z}_2$, where the generator $r \in \mathbb{Z}_4$ acts by parity on $\mathrm{SO}(2k)$ and $r^2 = -1 \in SO(2k)$. Now let $G’=(\mathrm{Spin}(2k) \rtimes R)/P$ be a cover of $G$ whose connected component is $G_0'=\mathrm{Spin}(2k)$. There is some element $r'$ of $R$ that projects to $r$, hence $r’$ acts on $\mathrm{Spin}(2k)$ by parity. If $k$ is odd, then $Z(G_0') = \mathbb{Z}_4$, and $(r’)^2$ must be one of the two elements of order 4 in $Z(G_0')$ to project to $(r)^2 = -1$. However, parity exchanges these two elements, so we find $(r’)^{-1} (r’)^2 r’ \ne (r’)^2$, which is a contradiction. The case of even $k$ is very similar.<|endoftext|> -TITLE: Is there a polytope with an essentially unique shape? -QUESTION [5 upvotes]: More percisely: - -Question: Is there a (convex) polytope that has a unique realization up to, say, projective transformations? - -I suppose I have to assume that it has more than $d+2$ vertices/facets if it is $d$-dimensional, to avoid some trivial cases such as simplices. -By realization I mean just any other polytope that has the same combinatorics as the original one, that is, the same faces being incident in the same way. -For example, the hexagon is not such a polytope, as there are many irregular hexagons that are not projective transformations of the regular hexagon. - -More generally, such a polytope cannot be simple of simplicial, as in these cases there are generic ways to move vertices or facets to change its shape while keeping the combinatorics. -I am not sure that projective transformations are the most general thing one wants to exclude to make this an interesting question. -I am open for other ideas. -Here is another similar question: - -Question II: Is there a centrally symmetric polytope that has a unique realization up to linear transformations? - -This time with $> d+1$ vertices/facets if it is $d$-dimensional. - -REPLY [2 votes]: What you are after seems to be projectively unique polytopes. And so the paper of Adiprasito and Ziegler may answer your question. Perhaps two papers in the references of that paper will give you many examples, as they contain operations that preserve this kind of uniqueness. - -A construction for projectively unique polytopes by Perles, MA and Shephard, GC -Constructions for projectively unique polytopes by McMullen, P. - -Regards, Guillermo<|endoftext|> -TITLE: All non-compact simply connected $2$-manifolds with boundary -QUESTION [8 upvotes]: There are two corresponding posts MSE and MSE by me without any answers. - -Problem: Let $\Sigma$ be a non-compact simply-connected $2$-dimensional manifold, -with boundary. Then, up to homeomorphism $\Sigma$ is of the form: delete a closed subset from -the boundary $\Bbb S^1$ of the closed unit disc. - -Motivation: The lemma 1.8., on page 30 of the book A Primer on Mapping Class Group by Benson Farb and Dan Margalit, says the following: - -Two simple closed curves on a surface having finitely many -intersection points and having no bi-gon, whenever lifted(assuming the -existence of liftings) to the universal cover intersects at most one -point. - -Now, the lemma has been proved for closed hyperbolic surfaces. One crucial step in proving this lemma is: there are two arcs of two liftings together bound a disc in the universal cover $\Bbb H$, which is possible by the plane Jordan curve theorem. In other words, as long as the universal cover is $\Bbb S^2$ or a convex subset of $\Bbb R^2$, the argument of the lemma 1.8. works fine. -My Thoughts: I am trying to use collaring the boundary of $\Sigma$ to get a non-compact simply connected surface without boundary. I know there is a technique for proving any non-compact simply connected surface without boundary is homeomorphic to $\Bbb R^2$, and this technique is straight forward, in the sense that it does not use the classification theory(considering genus, number of compact boundary components, orientability, isomorphic diagram) of all $2$-dimensional manifolds. Also, $\Sigma$ is contractible. -Any help in proving the problem will be appreciated. Thanks in advance. - -REPLY [10 votes]: Here is one proof, using the Uniformization Theorem. This proof will be easier in the setting of the "Primer" since the authors are considering universal covering spaces of complete hyperbolic surfaces with geodesic boundary. -Start with a simply-connected surface with boundary $S$ and let $DS$ denote the double of $S$ along its boundary. Then $DS$ admits an involution $\tau$ fixing $\partial S\subset DS$ pointwise. This all can be done smoothly. Now, put a $\tau$-invariant Riemannian metric on $DS$; this defines a conformal structure on $DS$ with respect to which $\tau$ is an antiholomorphic involution. Let $X$ denote the universal covering space of $DS$ with lifted conformal structure. Then $\tau$ lifts to antiholomorphic involutions on $X$. By the Uniformization Theorem, $X$ is conformal to the unit disk (which I will equip with the Poincare metric) or complex plane or $S^2$. I will consider the first case since the proof in the two other cases is similar but simpler. The surface $S$ lifts diffeomorphically to a subsurface $Y\subset X$ (since $S$ is simply connected). Each boundary component $c$ of $Y$ in $X$ is fixed by a lift $\sigma_c$ of $\tau$. Since $\sigma_c$ is an antiholomorphic involution of the unit disk, it is a hyperbolic isometry, hence, its fixed-point set is a geodesic in the hyperbolic plane ${\mathbb H}^2$. Thus, $Y$ is a closed convex subset in ${\mathbb H}^2$. Now, switch to the projective (Klein) model of the hyperbolic plane. Every convex subset of ${\mathbb H}^2$ then becomes a convex subset of the Euclidean plane. It is an elementary exercise to see that each compact convex subset (with nonempty interior) of the Euclidean plane is homeomorphic to the closed unit disk ${\mathbb D}$ in ${\mathbb R}^2$. Under this homeomorphism $cl_{{\mathbb R^2}}(Y)\to {\mathbb D}$, $Y$ maps to the complement to a closed subset of the boundary of ${\mathbb D}$. Since $Y$ is homeomorphic to your surface $S$, you get the statement you are after.<|endoftext|> -TITLE: Forms of ${\rm SL}(2)$ -QUESTION [9 upvotes]: I know all real forms of ${\rm SL}(2,{\Bbb C}$). They are ${\rm SL}(2,{\Bbb R})$ and ${\rm SU}(2)$. -Moreover, ${\rm SL}(2,{\Bbb R})$ is isomorphic to ${\rm SU}(1,1)$. Thus I can say that all real forms of ${\rm SL}(2,{\Bbb C})$ -are of the form ${\rm SU}(2,F_\lambda)$, where $F_\lambda$ is the diagonal Hermitian form on ${\Bbb C}^2$ -with matrix ${\rm diag}(1,\lambda)$, $\lambda$ taking the values 1 and $-1$. - -Question. Is it true that any ${\Bbb Q}$-form of ${\rm SL}(2,{\Bbb C})$ is isomorphic to ${\rm SU}(2,F_{K,\lambda})$, -where $F_{K,\lambda}$ is the diagonal Hermitian form on $K^2$ for some quadratic extension $K/{\Bbb Q}$ with matrix ${\rm diag}(1,\lambda)$, -for some $\lambda\in {\Bbb Q}^\times$ ? - -REPLY [14 votes]: $\DeclareMathOperator\SL{SL}\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\Br{Br}\DeclareMathOperator\U{U}\DeclareMathOperator\disc{disc}\DeclareMathOperator\Nm{Nm}\DeclareMathOperator\diag{diag}\DeclareMathOperator\SU{SU}\DeclareMathOperator\GL{GL}$The general theory tells us that forms of a reductive group $G$ are classified by $H^1$ of the automorphism group, which is a semidirect product $(\text{root datum automorphisms}) \ltimes (\text{inner automorphisms})$. For $\SL_2 / k$ the root datum has no automorphisms, so the classifying object is $H^1(k, \PGL_2(\overline{k})) \cong H^2(k, \smash{\overline{k}}^\times)[2] = \Br(k)[2]$. Hence, by standard results on the Brauer group, a form of $\SL_2$ over a number field is uniquely determined by the set of places at which it's locally non-split, and the only constraint on this set is that it must have even size. -Now, let's look at unitary groups $\U(V)$ over local fields. If $k$ is local and $K$ is a degree 2 étale ring extension of $k$, then $K$ is either $k \oplus k$ or a field extension. In the former case, any Hermitian space wrt $K/k$ is obviously split. In the latter case, the isomorphism class of $V$ is uniquely determined by the discriminant $\disc(V) \in k^\times / \Nm(K^\times)$. For your example, with matrix $\diag(1, \lambda)$, the discriminant is $-\lambda$. When $V$ is split, then one computes that $\SU(V) \cong \SL(2)$, as in the familiar argument for $k = \mathbf{R}$ (of course $\U(V)$ is not $\GL(2)$ unless $K = k \oplus k$!) -Conversely: it's non-obvious, but true, that if $V$ is non-split then $\SU(V)$ is also non-split. If it were split, then the Borel subgroup of $G$ would have to have a fixed point in $\mathbf{P}(V)$ by the Borel–Morozov theorem; and the restriction of the Hermitian form to this line has to be zero, hence $V$ is split. -Going back to the global case, if $K = k(\sqrt{\alpha})$, and $\beta = -\lambda$, we have shown that $\SU(F_{K,\lambda})$ is split locally at $v$ if and only if $(\alpha, \beta)_{k_v} = 1$, where $(-, -)_{k_v}$ is the Hilbert symbol. In other words, $\SU(F_{K,\lambda})$ is split locally at $v$ if and only if the quaternion algebra $D = (\alpha, \beta)_k$ splits there. So $\SU(F_{K,\lambda})$ is isomorphic to $(D^\times)^{\Nm = 1}$. -Since every quaternion algebra is $(\alpha,\beta)_{k}$ for some $\alpha$ and $\beta$, we conclude the answer to the question is "Yes: if $k$ is a number field, every $\overline{k}/k$-form of $\SL_2 / k$ is isomorphic to such an $\SU(2, F_{K, \lambda})$ for some quadratic extension $K/k$ and $\lambda \in k^\times$". -[PS: With a little more work, one should be able to write down explicitly an isomorphism from $\SU(F_{k(\sqrt{\alpha}),\lambda})$ to the norm 1 elements of $D^\times$ where $D = (\alpha, -\lambda)_k$, using the fact that $D$ is naturally a 2-dim'l vector space over $k(\sqrt{\alpha})$. But it's a Sunday morning so I'm going to be lazy and not do the exercise.]<|endoftext|> -TITLE: Approximating a real in the ground model -QUESTION [11 upvotes]: Let $\mathbb{P}$ be a proper notion of forcing, having the Sacks property. Suppose that $\dot{D}$ is a $\mathbb{P}$-name for an infinite subset of $\omega$. I'm looking for a set which approximates $\dot{D}$ both from above and below, that is: -Is there a set $A\subseteq\omega$ (in the ground model) and a $p\in\mathbb{P}$ such that (1) $p\Vdash\dot{D}\subseteq\check{A}$, and (2) for any finitely many elements $a_1,\ldots,a_n\in A$, there is a $q\leq p$ such that $q\Vdash a_1,\ldots,a_n\in\dot{D}$? -It might be useful to note that if $q_n$ is a decreasing sequence in $\mathbb{P}$ such that each $q_n$ decides whether $n\in\dot{D}$, then $A=\{n:q_n\Vdash n\in\dot{D}\}$ satisfies (2), though I don't think it satisfies (1) if $\Vdash\dot{D}\notin\mathbf{V}$. Meanwhile, the Sacks property ensures that there is a set in the ground model satisfying (1), but at least for the set you get by naively applying the Sacks property, it need not satisfy (2). - -REPLY [14 votes]: The answer is no. Here is a counterexample: For definiteness, let's work with $\mathbb P$ equal to Sacks forcing, though the proof works verbatim for any reasonable forcing whose generic can be understood as a real. Let $s$ be Sacks generic over $V$ and let $\dot{D}$ be the name for the set in the extension where for each $n \in \omega$ we let $2n \in D$ iff $s(n) = 1$ and $2n + 1 \in D$ iff $s(n) = 0$. So, for instance, if the first four bits of $s$ are, say, 0110 then $1, 2, 4, 7 \in D$ and $0, 3, 5, 6 \notin D$. Note that for each $n$ exactly one of $\{2n, 2n+1\}$ is in $D$. Also clearly we can read off $s$ from $D$ so it's not in the ground model. -Now suppose towards a contradiction that there is an $A \subseteq \omega$ satisfying your conditions for this $\dot{D}$. It can't be the case that there is an $n \in \omega$ so that $2n, 2n+1 \in A$ since that gives us two elements of $A$ for which no $q$ can force both of them to be in $\dot{D}$, contradicting (2) in your question. Also, for no $n$ can it be the case that neither $2n$ nor $2n+1$ is in $A$ since one of them is in $\dot{D}$ and hence in this case $A$ wouldn't cover $\dot{D}$. Therefore exactly one of $2n$, $2n+1$ is in $A$ for each $n$, but this means that $A$ must be equal to $D$, which is a contradiction.<|endoftext|> -TITLE: Does Aut(G) → Out(G) always split for a compact, connected Lie group G? -QUESTION [7 upvotes]: The outer automorphism group of a topological group $G$ is constructed by the short exact sequence -$$ -1\longrightarrow \operatorname{Inn}(G) \longrightarrow \operatorname{Aut}(G) \longrightarrow \operatorname{Out}(G) \longrightarrow 1. -$$ -This sequence does not always split, see Non-split Aut(G) $\to$ Out(G)?, for example for the discrete group $G = A_6$. -I am interested in the case where $G$ is a compact, connected Lie group. Does the sequence always split in this case? (If $G$ has a simple Lie algebra $\mathfrak{g}$ then I believe the answer is yes.) - -REPLY [5 votes]: Yes, $\operatorname{Aut}(G) \to \operatorname{Out}(G)$ always splits. The proof is just as in my answer to your question Classification of (not necessarily connected) compact Lie groups: regard $\operatorname{Aut}(G)$ as an extension of $\operatorname{Inn}(G) = G/\operatorname Z(G)$ by a discrete group $\operatorname{Out}(G)$, and lift $\operatorname{Out}(G)$ to $\operatorname{Aut}(G)$ as the automorphisms that preserve a pinning in the sense of that answer. (These are often called "diagram automorphisms".) Over in that other question we did not get an honest section of the component group inside the Lie group because you did not assume that the identity component was centreless, but since the adjoint group $\operatorname{Inn}(G)$ is centreless, everything is fine here.<|endoftext|> -TITLE: Divergent sums of reciprocals with unique factorization property -QUESTION [6 upvotes]: Let $A=\{a_n|n\in\mathbb{N}\}$ be a sequence of positive integers with the following properties: - -$1 -TITLE: Diophantine equation $3(a^4+a^2b^2+b^4)+(c^4+c^2d^2+d^4)=3(a^2+b^2)(c^2+d^2)$ -QUESTION [9 upvotes]: I am looking for positive integer solutions to the Diophantine equation $3(a^4+a^2b^2+b^4)+(c^4+c^2d^2+d^4)=3(a^2+b^2)(c^2+d^2)$ for distinct values of $(a,b,c,d)$. -There are many solutions with $a=b$ such as $(7,7,11,13)$ and $(13,13,22,23)$, and many solutions with $c=d$ such as $(3,5,7,7)$ and $(7,8,13,13)$, but it seems not so many with $4$ distinct values. The smallest distinct solution I know of is $(35,139,146,169)$. -Are there infinitely many solutions? Is there perhaps a parameterization of such solutions? -Of special interest are solutions where the quantity $(c^2+d^2)-(a^2+b^2)$ is a perfect square, such as $(a,b,c,d)=(323,392,407,713)$. Are there other solutions of this type? -This comes up in solving an unsolved problem in geometry that I can't discuss, but you might get your name on a paper if you can help. - -REPLY [3 votes]: $3(a^4+a^2b^2+b^4)+(c^4+c^2d^2+d^4)=3(a^2+b^2)(c^2+d^2)\tag{1}$ -We show above equation gives infinitely many integer solutions if c = b+d. -We get -$$a^2 = \frac{c^2+d^2-b^2}{2}+\frac{\sqrt{-3c^4+6c^2d^2+18b^2c^2-3d^4+18b^2d^2-27b^4}}{6}\tag{2}$$ -Hence we find the rational solutions of quartic curve for $c$ below $$v^2 = -3c^4+(18b^2+6d^2)c^2+18b^2d^2-27b^4-3d^4\tag{3}$$ -Let $c = b+d$, then RHS of equation $(3)$ becomes to $-12b^2(b^2-2bd-2d^2)$. -Parametric solution of $b^2-2bd-2d^2=-3u^2$ is given -$(b,d)= \left( \dfrac{-5-2k+k^2}{(k+2)(k-1)}, \dfrac{2(1+k+k^2)}{(k+2)(k-1)} \right)$. -Hence equation $(3)$ has a solution $(c,v)= \left( \dfrac{7+4k+k^2}{(k+2)(k-1)}, \dfrac{6(k^2+4k+1)(-5-2k+k^2)}{(k+2)^2(k-1)^2} \right)$. -k is arbitrary. -Then we get $a^2 = \dfrac{3(k^4+4k^3+4k^2+3)}{(k+2)^2(k-1)^2}$. -Hence we find the rational solutions of quartic curve -$$V^2 = 3(k^4+4k^3+4k^2+3)\tag{4}$$ -Quartic curve $(4)$ can be transformed to elliptic curve below. -$Y^2 = X^3 -156X + 560$ with rank 1 and generator is $P(X,Y)=(2,-16)$. -Hence this cuve has infinitely many rational points. -One of rational points is $2P(X,Y)=(\frac{65}{4}, \frac{-385}{8})$. -This point gives a solution $(a,b,c,d)=(1407459, 656995, 1391911, 2048906)$. -Thus equation $(1)$ gives infinitely many integer solutions. -Another solution of equation $(1)$ with $c\neq a\pm b, d\neq a\pm b$. -$a = 7n^4+100n^3+42n^2+100n+7$ -$b = 13n^4+28n^3+174n^2+28n+13$ -$c = 19n^4+112n^3+174n^2-56n+7$ -$d = 7n^4-56n^3+174n^2+112n+19$ -$n$ is arbitrary. -$n = 5$ is excluded. - n a b c d - [2 143 133 199 67] - [3 247 217 373 61] - [4 9271 8029 14599 1459] - [6 32791 29341 54751 3931] - [7 3367 3097 5761 433] - [8 9263 8773 16207 1339] - [9 481 469 859 79] - [10 175207 175693 318847 32539] - -Numerical solutions of equation $(1)$. -$(a,b,c,d)<1000$ -fg1: 1 ($c=a\pm b,d=a\pm b$) -fg2: s ($c^2+d^2-a^2-b^2$ is square number) -There are many solutions such as $c\neq a\pm b, d\neq a\pm b$. -Total number of solutions : $378$ -Case of $c=a\pm b,d=a\pm b: 274(72.5$ percecnt) -Others: $104(27.5$ percent) -There are two solutions such as $c^2+d^2-a^2-b^2$ is square number. -$(a,b,c,d)=(645, 323, 713, 407),(645, 392, 713, 407)$ -There is no other solution with $(a,b,c,d)<10000$ - a b c d fg1 fg2 - 5 3 7 2 1 - 8 7 13 1 1 - 16 5 19 11 1 - 24 11 31 13 1 - 31 24 43 11 - 33 7 37 26 1 - 35 13 43 22 1 - 39 16 49 23 1 - 43 35 62 13 - 45 32 67 13 1 - 49 39 74 11 - 51 40 79 11 1 - 56 39 67 31 - 56 9 61 47 1 - 57 35 77 23 - 57 55 97 2 1 - 63 17 73 46 1 - 77 40 103 37 1 - 80 19 91 61 1 - 85 11 91 74 1 - 88 65 133 23 1 - 91 69 139 22 1 - 95 24 109 71 1 - 97 73 122 49 - 105 104 181 1 1 - 112 75 163 37 1 - 115 56 151 59 1 - 120 13 127 107 1 - 120 23 133 97 1 - 143 133 199 67 - 143 25 157 118 1 - 145 119 229 26 1 - 152 65 163 103 - 152 35 143 133 - 152 147 247 23 - 152 35 143 133 - 155 87 193 77 - 155 144 259 11 1 - 160 87 217 73 1 - 161 15 169 146 1 - 161 151 266 13 - 165 56 199 109 1 - 168 85 223 83 1 - 169 31 161 151 - 175 32 193 143 1 - 187 93 247 94 1 - 192 133 283 59 1 - 195 29 211 166 1 - 203 72 247 131 1 - 205 163 247 122 - 208 17 217 191 1 - 209 136 301 73 1 - 217 95 277 122 1 - 221 64 259 157 1 - 224 31 241 193 1 - 231 185 361 46 1 - 240 229 341 109 - 247 105 313 142 1 - 247 168 271 163 - 247 217 373 61 - 253 240 427 13 1 - 259 176 379 83 1 - 261 19 271 242 1 - 272 105 337 167 1 - 273 152 373 121 1 - 279 40 301 239 1 - 280 111 349 169 1 - 280 133 337 157 - 280 201 259 241 - 287 240 457 47 1 - 288 35 307 253 1 - 291 115 314 199 - 299 205 439 94 1 - 301 275 499 26 1 - 304 161 409 143 1 - 315 88 367 227 1 - 320 21 331 299 1 - 323 37 343 286 1 - 325 193 407 157 - 325 264 511 61 1 - 333 115 403 218 1 - 336 215 481 121 1 - 343 255 473 122 - 344 305 539 59 - 351 329 589 22 1 - 352 123 427 229 1 - 357 80 403 277 1 - 368 297 577 71 1 - 368 335 473 217 - 369 175 481 194 1 - 385 23 397 362 1 - 387 208 523 179 1 - 391 129 469 262 1 - 399 41 421 358 1 - 400 377 673 23 1 - 407 48 433 359 1 - 416 235 571 181 1 - 423 280 613 143 1 - 424 395 559 241 - 425 184 541 241 1 - 433 217 559 218 - 437 88 487 349 1 - 440 43 463 397 1 - 440 189 559 251 1 - 441 319 661 122 1 - 451 104 511 347 1 - 456 25 469 431 1 - 459 245 619 214 1 - 473 135 553 338 1 - 475 141 559 334 1 - 477 403 763 74 1 - 480 371 739 109 1 - 493 312 703 181 1 - 495 448 817 47 1 - 504 409 581 337 - 513 200 637 313 1 - 517 203 643 314 1 - 520 341 751 179 1 - 520 147 607 373 1 - 527 385 793 142 1 - 528 47 553 481 1 - 533 27 547 506 1 - 536 465 781 169 - 539 240 691 299 1 - 544 531 931 13 1 - 545 441 806 139 - 551 265 721 286 1 - 552 145 637 407 1 - 559 56 589 503 1 - 559 240 671 329 - 560 429 859 131 1 - 561 464 889 97 1 - 575 49 601 526 1 - 583 320 793 263 1 - 589 336 811 253 1 - 589 555 946 91 - 595 549 991 46 1 - 600 217 733 383 1 - 611 120 679 491 1 - 613 245 598 473 - 616 29 631 587 1 - 616 159 709 457 1 - 616 559 889 229 - 621 104 679 517 1 - 624 325 851 301 - 627 413 907 214 1 - 637 155 727 482 1 - 645 323 713 407 s - 645 392 713 407 s - 649 455 961 194 1 - 665 336 671 481 - 665 193 767 473 - 667 165 763 502 1 - 675 53 703 622 1 - 679 151 722 539 - 680 427 967 253 1 - 689 231 829 458 1 - 695 488 817 403 - 696 385 949 311 1 - 704 295 889 409 1 - 705 31 721 674 1 - 713 232 853 481 1 - 715 272 883 443 1 - 721 344 923 379 - 725 112 787 613 1 - 728 55 757 673 1 - 728 165 823 563 1 - 728 703 937 469 - 735 64 769 671 1 - 744 169 869 581 - 759 305 949 454 1 - 767 360 997 407 1 - 775 177 877 598 1 - 776 489 869 481 - 795 136 871 659 1 - 800 33 817 767 1 - 805 189 869 626 - 825 248 973 577 1 - 832 183 937 649 1 - 840 59 871 781 1 - 845 328 763 713 - 883 480 893 637 - 899 61 931 838 1 - 901 35 919 866 1 - 907 155 938 767 - 912 427 973 613 - 921 91 913 851 - 935 72 973 863 1 - 936 775 989 739<|endoftext|> -TITLE: When a $C^*$-algebra is an ideal in its second dual? -QUESTION [7 upvotes]: I would like to know which $C^*$-algebras are ideals in their second duals? -There is a paper by S. Watanabe that claims in introduction that it is well known that a $C^*$-algebra is an ideal in its second dual iff it is a dual $C^*$-algebra. But I do not know what does he mean by term "dual $C^*$-algebra". -For the general case of normed algebras there is a criterion (see Banach algebras and the general theory of *-algebras. Volume I. Algebras and Banch algebras. Theodore W. Palmer, theorem 1.4.13): $A$ is a two-sided ideal in $A^{**}$ iff the maps $L_a:A\to A$, $b\mapsto ab$ and $R_a:A\to A$, $b\mapsto ab$ are weakly compact for all $a\in A$. -May be this criterion could be improved for the case of $C^*$-algebras. You may even assume that $A$ is commutative. - -REPLY [8 votes]: Warning: the following is just what I found from some work on MathSciNet, following Yemon's hint in the comments. It's not meant to be accurate historical notes. -A "dual" $C^\ast$-algebra is defined as follows. Let $A$ be an algebra and for a subset $M\subseteq A$ let $R(M) = \{ x\in A : Mx=\{0\}\}$; similarly define $L(M) = \{ x\in A : xM=\{0\}\}$. Then a $C^\ast$-algebra is dual if for each closed left ideal $I$ we have that $L(R(I))=I$. (The involution can be used to show that the analogous definition with right ideals gives the same notion). -An early paper which studied these is Kaplansky, The structure of certain operator algebras, see section 2. It seems that Berglund, Ideal $C^\ast$-algebras was the first to obtain the equivalence you seek (why is the Duke journal archive behind a paywall??) A short proof is in McCharen, A characterization of dual $B^{\ast}$-algebras These give the following: -Claim: A $C^\ast$-algebra $A$ is an ideal in its bidual if and only if $A$ is dual. -We now combine this with known characterisations of dual $C^\ast$-algebras: -Claim: A $C^\ast$-algebra $A$ is dual if and only if $A$ is isomorphic to a $C^\ast$-subalgebra of $K(H)$ for some $H$ if and only if $A$ is the $c_0$-direct sum of algebras of the form $K(H)$ for some (finite or infinite dimensional) $H$. -Thus $K(H)$ and $c_0$ really are the archetypal examples. These results are quoted in the paper of Kaplansky I linked above, and in Dixmier's book (English edition, section 4.7.20) though in both cases it seems only further references are given, not proofs. I am afraid that I don't know of a modern, self-contained treatment.<|endoftext|> -TITLE: Is there a canonical resource for mathematical Lorem ipsum text? -QUESTION [8 upvotes]: I'm designing a template intended for mathematical writing. I'd like to fill the template with Lorem ipsum text, but would like that text to showcase some of the common ways that mathematics is typeset. I could use some randomly generated text using either LaTeX's blindtext package or a random paper generator (although the phony formulas in those can get pretty crunchy). But is there some canonical mathematics text used for this purpose that has the same meaninglessness as Lorem ipsum? Or is there some classical text (language doesn't matter) that would be cool to use for this? - -REPLY [13 votes]: As a "showcase", and a test for fonts, I use fonttest.tex<|endoftext|> -TITLE: Why strong limit cardinals in the definition of condensed sets? -QUESTION [16 upvotes]: A condensed set à la Clausen–Scholze is, as far as I understand it, a small sheaf on the large site of profinite spaces. In Scholze's notes they are described as being objects of a category that is a large sequential colimit of toposes, each of which is $\mathrm{Sh}(*_{\kappa\text{-proét}})$, where the index $\kappa$ runs over the uncountable strong limit cardinals. The $\kappa$-proétale site $*_{\kappa\text{-proét}}$ consists of the profinite spaces whose underlying set is smaller than $\kappa$ (the topology is for now not important). The inclusions of sites induce fully faithful inverse image functors between the toposes, and the colimit defining the condensed sets is given by the colimit along these inverse image functors. -We can concretely describe this colimit as having objects pairs $(\kappa,F)$ consisting of an uncountable strong limit cardinal and a sheaf on $*_{\kappa\text{-proét}}$, where a map $(\kappa,F) \to (\kappa',G)$ is a map of sheaves on $*_{\lambda\text{-proét}}$ where $\lambda = \max\{\kappa,\kappa'\}$ where we have included the sheaf in the smaller topos into the larger topos (i.e. extended it to the larger site in the usual way). -Now I would like to know what goes wrong if I don't take this limit only over uncountable strong limit cardinals, but rather over all cardinals. (In this setup, a condensed set would be a pair consisting of an arbitrary cardinal $\mu$ and a sheaf on $*_{\mu\text{-proét}}$; morphisms are as above.) The strong limit cardinals just give a cofinal sequence, and it's not clear to me that the result is not equivalent. If there are specific properties of strong limit cardinals that are used, what are they? - -REPLY [8 votes]: Let me write down an answer to mark the question as answered. -As David points out, the definition would not change if one used all cardinals $\kappa$ instead of the strong limit ones, as the latter are cofinal in the former. -But strong limit cardinals are used to study the "individual layers" appearing in the definition : $\kappa$-condensed sets/abelian groups etc. -To prove that one can compare those to sheaves on $\kappa$-small profinite sets, or $\kappa$-small extremally disconnected sets, one uses that $\kappa$ is strong limit; see propositions 2.3, 2.5 of the notes and their proofs. -In turn, this comparison is very useful technically speaking, as sheaves on the site of $\kappa$-small extremally disconnected sets are just product preserving functors, and thus the category of sheaves enjoys very nice properties (with respect to co/limits for instance); and also the functors that connect these categories of sheaves are also particularly well-behaved, which allows one to make "local" computations of co/limits, which is again useful to prove specific things about their behaviour. For examples of this, see proposition 2.9 and theorem 2.2 of the notes. -As Peter pointed out in the comments, another thing that's good in $\kappa$-condensed abelian groups that doesn't work in general, but does if $\kappa$ is strong limit, is the fact that they are generated by compact projectives.<|endoftext|> -TITLE: Ternary sequences satisfying $ x_i + y_i = 1 $ for some $ i $ -QUESTION [12 upvotes]: Consider a set of strings $ {\mathcal S} \subset \{0, 1, 2\}^n $ satisfying the following two conditions: 1.) every string in $ {\mathcal S} $ has exactly $ k $ symbols from $ \{0, 1\} $ (i.e., $ \forall x=x_1 \cdots x_n \in {\mathcal S} \;\; |\{ i : x_i = 2 \}| = n - k $), and 2.) for every two strings $ x, y \in {\mathcal S} $ there exists a location $ i \in \{1, \ldots, n\} $ such that either $ x_i = 0, y_i = 1 $, or $ x_i = 1, y_i = 0 $ (in other words, such that $ x_i + y_i = 1 $). -Denote by $ S_{k,n} $ the maximum possible cardinality of a set satisfying these two conditions. Clearly, $ S_{k,n} \geq S_{k,k} = 2^k $, for any $ n \geq k $. My feeling is that $ S_{k,n} $ cannot be "much bigger" than $ 2^k $, regardless of $ n $ (in fact, I couldn't find an example with $ S_{k,n} > 2^k $). Can one derive an upper bound that would establish this? The precise statement I am interested in proving is the following: -$$ \lim_{k \to \infty} \frac{1}{k} \log_2 \max_{n \in \mathbb{N}} S_{k,n} \stackrel{?}{=} 1 .$$ - -REPLY [9 votes]: I claim that $S_{k,n}=2^k$ for all $n\geqslant k$. Moreover, $\sum 2^{-m_i}\leqslant 1$ if the set of strings satisfies this condition, and $m_i$ denotes the number of zeros/ones in $i$-th string. -Proof. Toss a coin for each location and consider the following events enumerated by your strings: if the string has 1/0 at some location, the corresponding coin must come out heads/tails correspondingly. No two events may occur together, thus the sum of their probabilities is at most 1.<|endoftext|> -TITLE: Elementary questions on the Freyd-Mitchell embedding theorem -QUESTION [7 upvotes]: I have a few elementary questions about the Freyd-Mitchell embedding theorem which I can't see answered elsewhere here and which all arise from some confusions I ran into. - -Can someone point out why (and where in the proof) exactly you need that the abelian category $\mathcal{A}$ is small? Is it to show that the functor category $\mathsf{Fun}(\mathcal{A},\mathsf{Ab})$ is Grothendieck? - -Does the FH embedding theorem also hold if $\mathcal{A}$ is only essentially small? I would say so but have seen this nowhere written out. If $\mathcal{A}$ is essentially small, there is an equivalence $F \colon \mathcal{A} \to \mathcal{A}'$ into a small category $\mathcal{A}'$. The category $\mathcal{A}'$ is automatically abelian and $F$ is an exact equivalence. Composed with the FH embedding of $\mathcal{A}'$ we get an exact embedding of $\mathcal{A}$ into a module category. Where's the mistake? - -This one really confuses me. I thought that an exact embedding $F \colon \mathcal{A} \to R\text{-}\mathsf{Mod}$ induces an isomorphism $\mathrm{Sub}(X) \to \mathrm{Sub}(F(X))$ between the lattices of subobjects of $X$ and $F(X)$, therefore maps simple objects to simple modules etc., and I can deduce the Jordan–Hölder theorem for a finite length object in $\mathcal{A}$ simply from the one for modules. But the comments here seem to indicate that this is wrong. Why is it so? I must be stupid here. - -REPLY [10 votes]: You need this to know that $Fun(A,Ab)$ is locally small; and also indeed to check that $Fun(A,Ab)$ has nice properties such as Grothendieck-ness, or the existence of injective cogenerators etc. In other words it's needed for most of the steps that lead to the reduction to the case of a small subcategory of a nice Grothendieck abelian category. - -Yes of course it works, and your argument shows that. I guess it's never written out because category theorists don't really care about the difference between small and essentially small. I would be surprised though if it were actually never mentioned. - -It doesn't induce an isomorphism, only an embedding: given a subobject $Y\to F(X)$, how do you cook up a $Z$ such that $F(Z) \cong Y$ ? For instance look at the exact embedding $\mathbb Q-Mod\to \mathbb Z-Mod$, it certainly does not map simples to simples. - -REPLY [9 votes]: Here is an explicit counterexample when $A$ is not assumed to be small. Any abelian category admitting an exact (fully faithful) embedding into $\text{Mod}(R)$ must be well-powered, meaning every object must have a set of subobjects (since the same is true in $\text{Mod}(R)$ and an exact embedding induces an embedding on posets of subobjects, but not, as Maxime points out, an isomorphism). There are abelian categories that are not well-powered; you can see some examples in this MO question. I particularly like Jeremy Rickard's example of the (still locally small!) category of eventually constant functors $\text{Ord} \to \text{Ab}$.<|endoftext|> -TITLE: Subtracting the weak limit reduces the norm in the limit -QUESTION [12 upvotes]: Question -Let $X$ be some reflexive Banach space. Suppose $x_n$ is some sequence in $X$ that weak converges to some $y \neq 0$. Is it the case that -$$ \limsup \|x_n - y\| < \limsup \|x_n\| ?$$ -Positive with Hilbert spaces -In the Hilbert space case this is true, as -$$ \langle x_n - y, x_n -y \rangle = \| x_n\|^2 + \|y\|^2 - \langle x_n, y\rangle - \langle y, x_n\rangle $$ -and by weak convergence the latter two terms both converges to $\|y\|^2$ and we get -$$ \limsup \|x_n - y\|^2 = \limsup \|x_n\|^2 - \|y\|^2. $$ -A negative example with a non-reflexive Banach space and weak* convergence -Let $X = L^\infty(\mathbb{R})$, and take $x_n = \mathbf{1}_{[-2,-1]} + \mathbf{1}_{[n,n+1]}$. Then $x_n$ weak* converges to $\mathbf{1}_{[-2,-1]}$ but $\|x_n - \mathbf{1}_{[-2,-1]}\|_{\infty} = 1 = \|x_n\|_\infty$. - -REPLY [11 votes]: The property you indicate is known as (strict) Opial’s Property (see https://en.m.wikipedia.org/wiki/Opial_property). It fails generally in reflexive spaces; in fact, it fails generally even for uniformly convex spaces where it is equivalent to Delta convergence (see https://en.m.wikipedia.org/wiki/Delta-convergence). Indeed, for $L^p[0,1]$, where $p\ne 2$, Opial’s Property fails. Using the duality pairing $\langle\cdot\,,\cdot\rangle\colon X\times X^*\to \mathbb{R}~or~\mathbb{C}$ and (the normalized) duality map $$J(x):=\{f\in X^*:\|f\|^2=f(x)=\|x\|^2\}\,,$$ Brailey Sims (https://www.mathshunter.edu.au/brailey/Research_papers/As%20Support%20Map%20Characterisation%20of%20the%20Opial%20Conditons.pdf) characterised spaces with the strict Opial Property as those spaces such that whenever $x_n$ converges weakly to $x\ne 0$, then $$\limsup_{\substack{n\to\infty\\j x_n\in J x_n}}\langle x\,,j x_n\rangle>0\,.$$ -(The $\limsup$ may be equivalently replaced with $\liminf$). From this it is easy to deduce that the sequence spaces $\ell^p$, where $1\le p<\infty$ have Opial’s Property.<|endoftext|> -TITLE: Given a complex vector bundle with rank higher than 1, is there always a line bundle embedded in it? -QUESTION [13 upvotes]: When I read the GTM 082, "The Splitting Principle of the complex vector bundle", I see that in the proof we split off one subbundle at a time by pulling back to the projectivization of a quotient bundle. I don't know why we need to pull this back to the projectivization of the quotient bundle. I mean if there is always a line bundle embeded in the given complex vector bundle, we can just split it on the original vector bundle, every time we take quotient of these two bundle, we get a quotient bundle with dimensional reduced by 1. So in the similar way we can conclude that every bundle can be decomposed into the direct sum of several line bundle. So is it true? Can we always find a line bundle embedded into a given complex vector bundle? - -REPLY [8 votes]: I think thinking in terms of classifying spaces will help clarify the situation. We know that a rank $n$ complex vector bundle $V$ on $X$ is the same thing as a homotopy class of maps $f:X\to BU_n$. A choice of decomposition $V\cong L\oplus W$ where $L$ is a line bundle corresponds to a lift of $f$ to $B(U_1\times U_{n-1})\cong BU_1\times BU_{n-1}$. The homotopy fiber of the map $BU_1\times BU_{n-1}\to BU_n$ is $U_n/(U_1\times U_{n-1})\cong \mathbb{CP}^{n-1}$. This is a non-trivial principal fibration and so it has no section. -In particular you can take the universal vector bundle on $BU_n$ (or if you want, its restriction to some finite skeleton given by a suitable complex Grassmannian) and this will give you a vector bundle that admits no line bundle. -Note that $BU_1\times BU_{n-1}$ is precisely the projectivization of the universal vector bundle on $BU_n$, so this is just another way of phrasing the splitting principle. -More precisely you can find a sequence of obstructions in $H^{i+1}(X;\pi_i\mathbb{CP}^{n-1})$ for $i\ge2$ and you can find a sub-line bundle if and only if these obstruction vanish. For example you can always find a such a line bundle if the (cohomological) dimension $X$ is less or equal than 2<|endoftext|> -TITLE: Toroidal alternating sign matrices -QUESTION [6 upvotes]: Consider $n\times k$ matrices with entries from $\{0,1,-1\}$ such that the sum in each row and each column is 0 and the non-zero numbers in each row/column alternate in sign (so, they alternate if we make a torus from an $n\times k$ table). Were such things studied? Enumerated? - -REPLY [3 votes]: Recall Kuperberg's proof of the ASM conjecture [arXiv:math/9712207] via a relation to a quantum-integrable two-dimensional lattice model from classical statistical mechanics, namely a special case of the six-vertex model with domain-wall boundaries. See also the excellent book by Bressoud, Proofs and Confirmations (The Story of the Alternating-Sign Matrix Conjecture). -The same mapping will relate the problem you're interested in to a special case of the six-vertex model with toroidal boundary conditions. Let me denote the three weights of the (symmetric) six-vertex model by $a,b,c$ as usual for this model. For ASM it suffices to consider the special case with vertex weights $a=b$ and $c$ free, which is also known as the F-model in the physics literature. Its partition function for toroidal boundaries is a generating function for $x$-enumerations, including $x=1$ (counting the number of ASM) and $x=2$ (for domino tilings), of the ASM you're after. -Its partition function was found exactly long ago by Lieb [PRL 18 (1967) 1046]. The main difference with the case of usual ASM (or domain-wall boundaries in terms of the six-vertex model) is that for finite size the result is not given in a closed form, but only up to solving systems of algebraic equations known as the "Bethe-ansatz equations". For sufficently low size this could be done numerically. The large-size asymptotics of the partition function, however, is known in closed form due to Lieb; and since then surely also including the leading finite-size corrections. (See also the rigorous work of Duminil-Copin and collaborators on related models using the Bethe ansatz.) -Since the model is invariant under (discrete) translations in either direction there is no phase separation, and there are no limit shapes. - -Edit. Let me outline how the problem of $x$-enumerating toroidal alternating-sign matrices, which I will abbreviate to tASM, is turned to that of solving coupled algebraic equations. -Step 1. Turn the problem into linear algebra. In physics this is called the transfer-matrix method. I also think that this is a computationally quite efficient way to find the number of tASM for sufficiently low $n,k$. Without loss of generality assume that $n\geq k$. -This goes as follows. Let me use the bijection that you mentioned between tASM and Eulerian (divergence-free) orientations of the square lattice with toroidal boundary conditions; the $\pm1$ correspond to the source/sink orientations around a vertex. -Consider the partition function (moment-generating function) -$$ Z = \sum_{C \,:\, \text{config}} \text{weight}(C) , \quad \text{weight}(C) = \prod_{v \, : \, \text{vertex}} \text{weight}(v) $$ -for your favourite choice of vertex weights, e.g. $\text{weight}(v) = 1$ if orientation around $v$ is Eulerian, $=0$ else; or e.g. a refinement that counts the number of signs. -$$ Z = \text{tr}[t^k] $$ -where $t$ is a $2^n \times 2^n$ matrix constructed as follows. Let $V = \mathbb{C} \, e_\uparrow \, \oplus \, \mathbb{C} \, e_\downarrow$ be the complex vector space with basis vectors $e_s$ labelled by the the two possible orientations of a single edge, $s \in \{\uparrow,\downarrow\}$. Its $n$-fold tensor product $W = V^{\otimes n}$ has basis $\{ e_{s_1} \otimes \dots \otimes e_{s_n} \}_{s_1,\dots,s_n}$ indexed by all possible configurations on $n$ edges, which we will think of as the $N$ vertical edges connecting two adjacent rows of vertices of the lattice. The transfer matrix is the linear operator $T \in \mathrm{End} \, W$ is defined as -$$\langle e_{s'_1} \otimes \dots \otimes e_{s'_n}, t \, e_{s_1} \otimes \dots \otimes e_{s_n}\rangle = \sum_{C} \prod_{v \, : \, \text{vertex} \,\in\, \text{row}} \text{weight}(v) $$ -where the sum is over all configurations of orientations for the horizontal edges connecting the row of vertices, with periodic boundary conditions in the horizontal direction. As my description indicates we think of this as follows. Fix a row of $n$ vertices in the lattice. The matrix entries defined above give the weight (probability up to normalisation) where the configuration $e_{s_1} \otimes \dots \otimes e_{s_n}$ on the row of vertical edges directly below our row of vertices is transferred to the configuration $e_{s'_1} \otimes \dots \otimes e_{s'_n}$ on the row of vertical edges directly above it. -Define the $\mathfrak{sl}_2$-weight (in the sense of representation, not probability, theory) of a configuration on a row of vertical edges to be the number of $1\leq i\leq n$ such that $s_i = \,\downarrow$. Then, as the orientation is Eulerian and using the horizontal periodicity, $t$ preserves the $\mathfrak{sl}_2$-weight. The nonzero entries of $t$ precisely correspond to transfering a configuration with a given $\mathfrak{sl}_2$-weight to any other configuration with that same $\mathfrak{sl}_2$-weight. Moreover, the above sum defining the matrix entry of $t$ contains two terms if $s_i = s'_i$ for all $i$ (namely: all horizontal edges have the same orientation) and one term otherwise. -Next, $t^2$ transfers a configuration through two rows of the lattice (composition corresponds to summing over all possible configurations on the intermediate vertical edges). Likewise, the matrix entries of $t^k$ gives the (probabilistic) weight of all possible configurations on an $n\times k$ lattice with prescribed configurations on the vertical edges at the bottom and the top, and periodic boundary conditions in the horizontal direction. Finally, its trace (over $W$) amounts to imposing periodic boundary conditions in the vertical direction as well to get $Z$. -(In practice one defines $t$ in turn by a similar slicing of the row, thinking of it as consisting of a successing of operators --- acting e.g. from left to right --- encoding the weights of a single vertex. All of this is usually done via graphical notation, see e.g. the left column on p5 of the paper [arXiv:1702.05474] that I wrote with Rick Keesman, take `staggering' therein to be zero.) -Step 2. Compute the eigenvalues of $t$. Consider the case where the vertex weights are invariant under reversing the orientations on all edges. Then there are three vertex weights $a,b,c$ to specify. Overall rescalings only give an overall factor for $Z$ so are irrelevant. This leaves a two-parameter family of six-vertex models. It is known that at fixed value of the combination -$$ \Delta := \frac{a^2 + b^2 - c^2}{2ab}$$ -the transfer matrices commute with each other at different values of the remaining parameter. This one-parameter family of commuting transfer matrices can be simultaneously diagonalised. By $\mathfrak{sl}_2$-weight conservation this may be done per fixed $\mathfrak{sl}_2$-weight. This is what can be achieved, as I mentioned above, via an ansatz for the form of the eigenvectors that is due to Bethe. If the $\mathfrak{sl}_2$-weight is $m$ then this involves $m$ free parameters that have to solve a set of coupled algebraic equations to guarantee that they are actually eigenvectors. Depending on the (probabilistic) weights the dominant eigenvalue, most relevant when $k$ grows, occurs either for $\mathfrak{sl}_2$-weight $m=0$ ('ferromagnetic') or instead at the 'equator' $m=n/2$ (if $n$ is even, 'antiferromagnetic'). In particular one can compute the dominant eigenvalue (via Perron--Frobenius) and its large-$n$ asymptotics. For the details see e.g. the recent preprint [arXiv:2012.11675] by Duminil-Copin et al.<|endoftext|> -TITLE: limit of vector sequence is in range of limit of matrix sequence -QUESTION [5 upvotes]: $\{A_k\}_{k=1}^{\infty}$ is a convergent sequence of real $n \times d$ matrices of rank $n$, and the limit $\lim_{k \to \infty}A_k := A_*$ also has rank $n$. -$\{x_k\}_{k=1}^{\infty}$ is a convergent sequence of real $d$ dimensional vectors, $\lim_{k \to \infty}x_k := x_*$. -We are given that $x_k \in \text{Range}(A_k^T)$ for all $k$. Is it true that $x_* \in \text{Range}(A_*^T)$? - -REPLY [3 votes]: Here is another way of seeing that the answer is "yes". -Recall that the orthogonal projection onto $\mathrm{Range}(A_k^T)$ is $P_k = A_k^T(A_kA_k^T)^{-1}A_k$. Since $A_k$ has full rank $n$ for each $k$, the inverse in the middle always exists, and since the limit $A_*$ also has full rank $n$ that inverse does not blow up (i.e., the singular values of $A_kA_k^T$ are bounded away from $0$). Thus $\lim_{k\rightarrow \infty}P_k = P_* = A_*^T(A_*A_*^T)^{-1}A_*$ is the orthogonal projection onto $\mathrm{Range}(A_*^T)$, so -$$ -\|P_*x_* - x_*\| = \lim_{k\rightarrow \infty} \|P_k x_k - x_k\| = \lim_{k\rightarrow\infty}0 = 0, -$$ -so $P_*x_* = x_*$, so $x_* \in \mathrm{Range}(A_*^T)$.<|endoftext|> -TITLE: Generalized cancelation properties ensuring a monoid embeds into a group -QUESTION [10 upvotes]: Context: an obvious necessary condition for a monoid to embed into a group (as submonoid) is to satisfy the left and right cancelation rules: -$$xy=xz \quad\Longrightarrow y=z;$$ -$$yx=zx \quad\Longrightarrow y=z.$$ -It's sufficient for commutative monoids, by an easy standard construction. However, in general it's known not to be sufficient, as already mentioned at MO (see this question and this question). The first such construction is due to Malcev. Malcev's proof (1936), as described in this 1969 paper by R. Johnson (Proc AMS, link with unrestricted access), consists in checking that a in a group, we have the (straightforward) "generalized cancelation" rule: -$$ea=db,eb=fa,ec=fb \quad \Longrightarrow \quad eb=dc$$ -Malcev's result consists then in constructing a cancelative monoid in which this rule fails (which is the less trivial part, and is not my point here). - -Motivated by the above, we can define a generalized cancelation rule as a rule of the form -$$w_1=w'_1,\dots,w'_n=w'_n \quad \Longrightarrow \quad w_0=w'_0$$ -where $w_i,w'_i$ are non-negative words in some countable alphabet. A monoid is said to satisfy this generalized cancelation rule if it satisfies the above implication for each replacement of the letters by monoid elements. -Let $\mathcal{G}$ be the set of generalized cancelation rules that are satisfied by all groups. -Clearly a monoid that embeds into a group, satisfies all rules in $\mathcal{G}$. One can first ask about the converse: if a monoid satisfies all rules in $\mathcal{G}$, does it embed into a group? -The answer is actually a trivial yes! Indeed, starting from such a monoid $M$, define the enveloping group $i:M\to G_M$ in the obvious way (presentations: generators = $M$, relators = monoid law). Then $i$ is injective: indeed every relation of the form $i(m)=i(m')$ can be interpreted as some generalized cancelation rule, and eventually implies $m=m'$. -At a formal level this hence provides a characterization of monoids embedding into groups. But it's hopelessly non-practical. My question is then: -Is there a finite set $\mathcal{F}\subset\mathcal{G}$ of generalized cancelation rules such that a monoid embeds into a group iff it satisfies all the rules in $\mathcal{F}$? - -REPLY [15 votes]: The answer is no. What you call a generalized cancellation rule is called a quasi-identity in universal algebra. Malcev proved in 1939 that there is no finite basis of quasi-identities defining group embeddable monoids or equivalently defining the quasi-variety of monoids generated by groups. -You can find details in Volume 2 of Clifford and Preston's classical text The Algebraic Theory of Semigroups. Malcev gave an infinite basis and Lambek another that has a geometric feel to it based on polyhedra. The precise theorem you want is Theorem 12.30 of Clifford and Preston Volume 2.<|endoftext|> -TITLE: CW-presentation of configurations of points in plane and space -QUESTION [6 upvotes]: I know from the the theory of Artin groups that, as the $K(\pi,1)$ conjecture is known for Braids group, that using Salvetti complexes we have a fairly explicit finite CW-complex presentation of the classyfing spaces of Braid groups, i.e. of the configuration space of $n$ unordered points in $\mathbb{R}^2$. (See for e.g. this survey paper on the $K(\pi,1)$ conjecture). If My understanding is correct (let me know if I got it wrong), the classyfing space of $B_n$ can be given as a CW-complex with a cell of dimension $k$ for each subsets of size $k-1$ of $\{1,\dots,n-1\}$. -I have two related question about this: - -First a reference request: Is there a references that explicitly describe these CW-complexes in the special cases of Braid groups without going though the general case of an Artin group ? or at least that spell out explicitly the description in the case of Braid groups ? - -Are there higher dimensional version of this ? i.e. "nice and explicit" CW-presentations of the configuration spaces of $n$ unordered points in $\mathbb{R}^d$ ? - -REPLY [2 votes]: Here is the original paper by Fox and Neuwirth: -Fox, R.; Neuwirth, L. The braid groups. Math. Scand. 10 (1962), 119–126. -I remember reading this in the late 1970's in grad school, and found it clear enough that it was obvious to me how to generalize this to get a CW complex on all of the configuration spaces $B(\mathbb R^n,k)$. -Then there is Jeff Smith's thesis, eventually published: -Smith, Jeffrey Henderson Simplicial group models for $\Omega^n \Sigma^n X$. -Israel J. Math. 66 (1989), no. 1-3, 330–350. -He gives an explicit simplicial $E_n$--operad. -This paper relates these: -Kashiwabara, Takuji On the homotopy type of configuration complexes. Algebraic topology (Oaxtepec, 1991), 159–170, Contemp. Math., 146, Amer. Math. Soc., Providence, RI, 1993. -Yes, all of these papers are pre ArXiv, but they should not be forgotten.<|endoftext|> -TITLE: Existence of normal microbundles -QUESTION [10 upvotes]: In the same paper where Milnor introduced the concept of microbundles, he gave the following definition. $M$ has a microbundle neighborhood in $N$ if there is a neighborhood $U$ of $M$ in $N$ and a retraction $r: U \to M$ so that $(M, U, i, r)$ is a microbundle (where here $i$ is the inclusion map). -In a remark, Milnor mentions that he does not know if every locally flat submanifold $M$ of a topological manifold $N$ has a microbundle neighborhood. It has been many years since that foundational paper, so I imagine that result is now known. Does anybody know the state of this? - -REPLY [10 votes]: Not all locally flat submanifolds have a normal microbundle, but they do stably. -Rourke-Sanderson prove that there is a PL embedding $S^{19} \times I \to S^{29}$ with no topological normal microbundle. -Milnor, in Microbundles 1, showed that every submanifold $M \subset N$ stabilizes to give $M \times 0 \subset N \times \Bbb R^q$ with a microbundle neighborhood for some large $q>0$. -Hirsch, in "On normal microbundles", gives relatively simple proofs of this fact and a few related facts. $q$ is quadratic in $\dim N$ in his result; I don't know whether or not it is known that the optimal $q(\dim M, \dim N)$ is quadratic in the inputs. -The first place I look for references to these sorts of foundational questions is Sander Kupers' notes on diffeomorphism groups. I found the Rourke-Sanderson reference at the beginning of Chapter 28. He cites Brown for the stability claim, but does not provide a reference.<|endoftext|> -TITLE: How many untilts? -QUESTION [10 upvotes]: I read the following passage in Endomorphisms of power series fields and residue fields of Fargues-Fontaine curves by Kedlaya-Temkin: -"One can construct many algebraic extensions of $\mathbb{Q}_p$ whose completions $K$ tilt to the completed perfect closure of a power series field over $\mathbb{F}_p$." -They then give the two classical examples of the $p$-adic completions of $\mathbb{Q}_p(p^{1/p^{\infty}})$ and $\mathbb{Q}_p(\zeta_{p^{\infty}})$. -This made me wonder: -Questions: (a) What other algebraic extensions of $\mathbb{Q}_p$ tilt to $\mathbb{F}_p((t^{1/p^{\infty}}))$? -(b) How many are there (up to isomorphism)? -To be precise: Question (b) asks for the cardinality of the set of untilts of $\mathbb{F}_p((t^{1/p^{\infty}}))$ which are completions of algebraic extensions of $\mathbb{Q}_p$. -I suspect that this question already has an answer in the works of Fargues–Fontaine but my scientific French is too poor to understand if this is the case. - -REPLY [15 votes]: This specific question is probably not addressed in the literature; let's try to figure it out! -Let $K$ be an algebraic extension of $\mathbb Q_p$ such that the tilt of $\widehat{K}$ is isomorphic to $\mathbb F_p((t^{1/p^\infty}))$. We can observe the following: - -Tilting preserves residue fields, so necessarily $K$ has residue field $\mathbb F_p$, i.e. is totally ramified over $\mathbb Q_p$. - -Tilting preserves value groups, so the value group of $K$ is isomorphic to $\mathbb Z[\tfrac 1p]$. In particular, there can only be a finite amount of ramification of degree prime to $p$. - - -In particular, $K$ is pro-$p$ and totally ramified of infinite degree over a finite totally tamely ramified extension $K_0$ of $\mathbb Q_p$. Conversely, all such $K$ have perfectoid completion $\widehat{K}$. -As SashaP comments below, these conditions are however not yet sufficient for $\widehat{K}^\flat$ to be isomorphic to $\mathbb F_p((t^{1/p^\infty}))$. For example, one can find such an extension $K/\mathbb Q_p$ for which $\mathrm{Gal}_K$ maps isomorphically to the tame quotient of $\mathrm{Gal}_{\mathbb Q_p}$; but the tame quotient of $\mathrm{Gal}_{\mathbb Q_p}$ cannot be isomorphic to $\mathrm{Gal}_{\mathbb F_p((t))}$. It seems to be an interesting question to isolate those $K$ which have tilt isomorphic to $\mathbb F_p((t^{1/p^\infty}))$! -One definitely gets examples by taking any tower $K_0=\mathbb Q_p$, $K_1$, $K_2$, ... such that each $K_{i+1}/K_i$ is a degree $p$ extension given by extracting a $p$-th root of a uniformizer. At each step, this gives finitely many distinct choices, so there are at least $2^{\aleph_0}$ such extensions. This is also an evident upper bound, so this gives an answer to b).<|endoftext|> -TITLE: Constructing a $0/1$ polytope from an abstract simplicial complex -QUESTION [7 upvotes]: Let us fix $\Delta$ a finite simplicial complex, and label the vertices of $\Delta$ as $\{1,2,\ldots,n\}$. For each $F\in \Delta$ let us consider the point in $\mathbb{R}^n$ given by: -$$e_F := \sum_{i\in F} e_i,$$ -where $e_i$ denotes the $i$-th canonical vector in $\mathbb{R}^n$. Let us define the polytope $\mathscr{P}(\Delta)$ as the convex hull in $\mathbb{R}^n$ of all these $e_F$ for $F$ ranging in $\Delta$. -Why is this construction interesting? Well, it contains two famous families of polyhedra: - -Independence matroid polytopes. This is easy to see, since the family of independent sets of a matroid is indeed a simplicial complex. -Chain polytopes of posets. This is a little bit harder, but still elementary: the vertices of the chain polytope of a poset $P$ are the antichains of $P$, and this defines a simplicial complex. - -I have two questions. Has this family of polytopes already been studied? Are there any other big (or interesting) family of polytopes that fall into this category? -As a side note: there is a nice formula for the volume of chain polytopes, I do not know if this is too naive, but maybe this could help to guess a "good" formula for the volume of independence matroid polytopes. - -REPLY [3 votes]: Another family within your class that has been studied is matching polytopes. Here for a graph $G = (V,E)$ we work in $\mathbb{R}^E$. The matching polytopes in then the convex hull of $e_M$ for matchings of the graph (not necessarily perfect). Since any subset of a matching is a matching it fits in the simplicial complex framework.<|endoftext|> -TITLE: Explicit description of the left adjoint $\mathfrak{F}$ to the relative nerve -QUESTION [7 upvotes]: $\newcommand{\op}{\mathrm{op}}\newcommand{\Un}{\mathrm{Un}}$Lurie's relative nerve functor $\mathrm{N}^{F}_{\bullet}$ is a simpler version of the unstraightening functor -$$\Un_\phi:\mathrm{sPSh}(\mathcal{C}_\bullet)\to\mathsf{sSets}_{/S}$$ -(or rather of its opposite; see HTT 3.2.5.1) which "unstraightens the homotopy coherence laws of a simplicial presheaf into a left fibration over $S_\bullet$" (I'm paraphrasing the nLab here) -Spelling out HTT 3.2.5.2, one sees that $\mathrm{N}^F_\bullet$ admits the following very simple description: it is the simplicial set where - -a vertex of $\mathrm{N}^F_\bullet$ is a pair $(A,x)$ with - -$A$ an object of $\mathcal{C}$; -$x$ a vertex of $F(A)$; - - -an edge of $\mathrm{N}^F_\bullet$ from $(A,x)$ to $(B,y)$ is a pair $(f,e)$ with - -$f:A\to B$ a morphism of $\mathcal{C}$; -$e: F(f)(x)\to y$ an edge of $F(B)$; - - -a $2$-simplex of $\mathrm{N}^F_\bullet$ is a $9$-tuple $(f,g,x,y,z,e_{01},e_{12},e_{02},\sigma)$ with - -$f: A\to B$ a morphism of $\mathcal{C}$; -$g: B\to C$ a morphism of $\mathcal{C}$; -$x$ a vertex of $F(A)$; -$y$ a vertex of $F(B)$; -$z$ a vertex of $F(C)$; -$e_{01}: F(f)(x)\to y$ an edge of $F(B)$; -$e_{12}: F(g)(y)\to z$ an edge of $F(C)$; -$e_{02}: F(g\circ f)(x)\to z$ an edge of $F(C)$; -$\sigma: e_{12}\circ F(g)(e_{01})\Rightarrow e_{02}$ a $2$-simplex of $F(C)$ as in the diagram -$$\require{AMScd} -\require{cancel} -\def\diaguparrow#1{\smash{\raise.6em\rlap{\!\!\!\!\!\!\scriptstyle #1} - \lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}}}} -\begin{CD} -&& F(g)(y)\\ -& \diaguparrow{F(g)(e_{01})} @VVe_{12}V \\ -F(g\circ f)(x) @>>e_{02}> z -\end{CD}$$ - - -and so on. - - -Now, in HTT 3.2.5.5, Lurie introduces a functor $\mathfrak{F}_X$ that is (FWIU) supposed to be a simpler analogue of the straightening functor $\mathrm{St}_\phi$. However, Lurie does so by invoking the adjoint functor theorem, similarly to the definition of $\Un_\phi$. -I gather that nevertheless those functors admit nice constructions (as shown for $\Un_\phi$ in Rezk's notes); so, my question is: - -Question: What is a concrete description of Lurie's $\mathfrak{F}_X$ functor? - -REPLY [8 votes]: There is indeed a simple, explicit description of the left adjoint to the relative nerve functor. This left adjoint sends a simplicial set over the category $C$, i.e. a morphism of simplicial sets $p \colon X \to C$, to the functor $C \to \mathbf{sSet}$ that sends an object $c \in C$ to the simplicial set $X/c$ defined by the following pullback square of simplicial sets: -$\require{AMScd}$ -\begin{CD} -X/c @>>> C/c\\ -@V V V @VV \mathrm{dom} V\\ -X @>>p> C -\end{CD} -A reference for this description of the left adjoint is the paper - -Heuts, Gijs; Moerdijk, Ieke. Left fibrations and homotopy colimits. Math. Z. 279 (2015), no. 3-4, 723--744. doi - -in which they denote this left adjoint by $r_!$. -Let me take this opportunity to remind everyone that the relative nerve construction appeared way back in the 1980's in §4 of the following paper written by the category theorist John Gray: - -Gray, John W. The representation of limits, lax limits and homotopy limits as sections. Mathematical applications of category theory (Denver, Col., 1983), 63--83, Contemp. Math., 30, Amer. Math. Soc., Providence, RI, 1984. doi<|endoftext|> -TITLE: What is the motivation for excellent rings? -QUESTION [8 upvotes]: First of all I am not formally educated in mathematics so pardon my ignorance if this is obvious and I am skipping something vital, but I am interested nonetheless in what the original motivation and applications are for excellent rings and by extension quasi-excellent rings conceptually. I know they were first described by Grothendieck in effort to describe phenomena with the resolutions of singularities, but I can't seem to find any specific conceptual applications they have. Every time I've tried researching this I get the definition of them without any further elaboration of why they are defined the way they are. If someone could describe their specific uses or provide a source to do so that would be very much appreciated. - -REPLY [5 votes]: (As the name suggests,) (quasi-)excellent rings are rings that are very well-behaved under various natural operations: localisation, finite type extension, formal completion (some hypothesis here), henselisation. One may well imagine that Grothendieck was looking for a property permanent under these operations. Excellent integral domains are also universally Japanese, which may be useful to know. -The first non-trivial example of an excellent ring is that of a complete noetherian local ring. The formal completion $\widehat{A}$ of a local ring $A$ may not inherit some properties of $A$, like reduceness or normality; it does when $A$ is excellent. -Now, my familiarity with excellent rings is rather superficial and I can never remember all the things that go into defining them. My go to place is Exposé I in Travaux de Gabber. There, in 2.11, Raynaud briefly mentions the link with the resolution of singularities: every characteristic zero quasi-excellent scheme $X$ admits a desingularisation à la Hironaka; conversely, if every integral scheme $Y$ finite over $X$ admits a resolution of singularities, then $X$ is quasi-excellent.<|endoftext|> -TITLE: A "proof" that all separately continuous maps on LF-spaces are continuous -QUESTION [5 upvotes]: Problem -Consider the locally convex spaces $C^\infty(\mathbb{R})$ and $C^\infty_c(\mathbb{R})$, the former equipped with its standard Fréchet topology, the latter equipped with the inductive limit topology given by $C^\infty_c(\mathbb{R}) = \varinjlim C^\infty_c([-n,n])$ where each $C^\infty_c([-n,n])$ is equipped with its Fréchet topology inherited from $C^\infty(\mathbb{R})$. This makes $C^\infty_c(\mathbb{R})$ into an LF-space. -Note that here we use the notation $C^\infty_c(K) = \{f \in C^\infty(\mathbb{R}) : \text{supp }f \subset K\}$ for compact sets $K$, as it is used in Treves. -Now, the map $C^\infty(\mathbb{R}) \times C^\infty_c(\mathbb{R}) \to C^\infty_c(\mathbb{R}), (f,g) \mapsto fg$ is separately continuous, but not jointly so (see Treves, Chapter 41). Hence, there must be something wrong with the very basic looking arguments in the following statement, and I'm curious if anyone can tell me which step is the problematic one. - -"Theorem": Every separately continuous bilinear map $\psi: C^\infty(\mathbb{R}) \times C^\infty_c(\mathbb{R}) \to C^\infty_c(\mathbb{R})$ is jointly continuous. - -"Proof": A linear map $E \to F$ from an LF-space $E = \varinjlim E_n$ to a locally convex space $F$ is continuous if and only if all restricted maps $E_n \to F$ are continuous (Treves, Proposition 13.1). -Hence separate continuity of $\psi$ implies separate continuity of the restricted maps -$$ \psi_n : C^\infty(\mathbb{R}) \times C^\infty_c([-n,n]) \to C^\infty_c(\mathbb{R}).$$ -But if $E,F$ are Fréchet and $G$ is locally convex, every bilinear separately continuous map $E \times F \to G$ is even jointly continuous. (Treves, Corollary to Theorem 34.1). Hence all $\psi_n$ are continuous. -But it is also true that if $E = \varinjlim E_\alpha$ and $F = \varinjlim F_\alpha$ are locally convex inductive limit spaces, and $G$ is any locally convex space, then a bilinear map $E \times F \to G$ is continuous if and only if the restrictions $E_\alpha \times F_\beta \to G$ are continuous for all $\alpha,\beta$ (Mallios, Chapter IV, Lemma 2.1). -Viewing $C^\infty(\mathbb{R}) = \varinjlim C^\infty(\mathbb{R})$ as a trivial inductive limit, this means that the continuity of the $\psi_n$ implies the continuity of $\psi$. This "proves" the statement. $\stackrel{?}{\square}$ -Notes -I came across (a more general version of) this statement and its proof in the paper "Cyclic-type cohomology of strict inductive limits of Fréchet algebras" by Lykova, Lemma 4.1. I don't want to appear like I'm calling them out, though, it may very well be that only my use of their statement is incorrect. -The books I cite, "Topological Vector Spaces, Distributions and Kernels" by Treves and "Topological Algebras: Selected topics" by Mallios seem to me very respected, and none of the statements seem obviously false to me, so I'm a bit confused here. -One suspicion I have is that treating $C^\infty(\mathbb{R})$ as a trivial inductive limit is the issue, but the proof of Lemma 2.1 in Mallios looks very straightforward, and doesn't seem to care whether the inductive limit is a strict one or not. - -REPLY [6 votes]: Already the first sentence in the ''proof'' is doubtful: The characterization of continuity of maps $f:\lim_\limits{n\to} E_n \to F$, that all restrictions to $E_n$ are continuous, holds for linear maps but there is no reason that this should be true (for colimits in the category of locally convex spaces) for bilinear maps. -This is not only doubtful but indeed wrong. For example, in an old article with the late Susanne Dierolf (Inductive limits of topological algebras. Linear Topol. Spaces Complex Anal. 3 (1997), 45–49) we constructed a strict locally convex inductive limit of Fréchet commutative algebras such that the multiplication is discontinuous (although the restrictions to the ''steps'' are continuous). I believe that Mallios claimed the contrary.<|endoftext|> -TITLE: Projections in infinite dimensional statistical manifolds -QUESTION [6 upvotes]: I'm struggling to understand the geometry of projection for infinite dimensional statistical manifolds. In finite dimensions, a strictly convex smooth function $F$ defines a Bregman divergence. From this function/divergence, an finite-dimensional information-geometric structure is defined, which is dually-flat. The Riemannian metric is obtained from the Hessian of the strictly convex function, and the rest of the information is obtained from partial derivatives (including the connection $\nabla^{F}$). In fact, there is a global coordinate system $[\theta]$ defined by the gradient $\nabla F$. In this case, it can be shown that unique projections from a point on the manifold to a submanifold exist, provided the submanifold is $\nabla$-flat (i.e. corresponds to a $\nabla$-affine subspace in the $[\theta]$-coordinate system). This, of course, broadly generalizes a projection to an affine subspace of Euclidean space. Does there exist a suitable analog for these facts in the infinite-dimensional information case? - -REPLY [3 votes]: Are you trying to construct an infinite-dimensional (Hilbert) manifold of probability measures on a fixed measurable space? Or you simply want to find a Bergman divergence analog in order to generalize a divergence-like metric in an infinite-dimensional manifold? It is not entirely clear in the OP what kind of analog you are looking for. -In a finite-dimensional statistical manifold, for example an exponential family with natural parameterizations, the manifold does come with a canonical connection as you mentioned. In the finite-dimensional situation, since we can take a parameterization, the geometry of the statistical manifold is embedded in the parameter space. The image of such a projection to a $\nabla$-flat submanifold would correspond to a sub-family of the exponential family, also a restricted collection of parameters that define this sub-family. -In an infinite-dimensional statistical manifold, assuming that you want to compare two points; you can simply calculate the norms of their difference and other kind of similarity measures are possible [Harandi et.al.]. If you are talking about infinite-dimensional statistical manifold that consists of probability measures that cannot be The inner product structure like we have for low-dimensional parameter space, however, does not always exist. In a special case pointed out by [Newton2], we can use an $\alpha$-divergence (which is a generalization to Bergman divergence, see this post from stat.SE) to partially describe a similar inner product-like structure for an infinite-dimensional statistical manifold. The asymptotic results from geometric perspective [Kass&Vos] may generalize into infinite-dimensional situation if you can find a "good finite-dimensional basis approximation". However, a strict inner product structure is not always possible in infinite-dimensional situation, therefore projections may not generalize. If it does generalize, its meaning would likely be a sub-family of probability measures. -In another direction, if you wish to construct a realization (or draw samples) from an inifinite-dimensional statistical manifold, the starting point would be Dirichlet processes wiki. And [Newton] proposed to use transformation (Fenchel–Legendre transform) approach to construct new infinite-dimensional statistical manifolds in a more abstract way. -Reference -[Newton2] Infinite-dimensional statistical manifolds based on a balanced chart, 2016. -[Harandi et.al.] Bregman Divergences for Infinite Dimensional Covariance Matrices,2014. -[Newton] An infinite-dimensional statistical manifold modelled on Hilbert space,2012. -[Kass&Vos] Geometrical Foundations of Asymptotic Inference, 1997.<|endoftext|> -TITLE: Non-set-theoretic consequences of forcing axioms -QUESTION [8 upvotes]: This article by Quanta Magazine states: - -... forcing axioms ... are workhorses that regular mathematicians “can actually go out and use in the field, so to speak,” ... - -What are some examples of uses and/or consequences of forcing axioms in "regular mathematics" (i.e. not set theory / logic)? - -REPLY [3 votes]: Farah's proof that all automorphism's of the Calkin algebra are inner under ZFC + Open coloring axiom,. The Calkin algebra is the quotient of the algebra of continuous linear operators on a separable Hilbert space by the ideal of compact operators (if I remember correctly this is the only closed ideal). It was known for a long time that you can construct an outer automorphism under ZFC + Continuum Hypothesis, but it wasn't known if CH is necessary for this. -I'm not sure if OCA is considered to be a forcing axiom, but it is a consequence of PFA, which certainly is. My somewhat vague understanding is that OCA is supposed to be a consequence of PFA which is both comprehensible and maybe even useful to non-logicians. In general, CH gives lots of isomorphisms between structures (so lots of automorphisms) and PFA makes things more rigid. Mathematicians like rigidity, so maybe they will like consequences of PFA. The stuff on automatic continuity of homomorphisms between Banach algebras (Kaplansky's conjecture) mentioned by Golshani fits in this picture. CH gives you discontinuous homomorphisms and PFA gives automatic continuity. -Farah's theorem descends from the earlier theorem of Shelah that is it consistent with ZFC that every automorphism of $\mathcal{P}(\mathbb{N})/\mathrm{Fin}$ is induced by an automorphism of $\mathcal{P}(\mathbb{N})$, here $\mathcal{P}(\mathbb{N})$ is the boolean algebra of subsets of $\mathbb{N}$ and $\mathrm{Fin}$ is the ideal of finite sets. Shelah's original argument is a complicated forcing, it was later proven from ZFC + OCA + Martin's axiom. Analogous results hold for other quotients of $\mathcal{P}(\mathbb{N})$ by other ideals, Farah has a nice survey. -Farah has other interesting applications of set theory to operator algebras. Some of this uses forcing axioms, and some of it doesn't. For example Farah and Hirshberg have constructed, under ZFC + a form of diamond, an inductive limit of matrix algebras which is not isomorphic to its opposite. The extra axiom they use is intermediate between CH and V = L, and I don't think it's considered to be a forcing axiom. -It seems that in general set theory has a lot to say about non-separable operator algebras. I think most operator algebraists now mainly work with separable algebras, probably partly to avoid set-theoretic issues, but there are certainly interesting non-separable algebras such as the Calkin algebra, and it is interesting to see which results from the separable case generalize to the non-separable case and which conjectures in the separable case fail in the non-separable case. My impression is that operator algebraists have been pretty open to input from logic. (Logicians are often very interested in applying logic to other areas, other areas have varied levels of interest in being applied to.)<|endoftext|> -TITLE: What is the endgoal of formalising mathematics? -QUESTION [29 upvotes]: Recently, I've become interested in proof assistants such as Lean, Coq, Isabelle, and the drive from many mathematicians (Kevin Buzzard, Tom Hales, Metamath, etc) to formalise all of mathematics in such a system. I am very sympathetic to some of the aims of this project, such as finding mistakes in published proofs, and definitively verifying complex and controversial proofs [Kepler conjecture (Hales), four-colour theorem (Appel/Haken), Fermat's last theorem (Wiles), ...]. -However, I'd like to understand more about the ultimate goals of this project, and the vision for achieving a computer formalisation of all of mathematics. I understand a good portion of core (undergraduate) mathematics has been formalised. Yet, the bulk of research-level mathematics remains unformalised, and it seems to me that we are currently creating new mathematics at a far greater rate than we are formalising mathematics. Are we always going to be "chasing our tails" here? What needs to change so that we can catch up and achieve a complete formalisation of mathematics? -Another issue is, of course, new mathematics is being created all the time. If we achieve complete formalisation, how do we sustain this? Will there be "professional formalisers" whose job it is to formalise the recent results of mathematicians? Will mathematicians themselves be expected to formalise their results as they are proved? What role will proof assistants play in 20 (or 50, or 100) years? - -REPLY [16 votes]: Other answers address several aspects of your question, but to add a point not mentioned yet — you write: - -it seems to me that we are currently creating new mathematics at a far greater rate than we are formalising mathematics. Are we always going to be "chasing our tails" here? What needs to change so that we can catch up and achieve a complete formalisation of mathematics? -Another issue is, of course, new mathematics is being created all the time. If we achieve complete formalisation, how do we sustain this? Will there be "professional formalisers" whose job it is to formalise the recent results of mathematicians? Will mathematicians themselves be expected to formalise their results as they are proved? - -Certainly, at the moment, formalising maths is very laborious and time-consuming. But I think most people who work the formalisation hope that it won’t stay so bad! It’s been getting steadily better as proof assistants and their libraries have progressed — “better” both in terms of how quickly experienced users can cover ground, and in terms of the learning curve for new users. This improvement hasn’t perhaps been as fast as we’d like, or might have hoped — but it’s been steady, and continuing. -If usability continues to improve, then it’s reasonable to imagine a future proof assistant with a role comparable to LaTeX today. Typesetting our work in LaTeX takes a bit of work, and has a bit of a learning curve — but most mathematicians accept the time and effort as worthwhile for the payoff of good typesetting. A very few mathematicians outsource this to professional typesetters; a larger minority leave it to their co-authors when possible. But it’s part of the expectations of the field, and there’s no problem of “tail-chasing” — no question of “can we typeset mathematics as fast as we can produce it?”, or of requiring major grant support to get work typeset. This is one possible future that I and some others in formalisation envisage, and that it seems reasonable to hope for in a timescale of a few decades. -One likely difference from LaTeX is that this could well arrive in some subfields of maths sooner than others, since (so far) some topics seem much more amenable to formalisation.<|endoftext|> -TITLE: Checking if Hochschild cohomology $\mathit{HH}^2(A)=0$ -QUESTION [5 upvotes]: I am trying to compute the Hochschild cohomology of a particular bound quiver path algebra. The quiver $Q$ consists of one vertex and four loops $x,y, h_1,h_2$, and the relations $I$ are generated by: - -All paths of length greater than 3. -All paths of length 3, except $yh_1x$ and $xh_2y$, and $yh_1x+xh_2y$. -All paths of length 2, except $yh_1, h_1x, xh_2,h_2y$. - -Basically, in this algebra I have $yh_1x=-xh_2y$, and only other nonzero paths are the subpaths of these 2. -I am interested in $\mathit{HH}^2(kQ/I)$. More specifically, I am interested in whether $\mathit{HH}^2(kQ/I)=0$ for some infinite field $k$. I couldn't find or come up with a direct way of computing it, and my attempt using GAP's QPA package ran into memory problems. So I was wandering what are the tractable ways to compute this cohomology or prove that is zero or non-zero, either on paper or using computer algebra. -GAP code: - -LoadPackage("qpa"); -Q := Quiver(1, [[1,1,"x"],[1,1,"y"],[1,1,"h_1"],[1,1,"h_2"]]); -R := PathAlgebra(Rationals,Q); -gens:= GeneratorsOfAlgebra(R); -x:=gens[2]; -y:=gens[3]; -h_1:=gens[4]; -h_2:=gens[5]; -relations :=[x^2,y^2,h_1^2,h_2^2,xy,yx,h_1h_2,h_2h_1,xh_1x,xh_1y,yh_1y,xh_2x,yh_2x,yh_2y,yh_1x+xh_2y,h_1xh_1,h_1xh_2,h_2xh_1,h_2xh_2,h_1yh_1,h_1yh_2,h_2yh_1,h_2yh_2]; -gb := GBNPGroebnerBasis(relations,R); -I:=Ideal(R,gb); -GroebnerBasis(I,gb); -A:=R/I; -M := AlgebraAsModuleOverEnvelopingAlgebra(A); -HH2 := ExtOverAlgebra(NthSyzygy(M, 1), M); - -REPLY [6 votes]: Of course Tyler Lawson's answer is the more conceptual and insightful one, but in case it is useful, I ran your GAP script on a machine with 64 GB of RAM, which turned out to be enough. If I understand the output of ExtOverAlgebra correctly, it seems that your $HH^2(A)$ group is a 138-dimensional vector space. -The GAP session is pasted below. Please let me know if I have misunderstood anything, including possibly the output of ExtOverAlgebra, which I never used before running your script. -gap> LoadPackage("qpa"); -───────────────────────────────────────────────────────────────────────────── -Loading GBNP 1.0.3 (Non-commutative Gröbner bases) -by A.M. Cohen (http://www.win.tue.nl/~amc) and - J.W. Knopper (J.W.Knopper@tue.nl). -Homepage: http://mathdox.org/products/gbnp/ -───────────────────────────────────────────────────────────────────────────── -───────────────────────────────────────────────────────────────────────────── -Loading QPA 1.30 (Quivers and Path Algebras) -by Edward Green (http://www.math.vt.edu/people/green) and - Oeyvind Solberg (https://folk.ntnu.no/oyvinso/). -Homepage: https://folk.ntnu.no/oyvinso/QPA/ -───────────────────────────────────────────────────────────────────────────── -true -gap> Q := Quiver(1, [[1,1,"x"],[1,1,"y"],[1,1,"h_1"],[1,1,"h_2"]]); - -gap> R := PathAlgebra(Rationals,Q); -]> -gap> gens:= GeneratorsOfAlgebra(R); -[ (1)*v1, (1)*x, (1)*y, (1)*h_1, (1)*h_2 ] -gap> x:=gens[2]; -(1)*x -gap> y:=gens[3]; -(1)*y -gap> h_1:=gens[4]; -(1)*h_1 -gap> h_2:=gens[5]; -(1)*h_2 -gap> relations :=[x^2,y^2,h_1^2,h_2^2,x*y,y*x,h_1*h_2,h_2*h_1,x*h_1*x,x*h_1*y,y*h_1*y,x*h_2*x,y*h_2*x,y*h_2*y,y*h_1*x+x*h_2*y,h_1*x*h_1,h_1*x*h_2,h_2*x*h_1,h_2*x*h_2,h_1*y*h_1,h_1*y*h_2,h_2*y*h_1,h_2*y*h_2]; -[ (1)*x^2, (1)*y^2, (1)*h_1^2, (1)*h_2^2, (1)*x*y, (1)*y*x, (1)*h_1*h_2, - (1)*h_2*h_1, (1)*x*h_1*x, (1)*x*h_1*y, (1)*y*h_1*y, (1)*x*h_2*x, - (1)*y*h_2*x, (1)*y*h_2*y, (1)*x*h_2*y+(1)*y*h_1*x, (1)*h_1*x*h_1, - (1)*h_1*x*h_2, (1)*h_2*x*h_1, (1)*h_2*x*h_2, (1)*h_1*y*h_1, (1)*h_1*y*h_2, - (1)*h_2*y*h_1, (1)*h_2*y*h_2 ] -gap> gb := GBNPGroebnerBasis(relations,R); -[ (1)*x^2, (1)*x*y, (1)*y*x, (1)*y^2, (1)*h_1^2, (1)*h_1*h_2, (1)*h_2*h_1, - (1)*h_2^2, (1)*x*h_1*x, (1)*x*h_1*y, (1)*x*h_2*x, (1)*x*h_2*y+(1)*y*h_1*x, - (1)*y*h_1*y, (1)*y*h_2*x, (1)*y*h_2*y, (1)*h_1*x*h_1, (1)*h_1*x*h_2, - (1)*h_1*y*h_1, (1)*h_1*y*h_2, (1)*h_2*x*h_1, (1)*h_2*x*h_2, (1)*h_2*y*h_1, - (1)*h_2*y*h_2 ] -gap> I:=Ideal(R,gb); -]>, - (23 generators)> -gap> GroebnerBasis(I,gb); - -gap> A:=R/I; -]/ -]>, - (23 generators)>> -gap> M := AlgebraAsModuleOverEnvelopingAlgebra(A); -<[ 14 ]> -gap> HH2 := ExtOverAlgebra(NthSyzygy(M, 1), M); -[ <<[ 602 ]> ---> <[ 784 ]>>, - [ <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>>, - <<[ 602 ]> ---> <[ 14 ]>>, <<[ 602 ]> ---> <[ 14 ]>> ], - function( map ) ... end ] -gap> Length(HH2[2]); -138<|endoftext|> -TITLE: low dimensional manifolds by gluing the boundary of a ball -QUESTION [6 upvotes]: Recall that one way of drawing closed 2-manifolds is to take a disk $D^2$, take a cellular decomposition of $\partial D^2$, pair the vertices in this cellular decomposition so that the pairing preserves edges, and then take $D$ together with this quotient of the boundary. -We can do this in other dimensions as well, for example in dimension 3, every closed 3-manifold can be obtained by a similar procedure where we take $B^3$, take a cellular decomposition of $\partial B^3$, pair the vertices of this cellular decomposition so that the pairing preserves edges and faces, and then look at the quotient of $B^3$ by this pairing. -Threlfall and Seifert did this for the Poincaré homology sphere (see for example here - which also contains a different such description due to Kreines). In fact, they take the cellulation of $\partial B^3$ to be the dodecahedron. Is there a complete (presumably rather short) list of all of the 3-manifolds obtained in such a way where the cellulation is a Platonic solid? $T^3$, $\mathbb{R}P^3$, and the Seifert-Weber space are other examples that come to mind. I'd guess that the Poincaré homology sphere is maybe the only homology sphere on that list. More generally, I'd like to look through a list of the 3-manifolds that occur in this way using simple cellulations. -This can also be done in a similar way in dimension 4 to produce all smooth closed 4-manifolds. Are there some nice pictures/examples of this being carried out somewhere? I'd love to see such pictures of $S^2 \times S^2, T^4, \mathbb{C}P^2,...$ - -REPLY [5 votes]: This closed orientable 3-manifolds obtained by gluing faces of the Platonic solids were classified by Everitt. -That is for regular polyhedra with equal dihedral angles, and the gluing is done geometrically. However, it is also possible to do the gluing topologically, and for that problem, I only have a partial answer. There are 3 closed orientable 3-manifolds obtained by gluing faces of the tetrahedron. They are $S^3$, $L(4,1)$, and $L(5,2)$. Explicit gluings can be seen in figure 2 of this paper of Jaco and Rubinstein. -There are 17 closed orientable 3-manifolds obtained by gluing faces of the octahedron, 13 of which are prime. They are listed in Proposition 4.2 of this paper by Heard, Pervova, and Petronio. -Presumably, the closed orientable 3-manifolds obtained from the cube have been enumerated, but I don't know of a reference. They include $\mathbb{R}P^3$, the 3-torus and at least 2 of the other closed orientable Euclidean 3-manifolds. I imagine there are a lot of 3-manifolds obtained from the dodecahedron and icosahedron, but I doubt that anyone has enumerated them all. -As for 4-manifolds, I will leave that for someone else to answer, except to note that there are no 4-manifolds obtained from a single pentachoron (4-simplex), since it has 5 tetrahedra in its boundary and this causes a parity issue.<|endoftext|> -TITLE: Poynting vector and differential forms -QUESTION [13 upvotes]: It is well known that electromagnetic field is a 2-form and Maxwell's equation can be reformulated in language of differential forms. -What is the Poynting vector in this language? - -REPLY [11 votes]: Too long for a comment. -It is worth mentioning the controversy that arose about the expression of the Poynting vector $P$, whether the Abraham form $E\times H$ or the Minkowski form $D\times B$ is valid. The notations are that the electro-magnetic field, a closed $2$-form in Minkowski space, has coordinates $(E,B)$. The vectors $(D,H)$ are given by -$$D=\frac{\partial L}{\partial E},\qquad H=-\frac{\partial L}{\partial B}$$ -where $L(E,B)$ is the density of the Lagrangian from which the Maxwell's equations derive. For instance, the standard equations follow from the choice $L=\frac12(|E|^2-|B|^2)$ (here the light speed is set to $1$). Other choices are possible, like that of Born-Infeld, which yield non-linear models. -The amazing resolution of the controversy is that both expressions equal each other if, and only if $L$ is Lorentz invariant. In mathematical terms, this means that $L$ is a function of the quantities -$$\frac12(|E|^2-|B|^2),\qquad E\cdot B$$ -only. -Going further, the Maxwell's equations imply that the following tensor is (row-wise) divergence free -$$T=\begin{pmatrix} -W & E\times H \\ -D\times B & --W_B\otimes B-W_D\otimes D+\sigma I_3 -\end{pmatrix}$$ -where -$$\sigma:=B\cdot W_B+D\cdot W_D-W.$$ -Hereabove $W:=D\cdot E-L(E,B)$ is the energy density (a partial Legendre transform of $L$), and subscripts are differentials. That ${\rm Div}_{t,x}T=0$ expresses the conservation laws of the energy and of the Poynting vector. The important point is that $T$ is a symmetric tensor, and this is ensured by the Lorentz invariant.<|endoftext|> -TITLE: "Laurent phenomenon"? -QUESTION [10 upvotes]: Define the recurrence -\begin{align*} -n(2n+x-3)u_n(x) -&=2(2n+x-2)(4n^2+4nx-8n-3x+3)u_{n-1}(x) \\ -&-4(n+x-2)(2n-3)(2n+2x-3)(2n+x-1)u_{n-2}(x) -\end{align*} -with initial conditions $u_0(x)=0$ and $u_1(x)=x+1$. -The subject of "Laurent phenomenon" was motivated by Somos sequences. In the same spirit, I ask: - -QUESTION. Is it true that each $u_n(x)$ is a polynomial in $x$? In fact, with positive integer coefficients. - -EXAMPLES. $u_2(x)=5x^2 + 13x + 6$ and $u_3(x)=22x^3 + 114x^2 + 164x + 60$ and -\begin{align*} -u_4(x)&=93x^4 + 814x^3 + 2367x^2 + 2606x + 840 \\ -u_5(x)&=386x^5 + 5140x^4 + 25030x^3 + 54500x^2 + 51024x + 15120. -\end{align*} - -REPLY [17 votes]: In fact, -$$ -u_n(x) = -{2}^{n-1}\prod _{k=0}^{n-1}(2x+2k+1) --{2\,n-1\choose n-1}\prod _{k=0}^{n-1}(x+k) , -\tag1$$ -which is a polynomial with integer coefficients. -P.S. the proof rests on a routine verification that (1) satisfies the given recurrence relation and initial conditions.<|endoftext|> -TITLE: Upper bounds for Bessel functions -QUESTION [6 upvotes]: Cosider the K-Bessel function $$K_\nu(x):= \frac\pi 2 \frac{I_{-\nu}(x)-I_\nu(x)}{\sin(\nu\pi)}.$$ -See also Watson, G. N., A treatise on the theory of Bessel functions., Cambridge: University Press, Chapter III, p.78. -This book contains many results on the asymptotic behaviour of functions of this kind. However, I am interested in upper bound estimates against simpler terms. Something similar is used in a another paper to show resolvent estimates by estimating K-Bessel functions against exponential and polynomial terms. At the moment I try to reproduce this estimates as a part of my research. However, no explanation on these kinds of estimated is provided and it falls short in providing the literature so I am looking for a reference on this topic. - -REPLY [4 votes]: $\newcommand{\om}{\omega}\newcommand{\al}{\alpha}\newcommand{\la}{\lambda}\newcommand{\Ga}{\Gamma}$Inequality (14) in the paper by Glushak and Pokruchin referred to in your comment is equivalent to -\begin{align*} - &\Big\|\int_0^\infty t^{\nu+n}K_{\nu-n+1}(t\sqrt\mu)Y_k(t)x\,dt\Big\| \\ - &\le M\|x\|\int_0^\infty t^{\nu+n-1/2}\exp((\om-\sqrt\mu)t)\,dt \tag{14} -\end{align*} -given that, by formula (3) in the paper, -\begin{equation} -\|Y_k(t)\|\le M\exp(\om t) \tag{0} -\end{equation} -for some real $\om\ge0$ and all real $t>0$. Here, $\nu=(k-1)/2$ (by p. 40 of the paper), $k\ge0$, and $n$ is a natural number. -One can easily see that inequality (14) is false in general (or, maybe, always). -However, one can save the conclusion (at the bottom of p. 42 of the paper) that -\begin{equation*} - \|R^k(\mu)\|\le M(k)/(\mu-\om_1)^n \tag{1} -\end{equation*} -for some real $M(k)>0$ not depending on $\mu$ or $n$, some real $\om_1$, and all real $\mu>\om_1$, which makes the Hille--Yosida theorem applicable. -Indeed, for real $u>0$ and $\al$ -\begin{equation*} - K_\al(u)=\int_0^\infty e^{-u\cosh z}\cosh\al z\,dz \tag{2} -\end{equation*} -by the second display under the picture in Section Modified Bessel functions: $I_\al$, $K_\al$. By rescaling, without loss of generality, $\|x\|=1$ and (0) holds with $M=1$. So, by (2), with -\begin{equation} - m:=n-\nu+1,\quad l:=2\nu+2=k+1\ge1,\quad\mu>4\om^2,\quad r:=\om/\sqrt\mu\in(0,1/2), -\end{equation} -the left-hand side of (14) is no greater than -\begin{align*} - &\int_0^\infty dt\, t^{\nu+n}e^{\om t} - \int_0^\infty dz\,e^{-t\sqrt\mu\,\cosh z}\cosh mz \\ - &=\int_0^\infty dz\,\cosh mz - \int_0^\infty dt\, t^{\nu+n}e^{-t(\sqrt\mu\,\cosh z-\om)} \\ - &=\Ga(\nu+n+1)\int_0^\infty \frac{dz\,\cosh mz}{(\sqrt\mu\,\cosh z-\om)^{m+l}} \\ - &=\frac{\Ga(\nu+n+1)}{(\sqrt\mu)^{m+l}}\int_0^\infty \frac{dz\,\cosh mz}{(\cosh z-r)^{m+l}} \\ - &\ll\frac{\Ga(\nu+n+1)}{(\sqrt\mu)^{m+l}}\int_0^\infty dz\,f(z)^m e^{-lz}, \tag{3} -\end{align*} -where $a\ll b$ means that $|a|\le Cb$ for some real $C>0$ not depending on $n$ or $\mu$ or $\om$ (as long as $r$ is small enough) and -\begin{equation*} - f(z):=\frac{e^z}{\cosh z-r}. -\end{equation*} -Note that $\max_{z>0}f(z)=(1-r^2)^{-1}$. So, the left-hand side of (14) is -\begin{align*} - &\ll\frac{\Ga(\nu+n+1)}{(\sqrt\mu)^{m+l}}\, (1-r^2)^{\nu-1-n} - \ll\frac{\Ga(\nu+n+1)}{(\sqrt\mu)^{n+\nu+3}}\, (1-r^2)^{-n}=:C_n(\mu). -\end{align*} -So, by the equality in formula (14) in the paper, for $\mu>\om_1:=1+4\om^2$, -\begin{align*} - \|R^k(\mu)\|&\ll\frac{C_n(\mu)}{2^{n+\nu-1}(n-1)!\Gamma(\nu+1)\mu^{n/2-k/4}} \\ - &\ll\frac{\Ga(\nu+n+1)(1-r^2)^{-n}}{2^{n+\nu-1}(n-1)!\Gamma(\nu+1)\mu^n} \\ - &\ll\frac{\Ga(\nu+n+1)(4/3)^n}{2^{n+\nu-1}(n-1)!\Gamma(\nu+1)\mu^n} \\ - &\ll\frac{M(k)}{(\mu-\om_1)^n}, -\end{align*} -which proves (1). - -Edit: In fact, $\max_{z>0}f(z)=2(1-r^2)^{-1}=:c>2$ -- I previously missed the factor $2$ here. So, the last integral in (3) is $\asymp c^m/\sqrt m$. Since the bounding was lossless (up to inessential constant factors) everywhere in the above reasoning, this appears to invalidate the entire result in the paper by Glushak and Pokruchin.<|endoftext|> -TITLE: Finding many subsets of $V_{\lambda+2}$ stable under $j:V\prec V$ -QUESTION [5 upvotes]: Working over $\mathsf{ZF}$ with an embedding $j:V\prec V$ with a critical point $\kappa$. Take $\lambda=\sup_{n<\omega}j^n(\kappa)$. (You may assume $\mathsf{DC}_\lambda$ if you need, but I am not sure it is consistent with a Reinhardt cardinal.) -My question is: is there a family $\mathcal{A}\subseteq V_{\lambda+2}$ such that - -$j(\mathcal{A})=\mathcal{A}$, $j(x)=x$ for all $x\in\mathcal{A}$, and -$\lvert\mathcal{A}\rvert\ge\lvert V_\lambda\rvert$? - -Note that if we take $\mathcal{A}=\{x\subseteq V_{\lambda+1}\mid x=j(x)\}$, then $j(\mathcal{A})\neq \mathcal{A}$ since $\alpha\in\mathcal{A}$ for all $\alpha<\kappa$. Hence $\mathcal{A}$ has to be a proper subset of $\{x\subseteq V_{\lambda+1}\mid x=j(x)\}$. -If it has a negative answer, what is the least $\alpha$ which ensures the existence of a family $\mathcal{A}\subseteq V_\alpha$ with the mentioned conditions? - -REPLY [9 votes]: There can be no such set $\mathcal A$. In fact, no set $\mathcal A$ with $j(\mathcal A) = \mathcal A$ and $j(x) = x$ for all $x\in \mathcal A$ can surject onto $\kappa$, and this does not require that the codomain of $j$ is $V$. The reason is that if $f :\mathcal A\to \kappa$ were a surjection, then $j(f) : \mathcal A\to j(\kappa)$ would be a surjection; but for all $x\in \mathcal A$, $j(f)(x) = j(f)(j(x)) = j(f(x)) = f(x)$, so $j(f) = f$, and in particular $\kappa\notin \text{ran}(j)$, a contradiction.<|endoftext|> -TITLE: Homotopy equivalent smooth 4-manifolds which are not stably diffeomorphic? -QUESTION [10 upvotes]: Recall that two 4-manifolds $M$ and $N$ are stably diffeomorphic if there exist $m,n$ such that -$$M \#_n (S^2 \times S^2) \cong N \#_n (S^2 \times S^2).$$ -That is, they become diffeomorphic after taking sufficiently many connected sums with $S^2 \times S^2$. -I am interested to find examples $M$ and $N$ which are homotopy equivalent $M \simeq N$, but where $M$ and $N$ fail to be stably diffeomorphic. -I know of two sources of examples of such manifolds. In Example 5.2.4 of -Topological 4-manifolds with finite fundamental group P. Teichner, PhD Thesis, University of Mainz, Germany, Shaker Verlag 1992, ISBN 3-86111-182-9. -Teichner constructs a pair of $M$ and $N$ where the fundamental group $\pi$ is any finite group with Sylow 2-subgroup a generalized Quaterion group $Q_{8n}$ with $n \geq 2$. -Another pair of $M$ and $N$ with fundamental group the infinite dihedral group was constructed in: -On the star-construction for topological 4-manifolds. P. Teichner, Proc. of the Georgia International Topology Conference 1993. Geom. top. AMS/IP Stud. Adv. Math. 2 300-312 A.M.S. (1997) -Are there any other known examples of this phenomenon? I have been unsuccessful in finding any others in the literature, but this is not my area of expertise. Are there any general results about when this can occur? - -REPLY [3 votes]: For orientable 4-manifolds, I believe you gave a complete list of the known examples. For non-orientable, the phenomenon does generalise. Kreck showed that for every 1-type $(\pi,w)$ with $\pi$ a finitely presented group and $w \colon \pi \to C_2$ a surjective homomorphism, there is a 4-manifold $M$ with $\pi_1(M) \cong \pi$ and orientation character $w$, such that $M\# K3$ and $M \#^{11} S^2 \times S^2$ are homeomorphic (so in particular homotopy equivalent) but not stably diffeomorphic. -@incollection {Kreck-84, -AUTHOR = {Kreck, M.}, -TITLE = {Some closed {$4$}-manifolds with exotic differentiable -structure}, -BOOKTITLE = {Algebraic topology, {A}arhus 1982 ({A}arhus, 1982)}, -SERIES = {Lecture Notes in Math.}, -VOLUME = {1051}, -PAGES = {246--262}, -PUBLISHER = {Springer, Berlin}, -YEAR = {1984} -}<|endoftext|> -TITLE: Real-rooted polynomials -QUESTION [7 upvotes]: I proposed this question at MO which was resolved neatly by Gerald Edgar in the form -$$ -u_n(x) = -{2}^{n-1}\prod _{k=0}^{n-1}(2x+2k+1) --{2\,n-1\choose n-1}\prod _{k=0}^{n-1}(x+k).$$ -Now that we confirmed that $u_n(x)$ are all polynomials. I would like to add a follow up: - -QUESTION. For $n\geq1$, are the roots of $u_n(x)$ real negative numbers? It seems to be true. - -REPLY [18 votes]: Yes, because $u_n(x)$ switches sign between each consecutive pair in -$x=0,-1,-2,-3,\ldots,1-n$, and $u_n(-n) = 0$. -In general, if $P,Q$ are continuous functions each with $n$ simple roots -in an interval $I$, and those roots interlace, then the same argument gives -at least $n-1$ roots of $P-Q$ in $I$, and then an extra root (or even two) -might be forced by the values of $P-Q$ at the endpoints of the interval. -Here $P$ and $Q$ are $2^{n-1} \prod_{k=0}^{n-1} \, (2x+2k+1)$ and -${2n-1 \choose n-1} \prod_{k=0}^{n-1} \, (x+k)$, and $I = (-\infty,0)$.<|endoftext|> -TITLE: Next steps for a Morse theory enthusiast? -QUESTION [22 upvotes]: I don't know if this question is really appropriate for MO, but here goes: I quite like Morse theory and would like to know what further directions I can go in, but as a complete non-expert, I'm having trouble seeing forward to identify these directions and where I should be reading. Below, I will mention my background and particular interests, then mention things that I've heard of or wondered about. I would appreciate references appropriate for my level, or even better, sketches of any historical or recent Morse-y trajectories. -I have read Milnor's Morse Theory and Lectures on the H-cobordism Theorem (the latter was the subject of my undergraduate thesis). I have also read a little bit about Morse homology. I think the issue is that my knowledge of Morse theory ends there, not only in detailed knowledge, but also in terms of themes and trajectories. That makes it difficult to know where to look next. My main interests (at the current time) are in differential topology and symplectic stuff. To give this question a reasonable range, here are a couple restrictions: - -This question concerns topics in "Morse theory" (in some broad sense), not applications of Morse theory to other things. I am definitely interested in those as well, but that list would be unending. In particular, I'm moving my toric curiosities to a different question. -I'm mainly interested in manifold-y stuff, as opposed to say, discrete or stratified Morse theory. -Restricting to finite dimensions is perfectly fine for this context. I am aware that there are Hilbert/Banach manifolds and such to be discussed, but I don't know anything about them. Perhaps I can't outlaw Floer theory entirely, but I'll just say that while I plan to learn about it eventually, I think it's beyond my present scope. - -Here are some specific things that I have wondered about: -Cohomology products: I imagine that for a Morse-Smale pair, the cup product (or its Poincaré dual) could be computed by intersection numbers of the un/stable manifolds, though I haven't read an account of this in detail. Near the end of Schwarz's Morse Homology (which I have not read), he defines the cup product in an analogous style to the usual singular cohomology construction. Perhaps most interesting are the products in Chapter 1 of Fukaya's "Morse Homotopy, $A^\infty\!$-Category, and Floer homologies." I have not read this yet, but hope to do so in the near-ish future. Are there any other major view of the cup product in Morse cohomology that I have missed here? -CW Structure: In Morse Theory, Milnor describes manifolds by adding cells and then sliding them around to get an actual CW structure (i.e. cells only attach to lower-dimensional cells). This is useful, but quickly leaves the manifold behind and just becomes a question about homotoping attaching maps. The un/stable manifolds add an important layer of detail about handle decompositions, but even with a Morse-Smale pair, the "attaching" maps notoriously fail continuity. Fixing this seems to be a finicky question and I'm not sure where the answer lies. If I understand correctly, this is related to compactifying moduli spaces of flow lines, so perhaps the answer can be found in Schwarz's book or Hutchings' notes? (Although a comment on this MO question purports that Hutchings' assertion is mis-stated.) Is a bona fide CW structure related to what Cohen-Jones-Segal were looking for in "Morse theory and classifying spaces"? (Yet again, I have not read, but I am intrigued and hope to.) -Finite volume flows: Another paper that I have been intrigued by, but have not read is Harvey and Lawson's "Finite volume flows and Morse theory." It seems like a beautiful way to circumvent the aforementioned issues of discontinuity and create a whole new schema of Morse theory in the process. However, reading it would probably involve learning about currents first… It seems very elegant in and of itself, but it might be interesting to know where this theory goes and what is being done with it, as motivation to learn the necessary background. -Cerf theory: I've heard a little bit about Cerf theory, but I can't really find any references on it (in English, since I don't speak French). As a way to understand the relationship between different handle decompositions, it seems like a very natural thing to pursue. Perhaps it's unpopular because of the difficulty/length of Cerf's paper? Or because it was later subsumed by the framed function work of Hatcher, Igusa and Klein (and maybe others, I just don't know anything about this area), as mentioned in this MO question? I really don't even know enough about this to ask a proper question, but I would love any suggestions for how to learn more. -Other: Any other major directions that you would suggest to a Morse theory enthusiast? - -REPLY [11 votes]: A recent breakthrough result which uses Morse theory in a substantial manner is Watanabe's disproof of Smale conjecture in dimension 4. In it, he provides a method to compute Kontsevich's configuration space integrals by counting certain broken flowlines for gradients of Morse functions. These Morse-theoretic invariants are used to prove that certain 4-dimensional disk bundles with trivialized are not trivial bundles. There is still much to do in developing the properties of these types of invariants, and in using them to detect non-trivial homotopy groups of the diffeomorphism groups of other manifolds.<|endoftext|> -TITLE: How can we be sure that results that rely heavily on extensive computations are correct? -QUESTION [19 upvotes]: Recently a ''bug'' was discovered in one of the most popular mathematics software, Wolfram Mathematica (see links here and here). It concerns the evaluation of the sum -$$ -\sum_{k=1}^{n-1} \frac{(-1)^{k-1}(k-1)!^2}{(n^2-1^2)\ldots(n^2-k^2)}, -$$ -a fairly straightforward computational exercise, as one would expect. Surprisingly, Mathematica incorrectly evaluates this sum to -$$ -\frac{1}{n^2}, -$$ -instead of the correct expression, which is -$$\frac{1}{n^2}-\frac{2(-1)^{n-1}}{n^2 \binom{2n}{n}}.$$ -Another user (in second link above) found that Maple 2020 also makes the same incorrect evaluation. -This raises the question whether we can trust widely used software like Mathematica and Maple with (much) more complex computational tasks, and in particular theorems and lemmas that appear in published literature that explicitly rely on large scale computations performed with such applications. -In some cases, the peer review process involves replicating computational results that appear in a manuscript, but this is (more often than not) not the case. Furthermore, it is not unlikely that reviewers will use the same software to double check these results as the author, thereby replicating the same mistake, or bug. -To what extent can we trust results that were obtained with the aid of extensive computations? At what point can we safely accept the ''truthfulness'' of a claim if its replication requires months (sometimes years) of number crunching performed by software that can make such elementary mistakes as above? - -REPLY [10 votes]: In the long run, the best we can probably do is to develop computer algebra packages whose computations are formally verified. To my knowledge, there has not been a lot of effort in this direction. Muhammad Taimoor Khan’s 2014 Ph.D. thesis, Formal Specification and Verification of Computer Algebra Software, provided a proof of concept; he implemented a fragment of Maple that he called MiniMaple and formally verified a particular package of routines for Gröbner basis computations. -At the present time, however, there seems to be little demand for computer algebra systems that meet the bar of formal correctness, in part because the sacrifice in computational speed for the sake of increased certainty does not appeal to most people. Thus, people who want their computations to be formally correct have to code them up on an ad hoc basis. -For example, the original Hales–Ferguson proof of the Kepler conjecture involved heavy computations that were carried out using traditional software packages such as CPLEX. When it came to producing a formally verified version of these computations for the Flyspeck proof, there was no royal road; the computations had to be completely re-programmed from scratch. I am not sure exactly what the slowdown factor was when passing from the traditional computation to the formally verified computation, but as reported in A formal proof of the Kepler conjecture, the formally verified versions of the computations took more than 5000 processor-hours to complete. Moreover, even Flyspeck has some potential loopholes; see Mark Adams's presentation on Flyspecking Flyspeck for an interesting discussion.<|endoftext|> -TITLE: Cohomology of resolution of singularity -QUESTION [5 upvotes]: If $X,Y$ are smooth projective schemes, then if we have a surjection $f:X\to Y$, we have an injective map on étale cohomology, or more generally on any Weil cohomology (see https://mathoverflow.net/q/172527). The proof of this statement uses Poincare duality on the target $Y$. -I would like to understand the cohomology of the resolution of singularity of a scheme $X$. If $\tilde{X}\to X$ is a resolution of singularites, then this is a surjective map (since it is a birational map between projective schemes). Can this be generalized to say that -$$H^*(X)\to H^*(\tilde{X})$$ -is injective? Since the ``usual" argument uses Poincare duality on the target $X$, and $X$ is non-smooth (otherwise, what's the point of the resolution of singularities), the usual argument fails. - -REPLY [3 votes]: Given a resolution $\pi:\tilde X \to X$, you can ask whether the pullback morphism $\pi^*:H^k(X) \to H^k(\tilde X)$ is injective for some (or all) $k$. As Donu points out, the mixed Hodge structure on $H^k(X, \mathbf C)$ would be pure which is a restrictive condition. One case where this is true for all $k$ is when $X$ has at worst quotient singularities by a result of Steenbrink. For orbifolds, $H^k(X) \cong IH^k(X)$ for all $k$, where $IH^k(X)$ is the intersection cohomology, so we have the injection again by David's comment. Another example for a specific $k$ is for rational singularities, i.e., $R^i\pi_*\mathcal O_{\tilde X} = 0$ for all $i > 0$. By playing around with the Leray spectral sequence, it follows that $\pi^*$ is injective for $k \le 2$ (although I think $H^1(X)$ is pure for normal singularities by the same logic). In general $H^2(X)$ and $IH^2(X)$ are not isomorphic for rational singularities. You can see this by taking a scheme with a isolated rational singularity which is not $\mathbf Q$-factorial. -However, there is always a map $H^2(X) \to IH^2(X)$ which will be injective if $H^2(X)$ carries a pure Hodge structure. More generally, any injection $H^k(X) \hookrightarrow H^k(\tilde X)$ will factor through the injection $IH^k(X) \hookrightarrow H^k(\tilde X)$ coming from the decomposition theorem (at least for proper schemes).<|endoftext|> -TITLE: Intersection map giving rise to Poincaré duality -QUESTION [10 upvotes]: Let $M$ be a smoothly triangulated compact $d$-dimensional manifold. Consider the subcomplex $C_*^{\pitchfork T}(M)$ of smooth singular chains which are transverse to the triangulation. An inductive chain homotopy construction establishes that these are quasi-isomorphic to all smooth, and thus all singular, chains. -Define the intersection map $I : C_n^{\pitchfork T}(M; R) \to C^{d-n}_\Delta(M; R)$ (the latter being simplicial cochains arising from the triangulation) by sending $\sigma : \Delta^d \to M$ to the cochain whose value on the an element of the triangulation whose characteristic map is $\iota : \Delta^{d-n} \to M$ is the count of the zero manifold given by the pullback of $\sigma$ and $\iota$. Here either $R$ is $\mathbb{Z}/2$ or $M$ must be oriented and the count is with the usual signs, and one uses some version (such as this) of transversality for manifolds with corners. -Fun exercise: with appropriate signs, $I$ is a map of chain complexes. (Hint: as in the proof that degree as defined by counting preimages is homotopy invariant, this relies on the classification of one-manifolds.) Poincaré duality implies that the domain and range of $I$ are quasi-isomorphic. -Question: why is $I$ a quasi-isomorphism? -I think I can prove this, but only in the mod-two setting, by using Thom's seminal work on bordism and Quillen's elementary approach to cobordism (just the definitions of his "elementary" paper - not the main results, which to me are quite deep despite the title of the paper). But there must be a more direct argument, which covers the oriented case as well, and it seems like this should be in the literature somewhere - from the 1940's maybe? -(Motivation: Greg Friedman, Anibal Medina and I have what we think is a new approach to questions such as Do chains and cochains know the same thing about the manifold? through vector field flows, and would like to build on existing knowledge of the interplay between intersection and duality.) - -REPLY [2 votes]: In a related and highly relevant comment thread, Mike Miller pointed me to this preprint of Lipyanskiy. I'm sure there are arguments which work, such as what Joshua and Dmitri and I discuss in the comments. But before I forget I'd like to point to Lipyanskiy's work, especially Section 12, and mark the question as answered.<|endoftext|> -TITLE: Error term in Davenport's sum $\sum_{n \leq x } \mu(n) \exp(2 \pi i \alpha n ) $ -QUESTION [9 upvotes]: Reference request: -Davenport proved that for every fixed $N>1 $ one has $$ \sup_{\alpha \in \mathbb R } \left | \sum_{1\leq n \leq x } \mu(n) \exp(2 \pi i \alpha n )\right | = O_N\left( \frac{x}{(\log x)^N}\right).$$ -Does anyone know if the error term has been improved to something like $O(x \exp(-c \sqrt { \log x } ))$, i.e. as the prime number theorem error? - -REPLY [13 votes]: If there's a bad Landau--Siegel zero $\mod q$, then the Möbius function behaves a lot like a primitive quadratic character $\mod q$, and then the exponential sum at $\alpha =a/q$ roughly has size $x/\sqrt{q}$ (as would be the case if $\mu(n)$ were replaced by a character). Siegel's theorem is what enables one to save arbitrary powers of $\log$ in Davenport's result. If you assume that there are no Siegel zeros, then one can get a refinement as asked. This was done by Hajela and Smith; and on GRH one can get a power saving. See the paper of Baker and Harman for details and references.<|endoftext|> -TITLE: What's a great christmas present for someone with a PhD in Mathematics? -QUESTION [92 upvotes]: Christmas is just around the corner and I haven't bought all the gifts for my family yet ( yeah, ) -My Dad has a PhD in Mathematics, he works in Graph theory and his thesis was about Quasiperiodic tilings. -What do you think would make a good gift for him? -I'll appreciate anything you could think of! -Thanks for reading, hope you have a great day . -p.s.: after reading all the tags in this website I think this is the right one for this kind of question? please correct me if I'm wrong! - -REPLY [3 votes]: A book of interviews of famous mathematicians could be good. I have in mind particularly More Mathematical People, which I've gotten a lot of mileage from here at MathOverflow. - -REPLY [3 votes]: A good red wine from Bodegas Langa (Spain)<|endoftext|> -TITLE: Properness for small forcings -QUESTION [8 upvotes]: It is easy to see a forcing of size $\aleph_1$ is proper if and only if is semiproper. I was wondering when such an equivalency holds between semi-proper and stationary-preserving forcings in $\rm ZFC$? Or consistently in a model where significant fragments of $\rm MM$ fail. - -REPLY [3 votes]: The answer is no, as the following upcoming work of Shelah and Usuba shows: -Theorem (Shelah-Usuba): -The following theories are equiconsistent with ZFC: -ZFC+CH+ “there is an $\omega _1$-stationary preserving $\sigma$-Baire poset of size $\aleph_1$ -​ which is not semiproper”. -ZFC+“Martin’s axiom for semiproper posets of size $\aleph_1$” + “there is an $\omega _1$-stationary preserving $\sigma$-Baire poset of size $\aleph_1$ which is not semiproper”. -ZFC+CH+“every $\omega _1$-stationary preserving $\sigma$-Baire poset of size $\aleph_1$ is semiproper”. -See $\omega_1$-Stationary preserving $\sigma$-Baire posets of size $\aleph_1$.<|endoftext|> -TITLE: Are the vertical sections of the Ackermann function primitive recursive? -QUESTION [16 upvotes]: The Ackermann function $A(m,n)$ is a binary function on the natural numbers defined by a certain double recursion, famous for exhibiting extremely fast-growing behavior. -One finds various slightly different formulations of the Ackermann function, with slightly different initial conditions. I prefer the following version: -\begin{eqnarray*} -A(0,n) &=& n+1 \\ -A(m,0) &=& 1, \quad\text{ except for }A(1,0)=2\text{ and }A(2,0)=0\\ -A(m+1,n+1) &=& A\bigl(m,A(m+1,n)\bigr). -\end{eqnarray*} -I like this version because it has the nice properties that -\begin{eqnarray*} - A(0,n) &=& n+1\\ - A(1,n) &=& n+2\\ - A(2,n) &=& 2n\\ - A(3,n) &=& 2^n\\ - A(4,n) &=& 2^{2^{\cdot^{\cdot^2}}}{}^{\bigr\}n}\\ -\end{eqnarray*} -In this way, the levels of the Ackermann function exhibit increasingly fast-growing behavior. -Each horizontal level of the Ackermann function, the function $A_m(n)=A(m,n)$ considered as a unary function with $m$ fixed, is defined by recursion using the previous level, and consequently all these functions are primitive recursive. -Furthermore, these functions are unbounded in the primitive recursive functions in the sense that every primitive recursive function is eventually bounded by some horizontal level $A_m$ of the Ackermann function. It follows from this that the diagonal of the Ackermann function $n\mapsto A(n,n)$ is not primitive recursive. -My question is about the sections of the Ackermann function taken on the other coordinate, the vertical sections of the Ackerman function, the functions $A^n(m)=A(m,n)$, for fixed $n$. -Question. Are the vertical sections of the Ackermann function $A^n$ each primitive recursive? -I don't expect the answer to depend on which particular version of the Ackermann function one uses, but just in case, let me mention another standard version, which has a smoother definition, which works better in many of the inductive arguments, even though the level functions are not as nice. -\begin{array}{lcl}\operatorname {A} (0,n)&=&n+1\\\operatorname {A} (m+1,0)&=&\operatorname {A} (m,1)\\\operatorname {A} (m+1,n+1)&=&\operatorname {A} (m,\operatorname {A} (m+1,n))\end{array} - -REPLY [22 votes]: No, already $A(n,3)$ is not primitive recursive. Let me use the essentially equivalent up-arrow notation: $A(n,m)=2\uparrow^{n-1}m$, and argue why $f(n)=2\uparrow^n 3$ is not PR. I claim $f(2n-2)\geq 2\uparrow^{n+1}n=A(n,n)$. $n\mapsto A(n,n)$ outgrows all $n\mapsto A(m,n)$, so by a classical argument it eventually outgrows all PR functions, and hence so will $f$ after we justify the claim. -It is enough to show that for $m\geq 3$ we have $2\uparrow^n m\geq 2\uparrow^{n-1}(m+1)$. We show this by induction on $m$. Indeed, $2\uparrow^n m=2\uparrow^{n-1}(2\uparrow^n(m-1))$, so it is enough to show $2\uparrow^n(m-1)\geq m+1$. This holds true for $m=3$ as $2\uparrow^n 2=4$ for all $n$, and for larger $m$ we have $2\uparrow^n(m-1)=2\uparrow^{n-1}(2\uparrow^n(m-2))\geq 2\uparrow^{n-1}m\geq m+1$. -Let me note that the bound $f(2n-2)\geq 2\uparrow^{n+1}n$ is far from optimal, in fact it should hold for $f(n+2)$, at least for large enough $n$. - -REPLY [20 votes]: If you don't mind not getting the best possible result, then the proof is very short. Indeed, -$$A(n,4)=A(n-1,A(n,3))$$ -so if you can prove that $A(n,3)\geq n-1$ then $A(n,4)\geq A(n-1,n-1)$, which we know doesn't grow primitive-recursively. But $A(n,3)$ grows much more quickly than $n-1$.<|endoftext|> -TITLE: Divisibility condition implies $a_1=\dotsb=a_k$? -QUESTION [10 upvotes]: Let $a_1, a_2, \dotsc, a_k$ be $k$ positive integers, $k\ge2$. For all $n\ge n_0$ there is a positive integer $f(n)$ such that $n$ and $f(n)$ are relatively prime and $a_{1}^{f(n)}+\dotsb+a_{k}^{f(n)}$ is a multiple of $a_{1}^n+\dotsb+a_{k}^n$. -Is it true that $a_1=a_2=\dotsb=a_k$ or is it possible to construct a counterexample? - -REPLY [7 votes]: Here's a a tweak of Seva's idea that gives a counterexample. Note that if $r$ is odd, then $2^{n}+1$ divides $2^{rn} + 1$. -Let $k = 6$, $a_{1} = 1$, $a_{2} = a_{3} = a_{4} = 2$, $a_{5} = a_{6} = 4$. Then $a_{1}^{n} + \cdots + a_{6}^{n} = 1 + 3 \cdot 2^{n} + 2 \cdot 4^{n} = (1+2^{n})(1+2^{n+1})$. -If $n$ is any positive integer, let $f(n) = 2n^{2} + n - 1$. We see that $f(n) \equiv -1 \pmod{n}$ and so $\gcd(n,f(n)) = 1$. Now, -$$a_{1}^{f(n)} + \cdots + a_{6}^{f(n)} = (1+2^{2n^{2} + n - 1})(1+2^{2n^{2} + n}).$$ -The second factor is $2^{n(2n+1)}+1$ and so is a multiple of $2^{n} + 1$. On the other hand, the first factor is $2^{(n+1)(2n-1)}+1$ and so it is a multiple of $2^{n+1} + 1$. Thus, $a_{1}^{f(n)} + \cdots + a_{6}^{f(n)}$ is a multiple of $a_{1}^{n} + \cdots + a_{6}^{n}$.<|endoftext|> -TITLE: Is every field the residue field of a discretely valued field of characteristic 0? -QUESTION [9 upvotes]: Let $k$ be a field of positive characteristic $p$. Is there necessarily a discrete valuation ring of characteristic $0$ with maximal ideal $(p)$ and residue field isomorphic to $k$? - -REPLY [10 votes]: Yes, by Hasse-Schmidt ("Die Struktur diskret bewerteter Koerper", Crelle's Journal, 1934) for any field $k$ of characteristic $p$ there exists a strict Cohen ring $A$, which is a Noetherian, complete local, discrete valuation ring with maximal ideal $pA$ and residue field $A/pA = k$. See also Mac Lane (Theorem 2 in "Subfields and automorphism groups of p-adic fields", Ann. of Math. 1939) and Cohen ("On the structure and ideal theory of complete local rings", Trans. AMS, 1946).<|endoftext|> -TITLE: Does ZF minus infinity imply collection? -QUESTION [10 upvotes]: In ZF with the axiom of infinity removed, is the axiom scheme of collection provable? -Note that Collection does follow from the axiom of Transitive Containment, which states that everything belongs to a transitive set. Mancini gave a model of ZF minus Infinity where Transitive Containment fails: see the end of Section 3 in -https://projecteuclid.org/euclid.ndjfl/1054837937 -Mancini's model validates Collection, so this does not answer my question. - -REPLY [11 votes]: ZF - Inf does imply Collection. Fix a set $X$ and a property $P$ (which can be formalized in terms of a formula and a parameter). Since we have separation, we may assume for all $x \in X,$ there is $y$ such that $P(x,y).$ Suppose $X$ is finite, with cardinality $n.$ A standard inductive argument shows there is a set $Y$ such that for all $x \in X,$ there is $y \in Y$ satisfying $P(x,y).$ -Now suppose $X$ is infinite. Then $\omega$ exists, by replacing every element of $\mathcal{P}_{\text{fin}}(X)$ with its cardinality. Thus ZF holds, and this instance of Collection is justified by the standard argument.<|endoftext|> -TITLE: Did Edward Nelson accept the incompleteness theorems? -QUESTION [13 upvotes]: Edward Nelson advocated weak versions of arithmetic (called predicative arithmetic) that couldn't prove the totality of exponentiation. Since his theory extends Robinson arithmetic, the incompleteness theorems apply to it. But if the incompleteness theorems are proven in theories stronger than those he accepts, he could presumably reject them. So my questions are first, did Nelson doubt either of the incompleteness theorems? And second, can the incompleteness theorems be proved in weak systems of arithmetic that don't prove the totality of exponentiation? -The closest thing I can find to an answer is an excerpt from his book Predicative Arithmetic, in which he says on page 81 "at least one of these two pillars of finitary mathematical logic, the Hilbert-Ackermann Consistency Theorem and Gödel's Second Theorem, makes an appeal to impredicative concepts." - -REPLY [5 votes]: Your second question has been properly answered by Emil Jerabek, I would say. Reading some of the comments, I feel I should write the following about your first question: -From talking to Ed Nelson and to people who knew him well, I can say that Ed Nelson has for a long time been firmly convinced that the exponential function somehow leads to inconsistency (and therefore PA is inconsistent). He has written about this at length and has pointed out some motivation for this view, like the Bellantoni-Cook characterisation of function complexity and his writings on predicativity. -Ed Nelson's deeper motivation for his view seems to have been the following: he had a feeling that somehow fixed point constructions (like an enumeration of all partial recursive functions or Goedel's incompleteness theorems) could be 'internalised' or 'made total' to produce a contradiction like '0=1'. Such a contradiction would only be possible given the exponential function. At the most fundamental level, Ed Nelson did not believe that the notion of completed infinite set was formally consistent.<|endoftext|> -TITLE: Can we make ZF − infinity + “all ordinals are finite” as strong as ZFC? -QUESTION [6 upvotes]: Consider the theory ZFfin := ZF − axiom of infinity + “all ordinals are finite” (i.e., every ordinal is zero or successor). -Of course, if we add the axiom of choice, this is not very interesting: it's basically first-order arithmetic in disguise. Basically what I'd like to know is whether we can make something combinatorially interesting out of ZFfin (without Choice). -Specifically: -Question: Can we find some axiom (or perhaps axiom scheme) T of a combinatorial flavor which makes ZFfin + T arithmetically at least as strong as ZFC (meaning that all arithmetical statements which are theorems of ZFC should also be theorems of ZFfin+T; of course, we also demand consistency of ZFfin+T)? -I don't have a precise definition of “of combinatorial flavor” but an example of something that is definitely not of that nature is the axiom scheme “all arithmetical theorems of ZFC are true”. I want something which looks like a large cardinal axiom, or a reflection scheme, or something set-theoretic like that, except that you can't do that as such because all ordinals are finite. -(If the question turns out to be too difficult, I might settle for other axioms which make ZFfin much stronger than arithmetic, without going as far as ZFC.) -Motivation: there are various axioms that can be added to the theory NFU (Quine's New Foundations with Urelements) to make it strong in various ways, and can be compared to the large cardinal axioms of ZFC. My question is whether analogous axioms can be found for ZFfin instead of NFU. - -REPLY [8 votes]: ZFfin implies every set is finite, and in particular choice holds. Suppose there is an infinite set $X.$ By replacing every element of $\mathcal{P}_{\text{fin}}(X)$ with its cardinality, we see $\omega$ exists, contradiction. So this theory is roughly arithmetic.<|endoftext|> -TITLE: $\ell^1$-norm of eigenvectors of Erdős-Renyi Graphs -QUESTION [13 upvotes]: Setting. Let $G(n,p)$ denote the usual Erdős-Renyi (random) graphs. For each such graph there is an associated Laplacian matrix $L = D - A$ where $D$ collects the degrees on the diagonal and $A$ is the adjacency matrix. This matrix has eigenvectors $v_1, \dots v_n$ (ordered with respect to increasing eigenvalue) and we assume the eigenvectors are all normalized in $\ell^2(\mathbb{R}^n)$. - -Localization. One natural question is whether any of these eigenvectors can be localized and this is often measured in the sense of $\| v_i\|_{\ell^{\infty}}$ being 'large', one entry being unusually big. One does not expect this to be the case and it's interesting to try to quantify this notion (and I think there are many papers about this). However, there is dual version: we know from Hölder that -$$\| v_i\|_{\ell^1} \leq \sqrt{n} \cdot \|v\|_{\ell^2} = \sqrt{n}$$ -with equality only attained for the flat vector. So the $\ell^1-$norm can also serve as a measure of localization: the more localized a vector is, the smaller it will be. -When I plot the value of $\|v_i\|_{\ell^1}$ for all the eigenvectors $i=1,2,\dots,n$, I get the following amusing curve. - -This picture shows $\|v_i\|_{\ell^1}$ for $i=1,\dots, 5000$ for a random realization of $G(n, p)$ for $n=5000$ and $p=0.4$. Knowing where in the spectrum an eigenvector lies seems to narrow down what its $\ell^1$-norm can be. There also seems to be a concentration phenomenon: the picture is pretty much the same for each random realization. Here's a second example for a random realization of $G(n, p)$ for $n=5000$ and $p=0.01$. - -The eigenvectors corresponding to the smallest eigenvalues seem to be localized in vertices that have unusually few neighbors and eigenvectors corresponding to the largest eigenvectors seem to be localized in vertices having unusually many neighbors — thus one would perhaps some concentration in the edges and that would also explain why the curve is largest in the middle and symmetric. But I am surprised to see such a nice curve that seems to go uniformly through the entire spectrum. -Question 1: Is this known or does it follow from known results? Is there a good heuristic why this might be true? -Another interesting question is about the maximum of $\| v_i\|_{\ell^2}$ which seems to be assumed somewhere in the middle. -Question 2: What can be said about -$$ 0 \leq \max_{1 \leq i \leq n}\frac{ \| v_i\|_{\ell^1}}{\sqrt{n}} \leq 1?$$ -Numerically it seems to be somewhere around 0.8. If we assume that the entries of a typical flat eigenvector behave like i.i.d. Gaussians, then a first guess would be that this ratio should be somewhere around $\mathbb{E} |X|$ where $X \sim \mathcal{N}(0,1)$. We have $\mathbb{E} |X| = \sqrt{2/\pi} \sim 0.79788$. Maybe a coincidence? -Background: A couple of years ago, Alex Cloninger and I played with the problem of trying to find structure in Laplacian eigenvectors. We found a method that worked pretty well in nice settings — we then tried to see what it would find on Erdős-Renyi random graphs (which seemed a natural test case: the eigenvectors should be mostly random and not particularly structured). Somehow our method picked out the edge and we didn't know why until we found this $\ell^1$ anomaly. It's mentioned in the paper (arXiv) but we never got a chance to ask a lot of people about it. - -REPLY [3 votes]: The following does not decide on whether the infimum is $0$, but it does give a -rough bound. -Let $v$ be a normalized eigenvector. Suppose $\|v\|_1\alpha$ t.b.d. By the no-gap delocalization property of Rudelson-Vershynin (see https://arxiv.org/pdf/1506.04012.pdf), with high probability, for al such vectors, -$$C(|I|/n)^{12}\leq \sum_{i\in I} |v_i|^2\leq n^{-\beta} \sum_{i\in I} |v_i|\leq n^{\alpha-\beta}.$$ -(I am simplifying a bit, out of laziness; in reality you have to make sure $I$ is not too small, but the argument still goes through. See the difference between Theorem 1.3 and 1.5 there.) -Therefore, -$$|I|\leq n\cdot (n^{(\alpha-\beta)})^{1/12}.$$ -Thus, if $\alpha<\beta$ then $|I|/n\to 0$. That is, the cardinality of the set of coordinates larger than $n^{-\beta}$ is larger than (say) $n/2$. But then, -$$n^{\alpha}\geq \sum_i |v_i|\geq \sum_{i\notin I} |v_i|\geq n^{1-\beta}/2.$$ -Thus, we get that $\alpha\geq 1-\beta$. Since $\beta$ is arbitrary subject to the constraint $\beta>\alpha$, it follows that $\alpha>1/2$, which contradicts $\alpha<1/2$. -Added in edit: I realize that I was writing about a matrix with independent centered entries; the Laplacian matrix does not fall in that framework, although the no gap delocalization for the normalized Laplacian is treated (with weaker bounds) in Eldan et als work, reference 9 in the cited paper of Rudelson-Vershynin. In particular, for $p\in (0,1)$ independent of $n$, it shows that for the second eigenvector (and some other near the edge), the $l_1$ norm is bounded below by $\sqrt{n}/(\log n)^C$. I suspect their method works for other eigenvectors.<|endoftext|> -TITLE: Chebyshev's other inequality -QUESTION [10 upvotes]: It is a simple fact, the granddaddy of correlation inequalities that if $f,g$ are monotone functions on $[0,1]$ then $$\int_0^1 f(x)g(x) dx \ge \int_0^1 f(x) dx \int_0^1 g(x) dx.$$ In their famous inequalities book, Hardy-Littlewood-Polya say this is due to Chebyshev and I even found an article calling this "Chebyshev's other inequality". But I haven't found a reference to a paper of Chebyshev where this appears. Does anyone know such a reference? (I want it because I am writing a book with a chapter on correlation inequalities). Many thanks. - -REPLY [9 votes]: The source for the inequality is -P.L. Chebyshev, On approximate expressions of some integrals in terms of others, taken within the same limits, Proc. Math. Soc. Kharkov 2, 93–98 (1882). -This article is online here. The relevant text is copied below, which presumably states $R_1>0$, hopefully someone who reads Russian can provide a translation. - Google translate suggests that $R_1=AB/12$ with $A,B$ the maximum of the derivatives of $du/dx$, $dv/dx$, which as noted by Emil Jeřábek cannot be quite correct.<|endoftext|> -TITLE: Uniqueness of infinite direct sum decomposition -QUESTION [5 upvotes]: A module $M$ over a ring $R$ is called semisimple if it admits a direct sum decomposition into simple modules. If $M$ admits a finite decomposition $M=\bigoplus_{i=1}^n S_i$ into simple $R$-modules $S_i$, then this decomposition is unique up to isomorphism and permutation of the factors because from such a decomposition one can cook up a composition series and then use the Jordan–Hölder theorem. -I would like to know: are infinite decompositions unique as well? -E.g. if $R$ is semisimple, then all $R$-modules are semisimple. Do they decompose uniquely? - -REPLY [7 votes]: Yes. A semisimple module $M$ is canonically isomorphic to -$$M \cong \bigoplus_i \text{Hom}_R(S_i, M) \otimes_{\text{End}(S_i)} S_i$$ -where $\text{Hom}_R(S_i, M)$ is what you might call the multiplicity space of the simple module $S_i$. It is naturally a module over the division ring $\text{End}(S_i)$, and from this expression it follows that the multiplicity of $S_i$ in any direct sum decomposition of $M$ into simples must be $\dim_{\text{End}(S_i)} \text{Hom}_R(S_i, M)$. (For this we need to know that $\text{Hom}_R(S_i, -)$ preserves infinite coproducts, but this follows from the fact that simple modules are cyclic and so in particular finitely generated.)<|endoftext|> -TITLE: Questions about $\text{Perf}(A)$ of dg algebra $A$ -QUESTION [5 upvotes]: [ALEXEY ELAGIN AND VALERY A. LUNTS, p.4.] Recall that triangulated category $\text{Perf}(A)$ is defined as the full thick triangulated subcategory of $D(A)$ generated by the dg $A$-module $A$. - - -[Kontsevich, Definition 1.] Dg algebra $A$ is called smooth if $A \in \operatorname{Perf}\left(A \otimes A^{o p}\right)$. It is compact if $\operatorname{dim} H^{\bullet}(A, d)<\infty$. This properties are preserved under the derived Morita equivalence. - -Could you please share your understanding of $\text{Perf}(A)$? - -What is the relation between $\text{Perf}(A)$ and perfect complex? -What good properties does it have? -Why do we use $\text{Perf}(A\otimes A^{op})$ to define "smoothness"? What is the geometrical motivation? - -Thank you very much! - -REPLY [6 votes]: As explained a little bit further in Elagin and Lunts' paper, the category $Perf(A)$ consists of the compact objects of $D(A)$, this is exactly what happens in the usual situation in algebraic geometry, for example for a nice space and certainly for a commutative ring, the perfect complexes as locally quasi-isomorphic to a bounded complex of free modules of finite type. The proof that these two notions are the same in D(R) can be found for example in the tag 07LT in the stacks project. -The so to say slogan of perfect complexes at least of good spaces is that they're finite, like the previous characterization as compact objects it is possible to show too that in a lot of situations of interest they are also the dualizable objects under the usual monoidal structure of $D(A)$. I won't recall what the definition of dualizable is ( but you can check the nlab entry ) but remember that finite dimensional vector spaces are what you must be thinking of. In this sense then perfect complexes are objects which are both topologically and algebraically finite. This means that perfect complexes are categorically closed under a lot operations you would expect f.d. vector spaces or compact (topological) spaces to be closed by. -These two nice properties are so that a lot of the geometry of the would-be space is reflected as properties of the category. As explained a bit by Kontsevich in the paper you list, a fundamental result is that of Bondal and Van der Bergh relating the category of perfect complexes of a separated scheme of finite type X with the category of perfect complexes of a single dg-algebra A. In particular if X is also smooth then the dg-algebra will be smooth in the sense of your question. This sort of characterization allows people to work purely axiomatically. -I hope this helps clarify some things and I apologize if my answer seems too vague. Perfect complexes appear in many ways and its a bit hard to list every good property and interesting result without knowing exactly from which angle the other person might be coming from.<|endoftext|> -TITLE: Space of simple closed curves in $S^2$ -QUESTION [5 upvotes]: I am curious about the topology of the space of simple closed curves in $S^2$. -The entire free loop space $LS^2$ admits an explicit description using the Morse theory of the energy functional for the round metric. The energy functional is Morse-Bott, with a sequence of critical manifolds corresponding to iterates of great circles. Topologically, this implies that $LS^2$ is homotopy equivalent to the complex constructed by starting with $S^2$ (the space of constant loops) and successively attaching copies of the unit disk tangent bundle of $S^2$ in some manner. -Given this context, my question can be phrased as follows. Let $\mathcal{S}S^2 \subset LS^2$ be the space of simple closed curves, i.e. injective continuous maps $S^1 \to S^2$. Does $\mathcal{S}S^2$ admit a similar nice description? -If the gradient flowline of the energy functional starting at any simple closed curve ended at a non-multiply-covered great circle, then it seems to me that this would give a retraction of $\mathcal{S}S^2$ onto the unit disk tangent bundle to $S^2$. However, I don't actually know if this is true or not. - -REPLY [5 votes]: Yes; I find this easier than the whole Morse theory package needed for $LS^2$. -Thm. The projection map $\text{Emb}(S^1, S^2) \to (TS^2 \setminus 0)$, given by $\gamma \mapsto \gamma'(0)$, is a weak homotopy equivalence. -A section sends a nonzero tangent vector $v$ above a point $p$ to the geodesic through $p$ with $\gamma'(0) = v$; it should be the case that $\text{Emb}(S^1, S^2)$ deformation retracts to the space of great circles (which is homeomorphic to $SO(3)$); I don't know a reference to the claim that the corresponding inclusion is a cofibration so that I can conclude. (Maybe there is an explicit deformation retraction from mean curvature flow.) -Pf. First observe that $\text{Diff}^+(S^2)$ acts transitively on $\text{Emb}(S^1, S^2)$ by the fact that the latter space is path-connected (roughly, the Schoenflies theorem) and the isotopy extension theorem. This (plus some general theorems about Frechet group actions) show that there is a fiber sequence $$\text{Diff}^+(S^2 \text{ rel } S^1) \to \text{Diff}^+(S^2) \to \text{Emb}(S^1, S^2).$$ The first space is the space of oriented diffeomorphisms which fix the equator pointwise. The same derivative map above defines a map of fiber sequences to $\{*\} \to TS^2 \setminus 0 \to TS^2 \setminus 0$ (sending a diffeomorphism $\varphi$ to $d\varphi_p(e_1)$, where $p$ is a point on the equator and $S^1$ the tangent vector corresponding to the equator. -The claim is that this the first space is contractible and the map $\text{Diff}^+(S^2) \to TS^2 \setminus 0$ is a weak equivalence. -Both of these claims follow from Smale's theorem $\text{Diff}(D^2 \text{ rel } \text{Nbhd}(\partial D^2)) \simeq \{*\}$ with a little massaging, where here my notation means "the space of diffeomorphisms of the disc which fix a neighborhood of the boundary". -First you show that $\text{Diff}(S^2 \text{ rel } S^1)$ deformation retracts onto the space of diffeomorphisms which are linear on an open nbhd of $S^1$ in a fixed trivialization of a tubular neighborhood of $S^1$, then use the fact that the space of oriented linear isomorphisms $NS^1 \to NS^1$ is contractible; this buys you that $\text{Diff}^+(S^2 \text{ rel } S^1)$ is equivalent to the space of diffeomorphisms which are constant on a neighborhood of the equator; now Smale's result applies. -For $\text{Diff}^+(S^2)$ itself, the argument is similar. The fiber of the map described above is the set of diffeomorphisms fixing a point and a tangent vector at that point; because the map is orientation-preserving, this leaves a contractible space of possible derivatives, and you can show this fiber deformation retracts onto the space of diffeomorphisms fixing a chosen point and with $d\varphi_x = \text{Id}$; then linearize, as above, so that this fiber is equivalent to the space of diffeomorphisms of $S^2$ which fix an open neighborhood of $p$ pointwise, which again is contractible by Smale.<|endoftext|> -TITLE: On a certain norm of the identity operator on $\mathbb R^2$ -QUESTION [5 upvotes]: $\newcommand\R{\mathbb R}\newcommand\Q{\mathcal Q}$For mutually orthogonal vectors unit vectors $a=[a_1,\dots,a_n]^T$ and $b=[b_1,\dots,b_n]^T$ in $\R^n=\R^{n\times1}$ (so that $n\ge2$) and for all $x=[x_1,x_2]^T\in\R^2$, let -$$\|x\|_{a,b}:=\sum_{i=1}^n|a_i x_1+b_i x_2|\quad\text{and}\quad \|x\|_1:=|x_1|+|x_2|.$$ -Let -$$N_{a,b}:=\max\{\|x\|_1\colon x\in\R^2,\,\|x\|_{a,b}\le1\},$$ -the norm of the identity operator on $\R^2$ with respect to the norms $\|\cdot\|_{a,b}$ and $\|\cdot\|_1$. -Note that $N_{a,b}=\sqrt2$ if $a=[1,1,0,\dots,0]^T/\sqrt2$ and $b=[-1,1,0,\dots,0]^T/\sqrt2$. - -Question: Is it always true that $N_{a,b}\le\sqrt2$, for all mutually orthogonal unit vectors $a$ and $b$ in $\R^n$? - -REPLY [7 votes]: Simply observe that -$$\|x\|_{a,b}=\|x_1a+x_2b\|_1\,.$$ -Thus, by orthogonality of $a,b$ and the easily-derived inequality $\|y\|_2\le\|y\|_1\le\sqrt{n}\|y\|_2$ for any $y\in\mathbb{R}^n$, we have -\begin{eqnarray*} -\|x\|_{a,b} & = & \|x_1a+x_2b\|_1 \\ - & \ge & \|x_1a+x_2b\|_2=\sqrt{x_1^2+x_2^2}=\|x\|_2 \\ -& \ge & \frac{1}{\sqrt{2}}\|x\|_1 -\end{eqnarray*} -for all $x\in\mathbb R^2$, and therefore $N_{a,b}\le\sqrt{2}$, as desired.<|endoftext|> -TITLE: Exchanging the components of a two-component link -QUESTION [5 upvotes]: Given a 2-component link in $S^3$ whose components are trivial knots, is it always possible to find a homeomorphism of $S^3$ that exchanges the components? -I guess the answer is "no" (but I could not find a counterexample), so here is a second question: -Which link-invariant could prevent the existence of such an exchanging homeomorphism? - -REPLY [6 votes]: Indeed, as you suspect, the answer is no. For instance, take the link $L$ obtained from the Hopf link by doing a $(2,1)$-cable of one component and a Whitehead double of the other. -A way of telling them apart is to look at the JSJ decomposition of the complement: each piece contains one component, and one component is Seifert fibred, while the other is hyperbolic. -A reflection of this fact is that if you twist each of the components with respect to the other, you obtain different knots. (By "twist" here I essentially mean "do $\pm1$-surgery along one component.) -I also think that the 2-variable Alexander polynomial (or branched covers, coloured signatures...) should do the trick.<|endoftext|> -TITLE: Studying higher categories from the bottom up -QUESTION [8 upvotes]: It is standard in category theory to study things 'from the top down' -- to study structured sets we use categories, to study structured categories we use bicategories, to study structured bicategories we use Gray-categories, and to study all of them at once we can use $\infty$-categories. -Unless I'm mistaken, some of Grothendieck's initial work on fibered categories was implicitly using bicategorical ideas (indexed categories) without the bicategorical vernacular, and the development of $2$-category theory allowed us to more succinctly express the ideas in play. -This has obviously been fruitful, but it seems counterintuitive to study simpler things using more complicated things. I can write down the definition of a set in a few seconds (primitive), the definition of a category in about a minute, the definition of a bicategory takes maybe a few minutes if rushed, and a tricategory with all of its coherence pasting diagrams explicitly written out would take 27 pages (if I remember correctly, I'm moving and the GPS book with the definition is still in the old house). Ideally, I would like to be able to study the more complicated things using conceptually simpler ones. - -Has there been any work done attempting to seriously study higher categorical notions using lower categorical ones? - -For example, we can consider the $1$-category of pseudonatural transformations and modifications (in full generality or between two fixed pseudofunctors) and study it using the machinery of $1$-category theory, perhaps gaining insight into pseudofunctor $2$-categories. We could also study the category of bicategories and pseudofunctors. These examples both require prior knowledge of the definitions involved in the higher stages of the categorical hierarchy though, which is not ideal. -There are also some negative results in this direction; for example, we can't form a category of tricategories and trihomomorphisms since composition of trihomomorphisms isn't associative on the nose; the best we can hope for is a bicategory (see page $4$ of the linked Garner/Gurski paper). This obstacle seems to imply that an approach based on this method would be limited to using levels just below the one we want to study. -Another type of example would be attempting to study the ingredients of higher categories at lower levels. For example, the definition of a bicategory uses categories, functors and natural transformations. While I'm not aware of a way to compile all three of these into one structure without using bicategories, we can collect the top two levels into a category ${\bf Fun}$ of functors and natural transformations -- functor categories between fixed categories $\mathcal{C}$ and $\mathcal{D}$ can be obtained as the full subcategory of ${\bf Fun}$ on functors whose domain is $\mathcal{C}$ and whose codomain is $\mathcal{D}$. -While the composition of functors and Godement product of natural transformations together don't form any kind of named structure I'm aware of on ${\bf Fun}$, if we restrict out attention to endofunctor categories $\mathcal{C}^\mathcal{C}$ for a fixed category $\mathcal{C}$ then composition and Godement products constitute a tensor product, turning $\mathcal{C}^\mathcal{C}$ into a strict monoidal category; whatever structure composition constitutes on ${\bf Fun}$, it's one one that restricts to a strict tensor product on certain 'well-behaved' full subcategories. -This can be seen as a structure 'endowed' on the endofunctor category by virtue of the fact that $\mathfrak{Cat}$ is a strict $2$-category, but we could also in theory make the observations about $\mathcal{C}^\mathcal{C}$ and ${\bf Fun}$ first and study them, gaining insight into horizontal composition of $2$-cells at the $2$-categorical level using $1$-categorical machinery. This is just a rough example of what I'm looking for, but it's the best I've come up with after a week or so thinking about it. Any assistance is appreciated. -Linked paper: arXiv:0711.1761v2 [math.CT] - -EDIT: For another example in the second vein, we have the following lemma about sets and functions that essentially yields the intuituion for the Yoneda lemma, and can be used to prove it. - -Lemma For sets $X,Y$, let $Hom(X,Y)$ denote the set of all functions from $X$ to $Y$. For any two functions $f:A\to B$ and $g:C\to D$, the following diagram commutes -where $g\circ$ and $\circ f$ denote postcomposition with $g$ and precomposition with $f$, respectively. - -We can thusly study a central piece of $1$-category theory using sets and functions ($0$-category theory). - -REPLY [8 votes]: Well, what is an $\infty$-category? Most people who study $\infty$-categories would, if pressed for an actual definition, say something like "a quasicategory" or "a complete Segal space". But any such definition consists of a collection of sets and functions between them satisfying certain axioms. So in that sense, we always study the more complicated $\infty$-categories using the simpler sets. Even homotopy type theory, which treats $\infty$-groupoids or $\infty$-categories as "synthetic" objects that aren't built out of sets, uses set-like structures at the meta-level, with definitional equalities between terms. -Quillen model categories, and related tools like fibration categories or cofibrant replacement comonads, are another way of studying $\infty$-categories using 1-categories. The Riehl-Verity followers who say "$\infty$-category" to mean "object of some $\infty$-cosmos" are using this approach, since an $\infty$-cosmos is a certain kind of fibration category. Homotopy categories and homotopy 2-categories have already been mentioned as another way to study higher-categorical things with lower-categorical ones; "derivators" are an enhanced sort of homotopy category that's especially good for this purpose. (I wrote a blog post about some of these approaches.)<|endoftext|> -TITLE: Computations in condensed mathematics, page 32-34 -QUESTION [10 upvotes]: I started reading the Lectures on Condensed Mathematics. I am looking at the material at page 32-34. I have three fundamental computation questions: - - -At the last line of pg 32 - it seems to imply that for finite sets $S$, $\Bbb Z[S] \simeq \underline{Hom}(C(S, \Bbb Z ), \Bbb Z) $? - - - -In pg 33 line 4 how is $Hom(C(S,\Bbb Z), \Bbb Z)$ identified with measure on $S$? - - - -In proof of Proposition 5.7, there was the following equivalence, - -$$ RHom(\prod_J \Bbb R, \Bbb Z) \simeq RHom_{\Bbb R} (\prod_J \Bbb R, R\underline{Hom}(\Bbb R, \Bbb Z) )=0$$ -why is this true and what exactly does $RHom_{\Bbb R}$ mean? Is there some adjunction happening here from $\Bbb Z$ modules of $Cond(Set)$ to $\Bbb R$-modules of $Cond(Set)$? - -I would appreciate if there are some related references for the general set up in 2 and 3. - -For 1, I would like to compute the $T$ points for $T$ extremely. disconnected. -What i don't see is an easy expression for lhs. -$$\Bbb Z[S] (T)= \bigoplus_{C(T,S)}\Bbb Z$$ -Conversely for rhs we have -$$\underline{Hom}(C(S,\Bbb Z), \Bbb Z)(T)=Hom(\Bbb Z[T] \otimes C(S,\Bbb Z), \Bbb Z) $$ -This doesn't seem an easy expression to handle too. - -REPLY [13 votes]: Correct, as both sides are the $S$-indexed direct sum of copies of $\mathbb{Z}$. For the LHS this holds by the universal property of $\mathbb{Z}[S]$, and for the RHS note that $C(S,\mathbb{Z}) = \prod_S \mathbb{Z} = \oplus_S \mathbb{Z}$ which lets you calculate. - -By definition, one could say. It's reasonable to define a $\mathbb{Z}$-valued measure on a profinite set to be an element of $Hom(C(S,\mathbb{Z}),\mathbb{Z})$. You could also define a measure to be a finitely additive assignment of an integer to each clopen subset of S. You can consider the indicator functions of clopen subsets to see the equivalence. - -$RHom_{\mathbb{R}}(M,N)$ means a complex calculating Ext's from M to N in the category of $\mathbb{R}$-modules in condensed abelian groups. The equality holds because both are the same as $RHom_{\mathbb{R}}(\mathbb{R} \otimes_\mathbb{Z}^L \prod_J \mathbb{R},\mathbb{Z})$ by some adjunctions. - - -There's also another way of looking at 3. Abstractly, the claim is that if $A$ is a condensed algebra and $N$ a condensed abelian group with $\underline{RHom}(A,N)=0$, then $RHom(M,N)=0$ whenever $M$ has an $A$-module structure. By adjunction, the hypothesis is equivalent to saying that $RHom(A\otimes_{\mathbb{Z}}^L M,N)=0$ for all condensed abelian groups $M$. If $M$ has an $A$-module structure, then $M$ is a retract of $A\otimes^L M$ by the action map $A\otimes^LM\rightarrow M$ in one direction and the unit $M=\mathbb{Z}\otimes^L M\rightarrow A\otimes^LM$. Thus we deduce $RHom(M,N)=0$ as claimed.<|endoftext|> -TITLE: Faithful traces on quasi-diagonal C*-algebras -QUESTION [5 upvotes]: Recall that a separable C*-algebra $A$ is quasi-diagonal if there are completely positive and contractive maps $\varphi_k \colon A \rightarrow M_{n(k)}$ such that $||\varphi_k(ab) - \varphi_k(a)\varphi_k(b)|| \rightarrow 0$ and $||\varphi_k(a)|| \rightarrow ||a||$ for every $a, b \in A$, where $M_{n(k)}$ denotes the algebra of $n(k) \times n(k)$ complex matrices. -The following ought to be known to experts, but I've had some issues trying to prove it. -Question: Let $A$ be a quasi-diagonal separable C*-algebra. Does $A$ admit a faithful trace? -Observe we do not assume that $A$ is simple. The idea would be to use an ultralimit of the cp maps $\varphi_k \colon A \rightarrow M_{n(k)}$ composed with the usual traces $tr_{n(k)} \colon M_{n(k)} \rightarrow \mathbb{C}$. This does not necessarily yield a faithful trace, though, as the traces $tr_{n(k)}(\varphi_k(a^*a))$ might be very small while $||\varphi_k(a^*a)||$ remains large. However, one could in some cases tweak $\varphi_k$ by adding a factor that enlarges the trace, maintains the asymptotic multiplicativity and the norm. The actual proof requires to do this everywhere at once, that is, for every possible element $a^*a$, and this seems impossible (or unreasonably hard). -Note, moreover, that there is no obvious K-theoretic obstruction, as every countable abelian group $K_0(A)$ embeds into $\mathbb{R}$. -Any help is greatly appreciated. - -REPLY [11 votes]: No, separable (unital) quasi-diagonal $C^\ast$-algebras do not necessarily admit a faithful tracial state. For instance, the $C^\ast$-algebra -\begin{equation} -A= \{ f\in C([0,1], \mathcal O_2) : f(0) \in \mathbb C 1_{\mathcal O_2}\} -\end{equation} -(where $\mathcal O_2$ is the Cuntz algebra with two canonical generators) is homotopic to $\mathbb C$, and hence is quasi-diagonal by the homotopy invariance of quasi-diagonality due to Voiculescu. However, any trace on $A$ vanishes on the ideal $C_0((0,1], \mathcal O_2)$, since this is purely infinite in the sense of Kirchhberg-Rørdam [Kirchberg, Eberhard; Rørdam, Mikael Non-simple purely infinite C∗-algebras. Amer. J. Math. 122 (2000), no. 3, 637–666.]. So the only tracial state on $A$ is the one factoring through evaluation at 0. -If you consider non-unital $C^\ast$-algebras, $C_0((0,1], \mathcal O_2)$ is an example of a separable quasi-diagonal $C^\ast$-algebra with no tracial states. -On the other hand, if you consider separable residually finite-dimensional (RFD) $C^\ast$-algebras, then they always admit a faithful tracial state: a separable $C^\ast$-algebra is RFD if it embeds into $\prod_{n\in \mathbb N} M_{k(n)}(\mathbb C)$ for some sequence $k(n)$ of natural numbers. As $\prod_{n\in \mathbb N} M_{k(n)}(\mathbb C)$ has a faithful tracial state, e.g. $\sum_{n\in \mathbb N} 2^{-n} \mathrm{tr}_{k(n)}$, so does any $C^\ast$-subalgebra.<|endoftext|> -TITLE: $\mathbb{Z}$-graded algebras and tensor products -QUESTION [7 upvotes]: Let $A = \bigoplus_{k \in \mathbb{Z}} A_k$ be a not necessarily commutative $\mathbb{Z}$-graded unital algebra over a field $\mathbb{K}$, and assume that it is strongly graded: -$$ -A_kA_l = A_{k+l}. -$$ -In general can it happen that the multiplication does not give an isomorphism -$$ -A_k \otimes_{A_0} A_l \simeq A_{k+l}? -$$ -The map will be surjective since we are assuming the gradind to be strong, but will it be injective? - -REPLY [8 votes]: No it cannot happen. -And not only for strongly $\mathbb{Z}$-graded rings; this is always the case for any strongly $G$-graded ring, where $G$ is a group. $A_k \otimes_{A_0} A_l \simeq A_{k+l}$ is an isomorphism of $A_0$-bimodules. -(See: Corollary 3.1.2, p.82, from Methods of Graded Rings).<|endoftext|> -TITLE: Is there a way to reconstruct the convolution $(f * g)(x)$ of $f$ with a Gaussian $g$ from sampled values, $(f*g)(a), a \in A$? -QUESTION [6 upvotes]: Suppose that $f: \mathbb{R} \to \mathbb{C}$ is a function which has support in $[-1,1]$. Let $g = g_\sigma$ be a centered Gaussian with variance $\sigma^2$. Is there a way to reconstruct the convolution of $f$ with $g$ (on whole $\mathbb R$) if only sampled values of this convolution are given, i.e. -$$ -\{ (f *g)(a) : a \in A \} -$$ -where $A \subset \mathbb R$ is a set such as a uniform grid, $A=c\mathbb Z$ for some $c>0$? Can we make some strong assumptions on $f$ such that there exists an explicit reconstruction? - -REPLY [2 votes]: I feel that a Fourier-based solution should be possible along these lines: Let $h(t)=\int f(x) \varphi (x-t) dx$ represent the convolution and $f(x)$ be sufficiently bounded. Using sampling rate $1$ for simplicity we have the exact result $$h(t) = \sum a(n) \frac{\sin \pi(t-n)}{\pi(t-n)}$$ where $$a(n) =\int h(t) \frac{\sin \pi(t-n)}{\pi(t-n)} dt.$$ If $h(t)$ was suitably band-limited one would actually have $a(n)=h(n)$. In fact, $h(n)$ are the only values given to us. The catch is that $h$ is not band limited although we do know that $f(x)$ is nicely bounded. Presumably if the sampling rate were increased (equivalent here to making $f(x)$ more bounded) the approximation involved in using $a(n)= h(n)$ would increase in accuracy. Since $h(t)$ is not band-limited, perfect reconstruction from the sampled $h(n)$ values may not be possible. On the other hand, the boundedness of $f$ and the fact that the Fourier transform of $\varphi$ is supported on all of $R$ gives some hope that the argument can be patched up to work. In particular, one could try to evaluate the integral for $a(n)$ by discretizing it given a sufficiently dense lattice for $h(t)$. As an aside, the moments solution that was proposed was extremely interesting. Can someone please try to complete the Fourier-based approach; it would be extremely worthwhile.<|endoftext|> -TITLE: Picard group of a cubic hypersurface -QUESTION [9 upvotes]: Consider the following cubic hypersurface in $\mathbb{P}^5$: -$$ -X = \{z_0z_3z_5-z_1^2z_5-z_0z_4^2+2z_1z_2z_4-z_2^2z_3 = 0\}\subset\mathbb{P}^5 -$$ -The singular locus of $X$ is the Veronese surface $V\subset X$. I would like to ask if it is known what is the Picard group of $X\setminus V$? -Thank you very much. - -REPLY [6 votes]: Another way to find $\mathrm{Pic}(X)$ is the following. Note that the cubic $X$ is the symmetric determinantal cubic and it has a resolution of singularities -$$ -\tilde{X} = \mathbb{P}_{\mathbb{P}^2}(S^2\Omega_{\mathbb{P}^2}(2)). -$$ -its explicit form implies that $\mathrm{Pic}(\tilde{X}) \cong \mathbb{Z} \oplus \mathbb{Z}$. Furthermore, the exceptional divisor of the contraction $\tilde{X} \to X$ is the subvariety -$$ -E = \mathbb{P}_{\mathbb{P}^2}(\Omega_{\mathbb{P}^2}(1)) -$$ -and its embedding into $\tilde{X}$ is the relative double Veronese embedding. Finally, it is easy to check that the class of $E$ in $\mathrm{Pic}(\tilde{X})$ is equal to -$$ -2H + 2h, -$$ -where $h$ is the hyperplane class of $\mathbb{P}^2$ and $H$ is the relative hyperplane class of $\mathbb{P}_{\mathbb{P}^2}(S^2\Omega_{\mathbb{P}^2}(2))$. -Therefore -$$ -\mathrm{Pic}(X \setminus V) = \mathrm{Pic}(\tilde{X} \setminus E) \cong \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}. -$$<|endoftext|> -TITLE: Can $Ord$ have nontrivial second-order elementary self-embeddings? -QUESTION [16 upvotes]: I forgot to mention originally: this was motivated by this old MSE question. -It's not hard to show in $\mathsf{ZFC}$ that there are nontrivial elementary embeddings from $(Ord; \in)$ to itself - or rather, it's not hard to write down specific formulas which $\mathsf{ZFC}$ proves define such maps. When we add common algebraic operations, things get messier but the picture seems to stay the same - e.g. the result continues to hold for $Ord$ equipped with ordinal addition, multiplication, and exponentiation in addition to $\in$. It seems like any "reasonably simple" expansion of $Ord$ has $\mathsf{ZFC}$-provable nontrivial elementary self-embeddings. -I'm interested in a particular precisiation of this intuition. Specifically, the following seems very close to the Kunen inconsistency but I can't quite get it there: - -Is it consistent that there is a nontrivial second-order elementary self-embedding $f:(Ord;\in)\rightarrow (Ord;\in)$? - -Note that the second-order quantifiers in this case range over subclasses of $Ord$ or its finite Cartesian powers, not just subsets (= bounded subclasses). So we really have to ask this within an appropriate class theory. Given that that may be a bit annoying to consider, here's a "set-ified" version of that question: - -Is it consistent with $\mathsf{ZFC}$ that there is an inaccessible $\kappa$ and a second-order nontrivial elementary embedding $f:(\kappa;\in)\rightarrow(\kappa;\in)$? - -(Here correspondingly the second-order quantifiers range over the $\mathcal{P}(\kappa^{n})$s for $n\in\omega$ as appropriate, so everything is nicely captured by $V_{\kappa+2}$: $f$ itself, or an equivalent coding thereof, lives in $V_{\kappa+1}$, the evaluation of each second-order formula over $V_{\kappa}$ is determined by $V_{\kappa+1}$, and $V_{\kappa+2}$ can express "is a second-order elementary embedding" directly.) -The difficulty I run into in getting a negative answer here by invoking the Kunen inconsistency is that even using second-order logic there doesn't seem to be a way to code $V$ into $Ord$ (or $V_\kappa$ into $\kappa$). Of course if $V$ is "sufficiently canonical" we can do this, e.g. the existence of such an $f$ (or such a $\kappa$ and $f$) clearly contradicts $\mathsf{V=L}$, but I don't see an argument for the general case. - -REPLY [17 votes]: Answering this question would either require refuting choiceless large cardinals or getting close to refuting Woodin's HOD Conjecture. -First, if choiceless cardinals are consistent, one cannot rule out the existence of a second-order elementary embedding. Corollary 4.8 in Usuba's "A note on Lowenheim-Skolem cardinals" states that if there is a proper class of Lowenheim-Skolem cardinals, one can force the Axiom of Choice by a homogeneous definable class forcing. (He doesn't say homogeneous, but it will be.) Therefore assume there is a $j : V\to V$ and there is a proper class of Lowenheim-Skolem cardinals. Let $\mathbb P$ be Usuba's forcing, and let $G\subseteq \mathbb P$ be a generic filter. Then $j\restriction \text{Ord}$ is second-order elementary in $V[G]$: if $\varphi(\vec v)$ is a second-order formula and $V[G]\vDash \varphi(\vec \alpha)$, then by homogeneity, $1_\mathbb P\Vdash \varphi(\vec \alpha)$, and so by the definability of $\mathbb P$ and the elementarity of $j$, $1_\mathbb P\Vdash \varphi(j(\vec\alpha))$, and hence $V[G]\vDash \varphi(j(\vec\alpha))$. -Second, assuming the HOD Conjecture and the existence of an extendible cardinal, one can rule out the existence of a nontrivial second-order elementary embedding of the ordinals. This follows from Theorem 3.47 of Woodin's paper "In search of Ultimate L." The reason is that any second-order elementary embedding $j : \text{Ord}\to\text{Ord}$ extends to an elementary embedding from the structure $(\text{HOD},T)$ to $(\text{HOD},T)$ where $T$ is the $\Sigma_2$ theory of $V$ with ordinal parameters. It is a prominent open question whether such an elementary embedding can ever exist. It is also open whether there can be an elementary embedding from $\text{HOD}$ to $\text{HOD}$ in the context of the HOD Conjecture. -So if strong choiceless large cardinals are consistent, so is a second-order embedding, while if the HOD Conjecture is true, second-order embeddings are ruled out by more compelling large cardinal axioms. The question is therefore just one aspect of one of the most important questions in set theory/large cardinals: to what extent can arbitrary sets be approximated by ordinal definable ones?<|endoftext|> -TITLE: Tiling a rectangle with all simply connected polyominoes of fixed size -QUESTION [15 upvotes]: For which values of $n$ can we tile some rectangle with one copy of each free simply-connected $n$-omino (that is, each polyomino with $n$ squares that has no holes)? -It appears that it is possible for $n=1$ (trivial), $n=2$ (trivial), $n=5$ (see here), and $n=7$ (see here); and impossible for $n=3$ (trivial), $n=4$ (by a parity argument), and $n=6$ (by a parity argument). Is it known to be possible or impossible for any $n\geq 8$? - -REPLY [12 votes]: Timothy Budd's construction for $n =20$ actually shows that it is impossible for $n \geq 18$. The following images show constructions for $18 \leq n \leq 23$. For each of them, the polyomino containing the red square must fill the cavity and there are not enough possible ways of attaching the remaining (at most $2$) squares to cater for all siblings. - -To go from $n$ to $n+4$ we can for instance add an additional column and increase the size of the "lump" by $2$.<|endoftext|> -TITLE: Rational sextic plane curves -QUESTION [14 upvotes]: In Coolidge's venerable Treatise on algebraic plane curves, Theorem 28 p. 392 states: - -Let $S=\{P_1,\ldots ,P_{10}\} \subset \mathbb{P}^2$ such that for any $i$, there is a sextic curve (edit: integral) singular along $S\smallsetminus P_i$; then there is a sextic curve singular along $S$. - -I am probably missing something, but it seems to me that the text before and after this statement has nothing to do with the Theorem. Does someone know a proof? - -REPLY [10 votes]: Let me try to reconstruct Coolidge's argument at pages 390-392. -Note: there is a misprint at line 4 of p. 392, where "with double points" should be "with triple points". - -Proposition 1 [Coolodge, p. 390-391]. Assume that $P_1, \ldots, P_8$ are points on the plane, not three on a line and not six on a conic. Let $\phi_1$, $\phi_2$ be a basis for the pencil of cubics through the $P_i$ and let $f$ be a non-degenerate sextic with the $P_i$ as double points. Then a point $P_9$ is a further double point of a non-degenerate sextic with double points at $P_1, \ldots, P_8$ if and only if it is a smooth point of the curve $$\frac{\partial(\phi_1, \, \phi_2, \, f)}{\partial(x_1, \, x_2, \, x_3)}=0.$$ -This curve has triple points at each of the points $P_1, \ldots, P_8$ and passes simply through the $12$ singular points of the nodal curves of the pencil of cubics spanned by $\phi_1$ and $\phi_2$. - - -Proposition 2 [Coolidge, Theorem 27, p. 391]. Assume that nine points of the plane are the base locus of a pencil of nodal sextics. Then there are precisely $12$ sextics in the pencil having a further double point. - -Using these results, we can now prove the - -Theorem. Assume that $P_1, \ldots, P_{10}$ is a set of ten points such that for any nine of them there is a non-degenerate sextic (and hence a pencil of non-degenerate sextics) having nodes there . Then there is a non-degenerate sextic having nodes at all of the ten points. - -Proof. Consider the nodal sextics through $P_1, \ldots P_8$. Since by assumption there is a nodal sextic through $P_1, \ldots, P_9$, we deduce that $P_9$ is a smooth point of the curve $C$ of degree $9$ and with triple points at $P_1, \ldots, P_8$ described in Proposition $1$. Analogously, consider the nodal sextics through $P_1, \ldots, P_7, \, P_9$. Then $P_8$ is a smooth point of the curve $D$ of degree $9$ and with triple points at $P_1, \ldots, P_7, \, P_9$. -The $12$ nodal points $Q_1, \ldots, Q_{12}$ of the $12$ singular curves of the pencil of nodal sextics through $P_1, \ldots, P_9$ (see Proposition 2) must belong to both $C$ and $D$. Moreover, since $$9\cdot 9 - 7 \cdot 3 \cdot 3 - 3 \cdot 1 - 1 \cdot 3 = 12,$$ the only intersection points of $C$ and $D$ are $P_1, \ldots, P_9$ and $Q_1, \ldots, Q_{12}$. -Finally, by assumption we also have pencils of nodal sextics through $P_1, \ldots, P_8, \, P_{10}$ and through $P_1, \ldots, P_7, \, P_9, \, P_{10}$. Then $P_{10}$ belongs to the intersection of $C$ and $D$, too. By the remark above, this means that $P_{10}=Q_i$ for some $i$. -Summing up, there is a curve in the pencil of nodal sextics through $P_1, \ldots, P_9$ which has a further node at $P_{10}$. This concludes the proof.<|endoftext|> -TITLE: Are all integers not congruent to 6 modulo 9 of the form $x^3+y^3+3z^2$? -QUESTION [10 upvotes]: Are all integers not congruent to 6 modulo 9 of the form $x^3+y^3+3z^2$ -for possibly negative integers $x,y,z$? -We have the identity $ (-t)^3+(t-1)^3+3 t^2=3t-1$. -The only congruence obstruction we found is 6 modulo 9. - -REPLY [3 votes]: We found another approach based on integral points on genus 0 curves. -Let $G(x,z,a_1,a_2,a_3)=(a_1 x+a_2)^3+(-a_1 x+a_3)^3+3 z^2$. -For fixed $n$ and $a_i$, $G=n$ is degree two genus 0 curve and it might -have infinitely many integral points, which gives infinitely many solutions -to the OP. -WolramAlpna on solve (10x-5)^3+(-10x+3)^3+3*y^2=25 over integers. -Related paper is Representation of an Integer as a Sum of Four Integer Cubes and Related Problems -Added 22 Dec 2020 -We can solve $n \equiv 1 \pmod{3}$ -The solutions of $(-x-1)^3+(x+2)^3 - 3y^2=(3N+1)$ -are $x=-1/2\sqrt{4y^2 + 4N + 1} - 3/2$. -When we express $4N+1$ as difference of two squares $4N+1=z^2-y'^2$ with $y'$ -even we have solution. -Currently the only unsolved case appears to be $n \equiv 0 \pmod{9}$.<|endoftext|> -TITLE: In what sense exactly are the Einstein metrics distinguished? -QUESTION [5 upvotes]: EDIT: In general relativity given a manifold $M$ one can consider a functional on (pseudo-) Riemannian metrics $g$ $$\int_M R\,\, dvol_g,$$ -where $R$ is the scalar curvature and $vol_g$ is the (pseudo-) Riemannian measure. The extremal metrics for this functional (solutions of the Euler-Lagrange equation) satisfy the so called Einstein equation. -As far as I heard the distinguished property of the above functional is that $R$ involves second derivatives of the metric, but the Euler-Lagrange equation is a differential equation not of the fourth order in the metric, but only of the second order in the metric. - -I am looking for a precise statement within what class of functionals on metrics the above property distinguishes the scalar curvature $R$. Say assume I have a generally covariant expression involving at most second order derivatives of the metric such that the Euler-Lagrange equation is also of second order in the metric. Does it imply that this expression is proportional to $R$? - -REPLY [8 votes]: If I understood your question correctly, the answer indeed is due to Lovelock. I think it's important to state all the hypotheses clearly, because they are not always reported accurately. -Theorem. (Lovelock, 1971) Given a metric $g_{ab}$ and a covariantly constructed symmetric 2-tensor $T^{ab}(g,\partial g, \partial^2 g)$ such that $\nabla_a T^{ab} = 0$, then necessarily the tensor $T^{ab} dvol_g = \frac{\delta L}{\delta g_{ab}}$ is the Euler-Lagrange derivative of the Lagrangian density $L = \sum_k \alpha_k R_{[a_1 a_2}{}^{a_1 a_2} \cdots R_{a_{2k-1} a_{2k}]}{}^{a_{2k-1} a_{2k}} dvol_g$, where $R_{abcd}$ is the Riemann curvature tensor, $[{\cdots}]$ denotes full antisymmetrization, and $\alpha_k$ are arbitrary constants (their values determine $T^{ab}$). -The number of terms in $L$ is finite in any given dimensions, because antisymmetrizing over more indices than the dimension always gives zero. In 4 dimensions, the only possibilities are $L = (\alpha_0 + \alpha_1 R + \alpha_2 R_{abcd} R^{abcd}) dvol_g$. -If $T^{ab} dvol_g = \delta L/\delta g_{ab}$, for any covariantly constructed Lagrangian $L$, then $T^{ab}$ is obviously symmetric and $\nabla_a T^{ab}=0$. So Lovelock's theorem actually classifies all Lagrangians that depend depend covariantly on the metric and Riemann curvature (including its derivatives) that have second order Euler-Lagrange equations, up to null Lagrangians, those whose Euler-Lagrange equations are identically zero. In fact, in 4 dimensions, a particular choice of the constants with $\alpha_2\ne 0$ gives such a null Lagrangian, the so-called Gauss-Bonnet term. So, among 4-dimensional Lagrangians, by adding a null Lagrangian, one can always reduce the desired variational principle to the Einstein-Hilbert form, $L= (\alpha_0 + \alpha_1 R) dvol_g$. - -Lovelock, D., The Einstein tensor and its generalizations, J. Math. Phys. 12, 498-501 (1971). ZBL0213.48801. - -REPLY [3 votes]: OK I think I understand the question now. -You want to look at Lovelock gravities. -These are the most general theories (not bothering with a list of qualifiers here, see wikipedia) that produce 2nd order EOMS. -For dimensions d=3 and d=4 the only Lovelock functional is indeed the Einstein-Hilbert one.<|endoftext|> -TITLE: Zermelo-Frankel set theory for algebraists -QUESTION [6 upvotes]: $\DeclareMathOperator\Var{Var}\DeclareMathOperator\CRings{CRings}\DeclareMathOperator\Grp{Grp}\DeclareMathOperator\Sets{Sets}$I'm not a logician/set theorist, and I have some questions on set theory and references that may seem "trivial" for experts. Still I ask the question – if you have references this would be interesting. -In algebraic geometry (See Hartshorne's book, Appendix A) the following theorem is proved: -Let $\Var(k)$ be the "category of non-singular quasi-projective varieties over an algebraically closed field $k$ and morphisms of varieties over $k$. This category is defined in Hartshorne's book. -Theorem 1.1. There is a unique intersection theory $A^*(X)$ for algebraic cycles on $X\in \Var(k)$ modulo rational equivalence satisfying the axioms A1–A7. -The axioms A1–A7 are listed on page 426-427 in the book. For a variety $X\in \Var(k)$ one defines -a commutative unital ring $A^*(X)$ – the Chow ring – and this construction is unique. There is only one way to do this, meaning there is a unique functor -$A^*(-) : \Var(k) \rightarrow \CRings$ -such that axioms A1–A7 hold. Here $\CRings$ is the category of commutative unital rings and maps of unital rings. -In algebra one defines a group $(G, \bullet)$ as a set $G$ with an operation $\bullet: G\times G \rightarrow G$ satisfying $3$ axioms: G1 Associativity, G2 existence of identity and G3 existence of inverse. One defines a morphism of groups and the "category of groups" $\Grp$. Clearly the category of groups $\Grp$ contains non-isomorphic groups, hence the axioms G1–G3 does not uniquely determine -one group. There are many different groups satisfying G1–G3. -In ZF set theory set theorists write down 9 axioms ZF1-ZF9, and these axioms determine $\Sets$ – the "category of sets". $\Sets$ is a category with "sets" as objects and "maps between sets" as morphisms. We would like the category $\Sets$ to be uniquely determined by the axioms ZF1–ZF9 similarly to what happens for the Chow ring. Is it? Is there a unique category $\Sets$ fulfilling the axioms ZF1–ZF9? If yes I ask for a reference. -For reference, Wikipedia has the page ZF set theory. - -REPLY [3 votes]: You have to specify what you mean by a category 'fulfilling a list of axioms of set theory', because axioms of set theory are usually statements about sets in some logical language. -One reasonable interpretation would be to use the internal logic of a category to form an internal logical language, then formulate the axioms of set theory in the internal language and ask if they're true in the ambient category. Set can trivially internally formulate and satisfy all of the set-theoretic axioms of whatever background set-theoretic universe we're in, so in particular if we're working in $ZF$ then Set will satisfy ZF1-ZF9. -It turns out that other categories do this as well, so Set is not unique in this regard as other answers state. Specifically, Mike Shulman wrote a nice paper exploring how to extend the usual internal logic of a pretopos to a more general interpretation called the stack semantics, which allows us to formulate axioms involving unbounded quantifiers. Using the stack semantics of a topos, we can formulate all the axioms of $ZF$ internally and ask if they're true in the ambient topos. -Collection and replacement are satisfied in the stack semantics of any topos, but the separation axiom requires the introduction of a new topos-theoretic axiom schema -- any topos satisfying this additional axiom schema is called an autological topos, and these are precisely the categories which internally have the full strength of $ZF$ set theory. -So it seems like the answer to your question is: - -No, Set is not the only category satisfying ZF1-ZF9 in the above sense, and the (bi)category of all categories satisfying ZF1-ZF9 would be precisely the (bi)category of autological toposes. - -If you are looking for a unique characterization of Set, it is the terminal object in the category of Grothendieck toposes and geometric morphisms. -As mentioned in Andrej Bauer's answer it is also the initial 'ZF-algebra'. For a summary on this view with definitions you can take a look at these notes on Algebraic Set Theory by Steve Awodey, or take a look at the classical reference on Algebraic Set Theory by Joyal and Moerdijk for a full explanation. -Linked paper: arXiv:1004.3802 [math.CT] - -REPLY [2 votes]: There is such a thing as Algebraic Set Theory where: - -models of set theory are simply algebras for a suitably presented algebraic theory and then many familiar set theoretic conditions (such as well-foundedness) are thereby related to familiar algebrauc ones (such as freeness) ... it was developed by Joyal & Moerdjik in 1988 and was first presented in detail in a book published in 1995 by them.<|endoftext|> -TITLE: Atiyah's proof of the moduli space of SD irreducible YM connections -QUESTION [15 upvotes]: In the paper "Self-duality in Four-dimensional Riemannian Geometry" (1978), Atiyah, Hitchin and Singer present a proof that the space of self-dual irreducible Yang-Mills connections is a Hausdorff manifold, and if it is not the empty set, then the dimension is given by -$$p_1(\text{Ad}(P))-\frac{1}{2}\dim G(\chi(M)-\tau(M))$$ -Where $\chi(M)$ is the Euler characteristic and $\tau(M)$ the signature. -EDIT: It turns out the original paper contained an error/typo. It should in fact be -$$2p_1(\text{Ad}(P))-\frac{1}{2}\dim G(\chi(M)-\tau(M))$$ -End of edit. -Although I would love to be able to understand the full paper, I am not in a position to be able to do so yet, I am only trying to understand the computation of this dimension, because I am interested in some applications of the Atiyah-Singer index theorem. -To compute this dimension, the following is utilised in the paper: -Let $D:\Gamma(V_-\otimes E)\to\Gamma(V_+\otimes E)$ be the Dirac operator for a spinor bundle with values in some auxiliary bundle $E$. By the index theorem, -$$\text{ind}(D)=\int_M\text{ch}(E)\widehat{A}(M)$$ -In dimension four, we have $\widehat{A}(M)=1-\frac{1}{24}p_1(M)$ (but where is this used?). For the proof, we take $E=V_-\otimes\text{Ad}(P)$. Then $\text{ch}(E)=\text{ch}(\text{Ad}(P))\text{ch}(V_-)$. So far, so good. I lose track in the following computation: -$$\text{ind}(D)=\int_M\text{ch}(\text{Ad}(P))\text{ch}(V_-)\widehat{A}(M)\\ -\color{red}{=p_1(\text{Ad}(P))+\dim G(\text{ind}(D'))}=\\ -p_1(\text{Ad}(P))-\frac{1}{2}\dim G(\chi-\tau)$$ -Where $D':\Gamma(V_+\otimes V_-)\to\Gamma(V_-\otimes V_-)$. I have been trying to find a result that explains the red coloured part of the equation, because this step seems completely non-trivial, and in spite of that, it is not elaborated upon within the paper at all, and I am not able to find any sources that explain this step. In Index of Dirac operator and Chern character of symmetric product twisting bundle the accepted answer seems to give an answer that goes some way towards explaining how this result is obtained, in a very particular case. However, I am not very experienced in this area and I don't know how to generalise the result to an arbitrary principal $G$-bundle. I am looking for an explanation of the above, whether someone is able to provide their own response or a reference. Either one would be greatly appreciated. - -REPLY [14 votes]: Hopefully I remember this well. My adviser explained this computation to me I don't even want to think how many years ago. -The deformation complex of the SD equation is $\DeclareMathOperator{\Ad}{Ad}$ -$$L=d_A^-\oplus d_A^*:\Omega^1\big(\, \Ad(P)\,\big)\to\Omega^2_-\big(\; \Ad(P)\;\big)\oplus \Omega^0\big(\;\Ad(P)\;\big). $$ -The dimension of the moduli space of self-dual connections is the index of this operator. $\DeclareMathOperator{\ind}{ind}$ $\DeclareMathOperator{\ch}{ch}$ $\DeclareMathOperator{\hA}{\widehat{A}}$This operator is obtained by twisting with $\Ad(P)$ the operator -$$ D=d^-+d^*:\Omega^1(M)\to \Omega^2_-(M)\oplus \Omega^0(M) $$ -This is the operator $D: \Gamma(V_+\oplus V_-)\to \Gamma(V_-\oplus V_-)$ in the paper you mentioned. -The Atiyah-Singer index theory shows that $\ind L$ is -$$\ind L= \int_M \big[\; \ch(\Ad(P)) \hA(X)\ch(V_-)\;\big]_4, $$ -where $[--]_4$ denotes the degree $4$ part of a non-homogeneous differential form. -We deduce -$$\ch(\Ad(P))=\dim G +\ch_2(\Ad(P))+\cdots = \dim G+p_1(\Ad(P))+\cdots, $$ -$$\ind L= \int_M \big(\; p_1(\Ad(P))+(\dim G)\rho_D\;\big) $$ -where the degree $4$ from $\rho_D= [\hA(X)\ch(V_-)]_4$ is the index density of $D$ appearing in the Atiyah-Singer index theorem -$$ -\ind D=\int_M \rho_D. -$$ -Thus -$$ -\ind L=\int_M p_1(\Ad(P))+\dim G\ind D= \int_M p_1(\Ad(P))+\dim G(b_1 -b_2^--b_0). $$ -Now express $(b_1-b_2^--b_0)$ in terms of the signature $\tau=b_2^+-b_2^-$ and the Euler characteristic $\chi=2b_0-2b_1+b_2^++b_2^-$.<|endoftext|> -TITLE: Is there a non-free group $G$ whose subgroups are all freely decomposable? -QUESTION [10 upvotes]: Suppose that $G$ is a group such that every subgroup $H \subseteq G$ (including $G$ itself) is either free or a non-trivial free product, i.e. $H = H_1 * H_2$ with $H_1, H_2$ both non-trivial. Is there an example of such a $G$ which is not free? -If $G$ is finitely generated then Grushko's theorem implies $G$ must be free, but the infinitely-generated case seems likely to have a non-free example. - -REPLY [19 votes]: Yes, there's an example. -Kurosh proved that the group $G$ with presentation -$$\langle (a_n)_{n\ge 0},(b_n)_{n\ge 1}\mid a_nb_na_n^{-1}b_n^{-1}=a_{n-1},\;\forall n\ge 1\rangle$$ -has the following properties: $G$ is torsion-free, isomorphic to $G\ast\mathbf{Z}$, all freely indecomposable subgroups of $G$ are cyclic, but $G$ is not free (it's easily checked not to be residually nilpotent: the intersection of the lower central series contains all $a_n$). -Reference: A. Kurosch. Zum Zerlegungsproblem der Theorie der freien Produkte. Mat. Sbornik 2 (44): 5, 995–1001, 1937. - -[Added] Context: -Kurosh asked in 1934 (Math Ann.) the (free) Zerlegungsproblem: does every group have a free product decomposition into freely indecomposable factor? -In the same 1934 paper he proved that if the answer is positive for a given group, then all such decompositions are isomorphic. This was improved by Baer and Levi (Comp. Math. 1936): on an arbitrary group, any two free product decompositions (with possibly decomposable factors) have a common refinement (in a suitable sense). Then in 1937 (Mat. Sbornik) Kurosh found the above example providing in general a negative solution to the Zerlegungsproblem. Later, Grushko (Mat. Sbornik 1940) and then independently B.H. Neumann (J. LMS 1943) proved that the Zerlegungsproblem has a positive answer for finitely generated groups, showing that the generating rank is additive (and not only sub-additive) under free products. - -Added remark: there's an embedding of the countable group $G$ into the 3-generator group with 1-relator presentation -$$\Gamma=\langle a,x,y\mid a={^x}\!a\;^y\!a\rangle\qquad (\text{where } {^x}\!a\;^y\!a=xax^{-1}yay^{-1}),$$ -given by $a_n\mapsto x^{-n}ax^{n}$, $b_n\mapsto x^{-n}yx^{n}$. I don't know if this 1-relator group has ever been studied specifically. -More precisely, $\Gamma$ is the ascending HNN-extension of $G$ associated with the injective endomorphism $i$ given by $a_n\mapsto a_{n+1}$, $b_n\mapsto b_{n+1}$ (note that $G=i(G)\ast\langle b_0\rangle$).<|endoftext|> -TITLE: Branching laws for smooth representations -QUESTION [8 upvotes]: Let $E / F$ be a quadratic extension of nonarchimedean local fields (characteristic 0 if it matters), and $\pi$ an irreducible infinite-dimensional smooth representation of $GL_2(E)$. Let $B$ be the upper-triangular Borel of $GL_2$. I'd like to know: do we always have -$$\operatorname{dim} Hom_{B(F)}(\pi, \mathbf{C}) = 1 ?$$ -I know how to construct a non-zero element in this space, so my concern is to prove its dimension is $\le 1$, i.e. to show that $(GL_2(E), B(F))$ is a Gelfand pair. -Here's why I think it should be true. Firstly, the analogous result is true if we replace the field extension $E$ with $F \oplus F$; this is the main result of the Harris--Scholl paper cited below. Secondly, it's true if $\pi$ is discrete-series: by Frobenius reciprocity we have $Hom_{B(F)}(\pi, \mathbf{C}) = Hom_{GL_2(F)}(\pi, Ind_{B(F)}^{GL_2(F)}(\mathbf{C}))$. The representation $Ind_{B(F)}^{GL_2(F)}(\mathbf{C})$ has a trivial subrepresentation and the quotient is the Steinberg representation $\mathrm{St}$, so we have a left-exact seq -$$0 \to Hom_{GL_2(F)}(\pi, \mathbf{C}) \to Hom_{B(F)}(\pi, \mathbf{C}) \to Hom_{GL_2(F)}(\pi, \mathrm{St}).$$ -From the results in Prasad's 1992 paper cited below, exactly one of $Hom_{GL_2(F)}(\pi, \mathbf{C})$ and $Hom_{GL_2(F)}(\pi, \mathrm{St})$ is 1-dimensional and the other is zero, so we are done. -However, if $\pi$ is a principal series, we have the same exact sequence but it can happen that $Hom_{GL_2(F)}(\pi, \mathrm{St})$ and $Hom_{GL_2(F)}(\pi, \mathbf{C})$ are both non-zero (this occurs iff $\pi$ is the normalised induction of a pair of characters of $E^\times$ which are distinct, and both trivial on $F^\times$). I tried to bash out this case via Mackey theory, but I couldn't get it to work. -Harris, M.; Scholl, A. J., A note on trilinear forms for reducible representations and Beilinson’s conjectures., J. Eur. Math. Soc. (JEMS) 3, No. 1, 93-104 (2001). Preprint version Published version -Prasad, Dipendra, Invariant forms for representations of $\text{GL}_2$ over a local field, Am. J. Math. 114, No. 6, 1317-1363 (1992). - -REPLY [5 votes]: You can approach the problem via the mirabolic subgroup $P_2(F)\subset B_2(F)$. First we can restrict to $\pi$ with central character trivial on $F^\times$. Then you want to know if in this situation $Hom_{P_2(F)}(\pi,1)$ is of dimension at most $1$. This is indeed the case for unitary representations for example by [Matringe, Pacific Journal 2014, Proposition 2.5]. In particular this seems to take care of the remaining case you had of representations of the form $\chi_1\times \chi_2$ with $(\chi_i)_{|F^\times}=1$, which are clearly unitary as both $\chi_i$'s have to be unitary in this case.<|endoftext|> -TITLE: If a subspace $F$ is contained in a subspace $G$, and $H$ is close to $G$, can we choose a subspace of $H$ close to $F$? -QUESTION [5 upvotes]: Let $E$ be a Banach space. Recall that the collection of all closed linear subspaces of $E$ can be turned into a metric space in a number of ways. In particular, consider the notion of a gap: if $G$ and $H$ are subspace of $E$, then -$$g(G,H)=\max\{\sup\limits_{g\in \partial B_{G}} d(g, H),~\sup\limits_{h\in \partial B_{H}} d(h, G)\},$$ -where by $\partial B_{G}$ and $\partial B_{H}$ we mean the intersection of the unit sphere with $G$ and $H$, respectively. Note, that $g(G,H)\le h(G,H)\le 2g(G,H)$, where $h(G,H)$ is the Hausdorff distance between $\partial B_{G}$ and $\partial B_{H}$. - -Let $F\subset G$ be subspaces of $E$ and let $\varepsilon>0$. Does there exist $\delta>0$ such that every subspace $H$ with $g(G,H)<\delta$ contains a sub-subspace $J\subset H$ with $g(F,J)<\varepsilon$? - -REPLY [11 votes]: The answer is "No". You can derive this from Lemma 5.9 and Proposition 5.3 in my paper Ostrovskiĭ, M. I. Topologies on the set of all subspaces of a Banach space and related questions of Banach space geometry. Quaestiones Math. 17 (1994), no. 3, 259–319. In that Lemma a collection of subspaces $G_\varepsilon$, $\varepsilon\in(0,1)$ is constructed in the space $X\oplus X/Y$ which converges to $Y\oplus X/Y$ with respect to gap, and is such that all $G_\epsilon$ are isomorphic to $X$. -We get a counterexample in cases where $X/Y$ does not admit an isomorphic embedding into $X$ (such examples are well-known, e.g. $X=\ell_1$ and $X/Y=c_0$). In fact, if there would be a subspace $U$ in $G_\varepsilon$, which is close to $X/Y$, since $X/Y$ is complemented in the whole space, by Berkson's proposition (see Proposition 5.3 in the paper mentioned above), this would imply that $U$ is isomorphic to $X/Y$, a contradiction.<|endoftext|> -TITLE: Is a Banach lattice isomorphic to a Hilbert space in fact a Hilbert lattice? -QUESTION [9 upvotes]: The title says it all: - -Let $E$ be a Banach lattice, which is isomorphic to a Hilbert space (as normed spaces). Is there an equivalent Hilbert norm on $E$, which still makes it a Banach lattice with respect to the original order structure on $E$? - -REPLY [4 votes]: Yes. This is a consequence of the fact that semi normalized unconditionally basic sequences in a Hilbert space are Riesz bases. For a theorem quoting proof of what you want, note that that the lattice structure on $E$ is order complete by reflexivity, and hence you have a family $E_a$ of finite dimensional sub lattices of $E$ that are directed by inclusion and whose union is dense. On each $E_a$ put a Hilbert lattice (with respect to the lattice structure on $E_a$ that is inherited from $E$) norm $\|\cdot\|_a$ that dominates the original norm and is no more than $C$ times the original norm. $E$ isomorphically embeds as a sublattice into an ultraproduct of $(E_a,\|\cdot\|_a)$, which is a Hilbert lattice.<|endoftext|> -TITLE: Is ${\bf Set}$ the terminal autological topos -QUESTION [9 upvotes]: An autological topos is a type of topos defined by Mike Shulman in his paper on stack semantics; specifically, they are toposes satisfying an additional topos theoretic axiom schema expressed in their internal stack semantics which gives their internal logics the full strength of $ZF$ set theory. In his own words: - -On the other hand, the stack-semantics version of [the] separation [axiom] seems to be a new topos-theoretic property. We call a topos with this property autological. Autology is also closely connected to the stack semantics: a topos is autological if and only if its stack semantics is representable, in the same sense that the usual Kripke-Joyal semantics is always representable. (This motivates the name: a topos is autological if the logic of its stack semantics can be completely “internalized” by representing subobjects.) Autology is also quite common and well-behaved; for instance, we will show in §9 that all complete topoi are autological, and in in [Shub] we will study its preservation by other topos-theoretic constructions (including realizability). - -Since Grothendieck toposes are complete, all Grothendieck toposes are autological.* Let $\mathfrak{Groth}$ denote the category of Grothendieck toposes and geometric morphisms, and $\mathfrak{aut}$ the category of autological toposes and geometric morphisms. - -${\bf Set}$ is terminal in $\mathfrak{Groth}$ which is a subcategory of $\mathfrak{aut}$ by the above, but is ${\bf Set}$ terminal in $\mathfrak{aut}$? - -This would seem like a more satisfying characterization since autological categories are precisely the categories with the internal logical strength of $ZF$, so the category of sets in $ZF$ should be universal in some sense among these categories. -EDIT: As Qiaochu Yuan suggests in the comments, we may want morphisms other than geometric morphisms in $\mathfrak{aut}$ in order to correctly characterize ${\bf Set}$ as universally modeling $ZF$ internally. -As a bonus question: - -How does this relate to the fact that ${\bf Set}$ is the initial $ZF$-algebra? Is it some sort of dual statement? - -*This is proven in IZF set theory in Mike's paper. -**This question occurred to me while trying to answer this question. -Linked paper: arXiv:1004.3802 [math.CT] - -REPLY [6 votes]: The answer to the original question is no. Indeed, there are autological toposes that do not admit any geometric morphism to $\rm Set$, such as realizability toposes and filterquotients. (As pointed out by მამუკა ჯიბლაძე in the comments, this is also true of the topos of finite sets and that of sets below some inaccessible cardinal, which are also autological, but unlike the former realizability toposes and filterquotients contain NNOs, and unlike the latter they require no hypotheses beyond ZFC.) -Switching to logical functors doesn't help matters. There are already Grothendieck toposes that don't admit any logical functor from $\rm Set$ (indeed, any topos admitting a logical functor from $\rm Set$ must be Boolean, since a logical functor preserves the isomorphism $1+1\cong\Omega$), while I believe the trivial topos is terminal among any kind of topos with logical functors. -In general, I don't think one can expect to uniquely characterize $\rm Set$ using any elementary property such as autology (even combined with initiality or terminality). I don't know a watertight argument against this myself, but maybe some set theorist can come up with one. Note that the initial ZF-algebra, like the terminal Grothendieck topos, has a non-elementary property of set-indexed completeness.<|endoftext|> -TITLE: Number of d-Calabi-Yau partitions -QUESTION [14 upvotes]: This problem arises from algebraic geometry/representation theory, see https://arxiv.org/pdf/1409.0668.pdf (chapter 2). -We call a partition $p=[p_1,...,p_n]$ with $2 \leq p_1 \leq p_2 \leq ... \leq p_n$ d-Calabi-Yau (for $d \geq 1$) when $n-d-1=\sum\limits_{i=1}^{n}{\frac{1}{p_i}}$. -For example for $d=1$ there are 4 1-Calabi-Yau partitions: [2,2,2,2],[3,3,3],[2,4,4] and [2,3,6]. For $d=2$ there are 18, see example 2.15. (b) in https://arxiv.org/pdf/1409.0668.pdf . - -Question 1: Is there a nice formula for the $d$-Calabi-Yau partitions for a fixed $d$? - -It seems hard to calculate the number for $d \geq 3$ even with a computer since a term $p_i$ can be pretty big. - -Question 2: For a given $d$, what is the maximal value a term $p_i$ can have? - -For $d=1$ it is 6 and for $d=2$ it is 42. - -REPLY [6 votes]: Following up on @pbelmans's answer and researching an article mentioned in OEIS A007018, I believe your Question 2 was answered just under 100 years ago by David Curtiss (On Kellogg's Diophantine problem, Amer. Math. Monthly 29 (1922) 380-387).* -Curtiss confirms Kellogg's conjecture that the maximum $x_i$ in any $$\frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n} = 1$$ is given by the sequence $u_1 = 1$ and $u_{k+1} = u_k(u_k+1)$ which begins 1, 2, 6, 42, 1806, 3263442. On p386 he explains, "But the value $u_n$ is actually attained by giving to the $x$'s the values $u_k+1$, so that $u_n$ is the maximum of $f_{n-1}(x)$." (That last expression is something he defined to simplify the proof.) Indeed, -\begin{gather*} -\frac{1}{1+1} + \frac{1}{2+1} + \frac{1}{6} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1, \\ -\frac{1}{1+1} + \frac{1}{2+1} + \frac{1}{6+1} + \frac{1}{42} = \frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{42} = 1, \\ -\frac{1}{1+1} + \frac{1}{2+1} + \frac{1}{6+1} + \frac{1}{42+1} + \frac{1}{1806} = \frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{43} + \frac{1}{1806} = 1, \dots -\end{gather*} -* There's also a sci.math.research thread from 1996 where a claim that Curtiss's proof is wrong is retracted, but the proof does seem to be difficult to follow. Gerry Myerson provided a reference to a simpler 1995 proof by Izhboldin and Kurliandchik. - -I'm including a comment to @katago here so that answers to both questions are in one place. -Looking through MathSciNet, the most recent upper bound may be Browning & Elsholtz 2011, something like $$(1.264085...)^{(5/12)^{2^{n−1}}}$$ where apparently the 1.26 is a known constant that arises in these problems. There are also lower bounds in the literature. -So the short answer to Q1 is no, at least not yet.<|endoftext|> -TITLE: Boardman-Vogt resolution of the little 2-cubes operad -QUESTION [5 upvotes]: If $\mathbf{P}$ is a (coloured) operad, one can build a topological operad $W(\mathbf{P})$ called the $W$-construction or the Boardman-Vogt resolution of $\mathbf{P}$. Let me denote the resulting map of operads $\varepsilon: W(\mathbf{P}) \to \mathbf{P} $. -My question is: if $\mathbf{P} = \mathbf{E}_2$ the little 2-cubes (or 2-discs) operad, how does $W(\mathbf{E}_2)$ look like? And even more of my interest: if $A$ is a $\mathbf{P}$-algebra, how does the $W(\mathbf{E}_2)$-algebra $\varepsilon^* (A)$ look like? -Edit: More concretely, how should $\varepsilon^* (A)$ be understood as an $\mathbf{E}_2$-algebra "up to homotopy"? - -REPLY [4 votes]: In the more general setting of a symmetric monoidal category $\mathsf{M}$ and a general colored operad $\mathsf{O}$, the structure of an $\mathsf{O}$-algebra $X$ regarded as a $\mathsf{WO}$-algebra is described explicitly in the book Homotopical Quantum Field Theory (called HQFT below), Corollary 7.2.9. The arxiv version is here. When $\mathsf{M}$ is a suitable category of topological spaces $\mathsf{Top}$, this pullback algebra structure is as Bertram Arnold described in the comments above, where you first forget about the edge lengths. In the general context of $\mathsf{M}$, the interval $[0,1]$ is replaced by a commutative segment $\mathsf{J}$ (Definition 6.2.1 in HQFT), which provides a notion of homotopy and comes with a counit map to the monoidal unit of $\mathsf{M}$. In $\mathsf{Top}$, this counit is $[0,1] \to *$. -To understand Corollary 7.2.9, you will first need to read Theorem 7.2.1 in HQFT, which explicitly describes the structure of a $\mathsf{WO}$-algebra using the formalism of trees and commutative segment. That theorem is one way to formally say that a $\mathsf{WO}$-algebra is an $\mathsf{O}$-algebra up to homotopy. In HQFT Definition 6.3.1, the Boardman-Vogt construction $\mathsf{WO}$ is defined as an entrywise coend indexed by a parameter category built from trees. You can apply that definition and the results mentioned above to the little 2-cube operad or any other operad. A number of examples can be found in HQFT.<|endoftext|> -TITLE: Multivariable higher-order chain rule -QUESTION [5 upvotes]: I am trying to understand the chain rule under a change of variables. Given a function $f : \mathbb R^n \rightarrow \mathbb R$ and a change of variables $G : \mathbb R^m \rightarrow \mathbb R^n$, what is the derivative -$\partial^\alpha ( f \circ G )$ -where $\alpha$ is a multiindex in the variables $x_1,\dots,x_m$ of degree $k$? We assume all necessary derivatives exist. References to the literature would be helpful too. I haven't found this general case treated in my sources. - -REPLY [8 votes]: You’re looking for the multivariate version of the formula of Faa di Bruno. -Addendum: As the OP notes, the version in Wikipedia is not in sufficient generality, since it takes $f:\mathbb R\to\mathbb R$. For a version that allows $f$ to depend on more than one variable, see for example the article: - -A Multivariate Faa Di Bruno Formula with Applications, M. Constantine and T. H. Savits, Trans. AMS, 348, Number 2, February 1996.<|endoftext|> -TITLE: Hasse invariant and the Clifford algbera -QUESTION [6 upvotes]: Let -$$q = a_1 x_1^2 + \cdots + a_n x_n^2$$ -be a quadratic form over some $p$-adic field $\mathbb{Q}_p$. We thus have its Hasse invariant -$$\mathcal{h}(q) = \prod_{1 \leq i < j \leq n} (a_i,a_j)_p \in \{\pm 1\},$$ -where $(a_i,a_j)_p$ is the usual Hilbert symbol. -Let $\mathcal{C}(q)$ be the Clifford algebra associated to $q$. If $n$ is even, this is a central simple algebra; in fact, it is a tensor product of quaternion algebras. It thus gives a $2$-torsion element in the Brauer group $\mathbb{Q}/\mathbb{Z}$ of $\mathbb{Q}_p$, i.e. an element $\mathcal{h}'(q) \in \{\pm 1\}$. -Question: How is $\mathcal{h}'(q)$ related to $\mathcal{h}(q)$? We should be able to express $\mathcal{h}'(q)$ in terms of $\mathcal{h}(q)$ along with $n$ and the discriminant, but I don't quite understand all the constructions well enough to do this. - -REPLY [2 votes]: You can find some information about this in Lam's book "Introduction to quadratic forms over fields," particularly in the 3rd chapter. I'll give the answer in a "field agnostic" way, not specifically for $p$-adic fields. -In general, the Hasse invariant is a slight modification of either the Brauer class of the Clifford algebra (if the form has even dimension), or the Brauer class of the even part of the Clifford algebra (if the form has odd dimension). If $h'$ denotes the class in the Brauer group of the Clifford algebra, or even Clifford algebra in the odd case (in either case, a central simple algebra), the relationship looks like this (taken from Lam, prop 3.20, and depending on the residue of $n$ modulo $8$): -if $n \equiv_8 1, 2$ then $h = h'$, -if $n \equiv_8 3, 4$ then $h = h' + \delta$, -if $n \equiv_8 5, 6$ then $h = h' + \gamma$, -if $n \equiv_8 7, 8$ then $h = h' + \delta'$, -where $\gamma, \delta, \delta'$ are quaternion algebras as follows: $\gamma = (-1, -1)$ is Hamilton's quaternion algebra, $\delta = (-1, -d)$ and $\delta' = (-1, d)$, and $d$ is the discriminant of the form $q$.<|endoftext|> -TITLE: Computations in condensed mathematics II, page 43 -QUESTION [5 upvotes]: This is a follow up on my previous question for Lectures on Condensed Mathematics. I am reading ahead at page 43. But it is not directly clear to me from the results that: - -How do we know $\Bbb Z[[T]], \Bbb R, \Bbb Z_p$ are solid modules? - -How did we obtain that $\Bbb R^{L \blacksquare}=0?$ - -For the last equation 6.4 to holds, does $-\otimes^{L\blacksquare}-$ commute with filtered limits in each variable? - - -Some elaborations would help. -I'd also like to know what are the formal aspects of the deductions (which I suppose is the great part this new category). - -Basically all I know which are solid: are - -those of the form $(-)^{L\blacksquare}$, -Those which are local objects. Characterization 5.8. -$\prod_I \Bbb Z$, the cpt. projective generators. - -REPLY [5 votes]: Let me basically repeat what R. says in the comments. -For 1, $\mathbb{Z}[[T]]$ and $\mathbb{Z}_p$ are solid becuase $\mathbb{Z}$ is solid and solid abelian groups are closed under all limits and colimits. But $\mathbb{R}$ is not solid, see 2. -For 2, the first point is that $\mathbb{R}$ is pseudocoherent as a condensed abelian group, i.e. the following equivalent conditions are satisfied: - -$Ext^i(\mathbb{R},-)$ commutes with filtered colimits for all i; -$\mathbb{R}$ admits a projective resolution where the terms are compact projective condensed abelian groups. - -(The equivalence of 1 and 2 is valid in any abelian category generated by compact projectives. It's a good exercise in homological algebra if you'd like to try.) -The pseudocoherence of $\mathbb{R}$ follows from the short exact sequence $\mathbb{Z}\rightarrow\mathbb{R}\rightarrow\mathbb{R}/\mathbb{Z}$. Indeed, $\mathbb{Z}$ is clearly pseudo-coherent (it is projective and compact) and the Breen-Deligne resolution implies that any compact abelian group is psuedo-coherent. Equivalent condition 1) shows that pseudocoherent modules have the 2 out of 3 property in short exact sequences, so this implies that $\mathbb{R}$ is pseudo-coherent. -The second point is that for a pseudocoherent condensed abelian group $M$ the derived solidification of $M$ identifies with $\underline{RHom}(\underline{RHom}(M,\mathbb{Z}),\mathbb{Z})$. Indeed, using condition 2 one reduces to checking that when $M=\mathbb{Z}[S]$ for $S$ extr. disconnected we have that $\underline{RHom}(\mathbb{Z}[S],\mathbb{Z})$ lives in degree $0$ and $\underline{RHom}(-,\mathbb{Z})$ on it is the derived solidification of $\mathbb{Z}[S]$. But these we verified in producing the solid theory (they follow from Specker's theorem and the definition of solidification). -Thus it suffices to see that $\underline{RHom}(\mathbb{R},\mathbb{Z})=0$. But this was verified in the proof of the existence of the solid theory. It follows from the Breen-Deligne resolution and the fact that $\mathbb{R}$ is contractible (so its cohomology with $\mathbb{Z}$-coefficients vanishes). -For 3, the answer is yes (Edit: no, I misread and thought the question was about filtered colimits. See comments) . For ordinary tensor product this is clear, and it follows for derived tensor product because filtered colimits are exact so they have vanishing higher derived functors.<|endoftext|> -TITLE: Recognizing algebraic independence among Schur polynomials -QUESTION [13 upvotes]: Given a set of integer partitions $\{\lambda_1, \lambda_2,\dots \lambda_n\}$. Are there combinatorial criteria for deciding whether the associated Schur polynomials $s_{\lambda_1}, s_{\lambda_2},\dots s_{\lambda_n}$ are algebraically independent? -If this is too hard, are there any surprising necessary or sufficient criteria? - -REPLY [4 votes]: I'll add further details to my comment and make it an answer. This will give algebraic relations between certain collections of six Schur polynomials. Hence, for algebraic independence it will give a necessary condition of avoiding these six Schur polynomials together. This relation comes from Bijective proofs for Schur function identities which imply Dodgson's condensation formula and Plücker relations by Fulmek and Kleber (see Theorem 2 of this paper). I will change notation from the linked paper to agree with the notation in the question. Take $a_1 \geq a_2 \geq \cdots \geq a_{r+1}$ if we let -$$\lambda_1 = (a_1, a_2, \dots, a_r)$$ -$$\lambda_2 = (a_2, a_2, \dots, a_{r+1})$$ -$$\lambda_3 = (a_2, a_3, \dots, a_r)$$ -$$\lambda_4 = (a_1, a_2, \dots, a_{r+1})$$ -$$\lambda_3 = (a_2-1, a_3-1, \dots, a_{r+1}-1)$$ -$$\lambda_4 = (a_1+1, a_2+1, \dots, a_r+1)$$ -then -$$s_{\lambda_1}s_{\lambda_2} = s_{\lambda_3}s_{\lambda_4} + s_{\lambda_5}s_{\lambda_6}.$$ -Letting $\lambda_1 = \lambda_2 = \cdots = \lambda_{r+1} = c$ we find that $\lambda_1 = \lambda_2 = (c)^r$, $\lambda_3 = (c)^{r-1}$, $\lambda_4 = (c)^{r+1}$, $\lambda_5 = (c-1)^r$, and $\lambda_6 = (c+1)^r$ which are all rectangle shapes recovering a result of Kirillov (which is cited in the linked paper).<|endoftext|> -TITLE: What nodes of a graph should be vaccinated first? -QUESTION [23 upvotes]: Consider a graph, choose some "p: 0 -TITLE: What is known in general about the liquid transfer problem? -QUESTION [10 upvotes]: In several puzzle books, I have seen the following kind of a problem: there are several containers that can hold up to certain amounts of liquid (these liquids are assumed to be infinitely divisible). Given certain initial amounts, it is asked whether a certain other configuration (often equal division among some of the containers) is possible to achieve. There are variations, such as being able to throw out some of the liquid or being near a source such as a river (that is infinite for all intents and purposes). -Here is a specific example from The Chicken from Minsk by Chernyak and Rose: - -Besides chess playing and problem solving, drinking is and always has been the most common form of recreation in Russia. Vassily has acquired a $12$-liter bucket of vodka and wishes to share it with Pyotir. However, all Pyotir has is an empty $8$-liter bottle and an empty $5$-liter bottle. How can the vodka be divided evenly? - -The condition is that, at each step, some container must either be emptied or filled to the brim; we cannot measure more precisely. One can identify at least three general problems, given the finite list of container sizes and an initial configuration: - -Given another configuration, produce a condition or algorithm for deciding whether it is achievable. -If a configuration is achievable, determine a way of showing how to go from the initial configuration to this one. -Count how many distinct achievable configurations exist. - -What is known about this problem, perhaps in complexity or computability theory if not in number theory? While I have seen instances here and there, I am unaware of a name for the general problem and so am unable to begin searching for its related literature. One trivial observation is that, if the container sizes and initial amounts in each container are all multiples of an integer, then all amounts in all individual containers in all achievable configurations must be multiples of this integer; the contrapositive might be useful. - -REPLY [15 votes]: These are also known as 'decanting problems' or water pouring puzzles. There is a list of literature references in that Wikipedia article. -They're quite popular among puzzling aficionados, and it should come as no surprise that our sister site Puzzling Stack Exchange has a question about this topic: A general solution to the decanting problem? (aka jug-pouring, water-pouring). In particular, your question 1. can be answered by checking if the greatest common divisor of all bottles divides the target amount. The GCD of 8 and 5 is 1, so you can basically achieve all volumes which are a multiple of 1 (liter). One of the answers provides a partial algorithm to solve your question 2., and the other one a program to generate the necessary steps.<|endoftext|> -TITLE: Are functor categories with triangulated codomains themselves triangulated? -QUESTION [20 upvotes]: I'm fairly confident that the following assertion is true (but I will confess that I did not verify the octahedral axiom yet): - -Let $T$ be a triangulated category and $C$ any category (let's say small to avoid alarming my set theorist friends). Then, the category of functors $C \to T$ inherits a natural triangulated structure from T. - -By "natural" and "inherits" I mean that the shift map $[1]$ on our functor category sends each $F:C \to T$ to the functor $F[1]$ satisfying $F[1](c) = F(c)[1]$ on each object $c$ of $C$; and similarly, distinguished triangles of functors -$$F \to G \to H \to F[1]$$ -are precisely the ones for which over each object $c$ of $C$ we have a distinguished triangle in $T$ of the form $$F(c) \to G(c) \to H(c) \to F[1](c).$$ -The main question is whether this has been written up in some standard book or paper (I couldn't find it in Gelfand-Manin for instance). Perhaps it is considered too obvious and relegated to an elementary exercise. Mostly, I am interested in inheriting t-structures and hearts from $T$ to functor categories $C \to T$, and would appreciate any available reference which deals with such matters. - -REPLY [13 votes]: I believe I have a simpler counterexample, which I learned from Paul Balmer's course on tensor-triangular geometry last spring: -Claim The arrow category $\mathcal{T}^{\bullet \to \bullet}$ of a triangulated category $\mathcal{T}$ never has any triangulated structure unless $\mathcal{T} = 0$. Actually, we don't even need $\mathcal{T}$ to be triangulated here: if $\mathcal{T}$ is any additive category such that $\mathcal{T}^{\bullet \to \bullet}$ is triangulated, then $\mathcal{T} = 0$. -Proof: Suppose $\mathcal{T}$ is an additive category such that $\mathcal{T}^{\bullet \to \bullet}$ is triangulated. Let $a$ be an arbitrary object in $\mathcal{T}$, with identity morphism $1_a : a \to a$. Let $t$ denote the unique morphism $a \to 0$. Then -$\require{AMScd}$ -\begin{CD} -a @>1_a>> a\\ -@V 1_a V V @VV t V\\ -a @>>t> 0 -\end{CD} -defines a morphism $\alpha : 1_a \to t$ in $\mathcal{T}^{\bullet \to \bullet}$. Note that $\alpha$ is an epimorphism. In any triangulated category, all epimorphisms are split, so let $\beta : t \to 1_a$ be a splitting of $\alpha$ (that is, $\alpha \circ \beta$ is the identity morphism of $t$). Then $\beta$ is a commutative diagram -\begin{CD} -a @>t>> 0\\ -@V f V V @VVs V\\ -a @>>1_a> a -\end{CD} -such that $1_a \circ f = 1_a$ (and $t \circ s = 1_0$). From this and the commutativity of the diagram, we see that $1_a = 1_a \circ f = s \circ t$ factors through $0$. Thus, $a = 0$. Since $a$ was arbitrary, $\mathcal{T} = 0$. -Edit: Of course we could make the statement even weaker: we only really needed that $\mathcal{T}$ has a zero object. But if $\mathcal{T}^{\bullet \to \bullet}$ is triangulated, then $\mathcal{T}$ must be additive, because it embeds as an additive subcategory of $\mathcal{T}^{\bullet \to \bullet}$ via $a \mapsto 1_a$.<|endoftext|> -TITLE: Find a combination of convex function so that it is positive -QUESTION [5 upvotes]: A student in my class asked me the following question, I did know what tools will be needed to attack it. But I found it is an interesting question. -Let $f_1,f_2$ be two convex functions on $[0,1]$ and let $\varepsilon>0$. For every $x\in[0,1]$，we have $\max\{f_1(x),f_2(x)\}>\varepsilon$. Does there exist a $\lambda\in [0,1]$ so that -$$ -\lambda f_1(x)+(1-\lambda)f_2(x)>0, \quad \forall x\in [0,1]? -$$ -Any hints are welcome! Thanks! - -REPLY [2 votes]: Here is the easiest explanation, two intersecting straight lines satisfy this property, and two intersecting straight lines are the worst-case among all convex functions when fixing the intersection point of the two functions. -And the following proof is an effort to make the above intuition rigorous and mathematical readable. - -The following argument only available for convex function with reasonable regularity -First, as point out by Loic Teyssier, we need to consider the case independently: when $f_1,f_2$ do not intersection. And when $f_{1}, f_{2}$ is not intersect and then we can take $\lambda=0$ or 1 to make $\lambda f_{1}(x)+(1-\lambda) f_{2}(x)=\max \left\{f_{1}, f_{2}\right\}$. -Second, as point out by Dieter Kadelka, we need to consider the endpoint $\{0,1\}$ separate because the prove of N-L formula relay the point $a,b$ is in some interval, in the following argument. This can be down use the convex property to gain a lower bound on $f_1(0),f_1(1),f_2(0),f_2(1)$. -Then, this is true and in fact, we can determine $\lambda$. We need the following 2 lemmas. -In the case $\exists \quad x_{0} \in[0,1] . \quad f_{1}\left(x_{0}\right)=f_{2}\left(x_{0}\right)>0$ - -$f$ is a continue couvex function on [0.1], then $f$ is lipschitz function. - - -$f$ is a lipschitz function. then $f^{\prime}(x) \quad$ exist a.e. in $[0,1]$ and $\forall \quad 0\leq a< b\leq 1$, -$$f(b)-f(a)=\int_{a}^{b} f^{\prime}(x) d x$$ -We can understand the previous N-L formula in distribution to avoid unnecessary trouble（The derivative does not exist, the left and right derivatives are not equal, and the derivative is zero but the function is a strictly monotonic function） - -then $\forall x \in[0.1]$, -$\begin{aligned} \lambda f_{1}(x)+(1-\lambda) f_{2}(x) &=\lambda\left(f_{1}\left(x_{0}\right)+\int_{x_{0}}^{x} f_{1}^{\prime}(t) d t\right)+(1-\lambda)\left(f_{2}\left(x_{0}\right)+\int_{x_{0}}^{x} f_{2}(t) d t\right) \\ &=\lambda f_{1}\left(x_{0}\right)+(1-\lambda) f_{2}\left(x_{0}\right)+\lambda \int_{x_{0}}^{x} f_{1}^{\prime}(t) d_{t}+(1-\lambda) \int_{x_{0}}^{x} f_{2}^{\prime}(t) d t \end{aligned}$ -Now $\quad$ WLOG $\quad$ we assame -$f_{1}(x)>0$ When $0 \leqslant x \leqslant x_{0}$, $f_{2}(x)>0 $ when $ x_{1} \leqslant x \leqslant 1$, and assume $f_{1}^{\prime}\left(x_{0}\right)>0. f_{2}^{\prime}\left(x_{0}\right)<0$. -When $ 1 \geqslant x>x_{0}$. Then, -$$ -\begin{array}{l} -\int_{x_{0}}^{x} f_{1}^{\prime}(t) d t \geqslant\left|x-x_{0}\right| \cdot f_{1}^{\prime}\left(x_{0}\right) \\ -\int_{x_{0}}^{x} f_{2}^{\prime}(t) d t \geqslant\left|x-x_{0}\right| \cdot f_{2}^{\prime}\left(x_{0}\right) -\end{array} -$$ -When $0 \leqslant x -TITLE: Which random variables can be written as the difference of two independent positive random variables? -QUESTION [6 upvotes]: Can we characterize random variables $X$ that satisfy -$$ -X\sim Y - Z -$$ -for two independent positive random variables $Y$ and $Z$? -Are $Y$ and $Z$ unique in some sense? -Can (one possible choice of) $Y$ and $Z$ be constructed (e.g. formulas for probability density or characteristic function, or sampling algorithms) when they exist? - -Possibly simpler question: -Which random variables $X$ satisfy -$$ -X\sim Y_1-Y_2 -$$ -for i.i.d. positive random variables $Y_1\sim Y_2\sim Y$? Since the characteristic function satisfies -$$ -\phi_X = \phi_{Y}\overline{\phi_{Y}} -$$ -we must have $\phi_{X}\geq 0$ -- is that sufficient? Is $Y$ unique in some sense? Can it be constructed? -For example, Laplace random variables satisfy $\phi_{X} = (1+x^2)^{-1}=(1+ix)^{-1}\overline{(1+ix)^{-1}}=\phi_{Y}\overline{\phi_{Y}}$ where $Y$ is exponential. Exponentials are positive of course, so we got lucky with this particular decomposition and can write $X=Y_1-Y_2$ as desired. Had we picked $(1+x^2)^{-1}=(1+x^2)^{-1/2}\overline{(1+x^2)^{-1/2}}$ this wouldn't have worked out. -This approach slightly generalizes to $\phi_{X}$ that are rational in $x^2$, but not at all (at least not obviously to me) to only slightly different examples like Linnik random variables, where $\phi_{X} = (1+|x|^{\alpha})^{-1}$, or to limits such as normal random variables, where $\phi_{X}=e^{-\sigma^2x^2}$. - -The only result I found that goes remotely in this direction was a theorem by Boas and Kac that positive definite functions with compact support have a convolution square root with half-length compact support. This has a support flavor, but a different one than I'm looking for. - -REPLY [3 votes]: As soon as $\phi(x)$ decays too rapidly, you are doomed. For example, if $\phi(x)=e^{-x^2}$, then $\psi(x)=Ee^{itY}$ would have to satisfy $|\psi(x)|=e^{-x^2/2}$, but now the rapid decay will make the distribution of $Y$ absolutely continuous with holomorphic density $f_Y$ (this is well known and easily proved since $f_Y(z)=\int_{-\infty}^{\infty}\psi(t) e^{-itz}\, dt$ converges for all $z\in\mathbb C$), so it's not possible to have $f_Y(y)=0$ for $y<0$. -To state this more formally, this argument shows that if $|\phi_X(x)|\lesssim e^{-c|x|}$ for some $c>0$ and $X= Y_1-Y_2$, with $Y_j$ independent and $Y_1\sim Y_2$, then the distribution of $Y_j$ is of the form $d\mu(y)= f(y)\, dy$ with a real analytic $f$. In particular $P(Y\in A)>0$ for any $A\subseteq\mathbb R$ of positive Lebesgue measure.<|endoftext|> -TITLE: Infinity-categorical analogue of compact Hausdorff -QUESTION [25 upvotes]: Recently I became through this mathoverflow question aware of the article Codensity and the ultrafilter monad by Tom Leinster. There he shows that the ultrafilter monad on the category $\mathrm{Set}$ arises from the adjunction -$$ \mathrm{Set} \rightleftarrows \mathrm{Fun}(\mathrm{FinSet}, \mathrm{Set})^{\mathrm{op}},$$ -where the left adjoint is given by the coYoneda-embedding (that it has a right adjoint follows either by a construction or the adjoint functor theorem). Moreover it is known that the category of compact Hausdorff spaces is monadic over $\mathrm{Set}$ and that the corresponding monad is the ultrafilter monad as well, exhibiting the category of compact Hausdorff spaces as algebras over this monad. -Moving to $\infty$-categories, it is natural to replace $\mathrm{Set}$ by the $\infty$-category $\mathcal{S}$ of spaces (or animae, as some call it). This has the sub-$\infty$-category $\mathcal{S}^{\mathrm{fin}}$ of finite spaces (i.e. the smallest finitely cocomplete subcategory containing the point). Using the coYoneda embedding and the adjoint functor theorem, we obtain again an adjunction -$$\mathcal{S}\rightleftarrows \mathrm{Fun}(\mathcal{S}^{\mathrm{fin}}, \mathcal{S})^{\mathrm{op}}.$$ -Can one describe the resulting monad and algebras over it? Is it a known $\infty$-category? Moreover, one might ask about its relation to to other $\infty$-categories, like profinite spaces or condensed spaces. -Edit: As Denis and Dustin pointed out, it is much more natural to replace $\mathrm{FinSet}$ by the $\infty$-category of $\pi$-finite spaces (instead of $\mathcal{S}^{\mathrm{fin}}$), i.e. spaces whose homotopy groups are concentrated in finitely many degrees and are finite there. - -REPLY [12 votes]: That's a good question! I think Barwick and Haine have thought much more about this, and maybe they already know the answer? What I say below is definitely known to them. Also beware that I've written the below in a stream of consciousness, not quite knowing where it will go when I started. -I'll write "anima" for what is variously called homotopy types/spaces/$\infty$-groupoids/..., and denote their $\infty$-category $\mathrm{An}$($=\mathcal S$). We can also consider the $\infty$-category $\mathrm{CondAn}=\mathrm{Cond}(\mathrm{An})$ of condensed anima (this is, by the way, also the animation of the category of condensed sets). If $X\in \mathrm{CondAn}$ is a condensed anima, then $\pi_0 X$ is a condensed set, and for any point $x\in X$, one can define homotopy groups $\pi_i(X,x)$ for $i\geq 1$, which are condensed groups (abelian for $i\geq 2$). Slightly more generally, if $S$ is any profinite set and $g: S\to X$ is any map, one can define a group object $\pi_i(X,g)\to S$ in condensed sets over $S$, whose fibre over any $s\in S$ is $\pi_i(X,g(s))$. Then a map of condensed anima is an equivalence if and only if it induces an equivalence on $\pi_0$ and all $\pi_i$ for $i\geq 1$ (at all base points, including profinite families of basepoints). -So, just like in a very very crude approximation an anima $X$ is something like the collection $\pi_0 X,\pi_1 X,\pi_2 X,\ldots$ of a set, a group, and abelian groups, a condensed anima is something like a collection of a condensed set, a condensed group, and condensed abelian groups. In particular, already $\pi_0 X$ can be an interesting topological space like a manifold, so a space. This is why we do not say "condensed space", as then it would seem like forgetting to condensed sets should forget the "space" structure, but it rather forgets "abstract homotopy" structure. -Now the following seems like the obvious "$\infty$-categorical compact Hausdorff spaces": -Definition. A condensed anima $X$ is "compact Hausdorff" if $\pi_0 X$ and all $\pi_i X$ for $i\geq 1$ are compact Hausdorff. -Recall here that compact Hausdorff spaces embed fully faithfully into condensed sets. The second statement means more precisely that for all profinite sets $S$ with a map $g: S\to X$, the group object $\pi_i(X,g)\to S$ in condensed sets over $S$ is compact Hausdorff. (This is a little stronger than only asking it at all fibres.) -So in this case $\pi_0 X$ is a compact Hausdorff space, $\pi_1 X$ is a compact Hausdorff group, and $\pi_2 X,...$ are compact Hausdorff abelian groups. -It turns out that there is a nice characterization of "compact Hausdorff" condensed anima. In fact, there is a general topos-theoretic notion of "coherent"="qcqs" objects. This is usually studied for $1$-topoi, but it generalizes easily to $n$-topoi. Basically, an object is quasicompact if any cover admits a finite subcover; it is quasiseparated if the diagonal is quasicompact; it is 2-quasiseparated if the diagonal is quasiseparated; etc.; and coherent = quasicompact and $n$-quasiseparated for all $n\geq 1$. Then coherent condensed sets are exactly compact Hausdorff spaces, and: -Proposition. Coherent condensed anima are exactly the "compact Hausdorff" condensed anima. -Note: In a $1$-topos, coherent objects often agree with the finitely presented objects, but this fails dramatically for $\infty$-topoi, where coherence and finite presentation are two quite different finiteness conditions. In the case of anima, coherence means finite homotopy groups, while finite presentation should mean generated under finite colimits from the point; these are very different notions. As already discussed in the comments, the "finite homotopy groups" condition seems more relevant for the question. -Now we have a good notion of "$\infty$-categorical compact Hausdorff spaces". The question however started from a different angle, namely as trying to describe it via a monad on anima. The good news is: -Proposition. Compact Hausdorff condensed anima are monadic over anima. -This can be deduced from Barr-Beck-Lurie, although it takes some work. -It remains to understand the monad (and see whether it can be described as a codensity monad). The monad takes an anima $X$ to $\lim_{X\to Y} Y$ where the diagram is over all maps from $X$ to a compact Hausdorff condensed anima $Y$: This computes the desired left adjoint. Assume for the moment that the diagram category was small; then this limit is still a compact Hausdorff condensed anima: The compact Hausdorff condensed anima are stable under all small limits, as they are stable under finite limits and all small products. Now the diagram category is not actually small, so one has to argue slightly more carefully to see the existence of the left adjoint. -If $X$ is actually a set, then one can show that the left adjoint is still the same as usual, given by the Stone-Čech compactification. This is the same as $\lim_{X\to Y} Y$ where we restrict $Y$ to be a finite set. Ultimately, the possibility to restrict $Y$ to finite sets here -- coming from the fact that the Stone-Čech compactification is totally disconnected, and totally disconnected compact Hausdorff spaces are pro-finite -- is what makes it possible to describe compact Hausdorff spaces in terms of the codensity monad for $\mathrm{FinSet}\hookrightarrow \mathrm{Set}$. -The first interesting new case is $X=K(G,1)$, for some discrete group $G$. Ignoring higher homotopy groups, we are then interested in the universal compact group $H$ with a map $G\to H$. In general, this is known as the "Bohr compactification" of $G$. If $G=\mathbb Z$, then we look for the free compact group on one generator. This is necessarily abelian, and then one can use Pontrjagin duality to actually determine this (I hope I didn't screw this up): Take $\prod_{\mathbb R/\mathbb Z}\mathbb R/\mathbb Z$, the product of $\mathbb R/\mathbb Z$ (as a discrete set) many copies of the circle $\mathbb R/\mathbb Z$, with its tautological "diagonal" element, and take the closed subgroup generated by this element. -What we see from the example is that already for the anima $X=K(\mathbb Z,1)$ (aka the circle), the monad takes an extremely complicated value (note that we were ignoring higher homotopy groups, but the computation of $\pi_1$ is correct), that in particular is not itself totally disconnected, and so cannot be written as a limit of finite anima. So I gather that these "$\infty$-categorical compact Hausdorff spaces" cannot be described in the way the question started. -This, then again, begs the question what algebras for the monad in the question are! -Well, I don't know the precise answer, but one can also consider "totally disconnected compact Hausdorff" condensed anima, asking now that all $\pi_i X$ are totaly disconnected compact Hausdorff. So $\pi_0 X$ is a profinite set, $\pi_1 X$ is a profinite group, and $\pi_2 X,\ldots$ are profinite abelian groups. -Proposition. "Totally disconnected compact Hausdorff condensed $n$-truncated anima" are equivalent to the Pro-category of $n$-truncated anima with finite homotopy groups. -One can also pass to the limit $n\to \infty$ in some sense, but has to be careful as this does not exactly commute with passage to Pro-categories. It is still true that any totally disconnected compact Hausdorff condensed anima $X$ maps isomorphically to the $\lim_{X\to Y} Y$ where $Y$ runs over anima with finite homotopy groups. -Now totally disconnected compact Hausdorff condensed anima are not monadic anymore over anima, but the forgetful functor still detects isomorphisms, and has a left adjoint, so gives rise to a monad on anima, and totally disconnected compact Hausdorff condensed anima embed fully faithfully into algebras over this monad. And this monad, by the last paragraph, can be identified with the codensity monad for the inclusion $\mathrm{An}^{\mathrm{coh}}\hookrightarrow \mathrm{An}$ of coherent anima (=anima with finite homotopy groups) into all anima. -So, if I'm not screwing this up, then the category of algebras over this monad is some kind of hull of totally disconnected compact Hausdorff condensed anima (including all geometric realizations that are split on underlying anima); this hull is contained in compact Hausdorff condensed anima. -In summary, if one takes "finite anima" in the question to mean "finite homotopy groups", then this gives rise to a monad whose algebras lie somewhere between totally disconnected compact Hausdorff condensed anima, and all compact Hausdorff condensed anima. I think they definitely include all those for which $\pi_0 X$ is arbitrary compact Hausdorff, but $\pi_i X$ for $i\geq 1$ is totally disconnected. -Hmm... OK, let me make the following: -Conjecture: Algebras over the codensity monad for $\mathrm{An}^{\mathrm{coh}}\hookrightarrow \mathrm{An}$ are exactly those compact Hausdorff condensed anima $X$ for which all $\pi_i X$ for $i\geq 1$ are totally disconnected. -I'm willing to conjecture this for the following reason: while one can obtain all compact Hausdorff spaces as quotients of profinite sets by closed equivalence relations, nothing like this happens for groups: a quotient of a profinite group by a closed equivalence relation is still a profinite group.<|endoftext|> -TITLE: Constant curvature metrics on complements of codimension 2 submanifolds -QUESTION [6 upvotes]: I apologize in advance for my lack of knowledge with respect to geometry and the literature - any pointers are welcome. Let $M^n$ be a closed smooth $n$-dimensional manifold. Call $M$ good if for every $c \in \mathbb{R}$ there exists a codimension 2 closed submanifold $X^{n-2}$ so that $M-X$ admits metrics $g$ of constant scalar curvature $c$. What are some necessary/sufficient conditions for $M$ to be good? -In dimension 3, $S^3$ admits such metrics for $c > 0$ with $X$ empty. Taking $X$ to be the unknot and removing it we have $\mathbb{R}^2 \times S^1$ which admits a flat metric, and for $c<0$, removing a point yields $\mathbb{R}^3$ and so any nonempty $X$ does the trick. By the Lickorish–Wallace theorem, this then extends to all such 3-manifolds. So all such 3-manifolds are good. -In dimension 2, by taking $X$ to be empty, 2 points, or a single point, depending on if $c$ is positive, zero, or negative, we see that $S^2$ is good. Positive genus surfaces presumably can not be good by some variation of Gauss-Bonnet. -What about in dimension 4? In dimension $n$? I would also be interested in hearing about the variation of the problem with constant sectional curvature metrics. -Edit: -Michael Albanese completely answers my original version of the question below. But I am also interested in the case where all of the metrics are complete. - -REPLY [7 votes]: Theorem 1.4 of Kazdan & Warner's Scalar Curvature and Conformal Deformation of Riemannian Structure states the following: - -Let $M$ be a noncompact manifold of dimension $\geq 3$ diffeomorphic to an open submanifold of some compact manifold $M_1$. Then every $K \in C^{\infty}(M)$ is the scalar curvature of some Riemannian metric on $M$. - -So for any closed smooth manifold $M$ of dimension at least three, and any closed smooth codimension two submanifold $X$, every smooth function arises as the scalar curvature of some metric on $M - X$. In particular, $M$ is good.<|endoftext|> -TITLE: What are the "simplest" polytopes with an automorphism group of $\mathrm M_{11} \hspace {-1.25pt} $? -QUESTION [6 upvotes]: Do any polytopes have an automorphism group of the smallest of the sporadic groups, the Matthieu group $\mathrm M_{11} \hspace {-1pt} $? Indeed, they must exist. What are the simplest such polytopes we can construct, where "simplest" can be interpreted in a number of equally valid ways such as possessing: - -The lowest dimensionality, -The fewest vertices and/or facets, -And/or particularly elegant faces and/or topological structure (the abstract polytope it generates)? - - -This exercise was inspired by a comment made by @Rudi_Birnbaum: - -I have found an interesting quote: "The Mathieu groups M11 and M22 are not automorphism groups of any polytopes.". Is this in contradiction to what has been written above? - -@S.Carnahan replied: - -@Rudi_Birnbaum I cannot read the article because of the paywall. However, given the name of the article, I suspect the author is restricting his view to regular polytopes, and didn't bother to include the word "regular" in that particular sentence. - -I then added: - -@Rudi_Birnbaum I couldn't find that sentence in the linked article, but it does appear in this one. S. Carnahan is correct that the author is discussing regular polytopes, but they are also "abstract", which refers to a very different concept. Of course, many important lattices such as the Leech lattice include $\mathrm M_{11}$ as a subgroup of their automorphism group. - - -In the answer, I will explain what the modifiers "regular" and "abstract" mean in this context. I will then fully characterize a specific non-regular, geometric (as opposed to abstract) polytope with $66$ vertices in $\mathbb Z^{11} \! $ which $\color {red} {\textbf{does not}}$ have an automorphism group of precisely $\mathrm M_{11} \hspace{-1pt} $. @M.Winter has shown that my solution is, in fact, incorrect. - -REPLY [4 votes]: $\color {red} {\textbf {WARNING}}$ -I believe @M.Winter has shown in the comments that the following answer is, in fact, $\color {red} {\textbf {incorrect}}$. It would appear that I have merely constructed a rectified $11$-simplex. I was misled by Magma's AutomorphismGroup(P) : TorPol -> GrpMat function, which only finds subgroups of $\mathrm {GL}_d(\mathbb Z)$. However, they also provided references showing that the feat is indeed possible. I will begin working on providing a correct answer tomorrow. -In the meantime, I invite others to try their hand at the problem. In general, simple polytopes, should be preferred, where "simple" can be interpreted in a number of equally valid ways. I provided some examples in the question. -Unless there are objections, I will leave this answer up, as I think it still contains some valuable information and valid ideas for how to go about fixing it. - -Regular and/or Abstract Polytopes -Regular polytopes possess the greatest degree of symmetry a polytope can have. Their automorphism group acts transitively on its faces of any dimension. This includes their vertices, edges, $\ldots$, and facets. The actual definition is even stricter than this, requiring that the automorphism group act transitively on its flags, but I won't get into that level of detail. -In our geometric setting, we are only concerned with real, Euclidean, convex regular polytopes. This rules out self-intersecting star polytopes, complex polytopes, tesselations of the hyperbolic plane, and infinite apeirotopes. Alongside regularity, these requirements are highly restrictive: - -In two dimensions, there are an infinite number of regular polygons. -In three dimensions, the regular polytopes are more commonly known as the five Platonic solids. -There are six four-dimensional regular polytopes, up to similarity. - - -In all higher dimensions $d > 4$, there are only three regular polytopes, up to similarity: - -The self-dual $d$-simplex, -The $d$-cube/measure polytope, -And its dual, the $d$-orthoplex/cross polytope. - - - -There is no regular geometric polytope with an automorphism group of $\mathrm M_{11} \hspace {-1pt} $. -Abstract polytopes can be generated from the more familiar geometric polytopes by "forgetting" the geometry of the polytope and retaining only the topological information regarding how its faces of various dimensions are connected to each other. One can also construct valid abstract polytopes that can not be "realized" as geometric polytopes such as the $11$-cell and the $57$-cell. -By modifying the definition slightly, the concept of regularity can be extended to abstract polytopes, and there are far more regular abstract polytopes than there are regular geometric polytopes. Hence, one can construct regular abstract polytopes with automorphism groups which do not apply to any of the regular geometric polytopes. -It is important to note that the automorphism groups of a geometric polytope and the abstract polytope generated from it need not be isomorphic. By forgetting the geometric information, the latter can become larger than the former. The former should always be a subgroup of the latter. -If my tentative understanding of the papers linked in the question and some of their references is correct, there are abstract polytopes with an automorphism group of $\mathrm M_{11} \hspace {-1pt}$, but none of them are regular. I will proceed to construct a non-regular geometric polytope with an automorphism group of $\mathrm M_{11} \hspace {-1pt} $. - -Construction of an $\mathrm M_{11} \! $ Polytope -$\mathrm M_{11} \! $ has irreducible $\mathbb Q$-linear representations of degrees $1$, $10$, $11$, $20$, $32$, $44$, $45$, and $55$. -The degree-$11$ representation, corresponding to the character $\chi_5 \hspace {-0.5pt} $, is particularly appealing because it's absolutely irreducible over the rationals, and its invariant form is just the identity matrix $I_{11} \hspace {-1pt} $. Thus, the lattice it generates is simply $\mathbb Z^{11} \! $. For these reasons, I will use it below. -It is worth noting, however, that the degree-$32$ representation, which corresponds to $\chi_6 + \chi_7 \hspace {-0.5pt} $ and is not absolutely irreducible over the rationals, generates a lattice with an automorphism group of $\mathbb Z_2 \! \times \mathrm M_{11} \hspace {-1pt} $. This is, I think, as close to an automorphism group of $\mathrm M_{11} \hspace {-1pt}$ as can be achieved, as lattices, by definition, must have inversion symmetry. -I begin by enumerating the orbits of the short vectors of $\mathbb Z^{11} \! $ under the degree-$11$ representation of $\mathrm M_{11} \hspace {-0.5pt} $: - -The $22$ norm-$1$ unit vectors fall into a single orbit. This is merely the $11$-orthoplex. -The $220$ vectors of (squared) norm $2$ fall into a single orbit. These can't have an automorphism group of just $\mathrm M_{11} \hspace {-1pt} $. -The $1320$ norm-$3$ vectors fall into two orbits, $O_1 \! \hspace {1pt} $ and $O_{-1}$, both of of size $660$. Each one is non-antipodal, in the sense that $v \in O_i \! \implies \! {-v} \notin O_i$. However, they form a pair, i.e. $v \in O_i \! \implies \! {-v} \in O_{-i}$. The representation doesn't act primitively on them, and it seems doubtful that such a large set of vectors could have an automorphism group of just $\mathrm M_{11} \hspace {-1pt} $. -The orbits of $5302$ vectors of norm $4$ include a scaled up orthoplex from norm $1$, two pairs of non-antipodal orbits of sizes $330$ and $1320$, and one antipodal orbit of size $1980$. The same doubtful comment from norm $3$ applies here as well. -There are $15\,224$ vectors of norm $5$. They include a pair of non-antipodal of orbits of size $66$. Notably, the representation acts primitively on these vectors, corresponding to the $S_5$ maximal subgroup of $\mathrm M_{11} \hspace {-1pt} $. We can choose one arbitrarily and proceed. - -(At norms $8$ and $11$, one finds two more pairs of orbits with primitive actions of sizes $165$ and $12$, respectively. The latter is just a simplex, but perhaps the former could be used to construct a polytope which actually has an automorphism group of $\mathrm M_{11} \hspace {-1pt} $.) -At this point, I noticed something quite astonishing: The union of these two orbits are precisely the weight-$5$ code words of the unextended ternary Golay code, where the identity element of $\mathbb F_3$ is mapped to $0$, one non-identity element to $+1$, and the other to $-1$. So we can just take a non-antipodal coset of these code words, such that they are invariant under the action of our degree-$11$ representation of $\mathrm M_{11} \hspace {-1pt} $, as the vertices of our polytope. Explicitly, they can be given by the cyclic permutations of the following six vectors: - - - - -$\pmb {v_1}$ -$\pmb {v_2}$ -$\pmb {v_3}$ -$\pmb {v_4}$ -$\pmb {v_5}$ -$\pmb {v_6}$ -$\pmb {v_7}$ -$\pmb {v_8}$ -$\pmb {v_9}$ -$\pmb {v_{10}}$ -$\pmb {v_{11}}$ -$\pmb {\sum v_i}$ - - - - -$+1$ -$+1$ -$+1$ -$\hphantom + 0$ -$+1$ -$\hphantom + 0$ -$\hphantom + 0$ -$+1$ -$\hphantom + 0$ -$\hphantom + 0$ -$\hphantom + 0$ -$+5$ - - -$+1$ -$\hphantom + 0$ -$-1$ -$+1$ -$-1$ -$-1$ -$\hphantom + 0$ -$\hphantom + 0$ -$\hphantom + 0$ -$\hphantom + 0$ -$\hphantom + 0$ -$-1$ - - -$-1$ -$-1$ -$+1$ -$\hphantom + 0$ -$\hphantom + 0$ -$-1$ -$+1$ -$\hphantom + 0$ -$\hphantom + 0$ -$\hphantom + 0$ -$\hphantom + 0$ -$-1$ - - -$+1$ -$-1$ -$\hphantom + 0$ -$-1$ -$\hphantom + 0$ -$+1$ -$\hphantom + 0$ -$-1$ -$\hphantom + 0$ -$\hphantom + 0$ -$\hphantom + 0$ -$-1$ - - -$-1$ -$\hphantom + 0$ -$\hphantom + 0$ -$-1$ -$-1$ -$\hphantom + 0$ -$+1$ -$+1$ -$\hphantom + 0$ -$\hphantom + 0$ -$\hphantom + 0$ -$-1$ - - -$+1$ -$\hphantom + 0$ -$+1$ -$-1$ -$\hphantom + 0$ -$\hphantom + 0$ -$-1$ -$\hphantom + 0$ -$-1$ -$\hphantom + 0$ -$\hphantom + 0$ -$-1$ - - - - -Magma manages to construct the polytope without too much effort, and after a few days of computation, $\color {red} {\textbf {fails to}}$ confirm that its automorphism group is indeed $\mathrm M_{11} \hspace {-1pt} $. - -Polytope Details -The polytope has a volume of -$$ \frac {9 \cdot 509} {32 \cdot 25 \cdot 7 \cdot 11} $$ -Remarkably, this fills only -$$ \frac {3^5 \cdot 509} {2^{11} \cdot 5^{13/2} \, \pi^5} \approx 5.649 \cdot 10^{-6} $$ -of the volume of the circumscribed ball $B^{11} \! \left ( \sqrt 5 \right ) $. -Its face structure is: - - - - -$\pmb d~~$ -$\textbf{Simplices}~~$ -$~~~\textbf{Volume}\\~~\textbf{(per face)}$ -$\hspace{0.5pt}\textbf{Rectified}~~\\\textbf{Simplices}~~$ -$~~~\textbf{Volume}\\~~\textbf{(per face)}$ - - - - -$0~~$ -$~~~~~~~~\hspace{3pt}66~~\\\text {vertices}~~$ - - - - - -$1~~$ -$~~\hspace{4pt}660~~\\\text {edges}~~$ -$~~\,\text {(length)}\\~~~~~\,\sqrt 6$ - - - - -$2~~$ -$~~~~~~~2200~~\\\text {triangles}$ -$~~\hspace{2pt}\text {(area)}\\~~\hspace{1pt}3 \sqrt 3/2$ - - - - -$3~~$ -$~~~~~~~~~\hspace{3pt}3960~~\\\text {tetrahedra}~~$ -$~\sqrt 3$ -$~~~~~~~~~~~495~~\\\text {octahedra}~~$ -$~~~~~\,4 \sqrt 3$ - - -$4~~$ -$5544~~$ -$~~~3 \sqrt 5/8$ -$792~~$ -$~~~~~~~~33 \sqrt 5/8$ - - -$5~~$ -$5544~~$ -$~~9 \sqrt 2/40$ -$924~~$ -$~~~~~117 \sqrt 2/20$ - - -$6~~$ -$3960~~$ -$~~3 \sqrt 7/80$ -$792~~$ -$~~~~~171 \sqrt 7/80$ - - -$7~~$ -$1980~~$ -$~3 \sqrt 6/280$ -$495~~$ -$~~~~~~~~~~\,9 \sqrt 6/7$ - - -$8~~$ -$660~~$ -$~~~27/4480$ -$220~~$ -$~~~~\,6669/4480$ - - -$9~~$ -$132~~$ -$\sqrt {30}/4480$ -$66~~$ -$251 \sqrt {30}/2240$ - - -$~~~~~~~~~10~~\\\text {(facets)}~~$ -$12~~$ -$~~\displaystyle \frac {3 \sqrt {11}} {44\,800} $ -$12~~$ -$~~~\displaystyle \frac {3039 \sqrt {11}} {44\,800} $ - - - - -$d$-simplices have $d + 1$ vertices, and rectified $d$-simplices (such as the rectified $10$-simplex) have $d(d + 1)/2$ vertices. -The automorphism group of the abstract polytope generated by this polytope is the symmetric group $S_{12} > \mathrm M_{11} \hspace {-1pt} $. It acts transitively on both of the types of facets listed in the bottom row of the table. -The inner products between any given vertex and the full set of vertices is the multiset $ \left \{5, 2^{20}, -1^{45} \right \} $. By taking both orbits and rescaling the vertices down to the unit sphere, we can create an antipodal $(11, 132, 2/5)$ spherical code. This is superior to the $(11, 78, 2/5)$ code listed on Dr. Neil Sloane's webpage. I was optimistic that it might even be optimal, but Prof. Henry Cohn very kindly calculated an even better $(11, 172, 2/5)$ code at my prompting. - -Further Extensions -Is this construction known? What about the dual of this polytope? It has only $24$ vertices, although they no longer lie on a single sphere. Do polytopes exist with automorphism groups matching any given finite group? -What about lattices? Is the $\mathbb Z_2 \! \times G$ factor necessary? If so, when? I found that it also occurred when constructing two $76$- and one $77$-dimensional lattices from the invariant forms corresponding to the (absolutely irreducible over the rationals) $\chi_4$, $\chi_5$, and $\chi_6$ characters of the sporadic pariah Janko $\mathrm J_1 \hspace {-1pt} $ group.<|endoftext|> -TITLE: Alternative proofs sought after for a certain identity -QUESTION [16 upvotes]: Here is an identity for which I outlined two different arguments. I'm collecting further alternative proofs, so - -QUESTION. can you provide another verification for the problem below? - -Problem. Prove that -$$\sum_{k=1}^n\binom{n}k\frac1k=\sum_{k=1}^n\frac{2^k-1}k.$$ -Proof 1. (Induction). The case $n=1$ is evident. Assume the identity holds for $n-1$. Then, -\begin{align*} \sum_{k=1}^{n+1}\binom{n+1}k\frac1k-\sum_{k=1}^n\binom{n}k\frac1k -&=\frac1{n+1}+\sum_{k=1}^n\left[\binom{n+1}k-\binom{n}k\right]\frac1k \\ -&=\frac1{n+1}+\sum_{k=1}^n\binom{n}{k-1}\frac1k \\ -&=\frac1{n+1}+\frac1{n+1}\sum_{k=1}^n\binom{n+1}k \\ -&=\frac1{n+1}\sum_{k=1}^{n+1}\binom{n+1}k=\frac{2^{n+1}-1}{n+1}. -\end{align*} -It follows, by induction assumption, that -$$\sum_{k=1}^{n+1}\binom{n+1}k\frac1k=\sum_{k=1}^n\binom{n}k\frac1k+\frac{2^{n+1}-1}{n+1}=\sum_{k=1}^n\frac{2^k-1}k+\frac{2^{n+1}-1}{n+1} -=\sum_{k=1}^{n+1}\frac{2^k-1}k.$$ -The proof is complete. -Proof 2. (Generating functions) Start with $\sum_{k=1}^n\binom{n}kx^{k-1}=\frac{(x+1)^n-1}x$ and integrate both sides: the left-hand side gives -$\sum_{k=1}^n\binom{n}k\frac1k$. For the right-hand side, let $f_n=\int_0^1\frac{(x+1)^n-1}x\,dx$ and denote the generating function -$G(q)=\sum_{n\geq0}f_nq^n$ so that -\begin{align*} G(q)&=\sum_{n\geq0}\int_0^1\frac{(x+1)^n-1}x\,dx\,q^n =\int_0^1\sum_{n\geq0}\frac{(x+1)^nq^n-q^n}x\,dx \\ -&=\int_0^1\frac1x\left[\frac1{1-(x+1)q}-\frac1{1-q}\right]dx=\frac{q}{1-q}\int_0^1\frac{dx}{1-(x+1)q} \\ -&=\frac{q}{1-q}\left[\frac{\log(1-(1+x)q)}{-q}\right]_0^1=\frac{\log(1-q)-\log(1-2q)}{1-q} \\ -&=\frac1{1-q}\left[-\sum_{m=1}^{\infty}\frac1mq^m+\sum_{m=1}^{\infty}\frac{2^m}mq^m\right]=\frac1{1-q}\sum_{m=1}^{\infty}\frac{2^m-1}m\,q^m \\ -&=\sum_{n\geq1}\sum_{k=1}^n\frac{2^k-1}k\,q^n. -\end{align*} -Extracting coefficients we get $f_n=\sum_{k=1}^n\frac{2^k-1}k$ and hence the argument is complete. - -REPLY [6 votes]: Here's a sketch of a proof of a generalization: -\begin{multline} -\quad -\sum_{k=1}^n\binom nk \frac{t^k}{k+a}\\ =\frac{1}{\binom{a+n}{n}}\sum_{k=1}^n \binom {a+k-1}{k-1} \frac{(1+t)^k-1}{k}. \quad -\tag {$*$} -\end{multline} -(This is a generalization of Terry Tao's generalization, which is the case $a=0$.) -We start with the identity -$$\sum_{k=0}^n \binom nk \frac{t^k}{k+a} = \frac {1}{a\binom{a+n}{n}}\sum_{k=0}^n \binom{a+k-1}{k} (1+t)^k.$$ -This is a special case of a well-known linear transformation for the hypergeometric series, the case $b=a+1$ of -$${}_2F_1(-n,a; b\mid -t) =\frac{(b-a)_n}{(b)_n}\,_2F_1(-n,a; 1-n-b+a\mid 1+t),$$ -where $(u)_n = u(u+1)\cdots (u+n-1)$, which can be proved easily in several ways. -Since $\frac{1}{a}\binom{a+k-1}{k} = \frac {1}{k}\binom{a+k-1}{k-1}$ for $k\ge 1$, we have -$$\sum_{k=1}^n\binom nk \frac{t^k}{k+a} - =\frac{1}{\binom{a+n}{n}}\sum_{k=1}^n \binom {a+k-1}{k-1} \frac{(1+t)^k-1}{k}+C$$ -where $C$ is a constant (as a polynomial in $t$). But $C=0$ since each summand has no constant term in $t$, and $(*)$ follows.<|endoftext|> -TITLE: Quantitative word problem for 3-manifold groups -QUESTION [8 upvotes]: The word problem for 3-manifold groups is solvable: given a based loop $\gamma$ in $M^3$ there is an algorithm to decide whether $\gamma$ bounds a disk. -What kinds of quantitative results are known about this problem? Specifically, what are the known best upper bound estimates for the time and space complexity of these algorithms? -Suppose we use a triangulation as a genie, are there any results like the following? If $(M,T)$ is a closed 3-manifold $M$ and a triangulation $T$ of $M$ then there exists a PL loop $\gamma$ in $M$ such that if $T'$ is a triangulation of $M$ with a 2-skeleton containing a disk bounding the loop then the Pachner distance between the two triangulations $T$ and $T'$ must be greater than $e^l$, where $l$ is the length of $\gamma$. - -REPLY [3 votes]: Moshe’s answer points out that one gets a quadratic algorithm for the word problem for subgroups of $GL_N(\mathbb{Z})$. I thought I would expand a comment on his answer into an answer, since I’m not aware of such a result strictly in the literature (even though it is relatively trivial, the estimate depends on recent advances). -Firstly, as you can find from the reference 3-Manifold Groups cited in Moshe’s answer, a compact orientable 3-manifold $M$ admitting a non-positively curved metric (which includes compact irreducible manifolds with non-empty boundary) admits a faithful representation of its fundamental group into $GL_N(\mathbb{Z})$ for some $N$ depending on $M$. This follows from results of Yi Liu in the graph manifold case and Przytycki-Wise in the “mixed manifold” case (see Cor. 1.2), where there is a hyperbolic piece of the JSJ decomposition (such irreducible manifolds always admit non-positively curved metric by a result of Bernhard Leeb). -The remaining graph manifold cases may have linear fundamental group, but there are certain restrictions. -In any case, one can show that finitely-generated subgroups of $GL_N(\mathbb{Z})$ have word problem which is $O(n\log^2 n)$ in the length of the word $n$. -Suppose that we are given $k$ $N\times N$ matrices $A_1,\ldots, A_k \in GL_N(\mathbb{Z})$ which -generate the group $G -TITLE: Regular or h-regular CW-complex structure for the Poincaré homology sphere -QUESTION [5 upvotes]: I am looking for a regular (the characteristic maps of the cells are homeomorphisms) or h-regular (the characteristic maps of the cells are homotopy equivalences) CW-complex structure for the Poincaré homology sphere. I would like to find a more economic one than the triangulation having f-vector: [16, 106, 180, 90]. I would like to minimize the number of cells. -Thanks in advance and any idea or potential reference is welcome! - -REPLY [2 votes]: Henrik Rüping's suggestion (in the comments) decomposes the Poincaré homology three-sphere as 12 pentagonal pyramids. The resulting face vector is -[0 + 12, 6 + 30, 10 + 20, 5 + 1] = [12, 36, 30, 6] -for a total of 84 cells. You can reduce the number of three-cells, at the cost of increasing the number of cells overall, as follows. In Rüping construction, join the pyramids in pairs (along their pentagonal face) to get six bi-pyramids. This has an face vector of -[6, 30, 30, 6] -but is no longer regular -- every bi-pyramid meets the central vertex twice. We can fix this by "blowing up" the central vertex to obtain a small dodecahedron. This truncates the apex and nadir of each bi-pyramid. Now the face vector is -[6 + 1, 30 + 12, 30 + 30, 6 + 20] = [7, 42, 60, 26] -with sum 135. At least the number of three-cells is smaller...<|endoftext|> -TITLE: Fourier cosine transform of $\frac{\sin(b\sqrt{a^2+x^2})}{a^2+x^2}$ -QUESTION [5 upvotes]: I've noticed it is not in Erdelyi or Gradshteyn, although the version with the sine replaced by a cosine is in Erdelyi (page 26 eq. 33). -I've tried using the substitution $x=a \sinh z$ to avoid a square root. -I would also like to know the Fourier sine transform of the sine and cosine form. - -REPLY [4 votes]: This is an integral in Oberhettinger's Table of Fourier Transforms (page 27, with $r=\sqrt{a^2+x^2}$). It is not a very helpful result, but it does indicate why a fully closed-form expression will not be forthcoming (you would need the indefinite integral of a Bessel function of argument $\sqrt{1-x^2}$). - -There is a typo in the formula, a numerical check does not match. For $yb.$$ -Notes: -• As a further check, note that the derivative with respect to $b$ gives formula 3.876.2 of Gradshteyn. -• The corresponding formulas in Erdelyi contain several typo's. These have been corrected in Gradshteyn, the formulas in question have a label $^6$. -• The integral also implies the identity -$$\int_0^1 K_0\left(\sqrt{1-t^2}\right)\,dt=\tfrac{1}{2e} \text{Ei}(1)-\tfrac{e}{2} \text{Ei}(-1).$$<|endoftext|> -TITLE: Is there a complete characterization of ordered fields without definable proper subfields? -QUESTION [5 upvotes]: $\mathbb{Q}$ has no proper subfields. As a result, all ordered fields elementarily equivalent to $\mathbb{Q}$ have no proper subfields which are first-order definable without parameters. And by the Tarski-Seidenberg theorem, real closed fields also have no proper subfields first-order definable without parameters. -My question is, are all the ordered fields of characteristic $0$ which have no proper subfields first-order definable without parameters known? Or failing that, are all the countable ordered fields of characteristic $0$ with this property at least known? - -REPLY [9 votes]: This is an interesting question. We know some things about this, but we do not have a characterization of fields with this property. As Wojowu says above the restriction to countable fields doesn't help, and I don't think that restricting to ordered fields helps either. This property implies that the field cannot define the subfield $\mathbb{Q}$, and of course we are particularly interested in definability of that subfield. -First note that an imperfect field defines the image of the Frobenius, so we can just consider perfect fields. I believe that basically all known model-theoretically tame perfect fields do not admit definable subfields, but I'm not sure how much of this has been written down. We don't know which fields are tame and which aren't. -As you point out, real closed fields have this property. This is because any infinite definable subset of $R$, for $R$ real closed, has nonempty interior. So any definable subfield must have interior, but it is easy to that there are no proper subfields with interior. The last claim holds for any topological field. So we can apply this argument to other topological fields, we just need every infinite unary definable to have interior. For example for $\mathbb{Q}_p$ this follows from a quantifier elimination result. Similar arguments should work over any local field of characteristic zero. I think it should also generalize to Henselian fields of characteristic zero, this should follow from quantifier eliminations for Henselian valued fields. So you cover things like $F((t))$ for any field $F$ of characteristic zero. To get something ordered you can take $F$ to be any ordered field. -Among the fields that people are usually interested in, there is a big divide between the large and non-large. Largeness is a field-theoretic notion introduced by Pop, it's supposed to be a field-arithmetic notion of tameness. Finite fields, number fields, and function fields are not large, most other interesting fields are. It's a theorem of Julia Robinson that any number field defines $\mathbb{Q}$. There are a number of results saying that various function fields define the field of constants. For example $\mathbb{R}(t)$ defines $\mathbb{R}$. I don't know what the state of the art is here, but for example we do know that if $F$ is an extension of $K$ with transcendence degree one, $K$ is algebraically closed in $F$, and the group of $n$th powers in $K$ has finite index for some $n \ne \mathrm{Char}(K)$, then $F$ defines $K$. See for example Fehm and Geyer's paper "A Note on Defining Transcendentals in Function Fields". -So it's worth restricting to perfect large fields. Arno Fehm has shown that a perfect large field cannot existentially define a proper subfield. Perfect large fields can define subfields, the fraction field of $\mathbb{R}[[x,y]]$ is large and defines $\mathbb{Q}$, this should also be orderable. We do not have a characterization of perfect large fields that do not define proper subfields. -If a field $K$ is perfect, large, and model complete (possibly after constants have been added to the language) then it cannot define any proper subfields - just because any definable set is existentially definable. For example, the theory of $n$-ordered fields has a model companion, the theory of $n$-ordered pseudo real closed fields, and these fields are model complete in the language of fields (without the orders!), so $n$-ordered pseudo real closed fields do not admit proper definable subfields. -I think it should be possible to show that the other known examples of model-theoretically tame perfect fields do not admit proper definable subfields, but I'm not sure how much is known/written down. -Edit: Fehm pointed out that Junker and Koeniggsmann show in the paper "Slim fields" that perfect PAC fields and characteristic zero Henselian fields do not define proper subfields. They study a class of fields they call "very slim", a very slim field does not define proper subfields. We don't have a classification of very slim fields, either.<|endoftext|> -TITLE: $m$-fold composite $p^{(m)}(x) \in \mathbb{Z}[x]$ implies $p(x) \in \mathbb{Z}[x]$ -QUESTION [27 upvotes]: Let $p(x)$ be a polynomial, $p(x) \in \mathbb{Q}[x]$, and $p^{(m+1)}(x)=p(p^{(m)}(x))$ for any positive integer $m$. -If $p^{(2)}(x) \in \mathbb{Z}[x]$ it's not possible to say that $p(x) \in \mathbb{Z}[x]$. -Is it possible to conclude that $p(x) \in \mathbb{Z}[x]$ if $p^{(2)}(x) \in \mathbb{Z}[x]$ and $p^{(3)}(x) \in \mathbb{Z}[x]$? -More general, suppose there exist positive integers $k_1 0$. Performing such choice for the polynomial $F_r(x)=f_rf(x)^r$, we get the index $rt$ with the valuation $v_r+rv_t$ (one way to see it is, again, via Newton polygons, but these inequalities can be easily written down explicitly). -Choose now an index $s$ minimizing $M=v_s+sv_t$, taking again the maximal index given a multiple choice. We have $M\leq v_t+tv_t<0$; therefore, $s>0$, since $v_0\geq0$. Then $M$ is the minimal valuation of a coefficient in all the $F_i$; moreover, such valuation appears in the coefficient of $x^{st}$ exactly once —that is, in $F_s(x)$. Therefore, in $f^{(2)}(x)$ the coefficient of $x^{st}$ also has valuation $M<0$.<|endoftext|> -TITLE: What would be a good introductory reference for learning jet-bundle theory? -QUESTION [5 upvotes]: I am interested in learning the theory of Jet bundles, and am aware of the standard reference "The geometry of jet bundles" by D. J. Saunders. However this appears to be a detailed book, suitable for those who wish to specialise in this area. Can somebody recommend a relatively more introductory book (for a reader who knows the necessary differential geometric pre-requisites for learning this subject, but has never encountered Jet bundles) ? Thanks so much ! - -REPLY [5 votes]: Two articles by A.M. Vinogradov provide a gentle introduction: -"Local symmetries and conservation laws", Acta Applicandae Mathematica volume 2, pages 21–78(1984) -"An informal introduction to the geometry of jet spaces", available here.<|endoftext|> -TITLE: The infimum of a gradient over the whole $\mathbb{R}^d$ -QUESTION [10 upvotes]: Let $\{f_k\}:\mathbb{R}^d\to\mathbb{R}$ be a sequence of $C^1$ functions which converges pointwise to 0. Is it true that -$$\lim_{k\to+\infty}\inf_{x\in\mathbb{R}^d}|\nabla f_k(x)|=0?$$ -If $d=1$ I believe the result is true, but what happens if d>1? I believe counterexamples can be constructed, maybe building a sequence of functions for which the modulus of the gradient remains bounded away from 0 but the modulus of each partial derivatives oscillates between 0 and a certain value, keeping the modulus of the whole gradient above the treshold. -I shall present the proof for the case $d=1$. -Take a compact set $A\subset \mathbb{R}$ which has non empty interior, then clearly -$$\limsup_{k\to+\infty}\inf_{x\in\mathbb{R}^d}|f^{\prime}_k(x)|\leq\limsup_{k\to+\infty}\inf_{x\in A}|f^{\prime}_k(x)|.$$ -Let us focus on the right hand side of the previous expression: suppose that there exists $l>0$ such that -$$\limsup_{k\to+\infty}\inf_{x\in A}|f^{\prime}_k(x)|\geq l.$$ -Then $\forall k$, due to compactness, there exists $x_k\in A$ such that -$$\limsup_{k\to+\infty}|f^{\prime}_k(x_k)|\geq l.$$ -But then, being $x_k$ a point of minimum, the following holds -$$\limsup_{k\to+\infty}|f_k^{\prime}(x)|\geq l\qquad\forall x\in A.$$ -Passing to a subsequence that I won't relabel, this implies that $\forall k\in\mathbb{N}$, either $f_k^{\prime}(x)\geq l/2$ for every $x\in A$, or -$f_k^{\prime}(x)\leq -l/2$ for every $x\in A$ due to the continuity of $f_k^{\prime}$. -Now take an interval $[x,y]\in A$ (with $x1$. - -REPLY [12 votes]: Here is the outline of the construction. -We shall start with considering a vertical half-strip going up, the left boundary of which will be called positive (shown in red) and the right boundary will be called negative (shown in blue) and carry out the following construction (generation 1 and generation 3 shown on the two next pictures; the rest of the strip work will be illustrated on generations 2): - -This was generation 1. - -And that is generation 3. How to proceed should be clear now (if not, look at the generation 2 below). -Notice that at generation $k$ we turn each side into a $3^{-k}$ dense tree and we have a winding strip between these trees one side of which is positive and the other one negative. Now expand each tree a bit into a domain with smooth boundary (the expansion can be made as small as we wish). - -Our function will be $\varepsilon=3^{-k}$ on the boundary of the red domain and $-\varepsilon=-3^{-k}$ on the boundary of the blue domain. -We can make a smooth foliation of the winding yellow strip with transversal lines of length about $3^{-k}$ as shown (the parts where the boundaries are true parallel straight lines are trivial). We can make a descent along each line with gradient about $1$ most of the time while near the endpoints we can declare any normal derivative larger than $1$ we want as long as it smoothly depends on the point (the exact value will be determined by what happens inside the domains). -Inside the domains we will draw the function level lines as follows (effectively removing one generation growth a time): - -The point is that we can move the level lines at speed $\le 1$ to keep the gradient above $1$, so we are forced to add $3^{-\ell}$ to the function when collapsing the level $\ell$ sprouts. Fortunately, the sum of the geometric progression is its largest term, so we keep the function at the level $3^{-\ell}$ around the part of the tree we added when going from generation $\ell-1$ to generation $\ell$. I'll call this part the $\ell$-th generation intervals. Note that they all are located on a fixed discrete grid of lines (I was too lazy to make that picture but I'll do it if somebody really wants it). -Generally the function increases in the red domain and decreases in the blue one according to the arrows below: - -Now comes some analysis. This function is of order $3^{-k}$ outside the blue and red domains that can be made of any width $\delta_k>0$ we want. However, if we keep the picture in place, the original trees will be bad. So we will shift it by some decreasing sequence $\mu_k>0$ in both vertical and horizontal directions demanding that the intervals $(\mu_k-4\delta_k,\mu_k+4\delta_k)$ are disjoint. Then for each level $\ell$, the $\delta_k$ neighborhoods of the, say, horizontal lines on which the $\ell$-th generation intervals lie are eventually disjoint (that happens as soon as $\mu_k$ gets too small to make it possible for a shift of a grid line to intersect another shift of a different grid line). Thus, for any fixed $\ell\ge 0$, each fixed point can lie only in finitely many $\delta_k$ neighborhoods of the shifted $\le \ell$-th generation intervals, i.e., eventually the function drops below $3^{-\ell}$ and stays there. -This gives you an example in a half-strip. What was crucial was that we can make these small shifts and that the boundary lines have the starting point. The possibility to shift is due to the foliation of the plane into vertical lines and the transversal foliation into the horizontal ones. So we need to make something like that but covering everything. The solution is the foliation into double spirals and the transversal foliation into rays from the origin: - -The spirals are unit width ones outside the disk of radius $1$ and logarithmic inside. This creates a nice coordinate system between the red and blue lines outside an arbitrarily small neighborhood of the origin and we can just replant our entire construction there. What to do near the origin? It turns out that we need to do nothing because the spirals themselves are already very close there (we need to gradually reduce the neighborhoods with $k$ to $0$, of course). So, if we stop the spirals at some point, extend them to narrow smooth domains and do the transversal foliation of the strip in between, we can just do the straight descent from red to blue as before without forcing any small values of the gradient: - -After just a turn or two, our coordinate system kicks in and the further expansion of the strip is of no concern. -The result is that every point except the origin is eventually covered by the coordinate system and our tree nonsense and also decides between which two rotations of the red/blue spirals it stays (so the coordinates get eventually consistent), so it is good in terms of convergence to $0$. But the origin is always good (if you look at out functions closely, you'll realize that they are actually odd). The escapes along the spirals are, of course, the same as along the boundary lines of the half-strip. -That is all (modulo minor boring formalities related to $C^1$-smoothness). -Describing pictures in words is not my strong side, but if you have trouble understanding anything, feel free to ask questions.<|endoftext|> -TITLE: The product of two Hausdorff measures -QUESTION [5 upvotes]: Let $(X,d_X)$ and $(Y,d_Y)$ be two compact metric space with Hausdorff dimensions $\dim_H(X)=n$ and $\dim_H(Y)=m$ and Hausdorff measures $\mathcal{H}^{n}$ and $\mathcal{H}^{m}$. -Assume that $\dim_H(X\times Y)=n+m$ for the cartesian product $(X\times Y, d)$ where $d=\sqrt{d_X^2+d_Y^2}$, then we have $(n+m)$-dimensional Hausdorff measure $\mathcal{H}^{n+m}$ on it. -Do we have $\mathcal{H}^{n+m}=\mathcal{H}^{n}\otimes\mathcal{H}^{m}$? -In particular, does the equation hold, if $Y$ is a smooth Riemannian manifold? - -REPLY [5 votes]: As noted in a comment, for Riemannian manifolds, the Hausdorff measures are equal to (up to a constant) the usual volumes. So this works. - -The metric case you mention can fail. There are metric spaces of Hausdorff dimension $1$ that are not "rectifiable". Every subset has either $\mathcal H^1(E) = 0$ or $\mathcal H^1(E) = \infty$. Discussion is in Chapter 3 of -Falconer, K. J., The geometry of fractal sets, Cambridge Tracts in Mathematics, 85. Cambridge etc.: Cambridge University Press. XIV, 162 p. (1985). ZBL0587.28004. - -And also note this is incorrect for general dimensions. Reference: third edition of -Falconer, Kenneth, Fractal geometry. Mathematical foundations and applications, Hoboken, NJ: John Wiley & Sons (ISBN 978-1-119-94239-9/hbk). xxx, 368 p. (2014). ZBL1285.28011. -Here are some things from Chapter 7. -Proposition 7.1 -If $E \subset \mathbb R^n, F \subset \mathbb R^m$ are Borel sets with $\mathcal H^s(E), \mathcal H^t(F) < \infty$, then -$$ -\mathcal H^{s+t}(E \times F) \ge c \mathcal H^s(E)\;\mathcal H^t(F) -\tag1$$ -where $c > 0$ depends only on $s$ and $t$. -The opposite inequality $\le$ can fail. -Example 7.8 -There exist sets $E, F \subset \mathbb R$ with $\dim_\mathcal H E = \dim_\mathcal H F = 0$ and $\dim_\mathcal H(F \times F) \ge 1$. -Falconer credits these results to: -Besicovitch, A. S.; Moran, P. A. P., The measure of product and cylinder sets, J. Lond. Math. Soc. 20, 110-120 (1945). ZBL0063.00354. -and -Marstrand, J. M., The dimension of Cartesian product sets, Proc. Camb. Philos. Soc. 50, 198-202 (1954). ZBL0055.05102. -plugs: -Those two papers are among those reprinted in -Edgar, Gerald A. (ed.), Classics on fractals, Reading, MA: Addison-Wesley Publishing Company. x, 366 p. (1993). ZBL0795.28007. -More on this from a student of mine: -Mullins, Edmond N., Jr, -Derivation bases, interval functions, and fractal measures. -Thesis (Ph.D.)–The Ohio State University. 1996. 97 pp. ISBN: 978-0591-18087-9 -(ProQuest LLC)<|endoftext|> -TITLE: Categorical presentation of direct sums of vector spaces, versus tensor products -QUESTION [6 upvotes]: My apologies in advance if this question is to vague, but here goes.... In the category of vector spaces, products are given by direct sums. In general category theory, the existence of products is a property of the category, there is no choice going on. On the other hand the tensor products of vector spaces is an examples of a monoidal structure. In general category theory, monoidal structures are chosen, there can exist more than one for any category. How is one to understand the distinction. Taking direct sums seems as "natural" as taking tensor products, so why is their general categorical interpretation so different? - -REPLY [22 votes]: One way to think about what the monoidal structure on vector spaces is doing is that it is telling us that vector spaces do not really form a category, or not "just" a category: they form a multicategory whose multimorphisms $V_1, \dots V_n \to W$ are given by multilinear maps $V_1 \times \dots \times V_n \to W$. We care a lot about multilinear maps and not just linear maps in practice so this is a very natural thing to do. -Multicategories are a strict generalization of monoidal categories; if the multihom functor $\text{Hom}(V_1, \dots V_n ; W)$ happens to be representable as a functor of $W$ for all $V_i$ then the representing object can be written $V_1 \otimes \dots \otimes V_n$ and this should define what is called an "unbiased" monoidal category (an axiomatization where we axiomatize all the $n$-fold tensor products at once rather than just binary ones). -In other words, the tensor product is describing an additional structure we care about in practice which is not captured by the category structure alone. Similar situations happen all the time, e.g. rings have additional structure given by multiplication which is not captured by addition alone (and it's not so bad to think of monoidal abelian categories, say, as categorified rings). What the category theory tells us is that the notion of direct sum can be defined using only the notion of linear map but the notion of tensor product requires the notion of multilinear map.<|endoftext|> -TITLE: Ramanujan's Master Formula: A proof and relation to umbral calculus -QUESTION [5 upvotes]: The Ramanujan's master theorem states that: -$$ -\int_0^{\infty}x^{s-1}\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}a_nx^ndx=\Gamma(s)a_{-s} -$$ -I found a really strange proof recently on a personal blog: -Define -$$ -\tau *a_n=a_{n+1} -$$ -we have -$$ -{\int_0^{\infty}x^{s-1}\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}a_nx^n\text{dx} \\=a_0\int_0^{\infty}x^{s-1}\sum_{n=0}^{\infty}\frac{(-1*\tau_a*x)^n}{n!}\text{dx} \\=a_0\int_0^{\infty}x^{s-1}\text{e}^{-\tau_ax}\text{dx} \\=\frac{a_0}{\tau_a^s}\int_0^{\infty}{(\tau_a x)}^{s-1}\text{e}^{-\tau_ax}\mathrm{d\tau_a x} =\frac{a_0}{\tau_a}\Gamma(s) \\=a_{-s}\Gamma(s) } -$$ -My question is, how can I understand the above proof? Can anyone give a rigorous explanation for the above proof? For example, I totally have no idea about the meaning of $\color{blue}{\mathrm{d\tau_ax}}$(in the sense of analysis). -Would it be related to Umbral calculus? -Also, I want some further readings if possible. - -REPLY [5 votes]: The RMF is definitely related to umbral calculus via the modified Mellin transform (MMT) pair and symbolic extension of the iconic Euler gamma function integral. The proof you copied? I don't know. The MMT pair allows for interpolation of the coefficients of generating functions, often directly connected to sinc and/or Newton interpolation. -First consider the MMT and its inverse -$$\tilde{f}(s) = MMT[f(x)] = \int_{0}^{\infty} f(x) \; \frac{x^{s-1}}{(s-1)!} \; dx$$ -$$f(x) = MMT^{-1}[\tilde{f}(s)] = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} \tilde{f}(s) \frac{x^{-s}}{(-s)!} \; ds .$$ -Then the RMF holds for a class of functions such that -$$f(x) = e^{-a.x} = \sum_{n \geq 0} \frac{(-a.x)^n}{n!} = \sum_{n=0} a_n \frac{(-x)^n}{n!} = - \sum_{n=0} \tilde{f}(-n) \frac{(-x)^n}{n!} \; ,$$ -that is, such that we may close the complex contour to the left (e.g., in the sense of the limit of a semicircle with its radius expanding to infinity) for $0 < \sigma < 1$ and $0 < x < 1$ when $F(s)$ has no singularities/poles within the contour. This rep allows an extension of the RMT (and the Mellin transform) to cases in which poles are present in $F(s)$ and other ranges of $x$. -Also note (see, e.g., Gelfand and Shilov's "Generalized Functions") the relation -$$D_x^{m+n+1} \; H(x) \frac{x^m}{m!} = H(x) \frac{x^{-n-1}}{(-n-1)!} = \delta^{(n)}(x)$$ -reflected in the two (of several) reps of the fractional differintegro op equivalent under analytic continuation -$$\frac{x^{\alpha-\beta}}{(\alpha-\beta)!} = \frac{d^{\beta}}{dx^\beta}\frac{x^{\alpha}}{\alpha!}=\int_{0}^{x}\frac{z^{\alpha}}{\alpha!}\frac{(x-z)^{-\beta-1}}{(-\beta-1)!} dz = \frac{1}{2\pi i} \oint_{|z-x|=|x|}\frac{z^{\alpha}}{\alpha!}\frac{\beta!}{(z-x)^{\beta+1}}dz ,$$ -with $H(x)$ the Heaviside step function. -So, under the conditions above, -$$\tilde{f}(-n) = \int_{0}^{\infty} f(x) \; \frac{x^{-n-1}}{(-n-1)!} \; dx = \int_{0}^{\infty} e^{-a. x} \; \delta^{(n)}(x) \; dx = a_n,$$ -and this suggests the analytic continuation and relation to umbral calculus -$$\tilde{f}(s) = \int_{0}^{\infty} f(x) \; \frac{x^{s-1}}{(s-1)!} \; dx = \int_{0}^{\infty} e^{-a.x} \; \frac{x^{s-1}}{(s-1)!} \; dx = (a.)^{-s} = a_{-s}.$$ -The iconic guiding example is the Euler gamma function integral rep with $(a.)^n = a_n = c^n$ -$$ (a.)^{-s} = a_{-s} = c^{-s} = F(s) = MT[f(x)= e^{-c\; x}] = \int_{0}^{\infty} e^{-c \; x} \; \frac{x^{s-1}}{(s-1)!} \; dx = \frac{1}{c^{s}}.$$ -Another useful example, which vividly illustrates the relation to the Appell Sheffer sequences of umbral calculus (of which the $x^n$ with e.g.f. $e^{x}$ is the basic example), is the integral rep for (what I call) the Bernoulli function, simply related to the Hurwitz zeta function and generalizing the Bernoulli polynomials, -$$ B_{-s}(z) = (B.(z))^{-s} = \int_{0}^{\infty} e^{-B.(z)t} \; \frac{t^{s-1}}{(s-1)!} \; dt $$ -$$ = \int_{0}^{\infty} \frac{-t}{e^{-t}-1} \; e^{-zt} \frac{t^{s-1}}{(s-1)!} \; dt = s \; \zeta(s,z)$$ -where the e.g.f. for the Bernoulli polynomials with $(b.)^n = b_n$ the Bernoulli numbers is -$$e^{B.(x)t} = e^{(b.+x)t} = e^{b.t} e^{xt} = \frac{t}{e^t-1} \; e^{xt}.$$ -Note that -$$B_n(z) = -n \; \zeta(1-n,z),$$ -$$B_n(1) = -n \; \zeta(1-n,1) =-n \; \zeta(1-n) (Riemann) = (-1)^n B_n(0) = (-1)^n b_n.$$ -Through this characterization, it is not too difficult to show that the Bernoulli function inherits all the elegant properties of a regular Appell sequence, such as $D_z \; B_{s}(z) = s \; B_{s-1}(z)$. - -Riemann knew all this stuff. Ramanujan intuited it. Hardy formalized it. I stumbled across it on a journey starting from the ladder ops of QM and a brief comment by my old math prof Stallybrass about the sequence $D^{m+n} H(x) \frac{x^m}{m!}$ in his integral transforms class an eon ago. -For application to defining fractional powers of operators, see my answer and comments therein to the MO-Q "What does the inverse Mellin transform really mean?" and several of my blog posts, such as "The Creation / Raising Operators for Appell Sequences." - -Added 2/5/21: -Other examples of interpolation of $a_n$ for the exponential generating funcrtion $g(t) = e^{a.t}$ from the MMT of $f(t) = g(-t) = e^{-a.t}$, or, conversely, surmising the MMT of $f(t)$ from the Taylor series coefficients of $g(t)$ via $a_n \; |_{n \rightarrow -s} =a_{-s} =\tilde{f}(s)$: -1) $\;g(t) = \cos(t) = \sum_{n \geq 0} \cos(\pi \frac{n}{2}) \; \frac{t^n}{n!}, $ -$\; \; \; \; \;f(t) = g(-t) = \cos(t) = \sum_{n \geq 0} \cos(\pi \frac{n}{2}) \; \frac{t^n}{n!} ),$ -$\; \; \; \; \;\tilde{f}(s) =\cos(\pi \frac{s}{2})$ for $0 < Re(s) < 1,$ -2) $\;g(t) = \sin(t)= \sum_{n \geq 0} \sin(\pi \frac{n}{2}) \; \frac{t^n}{n!},$ -$\; \; \; \; \;f(t) = g(-t) = \sin(-t) = \sum_{n \geq 0} \sin(-\pi \frac{n}{2}) \; \frac{t^n}{n!},$ -$\; \; \; \; \;\tilde{f}(s) =-\sin(\pi \frac{s}{2})$ for $-1 < Re(s) < 1,$ -3) $\;g(t) = \frac{1}{1-t} = \sum_{n \geq 0} \; n! \; \frac{t^n}{n!},$ -$\; \; \; \; \;f(t) = g(-t) = \frac{1}{1+t} = \sum_{n \geq 0} \cos(\pi n) \; n! \; \frac{t^n}{n!},$ -$\; \; \; \; \;\tilde{f}(s) =(-s)! $ for $0 < Re(s) < 1,$ -4) $\;g(t) = \frac{1}{1+t} = \sum_{n \geq 0} \cos(\pi n) \; n! \; \frac{t^n}{n!} ,$ -$\; \; \; \; \;f(t) = g(-t) = \frac{1}{1-t} = \sum_{n \geq 0} \; n! \; \frac{t^n}{n!},$ -$\; \; \; \; \;\tilde{f}(s)=\cos(\pi s) (-s)!$ for $0 < Re(s) < 1,$ -5) $\;g(t) = \ln(1-t) = \sum_{n \geq 0} \; -(n-1)! \; \frac{t^n}{n!} ,$ -$\; \; \; \; \;f(t) = \ln(1+t) = -\sum_{n \geq 0} \cos(\pi n) \; (n-1)! \; \frac{t^n}{n!},$ -$\; \; \; \; \;\tilde{f}(s) = -(-s-1)! $ for $-1 < Re(s) < 0,$ -6) $\;g(t) =\sum_{n \ge 0} \frac{x^n}{n!} \frac{t^n}{n!}, $ -$\; \; \; \; \;f(t) = J_0(2 \sqrt{xt}) =\sum_{n \ge 0} (-1)^n \frac{x^n}{n!} \frac{t^n}{n!},$ -$\; \; \; \; \;\tilde{f}(s) = \frac{x^{-s}}{(-s)!}$ for $0 < Re(s) < \frac{3}{4},$ -7) $\;g(t) = e^{-t^2} =\sum_{n \ge 0} \cos(\frac{\pi n}{2}) \; \frac{n!}{(\frac{n}{2})!} \; \frac{t^n}{n!}, $ -$\; \; \; \; \;f(t) = g(-t) = e^{-t^2},$ -$\; \; \; \; \;\tilde{f}(s) = \cos(\pi\frac{ s}{2}) \; \frac{(-s)!}{(-\frac{s}{2})!} = \frac{1}{2}\frac{(\frac{s}{2}-1)!}{(s-1)!} \;$ for $ Re(s) > 0.$ -8) $\;g(t) =\sum_{n \ge 0} \; \frac{a_{\bar{n}}\; b_{\bar{n}}}{c_{\bar{n}}} \; \frac{t^n}{n!} = F(a,b;c;t)$, the hypergeometric function, where, e.g., -$\; \; \; \; \;a_{\bar{n}} = \frac{(a+n-1)!}{(a-1)!}$, the rising factorial, -$\; \; \; \; \;f(t) = g(-t) = F(a,b;c;-t),$ -$\; \; \; \; \;\tilde{f}(s) = \frac{a_{-\bar{s}} \; b_{-\bar{s}}}{c_{-\bar{s}}},$ -see the Mellin-Barnes contour integral.<|endoftext|> -TITLE: Tree property at weak inaccessibles -QUESTION [6 upvotes]: Suppose $\kappa$ is a weakly inaccessible cardinal with the tree property. What can we say about the height of $\kappa$? Is it a weakly-hyper-Mahlo of some sort? Does it enjoy some kind of indescribability property? Of course it is weakly compact in $L$, but I am interested in what height properties we can say it has in a universe where $\kappa$ is not strongly inaccessible. - -REPLY [5 votes]: In his paper Boolean extensions which efface the Mahlo property William Boos proves the following consistency result: -Theorem. Assume GCH holds and $\kappa$ is weakly compact. Then there exists a cardinal preserving generic extension of the universe in which $\kappa$ is the least weakly Mahlo cardinal and the tree property holds at $\kappa$. -The proof of the theorem is very similar to that of Mitchell, the main difference is that at Mahlo cardinals below $\kappa$ he adds a club of singular cardinals into that cardinal. -The following is asked in the paper and is still open: -Question. Is it consistent that the tree property hold at the least weakly inaccessible cardinal?<|endoftext|> -TITLE: How to invoke constants badly -QUESTION [44 upvotes]: In a nice and witty lecture titled "how to write mathematics badly" (available on YouTube at https://www.youtube.com/watch?v=ECQyFzzBHlo&t=23s), Jean-Pierre Serre describes various ways in which a paper can be poorly/confusingly/inaccurately written. -Around min 34:00 in the previous link, he criticizes the use of the word "constant", in particular in inequalities. The example he provides is of the type: - -$$\|Af\|\le C\|f\|$$ for some constant $C$ - -where $A$ is a complicated operator depending on many parameters. In this case, he says, usually the only thing that the writer means is that $C$ does not depend on "some of the data" of the problem. He adds that this attitude "caused lots of mistakes". -What are examples of these mistakes? Has any significant piece of mathematics been rewritten or erased altogether because of some problem with proofs invoking "constants" too nonchalantly? - -REPLY [16 votes]: Edit: The original answer below refers to Nelson's attempt from 2011. Upon a cursory look at the afterword by Sam Buss and Terence Tao to Nelson's paper placed in arxiv in 2015 (after his death), it seems he later attempted to address the error referred to in the original answer below; it would be interesting to know what the experts think on how successful his efforts were or potentially can be. -Original Answer: Edward Nelson's recent project on finding inconsistency of arithmetic (which was the subject of a MathOverflow Question) might be pertinent. The error, discovered by Terence Tao, seems to be the dependence of a constant on the underlying theory that Nelson did not account for.<|endoftext|> -TITLE: Serre functor on the category $Perf(A)$, $A$ - k-algebra -QUESTION [7 upvotes]: Consider a finite-dimensional $k$-algebra $A$ of finite global dimension. Then it is known that the Serre functor on $D^b(mod-A)$ exists and is given by the Nakayama functor. The proof goes something like this: -The $k$-duality $(-)^*=R\underline{\text{Hom}}_k(-,k)$ gives an equivalence $D^b(mod-A)\to D^b(A-mod).$ The $A$-duality $(-)^\vee=R\underline{\text{Hom}}_A(-,A)$ maps the category of perfect complexes $Perf(A)$ to $Perf(A^{op})$, because $A^\vee$ is isomorphic to $A$ in $A-mod$. Then one proves that for $M\in Perf(A)$, $N\in D^b(mod-A)$ we have -$$Hom_{D(Mod-A)} (M,N)^*\cong Hom_{D(Mod-A)} (N,M^{\vee*}).$$ -Finally, we notice that for finite-dimensional algebras of finite global dimension $Perf(A)=D^b(A-mod).$ -But what if $\text{r.gl.dim}(A)=\infty$? The only reason I see why the above proof might not give us a Serre functor on $Perf(A)$ is that $(-)^*$ may not map $Perf(A)$ to $Perf(A^{op})$, since $A$ is not necessarily isomorphic to $A^*$ as a left $A$-module. In this case, is it enough to have $A\cong A^{op}$? - -REPLY [5 votes]: The assumption on the finite global dimension is just needed to have $Perf(A)=D^b(A-mod),$ which does not hold more generality. -It is indeed true that $Perf(A)$ has a Serre functor (given in the same way) when $A$ just is a Gorenstein algebra, that is the regular module $A$ has finite injective dimension as a left and right $A$-modules. -For example any selfinjective non-semisimple algebra (such as $K[x]/(x^n)$ for $n \geq 2$) is a Gorenstein algebra of infinite global dimension. -Good references are: -"Representation Theory: A Homological Algebra Point of View" by Zimmermann and the book by Happel on Triangulated Categories.<|endoftext|> -TITLE: What is the theorem of the highest weight used for? -QUESTION [6 upvotes]: $\DeclareMathOperator\End{End}$Over the past few months, I have taught myself the classification of reductive groups, and continued to non-abelian (as well as a small venture to non-compact) Harmonic Analysis. -I am now trying to put everything that I learn together into something coherent. Let's, therefore, take the case of compact real Lie groups. By Chevalley the category of compact real Lie groups is equivalent to the category of $\mathbb{R}$-anisotropic reductive linear algebraic groups whose connected components have $\mathbb{R}$-points. In particular, all of the representations of a compact real Lie group are algebraic. -Therefore, by the Theorem of the Highest Weight, we have a classification of the unitary dual (the set of irreducible unitary representations) of a real compact Lie group $G$. -From the Harmonic Analysis perspective: -$$L^2(G)\cong\bigoplus_{\pi\in\hat{G}} \End(\pi)(\cong \bigoplus_{\pi\in\hat{G}} \pi\otimes\pi^*).$$ -It remains, therefore, to choose a basis of each $\End(\pi)$, for each $\pi$ guaranteed by the Theorem of the Highest Weight. -I looked at the examples in Folland's book on Harmonic Analysis, and did not see any mention of the Theorem of the Highest Weight. This seemed to have been done completely ad hoc. -I was also told in an old question of mine to take a look at Zonal Spherical Functions, in the particular case that $K=1$. I must confess, I find Zonal Spherical Functions to be quite confusing. I suspect that the point is that these are the method by which Harish Chandra proved the Plancherel Theorem (the non-compact variant of Peter-Weyl), but it's not clear to me how to use this in practice. -But either way, I end up wondering: what is the point of the Theorem of the Highest Weight? Does it provide a basis for each $\End(\pi)$? If it does, then what is it? If it doesn't, and we end up using some other method for finding the basis for each $\End(\pi)$ (a method that appears to ad hoc show that we have exhausted all of the irreducible unitary representations), then what good is the Theorem of the Highest Weight? -To put it succinctly: what utility does the Theorem of the Highest Weight provide, and how, if at all, does it fit into the picture of finding a basis for each $\End(\pi)$? - -REPLY [7 votes]: $\DeclareMathOperator\SL{SL}$One direct response to the question of "what does the theorem of the highest weight give us?" is that the highest weight completely determines the eigenvalue by which Casimir acts on that irreducible. (And all of the center of the universal enveloping algebra, as well ….) (This illustrates a semi-tangible form of Harish-Chandra's isomorphism describing the structure of that center.) -The highest weight does also approximately (to my mind) tell how to find a spanning set for the irreds (in general). In small cases it can give a basis, but I think in general the specification of a basis is significantly subtler, … keywords "crystal basis" …. Names include Gelfand, Kashiwara, Lusztig, et al., and relatively recent results (perhaps showing my limited awareness here) from Brubaker, Bump, Friedberg … Chinta, Gunnells, … et al. -One "disappointing" result I do remember is that the ideal structure in Verma modules turns out to be more complicated than the most optimistic conjecture. So it is non-trivial to describe the linear dependence of images of the highest weight vector under lowering operators … unfortunately. For $\SL_2$ it certainly turns out well, and I think for $\SL_3$, but I believe already for $\SL_4$ there is a (complicated) counter-example …. -EDIT: to clarify, e.g., as prompted by @LSpice's remark, … given a choice of positive and negative roots, any repn (finite dimensional or not) with a (unique) highest weight vector is spanned by all the images of that vector under the "lowering operators", that is, the operators in the universal enveloping algebra coming from the negative root-spaces in the Lie algebra. -For $\SL_2$, it is easy and standard to see that the images of the highest weight vector under the (essentially unique) lowering operator are a basis … with the eventually-too-lower image being $0$. (The lowest weight vector ….) There are no non-obvious relations. -Even with $\SL_3$, it is non-trivial to see that the "reasonable" relations among images under negative root-space (lowering) operators are all there are. This amounts to something like showing that the only submodules of a Verma modular with given highest weight (Verma module being the universal module with given highest weight) are again Verma modules. This optimistic idea proves tooooo optimistic, with entailed complications. -EDIT_2: Jacques Dixmier's "Enveloping Algebras" (original in French, too) I believe gives a citation for the failure of submodules of Verma modules to be isomorphic to Verma modules, for $SL_4$.<|endoftext|> -TITLE: Elementary precise estimate of the covering number of euclidean balls by hypercubes -QUESTION [7 upvotes]: I am looking for a straightforward way to upper bound the covering number of a $d$-dimensional euclidean ball by $\ell_\infty$-balls of radius $\varepsilon$, which I will call cubes of sidelength $2\varepsilon$ for clarity. Let us denote this number by $\mathcal N(\varepsilon)$. -An elementary upper bound is to say $\mathcal N(\varepsilon) \leqslant 1 / \varepsilon^d$, as we can cover the whole cube of sidelength 2 with this many small cubes. -However, I know that the precise asymptotic behavior, up to multiplicative and additive constants, is the following -$$ -\log \mathcal N(\varepsilon) \approx \left\{\begin{split} -\frac{1}{\varepsilon^2} \log d \varepsilon^2\quad \text{if} \quad \varepsilon \geqslant 1/\sqrt{d}\\ -d \log \frac{1}{d\varepsilon^2} \quad \text{if} \quad \varepsilon \leqslant 1/\sqrt{d} \, . -\end{split}\right. \ -$$ -Therefore the easy upper bound of $d \log 1/ \varepsilon$ is loose when $\varepsilon$ is large. The change of regime happens when the diagonal of the smaller cubes become comparable to the radius of the larger ball. -The only proof I found, in online teaching notes [1], is convoluted. It consists in first bounding the covering number of $\ell_1$ balls by $\ell_2$ balls, thanks to Maurey's empirical method, and then to appeal to a duality result of [2]. -While the proof is elegant and sophisticated, the inconvenient is that it yields non-explicit constants. Also, it does not seem to take advantage of the ``easiness'' of the problem: covering with hypercubes should not be too hard as the cubes fit well together. I tried counting the number of cubes needed to cover the ball in the natural covering (cutting the cube of sidelength $2$ into small cubes of length $2\varepsilon$ ) but I have no idea how one would do so. For all I know, it might even be that the optimal covering has some overlap between the cubes. -EDIT: A nice improvement on the trivial bound by Rémi Peyre. -Considering the natural covering, it suffices to count the cubes that lie inside the ball of radius $1 + 2 \varepsilon \sqrt{d}$. Therefore -\begin{equation} -\mathcal N(\varepsilon) -\leq \frac{1}{2 \epsilon^d} \mathrm{Vol}\big(B(1 + 2\varepsilon\sqrt{d}) \big) -\leq \frac{1}{\sqrt{d\pi}} (2 \pi e)^{d/2} \Big( 1 + \frac{1}{2\varepsilon \sqrt{d}} \Big)^d \, . -\end{equation} -using a non-asymptotic version of Stirling. Taking logs we obtain something of order $d ( c + \log(1 + 1/(\varepsilon \sqrt{d}) )$. This is an improvement but is still far from the refined bounds above, when $\varepsilon \sqrt{d}$ is large. -[1] http://www.stat.yale.edu/~yw562/teaching/598/lec15.pdf -[2] Duality of metric entropy, S. Artstein, V. Milman, and S. J. Szarek -https://annals.math.princeton.edu/wp-content/uploads/annals-v159-n3-p07.pdf - -REPLY [3 votes]: Once you know the answer, the proof is a trivial induction on $d$. We will show that for some positive constants $A,B,K$ (to be chosen in the end) one can cover the ball of radius $r$ by -$$ -F(d,R^2)=\left(1+\frac{d}{R^2}\right)^{BR^2}+\left(\frac{AR^2}{d}\right)^{d/2}=F_1(d,R^2)+F_2(d,R^2). -$$ -cubes with sidelength $2K$ where $R^2=r^2+K^2$. If you are not too picky about constants, this is equivalent to what you have requested. -We will notice first of all that $F(d,R^2)=0$ for $RAR^2$, the comparison is trivial. Finally, if $R^2\le d\le AR^2$, we have at most $A^{AR^2/2}$ on the left and at least $2^{BR^2/2}$ on the right, so if $A^A\le 2^B$, we are fine. This tells the relation between $A$ and $B$ (both large) and $K$ is chosen last. -The end. -P.S. If you think a bit, you see that such induction is doomed to work: no fancy cover is possible because for the cubes of twice smaller size the covers of our slices are disjoint (the cube just cannot stretch enough in the vertical direction), so the stupid algorithm (cover slice by slice reducing the dimension by 1, then do the same for each slice of slice, etc.) is actually the best.<|endoftext|> -TITLE: rational points of a hyperelliptic curve of genus 3 -QUESTION [5 upvotes]: Let $K=\mathbb{Q}(\sqrt{-1}).$ I have the following hyperelliptic curve of genus 3: -$$ C : y^2 = (x^2-x+1)(x^6+x^5-6x^4 -3x^3+14x^2-7x+1) $$ -I want to find $C(K)$. My first attempt was to compute the Jacobian of the curve $J(\mathbb{Q})$. By Magma, $J(\mathbb{Q})$ has rank 0 and the torsion subgroup is isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/52\mathbb{Z}.$ (generated by the root of one of the factors on the right side of the hyperelliptic curve equation and the difference of two points at infinity.) One can see that the reduction of $J$ modulo 5 has this group type. -In order to find the rank of $J(K)$, I need to compute the rank of the quadratic twist $J^{-1}(\mathbb{Q})$. According to Magma, it is impossible to obtain an upper bound for the rank so I got unlucky. When the rank is positive, I don't think it is computationally feasible to extract $K$-points on the curve. Is there another method for finding rational points generally on a hyperelliptic curves of even degree and odd genus $3?$ Any help would be appreciated. - -REPLY [16 votes]: It turns out that $C(K) = C(\mathbb Q) = \{\infty_+, \infty_-, (0,1), (0,-1), (1,1), (1,-1)\}$. -To see this, consider a point $P \in C(K)$ and write $\bar{P}$ for its image under the nontrivial automorphism of $K$. Then $P + \bar{P}$ is an effective divisor of degree 2 on $C$, which is defined over $\mathbb Q$ (this is what Xarles is alluding to in his comment). So for any effective rational divisor $D$ of degree 2 on $C$, the linear equivalence class of $P + \bar{P} - D$ is a rational point on $J$. As you state correctly, $J(\mathbb Q)$ is finite and is isomorphic to ${\mathbb Z}/2 \times {\mathbb Z}/52$; it is generated by differences of rational points on $C$ together with the point of order 2 corresponding to the given factorization of the polynomial in the equation, which makes it easy to enumerate all these 104 points (in Magma, say). Magma represents these points as effective divisors in Mumford representation (minus a multiple of one of the points at infinity). So we look at all these points and check if these divisors are of degree 2 with points in the support defined over $K$. The only such divisors we find are sums of two rational points. This shows that there are no ``exceptional'' $K$-points, i.e., no $K$-points with $x$-coordinate not in $\mathbb Q$. If $x(P) \in \mathbb Q$ and $P \notin C(\mathbb Q)$, then $P + \bar{P}$ is in the class of pull-backs of a point on ${\mathbb P}^1$; such points all give the same point on $J$. To deal with these, we observe that such points correspond bijectively to rational points on the quadratic twist $C^{(-1)}$. However, there are no such points, since the polynomial on the right hand side of the equation of $C$ is always positive. This finishes the proof. -The business with the exceptional points is related to the fact that the map from the symmetric square $C^{(2)}$ (there is an unfortunate clash of notations with the quadratic twist) to $J$ is not injective: there is a $\mathbb P^1$ contained in $C^{(2)}$ coming from pulling back points on $\mathbb P^1$ under the hyperelliptic double cover, and this $\mathbb P^1$ gets contracted to one point, whereas the map is injective otherwise. -Since every quadratic point on $C$ gives rise to a rational point on $C^{(2)}$, the computation above actually shows that there are the following exceptional quadratic points on $C$. -$$ (\tfrac{1\pm\sqrt{-3}}{2}, 0), (\pm\sqrt{2}, \pm 4\sqrt{2}-5), - (\pm\sqrt{2}, \mp 4\sqrt{2}+5), - (\tfrac{2\pm\sqrt{2}}{2}, \tfrac{-5\mp4\sqrt{2}}{4}), - (\tfrac{2\pm\sqrt{2}}{2}, \tfrac{5\pm4\sqrt{2}}{4}), - (-1\pm\sqrt{2}, -11\pm8\sqrt{2}), (-1\pm\sqrt{2}, 11\mp8\sqrt{2}) -$$ -(The signs correspond in each pair of coordinates.)<|endoftext|> -TITLE: Size of largest square divisor of a random integer -QUESTION [12 upvotes]: Let $x$ be an integer picked uniformly at random from $1 \ldots N$. Write $x = r^2 t$ where $t$ is square-free. How does the expected value of $r$ scale with $N$? Is anything known about the variance of $r$? - -REPLY [23 votes]: The asymptotic frequency of square-free integers is known to be $6/\pi^2$, see [1]. -Denote by $P_n$ the uniform distribution on $[1,n]$ and by $E_n$ the corresponding expectation. Then -$$E_n(r)=\sum_{k \le \sqrt{n}} k P_n(r=k) \sim -\sum_{k \le \sqrt{n}} k \cdot \frac{1}{k^2} \cdot\frac{6}{\pi^2} \sim \frac{3}{\pi^2} \log(n) \,,$$ -where $A \sim B$ means that $A/B \to 1$ as $ n \to \infty$. -(In particular for $n=10^{10}$ the mean $E_n(r)$ is close to 7.) -Also, -$$E_n(r^2)=\sum_{k \le \sqrt{n}} k^2 P_n(r=k) \sim -\sum_{k \le \sqrt{n}} \frac{6}{\pi^2} \sim \frac{6\sqrt{n}}{\pi^2} \,, -$$ -so the variance of $r$ is asymptotic to $6\sqrt{n}/\pi^2$ as well. -[1] https://en.wikipedia.org/wiki/Square-free_integer - -REPLY [6 votes]: Just a little empirical data up to $N=10^{10}$, -three superimposed random runs. -Growing roughly linearly w.r.t. $\log_{10} N$ within that range.<|endoftext|> -TITLE: Heights of well-founded parts of models of $\mathsf{ZFC}$ -QUESTION [10 upvotes]: Work in the theory $\mathsf{ZFC}$ + "Every set is contained in some transitive model of $\mathsf{ZFC}$." -My question is the following: which ordinals are the heights of the well-founded parts of models of $\mathsf{ZFC}$? -For what follows, let $\mathsf{wfh}(M)$ denote the height of the well-founded part of $M$. -In the countable case, the answer is simple: a countable ordinal $\alpha>\omega$ is the height of the well-founded part of a (WLOG countable) model of $\mathsf{ZFC}$ iff $\alpha$ is admissible. The left-to-right direction holds since the well-founded part of any admissibile set is itself admissible; the interesting direction is right-to-left, where Barwise compactness comes into play. -In the uncountable case the left-to-right direction of the argument above still works, but the right-to-left direction breaks since we lose Barwise compactness. In fact, it's consistent with $\mathsf{ZFC}$ that there are admissible ordinals which are not of the form $\mathsf{wfh}(M)$ for any $M\models\mathsf{ZFC}$: - -Suppose $M\models\mathsf{ZFC}$ and $\mathsf{wfh}(M)=\omega_1^L$. Then we have (via mild abuse) that $L_{\omega_1^L}\subseteq M$. Since $L_{\omega_1^L}$ is locally countable, this means that $\omega_1^M$ must be ill-founded. Picking some ill-founded $M$-ordinal $\alpha<\omega_1^M$, we get in $M$ a bijection $\alpha\rightarrow\omega$ - which restricts externally to an injection $\omega_1^L\rightarrow\omega$. So we must have $\omega_1^L<\omega_1$. - -More generally, whenever $\kappa$ is an infinite cardinal such that $\kappa^+=(\kappa^+)^L$ we get that $\kappa^+\not=\mathsf{wfh}(M)$ for all $M\models\mathsf{ZFC}$ - just run the argument above with $\kappa^+$ in place of $\omega_1$ and "locally size-$\le\kappa$" in place of "locally countable." But this doesn't give a $\mathsf{ZFC}$ result since it's consistent that $L$ never computes successor cardinals correctly. So it's not even obvious to me that $\mathsf{ZFC}$ disproves "The possible values of $\mathsf{wfh}(M)$ are exactly the admissible ordinals." In particular note that it is consistent that $\omega_1$ is the height of the well-founded part of a model of $\mathsf{ZFC}$, since in fact it's consistent that $L_{\omega_1}\models\mathsf{ZFC}$ outright. - -REPLY [10 votes]: The answer to the question for ordinals of uncountable cofinality is provided by the following theorem, established by Magidor, Stavi, and Shelah, in their paper On the standard part of nonstandard models of set theory, J. Symbolic Logic 48 (1983), no. 1, 33–38. -Note that the ordinal defined in the question as $\mathsf{wfh}(M)$ (well-founded height of $M$) is referred to below as the standard part of $M$. -Theorem. For an ordinal $\alpha$ of uncountable cofinality the following are equivalent: -(A) $\alpha$ is the standard part of a nonstandard model of ZF. -(B) $\alpha$ is the standard part of a nonstandard model of KP. -(C) There exists $\gamma > \alpha$ such that $\gamma$ is a limit of ordinals which are models of ZF, and $\alpha$ has the tree property in $L_{\gamma}$. -(D) There exists $\gamma > \alpha$ such that $\gamma$ is a limit of admissible ordinals and has the $\alpha$ has the tree property in $L_{\gamma}$. -Regarding the case when $\alpha$ has countable cofinality: in the introduction to the aforementioned paper, the authors write the following (in what follows, Friedman's theorem refers to the fact that countable admissible ordinals coincide with the heights of well-founded parts of countable models of ZF). - -Friedman's theorem can be generalized to other ordinals, and the proof imitated, using some decomposition of $\alpha$ into small sets. (See [4].) All these generalizations handle just $\alpha$'s of cofinality $\omega$. - -In the above [4] refers to an abstract (published in the Notices of AMS, 1980), which later blossomed into another paper of Magidor, Shelah, and Stavi: -Countably decomposable admissible sets, Ann. Pure Appl. Logic 26 (1984), no. 3, 287–361.<|endoftext|> -TITLE: How to estimate the integral involving the distance function -QUESTION [6 upvotes]: Let $\Omega\subset\mathbb{R}^n$ be an open bounded domain with smooth boundary. Consider the following integral: -$$I(t)=\int_{\Omega}e^{-\frac{d^2(y,\partial\Omega)}{t}}{\rm d}y.$$ -My problem is how to calculate or estimate it (when $t\rightarrow 0^+$)? -We have tried the Coarea formula and got -\begin{align} -I(t)&=\int_{\Omega} e^{-\frac{d^2(y,\partial\Omega)}{t}}|\nabla d(y,\partial\Omega)|{\rm d}y\\ -&=\int_{0}^{+\infty} e^{-\frac{s^2}{t}}{\rm d}s\int_{\{x\in\Omega|d(x,\partial\Omega)=s\}}{\rm d}S_{n-1}(y)\\ -&=\int_{0}^{+\infty} e^{-\frac{s^2}{t}}|\{x\in\Omega|d(x,\partial\Omega)=s\}|_{n-1}{\rm d}s -\end{align} -where $|\cdot|_{n-1}$ is the $(n-1)$-dimensional Lebesgue measure. But we're stuck with how to estimate the measure. Maybe the condition is too weak? If you have some ideas you can share your answer even if you add some more conditions in $\Omega$. Thank you. - -REPLY [9 votes]: For small $t$ the dominant contribution comes from the boundary; starting from a point $y_0$ on the boundary and integrating inwards in a perpendicular direction we have $\int_{0}^\infty e^{-r^2/t}dr=\tfrac{1}{2}\sqrt{\pi t}$. Multiplication by the area $S$ of the boundary gives the estimate -$$I\rightarrow \tfrac{1}{2}S\sqrt{\pi t}.$$ -As a check, for a disc of radius $R$ one has -$$I=\pi ^{3/2} R \sqrt{t} \,\text{erf}(R/\sqrt{t})+\pi t \left(e^{-R^2/t}-1\right)$$ -which for small $t$ tends to $\pi R\sqrt{\pi t}+{\cal O}(t)$. -Similarly, for an $L\times L$ square one has -$$I=2 L \sqrt{\pi t} \,\text{erf}\left(\tfrac{1}{2}L/\sqrt{t}\right)+4 t \left(e^{-L^2/4 t}-1\right)\rightarrow 2L\sqrt{\pi t}+{\cal O}(t).$$ - -REPLY [8 votes]: The estimate you seek is reminiscent of H. Weyl's tube formula. I will give you some pointers referring for more details to section 9.3.5. of these lectures. -Denote by $r$ the distance to $\newcommand{\pa}{\partial}$ $\pa \Omega$ $\newcommand{\bn}{\boldsymbol{n}}$ and by $\bn$ the innner pointing unit normal of $\pa \Omega$. $\newcommand{\bp}{\boldsymbol{p}}$ There exists $r_0>0$ such that the map $\newcommand{\bR}{\mathbb{R}}$ -$$ -\pa \Omega \times [0,r_0)\ni(\bp,r)\stackrel{\Phi}{\longmapsto} \bp+r\bn\in \bR^n -$$ -is a diffeomorphism onto the region $\DeclareMathOperator{\dist}{dist}$ -$$A_{\rho_0}:=\big\{\; x\in \bar{\Omega};\;\;\dist(x,\pa \Omega)<\rho_0\;\big\}.$$ -Then -$$\int_{A_{\rho_0}} e^{-r^2/t} dx=\int_{\pa\Omega\times [0,r_0)}\Phi^*\big(e^{-\rho^2/t} dx\Big)=\int_{\pa\Omega\times [0,r_0)}e^{-\rho^2/t}\Phi^*(dx). -$$ -The pullback of the Euclidean volume form $dx$ via the map $\Phi$ is described explicitly in the above reference. It has the form $\newcommand{\eQ}{\mathscr{Q}}$ $\DeclareMathOperator{\tr}{tr}$ -$$ -\Phi^*(dx)= \eQ_\bp(r)dV_{\pa \Omega}(\bp) dr, -$$ -where, for each $\bp\in \pa \Omega$ $(r)$, $\eQ_\bp$ is a polynomial of degree $n-1$ in $r$ -$$ -\eQ_\bp(r)=\sum_{j=0}^{n-1}c_j(\bp) r^j=\det\big(1-r S_\bp\big), -$$ -where $S_{\bp}$ is the second fundamental form of the hypersurface $\pa \Omega$ at the point $\bp$ defined in terms of the inner normal $\bn$. More precisely if $(x^i)$ are local coordinates on $\pa \Omega$ near $\bp$, then -$$ -S_\bp(\pa_{x^i},\pa_{x^j})=\big(\; \bn(\bp),\pa^2_{x^ix^j}\bn(\bp)\;\big), -$$ -where $(-,-)$ denotes the canonical inner product on $\bR^n$. -Thus -$$\int_{A_{\rho_0}} e^{-r^2/t} dx=\int_0^{r_0} e^{-r^2/t}\left(\int_{\pa\Omega} \eQ_{\bp}(r)dV_{\pa \Omega}(\bp)\right) dr. $$ -The integral -$$ -K(r):=\int_{\pa\Omega} \eQ_{\bp}(r)dV_{\pa \Omega}(\bp) -$$ -appears in Weyl's tube formula and, more precisely $\newcommand{\bom}{\boldsymbol{\omega}}$ (see Eq. (9.3.18) in the above reference) -$$ -K(r)=\sum_{k=0}^{n-1} (-1)^{n-1-k}\bom_{n-k}r^{n-k}\mu_k(\Omega), -$$ -$$ -=\bom_0\mu_{n-1}(\Omega)r-\bom_1\mu_{m-2}(\Omega)r^2+\cdots +(-1)^{n-1}\bom_n\mu_0(\Omega)r^n, -$$ -where $\bom_m$ denotes the volume of the $m$-dimensional Euclidean unit ball and $\mu_k(\Omega)$ is the curvature measure of degree $k$. (You need to be careful about various sign conventions. In the above reference the second fundamental form is defined using the outer normal.)$\DeclareMathOperator{\vol}{vol}$ For example -$$ -\mu_{n-1}(\Omega)= \frac{1}{2} \vol_{n-1}(\pa\Omega), -$$ -$$ -\mu_{n-2}(\Omega)=\frac{1}{2\pi}\int_{\pa \Omega} \tr S_\bp dV_{\pa\Omega}(\bp), -$$ -where $\tr S_\bp$ is the mean curvature of $\pa\Omega$ at $\bp$. Also, $\mu_0(\Omega)$ is the Euler characteristic of $\Omega$. -The asymptotics of -$$ -J(t)=\int_0^{r_0} e^{-r^2/t}K(r) dr, -$$ -can be determined easily by making the change in variables $s=r^2/t$, $r=\sqrt{st}$ so that -$$ -J(t)=\frac{\sqrt{t}}{2}\int_0^{r_0^2/t} e^{-s}K(\sqrt{st}) s^{-1/2} ds -$$ -$$ -=\frac{1}{2}\sum_{k=0}^{n-1}(-1)^{n-1-k}t^{\frac{n-k}{2}}\mu_k(\Omega)\int_0^{r_0^2/t} e^{-s} s^{\frac{n-k}{2}-1} ds. -$$ -Observe that as $t\searrow 0$ -$$ -\int_0^{r_0^2/t} e^{-s} s^{\frac{n-k}{2}-1} ds=\;\underbrace{\int_0^{\infty} e^{-s} s^{\frac{n-k}{2}-1} ds}_{\Gamma\big( \frac{n-k}{2}\big)}\;+ O\big(t^{N}\big),\;\;\forall N>0. -$$ -Hence -$$ -J(t)=\frac{1}{2}\sum_{k=0}^{n-1}(-1)^{n-1-k}t^{\frac{n-k}{2}}\Gamma\Big(\;\frac{n-k}{2}\;\Big)\mu_k(\Omega) + O\big(t^{N}\big),\;\;\forall N>0. -$$ -Finally -$$ -|I(t)-J(t)|=\int_{\dist(x,\pa \Omega)>r_0} e^{-\dist(x,\pa\Omega)^2/t} dx -$$ -$$ -\leq e^{-r_0^2/t}\vol(\Omega)=O(t^{N}),\;\;\forall N>0. -$$ -Hence - -$$ I(t)=\frac{1}{2}\sum_{k=0}^{n-1}(-1)^{n-1-k}t^{\frac{n-k}{2}}\Gamma\Big(\;\frac{n-k}{2}\;\Big)\mu_k(\Omega) + O\big(t^{N}\big),\;\;\forall N>0.$$ - -The leading term of this yields the estimate indicated by Carlo Beenakker. -About the tube formula The tube formula for smooth domains or convex domains or more generally sets with positive reach are both special cases of the very general kinematic formulas. The Brunn-Minkowski formula is also a special case of the kinematic formula. - -REPLY [3 votes]: \begin{aligned} -& \int_{0}^{+\infty} e^{-\frac{s^{2}}{t}} \mathcal{L}_{n-1}(\{x \in \Omega \mid d(x, \partial \Omega)=s\}) d s \\ -=& \int_{0}^{+\infty} e^{-\frac{s^{2}}{t}} \mathcal{L}_{n-1}(\partial\Omega) d s+\int_{0}^{+\infty} e^{-\frac{s^{2}}{t}}\left(\mathcal{L}_{n-1}\left(\left\{x\in \Omega \mid d(x, \partial \Omega)=s\right)-\mathcal{L}_{n-1}(\partial \Omega)\right) d s\right. -\end{aligned} -Let us estimate the error term. It split into the estimate far from the boundary and the distortion estimate of the level set near the boundary. -The estimate far from the boundary: -$$ -\int_{0}^{+\infty} e^{-\frac{s^{2}}{t}} \mathcal{L}_{n-1}(\partial \Omega) d \delta=\frac{1}{2} \sqrt{\pi t} \mathcal{L}_{n-1}(\partial \Omega) -$$ -$$ -\int_{0}^{A} e^{-\frac{s^{2}}{t}} \mathcal{L}_{n-1}(\partial \Omega) d s=\frac{1}{2} \sqrt{\pi t} \mathcal{L}_{n-1}(\partial \Omega)+O\left(\sqrt{t} \cdot e^{-\frac{A^{2}}{t}} \mathcal{L}_{N-1}{(\partial \Omega)}\right) -$$ -The distortion estimate of the level set near the boundary: -Then estimate the distortion of $\mathcal{L}_{n-1}\left(\{x \in \Omega \mid d(x, \partial \Omega)=s\})-\mathcal{L}_{n-1}(\partial \Omega)\right.$ -$$ -\left.\mathcal{L}_{n-1}(\{x \in \Omega \mid d(x, \partial \Omega)=s\}\right)-\mathcal{L}_{n-1}(\partial \Omega)=O\left(s^{n-1}\right)=\frac{2 \pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)} s^{n-1}+o\left(s^{n-1}\right) -$$ -This estimate come from a similar way in proving Brunn-Minkowski expansion, which give the $\mathcal{L}_{n}(A+B_r)\sim \sum_{i=0}^n \lambda_i\mathcal{L}_{n-i}(A)\mathcal{L}_{i}(B_r)$ at least for $A$ is a convex domain. and drop the high order term $\mathcal{L}_{n-i}(A)\mathcal{L}_{i}(B_r), i\geq2$, which is minor when $t\to 0$. -Combine together: -Now a suitable choice of $s$ relies on $t$ will given a reasonable estimate for the error term., -$$E(t)=\int_{0}^{+\infty} e^{-\frac{s^{2}}{t}} \mathcal{L}_{n-1}(\{x \in \Omega \mid d(x, \partial \Omega)=s\}) d s-\frac{1}{2}\sqrt{\pi t}\mathcal{L}_{n-1}(\partial \Omega)$$ -In fact, a suitable $s_0$ is near the maximum of the minimum. -$$ -s_0=\max _{s} \min\left\{\frac{2 \pi^{\frac{1}{2}}}{\Gamma\left(\frac{n}{2}\right)} s^{n-1}+o\left(s^{n-1}\right), \quad \sqrt{t} \cdot e^{-\frac{s^{2}}{t}} \cdot \mathcal{L}_{n-1}(\partial \Omega)\right\} -$$ -I have not finished the calculation of the explicit order(the best choice of $s_0$), but I think this can give some good enough order of error term in the general case. -Then at least for $n\geq 3$, $\Omega\subset \mathbb{R}^n$, take $s_0\sim t^{\frac{1}{2}+\delta}, \delta>0$, we know $E(t)=o(t^{\frac{1}{2}+\delta^{'}}), \delta^{'}>0$. -In dimension 2 case it seems we even do not have $I \rightarrow \frac{1}{2} S \sqrt{\pi t}$ because the error term is out of control.<|endoftext|> -TITLE: Are there centrally-symmetric self-dual polytopes in dimension $d> 4$? -QUESTION [8 upvotes]: A convex polytope $P\subset\Bbb R^d$ is centrally symmetric if $-P=P$. It is self-dual (or better, self-polar?) if its polar dual $P^\circ$ is congruent to $P$, that is, there is a map $X\in\mathrm O(\smash{\Bbb R^d})$ with $\smash{P^\circ}=XP$. - -Question: Are there centrally symmetric self-dual polytopes in dimension $d>4$? - -Such exist in dimension $d=2$ and $d=4$: - -for $d=2$ we have the regular 2n-gons, -for $d=4$ we have the regular 24-cell. - -REPLY [5 votes]: There are centrally symmetric self-dual polytopes in every dimension. This follows from Proposition 3.9 in -Reisner, S., Certain Banach spaces associated with graphs and CL-spaces with 1- unconditional bases, J. Lond. Math. Soc., II. Ser. 43, No. 1, 137-148 (1991). ZBL0757.46030. -Moreover, in dimension $\geqslant 3$ the matrix $X$ can be chosen to be a permutation matrix. -Here is an example in dimension $3^d$ for every $d$. Start with Sztencel-Zaramba polytope $P$. This is the unit ball for the norm on $\mathbf{R}^3$ -$$ \|(x,y,z)\| = \max \left( |y|+|z|, |x|+\frac 12 |z| \right)$$ -whose dual norm satisfies -$$ \|(x,y,z)\|_* = \|(z,y,x)\|. $$ -We may now define inductively a sequence $\|\cdot\|_d$, which is norm on $\mathbf{R}^{3^d}$ (identified with $\mathbf{R}^{3^{d-1}}\times\mathbf{R}^{3^{d-1}}\times\mathbf{R}^{3^{d-1}}$). Chose $\|\cdot\|_1$ to be above norm, and use the recursive formula -$$ \|(x,y,z)\|_{d+1} = \|( \|x\|_d ,\|y\|_d , \|z\|_d )\|_1 .$$ -One checks by induction that there is a permutation matrix which maps the unit ball onto its polar. -To visualize the polytope $P$ you may use the Sage code -p1 = Polyhedron(vertices=[[0,1,1],[0,1,-1],[0,-1,1],[0,-1,-1],[1,0,1/2],[1,0,-1/2],[-1,0,1/2],[-1,0,-1/2]]) -p1.projection().plot()<|endoftext|> -TITLE: Density of Ramsey subsets of $\omega$ -QUESTION [5 upvotes]: For any set $X$ let $[X]^2=\{\{x,y\}:x\neq y \in X\}$. The starting point of this question is the following statement that follows from a more general theorem by Ramsey: - -If $\pi:[\omega]^2\to\{0,1\}$ is any map, then here is an infinite set $S\subseteq \omega$ such that the restriction $\pi|_{[S]^2}$ is constant. - -We call an infinite set $S$ with the above property a Ramsey set for the map $\pi:[\omega]^2\to\{0,1\}$. For $A\subseteq \omega$ we define its upper density $d(A)\in [0,1]$ by $$d(A) = \lim\sup_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}.$$ -Question. Given a map $\pi:[\omega]^2\to\{0,1\}$, is there necessarily a Ramsey set $S\subseteq \omega$ with $d(S)>0$? - -REPLY [8 votes]: Decompose $\omega$ into the disjoint union of the sets $I_k$ where $I_k=[k!,(k+1)!-1]$. Let $f(x,y)$ be 1 if $x,y$ are in distinct intervals, otherwise 0. It is easy to see that each homogeneous set for 1 is finite, for 0 has zero density. - -REPLY [6 votes]: If we define $\pi: [\mathbb N]^2 \rightarrow \{0,1\}$ randomly (say each $\pi(a,b)$ is determined by a coin flip), then almost surely there is no set $S$ with $d(S) > 0$ that is Ramsey for $\pi$. In fact, it is almost surely true that every $S$ with $d(S) > 0$ contains an induced isomorphic copy of the randomly colored infinite graph. -Even more: for a random coloring $\pi$ of $[\mathbb N]^2$, there is almost surely no set $S$ with $\sum_{n \in S \setminus \{0\}} \frac{1}{n} = \infty$ that is Ramsey for $\pi$. In fact, it is almost surely true that every such $S$ contains an induced copy of every coloring of every finite graph. (But in this case, "finite" cannot be improved to "infinite" as above.) -These results can be found in Section 2 of my paper "Which subsets of the infinite random graph look random?" (Mathematical Logic Quarterly 64 (2018), pp. 478-486), available here.<|endoftext|> -TITLE: Vectors with minimal Hamming weight in a rational vector space? -QUESTION [7 upvotes]: Suppose given $n\ge 1$ and a subspace $U$ in $\mathbb{Q}^n$. It is given as $\mathbb{Q}$-span of certain known vectors. -For $x \in U$, we let the Hamming weight of $x$ be the number of its nonzero entries. -Is it possible to find an element of minimal Hamming weight in $U\smallsetminus\{0\}$? Is there an algorithm? -(All I could find is concerned with the case of vector spaces over finite fields. Or, on the other hand, with vectors of minimal Euclidean length in lattices.) - -REPLY [3 votes]: Here is an Integer Linear Programming approach to this problem. -First, let's multiply given vectors by a suitable integer to make them all having integer components. Second, we notice that if certain rational coefficients deliver the minimum Hamming weight, then by scaling them, we can obtain integer coefficients delivering the same weight. Therefore, we can focus on the problem of finding an integer linear combination of integer vectors with the smallest nonzero number of nonzero components. -Let $v_1,\dots,v_m$ be given integer vectors. Let $c_i$ for $i\in\{1,\dots,m\}$ be integer variables corresponding to the coefficients in a linear combination. For each component $j\in\{1,\dots,n\}$, we further introduce two binary variables $p_j,q_j\in\{0,1\}$ and three inequalities: -$$\sum_{i=1}^m v_{ij} c_i \geq p_j - Mq_j,$$ -$$\sum_{i=1}^m v_{ij} c_i \leq -q_j + Mp_j,$$ -$$p_j + q_j \leq 1,$$ -where $M$ is a large positive constant (chosen empirically). The third inequality here restricts values to three possible cases: $(p_j,q_j)=(1,0)$ when the $j$-th component in the linear combination is positive; $(p_j,q_j)=(0,1)$ when the $j$-th component is negative; and $(p_j,q_j)=(0,0)$ when the $j$-th component is zero. Notice that the large value of $M$ when it comes with a nonzero coefficient makes the corresponding inequality silent (automatically satisfied). -Next, we exclude the zero linear combination by requiring -$$\sum_{j=1}^n p_j + q_j \geq 1.$$ -Finally, since $\sum_{j=1}^n p_j + q_j$, in fact, equals the weight of the linear combination, our objective is -$$\text{minimize}\quad \sum_{j=1}^n p_j + q_j.$$<|endoftext|> -TITLE: Local complexity of triangulations -QUESTION [13 upvotes]: According to the introduction in -Cooper, D.; Thurston, W. P., Triangulating 3-manifolds using 5 vertex link types, Topology 27, No. 1, 23-25 (1988). ZBL0656.57004. - -It is known that, for any dimension $n$, there is a finite set of link types such that every $n$-manifold has a triangulation in which the link of each vertex is in this set. - -(I assume the statement is about PL manifolds.) -What is a proof of or a reference to this known result? -Edit. There is a related discussion by Florian Frick here, but he only gets a bound on the valence of "ridges" (codimension 2 faces). I do not see how to generalize his sketch to faces of higher codimension. (Actually, I cannot follow his sketch.) - -REPLY [2 votes]: Theorem 3.37 of Florian Frick's thesis gives a sketch of the proof of statement in the question for smooth manifolds. At first sight, it only uses the Whitney embedding theorem and transversality, both of which are available for PL manifolds (since you get to pick the embedding you may assume it has a normal microbundle).<|endoftext|> -TITLE: Relative Entropy and p-norm -QUESTION [8 upvotes]: I asked this question on StackExchange but could not get any answer, therefore, I am posting it here. -I am currently reading the book "A Dynamical Approach to Random Matrix Theory". The authors introduce the notion of relative entropy and remark that relative entropy is a weaker measure of the distance between probability measures than the $L^p$ distance for any $p>1.$ The following inequality (authors call it 'elementary' and therefore leave the proof as an exercise) is mentioned as a justification for this claim: -$$\int f\log f\;d\mu\le 2\left[\int |f-1|^p\; d\mu\right]^{1/p}+\frac{2}{p-1}\int |f-1|^p\; d\mu,$$ -where $f\;d\mu$ and $\mu$ are a probability measure. -I have no idea how to get this inequality. I thought of using -$$\int f\log f\;d\mu= \int (f-1)\log f\;d\mu+ \int \log f\;d\mu,$$ -and then apply H"older's inequality to the first part but it doesn't seem to give me what I want. Any hint/help is appreciated. - -REPLY [16 votes]: The argument below is not very elegant,but it is, indeed, a standard exercise. Let $g=\max(f-1,0)$. We shall prove that -$$ -f\log f\le 2g+\frac 2{p-1}g^p\,. -$$ -The integration and Holder then give the result immediately -If $f<1$, there is nothing to prove ($LHS<0=RHS$). -If $0\le g\le 1$, then -$$ -f\log f=(1+g)\log(1+g)\le (1+g)g\le 2g. -$$ -Now assume that $g>1$. Let $g=e^t$, $t>0$. -Then -$$ -f\log f=(1+g)\log(1+g)\le 2g\log(2g)=2g(\log 2+\log g) -\\ -\le -2e^t(1+t)\le 2e^t+2te^t=2g+2\frac{(p-1)t}{p-1}e^t -\\ -\le 2g+2\frac{e^{(p-1)t}}{p-1}e^t=2g+2\frac{e^{pt}}{p-1}=2g+2\frac{g^p}{p-1}\,. -$$ -(we used that $e^u\ge u$ for all $u$ and, in particular, for $u=(p-1)t$).<|endoftext|> -TITLE: Different definitions of condensed sets -QUESTION [7 upvotes]: The $\kappa$-condensed sets are defined as the sheaves on the site of profinite spaces of cardinality less than $\kappa$ (with $\kappa$ an uncountable strong limit cardinal) with morphisms the continuous maps, and whose covers are finite collections of jointly surjective maps. -I understand that you get the same category of condensed sets if you instead take the larger category of compact hausdorff spaces of cardinality less than $\kappa$, or the smaller category of Stone-Cech compactifications of sets of cardinality less than $\kappa$. -What happens if you use the "same site" on the category of all topological spaces of cardinality less than $\kappa$? Does this also yield the condensed sets? If not, what goes wrong? - -REPLY [10 votes]: The question is not precise enough: it depends which topology you chose on the category of topological spaces. You will get the same category of sheaves if you are in a situation where Grothendieck's comparison lemma applies. -That is, you need to chose a topology on the category of all topological spaces, that induces the correct topology when restricted to compact spaces, and such that for each topological space X, the set of all maps $K \rightarrow X$ with $K$ compact are a covering. -There definitely is a (actually unique) topology that satisfies this, and others that don't, and this is a necessary and sufficient conditions to get the equivalence of the categories of sheaves. -The unique topology for which this is going to work can be described as follow: -A collection of maps $X_i \to X$ is a covering (generates a covering sieve) if and only for each profinite space K and each continuous map $K \to X$, there is a finite covering $K_j \to K$ by compact spaces $(K_j)_{j \in J}$ such that each map $K_j \to X$ factors through one of the $X_i$.<|endoftext|> -TITLE: Representing graphs as unions of trees -QUESTION [11 upvotes]: Suppose that $G$ is a simple connected (infinite) locally finite graph (i.e. each vertex has finite valence). Is $G$ a union of finitely many trees? If not, does it hold for graphs $G$ of bounded valence (i.e. there exists $d$ such that the valence of each vertex is $\le d$)? - -REPLY [14 votes]: (Unbounded valence.) No. Take disjoint complete graphs on $1,2,\ldots$ vertices and add some edges between them just to make this connected. Since you need at least $n/2$ trees to cover $K_n$, you can not cover by finitely many trees. - -(Bounded valence.) You may cover such graph by $d$ forests (even if multiple edges are allowed, but certainly no loops). Indeed, the graph is obviously countable. Let $v_1,v_2,\ldots,$ be its vertices. Assume that for each $n$ you may cover the subgraph on $\{v_1,\ldots,v_n\}$ by $d$ forests $L_1^n,\ldots,L_{d}^n$. For each edge $e$ denote $f_n(e)=s\in \{1,2,\ldots,d\}$ if $e\in L_s^n$. By diagonal argument, there exists a subsequence $(n_k)$ such that $f_{n_k}(e)$ is eventually constant for all edges $e$. Denote this constant $f_{\infty}(e)$ and denote $L_i^\infty=\{e: f_{\infty}(e)=i\}$, $i=1,\ldots,d$. These are our forests. - - -After this compactness abstract nonsense is said, we should prove a statement for a finite graph. There is a Nash-Williams criterion: a finite graph $G=(V,E)$ is a union of $m$ forests if and only if for any $U\subset V$ the number of edges joining vertices of $U$ is at most $m(|U|-1)$. If all degrees are at most $d$, this number of edges is at most $d|U|/2\leqslant d(|U|-1)$ when $|U|\geqslant 2$ (and there is nothing to check for $|U|=1$). -Now, if your trees are allowed to have common edges, just add edges to these forests to make them trees. If they must be disjoint, the statement is again false: take an infinite path $v_1v_2\ldots$ and replace each edge $v_{2i-1}v_{2i}$ by a 4-cycle $v_{2i-1}u_iv_{2i}w_i$. - -REPLY [6 votes]: If $G$ has colouring number or degeneracy at most $d$ (i.e there is a well=order $<^*$ such that for every vertex $v$ the number of neighbours $w$ of $v$ such that $w <^* v$ is less than $d$) then there is a cute argument (I think originally due to Erdős and Hajnal see this paper) that $G$ is the union of at most $d-1$ forests. -The idea is to rainbow colour for each $v$ the set $E_v$ of `backwards' edges at $v$, that is $E_v = \{ (u,v) \in E(G) \colon u <^* v\}$, with the colours $[d-1]$, which is possible since each $E_v$ has size at most $d-1$. We claim that each colour class is acyclic, and hence a forest. Indeed, if some colour class contains a cycle $C$, then since $C$ is finite it contains a largest vertex $v$ in the well-order. However then both edges in $C$ meeting $v$ must be in $E_v$, contradicting the fact that $C$ was monochromatic. -In fact, for infinite $\lambda$, having colouring number $\lambda^+$ is equivalent to being expressible as the union of $\lambda$ many forests, as shown in the same paper. The converse is not quite true for finite $\lambda$, although if you are expressible as the union of $\lambda$ many forests with $\lambda$ finite another nice compactness argument of Erdős and Hajnal shows that you have colouring number at most $2\lambda$. One should be able to build a counterexample to your other statement from this I expect, by exhibiting a locally finite graph with colouring number $\aleph_0$. For example a union of larger and larger complete graphs joined in a ray should probably work. -If you want the trees in your statement to be disjoint, then I don't know if there is a nice answer.<|endoftext|> -TITLE: Taylor expansion of the metric tensor in the normal coordinates -QUESTION [10 upvotes]: I am looking for a reference with a Taylor expansion of the metric tensor in the normal coordinates. -The coefficients should be written in terms of $\mathrm{Rm}, \nabla\mathrm{Rm}, \nabla^2\mathrm{Rm},\dots$ at the point. -It is easy to get using Jacobi equation, but I would prefer to have a reference (if it exists). -Comment. I need a tiny bit of this formula, namely the terms with the components of $\nabla^{n-2}\mathrm{Rm}$ in the coefficients for the monomials of degree $n$. - -REPLY [14 votes]: Using the reference https://arxiv.org/pdf/0903.2087.pdf, which agrees with https://arxiv.org/pdf/hep-th/0001078v1.pdf, which agrees with the reference U. Müller, C. Schubert and Anton M. E. van de Ven, J. Gen. Rel. Grav. 31 (1999) 1759-1768 [arXiv], one sees that the expansion in normal coordinates about a point $p\in M$ of a metric $g$ is given by -$$ -g_{ij} = g_{ij}(0) + \sum_{n\ge 2} \frac1{n!} C_{ijk_1\cdots k_n}\,x^{k_1}\cdots x^{k_n} -$$ -where $C_{ijk_1\cdots k_n}$ is the value at $p$ of a tensor of the form -$$ --\frac{2(n{-}1)}{n{+}1}\,\nabla_{k_3}\cdots\nabla_{k_n} R_{ik_1jk_2} + LOT_n -$$ -where $LOT_n$ is a polynomial in the curvature tensor and its derivatives of order strictly less than $n{-}2$. (Note that, in particular, $LOT_2 = 0$.) -Note: The first two references only verify the crucial coefficient $-2(n{-}1)/(n{+}1)$ as above for $n=2,3,4,5$, but, in any case, the coefficients for $n=2$ and $n=3$ are well-known. -Meanwhile, if one does not want to rely on these sources, one can note that these coefficients clearly cannot depend on the dimension of the underlying manifold, so it suffices to verify them in the simplest nontrivial case, i.e., when the dimension of the manifold is $2$. In this case, using the Gauss Lemma, a metric $g$ in geodesic normal coordinates $(x,y)$ centered on $p$ takes the form -$$ -g = \mathrm{d}x^2 + \mathrm{d}y^2 - + h(x,y)\bigl(x\,\mathrm{d}y-y\,\mathrm{d}x)^2, -$$ -where the function $h$ is arbitrary, subject to the condition that $(x^2{+}y^2)h(x,y)+1>0$. -Letting $r^2 = x^2 + y^2$ and letting $T$ be the radial vector field $x\,\partial_x + y\,\partial_y$, one computes the formula for the Gauss curvature of $g$ to be -$$ -K = -\frac{2(1+r^2h)(TTh) - r^2(Th)^2+2(5+3r^2h)(Th) + 8r^2h^2+12h}{4(1+r^2h)^2}. -$$ -Of course, one has the formula for the Riemann curvature tensor in the form -$$ -R = K\,(1{+}r^2h)\,(\mathrm{d}x\wedge\mathrm{d}y)\otimes(\mathrm{d}x\wedge\mathrm{d}y), -$$ -while the tensor $(1{+}r^2h)\,(\mathrm{d}x\wedge\mathrm{d}y)\otimes(\mathrm{d}x\wedge\mathrm{d}y)$ is parallel with respect to the Levi-Civita connection. Then, by assuming that $h$ vanishes to order $n{-}2\ge0$, one can easily verify that the above coefficient is correct.<|endoftext|> -TITLE: Can two versions of $\omega_1^{CK}(\mathsf{Ord})$ ever coincide? -QUESTION [8 upvotes]: The goal of this question is to fill in the gap in this old answer of mine. -For a transitive set $M$, thought of as an $\{\in\}$-structure, we define the following ordinals (this is not the notation used in the linked answer above, but on reflection I like it more, and to the best of my knowledge there is no standard notation): - -$\mathsf{Def}(M)$ is the supremum of the ordertypes of the well-orderings which are (first-order, with parameters) interpretable in $M$. - -$\mathsf{Ad}(M)$ is the height of the smallest admissible set with $M$ as an element. - - -Per the title, each of these can be thought of as a kind of "$\mathsf{Ord}$'s $\omega_1^{CK}$." Classically, $\omega_1^{CK}$ is defined as the smallest ordinal with no computable copy but is also the supremum of the definable well-orderings of $(\mathbb{N};+,\times)$ and the smallest admissible ordinal $>\omega$. -It's not hard to show that $\mathsf{Def}(A)\le\mathsf{Ad}(A)$ for all $A$, and equality does occur for some $A$ (e.g. $\mathsf{Ad}(L_\omega)=\mathsf{Def}(L_\omega)=\omega_1^{CK}$). However, $\mathsf{Def}(A)<\mathsf{Ad}(A)$ is also possible. The above-linked answer shows that we have $\mathsf{Def}(M)<\mathsf{Ad}(M)$ whenever $M\models\mathsf{ZFC}$ and $M^\omega\subseteq M$. My question is whether this closure condition is in fact needed: - -Is there any transitive $M\models\mathsf{ZFC}$ with $\mathsf{Def}(M)=\mathsf{Ad}(M)$? - -Note that since the definitions of $\mathsf{Def}$ and $\mathsf{Ad}$ are sufficiently simple, it's enough to ask whether there is a countable such $M$. - -REPLY [2 votes]: Every ordinal $α$ that is the least ordinal satisfying a given $Σ^1_1$ property (about $α$, equivalently, about $L_α$) is a Gandy ordinal, so $\mathsf{Def}(L_α)=\mathsf{Ad}(L_α)$. This includes the least $L_α$ satisfying ZFC, and even includes the least $L_α$ with $α$ an inaccessible cardinal in $L_{α^{+\text{CK}}}$ (but using $L_{α^{+\text{CK}}+1}$ would fail). -The proof of being Gandy is analogous to the proof of existence of recursive pseudowellorderings. Consider a theory $S$ including $\mathrm{ATR}_0$ and "ordinal $α$ satisfies a given $Σ^1_1$ property", with definable Skolem functions. Let $T$ be the tree of (essentially) finite partial $α$-models (as in $ω$-models) of $S$: A sequence of ordinals $<α$   $x_1,x_2,...,x_n$ is in $T$ iff there is no inconsistency proof shorter than $n$ for "$S$, symbols $x_1,x_2,...,x_n$, ordering of $x_1,x_2,...,x_n$, and $∃y<α \, φ(y) ⇒ φ(x_{⌈φ⌉})$ for the first $n$ one-variable formulas $φ$" ($⌈φ⌉$ is the Gödel number; $x_n$ is not used if $⌈φ⌉≤n$). Then the Kleene–Brouwer order of $T$ has a well-ordered initial segment of length $α^{+\text{CK}}$. The existence of the desired model ensures that the order is ill-founded, while its nonexistence in $L_{α^{+\text{CK}}}$ (since per assumption, $α$ is the least ordinal with the $Σ^1_1$ property, and we use $\mathrm{ATR}_0$) ensures that the well-founded length equals $α^{+\text{CK}}$. -In fact (not needed here), there is a recursive order-invariant linear order $f$ on $ℚ^{<ω}$ such that for all $α$ that are the least ordinal for some $Σ^1_1$ property, the initial well-founded portion of $f↾α^{<ω}$ has length $α^{+\mathrm{CK}}$ (by order invariance it does not matter how $α$ is order-preservingly embedded into $ℚ$). The construction is to interleave separate partial models as above for all variations of the theory $S$, except that we can skip variations at the cost of choosing a higher branch. -A related result is that for every finite $n$, $\mathsf{Def}(L_α)=\mathsf{Ad}(L_α)$ for the minimum $α$ such that $L_α$ satisfies ZFC and is correct about well-foundedness of $\mathbf{Σ}_n^{L_α}$ relations. The proof is as above except that we require that the partial models are correct about $Σ_{n+1}^{L_α}$ properties of the ordinals. This way a resulting model is correct about well-foundedness of relevant relations, and hence cannot be shorter than $α$. -As an aside, the results may seem surprising since working in ZFC or NBG, we can develop a theory of proper class well-orderings, and using replacement, failure of well-foundedness would be witnessed as a set. By diagonalization, increasing descriptive complexity gives strictly longer proper class well-orderings (at least for parameter-free complexity or if $V=HOD$), with 'longer' interpreted using embeddability into proper initial segments (which makes sense even though I think not all first-order definable class well-orderings are comparable using first-order definable classes). And for extensions of ZFC (including the replacement schema) using infinitary logic or constructible hierarchy above $V$, the increase in well-ordering lengths continues transfinitely with even the supremum of $L_\mathrm{Ord}(V)$ well-orderings not reaching the height of $\mathsf{Ad}(V)$ (assuming correctness about well-foundedness). But as soon as replacement (for countable sequences) fails (which presumably it would not for the 'true' $V$), in some cases, the supposed well-orderings may turn out to be an illusion.<|endoftext|> -TITLE: Lang's conjecture beyond the curve case -QUESTION [14 upvotes]: An algebraic variety $V$ is said to be of general type if it is of maximal Kodaira dimension. If $V$ is defined over a number field $K$, then one has the following conjecture due to Lang (Bombieri had made a similar conjecture in the case of surfaces; thus this conjecture is also known as the Bombieri-Lang conjecture): -If $V$ is a variety of general type defined over a number field $K$, then the set of $K$-rational points $V(K)$ of $V$ is not Zariski dense. -When $\dim V = 1$, this is Mordell's conjecture and was famously solved by Faltings in the 1980s. -Has there been any progress towards Lang's conjecture, in any situation beyond the curve case, since then? -"Progress" does not necessarily mean a proof, but even an outline of what a possible proof might look like or a program towards a proof would suffice. For example, Faltings, in his proof of Mordell's conjecture, had carried out a program that was already known at the time: given Parshin's trick, one shows that Mordell's conjecture is implied by Shafarevich's conjecture, and this is what Faltings actually proves. Is there anything like this beyond the curve case? - -REPLY [8 votes]: It may be instructive to think about what we really "know" in the case of projective surfaces. Let $k$ be an algebraically closed field of characteristic zero. -Let me first recall some basic definitions. -A projective (integral) variety over $k$ is of general type if some (hence any) desingularization of $X$ has big canonical bundle. Also, a projective variety is pseudo-Mordellic over $k$ if there is a proper closed subset $\Delta\subsetneq X$ such that, for every finitely generated subfield $K\subset k$ and every model $\mathcal{X}$ for $X$ over $K$, the set $\mathcal{X}(K)\setminus \Delta$ is finite. -If you can take $\Delta$ to be the empty set, then we say that $X$ is Mordellic over $k$. A projective variety $X$ is Mordellic over $k$ if and only if every subvariety of $X$ is pseudo-Mordellic over $k$. - - -Lang's conjecture (in its most optimistic form) is that a projective variety over $k$ is of general type if and only if it is pseudo-Mordellic over $k$ - - -Due to the birational invariance of both the notions of general type and pseudo-Mordellicity, you could as well stick to smooth projective varieties in this conjecture. -In the case of curves, Lang's conjecture boils down to showing that a projective curve is of general type if and only if it is Mordellic which is what Faltings achieved in 1983. -You ask whether Faltings's proof gives maybe a bit more than just the case of curves. Well, it certainly does. Namely, if $X$ is a projective variety over $k$ which admits a (quasi-)finite morphism to the moduli stack of principally polarized abelian varieties (ppav's), then every subvariety of $X$ is of general type (explanation omitted) and $X$ is Mordellic (by Faltings's theorem). Thus, any projective variety admitting a finite morphism to some moduli stack of ppav's will satisfy Lang's conjecture. Examples of such projective varieties are, for example, compact Shimura varieties of abelian type. -What about Faltings's 1992 result on closed subvarieties of abelian varieties? Let us assume that $X$ is a normal projective surface over $k$, and let $q=q(X)$ be the dimension of its Albanese variety. If $q>2$, then you can show that Lang's conjecture holds. That is: - - -Theorem. (Faltings) Assume that $X$ is a normal projective surface with $q>2$. Then $X$ is of general type if and only if $X$ is pseudo-Mordellic. - - -Sketch of proof. The fact that a pseudo-Mordellic surface is of general type follows from properties of K3 surfaces and the classification of surfaces; see Remark 7.8 in https://arxiv.org/abs/1909.12187 . Now, let $Y$ be the image of the Albanese map $X\to \mathrm{Alb}(X)$. Note that $Y$ is not an abelian variety, as this would contradict $q>2$. Note that $\dim Y = 1$ or $\dim Y =2$. If $\dim Y = 1$, then $Y$ is a hyperbolic curve (because it is not an abelian variety). The morphism $X\to Y$ is then a fibration of hyperbolic curves over a hyperbolic curve which readily implies that $X$ is pseudo-Mordellic. If $\dim Y=2$, use Faltings's 1991 theorem and Ueno's fibration theorem. QED -Now, this means that in the case of surfaces, it remains to prove Lang's conjecture whenever $q=0$, $q=1$, or $q=2$. However, if $X$ has a finite etale cover $X'\to X$ with $q(X') >2$, then $X'$ satisfies Lang's conjecture by the above. Since Lang's conjecture is "invariant under finite etale covers", it follows that $X$ also satisfies Lang's conjecture. (This is also what Noam Elkies is saying in his answer.) Thus, if $\dim X=2$, Lang's conjecture is really open if the augmented irregularity $\widehat{q}(X) $ is at most two. (The augmented irregularity is the supremum over $q(Y)$ as $Y$ runs over all finite etale covers of $X$.) -This suggests that the next "reasonable" case to study is that of surfaces with augmented irregularity equal to two. (The smaller the augmented irregularity, the further you are from any auxiliary abelian varieties. This makes life considerably harder.) -Using Stein factorization, in the case of surfaces with augmented irregularity two, one is reduced to studying finite ramified covers of abelian varieties. The next step we want to achieve is the following special case of Lang's conjecture: - - -Conjecture. Let $A$ be an abelian surface over a number field $K$ and let $X\to A$ be a finite ramified covering with $X$ a normal integral surface of general type over $K$. Then $X(K)$ is not dense. (Actually: we expect that $X_{\overline{K}}$ is pseudo-Mordellic, but peu importe.) - - -This non-density statement is not known. -Usually, Lang's conjecture is interpreted as suggesting that there are not so many points on a variety of general type. However, since $A(K)$ is dense (for some large enough number field $K$), one may also view Lang's conjecture in this specific situation as predicting that there are actually many rational points on an abelian variety. (That is, Lang predicts that there are so many rational points on an abelian variety that even after removing those coming from a ramified covering, the points remain dense.) -This brings me to my final point that some progress was made on this aspect of Lang's conjecture recently; see Theorem 1.3 in https://arxiv.org/abs/2011.12840 . A special case of that result reads as follows: - - -Theorem. Let $A$ be an abelian variety over a number field $K$ with $A(K)$ dense. Let $\pi:X\to A$ be a finite ramified morphism with $X$ a normal integral variety over $K$. Then $A(K)\setminus \pi(X(K))$ is dense in $A$. - - -This result says that there are "more" points in $A(K)$ than in $X(K$), as Lang predicted.<|endoftext|> -TITLE: Applications of equivariant homotopy theory in chromatic homotopy theory -QUESTION [9 upvotes]: I usually do computations in equivariant homotopy theory, but I would like to learn chromatic homotopy theory where one may use the equivariant techniques, e.g., slice spectral sequences, etc. -For this, I am looking for those papers which are dealt with the above kind of literature. -Any reference will be highly appreciated. - -REPLY [9 votes]: The canonical answer to this question is of course the celebrated solution by Hill, Hopkins and Ravenel to the Kervaire invariant one problem - -Hill, Michael A., Michael J. Hopkins, and Douglas C. Ravenel. "On the nonexistence of elements of Kervaire invariant one." Annals of Mathematics (2016): 1-262 (arXiv:0908.3724). - -Beside that amazing paper, equivariant homotopy theory can be used in the computation of Picard groups of some local categories. For example the following paper uses $C_4$-equivariant homotopy theory to study the Picard group of the $K(2)$-local category - -Beaudry, Agnes, Irina Bobkova, Michael Hill, and Vesna Stojanoska. "Invertible $ K (2) $-Local $ E $-Modules in $ C_4 $-Spectra." (arXiv:1901.02109) (2019). - -In both cases the intuition is that equivariant techniques are useful to run descent arguments along a Galois extension (typically by some small subgroup of the Morava stabilizer group). For example the slice spectral sequence allows one to resolve Borel $G$-spectra (i.e. "spectra with a $G$-action") by pieces that are not Borel anymore but which are "simpler" in some sense.<|endoftext|> -TITLE: What makes a Kähler manifold projective? -QUESTION [12 upvotes]: Let $X$ be a compact Kähler manifold, I know there are (at least?) 2 ways to make $X$ a projective manifold. - -(integral condition) If the Kähler class $[\omega]$ is integral, i.e., $[\omega]\in H^2(X,\mathbb Z)$, then $X$ is projective. -(Moishezon condition) If the Kähler manifold $X$ is also a Moishezon manifold, then $X$ is projective. - -In summary, we may write: - -Kähler+integral=projective, -Kähler+Moishezon=projective. - -And my question is that: what's the relationship between integral condition and Moishezon condition? -It includes: which one is stronger? Or whether one can deduced from the other？ Or are they actually equivalent? -Can we unify these 2 conditions together to get only one single condition? Or can we abstract the essence of both these 2 conditions? -I guess the Moishezon condition is another version of "integral" condition, this view comes from a post from @Gunnar Þór Magnússon - -A Moishezon manifold can now be characterized as a modification of a projective manifold, so there exists a generically 1-1 meromorphic map $X\to Y$, where $Y$ is projective. This is equivalent to $X$ admitting an integral Kähler current, which is roughly a closed (1,1)-form $T$ that has distribution coefficients (in local coordinates) whose cohomology class is integral. - -So can we unify these 2 conditions together to just one condition: $X$ admits an integral Kähler current? - Let me repeat my question again: what's the relationship between integral condition and Moishezon condition? -Added later: -I know $X$ admits an integral Kähler class if and only if $X$ has a positive line bundle, later I learned a compact complex manifold is Moishezon if and only if it carries a big line bundle. So in the view of "line bundles", we may write: - -Kähler+positive=projective -Kähler+big=projective - -And then it turns to the relationship between big and positive. -So what's the essence that makes a compact Kähler manifold projective? I apologize that my question seems a little bit vague, so any comment clarifying it is welcome! - -REPLY [12 votes]: If I understand the question correctly, I think that the answer is given by the main result in -S. Ji: Currents, metrics and Moishezon manifolds, Pac. J. Math. 158, No. 2, 335-351 (1993). ZBL0785.32011. -Essentially, the existence of a bimeromorphic modification that is projective gives a $d$-closed (1, 1)-current on our manifold (the pushforward of the Kähler form on the modification) that satisfies three properties. Conversely, the existence of such a current implies that the manifold is Moishezon. -Here is the rewriting in LaTex of the first page (warning: there is a misprint in the second line of the Introduction paragraph, the projective algebraic manifold is $\widetilde{M}$, not $M$). - -Abstract: A compact complex manifold $M$ is Moishezon if and only if there exists an integral closed positive $(1, 1)$-current $\omega$ such that $\omega \ge \epsilon\sigma$ and $\omega$ is smooth outside an analytic subvariety. -1. Introduction. Given a Moishezon manifold $M$, it is well known (cf. [Mo], [W]) that there is a bimeromorphic morphism $\pi : \widetilde M \to M$ such that the manifold $M$ is projective algebraic. Let $\tilde\omega$ be Kähler form on $\widetilde M$ with $[\tilde\omega] \in H^2(\widetilde M, \mathbf Z)$. Then the pushforward current $\omega = \pi_*\tilde\omega$ is a $d$-closed current on $M$ such that -(i) $\omega \in H^2(M, \mathbf Z)$; -(ii) $\omega$ is smooth on $M - S$, where $S$ is some proper analytic subset in $M$; -(iii) $\omega \ge \epsilon\sigma$ in the sense of currents, where $\epsilon > 0$ is some real number and $\sigma$ is a fixed positive definite $(1, 1)$-form (not necessarily $d$-closed) on $M$. -Conversely, we shall prove the following -Theorem 1.1. Let $M$ be a compact complex manifold of dimension $n$. Then $M$ is Moishezon if and only if there exists a $d$-closed $(1, 1)$-current $\omega$ on $M$ such that the conditions (i), (ii) and (iii) above are satisfied.<|endoftext|> -TITLE: What are some "good" examples of Kan simplicial manifolds? -QUESTION [6 upvotes]: According to the definition 1.1 of the paper Kan Replacement of simplicial manifolds by Chenchang Zhu https://arxiv.org/pdf/0812.4150.pdf, -A Kan simplicial manifold is a simplicial manifold $X$ such that for all $m \in \mathbb{N} \cup \lbrace 0 \rbrace $ and $0 \leq j \leq m$, the restriction map $Hom(\Delta^{m},X) \rightarrow Hom(\Lambda^{m}_{j}, X)$ is a surjective submersion. -I also encountered this notion in the definition 2.24 of the paper Higher Groupoid Bundles, Higher Spaces, and Self-Dual Tensor Field Equations by Branislav Jurco, Christian Samann, and Martin Wolf https://arxiv.org/pdf/1604.01639v2.pdf. and in the definition 1.2 of Integrating L∞-Algebras by Andre ́ Henriques (in the name of simplicial manifold satisfying Kan condition)https://arxiv.org/pdf/math/0603563.pdf. -But I could not find much good examples in each of the above 3 references and also anywhere else. I also could not construct one. -(Though I could find some examples like Cech $\infty$-groupoids and internal nerve of Lie groupoids in references Cech cocycles for differential characteristic classes – An ∞-Lie theoretic construction by Domenico Fiorenza, Urs Schreiber and Jim Stasheff and Kan Replacement of simplicial manifolds by Chenchang Zhu respectively. [Please check my 1st two comments for details]). -But it seems to me that, this notion is a very direct and natural generalisation of the notion Lie groupoid to Lie $\infty$-groupoid. (Though Lie $\infty$-groupoid is defined sometimes differently in some literatures). Though according to the discussion in https://ncatlab.org/nlab/show/Kan-fibrant+simplicial+manifold, it is not clear to me whether this notion is very useful or not from the perspective of homotopy theory, but the notion itself looks very elegant to me. -It would be very helpful for me if someone can suggest some interesting examples of Kan simplicial manifolds or suggest some literatures in this direction. -Thanks in advance. - -REPLY [6 votes]: Kan simplicial manifolds are in the same relation to differentiable ∞-stacks -(i.e., locally fibrant simplicial presheaves on the site of cartesian spaces and smooth maps) -as smooth manifolds are to sheaves of sets on the same site. -That is to say, Kan simplicial manifolds can be seen as the ∞-categorification of manifolds. -Some important examples include: - -Any ordinary manifold, interpreted as a constant simplicial object. - -The nerve of a Lie groupoid. In particular, the delooping of any Lie group, which represents principal bundles with this Lie group as a structure group. - -The Dold–Kan functor Γ applied to any nonnegatively graded chain complex of abelian Lie groups. - -In particular, applying Γ to the chain complexes U(1)[n], -we get the Kan simplicial manifold representing bundle (n-1)-gerbes. - -The nonabelian analogue of Γ applied to any crossed module whose two constituent groups are Lie groups -and the involved homomorphisms and actions are smooth. - -The nonabelian analogue of Γ applied to any (hyper)crossed complex whose constituent groupoids are Lie groupoids and the involved homomorphisms and actions are smooth. - -As a special case of the previous example, any simplicial Lie group is a Kan simplicial manifold.<|endoftext|> -TITLE: When are Fourier coefficients monotonic? -QUESTION [19 upvotes]: Given some sufficiently smooth function $f$ what conditions would be sufficient for its Fourier coefficients, as defined by -$$ -\hat{f}(n) := \int_{0}^{2\pi}\cos(nx)f(x)\ dx, \quad \text{for } n = 1,2,\ldots, -$$ -to be monotonic? Given the decay properties of Fourier coefficients, the monotonicity result would translate to -$$ -|\hat{f}(n)| \geq |\hat{f}(n+1)|, \quad n = 1,2,\ldots. -$$ -I haven't been able to find any literature regarding this and a result of this nature would be very interesting. - -REPLY [3 votes]: A comment on this problem would be that if $\hat f(n)$ are monotone ( here $f$ any continuous function, not necessarily odd or even, also I assume that $\hat f(n)$ is monotone not $|\hat f(n)|$ ) then one can assume that they are positive. -And if Fourier coefficients are real and positive then they must be absolutely convergent, that is $\{\hat f(n)\} \in l_1$. -This follows easily from property of Fejer's kernel, i.e. that it is positive operator -with integral 1: -$$ \sum_k (1-|k|/n)\hat f(k)exp(ikt) = \int F(t-s)f(s) \le \sup|f|$$ -so -$1/2 \sum_{k \in (-n/2, n/2)} \hat f(k)) \le \sup|f|$.<|endoftext|> -TITLE: Problem related to divisibility of even power sum -QUESTION [5 upvotes]: The question was posted in MSE(12/19/20)link, but gets no answer. Hence I'm posting in MO -Define $S_m(n)=1^m+2^m+\cdots+n^m$ - -Can it be shown that -$S_{2m}(uv)\equiv0\pmod{uv}\iff S_{2m}(u)\equiv0\pmod{u}$ and $S_{2m}(v)\equiv0\pmod{v}$ -Where $m,u,v$ are positive integer - -Example:$1|S_2(1)=1,5|S_2(5)=55,7|S_2(7)=140,25|S_2(25)=5525,35|S_2(35)=14910,49|S_2(49)=40425$. -(1) if $p$ prime, $p\nmid S_{2m}(p)$ then $p-1\mid 2m$ -Definition : For $m ≥ 0$, the $m$th power-sum denominator is the smallest positive integer $d_m$ such that $d_m · S_m(n)$ is a polynomial in $n$ with integer -coefficients. -The first few values of $d_m$ (see Sequence A064538) are -$d_m= 1, 2, 6, 4, 30, 12, 42, 24, 90, 20, 66, 24, 2730, 420, 90, 48, 510, . . . .$ -(2) Clearly $\gcd(u,d_{2m})=1\implies u\mid S_{2m}(u)$. -(3) $p\mid d_{m}\implies p\le m+1$ -I hope my following claim helps here: -• Consider $n\equiv 1\pmod2$ then $S_{2m}(n)\equiv0\pmod{n}\iff S_{2m}(\frac{n-1}2)\equiv0\pmod{n}$ -Source code Pari/GP -for(m=1,30,for(a=2,100,if(sum(i=1,a,i^(2*m))%a==0,print([2*m,a,sum(i=1,a,i^(2*m))])))) - -Proof for (1) -Proof for (3), see theorem 1 in this paper - -REPLY [3 votes]: If $a,b$ are coprimes then -$\color{cadetblue}{a|S_{2m}(a)\ \text{and} \ b|S_{2m}(b)\ \text{iff} \ ab|S_{2m}(ab)}$.....$(1)$ -$\mathcal Proof:$ Since, $S_{2m}(ab)=bS_{2m}(a) (\text{mod} \ a)$ and $S_{2m}(ab)=aS_{2m}(b) (\text{mod} \ b)$ and $(a,b)=1$, then surely $a|S_{2m}(a)$ and $b|S_{2m}(b)$ If $ab|S_{2m}(ab)$. -Conversely, if $a|S_{2m}(a)$ and $b|S_{2m}(b)$ and $S_{2m}(ab) \equiv r_{ab} (\text{mod} \ ab)$, then $a|r , b|r \Rightarrow lcm(a,b)=ab|r$ ( as $a,b$ are coprime) $\Rightarrow ab|S_{2m}(ab)$ -Now, $$S_{2m}(p^{k+1})=\sum_{n=0}^{p-1}\sum_{t=1}^{p^k} (np^k+t)^{2m}$$ -$\Rightarrow S_{2m}(p^{k+1}) \equiv pS_{2m}(p^k)+(\sum_{t=1}^{p^k} t^{2m-1})[(2m)\frac{p(p-1)}{2}p^k] \equiv pS_{2m}(p^k) (\text{mod}\ p^{k+1})$. -Hence, $\color{cadetblue}{p^{k+1}|S_{2m}(p^{k+1})\ \text{iff} \ p^{k}|S_{2m}(p^k)}$....$(2)$ -Let,$uv=\prod {p_i}^{a_i}$ and $uv|S_{2m}(uv)$, then definitely , ${p_i}^{a_i}|S_{2m}({p_i}^{a_i})$ from $(1)$.Then $(2)$ implies $p_i|S_{2m}(p_i)$ and ${p_i}^{k}|S_{2m}({p_i}^{k})$ for any, $k$. As these all are different primes, we get $u|S_{2m}(u)$ and $v|S_{2m}(v) \cdots \text{qed}$.<|endoftext|> -TITLE: Classes of birationally equivalent Calabi-Yau manifolds in the Grothendieck ring -QUESTION [6 upvotes]: Do birationally equivalent Calabi-Yau manifolds have the same classes in the Grothendieck ring of varieties? -Here a Calabi-Yau manifold is a smooth complex projective variety with trivial canonical bundle. -This is true for Calabi-Yau threefolds or holomorphic symplectic fourfolds. -Related conjectures: - -Birationally equivalent Calabi-Yau manifolds have equivalent derived categories. This is known for dimension three. - -Consider two derived equivalent Calabi-Yau manifolds which are simply connected. Then the difference of their classes in the Grothendieck ring of varieties is annihilated by some power of the class of the affine line. This is open even for K3 surfaces. - -Two derived equivalent smooth projective varieties have isomorphic Chow motives. This is known for dimension two. - -REPLY [7 votes]: This is not known. Motivic integration provides equality of classes of K-equivalent varieties (in particular, for birational with trivial canonical class) in the appropriate localization of the Grothendieck ring. This implies that birational Calabi-Yau varieties have equal Hodge numbers. -I believe the sharpest known result in this direction is Corollary 7.3 is in Kawamata's paper: https://arxiv.org/pdf/1710.07370.pdf.<|endoftext|> -TITLE: Resources for topos theory -QUESTION [10 upvotes]: I am trying to learn topos theory and I am finding a strong scarcity of resources. Is there any canonical textbook to refer someone to when learning this topic? -So far, I have only been able to find the following. - -The Stacks project (Part 1, Chapter 7 gives an introduction) -"Théorie des Topos et Cohomologie Etale des Schémas" by Grothendieck, Artin and Verdier (yes, this one is quite old and in French but I have found all the necessary topics together in a better way than any other resource known to me) - -In particular, I have only found a clear distinction between topology and pretopology in the second one, as well as clear guidance on how to work interchangeably between them. -Do people have any suggestions? Please list them below so other people trying to learn topos theory have a better idea of where to start! - -REPLY [4 votes]: Here are some references on Steven Vickers works to look at topoi from other "geometric" and logical sides. - -S. Vickers, Locales and Toposes as Spaces Chapter 8 in M. Aiello, I. Pratt-Hartmann and J. van Benthem (eds.), Handbook of Spatial Logics, 429–496. 2007 Springer, -S. Vickers, Toposes pour les vraiment nuls -S. Vickers, Sketches for arithmetic universes -S. Vickers, Arithmetic universes and classifying toposes - -I think it also worth to mention the following - -I. Moerdijk, Classifying Spaces and Classifying Topoi (surprisingly to me this one is available online as well) - -For some advanced matters I also suggest to read - -T. Streicher, Fibered Categories<|endoftext|> -TITLE: On a monotonicity property of Fourier coefficients of truncated power functions -QUESTION [7 upvotes]: Is it true that -$$a_{k,n}:=\int_0^{2\pi}x^k\cos(nx)\,dx$$ -is nonincreasing in natural $n$ for each $k\in\{0,1,\dots\}$? - -This question is related to this previous one. -Twice integrating by parts, one obtains the recurrence -$$a_{k,n}= \frac k{n^2} (2\pi)^{k - 1} - \frac{k (k - 1)}{n^2}\,a_{k-2,n}$$ -for $k\in\{2,3,\dots\}$, with $a_{0,n}=0=a_{1,n}$, which might be of use here. -We also have -$$a_{k,n}=\frac{(2 \pi )^{k+1}}{k+1}\;_1F_2\left(\frac{k}{2}+\frac{1}{2};\frac{1}{2},\frac{k}{2}+\frac{3}{2};-n^2 \pi ^2\right).$$ - -REPLY [8 votes]: I prefer to write $2\pi$ in the exponent and reverse the direction of the monotonicity (none of which really matters, of course), so we shall fix $k>1$ (real in general), $n\in\mathbb Z_+$ (that is essential) and define -$$ -a_n=\Re I_n, \qquad I_n=\int_0^1 (1-z)^ke^{2\pi in z}dz -$$ -(I immediately write the integral in the complex notation; the branch of the $k$-th power is positive on $(0,1)$; all our paths will be in the upper half-strip $0\le x=\Re z\le 1, y=\Im z\ge 0$). -We want to show that $a_n>0$ and $a_{n+1}4n$ and $k-1<4n$. We shall use different contours for these two cases. -Case 1: $k>4n$. Let $\rho=\frac {4n}{k}<1$. Consider the contour given by the equation $-k\alpha+2\pi n x=0$ (or, $x=\frac{2\alpha}{\pi\rho}$, $0\le\alpha\le\rho\frac{\pi} 2$) where the notation is clear from the following picture: - -The explicit equation for the curve (using $\alpha$ as the parameter) is -$$ -x=\frac{2\alpha}{\pi\rho},\quad y=\left(1-\frac{2\alpha}{\pi\rho}\right)\tan\alpha\,. -$$ -Differentiating, we get $\dot x=\frac{2}{\pi\rho}$, $\dot y=\left(1-\frac{2\alpha}{\pi\rho}\right)\frac 1{\cos^2\alpha}-\frac{2}{\pi\rho}\tan\alpha$, so for the direction $\theta$ of $dz$, we get -$$ -\tan\theta=\left(\frac{\pi\rho}2-\alpha\right)\frac 1{\cos^2\alpha}-\tan\alpha\,. -$$ -At the very least, we see that $-\alpha\le\theta\le\frac{\pi}2$. -Now let $F_n(z)$ be the integrand in $I_n$. We have $F_{n+1}(z)=e^{2\pi iz}F_n(z)$, so $|F_{n+1}(z)|=e^{-2\pi y}|F_n(z)|<|F_n(z)|$ -while $\arg F_{n+1}(z)=\arg F_{n}(z)+2\pi x=\arg F_{n}(z)+\frac{4\alpha}\rho$ on our contour. Note that $\arg F_{n}(z)=-k\alpha+2\pi n x=0$ on our contour, so we have -$$ -a_n=\int |F_n|\cos\theta|dz| -$$ -and -$$ -a_{n+1}=\int |F_{n+1}|\cos(\theta+\tfrac {4\alpha}\rho)|dz|\,. -$$ -Since $\cos\theta>0$ and $|F_{n+1}|<|F_{n}|$, all we need now is to show that -$$ -\cos(\theta+\tfrac {4\alpha}\rho)\le \cos\theta -$$ -If $-\alpha\le\theta\le 0$, we rotate by $\frac{4\alpha}\rho\in (4\alpha,2\pi)$, so we go beyond the symmetric angle $-\theta$ but stay short of the full revolution, so the inequality certainly holds. Now we need to consider $\theta>0$ only, in which case we want to show that we cannot rotate beyond $2\pi-\theta$, i.e., that $\theta+\frac{2\alpha}\rho\le\pi$ or -$$ -\theta\le \pi-\beta, \beta=\frac{2\alpha}\rho\,. -$$ -Note that $0\le\beta<\pi$ and if $\beta\le\frac\pi 2$, then there is nothing to prove because $\theta\le\frac\pi 2$. -So, assume $\beta>\frac \pi 2$. Then we need $\tan\theta\le\tan(\pi-\beta)$, i.e. (rewriting $\alpha$ in terms of $\beta$), -$$ --\tan(\tfrac\rho 2\beta)+\frac\rho 2(\pi-\beta)\frac 1{\cos^2\tfrac\rho 2\beta}\le\tan(\pi-\beta). -$$ -Bringing the terms on the left hand side to the common denominator, we get -$$ -LHS=\frac{\rho (\pi-\beta)-\sin(\rho\beta)}{2\cos^2\tfrac\rho 2\beta}\le -\rho \frac{(\pi-\beta)-\sin(\beta)}{2\cos^2\tfrac 1 2\beta} -\\ -\le -\frac{(\pi-\beta)-\sin(\beta)}{1+\cos\beta}= -\frac{(\pi-\beta)-\sin(\pi-\beta)}{1-\cos(\pi-\beta)} -$$ -because $\rho<1$, $\sin(\rho\beta)\ge \rho\sin\beta$ and $\cos \tfrac\rho 2\beta\ge \cos \tfrac 1 2\beta$. -Let now $\tau=\pi-\beta\in(0,\frac\pi 2)$. Our inequality becomes -$$ -\frac{\tau-\sin\tau}{1-\cos\tau}\le\frac{\sin\tau}{\cos\tau}\,, -$$ -that is -$$ -\tau\cos\tau\le\sin\tau\,, -$$ -which is classical. -This finishes Case 1. -Case 2: $k-1<4n$. This case is way more interesting. We will not be able here to join $0$ to $1$ directly and will make a detour via $+i\infty$. In this case we define $\rho=\frac{4n}{k-1}>1$ and use the same equation but with $0\le\alpha<\frac \pi 2$. Even as $\alpha\to\frac\pi 2-$, we shall have only $x\to \frac 1\rho<1$ while $y\to+\infty$ with the picture like this: - -Of course, once we escape to $+i\infty$, we have to come back from it, which we will do over the right boundary line $z=1+it$ of the strip. That the abscissa of the escape is different from that of the return is not a problem because $F_n,F_{n+1}$ decay to $0$ when we go up, so we may think that we join the paths high up by a horizontal interval and then take the limit. -We still have $\theta<\frac\pi 2$. I claim that now $\theta>0$. Indeed, $\tan\theta$ is given by the same formula (only now $\rho>1$!), so we have -$$ -\tan\theta=\left(\frac{\pi\rho}2-\alpha\right)\frac 1{\cos^2\alpha}-\tan\alpha\ge \left(\frac{\pi}2-\alpha\right)\frac 1{\cos^2\alpha}-\tan\alpha -\\ -=\frac{(\pi-2\alpha)-\sin(2\alpha)}{2\cos^2\alpha} -=\frac{(\pi-2\alpha)-\sin(\pi-2\alpha)}{2\cos^2\alpha}>0\,. -$$ -Note that now the argument regime of $F_n(z)dz$ is $\theta-\alpha$ because $-(k-1)\alpha$ cancels with $2\pi nx$. Thus, on the way up, we have -$$ -\int |F_n(z)|\cos(\theta-\alpha)|dz| -$$ -we look at the way down now matching to $z$ the point $z'=1+i\Im z$ (which is a legitimate parameterization). Notice that $|F_n(z')|=\sin^k\alpha|F_n(z)|$ and the argument of $F_n(z')dz'$ is constant and does not depend on $n$ (only the power factor matters) and $|dz'|=\sin\theta|dz|$. Thus, if we combine both paths, we shall get -$$ -\int_{\text{upward path}} |F_n(z)|[\cos(\theta-\alpha)-\lambda\sin\theta\sin^k\alpha]|dz| -$$ -where $\lambda\in[-1,1]$ is some constant (common for $n$ and $n+1$). -The key point now is that -$$ -\cos(\theta-\alpha)=\cos\theta\cos\alpha+\sin\theta\sin\alpha -\ge\sin\theta\sin\alpha\ge |\lambda\sin\theta\sin^k\alpha| -$$ -so the factor is, again, positive. -For $F_{n+1}$, we get -$$ -\int_{\text{upward path}} |F_{n+1}(z)|[\cos(\theta-\alpha+\tfrac{4\alpha}{\rho})-\lambda\sin\theta\sin^k\alpha]|dz| -$$ -again, and we see that everything boils down to the inequality -$$ -\cos(\theta-\alpha+\tfrac{4\alpha}{\rho})\le\cos(\theta-\alpha)\,. -$$ -If $\theta-\alpha>0$, then it is $\le\frac{\pi} 2-\alpha$ and the rotation is by at most $4\alpha\le 2\alpha+\pi$, so we get short of $\frac{3\pi}2+\alpha$, where we could have any chance to bring the cosine up again. Thus, we need to consider only the case $\theta-\alpha<0$. If $\rho\le 2$, then our rotation is by at least $2\alpha$ and at most $2\pi$, which diminishes the cosine of any angle between $-\alpha$ and $0$. Finally, if $\rho>2$, then we shall just show that we necessarily have $\theta>\alpha$ (the case we have considered already). Indeed, comparing the tangents, we see that we need just to prove the inequality -$$ -\left(\frac{\pi\rho}2-\alpha\right)\frac 1{\cos^2\alpha}\ge 2\tan\alpha\,. -$$ -Since $\rho>2$, that reduces to -$$ -\pi-\alpha>\frac\pi 2>1\ge \sin(2\alpha) -$$ -and the story is over. -The admissible contours are not unique, but the leeway in choosing them is rather small. I just posted the most elegant ones I could find. I hope it is all correct, but feel free to ask questions if something looks fishy or is just unclear.<|endoftext|> -TITLE: Rank of a finite group and its representations -QUESTION [5 upvotes]: $\DeclareMathOperator\Rep{Rep}\DeclareMathOperator\rank{rank}$Let $G$ be a finite group, and $C=\Rep(G)$ be the monoidal category of complex finite-dimensional representations of $G$. As $C$ is finite and semisimple, one can get all representations from $\oplus$ and a finite set $I$ of irreducible representations. By classical character theory, there is a (noncanonical) bijection between $I$ and $\mathrm{Conj}(G)$. In this thread, I hope to understand a bijection, if any, between both sides with the consideration of $\otimes$. -To be more precise, let $V$ be an irreducible faithful representation of $G$. Then every representation occurs as a submodule of $V^{\otimes n}$ for some $n$ (cf this and this), and vice versa! We then say that $V$ itself generates $C$ under $\otimes$ and Cauchy completion. However, not every group has an irreducible faithful representation. In the same post, we can see that this largely deals with the "rank" of the socle of $G$. -To summarize, define the rank, $\rank(G)$, to be the minimal number of elements needed to generate $\mathrm{socle}(G)$ under conjugation. Define the rank, $\rank(C)$, to be the minimal number of irreducible elements needed to generate $C$ under $\otimes$ and Cauchy completion. Then -$$ -\rank(G) = 1 \Leftrightarrow \rank(\Rep(G)) = 1 -$$ -Question -Does this equivalence generalize to -$$ -\rank(G) = n \Leftrightarrow \rank(\Rep(G)) = n, -$$ -for each natural number $n$? -(EDIT As Qiaochu pointed out in the comment, this is true for finite abelian groups by Pontrjagin duality.) - -REPLY [8 votes]: The answer to your question is yes and is the main theorem of the paper Žmudʹ, È. M. -On isomorphic linear representations of finite groups. -Mat. Sb. N.S. 38(80) (1956), 417–430. -It can be found in Theorem 5 on page 245 of Characters of Finite Groups. Part 1. by Berkovich and Žmudʹ. The theorem is phrased in a different, but equivalent way, and is proved in a very similar way to Gaschutz's theorem. -The theorem of Žmudʹ says that $G$ has a faithful representation with $k$ irreducible constituents if and only if the socle of $G$ can be generated as a normal subgroup by at most $k$ elements. In particular, the least number of normal generators of $\mathrm{socle}(G)$ coincides with the least number of irreducible constituents in some faithful representation of $G$. -It now suffices to observe $\mathrm{rank}(C)$ is exactly the minimal number of irreducible constituents in a faithful representation of $G$. Indeed, if $V$ is any faithful representation, then the Burnside theorem (or R. Steinberg's generalization) shows that every irreducible module is a direct summand in a tensor power of $V$ and so the irreducible constituents of $V$ generate $C$ under tensor product, direct sums and taking direct summands. On the other hand, if $\rho_1,\ldots, \rho_k$ are irreducible representations whose direct sum is not faithful, then $\ker \rho_1\cap\dots\cap \ker \rho_k$ acts as the identity on all modules in the subcategory generated by the corresponding simple modules under the operations of direct sum, tensor product and taking direct summands and so these irreducible representations cannot generate $C$. -Thus $\mathrm{rank}(G)=\mathrm{rank}(C)$<|endoftext|> -TITLE: Comparison of Information and Wasserstein Topologies -QUESTION [5 upvotes]: There are many possible metrics one can place on the space of Gaussian probability measures on $\mathbb{R}^n$, with strictly positive definite co-variance matrices. Let's denote this space by $X$. -I'm particularly interested in the information geometric one (using the Fisher-Rao-Riemann metric) and the one induced by restricting the Wasserstein $2$ metric from $\mathcal{P}_2(\mathbb{R}^n)$ to the subspace $X$. But how doe these compare? Most specifically, is the Wasserstein $2$-distance dominated by the Fisher-Rao metric's induced distance function? - -REPLY [2 votes]: Wasserstein distance has not good properties of invariances compared to Fisher Metric, that could be extended on convex cones by Jean-Louis Koszul tools (https://link.springer.com/chapter/10.1007/978-3-030-02520-5_12), and on homogeneous symplectic manifolds for Lie groups by Jean-Marie Souriau tools (https://www.mdpi.com/1099-4300/22/5/498). -These topics are developd in GSI conferences: www.gsi2021.org<|endoftext|> -TITLE: Do we lose any solutions when applying separation of variables to partial differential equations? -QUESTION [38 upvotes]: For example, consider the following problem -$$\frac{\partial u}{\partial t} = k\frac{\partial^2 u}{\partial x^2},\hspace{0.5cm} u(x,0)=f(x),\hspace{0.5cm} u(0,t)=0,\hspace{0.5cm} u(L,t)=0$$ -Textbooks (e.g., Paul's Online Notes) usually apply separation of variables, assuming that $u(x,t)=\varphi(x)G(t)$ without any explanation why this assumption can be made. -Do we lose any solutions that way given that there are functions of two variables $x$ and $t$ that are not products of functions of individual variables? -Separation of variables gives the following solution when we consider only boundary conditions: -$$u_n(x,t) = \sin\left(\frac{n\pi x}{L}\right)e^{-k\left(\frac{n\pi}{L}\right)^2t},\hspace{0.5cm}n=1,2,3,\dotsc.$$ -The equation is linear, so we can take a superposition of $u_n$: -$$u(x,t) = \sum\limits_{n=1}^{\infty}B_n\sin\left(\frac{n\pi x}{L}\right)e^{-k\left(\frac{n\pi}{L}\right)^2t}$$ -where $B_n$ are found from the initial condition: -$$B_n = \frac{2}{L}\int\limits_0^Lf(x)\sin\left(\frac{n\pi x}{L}\right)dx,\hspace{0.5cm}n=1,2,3,\dotsc.$$ -Are there solutions $u(x,t)$ that cannot be represented this way (not for this particular pde but in general)? What happens in the case of non-linear equations? Can we apply separation of variables there? - -REPLY [12 votes]: As you correctly noted, in the end we write our solution as a superposition of separable solutions, so the right question really 'can we express every solution to our PDE as a sum of separable solutions'? -A thorough answer to this question requires a little linear algebra. What we want to do is find a set of functions $\{\varphi_n(x): n \in \mathbb{N}\}$ so that for each time $t$ write our solution $f$ as $f = \sum_{n=0}^{\infty} \varphi_n(x) G_n(t)$ where the $G_n$ are just some coefficients which are allowed to depend on time. Not only does such a set of functions exists, we can actually find a set of these functions through the process of separation of variables. -Let's consider the heat equation again. When we separate variables, we reduce the situation to two ODEs: -$$G'(t) = EG(t), \varphi''(x) = \frac{E}{k}\varphi(x) $$ where $E$ is some unknown constant. -Remember that the differentiation is linear: that is, for functions $f$ and $g$ and constants $a,b$ we have $\frac{d}{dx}(af(x)+bg(x)) = a\frac{df}{dx} + b \frac{dg}{dx}$. What this means is that our two ODEs are eigenvalue problems: we have an eigenvalue problem for the operator $\frac{d}{dx}$ with eigenvalue $E$, and an eigenvalue problem for the operator $\frac{d^2}{dx^2}$ with eigenvalue $\frac{E}{k}$. -We need the eigenvectors of $\frac{d^2}{dx^2}$ (i.e. the solutions to our $\varphi$ ODE) to form a basis for our space of functions. Luckily, there is a theorem that does exactly this sort of thing for us. -Spectral Theorem: - -Let $V$ be a Hilbert space and $T: V \to V$ a (sufficiently nice) self-adjoint map. Then there exists an orthonormal basis for $V$ which consists of eigenvectors for $T$. - -In order to make sense of this, we need one final ingredient: an inner product. This is just something which generalises the familiar `dot product' in three dimensions. The inner product of two functions $f$, $g$ is a real number, defined as -$$\langle f,g\rangle := \int_{0}^{\infty} f(x)g(x) dx$$. -A basis of functions $\{f_n: n \in \mathbb{N}\}$ is called orthonormal if $\langle f_n, f_n \rangle = 1$ and $\langle f_n, f_m \rangle = 0$ when $n \neq m$. -Finally, we just need to check that the operator $\frac{d}{dx}$ is self-adjoint. What this means is that for any two functions $f$, $g$ we have that $\langle \frac{d^2 f}{dx^2},g\rangle = \langle f,\frac{d^2g}{dx^2} \rangle$. This can be done by integration by parts: -$$\int_{0}^{L} f''(x)g(x) dx = - \int_{0}^{L} f'(x)g'(x) dx = \int_{0}^{L} f(x)g''(x) dx$$ where we have thrown away the boundary terms because the boundary conditions tell us that they are zero. -Hence, the operator $\frac{d^2}{dx^2}$ is self-adjoint, and so the spectral theorem tells us that its eigenvectors form a basis for our function space, so for any given $t$ we can express any chosen function as $$f = \sum_{n=0}^{\infty} \varphi_n(x) G_n(t)$$ Thus, we haven't lost any solutions in that we can write the equation like this. -I have skipped a few technical issues here: I haven't told you what the Hilbert space is, and when I say 'any' function, I really mean 'any square-integrable' function. But I don't think these technicalities are important in the understanding. - -As a fun extra, now that we have our inner product, we can use it to simply derive the coefficients in our series solution. We write our solution as $$f(x,t) = \sum_{n=0}^{\infty} \varphi_n(t) G_n(x)$$ and now lets take the inner product of $f$ with the basis element $\varphi_n(x)$. This gives us -$$\langle f(x,0), \varphi_n(x) = \langle \sum_{k=0}^{\infty} \varphi_k(x) G_k(0), \varphi_n(x) \rangle = \sum_{k=0}^{\infty} G_k(0) \langle \varphi_k(x) , \varphi_n(x) \rangle = \sum_{k=0}^{\infty} G_k(0) \langle \varphi_k(x) , \varphi_n(x) \rangle $$ -Here we have interchanged integration and summation. Finally, the orthonormality of the basis $\{\varphi_k(x)\}$ means that all of the terms but one are zero, so we get $$ \langle f(x,0), \varphi_n(x) = G_n(0) $$ Recall that $G_n(t) = B_n e^{-k\frac{n\pi}{L}^2 t}$, so $B_n = G_n(0)$ and writing our inner product formula in terms of an integral, we get $$\int_{0}^{L} f(x,0) \varphi_n(x) dx = \int_{0}^{L} f(x,0) \sin(\frac{n\pi x}{L}) dx $$ which is our usual expression for the series coefficients!<|endoftext|> -TITLE: A combinatorics problem and the probability interpretation -QUESTION [6 upvotes]: For a gaussian vector variable $w\sim N(0,I_{n\times n})$, the moments of square norm are $\mathbb{E} \|w\|^{2 r} = \prod_{t=0}^{r-1} (n + 2 t)$. -Based on Isserlis' theorem, $\mathbb{E} \|w\|^{2 r}$ can also be evaluated as -$$\sum_{\pi\in \mathcal{P}([r]), |\pi|\leq n}\frac{n!}{(n-|\pi|)!}\prod_{p\in\pi}(2 |p|-1)!!$$ -where $\mathcal{P}([r])$ means all partitions on set $[r]=\{1,\dots,r\}$, $\pi$ is a partition, $p$ is one block in a partition, $|\pi|$ and $|p|$ are number of blocks and number of elements in a block. -Now consider a variant of the above problem. -$$\sum_{\pi\in \mathcal{P}([r]), |\pi|\leq n}\frac{n!}{(n-|\pi|)!}\prod_{p\in\pi}\frac{1}{2}~(2 |p|-1)!!$$ -The above formula only differs from moments of square norm of gaussian vector variable with a factor $\frac{1}{2}$. -Is there a similar finite product solution and probability interpretation for the above formula? - -REPLY [7 votes]: Fix $n$. Let -$$ G(x) = \sum_{i=0}^n \frac{n!}{(n-i)!}\frac{x^i}{i!} = (1+x)^n. $$ -Let -$$ F(x) = \sum_{j\geq 1}\frac 12 (2j-1)!!\frac{x^j}{j!} = - \frac{1}{2\sqrt{1-2x}}-\frac 12. $$ -By the Compositional Formula (Theorem 5.1.4 of Enumerative -Combinatorics, vol. 2), the number you want is $r!$ times the -coefficient of $x^r$ in -$$ G(F(x)) = \frac{1}{2^n}\left( - 1+\frac{1}{\sqrt{1-2x}}\right)^n. $$ -You can expand this by the binomial theorem and then expand each term -into a power series to get a formula for your number as a sum with $n$ -terms.<|endoftext|> -TITLE: Final project ideas - computational geometry -QUESTION [8 upvotes]: Next semester I am teaching (the programming part) of a course in Computational Geometry. There are 14 weeks of class, two "pure theory/blackboard" hours per week, one "theory/blackboard exercises" hour per week and one "computer lab" hour per week. -My intention for the "computer lab" sessions is that the students work (almost from day one) on a programming project that they can submit at the end of the course. My thoughts for the structure and evaluation of the projects are more or less the same than in this link. To sum up: students can either design a graphical applet that illustrates some algorithm, work on an open problem or develop an application of Computational Geometry to other areas. -However, I am not an expert in Computational Geometry and I would like you to give me some suggestions of interesting topics for the final projects. -Let me give some context for the level of the projects: - -The students are 4th-year (that's last year in Spain) undergraduates in Mathematics. Their programming background is a two-semester course on basic programming in Python and two one-semester courses on Numerical Analysis where they did some programming in Matlab. Some of them have a wider background and a small fraction of them are also last-year Computer Science undergraduates. - -The textbook we are going to follow is: A Short Course in Computational Geometry and Topology by H. Edelsbrunner. (The theory professor chose it). - - -In previous years this course has been Machine-Learning-oriented, so I might also suggest some projects related with ML. - -REPLY [6 votes]: Although The Open Problems Project has grown a bit -out-of-date, we recently moved it to github to improve updating. -Because it was "originally aimed to record important open problems of interest to researchers in computational geometry and related fields," it has a slant -appropriate for your purposes. -A typical problem is this, proposed by -R. Nandakumar: - -Problem 67: Fair Partitioning of Convex Polygons. - -      -Figure from: Bárány, Imre, Pavle Blagojević, and András Szűcs. "Equipartitioning by a convex 3-fan." Advances in Mathematics 223, no. 2 (2010): 579-593. -DOI.<|endoftext|> -TITLE: Maxwell equations as Euler-Lagrange equation without electromagnetic potential -QUESTION [15 upvotes]: In (mathematical) physics many equations of motion can be interpreted as Euler-Lagrange (EL) equations. The Maxwell equation for electromagnetic (EM) field (say in vacuum and in absence of charges) seems to me quite unusual in comparison to examples known in classical mechanics. This is a system of first order PDE on 6 components of EM field. To get the Lagrangian density, one takes the first pair of the Maxwell equations and deduces from it existence of electromagnetic potential. Substituting the potential into the second pair of Maxwell equations, one gets second order equations for the potential. They can be presented as EL-equations for the potential. -I am wondering if there is a way to present the Maxwell equations as an EL-equation in terms of electromagnetic field only rather than potential. -I think I can prove that this is impossible if one requires in addition that the Lagrangian density is quadratic in fields and their first derivatives and invariant under the Poincare group. -ADDED 1: By the Maxwell equations I mean -$$\operatorname{div}\vec B=\operatorname{div} \vec E=0,\, \operatorname{rot}\vec E=-\frac{1}{c}\dot{\vec B},\, \operatorname{rot}\vec B=\frac{1}{c}\dot{\vec E}.$$ -Thus I am looking for a Lagrangian depending on $\vec E,\vec B$ as independent fields and their derivatives such that the EL-equation is equivalent to all these equations. -ADDED 2: Let me reformulate my question on the language of some of the answers and comments below. The standard approach to interpret Maxwell equations as EL-equation in electrodynamics is as follows. One selects out of 8 Maxwell equations four equations and declares them to be constrains, and the other four - equations of motions. One considers the variations of the action functional $\int L$ only in the class of electromagnetic fields $(\vec B,\vec E)$ satisfying the constrains. Its extrema recover the four equations of motion. This separation of the Maxwell equations into two halves seems to me to be artifical and different from all other examples I know. I am wondering if there is a way to consider all 8 Maxwell equations on equal footing for this purpose. - -REPLY [2 votes]: The answer to your question is in the paper below: -Anthony Sudbery (York U., England) Jul, 1985 -A Vector Lagrangian for the Electromagnetic Field 1986 J. Phys. A: Math. Gen. 19 L33<|endoftext|> -TITLE: Reference request: proof-theoretic strength of $\mathsf{KP}$ with recursively large ordinals and $\mathsf{CZF}$ with large set axioms -QUESTION [8 upvotes]: Large set axioms are notions corresponding to large cardinals on constructive set theories like $\mathsf{IZF}$ or $\mathsf{CZF}$. The notion of inaccessible sets, Mahlo sets, and 2-strong sets correspond to inaccessible, Mahlo, and weakly compact cardinals on $\mathsf{ZFC}$. -(See Rathjen's The Higher Infinite in Proof Theory and Ziegler's thesis for the definition of notions I mentioned, and check the description added in the ) -It is known that the proof-theoretic strength (or consistency strength if you are not familiar with the proof-theoretic one) of $\mathsf{CZF}$ is the same as that of $\mathsf{KP}$, so we may expect the strength of $\mathsf{CZF}$ with large sets are quite weaker than that of $\mathsf{ZF}$. -In fact, Rathjen described the proof-theoretic strength of large sets over $\mathsf{CZF}$ in his paper The Higher Infinite in Proof Theory. Here is the result he stated: - -Theorem 6.9. $\mathsf{CZF}+\forall x\exists I (x\in I \land \text{$I$ is inaccessible})$ can be interpretable in $\mathsf{KP}+\forall \alpha\exists \kappa>\alpha(\text{$\kappa$ is recursively inaccessible})$. Moreover, the interpretation preserves the validity of $\Pi^0_2$-sentences. - - -Theorem 6.14. $\mathsf{CZF}+\forall x\exists M (x\in M \land \text{$M$ is Mahlo})$ can be interpretable in $\mathsf{KP}+\forall \alpha\exists \kappa>\alpha(\text{$\kappa$ is recursively Mahlo})$. Moreover, the interpretation preserves the validity of $\Pi^0_2$-sentences. - - -Theorem 6.20. $\mathsf{CZF}+\forall x\exists K (x\in M \land \text{$K$ is 2-strong})$ can be interpretable in $\mathsf{KP}+\forall \alpha\exists \kappa>\alpha(\text{$\kappa$ is $\Pi_3$-reflecting})$. Moreover, the interpretation preserves the validity of $\Pi^0_2$-sentences. - -However, Rathjen does not state the proof of his results in this paper. Combining some known results may result in a proof of the statement given above, but I have no clear idea which one would be relevant. - -Question. Where can I find a proof (or a rough sketch of a proof) of the statements described above? - - -Added on 1 April 2021: I realized that just referring to Ziegler's thesis for the definition of large set axioms is not correct, since the different definitions result in different proof-theoretic strength. -Ziegler defined inaccessible sets as transitive sets which satisfy the second-order Strong Collection and $\mathsf{CZF}$. Some other references (e.g., Rathjen's paper I cited) add an additional requirement that an inaccessible set must satisfy the regular extension axiom $\mathsf{REA}$, which claims that every set is contained in a regular set. Let me call the former weakly inaccessible sets, and the latter strongly inaccessible sets. -It turns out that the existence of weakly inaccessible sets does not strictly increase the proof-theoretic strength: - -Theorem. The theory $\mathsf{CZF}$ + "every set is contained in a weakly inaccessible set" has the same proof-theoretic strength with $\mathsf{CZF+REA}$. - -(See Theorem 4.7 of Rathjen's The Anti-Foundation Axiom in Constructive Set Theories for its proof.) -However, the proof-theoretic strength of $\mathsf{CZF}$ with a strongly inaccessible set exceeds that of $\mathsf{CZF+REA}$. In fact, it has the same proof-theoretic strength with $\mathsf{KP}$ with a recursively inaccessible ordinal. (Theorem 6.5 of Rathjen's Proof Theory of Constructive Systems: Inductive Types and Univalence.) -Note that both weakly and strongly inaccessible sets are just $V_\kappa$ for an inaccessible cardinal $\kappa$ over $\mathsf{ZFC}$. There may be a difference between them over $\mathsf{ZF}$ since strongly inaccessible sets would think there is a proper class of regular cardinals. I do not think there is any direct relation between these two kinds of sets and weakly inaccessible cardinals. -Rathjen defined his inaccessible sets as strongly inaccessible sets in his papers, so we have to understand inaccessible sets in that sense. - -REPLY [2 votes]: Rathjen stated rough sketch of proof that intuitionistic theory (e.g. $\mathsf{CZF}+(\forall x)(\exists I)[x\in I\land I \text{ is inaccessible}]$) has at least the strength of the classical one (e.g. $\mathsf{KP}\omega+(\forall\alpha)(\exists\kappa)[\alpha<\kappa\land\kappa\text{ is recursively inaccessible}]$) in The Realm of Ordinal Analysis. - -The proof that the intuitionistic theory has at least the strength of the classical -one requires an ordinal analysis of the classical theories as given in [73, 74, -82] and a proof of the well-foundedness of the pertinent ordinal representation -system in the intuitionistic theory. - -From this description, Rathjen's proof is probably following. -Let $T_i$ be intuitionistic theory, $T$ be classical theory corresponding to intuitionistic one, and $\langle A,\prec\rangle$ be primitive recursive well-ordering which has proof-theoretic ordinal of $T$. - -Prove well-foundedness of $\prec\restriction$ in $T_i$ by using higher infinites where $\prec\restriction$ is proper initial seqment of $\prec$. -Embed $T$ into infinitary derivation $\mathrm{RS}$ and prove "$T\vdash\varphi$ implies there exists $\alpha\in A$ such that $\mathrm{RS}$-cutfree proof of $\varphi$ whose height bounded by $\alpha$ for all $\Sigma_1^{L_{\omega_1^\mathrm{CK}}}$-sentence $\varphi$" by using cut elimination and collapsings in $T_i$. -Prove "If there is $\mathrm{RS}$-cutfree proof of $\varphi$ whose height bounded by $\alpha\in A$, then $\varphi$ is true for all $\Pi^0_2$-sentence" by using transfinite induction up to $\alpha$ and $\Pi^0_2$-partial truth definition in $T_i$.<|endoftext|> -TITLE: Potential automorphy vs. automorphy -QUESTION [7 upvotes]: What is the geometric meaning of potential automorphy, and conceptually why is it so hard to go from potential automorphy to automorphy on the nose? -Is there an obstruction to descent from potential automorphy to automorphy and does it lie in some Galois cohomology group? Is there an example where a variety is potentially automorphic, but not automorphic? - -REPLY [11 votes]: $\DeclareMathOperator\GL{GL}$There is no obstruction to descent from potential automorphy to automorphy lying in some cohomology group. Conjecturally, potential automorphy should be equivalent to automorphy. -To see why, let us take as our primary object that can be automorphic or not, Galois representation instead of varieties or motives (to fix ideas). Specifically, we consider continuous irreducible representations $r:G_K \rightarrow \GL_n(\bar{\mathbf Q}_p)$, where $K$ is a number field, $G_K$ its absolute Galois group, and $p$ a prime number. Let us recall the following "folklore" conjecture: - -Conjecture (Fontaine–Mazur + Langlands). A Galois representation $r$ of $G_K$ as above is automorphic if and only if $r$ is unramified at almost all places of $K$ and potentially semi-stable at every place of $K$ dividing $p$. - -Here automorphic means that $r$ has the same $L$-factors at almost all primes as a cuspidal automorphic algebraic representation $\pi$ of $\GL_n(\bf{A}_K)$. -Now fix $K'$ a finite extension of $F$. It is almost obvious that the conditions after the "if and only if" -are satisfied for $r$ if and only if they are satisfied for $r_{\vert G_{K'}}$. Thus automorphy for $r$ should be the same as automorphy for $r_{\vert G_{K'}}$. -In other words, automorphy is conjecturally the same as potential automorphy. -Now why is it not possible to prove on the nose that if $r_{G_K'}$ is automorphic, that is associated with an automorphic representation $\pi'$ of $\GL_n({\bf A}_{K'})$, then $r$ itself is automorphic? Because that would mean to "descend" $\pi'$ into s suitable automorphic representation $\pi$ of $\GL_n(\bf{A}_K)$, -and this question, called "automorphic descent", is very hard. It is part of the extremely hard Langlands functoriality, and has been solved so far only in the case $K'/K$ solvable (by Langlands for $n=2$, Arthur–Clozel for general $n$).<|endoftext|> -TITLE: Two knots with same Dowker-Thistlethwaite code have isomorphic knot groups? -QUESTION [5 upvotes]: Given a (tame) knot diagram, one derives the Dowker-Thistlethwaite code by travelling around the knot and numbering each crossing 1,2,3,.... A negative sign is given to an even number if you cross on an underpass with that even number. Each crossing gets an even-odd pair of numbers, and we sort the pairs by the odd numbers. The resulting even number sequence is the Dowker-Thistlethwaite code of the knot. -This code determines a prime knot uniquely but there may be multiple composite knots with the same code. - -Question: Given any two knots with the same Dowker-Thistlethwaite code, are their knot groups isomorphic? - -I believe the answer is yes because of the comment on Wikipedia that two knots with the same code will be almost the same except for a reflection or a reflection of the way the connected sum is taken in a prime knot decomposition. Therefore one can define an isomorphism of fundamental groups easily enough. However there does not seem to be a reference for this fact either, and Dowker and Thistlethwaite (at least to me) don't seem to talk about this in their paper [1]. However, I'm mainly an algebraist and my algebraic topology expertise is not as thorough as I'd like, so I would appreciate some help. -[1] Dowker, C. H.; Thistlethwaite, Morwen B. Classification of knot projections. Topology Appl. 16 (1983), no. 1, 19--31. MR0702617 - -REPLY [2 votes]: You are correct. The DT code gives a composite diagram if and only if the permutation splits into two sub-permutations. So, if the code does not split, then it specifies a prime diagram which we can reconstruct up to taking a mirror image. -This is discussed (briefly) in Section 2.2 of "The Knot Book" by Colin Adams.<|endoftext|> -TITLE: On the definition of the etale site of an adic space -QUESTION [9 upvotes]: I have a question related to the definition of the etale site of an adic space. As a reference, I am using Huber's book "Etale Cohomology of Rigid Analytic Varieties and Adic Spaces". -First of all, to define the etale site of an adic space, we only consider adic spaces that are locally of the form $Spa(A,A^+)$ where $A$ has one of the following properties -i) $\hat{A}$ is discrete -ii) $A$ is strongly Noetherian Tate -iii) $\hat{A}$ is has a Noetherian ring of definition over which it is finitely generated. -Then the etale site of $X$ is defined in the same way as with schemes (namely by considering surjective families of etale morphisms). Why do we restrict ourselves only to adic spaces of the above form? Could we not, for example, consider stable adic spaces instead? Let me explain that. I give the definitions, as I haven't seen those being widely used. Those are taken from Wedhorn's notes, found here https://arxiv.org/pdf/1910.05934.pdf. -Definition 1 A Huber ring $A$ is called stably sheafy if every $\hat{A}$-algebra topologically of finite type is sheafy, i.e given such an algebra $B$ and any subring of integral elements $B^+$, $(B,B^+)$ is sheafy. -Definition 2 A stable adic space $X$ is an adic space that is covered by affinoid adic spaces $Spa(A,A^+)$ where $A$ is stably sheafy. -Now consider only adic spaces that are stable. This includes the adic spaces that Huber considers in his book. Then the category of adic spaces that are etale over $X$ (a stable adic space) is well defined. Moreover if we define $X_{et}$ in the same way as above, everything seems to work well (for example fiber products exist under the same assumptions as in Huber's book). So what is it (if there is something) that restricts us to work only in the case Huber treats in his book? Perhaps there are properties that we want the etale site to have and fail in this more general setting. Let me suggest one, that I cannot see if it is true or not. -Question 1 Let $f:X\rightarrow Y$ be an etale morphism of affinoid stable adic spaces. Is $\mathcal{O}_Y(Y)\rightarrow\mathcal{O}_X(X)$ flat? -If this is true, I expect the proof to be hard. The analogous statement ,in the usual setting, is lemma 1.7.6 in Huber's book. In any case, regardless whether we can work in this more general setting or not, I would like to see an answer related to question 1. -I haven't thought much of the latter and I haven't checked every detail to make sure everything works fine (but I hope it does). Any answer is very much appreciated. - -REPLY [13 votes]: Great question! -The short answer is that Huber simply wanted to be in a setting where everything is (stably) sheafy, and so put some assumptions ensuring this. Note that Huber's work remained somewhat obscure for some time, so he probably didn't want to further scare people by allowing the most general non-noetherian case! Also, the detailed analysis of sheafyness that happened in recent years wasn't known then. In particular, it was not known that there are interesting large classes of stably sheafy spaces outside of the cases considered by Huber. -On the other hand, it turns out that for the definition of the étale site, sheafyness is not really important. See, for instance, Chapter 9 (Definition 9.1.2) of Kedlaya-Liu 1, https://arxiv.org/abs/1301.0792, for a definition of the étale site for general preadic spaces (that is allowing non-sheafy Huber pairs). In www.math.uni-bonn.de/people/scholze/EtCohDiamonds.pdf (see especially Section 15), I also discuss étale sites of preadic spaces (in the analytic case). Mostly all basic properties of the étale site, especially pertaining to étale cohomology, extend to this more general setting. One important property is that étale maps of (qcqs) adic spaces admit "noetherian approximation", so one can often formally reduce to the cases considered by Huber. -Re your comment: There has recently been a large interest in studying the étale site of adic spaces not falling into the settings considered by Huber; the prototypical case is that of perfectoid spaces. -Incidentally, in the case of perfectoid spaces, we very much expect the answer to your question to be negative: Already the map -$$ -\mathbb C_p\langle T^{1/p^\infty}\rangle\to \mathbb C_p\langle (\tfrac Tp)^{1/p^\infty}\rangle -$$ -corresponding to an inclusion of perfectoid discs is probably not flat (although, I'm embarassed to say, we still don't know this??).<|endoftext|> -TITLE: Do these properties of a countable abelian group guarantee a Prüfer subgroup? -QUESTION [9 upvotes]: Suppose $(G,+)$ is a countable abelian group and $p$ is a prime number such that: - -The subgroup $pG$ has finite index in $G$, and -For every $n \in \mathbb{N}$, $G$ contains an element of order $p^n$. - -Must $G$ contain a subgroup isomorphic to the Prüfer $p$-group? -I have tried carrying out a pigeonhole-type argument as follows. Let $x_n$ be an element of order $p^n$. To produce a Prüfer subgroup, it suffices to show that infinitely many of the cyclic groups $\langle x_n\rangle$ are nested. Let $G_n = \langle x_1, x_2, \dots, x_n\rangle$. Since $G_n$ is a finite abelian group, it can be written in the form $G_n = \bigoplus_{i=1}^{k_n}{\mathbb{Z}/p^{r_i}\mathbb{Z}}$. If $k_n$ is bounded, then there must be an infinite sequence of nested cyclic groups and hence a Prüfer group. One many be inclined (as I was) to use condition (1) to show that $k_n$ is bounded, since $[G_n : pG_n] = p^{k_n}$. However, $[G_n : pG_n]$ is not bounded by $[G : pG]$, so this argument does not work. For example, the divisible group $G = \bigoplus_{n=1}^{\infty}{\mathbb{Z}[p^{\infty}]}$ contains $\bigoplus_{n=1}^{\infty}{\mathbb{Z}/p^n\mathbb{Z}}$ as a subgroup. -I don't see any immediate remedy to this argument, but I also don't see any simple candidates for counterexamples. My hope is that another approach can be used to find a Prüfer subgroup. - -REPLY [10 votes]: Yes, it must. And $G$ doesn't need to be countable. -Let $H$ be the $p$-primary component of the torsion subgroup of $G$. Then the natural map $H/pH\to G/pG$ is injective, so $H$ also satisfies (1), and clearly satisfies (2). So, replacing $G$ by the subgroup $H$, we shall assume that $G$ is a $p$-group. -Let $X$ be a finite subset of $G$ that generates $G/pG$, so -$$G=\langle X\rangle + pG.$$ -For some $n$, $p^nx=0$ for all $x\in X$, and so -$$p^nG=p^n\langle X\rangle + p^{n+1}G=p^{n+1}G.$$ -So $p^nG$ is divisible, and by (2) is nonzero. And a divisible abelian $p$-group is a direct sum of copies of the Prüfer $p$-group.<|endoftext|> -TITLE: Multicategories vs Categories -QUESTION [6 upvotes]: One of the initial motivating factors for learning category theory, besides needing it for my work, was the idea that almost all mathematical notions I would encounter could be understood using categories one way or another. -That’s largely been borne out at the $1$-categorical level, and (almost?) completely vindicated at the $\infty$-categorical level, but I keep encountering statements about multicategories that make me feel like I might still be missing out on some ‘big picture’ understanding of the sort typically furnished by categories. -For a specific example I recently came across an MO question about direct sums/tensor products of vector spaces, and the answer by Qiaochu Yuan seemed to essentially assert that although we can understand what’s happening in terms of categories the most natural view is furnished by multicategories, and further the notion of a monoidal category is strictly generalized by the notion of a multicategory in a satisfying way. Monoidal categories are also generalized by bicategories in a satisfying way though, so my first question is: - -Do multicategories generalize categories in a way that bicategories don’t? - -This is kind if vague, but returning to the example above I would ask if we can understand the situation involving vector spaces using bicategories to clarify things instead of multicategories. -If the answer to the first highlighted question is yes, - -Is there a way to recapture the additional understanding imparted by multicategories using higher categories? - -If not, then I would ask if a theory of higher multicategories exists and if the additional work of learning it over higher category theory is worth the understanding payoff. -If the answer to the first highlighted question is no, I am happy to stick with higher categories for now — I have a bonus question though: - -For those familiar with it, does the theory of augmented virtual double categories have any significant ‘big picture understanding’ advantages over the theory of bicategories? What about compared to higher categories? - -I am immediately attracted to the fact that the collection of all large categories (not even locally small), functors and natural transformations form an augmented virtual double category (what a mouthful), but is there any other nice conceptual payoff for the leap from categories to augmented virtual double categories? - -REPLY [6 votes]: Multicategories and bicategories, to me, are first of all completely -orthogonal generalisations of monoidal categories, with virtual -double categories as a common generalisation of multicategories and -(strict) $2$-categories (they are to multicategories as categories are -to monoids, or $2$-categories to monoidal categories). As for -generalising simply categories, they are even more different, as -multicategories are still a strictly associative structure while -coherence issues appear for bicategories. -So, to your second highlighted question, I would say that the answer -is no. A way to more precisely understand the difference, and a -positive answer to your first question, lies in your final (bonus) -question. -In addition to the aforementioned algebraic aspect, double categories -have a large conceptual advantage over bicategories for formal -category theory; in short, if you try to treat the objects of an -arbitrary $2$-category as abstract categories, you will be lacking a -lot of elements (the Yoneda structure coming from profunctors a.k.a. -bimodules) to speak about limits in them, while double categories (at least -the ones equipping their vertical category with proarrows) will give -you enough. Virtual double categories are just the relevant -generalisation for when bimodules do not compose, and the augmented -version deals with the case lacking identity bimodules (e.g. for -non-locally small categories). This is, in my opinion, the main conceptual -payoff for (augmented) virtual double categories. -To finish, two technical generalisations, the latter of which was your -second-and-a-half question: - -Virtual double categories (though not the augmented ones, as far as -I know), are an example of "generalised multicategories", something -that can be defined relative to any monad acting on a virtual proarrow equipment. For this, see Leinster's book suggested in varkor's -comment -(chapters 4 and 5) or the more general and recent A unified framework for generalized -multicategories by Cruttwell and -Shulman. - -There does exist a higher-categorical version of multicategories and -even virtual double categories, defined (as "generalised non-symmetric -$\infty$-operads") and put to great use (precisely to unify -$\infty$-multicategories and double $\infty$-categories) in Gepner–Haugseng's Enriched -$\infty$-categories via non-symmetric -$\infty$-operads -and follow-ups. A more general form, close to the generalised -multicategories, is offered by Chu–Haugseng's formalism of algebraic -patterns developed in Homotopy-coherent algebra via Segal -conditions (see in particular -section 9 with ex. 9.8). For the moment this is only used for algebraic aspects rather than formal (higher) category theory, but I would argue it is definitely worth learning (not over higher category theory, but as an extension of it).<|endoftext|> -TITLE: Function spaces satisfying $\mathcal{F}(M\times N)\simeq\mathcal{F}(M)\otimes\mathcal{F}(N)$ -QUESTION [8 upvotes]: Let $M \mapsto\mathcal{F}(M)$ be a map associating topological vector spaces of some type (that I will call "function spaces") to geometric spaces $M$ of some type. -For $M$, I'm mostly thinking of manifolds with some additional structure, or locally compact topological spaces. $\mathcal F$ may or may not be a functor in some way, though it's better if it's a contravariant functor. I'm mostly interested in the case where $\mathcal{F}(M)$ is a usual function space such as $L^p(M)$, $W^{k,p}(M)$, $\mathrm{Meas}(M)$, like in this question, and this one. I want the function spaces of the form $\mathcal{F}(M)$ to have some completed tensor product $\otimes$. -Question 1: When does it happen that $\mathcal{F}(M\times N)\simeq\mathcal{F}(M)\otimes\mathcal{F}(N)$ and when does it fail and how badly? -The above tensor property, when $\mathcal F$ is a functor, would be better intended to hold naturally, i.e. $\mathcal F$ is to be a monoidal functor from spaces with their Cartesian product $\times$ to function spaces with $\otimes$, but the emphasis is not on the categorical aspect. -Edit: I'm aware that, as Nik Weaver points out in the comments, I can't expect to get a completely general answer. Rather, the question (which I find very natural) should be intended in "community wiki" style, i.e. partial contributions are ok. - -REPLY [4 votes]: In the theory of stereotype spaces there is a series of natural functors that satisfy this identity with one of the two main tensor products (the so-called ``injective stereotype tensor product'' $\odot$): -$$ -{\mathcal F}(M\times N)\cong {\mathcal F}(M)\odot {\mathcal F}(M) -$$ -In particular, this holds for - -the stereotype algebras ${\mathcal C}$ of continuous functions on paracompact locally compact spaces: -$$ -{\mathcal C}(M\times N)\cong {\mathcal C}(M)\odot {\mathcal C}(M) -$$ - -the stereotype algebras ${\mathcal E}$ of smooth functions on smooth manifolds: -$$ -{\mathcal E}(M\times N)\cong {\mathcal E}(M)\odot {\mathcal E}(M) -$$ - -the stereotype algebras ${\mathcal O}$ of holomorphic functions on Stein manifolds: -$$ -{\mathcal O}(M\times N)\cong {\mathcal O}(M)\odot {\mathcal O}(M) -$$ - -the stereotype algebras ${\mathcal P}$ of polynomials (= regular functions) on affine algebraic manifolds: -$$ -{\mathcal P}(M\times N)\cong {\mathcal P}(M)\odot {\mathcal P}(M) -$$ - - -(The cases of ${\mathcal E}$ and ${\mathcal O}$ are just reformulations of the classical results of functional analysis, and the whole picture is stated here. This is closely connected with the constructions of group algebras in analysis.)<|endoftext|> -TITLE: Can one reuse positive referee reports if paper ends up being rejected? -QUESTION [26 upvotes]: Context: A submission to a very good generalist journal X received one positive referee report recommending publication and two shorter opinions which both deemed the paper a solid and valuable contribution and thus worthy of publication but perhaps not a priority given the backlog this journal has, thus only weakly recommending publication. Given the fairly high standing of the journal this logically resulted in a rejection. -Questions: - -Would it be appropriate upon resubmission to a new journal Y to inform the editor of Y of the existing detailed referee report at X? (To be clear: I do not mean sending them the actual report, just making them aware of its existence) - -Upshot:the referee's time is not wasted, in particular as the detailed referee report from X is the most thorough report I have ever received (out of over 100), the paper is quite technical in places and the expert referee has spent considerable amount of time on it. -Drawback: The paper starts on the bad footing of being a rejected one, but I think this should not be a big problem if Y is e.g. a specialist journal or if Y is deemed not as "highly ranked" as X. Moreover during the review process at X the referee asked for some clarifications and suggested some improvements which has led to a better paper. I think that even when resubmitting elsewhere we should keep the thank you to an anonymous referee for suggested improvements which already shows the paper was rejected, thus the drawback is there anyway. - -If the answer is yes to 1) should one ask the editor of X beforehand whether it is ok to pass on their contact details to potentially transfer the referee report to the editor at Y? - -Personally I think the answer is "Yes" to both questions but hearing other people's perspective is very interesting. - -REPLY [4 votes]: The following might be useful to have in mind: -if your paper is rejected at journal J1 and submitted at J2, the information at the J2 editorial board that it was rejected at J1 might be considered as negative, especially if the editorial board of J2 considers J1 as being of equal or lower level (which might not coincide with your own appreciation, or of J1's editorial board or referee in case they recommended J2 as alternative — keep in mind that the appreciation of the journal level varies according to subtopics and countries).<|endoftext|> -TITLE: Laplacian on manifolds and random matrix theory -QUESTION [8 upvotes]: Let $M$ be a compact Riemannian manifold with a metric $g$, and consider the spectrum of the Laplacian operator $\Delta$. -What is known about the relationship between this spectrum and random matrix theory? -In posing this question, I am imagining that the metric $g$ is drawn randomly from a suitable distribution. I am agnostic as to how this is done, but since the space of metrics on $M$ is somewhat unwieldy it may be simpler to consider special finite-dimensional spaces of $g$'s. -To be concrete, we could for instance imagine a two-dimensional Riemann surface of genus greater than one equipped with a uniformly negatively curved metric. Such metrics come in finite-dimensional families and it is natural to imagine drawing the metric from this set. -(Clearly in the genus one case the spectrum on a flat torus is not random, so I am also imagining that the topology of $M$ is suitably generic.) -In higher dimensions there are also sometimes natural finite-dimensional families of metrics, e.g. on Calabi-Yau manifolds and I am also interested to know what generic features of the spectra are known here as well. -As some physics motivation, if you consider a quantum particle moving on such a manifold, the energy levels are controlled by this spectrum. If the system is sufficiently generic, one expects chaotic behavior and hence some type of random matrix universality in properties of the spectrum. -Thanks for any answers or links to relevant literature! - -REPLY [4 votes]: There are quite a few connections. I will mention a result of mine where the connection is explicit and essential. Fix the metric $g$. Set $m=\dim M$ and assume that ${\rm vol}_g(M)=1$. -Denote by $\DeclareMathOperator{\spec}{spec}$ $\newcommand{\bR}{\mathbb{R}}$ $\spec(\Delta)$ the spectrum of $\Delta$ -$$ -\spec(\Delta)= \big\{\, 0=\lambda_0<\lambda_1\leq \lambda_2\leq \cdots \,\big\}, -$$ -where each eigenvalue is repeated according to its multiplicity. -Fix an orthonormal eigenbasis $(\Psi_k)_{k\geq0}$ of $L^2(M,g)$ -$$ -\Delta \Psi_k = -\lambda_k\Psi_k,\;\;\forall k. -$$ -Pick a nonnegative even Schwartz function $w:\bR\to[0,\infty)$. (E.g. $w(x)=e^{-x^2}$) Fix $\newcommand{\ve}{{\varepsilon}}$ $\ve>0$ and set $w_\ve(x)=w(\ve x)$. Consider the random Fourier series -$$ -U^\ve=\sum_{k\geq 0} X_k w_\ve\big(\sqrt{\lambda_k}\big)\Psi_k, -$$ -where $X_k$ are independent standard normal random variables. -Random matrices appear when you study the distribution of critical points and critical values of the random function $U^\ve$ as $\ve\searrow 0$. $\DeclareMathOperator{\Hess}{Hess}$ $\newcommand{\bp}{\boldsymbol{p}}$ -The Hessian of $U^\ve$ at a point $\bp\in M$ is a random matrix $\Hess_\bp^\ve$. Suitably rescaled it converges in distribution to a classical random matrix of the form $\newcommand{\one}{\boldsymbol{1}}$ -$$ X_m\one +A_m $$ -where $X_m$ is a normal random variable with mean zero and variance depending only on $m$ and $w$ and $A_m$ belongs to GOE, the Gaussian Orthogonal Ensemble. -The critical values of $U^\ve$ are with high confidence distributed in an interval of the form $[\ve^{-m},\ve^{-m}]$. -If you denote by $\spec(U^\ve)$ the set of critical values of $U^\ve$ then we obtain a random measure on $\bR$ -$$ -\mu^\ve=\sum_{c\in\spec(U^\ve)}\delta_{\ve^mc}. -$$ -Its expectation $\newcommand{\bE}{\mathbb{E}}$ $\bar{\mu}^\ve:=\bE[\mu^\ve]$ is a deterministic measure on $\bR$. Its mass -$$ -N^\ve:=\bar{\mu}^\ve[\bR] -$$ -is the expected number of critical points. -Suppose that $A_{m+1}$ is belongs to the GOE of symmetric matrices of dimension $(m+1)\times (m+1)$. We obtain similarly a random\probability measure -$$ -\sigma_{m+1}=\frac{1}{m+1}\sum_{\lambda\in \spec{A_{m+1}}\delta_\lambda. -$$ -Its expectation $\bar{\sigma}_{m+1}=\bE[\sigma_{m+1}]$ bis a deterministic probability measure on $\bR$ that has a rather explicit, albeit complicated, description. -In the paper I mentioned I show that -$$ -\bar{\mu}^\ve\to Z_m e^{-b_mx^2}\bar{\sigma}_{m+1} -$$ -weakly in the sense of measures as $\ve\searrow 0$. -I am not very precise here because this is true up to certain very explicit rescalings depending only on $m$ and $w$. The constants $Z_m$ and $b_m$ are also explicit and depend only on $m$ and $w$. -The normalizing constant $Z_m$ ensures that the right-hand-side above is indeed a probability measure.<|endoftext|> -TITLE: Optimal $\delta$ for Gromov's $\delta$-hyperbolicity of the hyperbolic plane -QUESTION [10 upvotes]: What is the minimal $\delta$ such that the hyperbolic plane is $\delta$-hyperbolic, in the sense of the four point definition of Gromov? -Four point definition of Gromov: A metric space $(X, d)$ is $\delta$-hyperbolic if, for all $w, x, y, z \in X$, -$$ d(w, x) + d(y, z) \leq \text{max}\{d(x, y) + d(w, z), d(x, z) + d(w, y) \} +2\delta. $$ -Empirically, the minimal value seems to be approximately $0.693$. -There is a related question, but this concerns the optimal $\delta$ in the $\delta$-slim definition. While this implies a bound on the $\delta$ of the four point definition, it hasn't yet helped me to derive the minimal value. -Any help (or a reference) would be greatly appreciated! - -REPLY [6 votes]: Indeed, the hyperbolic plane is $\log(2)$-hyperbolic (with the 4-point definition of hyperbolicity) and this is the optimal constant. The result is nontrivial and first appeared as Corollary 5.4 in -Nica, Bogdan; Špakula, Ján, Strong hyperbolicity, Groups Geom. Dyn. 10, No. 3, 951-964 (2016). ZBL1368.20057.<|endoftext|> -TITLE: Example of ODE not equivalent to Euler-Lagrange equation -QUESTION [17 upvotes]: I am looking for an explicit (preferably simple) example of an ODE with time-independent coefficients in $\mathbb{R}^3$ such that there does not exist an Euler-Lagrange equation -$$\frac{\partial L}{\partial q^i}=\frac{d}{dt}\frac{\partial L}{\partial \dot q^i}, \, \, i=1,2,3,$$ -with the same solutions. -I prefer to distinguish two cases: either $L$ depends explicitly on time or not. - -REPLY [2 votes]: If you're willing to consider PDEs, I coauthored an explicit example of a PDE in a gauge field $A$ that fails to be the Euler-Lagrange equation arising from any gauge-invariant lagrangian involving $A$ alone in arXiv:1501.07548. -The PDE in question is formula (12), which is a deformation of the (topologically massive, for $\mu\neq 0$) 3D Yang-Mills equation of motion. The relevant term is the one involving the mass parameter $m$; for $m\to\infty$, we find (topologically massive) Yang-Mills. To avoid reproducing most of that (short) paper here, I will only sketch the proof that this is not an EL equation for any gauge-invariant lagrangian: the PDE is gauge invariant under standard Yang-Mills type gauge transformations, but fails to satisfy a Noether identity. However, gauge-invariant PDEs arising as EL equations always satisfy Noether identities; contradiction.<|endoftext|> -TITLE: Nice way of referring to “some cardinal that always exists” -QUESTION [5 upvotes]: My apologizes if this question is not MO-appropriate. -Often I find myself wanting to state a hypothesis of a consistency result that can hold for an “aribtrary” regular cardinal. Only, it’s not really arbitrary, but it can be any regular cardinal below some large cardinal involved in the hypothesis. I’d like to state the result succinctly in a form like, “If ZFC is consistent with a measurable cardinal, then for every accessible regular cardinal $\mu$, it is consistent that $\varphi(\mu)$.” But this doesn’t really make sense because the accessible cardinals are not terms in the formal language. I can revert to talking about $\aleph_n$ for finite $n$, or $\aleph_\alpha$ for $\alpha$ a recursive countable successor ordinal, but this seems to lose the generality. If I try talking about “definable cardinals,” then I run into the problem of non-absolute definitions. -Does anyone have any preferred elegant/eloquent general ways to address this? If possible, I’d prefer to talk about consistency results rather than describe the cardinals in terms of the relation between some models. -EDIT: Does the following notion make sense? Suppose $\varphi(x)$ is a formula in the language of set theory. We say $\kappa_\varphi$ is an absolute cardinal if ZFC proves that there is a unique cardinal $\kappa$ such that $\varphi(\kappa)$, denoted $\kappa_\varphi$, and for any two models $M,N$ of ZFC with the cardinals of one being an initial segment of the cardinals of the other, $\kappa_\varphi^M =\kappa_\varphi^N$. Does this notion nicely capture examples of consistency results like I mentioned? For instance, if ZFC+inaccessible is consistent, then for any absolute successor cardinal $\kappa_\varphi$, it is consistent that there are no Kurepa trees on $\kappa_\varphi$. Are there examples that make this a bad definition? - -REPLY [6 votes]: Unless I'm misreading, there's an implicit identification in the question of "cardinal that always exists" with "cardinal below the first [large cardinal]." I don't really buy such an identification; the answer below treats the second notion, that is, the question of whether we can talk about "each cardinal below the first [large cardinal]" in a model-free way. This may not be relevant to the OP after all, in which case I'll delete this. - -I don't think there is one, at least not without a lot of circumlocution - and I don't think there should be. -To me there's a pretty fundamental ambiguity when one says anything like - -Assuming Con($\mathsf{ZFC}$ + a measurable), for every uncountable regular cardinal $\kappa$ below the first measurable it is consistent that $\kappa$ has property $P$ - -in the first place. Since in general there's no way to refer to a single arbitrary cardinal in the language of set theory alone, I'm not sure how to make sense of consistency in this context. My best interpretation of "consistent" in this context is in the sense of outer models, so that I'd interpret the above as saying: - -Suppose $M$ is a model of $\mathsf{ZFC}$ + a measurable. Let $X$ be the set of uncountable regular cardinals below the least measurable, in the sense of $M$. Then there is a family $(M_\kappa)_{\kappa\in X}$ such that each $M_\kappa$ is an outer model of $M$ satisfying "$\kappa$ has property $P$." - -However, this may not be what's intended - e.g. maybe we should restrict attention to countable $M$, or well-founded $M$, or both. Conversely, maybe we want to say more about those outer models - maybe each $M_\kappa$ is a set forcing extension of $M$ (even "uniformly" in an appropriate sense), or maybe each $M_\kappa$ still satisfies "$\kappa$ is a regular cardinal below the least measurable." -So the use of proof-theoretic language here really seems to obscure the actual meaning of the statement in question, to the point that genuine information is lost. Honestly I prefer it when authors don't use this sort of phrasing at all (or if they do, I prefer it if they give their notion of "consistency" a very precise definition somewhere else in the paper).<|endoftext|> -TITLE: How can I transform $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^k\sin(\pi rn)}$ into a modular form? -QUESTION [9 upvotes]: Let -$$f_k(z)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^k\sin(\pi zn)}$$ -be a family of holomorphic functions on the upper-half plane $\mathbb{H}=\{a+bi|b>0\}$ for each odd natural number $k$. These functions clearly satisfy the periodicity condition $f_k(z+2)=f_k(z)$, but more surprisingly they also satisfy the formula -\begin{equation} -z^{\frac{1-k}{2}}f_k(z)+z^{\frac{k-1}{2}}f_k\left(\frac{1}{z}\right)=\sum_{j=0}^{k+1}a_{k,j}z^{\frac{k+1}{2}-j}\tag{1} -\end{equation} -where $a_{k,j}$ are given by -\begin{equation} -a_{k,j}:=\frac{(-1)^{\frac{k+1}{2}}(2\pi)^k}{(k+1)!}{k+1\choose j}B_j\left(\frac{1}{2}\right)B_{k+1-j}\left(\frac{1}{2}\right)\tag{2} -\end{equation} -where $B_j(\cdot)$ are the Bernoulli polynomials. I discovered and proved this formula a year ago and didn't think much of it, but recently I have begun thinking about there recurrences in the context of modular/cusp forms and $L$-functions. More specifically, these two formulas seem quite similar to those of a cusp form of weight $1-k$ over the congruence subgroup $\Gamma(2)$. The match isn't perfect because of the extraneous polynomials that pop up on the LHS of (2), but it still makes one wonder. -Another piece of evidence that the functions $f_k(\cdot)$ are connected to modular forms is that they naturally induce $L$-functions in the same way that modular forms do. In this context I mean that if a modular form $g$ of weight $k$ has $q$-expantion $g(z)=\sum_{n}a_n n^{\frac{k-1}{2}}q^n$ then it induces the $L$-function $L(g,s)=\sum_{n}a_n n^{-s}$. The $L$-function attached to $f_k(z)$ is -\begin{align} -L(f_k,s)&=2i\frac{1-2^{1-\frac{k}{2}-s}}{1-2^{-\frac{k}{2}-s}}\zeta\left(s-\frac{k}{2}\right)\zeta\left(s+\frac{k}{2}\right)\\ -&=2i\frac{1-2^{1-\frac{k}{2}-s}}{1-2^{-\frac{k}{2}-s}}\prod_{p}\frac{1}{1-\left(p^{\frac{k}{2}}+p^{-\frac{k}{2}}\right)p^{-s}+p^{-2s}} -\tag{3}\end{align} -Making $L(f_k,s)$ a degree 2 $L$-function with a critical line $\Re(s)=\frac{1}{2}$. $L(f_k,s)$ is not of the Selberg class of $L$-functions since there is no function $\gamma_k(s)$ such that $\Lambda_k(s):=\gamma_k(s)f_k(s)$ satisfies $\Lambda_k(1-s)=\Lambda_k(s)$. -I suspect that the fact that the $L$ function generated is "almost" of the Selberg class corresponds to the fact that $f_k(z)$ is "almost" a cusp form. My overall goal with connecting $f_k(z)$ to modular forms would be to learn something non-trivial about the cosecant function $\frac{1}{\sin(\pi zr)}$ or to use the fact that the vector space of cusp forms $S_k(\Gamma(2))$ is finite dimensional and has a simple basis to find a new way to compute -$$\sigma_k^{(o)}(n):=\sum_{\substack{d|n \\ d\equiv 1\mod{2}}}d^k$$ -since the $q$-expantion of $f_k(z)$ is -$$f_k(z)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\sigma_k^{(o)}(n)}{n^k}q^n$$ - -REPLY [4 votes]: The generalized Eisenstein series -\begin{equation} -\hat{E}_{k,\chi,\psi}(z):=\sum_{n=1}^{\infty}\left(\sum_{d|n}\psi(d)\chi(n/d)d^{k-1}\right)q^n\tag{1} -\end{equation} -Is a eigenform over the space $M_k(RL,\chi\psi)$ for characters $\psi$ and $\chi$ with conductors $L$ and $R$ respectively, whenever $L>1$ and $\psi(-1)\chi(-1)=(-1)^k$. More generally, for any $t>0$ the function $\hat{E}_{k,\chi,\psi}\left(z^t\right)$ is a modular form over $M_k(RLt,\chi\psi)$ according to this online textbook, Theorem 5.8. -We now set $\psi=1$ to be the trivial character and $\chi_=\chi_2$ to be the unique character modulo $2$. This means that $L=2>1$ and $\psi(-1)\chi(-1)=1$ so the conditions of (1) are satisfied if $k$ is even, making $\hat{E}_{k,\chi_2,1}(z)$ a modular (eigen) form. Now, if $k'$ is any odd integer we can let $k=1-k'$ be an even integer, so -\begin{equation} -\hat{E}_{1-k',\chi_2,1}(z)=\sum_{n=1}^{\infty}\left(\sum_{d|n}\chi_2(n/d)d^{-k'}\right)q^n -\end{equation} -is a modular form of degree $1-k'$ over $\Gamma_1(2)$. Since $\sum_{d|n}\chi_2(n/d)d^{-k'}=\frac{\sigma^{(o)}_{k}(n)}{n^k}$, this means that -\begin{equation} -\hat{E}_{1-k,\chi_2,1}(z)=\sum_{n=1}^{\infty}\frac{\sigma^{(o)}_{k}(n)}{n^k}q^n\tag{2} -\end{equation} -Relating this to $f_{k}(z)$ is now a straightforward task. Namely, we see that -\begin{align*} -f_k(2z+1)&=2i\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\sigma_k^{(o)}(n)}{n^k}\exp\left(\frac{2\pi in(2z+1)}{2}\right)\\ -&=-2i\sum_{n=1}^{\infty}\frac{\sigma_k^{(o)}(n)}{n^k}\exp\left(2\pi inz\right)\\ -&=-2i\hat{E}_{1-k,\chi_2,1}(z) -\end{align*} -Thus, the transformation $f_k(2z+1)i/2$ turns $f_k(z)$ into a eigenform. Working out that $f_k(2z+1)i/2$ is an eigenform just from the formulae given in the problem statement seems hard, but also I am not aware of any general formula for Eisenstein series which yields the formulae given in the problem statement. -I find this stuff absolutely fascinating, and I'll be looking more into this and undoubtedly coming up with lots more fun stuff, so if anyone actually ends up reading this and cares about I'm writting then I will add more answers/edit this answer as time goes on to make the solution more complete. Thank you for your time.<|endoftext|> -TITLE: Is this set theory equivalent to ZFC? -QUESTION [11 upvotes]: Consider a variant of set theory with these axioms: - -Extensionality, -Regularity (foundation), -Separation, -Powerset, -Axiom of Choice, and -Transitive closure of a set-like relation is set-like. Update: This did not exactly represent what I had in mind, so the corrected version is given on the next line, and its precise formalization is given below. Sorry for my mistake and ensuing confusion. -The transitive closure of any set under a set-like relation is a set. - -Note that it does not explicitly postulate Pairing, Union, Infinity and Replacement. -Question: Is this set theory equivalent to $\mathrm{ZFC}$? - -Detailed explanation and formalization: - -We use symbol $\prec$ to represent a binary relation. In general, it is a definable class relation, that is a first-order formula with 2 free variables (and, possibly, additional parameters). As usual, we write $a\prec b$ to represent $\prec\!(a,b),$ and we assume that all bound variables in any formula are automatically renamed before a substitution to avoid variable name conflicts that would change its meaning. -We use “is a set” and “exists” as synonyms; “sethood” and “existence” are also synonyms. -We write $a\prec b\prec c$ to represent $a\prec b\land b\prec c$. This notation can also mix several different relation symbols, e.g. $a\prec b\in c$. -When we say that a relation $\prec$ is “set-like”, we mean -$$\color{green}{\forall x\,\exists y\,\forall z\left(z\prec x\;\Rightarrow\; z\in y\right)}.$$ -When we say that “$w$ is a superset of the transitive closure of $s$ under the relation $\prec$”, we mean -$$\color{maroon}{s\subseteq w\,\land\,\forall u\,\forall v\left(u\prec v\in w\;\Rightarrow\; u\in w\right)}.$$ -We also may rephrase it as “the transitive closure of $s$ under $\prec$ is a subset of $w$” or simply “the transitive closure of $s$ under $\prec$ is a set”. At this point, we do not need to define what “the transitive closure” exactly is, because we are only interested in asserting its sethood, so existence of any its superset $w$ is sufficient for our purposes. I suppose that, when the need arises, “the transitive closure” can be defined as the smallest such set, and can be carved out of its superset using Separation. -Our last axiom asserts that, provided $\prec$ is a set-like relation, the transitive closure of any set $s$ under that relation $\prec$ is a set. It can be formalized using the following axiom schema where $\prec$ ranges over all binary relations: -$$\left(\vphantom{\Large|}\color{green}{\forall x\,\exists y\,\forall z\left(z\prec x\,\Rightarrow\,z\in y\right)}\right)\,\Rightarrow\,\forall s\,\exists w\!\left(\vphantom{\Large|}\color{maroon}{s\subseteq w\,\land\,\forall u\,\forall v\left(u\prec v\in w\,\Rightarrow\,u\in w\right)}\right)\!.$$ - -REPLY [12 votes]: Accepting the convention that it is a logical axiom that the universe is nonempty, the answer is yes. We will formalize the transitive closure axiom schema (TC) as follows: for any definable (with parameters) binary relation $R,$ if for all $x,$ $\{y: y R x\}$ is a set, then for all $x,$ there is a set $T$ such that $x \in T$ and $T$ is closed downwards under $R.$ (*) Of course, this can only be weaker than asserting the existence of a minimum such $T.$ -For efficiency, we will prove Pairing, Union, Infinity, and Replacement from Extensionality, Separation, and TC. -Pairing: We first note that $\emptyset$ exists by applying separation to an arbitrary set. Next, for all $x,$ $\{x\}$ exists by applying TC to $x$ and the empty relation. Finally, for all $x, y,$ we get $\{x,y\}$ by applying TC to $x$ and the relation defined by $a R b$ iff $b = x$ and $a=y.$ -Union: Fix a set $S.$ By Separation and Russell's paradox, there is $x \not \in S.$ Define $R$ by $a R b$ iff $b = x$ and $a \in S$ or $b \in S$ and $a \in b.$ Then we get $\bigcup S$ by applying TC to $x$ and $R.$ -Infinity: Define a relation $R$ by $a R b$ iff $a$ and $b$ are natural numbers and $a=b+1.$ Then $\omega$ exists by applying TC to $\emptyset$ and $R.$ -Replacement: Fix a set $S$ and a definable function $F.$ Fix $x \not \in S.$ Define $R$ by $a R b$ iff $b=x$ and $a \in S$ or $b \in S$ and $a = F(b).$ Then we get $F"S$ by applying TC to $x$ and $R.$ -(*) Note that my formulation of TC only makes sense under the convention that the transitive closure of a relation is reflexive. Without this convention, then it's not clear we can prove the existence of $\{x\}$ from the axioms I specified. Of course, we can prove it exists from Separation and Power Set, which is included in the axioms listed in the question, but that feels overpowered for our purposes. -Edit: The question was updated with the intended formalization of the transitive closure schema. My TC here follows from Vladimir's version plus existence of $\{x\}$ for all $x,$ and the latter follows from Separation and Power Set.<|endoftext|> -TITLE: Does the random graph interpret the random directed graph? -QUESTION [8 upvotes]: The random graph is the Fraisse limit of the class of finite graphs, the random directed graph is the Fraisse limit of the class of directed graphs, a directed graph is just a set with a binary relation. -It's easy to see that the random directed graph interprets the random graph, in fact the second is a reduct of the first. I am curious to know about the other direction. I don't see an easy way to do it, or an easy way to rule it out. More generally, I would be curious have pointers to research on when countable homogeneous structures interpret other countable homogeneous structures. I don't know if anyone has thought about this, I'm not very familiar with that terrain. -Motivation: I am doing some work with a weak notion of interpretability. I show that the random graph weakly interprets the random directed graph, and I use this as a fairly important lemma. So I'm curious to know if anything is known/obvious to experts about actual interpretations. If there is an actual interpretation then it would be a bit silly to spend a page constructing a weak interpretation. -I am actually thinking about the random $k$-ary hypergraph and the random $k$-ary relation. - -REPLY [7 votes]: No, the random graph cannot interpret the random binary relation. -I’ll just answer the title question with the goal of illustrating a technique; I haven't considered $k$-ary structures. The approach is to use the property of least supports to set up a counting argument. Some equivalences are discussed at Least supports and weak elimination of imaginaries. Combined with $\omega$-categoricity we get a useful classification of all definable equivalence relations. I’ll use the following result with $M$ being the random graph. - -Fact. [1, Lemma 2.7(ii)] If $X\subseteq M^n$ is invariant, there is a unique smallest set $D\subset M$ such that $X$ is $D$-invariant. - -“$D$-invariant” means $X$ is preserved setwise by all automorphisms that fix $D$ elementwise. “Invariant” means $D_0$-invariant for some finite $D_0.$ Since $M$ has an $\omega$-categorical theory, invariant is the same as definable, and $D$-invariant is the same as $D$-definable. The proof in [1] uses the free amalgamation property. I’m not sure whether this special case is obvious or not. -Fix: - -a finite set $M_0\subset M$ -an $M_0$-definable set $X\subseteq M^n$ -an $M_0$-definable equivalence relation ${\sim}\subseteq X\times X,$ and -an $M_0$-definable relation $R\subseteq X\times X$ that respects ${\sim}$ i.e. it’s the preimage of a relation $R/{\sim}\subseteq (X/{\sim})\times(X/{\sim}).$ - -We aim to show that $(X/{\sim}, R/{\sim})$ is not isomorphic to the random binary relation. -For each complete $n$-type $p$ over $M_0$ let $X_p=\{x\in X:M\Vdash p(x)\}.$ Let $P=\{p:X_p\neq\emptyset\}.$ Pick representatives $x_p\in X_p$ for each $p\in P.$ Let $x_p/{\sim}$ denote the ${\sim}$-equivalence class of $x_p.$ By the Lemma, for each $p$ there is a unique smallest set $D$ such that $x_p/{\sim}$ is $D$-definable. We must have $D\subset \operatorname{rng}(x_p).$ Let $I_p\subset \{1,\dots,n\}$ be a minimal set of indices such that $\operatorname{rng}(x|I_p)=D$ - this step is necessary because $x_p$ might have repeated components. Let $\hat{p}$ denote the type of $x|I_p$ over $M_0,$ implicitly reindexing if necessary, or allowing types to use free variables $x_i,i\in I_p$ instead of the usual contiguous $x_1,\dots,x_{|I_p|}.$ Note that $I_p$ and $\hat p$ do not depend on the choice of $x_p\in X_p.$ -By uniqueness, whenever $x,y\in X_p$ have different supports, i.e. $\operatorname{rng}(x|I_p)\neq \operatorname{rng}(y|I_p),$ then $x\not\sim y.$ So $x/{\sim}$ is coded by $x|I_p$ modulo perhaps a permutation group acting on the indices of $I_p.$ -For each $p,q\in P$ define $R_{p,q}$ on vectors $\hat{x}\in X^{I_p}$ and $\hat{y}\in X^{I_q}$ by -$$R_{p,q}(\hat x,\hat y)\iff (\exists x\in X_p)(\exists y\in X_q)(x\supseteq \hat x \wedge y \supseteq \hat y \wedge R(x,y))$$ -Then for $x\in X_p$ and $y\in X_q$ we have $R_{p,q}(x|I_p,y|I_q)\iff R(x,y)$ because the choice of extension doesn’t affect the equivalence classes: by definition of $I_p$ any automorphism fixing $x|I_p$ will fix $x/{\sim},$ and similarly for $y|I_q.$ The relation $R_{p,q}$ is $M_0$-definable. -Let $d=\max_{p\in P}|I_p|.$ If $d=0$ then $X/{\sim}$ is finite, which is absurd. The $d=1$ case could be handled by symmetry, but the following counting argument happens to cover this case too. So assume $d\geq 1.$ -Pick $N$ large enough that $$4^{N^d}>|P|(N^d+2^{Nd})^d.$$ -Pick $p$ with $|I_p|=d.$ Use the extension property to pick $dN$ distinct vertices $x_{i,j}$ where $i\in I_p$ and $1\leq j\leq N,$ satisfying $M\Vdash \hat p((x_{i,{j_i}})_{i\in I_p})$ for each $j: I_p\to \{1,\dots,N\}.$ In graph theory terms we’re blowing up each vertex $(x_p)_i$ into $N$ vertices. It doesn’t matter whether there are edges between $x_{i,j}$ and $x_{i,k}$ for the same $i.$ -For each $q\in P$ I estimate that there are at most $(N^d+2^{Nd})^d$ types over $M_0$ realized by tuples of the form $((x_{i,j})_{i\in I_p,1\leq j\leq N},y)$ with $y\in X^{I_q}$ and $M\Vdash \hat{q}(y).$ This is because the type of the tuple is determined by the binary relations, and the only freedom is how the $1$-type of each $y_k$ over $M_0$ extends to a $1$-type over $M_0\cup \{x_{i,j}:i\in I_p, 1\leq j\leq N\}.$ There are at most $N^d+2^{Nd}$ such extensions: either $y_k=x_{i,j}$ for some $(i,j),$ or else there are the $2^{Nd}$ choices of whether or not each $\{x_{i,j},y_k\}$ is an edge. -Each $y\in X$ determines values $z_y(j)\in \{1,2,3,4\}$ for each $j: I_p\to \{1,\dots,N\},$ according to whether $R_{p,q}(x_{i,j_i},y)$ and $R_{q,p}(y,x_{i,j_i})$ are false/false, false/true, true/false or true/true. By the estimate in the previous paragraph, there are at most $|P|(N^d+2^{Nd})^d$ distinct functions of the form $z_y.$ So there is some $z\in 4^{N^{I_p}}$ not equal to $z_y$ for any $y.$ In terms of the original interpretation, we can pick $\xi^j\in X_p$ with $\xi^j_i=x_{i,j_i}.$ The equivalence classes $\xi^j/{\sim}$ are distinct because they have different supports. The extension property for the random binary relation requires that there exists $y\in X$ such that $R(\xi^j, y)$ and $R(y,\xi^j)$ are false/false etc according to $z(j).$ But then $z=z_y,$ contradicting the choice of $z.$ -[1] Macpherson, Dugald; Tent, Katrin, Simplicity of some automorphism groups., J. Algebra 342, No. 1, 40-52 (2011). ZBL1244.20002.<|endoftext|> -TITLE: how to compute Hilbert class field of $\Bbb Q(\zeta_{23})$? -QUESTION [8 upvotes]: I want to construct the Hilbert class field of $K=\Bbb Q(\zeta_{23}).$ I have no clue how to construct it except that I know that $[H(K):K]=3$ from Sage. Any references or comments are appreciated. - -REPLY [18 votes]: The cyclotomic field $K=\mathbb{Q}(\zeta_{23})$ contains the quadratic field $F=\mathbb{Q}(\sqrt{-23})$, and $F$ has class number $3$ (it is the first quadratic field, when those are ordered by absolute value of the discriminant, that has class number divisible by $3$). Thus, $F$ has an unramified cubic extensions $H$, and since the degree of $K$ is coprime with $3$, the extensions $H$ and $K$ are disjoint, so that the compositum $HK$ is an unramified cubic extension of $K$. Thus, your problem is reduced to finding the Hilbert class field of $F$. Magma (and presumably also Sage?) will just give it to you. It is the splitting field over $\mathbb{Q}$ of the cubic polynomial $x^3-x+1$. -In summary, the Hilbert class field of $K$ is obtained by adjoining to $K$ a root of $x^3-x+1$.<|endoftext|> -TITLE: Non-conjugate subgroups that are conjugate in complexification -QUESTION [12 upvotes]: In trying to come up with a counter-example in my line of research, I would like to find an example as follows: -$G$ is a semisimple Lie group with complexification $G^{\mathbb{C}}$. $H_1, H_2 \subseteq G$ are subgroups that are not conjugate as subgroups of $G$ but are conjugate as subgroups of $G^{\mathbb{C}}$. -Here is why I suspect this might be possible: we can certainly find pairs of elements that are conjugate in $G^{\mathbb{C}}$ but not in $G$. For instance, with $G = \text{SL}(2, \mathbb{R})$, $G^{\mathbb{C}} = \text{SL}(2,\mathbb{C})$ and $x = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $y = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$ then $x$ and $y$ are conjugates in $G^{\mathbb{C}}$ as $x$ is simply the Jordan canonical form of $y$ but they are not conjugate in $G$ as any $z \in \text{GL}(2, \mathbb{R})$ with $z^{-1} y z = x$ has negative determinant. -Unfortunately the example $G = \text{SL}(2, \mathbb{R})$ does not work at the level of subgroups: the two $2$-dimensional subgroups of $\text{SL}(2, \mathbb{R})$ are not equivalent in $\text{SL}(2, \mathbb{C})$ and the one-parameter subgroups are classified by the traces of their elements and are therefore equivalent in $\text{SL}(2, \mathbb{C})$ if and only if they are in $\text{SL}(2, \mathbb{R})$. - -REPLY [14 votes]: Define $M_s=\begin{pmatrix}0 & 0 & 1 & 0\\ 0 & 0 & 0 & s\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}$, and $G_s=\exp(\mathbf{R}M_s)\subset\mathrm{SL}_4(\mathbf{R})$. Then $G_1$ and $G_{-1}$ are not conjugate in $\mathrm{SL}_4(\mathbf{R})$ while they are conjugate in $\mathrm{SL}_4(\mathbf{C})$. - -$M_1$ and $M_{-1}$ are conjugate in $\mathrm{GL}_4$. Since over $\mathbf{C}$ any two conjugate matrices are conjugate by a matrix of determinant one (multiply the conjugating matrix by a suitable scalar). So $M_1$ and $M_{-1}$ are conjugate in $\mathrm{SL}_4(\mathbf{C})$. -  -Argument for non-conjugation over $\mathbf{R}$: one has to check that $M_1$ is conjugate to no scalar multiple of $M_{-1}$: -Let $M_E$ be the block matrix $\begin{pmatrix}0& E\\ 0 & 0\end{pmatrix}$, for $E$ some square matrix. Suppose $E,F$ invertible. If $UM_EU^{-1}=M_F$, $U$ preserves the common kernel of $E$ and $F$, is block upper-triangular, say of the form $\begin{pmatrix} A & C \\ 0 & B\end{pmatrix}$. The conjugacy then writes as $AE=FB$. So $\det(A)\det(E)=\det(F)\det(B)$. If $U$ has determinant 1, then $\det(B)=\det(A)^{-1}$. So $\det(F)\det(E)^{-1}=\det(A)^2$ is a square. In the present case, $E=I_2$ and $F=\lambda\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}$ has negative determinant so $\det(F)\det(E)^{-1}=-\lambda^2$.<|endoftext|> -TITLE: Arithmetic progressions in inverse image of totient function -QUESTION [7 upvotes]: I noticed on the OEIS that there are various sequences (e.g. A050515-A050520) that describe arithmetic progressions whose totients are all equal. For example, we have -$$\varphi(\{1,2\}) = 1$$ -$$\varphi(\{8,10,12\}) = 4$$ -$$\varphi(\{72,78,84,90\}) = 24$$ -$$\varphi(353640 + [0,4]\cdot 210) = 80640$$ -$$\varphi(583200 + [0,5]\cdot 30) = 155520$$ -$$\varphi(1158419010 + [0,6]\cdot 210) = 264781440$$ -(The notation above is somewhat sloppy; I'm trying to say that Euler's totient function -equals the right-hand side when evaluated on anything in the set.) Is there a simple reason to believe that this pattern should continue? In other words, for any $k$, does there exist an arithmetic progression $P$ of length $k$ such that $\varphi$ is constant on $P$? -A lemma of H. Gupta [Indian J. Pure Appl. Math, 12(1) (1981)] states that a solution to $\varphi(x) = n$ cannot exceed $A(n)$ where -$$A(n) = n\prod_{p\backslash n} {p\over p-1}$$ -(the product is taken over all primes $p$ dividing $n$), so my first thought was that some kind of density argument could be made. But it actually does not seem like this is the reason for the arithmetic progressions popping up. Indeed, the ratio -$${|\varphi^{-1}(n)|\over A(n)}$$ -does not seem to have a positive $\limsup$ as $n$ gets large. One observation is that if $\varphi$ is constant on an arithmetic progression $P$, it is also constant on any dilate $m\cdot P$ of $P$, and some of these dilates might be contained in larger progressions on which $\varphi$ is also constant. - -REPLY [5 votes]: We have a conditional result. -See Theorem 4 of S. W. Graham, J. J. Holt, C. Pomerance, On the Solutions to $\phi(n) = \phi(n + k)$, -Number Theory in Progress, eds. K. Gyory, H. Iwaniec, and J. Urbanowicz, eds., -Vol. 2 (de Gruyter, Berlin, 1999), pp. 867-882. -Theorem 4 in above: -Suppose that $j$, $j+k$, $\ldots$, $j+qk$ all have the same prime factors. Define $B=\prod_{i=0}^q (j+ik)$. For $i=0,\ldots q$, let -$$ -b_i=\frac B{j+ik}, \ g=\mathrm{gcd}(b_0,b_1,\ldots, b_q), -$$ -$$ -a_i=\frac{b_i}g=\frac B{(j+ik)g}. -$$ -Suppose that for some positive integer $r$, -$$ -a_0r+1, a_1r+1,\ldots, a_qr+1 -$$ -are all primes that do not divide $j$. If $n=j(a_0r+1)=\frac{Br}g+j$, then -$$ -\phi(n)=\phi(n+k)=\ldots=\phi(n+qk).$$ -The first requirement $j, j+k, \ldots, j+qk$ all have the same prime factors is satisfied with $j=(q+1)\#$ and $k=j$, where $N\#=\prod_{p\leq N} p$ is the primorial. -The second requirement $a_0r+1, \ldots, a_qr+1$ are all primes, is nontrivial. Currently, this can be achieved conditionally on Prime $k$-tuples Conjecture.<|endoftext|> -TITLE: Modularity of higher genus curves -QUESTION [10 upvotes]: The modularity conjecture for elliptic curves over number fields is well known, and indeed, is a theorem for all elliptic curves over $\mathbb{Q}$, and at least potentially, over any CM field. -What is the precise statement of the conjecture for higher genus curves? What are the modular/automorphic forms we expect to correspond to Galois representations realized in the $l$-adic cohomology of a smooth projective algebraic curve of genus $g$ > 1 over a number field $F$ via equality of $L$ functions? What is the state of progress towards the conjecture? References would be very welcome. - -REPLY [2 votes]: Following the suggestion of Faris I looked at Abelian Surfaces over Totally Real Fields are Potentially Modular by Boxer, Calegari, Gee & Pilloni, whose section 1.4.1 discusses the modularity conjecure for higher genus curves and points to On the Langlands Correspondence for Symplectic Motives by Benedict Gross. -Gross constructs a new form in the generic cuspidal automorphic representations of split orthogonal groups $SO_{2g+1}$ which are conjecturally attached by global Langlands correspondence to discrete symplectic motives over $\mathbb{Q}$ of rank $2g$. Some of these motives are $H^1(.)(1) = H_1(.)$ of genus $g$ curves over $\mathbb{Q}$, and more generally of polarized abelian varieties of rank $g$ over $\mathbb{Q}$ whose endomorphism rings are an order in a product of totally real fields. The $l$-adic realizations of these motives are Galois representations $G_\mathbb{Q} \rightarrow GSp_{2g}(\mathbb{Q}_l)$. The Langlands dual of $GSp_{2g}$ is $GSpin_{2g+1}$, interpreted here as a split orthogonal group. -The (conjecturally automorphic) representation attached to such a motive is put together from explicit local representations at all places of $SO_{2n+1}$. The explicit new form is built as restricted tensor product of local forms. -Results in the converse direction - attaching a symplectic motive to an automorphic representation of $SO_{2n+1}$ - appear in Potential Automorphy and Change of Weight by Barnet-Lamb, Gee, Geraghty, and Taylor. -Please feel free to strengthen this answer with more details and generalizations to other number fields.<|endoftext|> -TITLE: Intersection point of three circles -QUESTION [11 upvotes]: Can you provide a proof for the following proposition: - -Proposition. Let $\triangle ABC$ be an arbitrary triangle with orthocenter $H$. Let $D,E,F$ be a midpoints of the $AB$,$BC$ and $AC$ , respectively. Let $A'$ be a reflection of the point $A$ with respect to the point $E$ , $B'$ reflection of the point $B$ with respect to the point $F$ and $C'$ reflection of the point $C$ with respect to the point $D$. Consider the three circles $k_1,k_2,k_3$ defined by the points $AHA'$ , $BHB'$ and $CHC$' , respectively. I claim that $k_1$,$k_2$ and $k_3$ meet at a common point $P$. - - -GeoGebra applet that demonstrates this proposition can be found here. - -REPLY [11 votes]: The points $A,B,C$ are midpoints of the sides of $\triangle A'B'C'$, thus $H$ is the centre of the circumcircle $\omega$ of $\triangle A'B'C'$. Make an inversion with respect to $\omega$. The point $A$ maps to the intersection point $A_1$ of the tangents at $B',C'$ to $\omega$ (these tangents are the images of the circles $HB'A$, $HC'A$). -So, in triangle $A_1 B_1 C_1$ we join the vertices with the points of tangency of the incircle (or excircle) with the respective sides, and should prove that such three lines are concurrent. This is well known and follows from Ceva theorem, for example.<|endoftext|> -TITLE: Rank of jacobians of twists of hyperelliptic curves of genus one -QUESTION [5 upvotes]: For $a,b \in \mathbb{Z}$ we define the binary quartic form -$$\displaystyle F_{a,b}(u,v) = a(u^2 - v^2)^2 + 4bu^2 v^2.$$ -We shall assume throughout that the discriminant -$$\Delta(F_{a,b}) = 4096a^2 b^2 (a-b)^2$$ -of $F_{a,b}$ is non-zero; that is, the form $F_{a,b}$ is non-singular. Consider the twist family of genus one curves given by -$$\displaystyle d z^2 = F_{a,b}(u,v), d \in \mathbb{Z}.$$ -Is it known that for infinitely many $d$, the curve given above has a rational point and has jacobian whose rank is at least one (so that the curve is isomorphic to its jacobian, which has positive rank)? -The Jacobian of the curve $z^2 = F_{a,b}(u,v)$ (i.e., when $d = 1$) is given by -$$y^2 = x^3 - \frac{16(a^2 - ab + b^2)}{3} - \frac{64(a+b)(2a-b)(a-2b}{27} $$ -$$= \left( x - \frac{4b - 8a}{3}\right)\left( x - \frac{4a - 8b}{3}\right)\left( x - \frac{4b + 4a}{3}\right),$$ -which is an elliptic curve with full rational 2-torsion. Therefore it is amenable to the techniques introduced by A. Smith (https://arxiv.org/abs/1702.02325), so we know that for 100% of quadratic twists of this curve the $2^\infty$-Selmer rank is at most one. In particular, ordering the twists by $|d| \leq X$, that only $o(X)$ of such twists will have rank at least two. -This is a stronger conclusion that can be reached for any large family of elliptic curves, since in that case we only have control of the sizes of $p$-Selmer groups for $p = 2,3,5$ on average due to work of Bhargava and Shankar. However in this setting there is work due to Bhargava and Skinner, and subsequently Bhargava-Skinner-Zhang, which shows that in fact a positive proportion of curves will have positive rank in any large family. - -REPLY [2 votes]: The answer is yes, and it's fairly elementary. By the usual 2-descent, the curve $C$ gives a class $c$ in $H^1(\mathbb{Q},E[2])$, where $E$ is the Jacobian you wrote down. As you vary $d$, the groups $H^1(\mathbb{Q},E_d[2])$ are canonically isomorphic, and $c$ is also the class of $C_d$. To answer your question, you should condition on the possibility that $c$ "comes from" the 2-torsion subgroup of $E$ via $E(\mathbb{Q})/2E(\mathbb{Q}) \to H^1(\mathbb{Q},E[2])$. I don't know what the conditions on $a,b$ for this to happen are, but it won't matter. -If $C$ doesn't come from 2-torsion, and if $C_d$ has a rational point, then automatically $E_d$ has rank at least one (by descent). And as Alex B. says, it's easy to see that there are infinitely many squarefree $d$ for which this happens (his argument doesn't quite show this, but it's not hard to fiddle with congruence conditions to produce infinitely many squareclasses). -If $C$ does come from 2-torsion, then $C_d$ has a rational point for all $d$, but there is no guarantee that the rank is at least 1. So your question reduces to asking about the ranks of the twists of $E$. But now just apply the argument from the previous paragraph to some other class in $H^1(\mathbb{Q},E[2])$ represented by a binary quartic form (there are infinitely many of these).<|endoftext|> -TITLE: On the finiteness of an Auslander-Reiten component -QUESTION [6 upvotes]: I am reading a paper called A NOTE ON THE RADICAL OF A MODULE CATEGORY by -CLAUDIA CHAIO AND SHIPING LIU. This is Theorem 2.7: -And this is part of it's proof, in which the direction (2) $\Rightarrow $ (1) is shown. - -$\iota_S: S \rightarrow I(S)$ denotes the injective envelope and $\pi_S: P(S) \rightarrow S$ denotes the projective cover of a simple module $S$. Also dp$(f)$ denotes the depth of $f$. -As far as I know a quiver is locally finite means iff between each two vertices there is only a finite number of arrows between them -I really don't understand why it follows that the Auslander-Reiten component $\Gamma$ is finite. Also why does $\Gamma$ contain at most finitely many indecomposable injective modules? -Can anybody help with his? Any help is highly appreciated! - -REPLY [2 votes]: Since $A$ is an Artin algebra, it has only finitely many indecomposable injective modules in total (up to isomorphism), so there are finitely many in $\Gamma$. -In a locally finite quiver, given any $d\geq0$ and vertex $v$, there are finitely many paths ending at $v$ and having length at most $d$, because there are finitely many choices for each arrow. In particular, $v$ is reachable from only finitely many vertices by such paths. -This means that there only finitely many vertices of $\Gamma$ from which one of the finitely many indecomposable injectives may be reached via a path of length at most $d$, for any fixed $d>0$. Since the proof exhibits an $r$ such that every vertex of $\Gamma$ has a path of length at most $r$ to some indecomposable injective, it follows that $\Gamma$ has finitely many vertices.<|endoftext|> -TITLE: Ehrhart period collapse for $123\ldots k$-avoiding Birkhoff polytope? -QUESTION [14 upvotes]: For $1 \leq r \leq n$, let $\mathcal{B}^n_r$ denote the polytope of all real matrices -$$ \pi = \begin{pmatrix} \pi_{1,1} & \pi_{1,2} & \cdots & \pi_{1,n} \\ -\pi_{2,1} & \ddots & \cdots & \pi_{2,n} \\ -\vdots & \ddots & \ddots & \vdots \\ -\pi_{n,1} & \cdots & \cdots & \pi_{n,n} \end{pmatrix} \in \mathbb{R}^{n\times n}$$ -for which - -all entries are nonnegative: $\pi_{i,j}\geq 0$; -the sum along any row or column is equal to one: $\sum_{j}\pi_{i,j}=1$ for all $i$; $\sum_{i}\pi_{i,j}=1$ for all $j$. -the sum along any upper-left to lower-right chain of entries is at most $r$: $\sum_{k} \pi_{i_k,j_k} \leq r$ for all $(i_1,j_1)<(i_2,j_2) < \cdots < (i_m,j_m)$, where $(i,j) < (i',j')$ means $i\leq i'$, $j\leq j'$ with at least one of these inequalities being strict. - -By definition $\mathcal{B}^n_n$ is the Birkhoff polytope of doubly-stochastic matrices. In general $\mathcal{B}^n_r$ is a subpolytope of the Birkhoff polytope. -It is well-known that $\mathcal{B}^n_n$ is the convex hull of all permutation matrices, and so in particular $\mathcal{B}^n_n$ is an integral polytope. But this is not true of the $\mathcal{B}^n_r$ in general: for instance, -$$ \begin{pmatrix} 0.5 & 0 & 0.5 \\ 0 & 1 & 0 \\ 0.5 & 0 & 0.5 \end{pmatrix}$$ -is a vertex of $\mathcal{B}^3_2$. The vertices of $\mathcal{B}^n_r$ which are integral are precisely the permutation matrices of $123...r+1$-avoiding permutations. -For a polytope $\mathcal{P} \subseteq \mathbb{R}^n$ I use $L(\mathcal{P};t)$ to denote the Ehrhart function which at nonnegative integers $t$ counts the number of lattice points of $t\mathcal{P}$: -$$ L(\mathcal{P};t) := \# (t\mathcal{P}\cap \mathbb{Z}^n).$$ -Since $\mathcal{B}^n_r$ is not an integral polytope, but it is a rational polytope, its Ehrhart function $L(\mathcal{B}^n_r;t)$ is a priori only a quasipolynomial in $t$. -Question: Is $L(\mathcal{B}^n_r;t)$ in fact always an honest polynomial in $t$? -I have verified this for $1\leq r \leq n \leq 5$ via Sage. The computation that took the longest was $L(\mathcal{B}^5_2;t)$ which is equal to -(5959/249080832000) * (t + 1) * (t + 2) * (t + 3) * (t + 4) * (t^12 + 30*t^11 + 2534915/5959*t^10 + 22404750/5959*t^9 + 137606217/5959*t^8 + 620455590/5959*t^7 + 2117653385/5959*t^6 + 5561311650/5959*t^5 + 11311600324/5959*t^4 + 17737953240/5959*t^3 + 21126074400/5959*t^2 + 18162144000/5959*t + 10378368000/5959) -$\mathcal{B}^5_2$ has over 3000 vertices. The other data is available upon request. Probably someone with better computer skills than me can produce more data. -This phenomenon whereby an Ehrhart quasipolynomial has a smaller period than a priori predicted, or in extreme cases is in fact an honest polynomial, is called Ehrhart period collapse. It has apparently attracted some attention but remains mysterious. -Even if the above question has a negative answer, I'd still be interested in what could be said about the integer points $t\mathcal{B}^n_r\cap \mathbb{Z}^{n\times n}$. -P.S., Happy New Year's and here's to a better 2021! - -REPLY [4 votes]: The Ehrhart function $L(\mathcal B^n_r;t)$ is an honest polynomial. We will show this by following Per's suggestion and proving that it coincides with the Ehrhart function of a certain Gelfand-Tsetlin polytope. The following two steps can probably be combined into one, but I thought it was natural to think of it this way. -Step 1: For every $k\in \mathbb N$, the lattice points of $k\mathcal B_r^n$ are in bijection with the set of arrays $\tau\in \mathbb Z_{\geq 0} ^{n\times n}$ that are plane partitions, meaning that $i\geq i', j\geq j'$ implies $\tau_{i,j}\le \tau_{i',j'}$, that also satisfy $kr\geq \tau_{1,1}$ and for each $-n\le s\le n$ we have the traces -$$tr_s(\tau)=\sum_{i}\tau_{i,i+s}=k(n-|s|).$$ -Proof: This bijection is simply RSK applied to the array $\pi$. The entry $\tau_{1,1}$ records the length of longest increasing subsequence in the biword recording $\pi$, which is exactly the largest possible sum of $\pi_{i,j}$'s along an upper left-lower right chain. This is where the restriction $kr\geq \tau_{1,1}$ comes from. -Let's introduce the partitions $\lambda^n_r=\{r^n0^n\}$ and $\mu^n_r=\{(r-1)^n1^n\}$. -Step 2: The plane partitions from the previous lemma are in bijection with the lattice points in the Gelfand-Tsetlin polytope $GT(k\lambda^n_r, k\mu^n_r)=kGT(\lambda^n_r, \mu^n_r)$, therefore they are counted by the Kostka number $K(k\lambda^n_r, k\mu^n_r)$. -Proof: If we write a Gelfand-Tseltin pattern with top row given by $k\lambda^n_r=\{kr,kr,\dots, kr, 0, 0,\dots, 0\}$ we can erase the leftmost entries that are forced to be $=kr$ and the rightmost entries that are forced to be $=0$ to be left with a ($45^{\circ}$ rotated) $n\times n$ array. This array is a plane partition with largest part $\le kr$ and the appropriate traces. -For $k\mathcal B^{4}_2$ this bijection looks like simply selecting the array of $*$'s. -$$\begin{matrix} -2k && 2k && 2k && 2k && 0 && 0 && 0 && 0\\ -& 2k && 2k && 2k && * && 0 && 0 && 0 \\ -&& 2k && 2k && * && * && 0 && 0\\ -&&& 2k && * && * && * && 0 \\ -&&&& * && * && * && * \\ -&&&&& * && * && * \\ -&&&&&& * && * \\ -&&&&&&& * -\end{matrix}$$ -Combining both steps gives $L(\mathcal B^n_r ;t)=L(GT(\lambda^n_r, \mu^n_r);t)$ which implies the desired polynomiality. It is also expected that positivity of coefficients holds, but I believe it is not known yet.<|endoftext|> -TITLE: Example of an invariant metric on a nilpotent group which is not asymptotically geodesic -QUESTION [5 upvotes]: Let $X$ be a metric space. We say that $X$ is asymptotically geodesic if for all $\epsilon > 0$, there exists $R > 0$ such that, for all $x,y \in X$, there exists some finite sequence of points $p_0=x,p_1,p_2,...,p_N = y \in X$ such that each $d(p_{i-1},p_i) \le R$ and -$$ \sum_{i=1}^N d(p_{i-1},p_i) \le (1 + \epsilon)d(x,y). $$ -I came across this definition while studying finitely generated nilpotent groups and their scaling limits. Pansu proved in "Croissance des boules et des geodesiques fermees dans les nilvarietes" that, if a finitely generated (virtually) nilpotent group $\Gamma$ is endowed with a left-invariant metric $d$ such that $(\Gamma,d)$ is asymptotically geodesic, then the sequence of metric spaces $(\Gamma,\frac{1}{n}d)$ has a Gromov-Hausdorff limit (which he describes). -However, I've been having a lot of difficulties coming up with examples of metrics on nilpotent groups which aren't asymptotically geodesic, particularly under the extra assumption that the metric in question is bi-Lipschitz to a word metric on $\Gamma$. -I believe that it follows pretty quickly from Fekete's lemma that any invariant metric on $\mathbb{Z}$ which is bi-Lipschitz to the standard metric is asymptotically geodesic, and, unless I made a mistake, the same is true for $\mathbb{Z}^d$ by a more involved but fairly similar argument. But the property of being asymptotically geodesic is certainly not preserved by bi-Lipschitz maps a priori, and it seems implicit in the literature that there should be some metrics bi-Lipschitz to word metrics which are not asymptotically geodesic. (For instance, in Benjamini and Tessera, "First passage percolation in nilpotent Cayley graphs and beyond", one of the key ingredients in a proof is showing that a certain invariant metric is asymptotically geodesic, even though that metric is assumed to be bi-Lipschitz to a word metric). -So my question is: -(1) What is an example of a left-invariant metric on a finitely generated nilpotent group which is bi-Lipschitz to a word metric but not asymptotically geodesic? -As it happens, I actually have had difficulty constructing such a metric even if I drop the assumption that it is bi-Lipschitz to a word metric, so I would also be interested in the weaker question: -(2) What is an example of a left-invariant metric on a finitely generated nilpotent group which is not asymptotically geodesic? -And even though I'm interested primarily in the case of nilpotent groups, I'm also generally trying to understand this property, so an answer to the even weaker question might help too: -(3) What is an example of a left-invariant metric on a (finitely generated?) discrete group which is not asymptotically geodesic? - -REPLY [2 votes]: This answer seems to essentially be what @YCor was going for in his comments, but I'll give very explicit examples, since at the level he described the answer, there were some details that needed to be checked that weren't immediately clear to me. (I hope this is an appropriate situation to answer my own question.) -First, without any requirement of being bi-Lipschitz to a word metric, @YCor's first comment suggests a very easy example on $\mathbb{Z}$, that is, -$d(m,n) = \sqrt{|n-m|}$. It's helpful to remember that the square root of any metric is a metric, and taking the square root is a nice way to ruin asymptotic geodesicity. Anyone reading this should work out why asymptotic geodesicity fails in this example in order to understand the next one. -Now, for an example of a metric which is bi-Lipschitz to a word metric but not asymptotically geodesic (suggested by @YCor's second comment). -Take $\Gamma$ to be the Heisenberg group -$$ \Gamma := \langle X, Y, Z | [X,Y]=Z, [X,Z]=[Y,Z]=1\rangle.$$ -It's not too hard to show that each element of $\Gamma$ can be written uniquely -as $X^k Y^l Z^m$, and so we have a bijection $X^k Y^l Z^m \leftrightarrow (k,l,m)$ -between $\Gamma$ and $\mathbb{Z}^3$. -It's also a standard exercise to show that, for some constants $0 0$, -the function $\max(|k|,|l|,\epsilon\sqrt{|m|})$ is also bi-Lipschitz, and the smaller $\epsilon$ is, the easier it is to confirm that this function is subadditive and that it is not asymptotically geodesic. Here the heuristic to see that this metric is not asymptotically geodesic is the following: one cannot get an approximate geodesic from the identity to $Z^M$ by traveling along powers of $Z$ for the same reason that the square root of the standard metric on $\mathbb{Z}$ is not asymptotically geodesic. The paths which should approximate geodesics should come from words in $X$ and $Y$, but the length of such paths will be roughly some constant (independent of $\epsilon$) times $\sqrt{M}$, while the prescribed distance from the identity to $Z^M$ is -$\epsilon \sqrt{M}$.<|endoftext|> -TITLE: Quillen, Merkurjev and Suslin results about K2 of a conic -QUESTION [9 upvotes]: Let $X$ be a conic without rational points over a field $F$ and $Q$ its associated quaternion algebra. The paper -https://www.math.ucla.edu/~merkurev/papers/residue.pdf -presents a proof of the exactness of the sequence -$$(1) \qquad K_2(F) \to K_2(F(X)) \to \sum_x K_1(F(x))$$ -where $F(X)$ is the field of functions of $X$ and the sum is over all points $x$ of the conic, $F(x)$ being the residual field at the point $x$. This proof is elementary and involves computations in the quaternion algebra $Q$. The statement also is elementary, involving only $K_2$ of some fields. -As explained in this paper, the fact that (1) is exact was first discovered by Suslin 1982, by a more abstract argument using results of Quillen that -$$K_2(X) \to K_2(F(X)) \to \sum_x K_1(F(x))$$ -is exact, and the fact that $K_2(X)$ is the direct sum of $K_2(F)$ and $K_2(Q)$. -My question: is it possible to see this elementary proof as an "unfolding" of the abstract proof? Or to explain the connections between these two proofs? - -REPLY [11 votes]: No, the two proofs are quite different.<|endoftext|> -TITLE: Intuitive explanation why "shadow operator" $\frac D{e^D-1}$ connects logarithms with trigonometric functions? -QUESTION [17 upvotes]: Consider the operator $\frac D{e^D-1}$ which we will call "shadow": -$$\frac {D}{e^D-1}f(x)=\frac1{2 \pi }\int_{-\infty }^{+\infty } e^{-iwx}\frac{-iw}{e^{-i w}-1}\int_{-\infty }^{+\infty } e^{i t w} f(t) \, dt \, dw$$ -The integrals here should be understood as Fourier transforms. -Now, intuitively, why the following? -$$\left.\frac {D_x}{e^{D_x}-1} \left[\frac1\pi\ln \left(\frac{x+1/2 +\frac{z}{\pi }}{x+1/2 -\frac{z}{\pi }}\right)\right]\right|_{x=0}=\tan z$$ -There are other examples where shadow converts trigonometric functions into inverse trigonometric, logarithms to exponents, etc: -$$\left.\frac {D_x}{e^{D_x}-1} \left[\frac1{\pi }\ln \left(\frac{x+1-\frac{z}{\pi }}{x+\frac{z}{\pi }}\right)\right]\right|_{x=0}=\cot z$$ - -REPLY [11 votes]: The op $$T_x = \frac{D_x}{e^{D_x}-1} = e^{b.D_x},$$ -where $(b.)^n = b_n$ are the Bernoulli numbers, is (mod signs) often referred to as the Todd operator (maybe originally given that name by Hirzebruch, who used it to construct his Todd characteristic class). -It has a discretizing (or derivational) property that can be expressed in the following useful ways -$$f(x) = T_x T_x^{-1} f(x) = \frac{D}{e^D-1} \frac{e^D-1}{D} f(x) = T_x \int_{x}^{x+1} f(t) dt$$ -$$ = e^{b.D} \;\int_{x}^{x+1} f(t) dt = \int_{b.+x}^{b.+x+1} f(t) dt =\int_{B.(x)}^{B.(x)+1} f(t) dt$$ -$$ = F(B.(x)+1) - F(B.(x)) = F(B.(x+1)) - F(B.(x)) = D_x \; F(x),$$ -where -$$B_n(x) = (b.+x)^n = \sum_{k=0}^n \binom{n}{k} \; b_k \; x^{n-k}$$ -are the celebrated Appell Bernoulli polynomials, with the e.g.f. $e^{B.(x)t}= e^{(b.+x)t} = \frac{t}{e^t-1}e^{xt}$, and $F(x)$ is the indefinite integral/primitive of $f(x)$. The last equality illustrates the derivational property of the Bernoulli polynomials and completely defines them. -This leads to -$$\sum_{k=0}^n f(x+k) = T \; \int_{x}^{x+n+1} f(t) dt $$ -$$ = e^{b.D} \; \int_{x}^{x+n+1} f(t) dt = \int_{B.(x)}^{B.(x+n+1)} f(t) dt$$ -$$ = F(B.(x+n+1)) - F(B.(x)),$$ -and, in particular, the string of relations -$$\sum_{k=0}^n (x+k)^s =T_x \; \int_{x}^{x+n+1} t^{s} dt $$ -$$= e^{b.D} \int_{x}^{x+n+1} t^{s} dt = \int_{B.(x)}^{B.(x+n+1)} t^s dt$$ -$$ = T_x \; \frac{(x+n+1+)^{s+1} -x^{s+1}}{s+1} = e^{b.D} \frac{(x+n+1+)^{s+1} -x^{s+1}}{s+1}$$ -$$ = \frac{(B.(x+1+n))^{s+1} -(B.(x))^{s+1}}{s+1} = \frac{B_{s+1}(x+1+n) - B_{s+1}(x)}{s+1}$$ -$$ = \sum_{k=0}^n \frac{B_{s+1}(x+1+k) - B_{s+1}(x+k)}{s+1}$$ -$$ = \sum_{k=0}^n \frac{(B.(x+1+k))^{s+1} - (B.(x+k))^{s+1}}{s+1}$$ -$$ = \sum_{k=0}^n D_x \; \frac{(x+k)^{s+1}}{s+1}.$$ -If you appropriately take the limit $s \to -1$, you arrive at a relation to the natural logarithm from whence, along with the series expansions of the trig functions in Terry Tao's answer, you can tease out your particular formulas. -For a more sophisticated illustrative application of the discretizing formula, see Eqn. 1, "the Khovanskii-Pukhlikov formula, the combinatorial -counterpart to the Hirzebruch-Riemann-Roch formula (HRR) for a smooth -toric variety X with a very ample divisor D ... " on page 2 of the "$T_y$-operator on integrals over lattice polytopes" by Goda, Kamimura, and Ohmoto. -Note also the umbral inverse sequence to the Bernoulli polynomials, the Appell power polynomials -$$\hat{B}_n(x) = \frac{(x+1)^{n+1}-x^{n+1}}{n+1},$$ -with the .e.g.f. $\frac{e^t-1}{t}\; e^{xt}$, is defined also by the umbral compositional inversion -$$B_n(\hat{B}.(x)) = x^n = \hat{B}_n(B.(x)),$$ -so the - -derivational property of the Appell Bernoulli polynomials - -$$ \frac{(B_.(x)+1)^{n+1}}{n+1} - \frac{(B.(x))^{n+1}}{n+1} = \frac{(b.+x+1)^{n+1} - (b.+x)^{n+1}}{n+1}$$ -$$ = \frac{B_{n+1}(x+1) - B_{n+1}(x)}{n+1} = \hat{B}_n(B.(x)) = x^n = D \; \frac{x^{n+1}}{n+1},$$ - -reciprocal relationship of the defining e.g.f.s of the moments of the inverse pair of Appell polynomial sequences - -$$B(t) =e^{b.t}= \frac{t}{e^t-1},$$ -$$\hat{B}(t) = e^{\hat{b}.t}=\frac{e^t-1}{t}, $$ - -reciprocity of the dual ops - -$$T= B(D) = \frac{D}{e^D-1} = e^{b.D},$$ -$$T^{-1}= \hat{B}(D) = \frac{e^D-1}{D} = e^{\hat{b}.D},$$ - -dual polynomial generating properties of the ops - -$$T \; x^n = \frac{D}{e^D-1} \; x^n = e^{b.D} \; x^n = (b. + x)^n = B_n(x), $$ -$$ T^{-1} \; x^n = \frac{e^D-1}{D} \; x^n = e^{\hat{b.}D} x^n = (\hat{b.}+x)^n = \hat{B}_n(x),$$ - -umbral compositional inverse relationship of the dual sets of polynomials - -$$ B_n(\hat{B}.(x)) = T^{-1} \; T \; x^n = x^n = T \; T^{-1} \; x^n = \hat{B}_n(B.(x)),$$ - -and the discretizing property of the Todd operator - -$$ x^n = T \; T^{-1} x^n = T \; \int_{x}^{x+1} t^n \; dt$$ -$$ = T \frac{(x+1)^{n+1} - x^{n+1}}{n+1}$$ -$$ =\frac{(B.(x)+1)^{n+1} -(B.(x))^{n+1}}{n+1} = \hat{B}_n(B.(x))$$ -are all intimately (and productively) interlinked, different facets of an Appell duality, and can be generalized via the Mellin transform. (The inverse $T^{-1}$ to the Todd operator $T$ is called the Bernoulli operator in "Recent Contributions to the Calculus of Finite Differences: A Survey" by Loeb and Rota.) -This isn't the whole story--the relationships run even deeper through a Weyl algebra, Graves/Lie/Pincherle commutator, and ladder ops--but this perspective already leads to fruitful further exploration. For example, we obtain to boot in the limit as $n \to +\infty$ for the discretizing sum a modified Hurwitz zeta function as the generalization (interpolation) of the Bernoulli polynomials, -$$ B_{-s}(x) = s \; \zeta(s+1,x),$$ -which inherits the properties of an Appell sequence of polynomials. - -The 'shadow' equation is somewhat restrictive since it assumes the FT of $f(x)$ exists, which is not a necessary condition for the discretizing property to apply; e.g., note the similar Laplace transform Abel-Plana formula. -With a different normalization for the FT, -$$FT(f(x)) = \tilde{f}(\omega) = \int_{-\infty}^{\infty} e^{-i 2\pi \omega x} f(x) \; dx,$$ -and -$$f(b.+x) = e^{b.D_x} f(x) = \frac{D_x}{e^{D_x}-1} \; f(x) = \frac{D_x}{e^{D_x}-1} FT^{-1}[\tilde{f}(\omega)]$$ -$$ = \frac{D_x}{e^{D_x}-1} \; \int_{-\infty}^{\infty} e^{i 2\pi \omega x} FT[f(x)] \; d\omega = \int_{-\infty}^{\infty} e^{i 2\pi \omega x} \frac{i 2\pi \omega}{e^{i 2\pi \omega}-1} FT[f(x)] \; d\omega. $$ -Characterizing the action of the Todd operator using rather the Mellin transform interpolation a la Ramanujan/Hardy, gives an alternate, constructive route to the Hurwitz zeta function: -$$ B_{-s}(z) = (B.(z))^{-s} = (b.+z)^{-s} = e^{b.D_z} \; z^{-s}$$ -$$ = e^{b.D_z} \int_{0}^{\infty} e^{-zt} \; \frac{t^{s-1}}{(s-1)!} \; dt$$ -$$ = \int_{0}^{\infty} e^{-(b.+z)t} \; \frac{t^{s-1}}{(s-1)!} \; dt$$ -$$ =\int_{0}^{\infty} e^{-B.(z)t} \; \frac{t^{s-1}}{(s-1)!} \; dt $$ -$$ = \int_{0}^{\infty} \frac{-t}{e^{-t}-1} \; e^{-zt} \frac{t^{s-1}}{(s-1)!} \; dt = s \; \zeta(s+1,z).$$ -A series expansion for the Appell Bernoulli function for all real or complex $s$ and real or complex $z$ with $|z-1| < 1$ is given by the umbral binomial expansion -$$s \; \zeta(s+1,z) = B_{-s}(z)$$ -$$ = (b.+z)^{-s} = (b. + 1 - 1 + z)^{-s} = (B.(1)+z-1)^{-s}$$ -$$ = \sum_{n \geq 0} \binom{-s}{n} B_{-s-n}(1) \; (z-1)^n = \sum_{n \geq 0} \binom{-s}{n} (s+n) \; \zeta(s+n+1) \; (z-1)^n$$ -where -$$(b.+1)^{-s} = (B.(1))^{-s} = B_{-s}(1) = s \; \zeta(s+1,1) = s \; \zeta(s+1)$$ -with $\zeta(s)$, the Riemann zeta function.<|endoftext|> -TITLE: How to compute Hilbert class field of $\Bbb Q(\zeta_{31})$? -QUESTION [6 upvotes]: I tried constructing the Hilbert class field of $\Bbb Q(\zeta_{31})$ by imitating one of the problems from MO. I failed miserably as a quadratic field inside the cyclotomic field $\Bbb Q(\zeta_{31})$ has a class number different from $\Bbb Q(\zeta_{31})$. -Thanks for your help in advance. - -REPLY [10 votes]: This is a fun question, and I had already been thinking of making some comments on this in the question you link to. Apologies in advance for the long post. -Actually, the quadratic subfield of $\mathbb{Q}(\zeta_{31})$ is $\mathbb{Q}(\sqrt{-31})$, and has class number $3$. It is the second quadratic field, with respect to absolute value of discriminant, whose class number is divisible by $3$. However, that's not good enough this time for constructing the Hilbert class field of $\mathbb{Q}(\zeta_{31})$, since the latter has class group that is cyclic of order $9$. Moreover, the argument I gave in the question you link to for disjointness of the Hilbert class field of the quadratic and the cyclotomic field no longer works, because the degree of $\mathbb{Q}(\zeta_{31})$ is divisible by $3$. -First, let us get the second "obstacle" out of the way and give a better proof that the Hilbert class field $H$ of $F=\mathbb{Q}(\sqrt{-31})$ is disjoint from $K=\mathbb{Q}(\zeta_{31})$: just by thinking about how the Galois group of $\mathbb{Q}(\sqrt{-31})$ acts on elements, and hence on ideals, you will see that it acts by inversion on the class group of $\mathbb{Q}(\sqrt{-31})$. The way class field theory works, you can deduce that $H$ is Galois over $\mathbb{Q}$, the subgroup ${\rm Gal}(H/F)\cong \mathbb{Z}/3\mathbb{Z}$ is normal, and the quotient ${\rm Gal}(F/\mathbb{Q})$ acts on it by multiplication by $-1$, so that ${\rm Gal}(H/\mathbb{Q})$ is isomorphic to $S_3$. In particular, it is non-abelian, so $H$ cannot be contained in $K$, since its Galois group over $\mathbb{Q}$ is cyclic. -This tells you that the Hilbert class field of $F$, which one can compute to be obtainable by adjoining a root of $x^3+x+1$, gives you a piece of the Hilbert class field of $K$. But because this time the class group of $K$ is cyclic of order $9$, it does not give you everything. -Let $H_2$ denote the Hilbert class field of $K$. Before proceedings, let us think about the structure of $G={\rm Gal}(H_2/\mathbb{Q})$ (the fact that $H_2$ is Galois over $\mathbb{Q}$ again follows from general class field theory yoga). $G$ has a normal subgroup $N={\rm Gal}(H_2/K)$ that is cyclic of order $9$, and the quotient $G/N$ is cyclic of order $30$. First, I claim that this extension splits, so that $G$ is a semi-direct product $G\cong \mathbb{Z}/9\mathbb{Z}\rtimes \mathbb{Z}/30\mathbb{Z}$. In the case of $H/\mathbb{Q}$ we could see this simply by observing that the order of the normal subgroup was coprime to its index, and invoking Schur-Zassenhaus. In the current situation, that argument does not work, so instead we will exhibit a subgroup of $G$ that is cyclic of order $30$ and intersects $N$ trivially. There is a standard trick to this: take inertia at $31$. Let me call it $I$. It must be cyclic of order $30$, because $K$ is totally ramified at $31$ — we are using the fact that $\mathbb{Q}$ has no extensions that are unramified at all finite places — and it intersects $N$ trivially, since $H_2/K$ is everywhere unramified, while the extension cut out by $I$ is totally ramified at $31$. -Having established that $G = N\rtimes I$, we will have the complete structure of $G$ once we know how $I\cong G/N$ acts on $N$ by conjugation, or equivalently how ${\rm Gal}(K/\mathbb{Q})$ acts on the class group of $K$. The automorphism group of $N$ is cyclic of order $6$, and I claim that the image of $G/N$ in that automorphism group is everything. This will determine the whole group, since $G/N$, being cyclic of order $30$, has a unique quotient of order $6$. We already know that multiplication by $-1$ is in that image, because of what we said about the Galois group of $H$. Now, if $I\to {\rm Aut}N$ did factor through the quotient of order $2$, then the subgroup of $I$ that is cyclic of order $15$ would act trivially on $N$, and therefore would be normal in all of $G$. Moreover, it is contained (with index $2$) in the inertia subgroup $I$. It follows that its fixed field would be a Galois extension of $\mathbb{Q}$ that is unramified everywhere over $F$, and of degree $9$ over $F$. But wait, we said that $F$ only has class number $3$, not $9$, so this is impossible. It follows that $G/N\to {\rm Aut} N$ does not factor through a quotient of order $2$, but through a quotient of order $6$, i.e. is surjective. -This, finally, gives you a hint on how to find the Hilbert class field of $K$: applying all the same reasoning, we know in advance that the subgroup of $I$ that is cyclic of order $5$ (rather than $15$) will be normal in $G$, and that its fixed field will be an extension that is unramified of degree $9$ over the subfield of $F$ that is fixed by the subgroup of order $5$ inside ${\rm Gal}(F/\mathbb{Q})$. Now that you know in advance that this will succeed, you can fire up the computer and just wait for a few minutes: let $L$ be the subfield of $\mathbb{Q}(\zeta_{31})$ of degree $6$ over $\mathbb{Q}$. Magma will tell you that its class group is cyclic of order $9$ (we already knew this from our group theoretic considerations!) and will, after a few minutes, spit out a slightly horrendous looking polynomial of degree $9$ over $L$ whose root generates the Hilbert class field of $L$: -$$ -x^9 + \tfrac{1}{256}(351\alpha^4 + 22842\alpha^2 + - 999)x^7 + \tfrac{1}{256}(-9585\alpha^4 + 33210\alpha^2 + 567)x^6 + - \tfrac{1}{256}(56133\alpha^4 + 756702\alpha^2 + 26973)x^5 + \tfrac{1}{128}(-14096673\alpha^4 - - 289073286\alpha^2 - 9985113)x^4 + \tfrac{1}{256}(837980397\alpha^4 + 2627921070\alpha^2 - + 89938917)x^3 + \tfrac{1}{64}(-525358953\alpha^4 + 150497910738\alpha^2 + - 5208850071)x^2 + \tfrac{1}{128}(500734949193\alpha^4 - 5434147475802\alpha^2 - - 188657086959)x + \tfrac{1}{128}(329428602877167\alpha^4 + 3754393943660730\alpha^2 + - 129532294910295), -$$ -where $\alpha\in L$ has minimal polynomial -$$x^6 + 93x^4 + 899x^2 + 31$$ -over $\mathbb{Q}$. -Its compositum with $K$, obtained by adjoining to $K$ a root of the same polynomial, must then be the Hilbert class field of $K$. -Edit: Franz Lemmermeyer has found a much nicer polynomial that generates the same field over $L$, and therefore the same field over $K$: -$$ -x^9 - x^7 - 2x^6 + 3x^5 + x^4 + 2x^3 - x^2 + x - 3. -$$<|endoftext|> -TITLE: Riemann's attempts to prove RH -QUESTION [26 upvotes]: I read somewhere that Riemann believed he could find a representation of the zeta function that would allow him to show that all the non-trivial zeros of the zeta function lie on the critical line. I am wondering, then, is there any record of his attempts to prove RH? - -REPLY [27 votes]: The short answer is no. If anyone were aware of such a record, it would surely have been Carl Siegel, who undertook a careful study of Riemann’s unpublished notes. However, Siegel wrote: - -Approaches to a proof of the so-called “Riemann hypothesis” or even to a proof of the existence of infinitely many zeros of the zeta function on the critical line are not included in Riemann’s papers. - -Riemann himself, in his paper on the zeta function, said only that he made “some fleeting, vain attempts” (einigen flüchtigen vergeblichen Versuchen) to prove (what we now call) the Riemann hypothesis, and gave no indication that he recorded these attempts. -Having said that, I want to mention that there is some interesting speculation in Chapter 7 (on the Riemann–Siegel formula; see especially section 7.8) in H. M. Edwards's book Riemann’s Zeta Function about what Riemann’s train of thought might have been.<|endoftext|> -TITLE: Cancellation property for commutative monoid -QUESTION [5 upvotes]: Let $(M,+,e)$ be a commutative monoid with unit $e$. An element $a\in M$ is called cancellative element if -for any $b,c \in M$ such that $a+b=a+c$ implies that $b=c$. -Let $(\mathbf{N},+,0)$ the commutative monoid of natural numbers. -suppose that - -we have two morphisms of monoids $f:(\mathbf{N},+,0)\rightarrow (M,+,e)$ and $g:(M,+,e)\rightarrow (\mathbf{N},+,0) $ such that $g\circ f= id$. -The monoid $(M,+,e)$ is torsion-free. - -My question is the following: is the element $a=f(1)$ automatically a cancellative element in $(M,+,e)$ ? -Edit: By torsion-free I do mean that there does not exist a natural number $n>0$ and some element $x\in M-\{e\}$ such that $n x=e$. - -REPLY [7 votes]: The answer is no. Let $U=\{0,1\}$ under multiplication. Let $P$ be the semigroup of positive integers under $+$. Consider $S=P\times U$, the direct product and let $M=S\cup \{I\}$ where $I$ is an adjoined identity. Then $M$ is torsion-free, there is a homomorphism $f\colon \mathbf N\to M$ given by $f(0)=I$ and $f(n)=(n,0)$ for $n>0$ and $g\colon M\to \mathbf N$ with $g(I)=0$ and $g(n,x)=n$ for $n>0$ and $x\in \{0,1\}$ and clearly $gf=1$ but $(1,0)$ is not cancellable. - -REPLY [5 votes]: Consider the monoid $M=\mathbb{N}\times \{0,1\}$ where $$(n,a)*(m,b):=(n+m, a\cdot b).$$ -The unit element is $e:=(0,1)$. Note that this monoid is torsion free. Now consider the maps -$$g:(M,*,e)\rightarrow (\mathbb{N}, +,0), g(n,a)=n$$ -and $f: (\mathbb{N}, +,0) \rightarrow (M,*,e)$ such that $f(0)=e$ and $f(n)=(n,0)$. Then we have $g\circ f=id$, but $f(1)=(1,0)$ is not cancellative as -$$ (1,0)*(0,0)=(1,0)=(1,0)*(0,1).$$<|endoftext|> -TITLE: Euler's Master's Thesis -QUESTION [36 upvotes]: At the age of 16, Leonhard Euler defended his Master's Thesis, where he discussed and compared Descartes' and Newton's approaches to planet motion. I don't know anything else about it. In particular, I don’t know what position the young Euler supported. - -Is there any modern account of this dissertation? In English or French? - -Edit. The only source I know of is of dubious value, to say the least. On the occasion of a celebration of Euler's tri-centenary, I was offered a comics, authored by Andreas K. & Alice K. Heyne (illustrations by Elena S. Pini). There is a lot of good to say about it. But on page 10, block 3.1, Euler is concluding is defense with the words ... so the planets are dragged along by aether vortices. I wonder whether the authors have any source to support this citation. -By the way, I wish to mention that it took a considerable time for Newton's theory of gravitation to be accepted in France, and more generally in continental Europe. Descartes' reputation was so high that any contradiction to his writings was a priori rejected. The Principia were published in 1687, but they penetrated the French scientific community only once Emilie du Chatelet translated them circa 1745, on Voltaire's request. - -REPLY [8 votes]: As Franz says, it is impossible to know for definite which view the young Euler supported. -However, Newton had already disposed of the Cartesian vortex theory in the Principia which was published almost $40$ years before. I presume that Euler must have supported the view of Newton as Euler had almost certainly read the Principia by that age. I don't know how the history goes though, or how long it took for certain people to accept that the model of Descartes was definitely incorrect. -I suspect that at first there might have been quite a lot of resistance to Newton's views, but if the view of Descartes was still regarded as a tenable one in 1724, that is very interesting to me.<|endoftext|> -TITLE: Condensed criterion for sheafiness of adic spaces -QUESTION [22 upvotes]: Multiple times in talks about condensed mathematics (e.g. the Masterclass talks, Clausen's RAMpAGe talk), it is stated that the derived structure sheaf given by the condensed formalism "fixes" the non-sheafiness of non-sheafy adic spaces, and if the space is sheafy then "rational localizations are what you would expect" and the structure sheaf should be "concentrated in degree zero" in some infinity-categorical sense. -I am wondering about the opposite direction: could the condensed formalism be used as a criterion (or even characterization) of when an adic space is sheafy? -If the derived nature of the condensed formalism truly "fixes" the structure sheaf, then it seems to me like there should be some precises statement of the following form: -Conjecture. If the "condensed version" of the Huber pair $(A,A^+)$ is "concentrated in degree zero", then $(A,A^+)$ is sheafy. -Does such a statement exist? Is it trivial once one understands the construction? Or does it take work to prove, or is it even open? - -REPLY [25 votes]: Thanks for the question! One interpretation of the conjecture is true. Let me elaborate. The following results are kind of implicit in some discussion towards the end of www.math.uni-bonn.de/people/scholze/Analytic.pdf (see especially Proposition 13.16, Propositon 14.7, and some surrounding discussion), although some relevant computations are not explicitly done there, but in the Master Thesis of Grigory Andreychev (I hope it will be public soonish): - -To any Huber pair $(A,A^+)$, that is a pair of a certain kind of (always assumed complete here) topological ring $A$ together with an open and integrally closed subring $A^+\subset A$ consisting of powerbounded elements, one can associate an analytic ring $(A,A^+)_{\blacksquare}$. This is Proposition 13.16. Analytic rings are pairs of a condensed ring $B$ together with a notion of "completeness" for condensed $B$-modules: See Lecture 7 of www.math.uni-bonn.de/people/scholze/Condensed.pdf for the "non-animated" version (and Lecture 12 of Analytic.pdf for a more general version). As remarked in Remark 13.17, Andreychev has proved that this has the property that $(A,A^+)_{\blacksquare}[S]$ is concentrated in degree $0$ for any profinite set $S$, so it's an analytic ring in the "non-animated" sense (where everything is a usual condensed ring, and condensed module). - -This functor from Huber pairs to analytic rings is fully faithful. Again, this is part of Proposition 13.16. It is in this sense that the analytic geometry defined in these lectures extends the category of adic spaces. - -The functor $(A,A^+)\mapsto (A,A^+)_{\blacksquare}$ always has underlying condensed ring the condensed ring $\underline{A}$ corresponding to the topological ring $A$, but the notion of completeness for modules depends on the subring $A^+\subset A$. From this perspective, the seemingly obscure conditions on $A^+$ have a very natural interpretation, see Remark 13.18: Talking about such subrings of $A$ is just one way to talk about the associated analytic ring structures (other subrings of $A$ can also lead to such analytic ring structures, but for all such analytic ring structures there would be a different choice of subring that satisfies the conditions imposed on $A^+$). - -The functor $(A,A^+)\mapsto (A,A^+)_{\blacksquare}$ extends to more general pairs of a condensed animated ring $A$ (where "animated" = "simplicial, up to homotopy"), satisfying some conditions, and equipped with a similar subalgebra $A^+\subset \pi_0 A$. (Technically, it is enough if $A$ is nuclear over $\mathbb Z[[X_1,\ldots,X_n]]$ for some $n$.) - -For any rational open subset $U\subset \mathrm{Spa}(A,A^+)$, one can naturally define a condensed animated $A$-algebra $\mathcal O_X(U)$ together with a subalgebra $\mathcal O_X^+(U)\subset \pi_0 \mathcal O_X(U)$, fitting into the class implicit in 4). In particular, there is an associated analytic ring $(\mathcal O_X(U),\mathcal O_X^+(U))_{\blacksquare}$. This is the localization of the analytic ring $(A,A^+)_{\blacksquare}$ to the preimage of $U$ in $\mathrm{AnSpec}((A,A^+)_{\blacksquare})$ via the map of Proposition 14.7. - -The association $U\mapsto \mathcal O_X(U)$ defines a sheaf of condensed animated $A$-algebras on $\mathrm{Spa}(A,A^+)$, for any Huber pair $(A,A^+)$. This is a special case of Proposition 12.18. - -If $(A,A^+)$ satisfies any of the classical criteria of being sheafy, or generally if $A$ is Tate and sheafy, then $\mathcal O_X(U)$ is concentrated in degree $0$ and comes from the usual structure sheaf of Huber rings. This is Proposition 14.7. (The general Tate case uses some results of Kedlaya. It may be that by recent work of Ramero http://math.univ-lille1.fr/~ramero/CoursAG.pdf, Chapter 12, who extends some of Kedlaya's work to the case of general Huber rings, simply asking $(A,A^+)$ to be sheafy is enough.) - -Now conversely, if the sheaf $\mathcal O_X(U)$ of condensed animated $A$-algebras happens to be concentrated in degree $0$, and also to be quasiseparated, then it agrees with the (presheaf of condensed rings associated to the) presheaf of Huber rings defined by Huber, which is thus sheafy. - - -That final point 8) is, I think, a partial answer to your question. At least if $A$ is Tate, it shows that sheafyness is equivalent to the assertion that for all rational $U\subset \mathrm{Spa}(A,A^+)$, the condensed animated $A$-algebra $\mathcal O_X(U)$ is concentrated in degree $0$ and quasiseparated. One could wonder whether the "... and quasiseparated" ending is necessary; my gut feeling is that it is necessary. -Upshot: Huber was working in the context of complete topological rings (satisfying some conditions), and had to insist that his structure presheaf stays in the same realm. If you allow yourself more flexibility, in particular a notion of completeness that does not entail separatedness, and a formalism of topological rings that allows for higher homotopy groups (i.e., animated rings), then one can define a better version of his presheaf, that is always a sheaf. (See the beginning of Lecture 11 in Analytic.pdf, especially page 73, for some intuitive explanations.) Then sheafyness of Huber pairs is simply the question whether this more general construction stays in the classical realm, which fortunately happens so far in all cases of interest. I should mention that Bambozzi-Kremnizer have recently reached similar results, using different foundations, see arXiv:2009.13926. (In their approach, the role of $A^+$ is less clear.) -Edit: Let me actually be much more explicit about all of this. Consider a rational subset $U\subset \mathrm{Spa}(A,A^+)$, explicitly given as the locus $U=\{|f_1|,\ldots,|f_n|\leq |g|\neq 0\}$, for $f_1,\ldots,f_n,g\in A$ generating an open ideal. In that case, Huber's ring $\mathcal O_X^H(U)$ is given by -$$ -A\langle T_1,\ldots,T_n\rangle[\tfrac 1g]/\overline{(f_1-gT_1,\ldots,f_n-gT_n)}. -$$ -Intuitively, this is just expressing that on this subset $g$ is invertible, and $T_i=\tfrac{f_i}g$ has absolute value $\leq 1$, so we can allow all convergent power series in the $T_i$. To stay in the setting of complete topological rings, it is necessary to take the quotient by the closure of this ideal. -On the other hand, a theorem of Kedlaya (https://kskedlaya.org/papers/aws-notes.pdf, Theorem 1.2.7) shows that if $\mathcal O_X^H$ is a sheaf and $A$ is Tate, then it is not necessary to take the closure. Moreover, his results show that the sequence $(f_1-gT_1,\ldots,f_n-gT_n)$ is (Koszul-)regular. In other words, in this case $\mathcal O_X^H(U)$ is computed by the Koszul complex -$$ -A\langle T_1,\ldots,T_n\rangle[\tfrac 1g]/^{\mathbb L}(f_1-gT_1,\ldots,f_n-gT_n), -$$ -where I write, for an $A$-module $M$ and elements $a_1,\ldots,a_n\in A$, -$$ -M/^{\mathbb L}(a_1,\ldots,a_n) -$$ -for the (homological) Koszul complex -$$ -0\to M\xrightarrow{(a_1,\ldots,a_n)} M^n\to \ldots\to M^n\xrightarrow{(a_1,\ldots,a_n)} M\to 0. -$$ -What condensed mathematics allows you to do is to consider the derived quotient -$$ -A\langle T_1,\ldots,T_n\rangle[\tfrac 1g]/^{\mathbb L}(f_1-gT_1,\ldots,f_n-gT_n) -$$ -as the correct answer in general. For this, you have to consider it simultaneously as endowed with (something like) a topology, as a complex, and as a (commutative) algebra. Thus, you have to mix higher (i.e. homotopical) algebra with topology, and this is what condensed mathematics can easily accomodate. In fact, these are just the condensed animated rings I've been talking about above. (Animated rings are, at least in characteristic 0, just the "commutative algebras in the derived category", correctly understood; making them condensed amounts to putting something like a topology on them, in particular on their homotopy groups.) -In particular, what my answer above means is the following, at least if $A$ is Tate: $\mathcal O_X^H$ is a sheaf if and only if for all $U$ as above, the sequence $(f_1-gT_1,\ldots,f_n-gT_n)$ is Koszul-regular and generates a closed ideal. This is in fact already a theorem of Kedlaya, I believe.<|endoftext|> -TITLE: A non-Abelian de Rham complex? -QUESTION [18 upvotes]: This question is inspired by this physics stack exchange post, which is recent and has not received an answer yet, nontheless I feel that there is a better way to ask this question here with a larger scope than the OP did there. - -At first I am primarily interested in the local question so let $X$ be an $n$ dimensional smooth real manifold that if necessary might be assumed to be contractible or otherwise topological trivial. -Let us fix a Lie algebra $\mathfrak g$, let $\Omega^k(X,\mathfrak g)$ denote the module of smooth $\mathfrak g$-valued $k$-forms. -The following is known. Let us define $C:\Omega^1(X,\mathfrak g)\rightarrow \Omega^2(X,\mathfrak g)$ by $$ C(\omega)=d\omega+\frac{1}{2}[\omega\wedge\omega], $$ and call $C$ the curvature operator. It is known that at least locally $C(\omega)=0$ if and only if there is a function $f\in C^\infty(X,G)$ (where $G$ is a/the Lie group associated with $\mathfrak g$) such that $$ \omega=f^\ast\Xi_G\equiv \Delta_Gf, $$ where $\Xi_G\in\Omega^1(G,\mathfrak g)$ is the Maurer-Cartan form of $G$. I have taken the liberty of using $\Delta_G$ for this operation (nonstandard notation) and iirc this is sometimes referred to as the Darboux-derivative (treated in Sharpe for example). -Now, let $F=C(\omega)=d\omega+\frac{1}{2}[\omega\wedge\omega]\in\Omega^2(X,\mathfrak g)$. Such curvature forms satisfy the (differential) Bianchi identity $$ d_\omega F=dF+[\omega\wedge F]=0. $$ -It this seems that one may define a cochain complex-like structure $$ 0\longrightarrow C^\infty(X,G)\longrightarrow^{\Delta_G}\Omega^1(X,\mathfrak g)\longrightarrow^{C}\Omega^2(X,\mathfrak g)\longrightarrow^{d_\omega}\Omega^3(X,\mathfrak g) ... $$ -in that the composition of two subsequent arrows always return $0$, however this is not a true cochain complex because for example $C^\infty(X,G)$ is not an Abelian group (if $G$ is nonabelian). - -What I am primarily interested in is whether it is possible to define such a sequence rigorously in some way which satisfies a local exactness property. This sequence is locally exact at $\Omega^1(X,\mathfrak g)$, which can be verified using the Frobenius integrability theorem, but - in my opinion - a more interesting question is the $\Omega^2(X,\mathfrak g)$. -Specifically what is the necessary and sufficient (local) condition for a Lie algebra valued $2$-form to be the curvature form of a connexion? Even if the Bianchi identity is a complete local integrability condition, it is formulated in terms of $2$-forms so Frobenius' theorem does not apply, moreover in order to calculate the covariant exterior derivative of the curvature $2$-form one already needs to know the connection form as well. -If a necessary and sufficient condition can be given, is there an explicit "homotopy operator" that allows one to construct a connection form from a given curvature form? - -Remark:* In $\dim X=3$ the curvature form is the Euler-Lagrange form of the Chern-Simons Lagrangian, so I guess methods of the variational bicomplex could be used to attack this problem. As far as I see this only works in three dimensions though. - -REPLY [8 votes]: Unfortunately, things are subtle and terrible (at least, compared to the abelian case). I wrote a bit about this in an unpublished article. -One of the surprising, terrible features is that you can have two connections ω,η such that (1) curv ω = curv η, yet (2) ω and η are not gauge equivalent. Wu and Yang discussed this problem, and I think also were responsible for calling it the "field copy problem" (or maybe the name came from Mostow, who also wrote on it). It is quite weird when you sit down and think about it in terms of Yang-Mills field theory. From that perspective, you're looking at a connection whose dynamics is determined by a Lagrangian that only involves the curvature. The field copy problem shows that with nonabelian gauge groups, you can have two distinct connections that minimize the same Yang-Mills functional. In other words, observing the curvature form doesn't tell you enough to reconstruct the connection up to gauge equivalence, even in the special case of Yang-Mills connections! -Len Gross analyzed the field copy problem in "A Poincaré Lemma for Connection Forms", where he found an alternative collection of observables that suffices to reconstruct the connection up to gauge equivalence. I tried to make sense of his results using 2-categories in the article I linked above. Ultimately my research went in a different direction and I never got to follow through with this thread in the way I would have liked. -There is also an issue with the next term of the exact-ish sequence, too (your $d_\omega$): the Bianchi identity sure looks like it should be the map to use there, as you propose. Ok, so we'll differentiate the 2-form with respect to the connection... but wait! We don't even have a connection form to put our hands on at that point! There is no $\omega$ to use!<|endoftext|> -TITLE: Countable open covering of normal space -QUESTION [5 upvotes]: I read the following claim in Z.Frolik's article "A generalization of realcompact spaces" on page 135. -Two subset $M$ and $N$ of a space $X$ are called completely seperated if there exists a real valued continuous function $f$ on $X$ with $f(M)\subset \{0\}$ and $f(N)\subset\{1\}$. -Claim: Let $X$ be a normal space. For every countable open covering $\mathfrak{U}$ of $X$, there exists a countable open covering $\mathfrak{B}$ of $X$ such that for every $B$ in $\mathfrak{B}$ there exists an $A$ in $\mathfrak{U}$ such that $B$ and $X-A$ are completely seperated. -I didn't show the proof of the claim. - -REPLY [3 votes]: The claim is false it would imply that normal spaces are countably paracompact and hence that normality of $X$ would imply normality of $X\times[0,1]$. The latter is not the case, see Mary Ellen Rudin, A normal space $X$ for which $X\times I$ is not normal, Fundamenta Mathematicae, 73 (1971/72), 179-186. -To show that the property in the claim implies countable paracompactness we use Theorem 5.2.1 in Engelking's General Topology. Let $\{U_n:n\in\omega\}$ be an increasing open cover; we need to find open $O_n$ such that $\operatorname{cl}O_n\subseteq U_n$ for all $n$ and $\bigcup_nO_n=X$. Take an open cover $\{V_m:m\in\omega\}$ as in the claim; hence for every $m$ an $n$ such that $\operatorname{cl}V_m\subseteq U_n$. Now define $O_n=\bigcup\{V_m:m\le n$ and $\operatorname{cl}V_m\subseteq U_n\}$; then $\operatorname{cl}O_n\subseteq U_n$ for all $n$ and the $O_n$ form a cover (if $x\in V_m$ and $\operatorname{cl}V_m\subseteq U_n$ then $x\in O_n$).<|endoftext|> -TITLE: A simple oscillatory integral with a non-smooth phase -QUESTION [5 upvotes]: Let $\phi\in C_c^\infty(\mathbb{R})$ be an even function such that $\chi_{(-1/2,1/2)}\le\phi\le \chi_{(-1,1)}$, where $\chi_{(a,b)}$ stands for the indicator function of the interval $(a,b)$. For $\lambda>0$ consider the oscillatory integral -$$ -I(\lambda)=\int_\mathbb{R} \phi(x)\, \exp \left(i\lambda(x+\epsilon|x|^{\sqrt{2}})\right)\, dx, -$$ -with some fixed (very small) positive constant $\epsilon$. -My question is: what is the asymptotic behavior of this integral as $\lambda\rightarrow \infty$? I can show, by essentially doing careful integration by parts, that the upper bound is $\lesssim \lambda^{-\sqrt{2}}$, but I wonder whether $\lambda^{-\sqrt{2}}$ is also a lower bound? -Note, that if the exponent $\sqrt{2}$ is replaced by $2k$ for some positive integer $k$, then the integral decays like $\lambda^{-M}$ for any $M>0$ due to the non-stationary phase estimate (the derivative of the function $x+\epsilon x^{2k}$ is $\gtrsim 1$). -I would appreciate any hints on how to approach this problem. - -REPLY [8 votes]: $\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}\newcommand{\ep}{\epsilon}\newcommand{\tI}{\tilde I}$ -Take any $a\in(1,2)$ and then any nonzero $\epsilon\in(-1/a,1/a)$. Then -\begin{align*} - I(t)&:=\int_\mathbb{R} \phi(x)\, \exp(it(x+\epsilon|x|^a))\, dx \\ -&\sim\frac{2\epsilon\,\Gamma(a+1)}{it^a}\,\sin\frac{\pi a}2 \tag{1} -\end{align*} -as $t\to\infty$. This asymptotics does not depend on $\phi$, as long as -\begin{equation*} - \phi\in C_c^\infty(\mathbb{R}) \tag{2} -\end{equation*} -is a (not necessarily even) function such that -\begin{equation*} -\chi_{(-1/2,1/2)}\le\phi\le \chi_{(-1,1)}. \tag{3} -\end{equation*} -Indeed, let -\begin{equation*} - g(x):=x+\ep|x|^a. -\end{equation*} -Then $g'(x)=1+\ep ax^{[a-1]}$ for real $x\ne0$, where $x^{[c]}:=|x|^{c-1}x$. Therefore and because $|\ep|<1/a$, we see that $2\ge g'\ge1-|\ep|a>0$ on $(-1,1)$. So, there is a unique inverse $h$ of function $g$ on $(-1,1)$ such that for all $x\in(-1,1)$ and all $y\in(g(-1),g(1))$ we have -\begin{equation*} - y=g(x)\iff x=h(y). \tag{3.5} -\end{equation*} -Also, $g(-x)<00$. -Next, in view of (3), -\begin{equation*} - I_1=I_2+\tI_2, \tag{6} -\end{equation*} -where -\begin{align*} - I_2&:=\int_{g(-1/2)}^{g(1/2)} dy\,e^{ity}h''(y), \\ - \tI_2&:=\int_{g(S)} dy\,e^{ity}h''(y). -\end{align*} -Note that $h''\in C^1(g(S))$. So, integrating by parts, we have -\begin{equation*} - \tI_2\ll\frac1t. \tag{7} -\end{equation*} -Writing, for brevity, $x$ for $h(y)$ (cf. (3.5)), we have -\begin{equation*} - h'(y)=\frac1{1+\ep ah(y)^{[a-1]}}=\frac1{1+\ep ax^{[a-1]}} -\end{equation*} -and hence -\begin{align*} - h''(y)&=-\frac{\ep a(a-1)|x|^{a-2}}{(1+\ep ax^{[a-1]})^2},\\ h'''(y)&\ll|x|^{a-3}+|x|^{2a-4}\ll|x|^{a-3}\ll|y|^{a-3} \tag{8} -\end{align*} -for $|x|\le1/2$, that is, for $y\in[g(-1/2),g(1/2)]$. Next, for $y=g(x)\in[g(-1/2),g(1/2)]$ we have $y=x(1+\ep x^{[a-1]})$ and hence -\begin{align*} - x=h(y)&=\frac y{1+\ep x^{[a-1]}} \\ - &=\frac y{1+\ep y^{[a-1]}(1+O(|y|^{a-1}))} \\ - &=\frac y{1+\ep y^{[a-1]}} \, (1+O(|y|^{2a-2}) \\ - &=y\, (1+O(|y|^{a-1})) -\end{align*} -and -\begin{align*} - h''(y)&=-\frac{\ep a(a-1)|h(y)|^{a-2}}{(1+\ep ah(y)^{[a-1]})^2}\\ - &=-\ep a(a-1)|y|^{a-2}[1+O(|y|^{a-1})] \\ - &=-\ep a(a-1)|y|^{a-2}+O(|y|^{2a-3}). \tag{9} -\end{align*} -Further, -\begin{equation*} - I_2=I_3+\tI_3, \tag{10} -\end{equation*} -where -\begin{align*} - I_3&:=\int_{|y|\le\de} dy\,e^{ity}h''(y), \\ - \tI_3&:=\int_{[g(-1/2),g(1/2)]\setminus[-\de,\de]} dy\,e^{ity}h''(y), \\ - \de&:=t^{-3/4}. -\end{align*} -Integrating by parts and using (8) and (9), we have -\begin{align*} - \tI_3&\ll\frac{|h''(\de)|+|h''(-\de)|+O(1)}t - +\frac1t\, - \int_{|y|>\de} dy\,|y|^{a-3}, \\ - &\ll \frac{\de^{a-2}}t=t^{1/2-3a/4}=o(t^{1-a}). -\end{align*} -Using (9) again, we have -\begin{equation*} - I_3=-\ep a(a-1)I_4+O(\tI_4), \tag{11} -\end{equation*} -where -\begin{align*} - I_4&:=\int_{|y|\le\de} dy\,e^{ity}|y|^{a-2}, \\ - \tI_4&:=\int_{|y|\le\de} dy\,|y|^{2a-3}. -\end{align*} -Next, -\begin{equation*} - \tI_4\ll\de^{2a-2}=t^{-(2a-2)3/4}=o(t^{1-a}), \tag{12} -\end{equation*} -\begin{align*} - I_4&=t^{1-a}\int_{|z|\le t\de} dz\,e^{iz}|z|^{a-2} \\ - &=t^{1-a}\int_{|z|\le t^{1/4}} dz\,e^{iz}|z|^{a-2} \\ - &\sim t^{1-a}\int_\R dz\,e^{iz}|z|^{a-2} - =2 t^{1-a}\Gamma(a-1)\sin\frac{\pi a}2. \tag{13} -\end{align*} -Collecting the pieces (4)--(7) and (10)--(13), we get the result.<|endoftext|> -TITLE: Spaces of solutions to algebraic linear differential equations -QUESTION [7 upvotes]: What is the name of the function space formed by solutions to algebraic linear differential equations? Where can I find a discussion of its properties? -By an algebraic linear differential equation I mean a linear partial differential equation in $n$ variables whose coefficients are polynomials in those variables over a field $k$. The solutions of these equations obviously include all the polynomials, but also non-polynomial functions, e.g., the exponential function. -These equations and their generalizations are studied in the theory of algebraic $D$-modules, but literature on $D$-modules moves quickly to sheaves of differential operators on algebraic varieties without answering this basic question for the simplest affine case. The focus seems to be more on the properties of the ring of differential operators and their modules, rather than of their solutions, but I may be wrong. -Even just for $k = \mathbb{C}$ or $\mathbb{R}$ where can I find a description of the space of solutions of all algebraic differential equations in $n$ variables including its algebraic and functional-analytic properties? - -REPLY [3 votes]: Further to Sam Gunningham's comment, Frédéric Chyzak's thesis, Fonctions holonomes en calcul formel appears to throw some light on this question.<|endoftext|> -TITLE: Monoidal categories whose tensor has a left adjoint -QUESTION [16 upvotes]: Is there a name for monoidal categories $(\mathscr V, \otimes, I)$ such that $\otimes$ has a left adjoint $(\ell, r) : \mathscr V \to \mathscr V^2$? Have they been studied anywhere? What are some interesting examples? -A couple of remarks: when $I : 1 \to \mathscr V$ has a left adjoint, then $\mathscr V$ is semicartesian, i.e. the unit is terminal. When $\otimes$ has a left adjoint, which is furthermore the diagonal $\Delta : \mathscr V \to \mathscr V^2$, then $\mathscr V$ has binary products. - -I'll unwrap the definition here to make the structure more explicit. Let $(\mathscr V, \otimes, I)$ be a monoidal category. $\otimes$ has a left adjoint if we have the following. - -endofunctors $\ell : \mathscr V \to \mathscr V$ and $r : \mathscr V \to \mathscr V$; -for every pair of morphisms $f : \ell(X) \to Y$ and $g : r(X) \to Z$, a morphism $\{f, g\} : X \to Y \otimes Z$; -for every morphism $h : X \to Y \otimes Z$, morphisms $h_\ell : \ell(X) \to Y$ and $h_r : r(X) \to Z$, - -such that, for all $x : X' \to X$, $y : Y \to Y'$ and $z : Z \to Z'$, we have -$$y \otimes z \circ \{ f, g \} \circ x = \{ y \circ f \circ \ell(x), z \circ g \circ r(x) \}$$ -$$\{ h_\ell, h_r \} = h$$ -$$\{ f, g \}_\ell = f$$ -$$\{ f, g \}_r = g$$ - -Edit (2021-09-27): these are map pseudomonoids in $\mathrm{Cat}^{\mathrm{co}}$ in the terminology of Day–McCrudden–Street's Dualizations and Antipodes. In Kan extensions and cartesian monoidal -categories, Street defines a monoidal $\mathcal V$-category to be cartesian when it satisfies this condition. This terminology is justified at least when $\mathcal V = \mathrm{Set}$ by Tim Campion's answer. Preferably one would show it is correct whenever $\mathcal V$ has finite products. - -REPLY [19 votes]: Just to clean up the $\epsilon$ of room left after Qiaochu's answer -- we can get rid of the extra hypotheses. I'll write $I$ for the monoidal unit and $1$ for the terminal object. -Assume that $(\ell,r) \dashv \otimes$. Then the natural isomorphisms $A \cong I \otimes A \cong A \otimes I$ give rise, by adjunction, to maps $\ell A \to I$ and $r A \to I$, natural in $A$. We also have a unit map $A \to (\ell A) \otimes (r A)$, natural in $A$. Tensoring and composing, we get a map $A \to (\ell A) \otimes (r A) \to I \otimes I \cong I$, natural in $A$. That is, we have a cocone (with vertex $I$) on the identity functor for $V$. It follows that in the idempotent completion $\tilde V$ of $V$, there is a terminal object (which must be a retract of $I$). -Now, the idempotent completion $\tilde V$ again has a monoidal structure $\tilde \otimes$ with a left adjoint $(\tilde \ell, \tilde r)$. So the first part of Qiaochu's Eckmann-Hilton argument can be run in $\tilde V$: $I = I \otimes I = (I \times 1) \otimes (1 \times I) = (I \otimes 1) \times (1 \otimes I) = 1 \times 1 = 1$ (in the third expression, the products exist trivially, and in the fourth the product exists because $\otimes$ preserves products). That is, we must have $I_{\tilde V} = 1_{\tilde V}$. But $I_{\tilde V}$ is the image of $I_V$ in $\tilde V$, and the inclusion into the idempotent completion reflects terminal objects. Therefore $V$ has a terminal object, and $1_V = I_V$. -Then, as observed in the comments above, the second part of Qiaochu's Eckmann-Hilton argument can be run in $V$: $A \otimes B = (A \times 1) \otimes (1 \times B) = (A \otimes 1) \times (1 \otimes B) = A \times B$ (in the second expression, the products exist trivially, and in the third the product exists because $\otimes$ preserves products). That is, binary products exist in $V$ and agree with $\otimes$. In fact, the identity functor is an oplax monoidal functor from $(V,\otimes)$ to $(V,\times)$, which the argument shows is actually strong monoidal. Thus $(V,\otimes) \simeq (V,\times)$ as monoidal categories.<|endoftext|> -TITLE: When is the set of measurable functions a vector space? -QUESTION [8 upvotes]: I know this is not a research question, but I searched somewhat thoroughly and could not find the exact answer I want. But I've always wondered the following: suppose that $(X,\mathcal{M})$ is a measurable space and $Y$ is a real topological vector space equipped with the Borel $\sigma$-algebra $\mathcal{B}$. Let -$$L^0(X,Y):=\{f:X\to Y\;\mid\;f \text{ is measurable}\}.$$ -When is $L^0(X,Y)$ a real vector subspace of $Y^X$? In other words, what are "minimal" assumption needed on $X$ and $Y$ so that measurable functions form a vector space? -Point 1: For example in the proof of the case when $Y=\mathbb{R}$, if $f,g$ are Borel measurable functions then we use the following equality -$$\{ f+g < b\} = \bigcup_{r\in\mathbb{Q}} - \{f< r\} \cap \{g< b-r\}.$$ -to show that $f+g$ is measurable. So we have used the following assumptions - -$Y$ is ordered. -$Y$ has a countable dense set w.r.t that order. - -How much can this argument be generalized? -Point 2: When $Y$ is a Banach space, I know Bochner spaces come into play. Is there any result regarding the original question in this case? - -REPLY [8 votes]: The usual thing to do, even when $Y$ is a Banach space, is to define "measurable" in such a way that it works. (A while back I posted a counterexample to the general case. See below.) -Bochner measurable, meaning there exist simple functions $f_n$ that converge a.e. to $f$. "Simple" functions have finite range. In case (i) $X$ is any measurable space and $Y$ is a separable Banach space or in case (ii) $X$ is a perfect measure space and $Y$ is any Banach space, then Bochner measurable is the same as $f$ is "measurable" from $\mathcal M$ to the Borel sets in $Y$. But in general, Bochner measurability is the useful notion. -OR -(works for $Y$ a locally convex space) Weakly measurable, meaning $\phi \circ f$ is measurable for all continuous linear functionals $\phi : Y \to \mathbb R$. - -The mentioned counterexammple, is part of my answer HERE -It provides two measurable functions $f,g : \Omega \to B$ with $f+g$ not measurable. -$\Omega = T \times T$ where $T$ has power $2^{\aleph_0}$ and and $\Omega$ has the product sigma-algebra $\mathcal F$, with $\mathcal P(T)$ in each factor. -$B = l^2(T)$, a non-separable Hilbert space with orthonormal basis $\{e_t: t \in T\}$. We use the sigma-algebra $\mathcal B$ of Borel sets for the norm topology. -The definitions are $f\big((u,v)\big) = e_u$ and -$g\big((u,v)\big) = -e_v$. Then $f,g$ are $\mathcal F$ to $\mathcal B$ measurable, but $f+g$ is not. Details in the link.<|endoftext|> -TITLE: Are equivalences of categories stable under filtered colimits? -QUESTION [13 upvotes]: Let $D \colon \mathbf{J} \to \mathbf{Cat}$ be a filtered diagram of categories and functors. It has a colimit $\mathbf{C} = \mathrm{colim}\;D$. If you replace the diagram by a naturally isomorphic one $D' \colon \mathbf{J} \to \mathbf{Cat}$, then the colimit -$\mathbf{C'} = \mathrm{colim}\;D'$ is isomorphic to $\mathbf{C}$. What happens if you replace the diagram by a naturally equivalent one $D'' \colon \mathbf{J} \to \mathbf{Cat}$, is the colimit $\mathbf{C''} = \mathrm{colim}\;D''$ still equivalent to $\mathbf{C}$? -(I'm interested in $\mathbf{Cat}$, or in fact the category $\mathbf{MonCat}$ of monoidal categories, but suspect that there is a general statement in weak 2-categories. Any references are appreciated.) - -REPLY [6 votes]: With regards the question about MonCat, I will assume that you are taking strong monoidal functors as morphisms. (If you are dealing with strict morphisms things are straightforward.) -Then MonCat is an accessible category with filtered colimits and U:MonCat--->Cat preserves them. Since, viewing it as a 2-functor, it also reflects equivalences this reduces the MonCat question to the Cat one. -More generally, if T is a finitary flexible (aka cofibrant) 2-monad on Cat, the 2-category of algebras and pseudomaps T-Alg is accessible with filtered colimits and U:T-Alg-->Cat preserves them. This is a special case of Corollary 7.3 of my paper https://arxiv.org/abs/2003.06375. Flexibility or cofibrancy of the 2-monad means that it describes categorical structures with no equations between objects (like monoidal categories but not strict monoidal categories). For strict structures like strict monoidal categories, it is not true. -As to why filtered colimits of equivalences are equivalences in Cat, the approach via combinatorial model categories described by Simon Henry is a good general approach. An argument avoiding model categories, though similar in spirit, is to observe that, viewed as objects of the arrow category of Cat, equivalences are finitary injectives - see Diagram 2.3 of https://arxiv.org/abs/1712.02523 - and then use that finitary injectivity classes are always closed under filtered colimits, which is easy to prove (this is an easy part of the proof of Prop 4.7 of Locally presentable and accessible categories.) -P.S. Perhaps the original result of this nature is Lemma 5.4.9 of "Accessible categories .." by Makkai and Pare which shows that filtered colimits are bicolimits in Cat. This implies the property about equivalences.<|endoftext|> -TITLE: Euclid-style proof of Dirichlet’s theorem on primes in certain arithmetic progression -QUESTION [14 upvotes]: The well-known theorem of Dirichlet on primes in arithmetic progression states that given coprime natural numbers $a\le q$, there are infinitely many prime numbers congruent to $a\pmod q$. The standard proof is via analytic number theory; however, the special case $a=1, q=2$, is a celebrated theorem of Euclid, whose proof was essentially extended by Euler using cyclotomic polynomials to all cases $a=1$. This elementary algebraic approach — let’s call it Euclid-style proof — of using polynomials to resolve special cases of Dirichlet’s theorem is known to be impossible only if $a^2\not\equiv 1\pmod q$, which is (sometimes) known in the field as Schur-Murty Impossibility Theorem (after Issai Schur and Ram Murty). My question is - -Has anyone been able to give a generic Euclid-style argument for every case $a^2\equiv1\pmod q$ (or any partial results to that effect)? - -(Precisely, by “Euclid-style”, we mean the existence of a polynomial $p$ whose prime factors over (a sequence of) natural number arguments contains a prime congruent to $a\pmod q$; the simplest, non-trivial and well-known examples are $p(x)=4x-1$ for $a=3\,,q=4$, and $p(x)=4x^2+1$ for $a=1\,,q=4$.) - -REPLY [14 votes]: A proof of the construction of a polynomial, in English, is in the paper of Murty and Thain, Primes in Certain Arithmetic Progressions (Funct. Approx. Comment. Math. 35 (2006) pp. 249-259, doi:10.7169/facm/1229442627). See Section 2, which builds up to the Euclid-style proof as Theorem 6. In the proof is a typographical error: the defining displayed formula for $f(x)$ should have $f(x)$, not $f(x)^2$, on the left side of the formula. -The proof is illustrated afterwards for $p \equiv 4 \bmod 15$, but there is another typographical error: $f(x)$ here is the minimal polynomial of $\zeta + \zeta^{4}$ over $\mathbf Q$, which is $x^4 - x^3 + 2x^2 + x + 1$; it is written there with the linear term $x$ missing. -The basic idea of the proof can be described briefly if you know Galois theory. Let $\zeta_m$ be a root of unity of order $m$, such as $e^{2\pi i/m}$. When $a \bmod m$ has order $2$, $\{1,a\}$ is a subgroup of $(\mathbf Z/(m))^\times$, and ${\rm Gal}(\mathbf Q(\zeta_m)/\mathbf Q) \cong (\mathbf Z/(m))^\times$ in a standard way by Galois theory. Therefore the subfield of $\mathbf Q(\zeta_m)$ fixed by $\{1,a\bmod m\}$ using Galois theory is a field $L$ of degree $\varphi(m)/2$ over $\mathbf Q$. Write $L = \mathbf Q(\eta)$ for some number $\eta$ and take as $f(x)$ the minimal polynomial of $\eta$ over $\mathbf Q$. We'd like $f(x)$ to be in $\mathbf Z[x]$, so let $\eta$ be an algebraic integer generating $L$. By a careful choice of $f(x)$, every prime factor of $f(n)$ for for $n \in \mathbf Z$ is either a factor of ${\rm disc}(f)$ or is congruent to $1$ or $a \bmod m$. They use $f(x)$ to give a Euclid-style proof that there are infinitely many primes $p \equiv a \bmod m$ assuming there is at least one such prime. If there are only finitely many primes congruent to $a \bmod m$ then they use that finiteness to get a contradiction using the Chinese remainder theorem, so there are infinitely many such primes. -As an example, if $a = 4$ and $m = 15$ then $\zeta_{15} + \zeta_{15}^4$ generates the subfield of $\mathbf Q(\zeta_{15})$ fixed by $\{1,4\bmod 15\}$, and the minimal polynomial of $\zeta_{15} + \zeta_{15}^4$ over $\mathbf Q$ is the polynomial I mentioned earlier: $x^4 - x^3 + 2x^2 + x + 1$. With this polynomial you can give a Euclid-style proof that there are infinitely many primes $p \equiv 4 \bmod 15$. -It is natural to think that you should always be able to use $\eta = \zeta_m + \zeta_m^a$, since that sum definitely is fixed by $\{1, a\bmod m\}$. But watch out: we need to make sure $\zeta_m + \zeta_m^a$ is not fixed by anything else in ${\rm Gal}(\mathbf Q(\zeta_m)/\mathbf Q)$ in order to know it generates the subfield fixed by $\{1,a \bmod m\}$ rather than a smaller field. -Example 1. When $a = -1$ this always works: $\mathbf Q(\zeta_m + \zeta_m^{-1})$ has degree $\varphi(m)/2$ over $\mathbf Q$. -Example 2. When $m = 8$ and $a$ is $3$, $5$, and $7$, this idea works with $3$ and $7$ but there's a problem with $5$. The quadratic subfields of $\mathbf Q(\zeta_8)$ are $\mathbf Q(\sqrt{2})$, $\mathbf Q(\sqrt{-2})$, and $\mathbf Q(i)$, and $\zeta_8 + \zeta_8^3$ has minimal polynomial $x^2 + 2$, $\zeta_8 + \zeta_8^7$ has minimal polynomial $x^2 - 2$, but $\zeta_8 + \zeta_8^5$ is $0$, so it doesn't generate the subfield $\mathbf Q(i)$ fixed by $\{1, 5 \bmod 8\}$. -Example 3. For $4 \mid m$ and $a = 1 + m/2$, -$$ -a^2 = 1 + m + m(m/4) \equiv 1 \bmod m -$$ and $\zeta_m + \zeta_m^{1+m/2} = \zeta_m - \zeta_m$ is $0$, so $\zeta_m + \zeta_m^{1+m/2}$ doesn't generate the subfield of $\mathbf Q(\zeta_m)$ fixed by $\{1, 1+m/2 \bmod m\}$. Example 2 is the case $m = 8$ (so $a = 1 + m/2 = 5$). -That is why in the proof of Theorem 6 in Section 2 of Murty and Thain's paper, their chosen generator for the field $L$ fixed by $\{1, a \bmod m\}$ is not $\zeta_m + \zeta_m^a$ (it doesn't always work!), but $h_u(\zeta_m) := (u - \zeta_m)(u - \zeta_m^a)$ for an integer $u$ and they show $h_u(\zeta_m)$ has degree $\varphi(m)/2$ for all but finitely many integers $u$. Therefore you can take $\eta = h_u(\zeta_m)$ for all but finitely many integers $u$.<|endoftext|> -TITLE: Cohomology of doubly pinched torus via spectral sequences -QUESTION [5 upvotes]: Let $f:T^2\to Y$ be a resolution of singularities where $Y$ is a torus with two "pinched" points (or, if you prefer, two copies of $\mathbb{P}^1$ meeting at two points). I'm interested in using the Leray spectral sequence to calculate the cohomology of the constant sheaf on $Y$ . My goal is to better understand spectral sequences and this looks like a nice example to me. -To begin with, $Rf_*\mathbb{Q}_X$ has cohomology sheaves $R^0f_*\mathbb{Q}_X = \mathbb{Q}_Y$ and $R^1f_*\mathbb{Q}_X = \mathcal{S}_{\{a,b\}}$, the skyscraper sheaf with stalk $\mathbb{Q}$ at the pinched points $a,b$. -The $E_2$ page is -\begin{align*} -\begin{matrix} -H^2(Y; \mathbb{Q}_Y) & 0 & 0\\ -H^1(Y; \mathbb{Q}_Y) & 0& 0\\ -H^0(Y; \mathbb{Q}_Y) & H^0(Y;\mathcal{S}_{\{a,b\}})=\mathbb{Q}^{\oplus 2} & 0 -\end{matrix} -\end{align*} -and the $E_3 (=E_\infty)$ page is -\begin{align*} -\begin{matrix} -H^2(Y; \mathbb{Q}_Y)/ \text{im}(d_2^{0,1}) & 0 &0 \\ -H^1(Y; \mathbb{Q}_Y) & 0& 0\\ -H^0(Y; \mathbb{Q}_Y) & \text{ker}(d_2^{0,1}) & 0 -\end{matrix} -\end{align*} -Using the fact that the cohomology $H^n(T^2)$ is filtered by these objects, we find $H^0(T^2) = E_\infty^{0,0} = E_2^{0,0} = H^0(Y)$ and which obviously makes sense, as well as filtrations $$E_3^{0,1}\hookrightarrow H^1(X)$$ where $H^1(X) / \text{ker}(d^{0,1}_2) = E^{1,0}_3 = H^1(Y)$ and $$E_3^{0,2}\hookrightarrow (?) \hookrightarrow H^2(X)$$ where $H^2(X) / (?) = E^{2,0}_3 = H^2(Y)/\text{im}(d^{0,1}_2)$ and $(?)/E^{0,2}_3 = E^{1,1}_3$. Since $E^{1,1}_3 = 0 = E^{0,2}_3$, this implies $H^2(Y)/\text{im}(d^{0,1}_2) = H^2(X)$. -How can I get my hands on the differential $d_2^{0,1}$ in order to finish this calculation? - -REPLY [3 votes]: I think the most direct way to figure out the mystery differential is using the edge map -$$ \mathbf Z^2 \cong H^1(X,\mathbf Z) \to H^0(Y,R^1 f_\ast \mathbf Z) \cong \mathbf Z^2.$$ -Let's first think about how this map is defined: a class in $H^q(X)$ restricts to a class in $H^q(F_x)$ for each fiber $F_x$, and $H^q(F_x)$ is the stalk of $R^q f_\ast \mathbf Z$ at $x$. These classes are in fact compatible, i.e. come from a global section, which means that we get an element of $H^0(Y, R^q f_\ast \mathbf Z)$ from an element of $H^q(X)$. Now $H^1(X)$ has rank $2$, spanned by a meridian and a longitude. If I model $Y$ as the result of collapsing two disjoint meridian circles, then the longitude class restricts to zero in each fiber and the meridian circle restricts to the same generator in both special fibers. In particular the cokernel of the edge map is isomorphic to $\mathbf Z$, and the rank of the mystery differential is $1$. -(Also, I think you wrote your edge maps the wrong way around, possibly because your spectral sequences look transposed from how they're usually written.) -As a sanity check one can also argue geometrically that $Y \simeq S^2 \vee S^2 \vee S^1$, which gives the same cohomology groups. Indeed we can construct a CW decomposition of $Y$, taking the 1-skeleton to be a circle, and gluing on two $2$-cells. But both attaching maps are nonsurjective and therefore nullhomotopic.<|endoftext|> -TITLE: Non-linear hyperbolic PDE -QUESTION [6 upvotes]: I have the following PDE in two dimensions -$$ -2\partial_x\partial_y\sqrt{1-u^2}+\left(\partial^2_x-\partial^2_y \right)u=0, -$$ -with $u=u(x,y)$ with values between $-1$ and $1$, or alternatively -$$ -2\partial_x\partial_y\sin2\theta(x,y)+\left(\partial^2_x-\partial^2_y \right)\cos2\theta(x,y)=0, -$$ -with real $\theta(x,y)\sim\theta(x,y)+2\pi$, on some domain of the plane. Now, numerically I can obtain the solutions very quickly specifying some domain and an initial Cauchy line (as the equation hyperbolic), but I wish to have a deeper understanding of the solutions, so I'd like to see if there's a way to obtain analytic solutions. For instance, I know that $u=\cos(2\arctan(y/x))$ and $\theta(x,y)=\arctan(y/x)\pm1/2\arccos(c_1+c_2/(x^2+y^2))$, with $c_1, c_2$ some reals constants, are analytic, particular solutions, which strongly suggests that some general solution with arbitrary constants is plausible. -The problem is encountered in the context of elasticity of thin sheets. A so-called director field is imprinted on a thin elastic sheet, and it generates curvature upon a process called activation [1]. The director field $\theta(x,y)$ will induce a Riemannian metric on the new, curved sheet -$$ -g(x,y)=R[\theta(x,y)]diag(\lambda_1,\lambda_2)R[\theta(x,y)]^T, -$$ -with $R[\theta(x,y)]$ a $2\times2$ rotation matrix and $\lambda_1,\lambda_2$ some positive, known constants. Now, the aforementioned metric has a Gaussian curvature proportional to the equation written before, and the question I'm addressing is, for which $\theta(x,y)$s the generated curvature is zero ?, except for possibly isolated points where it may diverge. Now, the solutions I wrote before correspond to cones, but there should be more analytic solutions. -Any ideas ? Have you seen this equation or someone similar before ? -Thank you so much. -[1] Mostajeran, Cyrus; Warner, Mark; Ware, Taylor H.; White, Timothy J., Encoding Gaussian curvature in glassy and elastomeric liquid crystal solids, Proc. R. Soc. Lond., A, Math. Phys. Eng. Sci. 472, No. 2189, Article ID 20160112, 16 p. (2016). ZBL1371.82141. - -Edit: -Here is a summary of the solution from Robert Bryant's great answer, in a language more familiar to physicist. -Consider the (in general muliply-)connected domain $\mathscr{{W}}\subseteq\mathbb{R}^{2}$, with Cartesian coordinates $u$ and $v$, and the function $f:\mathscr{{W}}\rightarrow\mathbb{R}$ that solves $\frac{\partial^{2}f(u,v)}{\partial u\partial v}=f(u,v)$, -with $f(u,v)$ and $\partial f(u,v)/\partial v$ non-vanishing. This is a linear hyperbolic equation, so it's Cauchy problem is always well defined on $\mathscr{{W}}$, and the space of solutions is always non-empty. -The 1-forms -$$ -\alpha_{1} \equiv f\cos\left(u-v\right)\mathrm{d} u+\frac{\partial f}{\partial v}\sin\left(u-v\right)\mathrm{d}v -,\:\:\alpha_{2} \equiv f\sin\left(u-v\right)\mathrm{d} u-\frac{\partial f}{\partial v}\cos\left(u-v\right)\mathrm{d}v, -$$ -are closed. Therefore we can write locally $\mathrm{d} x =\alpha_{1},\: -\mathrm{d} y =\alpha_{2},$ for some functions $x$ and $y$ on $\mathscr{{W}}$. We define the function $(x,y):\mathscr{{W}}\rightarrow\mathbb{R}^{2}$ and the domain $\mathscr{{Z}\mathbb{\subseteq R}}^{2}$ as the image of $(x,y)$, i.e., $(x,y)\left(\mathscr{{W}}\right)=\mathscr{{Z}}$. We can use $x$ and $y$ as coordinates of $\mathscr{{Z}}$, as by definition they cover the latter completely. The function $u-v:\mathscr{{W}}\rightarrow\mathbb{R}$ can be pulled through $\mathscr{{Z}}$, that is $u-v=\theta\circ(x,y)$, with $\theta:\mathscr{{Z}}\rightarrow\mathbb{R}$ a function defined by the previous relation. -Inverting the definitions for $u$ and $v$ as functions of $x$ and $y$ we can write -$$\frac{\partial}{\partial x} =\frac{1}{f}\cos\left(u-v\right)\frac{\partial}{\partial u}+\frac{1}{\frac{\partial f}{\partial v}}\sin\left(u-v\right) -\frac{\partial}{\partial v}\\ -\frac{\partial}{\partial y} =\frac{1}{f}\sin\left(u-v\right)\frac{\partial}{\partial u}-\frac{1}{\frac{\partial f}{\partial v}}\cos\left(u-v\right)\frac{\partial}{\partial v}, -$$ -and it's just a matter of patience to verify that -$$ -\begin{align} -&2\frac{\partial^{2}}{\partial x\partial y}\sin2\theta+\left(\frac{\partial^{2}}{\partial x^{2}}-\frac{\partial^{2}}{\partial y^{2}}\right)\cos2\theta\\ - &=\Bigg\{ \left[\frac{1}{f}\cos\left(u-v\right)\frac{\partial}{\partial u}+\frac{1}{\frac{\partial f}{\partial v}}\sin\left(u-v\right)\frac{\partial}{\partial v}\right]^{2}\\ -&\:\:\:\:-\left[\frac{1}{f}\sin\left(u-v\right)\frac{\partial}{\partial u}-\frac{1}{\frac{\partial f}{\partial v}}\cos\left(u-v\right)\frac{\partial}{\partial v}\right]^{2}\Bigg\} \cos2\left(u-v\right)\\ - &+2\left[\frac{1}{f}\cos\left(u-v\right)\frac{\partial}{\partial u}+\frac{1}{\frac{\partial f}{\partial v}}\sin\left(u-v\right)\frac{\partial}{\partial v}\right]\\ -&\:\:\:\times\left[\frac{1}{f}\sin\left(u-v\right)\frac{\partial}{\partial u}-\frac{1}{\frac{\partial f}{\partial v}}\cos\left(u-v\right)\frac{\partial}{\partial v}\right]\sin2\left(u-v\right)\\ - &=\frac{4\cos^{2}(u-v)}{f\left(\frac{\partial f}{\partial v}\right)^{2}}\left(f(u,v)-\frac{\partial f}{\partial u\partial v}\right)\\ - &=0. -\end{align} -$$ -The particular solutions mentioned before are obtained with $f(u,v)=e^{\alpha u+v/\alpha}$, with some constant $\alpha$. But of course any other $f(u,v)$ will generate a solution. The most general real, separable solution is -$$ -f(u,v)=\int_{-\infty}^{\infty}\mathrm{d}\rho\:C(\rho)\:e^{\rho u+\rho^{-1}v}, -$$ -for some arbitrary kernel $C(\rho)$. So one can classify arbitrarily many solutions by $C(\rho)$. From a calculative standpoint, once chosen some $f(u,v)$ the problem is that in general it's difficult to solve the algebraic system to write down explicitly $u(x,y)$ and $v(x,y)$, so one can say that the non-linear differential equation in two variables was transformed into a problem of two non-linear algebraic equations. - -REPLY [7 votes]: As I understand it, the equation you are imposing on the function $\theta(x,y)$, defined on a domain $D\subset\mathbb{R}^2$ in the $xy$-plane is that, for some positive constants $\lambda_1\not=\lambda_2$, the metric -$$ -g = \lambda_1\,(\cos\theta(x,y)\,\mathrm{d}x+\sin\theta(x,y)\,\mathrm{d}y)^2 -+ \lambda_2\,(\sin\theta(x,y)\,\mathrm{d}x-\cos\theta(x,y)\,\mathrm{d}y)^2 -$$ -should be flat, i.e., that there should exist functions $p,q:D\to\mathbb{R}^2$ so that $g = \mathrm{d}p^2 + \mathrm{d}q^2$. In other words, the Jacobian matrix of the mapping $f = (p,q):D\to\mathbb{R}^2$ should have distinct, constant singular values. -Now, this is exactly (a local version of) the question posed in Are all maps $\mathbb{R}^2 \to \mathbb{R}^2$ with fixed singular values affine? In my answer to that question, I showed that any sufficiently differentiable solution $f$ defined on all of $D = \mathbb{R}^2$ must, in fact, be affine (equivalently, that $\theta$ must be constant), and I gave a formula for the local solutions that satisfy a non-degeneracy condition that shows how to reduce this problem locally to the hyperbolic linear equation $f_{uv} = f$ on an auxilliary domain $D'$ in the $uv$-plane plus a couple of 'quadratures' (i.e., writing an explicit closed $1$-form as the differential of a function). -I won't reproduce the analysis here, I'll just give the solution described there: Let $D'$ be a simply-connected domain in the $uv$-plane and let $f$ be a function on $D'$ such that $f_{uv} = f$ while $f$ and $f_v$ are nonvanishing. (It's easy to write many explicit solutions to this equation by separation of variables.) One easily sees that, when $f_{uv}=f$, the $1$-forms -$$ -\begin{aligned} -\alpha_1 &= \cos(u{-}v)\,f\,\mathrm{d}u +\sin(u{-}v)\,f_v\,\mathrm{d}v\\ -\alpha_2 &= \sin(u{-}v)\,f\,\mathrm{d}u -\cos(u{-}v)\,f_v\,\mathrm{d}v -\end{aligned} -$$ -are closed, and hence one can write $\alpha_1 = \mathrm{d}x$ and $\alpha_2 = \mathrm{d}y$ for some functions $x$ and $y$ on $D'$. Suppose that $(x,y):D'\to \mathbb{R}^2$ is one-to-one and, hence, a diffeomorphism, and let $\theta:(x,y)(D')=D\to\mathbb{R}$ be the function that satisfies $\theta(x,y) = u-v$. Then $\theta$ satisfies the given equation, as follows from the Chain Rule. This is the 'general' local smooth solution, i.e., every smooth solution for which $\mathrm{d}(\cos\theta(x,y)\,\mathrm{d}x+\sin\theta(x,y)\,\mathrm{d}y)$ and $\mathrm{d}(\sin\theta(x,y)\,\mathrm{d}x-\cos\theta(x,y)\,\mathrm{d}y)$ are nonvanishing can be written in the above form locally. (Of course, if one imposes the condition that one of these $2$-forms vanishes identically, the problem is even more easily solved, and one can see the formula for those solutions in the solution that I referenced above.)<|endoftext|> -TITLE: Is Collection really stronger than Replacement? -QUESTION [8 upvotes]: The two powerhouse schemata of set theory are Replacement and Collection: - -Replacement. For every definable function $f$ and every set $x$, $f"x$ is a set. -Collection. For every definable relation $R$ and every set $x$, there is a set $y$ such that for every $u\in x$ there is $v\in y$ such that $u\mathrel{R}v$. - -Easily, Collection implies Replacement, and assuming $\sf ZF$ we can prove the converse as well. If we omit the Power Set axiom, then the reverse implication no longer holds, and Collection is a strictly stronger schema than Replacement. -But since $\sf ZF$ without Power Set is strictly weaker, consistency-wise, than $\sf ZF$ itself, it raises the following question: - -Does $\sf\operatorname{Con}(ZF(C)-)\to\operatorname{Con}(ZF(C)^-)$? - -(Here $\sf ZF-$ is $\sf ZF$ without Power Set, but with Replacement, and $\sf ZF^-$ is the same theory with Replacement replaced by Collection. The C stands for the Well-Ordering Theorem.) - -REPLY [10 votes]: The theories are equiconsistent and have the same strength as second order arithmetic $\text{Z}_2$. Since we have an $L$-definable well-ordering of the constructible universe $L$, replacement implies collection and ZFC\P in $L$.<|endoftext|> -TITLE: Kähler differentials on an Artinian local ring -QUESTION [5 upvotes]: Suppose $R$ is a commutative Artinian local ring over an algebraically closed characteristic 0 field $k$. Suppose $f\in R$ is such that $df=0$ (in the sense that the element $df$ vanishes in the module of Kähler differentials). Is $f$ necessarily in $k$? - -REPLY [2 votes]: I finally remembered the example (though not the reference). Take $f=x^2y^2+x^5+y^5\in R=\mathbb{C}[[x,y]]$. Then $f_x,f_y$ form a regular sequence in $R$ and thus $R/I$ where $I=(f_x,f_y)$ is an Artin local ring. One checks $f\not\in I$. Thus, $df=0\in\Omega^1_{R/I}$, but $f\neq 0$ in $R/I$.<|endoftext|> -TITLE: Did human computers use floating-point arithmetics? -QUESTION [11 upvotes]: Before the proliferation of computers in the 1950s, did human computers use floating-point formats for their computations? -Floating-point calculation was reportedly implemented already in the 1910s (Wikipedia), so one might assume the idea must have been in circulation way earlier than that. However, I haven't found any information how people were computing back then. Since most human computer work must have been numerical, they may have used floating-point calculations. - -REPLY [23 votes]: In the field of hydrodynamics the first calculation by a human computer was carried out around 1920 for a project to transform an open sea into a closed lake, with the aim to protect Holland from flooding. The physicist Hendrik Lorentz headed a task force to calculate the effect of the dike on the tidal flow. -The human computer was the hydraulic engineer Jo Thijsse. Lorentz commented as follows on this work: - -The numerical calculations were so lengthy, that we came close to the -ultimate limit of what can be done in this way. I myself had no part -in this. I did try once or twice to set up and work out such a -calculation, but then it would turn out that I had made a mistake, so -that it had to be done all over again by others. - -The hydrodynamic calculation involved quantities differing by many orders of magnitude, and so necessarily required floating point arithmetic. Here is one page from their report, showing the exponential notation.<|endoftext|> -TITLE: Magic behind idempotent-complete categories a.k.a. why (sometimes) be Karoubian is sexier than be Abelian -QUESTION [10 upvotes]: It is well know that Karoubian categories (also called idempotent-complete categories) are living between additive and -Abelian categories. While one of the most famous advantages to -work with abelian cetegories is that they are closed under -building kernels and cokernels of arbitrary morphism, the -Karoubian categories have a slightly weaker property that only -idempotent morphisms (i.e. $p$ with $p^2=p$) from there share -this property. -Nevertheless it seems that in diverse constructions Karoubian categories -or Karoubian envelopes of additive categories provide a more -natural setting than Abelian categories. -I hope my question not becomes too broad: What is the philosophical -meaning behind Karoubian categories or say in simpler words why in some constructions they are more prefered (e.g. -category of pure motives and in K-theory) in contrast to say -at first glance more 'flexible' abelian categories? -My natural guess is that if we think about the construction -of of the category of pure motives we start with the category -of smooth varieties over a base field and pass after -application of this magic Karoubian completion functor to -idempotent-complete categories. Since it's not abelian it contains -by definition only kernel and cokernels of idempotent morphisms -but that's all we need there to proceed the constrution. -This lead me to conjecture that the main advantage of -Karoubian categories in contrast to Abelian categories mights show when -one have to perform a construction where one starts with -a certain preadditive category but the construction requireres -a category having at least some kernels and cokernels. -Now one can pass canonically to the Karoubian completion or -extend the initial category to an Abelian category. But exactly here I see -an obstacle with the secound and at first glance more 'natural' -approach: -Does there always a way to embedd a preadditive category in an abelian -category? If yes, seemingly the disadvantage of this approach -seems to be that this Abelian category is much harder to control, -while the Karoubian completion is constructed quite canonically and -behaves more 'similar' to initial preadditive category. -Questions: Is what I tried to sketch above exactly the motivation -why Karoubian categories are in some constructions more prefered -then Abelian categories? -Are there more reasons making Karoubian categories 'interesting'? -Is there any intuition or important example one should have in -mind how to think about Karoubian categories? - -REPLY [2 votes]: I'd like to comment that idempotent completion is sometimes the right thing to do 'by design'. (I don't have enough rep to just leave a comment though.) I'm reminded of the category of motives, in which you would like to have some kind of 'cellular decomposition' of your varieties, the archetypical example being that one would like the motive $[\mathbb{P}^1]$ to split as $[\mathbb{A}^1] + [\text{pt}]$ or something along those lines. Idempotent completion does exactly that: it forces the idempotent map $\mathbb{P}^1 \to \operatorname{Spec} k \to \mathbb{P}^1$ to have a kernel, which leads to the desired decomposition.<|endoftext|> -TITLE: Action of a dual Hopf algebra on a factor -QUESTION [5 upvotes]: Suppose that a finite-dimesnional Hopf $C^*$-algebra $H$ acts on a type $II_1$ factor $N$ minimally (that is, $N^{\prime}\cap (N\rtimes H)=\mathbb{C}$). Is it true that there always exists a minimal action of the dual Hopf algebra $H^*$ on $N$? - -REPLY [8 votes]: No, there might be no minimal action at all of $H^*$ on $N$. By Theorem A in this paper of Falguières and Raum (see also this paper from which other examples may be deduced), for any rigid C$^*$-tensor category $\mathcal{C}$ with finitely many irreducible objects, there exists a $II_1$ factor $N$ such that the bimodule category of $N$ is isomorphic with $\mathcal{C}$. In particular, given any finite, nonabelian group $G$, you can find a $II_1$ factor $N$ and an outer action $\alpha$ of $G$ on $N$ such that the only irreducible finite index $N$-$N$-bimodules are $1$-dimensional and given by the automorphisms $(\alpha_g)_{g \in G}$. Letting $H$ be the Hopf algebra associated with $G$, there is no minimal action of its dual $H^*$ on the $II_1$ factor $N$. Indeed, since $G$ is nonabelian, $G$ has irreducible representations of dimension larger than one and a minimal action of $H^*$ on $N$ would give irreducible finite index $N$-$N$-bimodules of dimension larger than $1$.<|endoftext|> -TITLE: Any shortcuts to understanding the properties of the Riemannian manifolds which are used in the books on algebraic topology -QUESTION [14 upvotes]: I'm now attending a reading seminar on the algebraic topology. -The seminar treats the book of Bott & Tu (Differential Forms in Algebraic Topology) and Milnor (Characteristic Classes). -In those books, theorems on the Riemannian manifolds are frequently just mentioned and used. -To mention some examples - -Riemannian manifold has a good cover. - -Exponential Map is used to find a tubular neighborhood for a pair of manifolds. (where one is a submanifold of the other ) and its properties are used in computations on the dual cohomology class and the diagonal cohomology class. - -Argument in a proof which states that we can reduce the general case to a local open submanifold with the Euclidean standard metrics. - - -and maybe more. -When I browse books on differential geometry or Riemannian manifolds, I get the feeling that I cannot avoid studying the standard materials like the connection, tensors... -But I have no time to study all that materials. -Is there some shortcut to understand those materials (at least for good manifolds) without studying all the details of these differential materials? (Maybe is there some axiomatic approach?) -Any suggestions on the references are welcome. -Thank you very much! - -REPLY [4 votes]: I just searched "Riemann" in Bott-Tu and scanned through most occurrences. Proving the existence of a good cover requires the fact that each point in a Riemannian manifold has a geodesically convex neighborhood. You can probably find the proof of this in most textbooks on Riemannian geometry. You can also safely just assume the existence of a good cover without learning the proof. -After that, all uses of Riemannian manifolds appear to rely on only the definition of a Riemannian metric and nothing more. Neither the Levi-Civita connection nor the curvature tensor are ever needed. Also, the book provides full details of any proof that uses a Riemannian metric. -Here are the uses of either a Riemannian metric or its equivalent on a vector bundle that I found: - -Two consist of the reduction of the tangent bundle into a sphere bundle and the reduction of the frame bundle into an orthonormal frame bundle. -The construction of the global angular form -The definition of a "radial function" -The definition of the gradient of a function<|endoftext|> -TITLE: Are we sure the calculus of inductive constructions and ZFC plus countably many inaccessible cardinals are equiconsistent? -QUESTION [17 upvotes]: This answer says, - -IIRC, the calculus of inductive constructions is equi-interpretable with ZFC plus countably many inaccessibles — see Benjamin Werner's "Sets in types, types in sets". (This is because of the presence of a universe hierarchy in the CIC.) - -But, I read "Sets in types, types in sets" and discovered that the book does not prove this statement. It only conjectures the strength of CIC. -Has "CIC and ZFC + countably many inaccessible cardinals are equiconsistent" been proven or disproven? - -REPLY [5 votes]: The situation is a bit subtle. One can interpret CIC in any model of ZFC with infinitely many inacessibles. However, interpreting ZFC in CIC is more subtle. First one needs to assume the law of excluded middle and choice in CIC (and perhaps quotient types depending on how smooth we want things to work). These are very strong assumptions and they increase the consistency strength over plain CIC, which appears to be much weaker. -Once excluded middle and choice are assumed, within each universe level $\mathcal{U}_1,\mathcal{U}_2,\ldots$ of CIC we can construct a model of ZFC. This constructs a chain $V_1 \subseteq V_2 \subseteq \cdots$ of models of ZFC where each is an end extension of the previous, up to canonical isomorphism. Thus, $V_2$ has at least one inaccessible, $V_3$ has at least two inaccessibles, and so on. Thus CIC with choice proves the consistency of ZFC + there are $k$ inaccessibles for any standard $k$. Note that I've been careful not to associate universe levels with natural numbers. Indeed, models of CIC and ZFC can have nonstandard natural numbers. However, universe levels are syntactic objects and therefore always standard. -So, the consistency of CIC with choice and excluded middle implies the $\Pi_1$ statement $$\forall k\,\operatorname{Con}(ZFC + k\text{-many inaccessibles})\tag{*}.$$ -This is strictly weaker than the consistency of ZFC with infinitely many inaccessibles. However, $(*)$ is actually enough to construct a full model of CIC! By compactness, we can construct a model $V$ of ZFC with $k$ inaccessibles, where $k$ is nonstandard. In this model, we can list the first standardly many inaccessibles as $\kappa_1 < \kappa_2 < \cdots$. This hierarchy allows us to construct a corresponding sequence of universes for a model of CIC. Note that this is not an interpretation per se since we can only witness externally that $k$ is nonstandard. -Nevertheless, from the above it follows that the consistency strength of CIC with excluded middle and choice is exactly $(*)$ and therefore strictly weaker than ZFC with infinitely many inaccessibles.<|endoftext|> -TITLE: Elementary proof that $\mathbb{R}^3 \setminus \{p_1,\dots,p_n\}$ is not homeomorphic to $\mathbb{R}^3$ -QUESTION [13 upvotes]: I was wondering if there were a proof of the fact that $$\mathbb{R}^3 \setminus \{p_1,\dots,p_n\} \: \text{is not homeomorphic to} \: \mathbb{R}^3$$ -for every $n \geq 1$ -that does not use cohomology or higher homotopy groups techniques (of course the fundamental group is allowed). -Thanks in advance - -REPLY [41 votes]: The fundamental group of the one-point compactification of $\mathbf R^3 \setminus \{p_1,\ldots,p_n\}$ is a free group on $n$ generators, for any $n \geq 0$.<|endoftext|> -TITLE: Certain interpolation property of von Neumann algebras -QUESTION [6 upvotes]: Von Neumann algebras have the following form of interpolation property: let $(x_n)_n$ and $(y_n)$ be increasing and decreasing, respectively, sequences of self-adjoint elements in a von Neumann algebra $M$ such that $x_n \leqslant y_n$ for all $n$. Then there is $z$ such that $x_n \leqslant z \leqslant y_n$ for all $n$. -Can it be strengthened to the following statement. Let $A$ and $B$ be two countable sets of selfadjoint elements with $A \leqslant B$ (each element of $A$ is dominated by each element of $B$). Is there $z$ such that $A \leqslant \{z\} \leqslant B$? -It is true in the commutative case but I am not sure if it works for matrices though. - -REPLY [6 votes]: It does not hold for matrices. Let $P_1 := \left[\begin{array}{cc} 1& 0 \\ 0& 0 \end{array}\right]$, $P_2:= \left[\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right]$, and $X:= \left[\begin{array}{cc} 3 & 2,5 \\ 2,5 & 4\end{array}\right]$, and take the sets $A:=\{P_1, P_2\}$ and $B:=\{X, Id\}$. Both $X$ and $Id$ are upper bounds for the set $A$, so the assumptions are satisfied. But any self-adjoint matrix $z$ that is an upper bound for $A$ and a lower bound for $Id$ has to be equal to the identity: indeed, its diagonal entries have to be equal to $1$, and therefore its off-diagonal entries have to be zero. But identity is not a lower bound for $X$, so there is no $z$ separating $A$ and $B$ at all.<|endoftext|> -TITLE: Angle between geodesics at different fixed points of a Riemannian isometry -QUESTION [5 upvotes]: Suppose I have an isometry $f$ of a Riemannian manifold $(M,g)$. Suppose further that $p$ and $q$ are fixed points of $f$. If $\gamma$ is a geodesic segment from $p$ to $q$, then so is $f(\gamma)$. -Let $\theta_p$ be the angle between $\gamma$ and $f(\gamma)$ at $p$ and $\theta_q$ be the angle between $\gamma$ and $f(\gamma)$ at $q$. Must we always get $\theta_p=\pm\theta_q$? -I’ve been trying to come up with a simple counterexample, but my inability to find one is starting to convince me that it might actually be true. - -REPLY [9 votes]: No. There is in fact a very tangible model that gives a counterexample. - -Take a piece of paper. Draw a straight line parallel to one of the edges. -Roll the paper up into a cone, with the vertex on the edge that you chose. -The straight line that you drew earlier is now a geodesic $\gamma$ on your Riemannian manifold (the cone). -$\gamma$ can be observed to intersect itself once, call the point $p$. On one side $\gamma$ forms a loop. On the other side of $p$ the two branches of $\gamma$ fly off to infinity. Forget about those two branches. -Start from $p$, travel along the loop part of $\gamma$. Call the half-way point $q$. -Your cone has a reflection symmetry that preserves the straight lines that connect $p$ to the vertex, and from vertex to $q$ that swaps the two halves. Call this isometry $f$. -$\gamma$ and $f(\gamma)$ are parallel at $q$. They are not at $p$ (with angle related to the defect angle of your cone). - -If you don't have a piece of paper handy, here's what the result looks like: - -N.B.: You can make the manifold smooth and complete by cutting off the vertex and replace by a small smooth cap. You can make it compact by putting a cap on the other end. In both cases you can make it so that the new pieces also respect the reflection symmetry.<|endoftext|> -TITLE: Looking for a BAMS text about the group with commutation relations defined using meaningful words -QUESTION [5 upvotes]: What I definitely remember is that I saw a description of the following in the Bulletin of the American Mathematical Society, sometime in eighties (or maybe nineties?) -One considers the group generated by letters of the Latin alphabet, with the following relations: $W(a,b,...,y,z)=W'(a,b,...,y,z)$ for any words that are permutations of each other and moreover each of them is a meaningful word in English. For example, $name=mane=mean=amen$ in this group. -The question was whether this group is commutative. -Does anybody remember it? Or maybe knows the answer? -Slightly later -Neil Strickland points out that it is Monthly (1986) rather than BAMS. -So formally he answered the question, except that there is another thing I did remember incorrectly: the question there was what is the center of this group! -So, what is the center of this group? - -REPLY [5 votes]: I checked that in /usr/dict/words there are a handful of words in which q appears without a following u, but none of them are free of repetition and have any anagrams. This means that we have a surjective homomorphism to the free group $\langle q,u\rangle$ sending all other letters to the identity, so $q$ and $u$ do not commute, as Jeremy suggested. -[UPDATE] My original answer assumed that we were only using words with no repeated letters, in which case Jeremy's comment would not be applicable. I now see that this restriction is not in fact in the original problem. Nonetheless, I have now run a brute-force check including words with repeated letters and reached the same conclusion. If you take any word that possesses anagrams and map it to $\langle q,u\rangle$ you get either a power of $u$ or the element $qu$.<|endoftext|> -TITLE: What is the point of reading classics over modern treatments? -QUESTION [25 upvotes]: There seems to be a good number of mathematicians who recommend reading "classic" works in a given field (where the term "classic" is in the sense defined below). Indeed, there are many well written classic texts, for instance - -anything by Milnor, including some papers like "Group of Homotopy Spheres, I." -Weyl, The concept of a Riemann surface -Pontryagin, Topological Group - -By "classic" text I loosely mean an old text written by the person who came up with the theory/idea or at least has made a major contribution somewhere very close to the subject of the book. -From a practical viewpoint, there are both advantages and disadvantages for reading the classics. Here are some that come to mind: -Pros: - -Reading the text gives a view of how the theory was originally created. (Thus by observing this, one may try to emulate the master's creative mindset.) - -By knowing that one is learning the theory "from the horse's mouth," one can have a peace of mind. - - -Cons: - -The terminology may be archaic. (e.g. A Course in Pure Mathematics by Hardy) -The exposition may be presented in a cumbersome manner in view of modern machinery. - -In particular, the "Cons" listed above result in more work and time spent on understanding the material than when reading a more modern textbook. -Edit: Per helpful comments (by @Geoff Robinson and @Francesco Polizzi) I am going to change my question to the following: -My Question. What are things one might learn from reading the classics that one would not gain from modern treatments? - -REPLY [2 votes]: Some "worked examples" where re-reading the classics has proved worthwhile: - -Darboux's method for solving second order partial differential equations (per Goursat's 1896 book Leçons sur l'intégration des équations aux dérivées partielles du second ordre, à deux variables indépendantes), combined with its group-theoretic formulation of Vessiot (1939, 1942), has been extended by Anderson, Fels, and Vassiliou (as explained by Anderson in https://youtu.be/KKljvg3yK1U). - -Lie's ideas about using continuous groups as a tool to solve -differential equations were revived and extended in the latter half -of the 20th century - refer the CRC Handbook of Lie Group Analysis -of Differential Equations, edited by N. H. Ibragimov - -Cartan's (1908) method of equivalence has proved to be valuable in solving a number of practical problems (per Shadwick's recent lecture - https://youtu.be/f3ATo79f1QA). See also Raouf Dridi's thesis Utilisation de la méthode d'équivalence de Cartan dans la construction d'un solveur d'équations différentielles, https://tel.archives-ouvertes.fr/tel-00264288. - -The work of Drach (1898) and Vessiot (1946) has been revived, corrected and incorporated into the nonlinear differential Galois theory of Umemura and Malgrange. (See also Pommaret's 1983 book Differential Galois Theory.)<|endoftext|> -TITLE: Rings where all indecomposable projective modules are finitely generated -QUESTION [10 upvotes]: Let $X$ be the class of (unital, associative and not necessarily commutative) rings $R$ where every indecomposable projective $R$-module is finitely generated. - -Question 1: Is there a nice equivalent characterisation when a ring is in $X$? - - -Question 2: $X$ should contain for example all Artin algebras. Does it also contain right artinian rings and if yes, is there an easy argument? - -REPLY [5 votes]: $X$ is contains semiperfect rings (hence right artinian rings) by two facts you can find in Lam. First every simple module has a projective cover. Second if $P$ is projective then $PJ\subsetneq P$ where $J$ is the radical. This latter fact is obvious if $J$ is nilpotent like for right Artinian rings and semiprimary rings but the proof for semiperfect rings is nontrivial. Any way once you know that $P/PJ$ is non-zero it has a simple quotient since $R/J$ is semisimple. That projective cover of this simple is of the form $eR$ with $e$ primitive and by definition of the projective cover, $P$ maps onto $eR$. Since this splits and $P$ is indecomposable $P\cong eR$. -I don't know more general results.<|endoftext|> -TITLE: Do $\mathbb{R}^n$ bundles have unit sphere bundles? -QUESTION [12 upvotes]: Recall there are multiple ways to define the unit sphere bundle of a vector bundle. One is by constructing a fiberwise vector space metric and declaring the sphere bundle to have fibers the unit spheres in each of the vector space fibers. The other way is to use the equivalence of vector bundles and principal $O(n)$ bundles and then since $O(n)$ acts faithfully on $S^{n-1}$ we may replace the fiber to obtain a sphere bundle. -There is, however, a distinction between vector bundles and $\mathbb{R}^n$ bundles, that is fiber bundles with fiber $\mathbb{R}^n$ and structure group $Homeo(\mathbb{R}^n)$. It is not always possible to assign a coherent vector space structure to the fibers to make it a vector bundle. Is there a notion of an associated sphere bundle in this context? -Now I tend to just believe fiberwise constructions are always possible, so I would believe someone if they said we could pick a metric on the fibers of any $\mathbb{R}^n$ bundle and define its unit sphere bundle in the same way as for vector bundles. But on the principal bundle side of things, this seems to me to be asserting that $Homeo(\mathbb{R}^n)$ has a subgroup $H$ so that the inclusion $H \rightarrow Homeo(\mathbb{R}^n)$ is a weak equivalence, and $H$ preserves the unit sphere $S^{n-1}$ while acting faithfully on it. This seems very difficult to believe. -I will add that this is an unstable question. We can always form the fiberwise one point compactification of a $\mathbb{R}^n$ bundle since homeomorphisms extend to the one point compactification. If the unit sphere bundle exists, it's fiberwise suspension should coincide with this sphere bundle. - -REPLY [14 votes]: It is not true in general that there is a subgroup $H$ of $Homeo(\mathbb{R}^n)$ such that the inclusion is a homotopy equivalence and $H$ preserves the unit sphere. If there was, then $H$ would also preserve the unit disk and thus every topological $\mathbb{R}^n$-bundle would contain a $D^n$-bundle. But it is not the case by a result of Browder, Theorem 1 of Open and Closed Disk Bundles.<|endoftext|> -TITLE: profinite completion and linear representations of finitely presented groups -QUESTION [6 upvotes]: Let $G$ be a finitely presented group. It is clear that if the profinite completion $\widehat{G} $ of $G$ is finite, then any finite dimensional complex linear representation $\rho: G\to \text{GL}(m, \mathbb{C})$ is finite, i.e., $\rho(G)$ is a finite group. For the converse, is there a counterexample for such $G$ such that any finite dim representation $\rho$ is finite, but $\widehat{G}$ is a infinite group? - -REPLY [10 votes]: Yes, there exists such a finitely presented group. -Let $\Gamma$ be a cocompact arithmetic lattice in a product of $\ge 2$ rank 1 groups simple groups (with trivial center) over locally compact fields of finite characteristic. So $\Gamma$ is finitely presented (it is even CAT($0$)). By Malcev, $\Gamma$ is residually finite. By Kazhdan-Margulis, $\Gamma$ is just-infinite: all normal subgroups in $\Gamma$ except $\{1\}$ have finite index. By Venkataramana's superrigidity, every finite-dimensional complex representation of $\Gamma$ has a finite image. - -My initial answer was conditional answer: -Let $\Gamma$ be a cocompact lattice in $\mathrm{PSp}(n,1)$, $n\ge 2$. This is a non-elementary hyperbolic group. Hence it is known that for many $x\in\Gamma-\{1\}$, the quotient $\Lambda_x=\Gamma/\langle\!\langle x\rangle\!\rangle$ is non-elementary hyperbolic. -It is also known (superrigidity) that every linear proper quotient of $\Gamma$ is finite. Hence, every linear quotient of $\Lambda_x$ has a finite image. Accordingly - -either $\Lambda_x$ has an infinite profinite completion, hence answers your question -or $\Lambda_x$ has finite profinite completion, and hence there exists a non-residually-finite hyperbolic group. The existence of such a group is a famous open problem. - -I'm not sure that any well-accepted conjecture predicts which of these properties holds (this might even depend on $x$).<|endoftext|> -TITLE: Can the category of S-local objects be reflective but not a localization by S? -QUESTION [12 upvotes]: This is cross-posted from MSE (and substantially re-written) after receiving no answers. -Suppose $\mathcal C$ is a category and $S \subseteq \operatorname{Mor}(\mathcal C)$ is some collection of morphisms in $\mathcal C$. Let $\operatorname{Loc}(S) $ be the full subcategory of $\mathcal C$ on the $S$-local objects. Then we have -$$ -\operatorname{Loc}(S) \stackrel{\iota}{\hookrightarrow} \mathcal C \stackrel{P}{\twoheadrightarrow} \mathcal C[S^{-1}], -$$ -where $\iota$ is the inclusion and $P$ is the canonical functor to the localization of $\mathcal C$ by $S$. We can ask about two possibly-existing objects: - -a left adjoint $L$ to $\iota$; -a right adjoint $r$ to $P$. - -In fact, (2) "subsumes" (1), in the following sense: If such an $r$ exists, then there is an equivalence of categories $j : \mathcal C[S^{-1}] \to \operatorname{Loc}(S)$ such that $r \cong \iota \circ j$. This implies that $j \circ P \vdash \iota$, so we have our $L$. Suppressing $j$, we can summarize as "if $P$ has a right adjoint, it must be $\iota$" or "if (2) exists, it is $\iota$, and hence (1) does too, and is $P$". -It's a bit tempting (at least to me) to try to extract something symmetrical like "if (2) exists, it must be $\iota$, and if (1) exists, it must be $P$" from this last summary, or maybe even "(1) exists iff (2) does", but neither of these actually follows from the statements above—to show them, we'd need a separate result about (1) "subsuming" (2) in the manner above. -So I'd like to know: Are the following equivalent statements true? - - -If $L \vdash \iota$, then there is an equivalence of categories $j : \mathcal C[S^{-1}] \to \operatorname{Loc}(S)$ such that $L \cong j \circ P$. -If $\iota$ has a left adjoint, then $P$ has a right adjoint. -If $L \vdash \iota$, then $L$ exhibits $\operatorname{Loc}(S)$ as a localization of $\mathcal C$ by $S$. - - -I suspect the answer is "not necessarily", because the nLab page on reflective localizations lists two facts which would clearly be implied by the third statement above but which do not clearly imply it. In particular, it says that $L$ will exhibit $\operatorname{Loc}(S)$ as a localization of $\mathcal C$ by the class of morphisms $L$ inverts—but these morphisms are in general a strict superclass of $S$. It also says that $L$ will have a universal property similar to the one for a localization by $S$, but which only applies to left adjoint functors that invert $S$. - -REPLY [14 votes]: Not in general, no - there must be some additional conditions on $S$, such as a saturation condition. -Consider for instance the presentable case. Then if $S$ is small, $Loc(S) $ is always reflexive, and is always a localization of $C$ at the saturated class generated by $S$, but there is no reason to expect it to be the localization at $S$. -What's happening is that the class of arrows that are inverted by the left adjoint $C\to Loc(S)$ is always closed under colimits in $C$ (let $x$ be $S$-local, then the collection of arrows $f:y\to z$ such that $\hom(f,x)$ is an isomorphism is closed under colimits), so if $S$ is too small, they won't all be inverted by $C\to C[S^{-1}]$. -What you get in the presentable situation is exactly what you describe (because out of a presentable category, left adjoint = colimit preserving, you can replace "left adjoint" by "colimit preserving"): $C\to Loc(S)$ is the universal colimit preserving functor out of $C$ that inverts $S$. -It would be nice to have an explicit example here but of the top of my head I don't have one - a strategy to find one would be to find a functor $C\to D$ which is very very far from preserving colimits, but inverts $S$ nonetheless. Then there's a hope that it would invert $S$ but not the saturated class it generates, which would imply $C[S^{-1}]\not\simeq Loc(S)$ -However, in general, model categories will provide good examples for another phenomenon : indeed for a model category $C$ with weak equivalences $W$, we have a very nice control over $C[W^{-1}]$ (this is why they were invented), but $Loc(W)$ will rarely be anything interesting, in fact it will often be trivial (in an appropriate sense), and so the inclusion does have a left adjoint, but $C[W^{-1}]$ is far from trivial. -Take for instance $C$ to be chain complexes over a ring $k$, and $W$ to be the class of quasi-isomorphisms (morphisms that induce an isomorphism in homology); then $C[W^{-1}]$ is very well known and well studied, it's $D(k)$, the derived category. In particular it is nonzero. -I claim that $Loc(W)$ consists only of the $0$ chain complex (in particular it is the trivial category $*$, and so it is easy to check that the inclusion $Loc(W)\to C$ does have a left adjoint). Let $C\in Loc(W)$ -Now consider the following chain complex : $D^n(k)= \dots \to 0\to k\to k\to 0\to \dots$ with only an identity arrow $k\to k$, the second $k$ in position $n-1$ and everything else $0$. Then the only map $D^n(k)\to 0$ is a quasi-isomorphism, so $0\to \hom(D^n(k),C)$ is an isomorphism. However $\hom(D^n(k),C)\cong C_n$ because an arrow $D^n(k)\to C$ consists of $f:k\to C_n$ and $g:k\to C_{n-1}$ such that $\partial_n f = g$, that is, an element $x\in C_n$ and its image under $\partial_n$, so in the end just an element $x\in C_n$. -It follows that $C_n=0$ so the claim now follows. -ADDED : Here's an explicit example of the first phenomenon : take the category of abelian groups, and the set $S$ to be the set of all $\mathbb Z/n\to 0$, where $n\geq 1$. Then $Loc(S)$ is the class of torsion-free abelian groups, and you have the classical unit $A\to A/tors(A)$ which shows that it is reflexive (although again it is a general fact, as $Ab$ is presentable). However, if you "naively" invert $S$, you don't get this. For instance, consider the $p$-Prüfer group $\mathbb Z/p^\infty = \mathrm{colim}_n \mathbb Z/p^n$. This clearly dies in $Loc(S)$, but I claim it doesn't in $Ab[S^{-1}]$. -Indeed, consider the following category $C$ : objects are abelian groups, and $\hom(A,B)$ is the quotient of $\hom_{Ab}(A,B)$ by the following subgroup : $f$ is in the subgroup if it factors through some finite abelian group. It's not hard to see that it does indeed define a category with the obvious composition law; and the functor $Ab\to C$ inverts $S$ (because in $\hom(\mathbb Z/n,\mathbb Z/n)$, $id - 0 \in$ the subgroup). So it factors as $Ab[S^{-1}]\to C$. But now $\mathbb Z/p^\infty$ is not isomorphic to $0$ in this new category, so it cannot be isomorphic to $0$ in $Ab[S^{-1}]$. -So what is happening here is exactly that we have this colimit of maps in $S$ ($\mathbb Z/p^n\to 0$) which is not in $S$ anymore. It must therefore be inverted by $C\to Loc(S)$, but it will not (in general) be inverted by $C\to C[S^{-1}]$, especially in cases where that colimit is "formal" (which it often is in presentable situations, almost by definition)<|endoftext|> -TITLE: Compact-open topology and Delta-generated spaces -QUESTION [7 upvotes]: Consider the set of continuous maps $C^0([0,1],[0,1])$ equipped with the compact-open topology. It is metrisable, and therefore sequential. It is also a k-space: see http://neil-strickland.staff.shef.ac.uk/courses/homotopy/cgwh.pdf Proposition 1.6. The proof relies on the facts that every k-closed subset is in particular $\overline{\mathbb{N}}$-closed where $\overline{\mathbb{N}}$ is the one-point compactification of $\mathbb{N}$, that every $\overline{\mathbb{N}}$-closed subset is sequentially closed, and therefore the kelleyfication functor adds no open subsets in the topology. Since $\overline{\mathbb{N}}$ is not $\Delta$-generated (its $\Delta$-kelleyfication is the discrete space $\overline{\mathbb{N}}^\delta$), the preceding proof does not work for $\Delta$-generated spaces. - -I am (almost) sure that $C^0([0,1],[0,1])$ is not $\Delta$-generated -and I would appreciate to see a proof. - -Motivation: This question is important for me because I am trying to understand specific things about the topology of the space of execution paths of a cellular multipointed $d$-space having a finite number of cells (in the sense of https://arxiv.org/abs/1904.04159). And the space above appears everywhere. - -REPLY [11 votes]: The mapping space $C([0,1],[0,1])$ in the compact-open topology is in fact $\Delta$-generated. -The reason for this is that every locally path-connected first-countable space is $\Delta$-generated. This was proved by Christensen, Sinnamon, and Wu in Proposition 3.11 of their paper The D-Topology for Diffeological Spaces, Pacific J. Math., 272, (2014). As has already been noted, $C([0,1],[0,1])$ is metrisable, and hence first-countable. In addition it's not difficult to see that it is also locally path-connected (in fact it is locally contractible in a strong sense).<|endoftext|> -TITLE: Hahn’s theorem on ordered fields -QUESTION [7 upvotes]: There is a theorem attributed to Hahn that every ordered field $F$ containing $\mathbb R$ is a subfield of a formal power series field $\mathbb R[[X^\Gamma]]$, where $\Gamma$ is an ordered abelian group. Can you give a nice reference in English for a proof of this theorem? Or if it is not too hard, please sketch a proof below. - -REPLY [11 votes]: This theorem, which extends Hahn's embedding theorem for ordered abelian groups to ordered fields, has a complicated history that makes it difficult to attribute it to any single author. However, by the early 1950s, as a result of the work of Kaplansky (1942), it appears to have assumed the status of a ``folk theorem'' among knowledgeable field theorists, with numerous proofs published thereafter. For an in-depth history of the embedding theorem along with numerous references to proofs, see my paper: -Hahn’s Über die Nichtarchimedischen Grössensysteme and the Development of the Modern Theory of Magnitudes and Numbers to Measure Them. In:From Dedekind to Gödel: Essays on the Development of the Foundations of Mathematics, edited by J. Hintikka, Kluwer Academic Publishers, Dordrecht, 1995, pp. 165-213. -https://www.researchgate.net/publication/325019391_Hahn's_Uber -Edit: -Strictly speaking, the theorem I am referring to is much deeper than the embedding theorem stated in the question, though I suspect it is the one that is intended. A weak formulation of the ordered-field-theoretic generalization of Hahn's Embedding Theorem asserts: If $K$ is an ordered field and $\Gamma$ is its ordered Abelian group of Archimedean classes, then there is an embedding of $K$ into the Hahn field $\mathbb{R}[[X^\Gamma ]]$. For increasingly stronger versions of the theorem, see the paper cited above.<|endoftext|> -TITLE: Counting monomials in product polynomials: Part I -QUESTION [9 upvotes]: This question is motivated by recent work of R P Stanley, Theorems and conjectures on some rational generating functions. Consider the polynomials -$$P_n(x)=\prod_{i=1}^{n-1}(1+x^{3^{i-1}}+x^{3^i}).$$ -Define the sequence $a_n$ to count the number of monomials of $P_n(x)$. For example, -\begin{align*} P_2(x)&=x^3 + x + 1 \qquad \qquad \qquad \qquad \qquad \qquad\qquad \,\,\implies \qquad a_2=3, \\ -P_3(x)&=x^{12} + x^{10} + x^9 + x^6 + x^4 + 2x^3 + x + 1 \qquad \implies \qquad a_3=8. \end{align*} -Recall the Fibonacci numbers $F_1=F_2=1$ and $F_{n+2}=F_{n+1}+F_n$. - -QUESTION. Is it true that $a_n=F_{2n}$? How does "ternary expansion" relate to Fibonacci? - -REPLY [13 votes]: Here is another argument. We have -$$ P_{n+1}(x)=(1+x+x^3)P_n(x^3). $$ -Now $P_n(x^3), xP_n(x^3)$, and $x^3P_n(x^3)$ all have $a_n$ monomials. If -a monomial $x^i$ appears in more than one of them, then it must appear -in $P_n(x^3)$ and $x^3P_n(x^3)$, but not $xP_n(x^3)$ (by considering -exponents mod 3). Thus we need to subtract off the number of -monomials $x^i$ that appear in $P_n(x)$ such that $x^{i+1}$ also -appears. By the uniqueness of the ternary expansion, the monomials with -such $x^i$ or $x^{i+1}$ are those appearing in -$(1+x)P_{n-1}(x^3)$. There are $2a_{n-1}$ such monomials, occurring -in pairs $x^i$ and $x^{i+1}$. Hence $a_{n+1}=3a_n-a_{n-1}$, the -recurrence satisfied by $F_{2n}$ (and with the correct initial -conditions).<|endoftext|> -TITLE: Generalizing the Pfaffian: families of matrices whose determinants are perfect powers of polynomials in the entries -QUESTION [14 upvotes]: Let $n$ be a positive integer, and let $M = (m_{ij})$ be a skew $2n \times 2n$ matrix. That is, we have $m_{ij} = -m_{ji}$ for $1 \leq i, j \leq 2n$. Then it is well-known that -$$\det M = p(M)^2,$$ -where $p$ is a polynomial in the entries $m_{ij}$. The polynomial $p(M)$ is called the Pfaffian of $M$. -Is there a generalization of this? That is, is there a natural family of $kn \times kn$ matrices whose determinants are perfect $k$-th powers of polynomials in the entries? - -REPLY [13 votes]: A good class of examples of this is given by Clifford algebras: If $V$ is a real vector space with endowed with a quadratic form $q:V\to\mathbb{R}$, the algebra $Cl(q)$ is the algebra generated by the elements of $V$ subject to the multiplication rule $x^2 = -q(x)$. If $M$ is a $Cl(q)$-module, say $M\simeq\mathbb{R}^m$, then we have an inclusion $V\hookrightarrow\mathrm{End}(M)$ and the characteristic polynomial of $x\in V\subseteq\mathrm{End}(M)$ is easily seen to be $(t^2+q(x))^{m/2}$, so we have -$$ -\det(x) = q(x)^{m/2} -$$ -for all $x\in V$. -For example, if $V$ is $\mathbb{R}^8$ with its standard Euclidean quadratic form $q$, then $Cl(q)$ is isomorphic to $\mathrm{End}_{\mathbb{R}}(\mathbb{R}^{16})$, so we can take $M=\mathbb{R}^{16}$ (and every $Cl(q)$-module is $\mathbb{R}^{16k}$ for some integer $k$). Thus, in this case, we have $\det(x) = p(x)^8$ where $p(x) = |x|^2$ for all $x\in V$. -In general, when $V\simeq\mathbb{R}^n$ and $q_n:V\to\mathbb{R}$ is nondegenerate, the dimension of a minimal nontrivial $Cl(q_n)$-module grows (roughly) exponentially with $n$, so the minimal $m$ grows exponentially with $n$. This shows that there are nontrivial 'irreducible' examples with $\det(x) = p(x)^k$ for $k$ arbitrarily large and that there is no bound on the possible dimension $n$ of the subspace $V\subset\mathrm{End}(M)$. -Remark: Given a linear subspace $V\subset\mathrm{End}(\mathbb{R}^{m})$ such that there exists a polynomial $p:V\to\mathbb{R}$ and an integer $k = m/\deg(p)>1$ such that $\det(x) = p(x)^k$, we say that the pair $(V,\mathbb{R}^m)$ is irreducible if there is no nontrivial subspace $M\subset\mathbb{R}^m$ such that $x(M)\subset M$ for all $x\in V$ and $\det(x_{|M}) = p(x)^j$ for all $x\in V$, where, necessarily, $j = (\dim M)/\deg(p)$. -The interesting problem for linear subspaces $V\subset\mathrm{End}(\mathbb{R}^m)$ on which the $\det$-function is a higher power of a polynomial on $V$ is to classify the irreducible ones of maximal dimension for a given $m$.<|endoftext|> -TITLE: Automorphisms over finite field that do not lift to an automorphism in characteristic zero -QUESTION [8 upvotes]: My main question is the following: is there an automorphism of the affine space $\mathbb{A}^n$ (automorphism of an algebraic variety) defined over a finite field which does not lift to an automorphism defined over a field in characteristic zero? -Let us first consider finite fields of the form $\mathbb{F}_p$ for some prime $p$. If you take an automorphism $f$ of the affine space that is defined over $\mathbb{Q}$ and such that $p$ does not divide any of the denominators of $f$ and of $f^{-1}$, you may consider the restriction of $f$ and $f^{-1}$ to automorphisms defined over $\mathbb{F}_p$, by considering the coefficients modulo $p$, and get automorphisms defined over $\mathbb{F}_p$. Is every automorphism of $\mathbb{A}^n$ over $\mathbb{F}_p$ obtained by this way? (one can do similar constructions for other finite fields and other fields of caracteristic zero). -For each prime $p$, every linear automorphism of $\mathbb{A}^n$, given by $(x_1,\ldots,x_n)\mapsto (a_{11}x_1+\cdots +a_{1n}x_n,\ldots,a_{n1}x_1+\cdots +a_{nn}x_n)$ for some matrix $(a_{ij})\in \mathrm{GL}_n(\mathbb{F}_p)$ comes from an element of $\mathrm{GL}_n(\mathbb{Q})$ whose entries are integers and whose determinant is not divisible by $p$. Its inverse in $\mathrm{GL}_n(\mathbb{Q})$ has thus all denominators that are not multiple of $p$. Hence, every linear automorphism of $\mathbb{A}^n$ over $\mathbb{F}_p$ comes from a linear automorphism of $\mathbb{A}^n$. -Similarly, if you take an elementary automorphism of $\mathbb{A}^n$ defined over $\mathbb{F}_p$, i.e. given by -$$(x_1,\dots,x_n)\mapsto (x_1+a(x_2,x_3,\ldots,x_n),x_2,\ldots,x_n)$$ -for some polynomial $a\in \mathbb{F}_p[x_2,\ldots,x_n]$, it comes from an automorphism of $\mathbb{A}^n$ defined over $\mathbb{Q}$ (here even over $\mathbb{Z}$). In particular, every "tame" automorphism (generated by linear and elementary automorphisms) of $\mathbb{A}^n$ over $\mathbb{F}_p$ comes from a tame automorphism of $\mathbb{A}^n$ over $\mathbb{Q}$. In dimension $n=1$ and $n=2$, every automorphism is tame, so every automorphism of $\mathbb{A}^n$ over $\mathbb{F}_p$ comes from an automorphism defined over $\mathbb{Q}$. The question is then more interesting for $n\ge 3$. It would give examples of non-tame automorphisms, a fact not know until now in positive characteristic (see the famous article Shestakov and Umirbaev - The tame and the wild automorphisms of polynomial rings in three variables for examples in characteristic zero). -This seems hard to answer in general, but can we then replace the variety $\mathbb{A}^n$ by an affine (or projective) algebraic variety, defined over $\mathbb{F}_p$ and find some automorphisms over $\mathbb{F}_p$ that do not come from automorphisms in characteristic zero? -EDIT: Will Sawin gave a nice answer for elliptic curves, where some given automorphism do not lift to $\mathbb{Q}$. Are there examples where some given automorphisms do not lift to any field of caracteristic zero? - -REPLY [8 votes]: Some simple examples are provided by affine elliptic curves. -Recall that a smooth genus zero curves $E$ over a field $k$ with a rational point $P$ has a natural group structure, where $P$ is the identity. Thus automorphisms of $E$ fixing $P$ are automorphisms of the group. -Automorphisms of the affine curve $E -\{P\}$ extend to automorphisms of $E$ fixing $P$, since affine curves have a canonical projective closure. -A lot is known about automorphisms of elliptic curves (as groups), and we can use this to construct examples, depending on how the problem is formulated. -The main subtlety in the formulation is that an elliptic curve over $\mathbb F_p$ can have many lifts to $\mathbb Q$. Thus whether an automorphism lifts might depend on the choice of a lift of the curve. -However, it is possible to construct automorphisms that do not lift regardless of the lift of a curve. This is because, over $\mathbb Q$, every elliptic curve has automorphism group of order $2$ – the automorphisms have finite order because they fix a point on a genus $1$ curve, and they act faithfully on the tangent space at the identity, which is $\mathbb Q$, but the only finite order elements of $\mathbb Q^\times$ are $\pm 1$. -However, the curve defined by $y^2=x^3-x$ has an automorphism of order $4$ over $\mathbb F_p$ for any $p \equiv 1 \mod 4$ and the curve defined by $y^2 = x^3-1$ has an automorphism of order $6$ over $\mathbb F_p$ for any $p \equiv 1 \mod 6$. -You could instead ask for lifts to any field of characteristic $0$, rather than just $\mathbb Q$. In this case, we would still have examples as long as you let me choose the lifts: choosing a lift of $y^2=x^3-1$ with nonzero $j$-invariant would guarantee the order $4$ automorphism doesn't lift, and similarly for $y^2=x^3-x$ and $j\neq 1728$. -If you merely want there to exist a lift of a curve to characteristic $0$ where all the automorphisms lift, there are still counterexamples. Supersingular curves in characteristic $2$ and $3$ have $24$ and $12$ automorphisms (over an algebraically closed field), respectively, which is more than elliptic curves can have in characteristic zero over an algebraically closed field. -However, if you wanted to find examples of a variety with some fixed automorphism that don't lift to any lift of that variety over any characteristic zero field, I believe you would have to look beyond elliptic curves.<|endoftext|> -TITLE: Action of the symmetric group $S_3$ on an elliptic curve $E$ defined over $\mathbb{Z}$ -QUESTION [5 upvotes]: I came up with the following question on a facebook group: find the positive integer solutions of the equation $$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=4$$ -Now clearly this is very difficult, indeed it is equivalent to find the integer solutions of an elliptic curve $E$ defined over $\mathbb{Z}$, in particular $E$ is the smooth plane cubic curve in $\mathbb{P}^2$ given by the equation $$E=(x^3 - 3 x^2 y - 3 x^2 z - 3 x y^2 - 5 x y z - 3 x z^2 + y^3 - 3 y^2 z - 3 y z^2 + z^3=0)$$ -Anyway I started playing with this: since $E$ is symmetric with respect to the permutations of the variables $x,y,z$ we have an action $$S_3 \times E \rightarrow E \\ (\sigma,[x,y,z]) \rightarrow [\sigma(x),\sigma(y),\sigma(z)]$$ where I've identified the set $\{x,y,z\}$ with the set $\{1,2,3\}$. -We can form the quotient of $E$ with respect to this action yielding a map $$f: E \rightarrow \mathbb{P}^1$$ of degree $6$. By Hurwitz's formula we have that the degree of the ramification divisor $R$ is $12$ and since the ramification points consists in those points with $2$ coordinates equal we have $3$ ramification points of order $4$. -My questions now are the following: - -From this description are we able to say that $E$ has infinite integer solutions, maybe noting that this type of elliptic curve is particular? -It seems to me that $S_3 \subset Aut(E)$ but searching on the internet I found that the automorphism group of elliptic curve can only be of type $\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/4\mathbb{Z}$ and $\mathbb{Z}/6\mathbb{Z}$. What I'm missing? - -I'm sorry if I said something wrong, I'm not an expert on this field. Thanks in advance for the help. - -REPLY [12 votes]: Technically speaking, an elliptic curve is a genus 1 curve with a choice of rational point. The automorphism group of an elliptic curve is the subgroup of automorphisms of the genus 1 curve that fix that rational point. So there is no contradiction with the facts you looked up, it just means that no point is fixed by all of $S_3$, and indeed (1:1:1) is not on the curve. -You can check in sage or Magma that this genus 1 curve does have infinitely many integral points (which are the same as the rational points, since the curve is projective). I chose $P = (1:-1:0)$ to be the identity. Then the elliptic curve has rank 1 and the group of rational points is generated by $Q = (4:-1:11)$ and the 6 "obvious" points with one coordinate equal to 0, which are the six torsion points. -This answers your two questions, but doesn't answer the original question from the facebook group because $Q$ doesn't have positive coordinates. I found that the point $9Q$, or explicitly, -$$ (154476802108746166441951315019919837485664325669565431700026634898253202035277999: - 36875131794129999827197811565225474825492979968971970996283137471637224634055579: - 4373612677928697257861252602371390152816537558161613618621437993378423467772036) $$ -does have positive coordinates. There are several other such points, enough to make me think that there are infinitely many, but I don't know how to prove it. -To do these computations yourself in sage, use the EllipticCurve_from_cubic function.<|endoftext|> -TITLE: Counting monomials in product polynomials: Part II -QUESTION [11 upvotes]: Encouraged by the responses to my earlier MO question, here is a follow up and upgraded quest. -Let $e\geq2$ be an integer. Define the polynomials -$$P_{n,e}(x)=\prod_{i=1}^{n-1}\left(1+x^{e^{i-1}}+x^{e^i}+\cdots+x^{e^{i+e-3}}\right).$$ -Denote the number of monomials in $P_{n,e}(x)$ by $a_{n,e}$. - -QUESTION 1. Is this true? -$$\sum_{n\geq1}a_{n,e}\,y^n=\frac{y}{\sum_{k=0}^{\lfloor\frac{e+1}2\rfloor}(-1)^k\binom{e+1-k}k\,y^k}.$$ -QUESTION 2. Emeric Deutsch interprets (for instance) the sequence $a_{n,3}$ as "number of walks of length $2n+1$ in the path graph $P_4$ from one end to the other one" (see on OEIS). Does this approach apply for all other $a_{n,e}$? - -REMARK. The case $e=3$ recovers my earlier request that $a_{n,3}=F_{2n}$ since -$$\frac{y}{1-3y+y^2}=\sum_{n\geq1}F_{2n}\,y^n.$$ -The case $e=2$ is trivially $a_{n,2}=2^{n-1}$ or $\sum_{n\geq1}a_{n,2}\,y^n=\frac{y}{1-2y}$. - -REPLY [5 votes]: The answer to Question 1 is positive. In Question 2 it is true that -Claim 1. $a_{n,e}$ equals to the number of walks of length $e+2(n-1)$ in the path graph $P_{e+1}$ from one end to the other one. -I start with general reformulations, then prove Claim 1, then deduce the generating function for $a_{n,e}$ (Question 1). - -Denote $V_i:=\{0,i,i+1,\ldots,i+e-2\}$; $f(0)=0$ and $f(i)=e^{i-1}$ for $i>0$. Any monomial in $P_{n,e}(x)$ have the form $x^N$ for $N=\sum f(c_i)$ for a certain choice $c_i\in V_i$ for all $i=1,2,\ldots,n-1$. The sum $\sum f(c_i)$ is the linear combination of powers of $e$ with non-negative integer coefficients not exceeding $e-1$. Thus such sums are in 1-to-1 correspondence with the multisets $\{c_1,\ldots,c_{n-1}\}$. - -Any fixed multiset $C=\{c_1,\ldots,c_{n-1}\}$ has the unique normal form: a sequence $(b_1,\ldots,b_{n-1})\in V_1\times V_2\times\ldots \times V_{n-1}$ such that - - -(i) $\{b_1,\ldots,b_{n-1}\}=C$; -(ii) if $b_j=0$ and $b_{j+1}>0$, then $b_{j+1}=j+e-1=\max(V_{j+1})$ (any 0 is followed by 0 or the maximum); -(iii) if $b_i>0$, $b_j>0$ and $i -TITLE: Restricted notions of set-theoretic geology -QUESTION [6 upvotes]: We say that $W$ is a ground of the universe $V$ if $W$ is a model of ZFC and there is a poset $P\in W$ such that $W[G]=V$ for some $G$ which is $P$-generic over $W$. The Ground Axiom ($\text{GA}$) asserts that $V$ has no nontrivial grounds while $\text{DDG}$ is the statement: for all grounds $W_1$, $W_2$ of $V$ there is a ground $U$ (of $V$) contained in $W_1\cap W_2$. -It has already been proven that $\text{GA}$ can be forced (by means of a class forcing notion) and that $\text{DDG}$ is a theorem of ZFC. I'm interested in restricting these notions. For example, $\text{GA}_{\text{$\sigma$-closed}}$ is the assertion that the universe is not a set-forcing extension of an inner model $\sigma$-closed forcing notion. -Similarly $\text{DDG}_{\text{$\sigma$-closed}}$ is the statement: for all $\sigma$-closed grounds $W_1$, $W_2$ of $V$ there is a $\sigma$-closed ground $U$ (of $V$) contained in $W_1\cap W_2$. Of course, $\text{GA}\rightarrow \text{GA}_{\text{$\sigma$-closed}}$. How about the converse? That is, are there models of ZFC satisfying $\neg\text{GA}+\text{GA}_{\text{$\sigma$-closed}}$ or $\neg\text{GA}+\text{GA}_{\text{ccc}}$? This question arises from Reitz - The ground axiom and when it was published the question was still open. -Moreover, I'm not able to exhibit models in which - -$\text{DDG}_{\text{$\sigma$-closed}}$ fails, -$\text{DDG}_{\text{ccc}}$ fails. - -The last two questions are probably easier. - -REPLY [8 votes]: If you add a Cohen real to $L$ (or any model of the Ground Axiom), you get a model of the countably closed Ground Axiom, since all intermediate extensions are grounds for Cohen forcing. On the other hand, if you add a Sacks real to $L$, you obtain a model with no ccc grounds since Sacks forcing is not ccc and there are no intermediate extensions. So the Sacks extension is a model of the ccc Ground Axiom but not the Ground Axiom. -A counterexample to the countably closed DDG can be obtained by adding mutual generics $G$ and $H$ for $\text{Col}(\omega_1,\mathbb R)$ to $L(\mathbb R)$ when $L(\mathbb R)$ is a model of DC but not AC. Then $L(\mathbb R)[G\times H]$ has $L(\mathbb R)[G]$ and $L(\mathbb R)[H]$ as countably closed grounds, but their intersection is $L(\mathbb R)$, and moreover any countably closed ground of the two models would contain all the reals, and hence be $L(\mathbb R)$. Since AC fails in $L(\mathbb R)$, it doesn't count as a ground. (Edit: Actually I am remembering now that this example was pointed out by Gunter Fuchs at the inner model theory meeting in Girona in 2018 after I had gone overboard using $\mathbb P_\text{max}$.) -To make Asaf happy, if possible, I also point out that assuming ZFC, the countably distributive DDG is true for ZF grounds, though the example above shows it fails for ZFC grounds. (Edit: I mean the intersection of any two ZF grounds closed under countable sequences contains a ZF ground closed under countable sequences. But this need not be a ground of the ZF grounds we started with.) If $M_0$ and $M_1$ are ZF grounds, there is a ZFC ground $N$ contained in both of them by a result of Usuba. Thus $N(\text{Ord}^\omega)$ is a common inner model of $M_0$ and $M_1$ that is closed under $\omega$-sequences. Now to show that $N(\text{Ord}^\omega)$ is a ground, it suffices by a theorem of Grigorieff to show $N(\text{Ord}^\omega) = N(X)$ for some set $X$. Let $\delta$ be large enough that $N$ has the $\delta$-cover property, and we claim $N(\text{Ord}^\omega) \subseteq N(\delta^\omega)$. To see this, suppose $\sigma$ is a countable set of ordinals. Then there is a set of ordinals $\tau\in N$ covering $\sigma$ of cardinality less than $\delta$. Let $f : \gamma\to \tau$ be the increasing enumeration of $\tau$. Then $\gamma < \delta$, so $\bar \sigma = f^{-1}[\sigma]\in N(\delta^\omega)$. Therefore $\sigma = f[\bar \sigma]\in N(\delta^\omega)$. -I couldn't figure out a counterexample to ccc DDG, though, but I don't doubt that there is one.<|endoftext|> -TITLE: Prove or disprove that $\sup_{n\in\mathbb{N}}\left|\sum_{\substack{d|n \\d -TITLE: Are such functions differentiable? -QUESTION [23 upvotes]: In my recent researches, I encountered functions $f$ satisfying the following functional inequality: -$$ -(*)\; f(x)\geq f(y)(1+x-y) \; ; \; x,y\in \mathbb{R}. -$$ -Since $f$ is convex (because $\displaystyle f(x)=\sup_y [f(y)+f(y)(x-y)]$), it is left and right differentiable. Also, it is obvious that all functions of the form $f(t)=ce^t$ with $c\geq 0$ satisfy -$(*)$. Now, my questions: -(1) Is $f$ everywhere differentiable? -(2) Are there any other solutions for $(*)$? -(3) Is this functional inequality well-known (any references -(paper, book, website, etc.) for such functional inequalities)? -Thanks in advance - -REPLY [42 votes]: Replace $x$ with $x+y$ to get -$f(x+y)\ge f(y)(1+x)$ or $f(x+y)-f(y)\ge xf(y)$. -Replace $y$ with $x+y$ and then interchange $x$ and $y$ to get $f(x+y)-f(y)\le xf(x+y)$. -Together, -$$ -xf(y)\le f(x+y)-f(y)\le xf(x+y). -$$ -Dividing by $x$ and taking the limit as $x\to0$ implies that $f$ is differentiable with $f'=f$. - -REPLY [27 votes]: For any $x$ and for sufficiently large $n$ such that $1+x/n>0$, it holds that -\begin{align} -f(x) &\ge f\left (\frac{(n-1)x}n \right) (1+x/n)\\ -&\ge f\left (\frac{(n-2)x}n \right)(1+x/n)^2 \\ -&\ge \cdots \ge f(0) \left(1+ \frac x n\right)^n. -\end{align} -by substituting $(x,y)=(x,(n-1)x/n), ((n-1)x/n, (n-2)x/n),...$ in the given equation. -In other words, -$$ -f(x) \ge \lim_{n\rightarrow \infty} f(0) \left(1+ \frac x n\right)^n = f(0)\cdot e^x. -$$ -On the other hand, for any $y$ and for sufficiently large $n$ such that $1-y/n>0$, we can similarly get the following inequality. -\begin{align} -f(y) &\le f\left( \frac{(n-1)y} n\right) / (1-y/n)\\ -&\le \cdots \le f(0)/(1-y/n)^n. -\end{align} -It implies $f(y)\le f(0) \cdot e^y$. Combining these inequalities, we get that $f(x)=f(0) \cdot e^x$ is the only solution as you wanted.<|endoftext|> -TITLE: Homological stability and Waldhausen A-theory -QUESTION [12 upvotes]: $\DeclareMathOperator{\Diff}{Diff}$ -From the work of Galatius - Randall-Williams and Berglund - Madsen we have homological stability (with respect to g) of $B\Diff_\partial (W_{g,1})$ and rational homological stability of $B\widetilde\Diff_\partial(W_{g,1})$, where $W_{g,1}$ is $(\#_g S^d \times S^d ) \setminus D^{2d}$ and $\widetilde\Diff$ denotes block diffeomorphisms (essentially an analogue of diffeomorphisms where path components are naturally pseudoisotopy classes). -By analyzing the Serre spectral sequence for the fibration $\widetilde\Diff_\partial(W_{g,1}) / \Diff_\partial (W_{g,1}) \rightarrow B\Diff_\partial (W_{g,1}) \rightarrow B\widetilde\Diff_\partial(W_{g,1}) $ we deduce from the aforementioned homological stability results, that $H_*(\widetilde\Diff_\partial(W_{g,1}) / \Diff_\partial (W_{g,1});\mathbb Q)_{\pi_0(\widetilde\Diff_\partial(W_{g,1}))} $ has stability with respect to $g$. By pseudo-isotopy implies isotopy and some of Wall's work on highly connected manifolds, one could replace $\pi_0(\widetilde\Diff_\partial(W_{g,1}))$ by a certain arithmetic group $\Gamma$ if she wished. -It is known that $\widetilde\Diff_\partial(W_{g,1}) / \Diff_\partial (W_{g,1})$ is related to Waldhausen A-theory, see Weiss' and Williams' Automorphisms of manifolds and algebraic K-theory: I. Is there a known proof using A-theory that the coinvariants of these homology groups have stability? - -REPLY [9 votes]: I don't think that you can deduce homological stability of the coinvariants from the Serre spectral sequence as you suggest. But this precise situation was studied in my paper "An upper bound for the pseudoisotopy stable range", which may be useful. -To answer your last question, A-theory only describe these groups in a range of degrees depending on the dimension of the manifold, so could not be used to prove homological stability of these coinvariants in a range of degrees tending to infinity. In the range of degrees in which this approximation applies the coinvariants are in any case zero, so do indeed stabilise.<|endoftext|> -TITLE: What is the meaning of this coboundary homomorphism for group hypercohomology? -QUESTION [5 upvotes]: $\require{AMScd}$ -Let $\Gamma=\{1,\gamma\}$ be a group of order 2. -In my problem from Galois cohomology of real reductive groups I came to a commutative diagram of $\Gamma$-modules -(abelian groups with $\Gamma$-action) -\begin{equation*}%\label{e:cd} -\begin{CD} -1 @>>>Q_1 @>>>Q_2 @>>> Q_3 @>>> 1 \\ -@. @VV{\rho_1}V @VV{\rho_2}V @VV{\rho_3}V \\ -1 @>>> X_1 @>>> X_2 @>>> X_3 @>>> 1 \\ -@. @VV{\alpha_1}V @VV{\alpha_2}V@VV{\alpha_3}V \\ -1 @>>> P_1 @>>> P_2 @>>> P_3 @>>> 1 \\ -\end{CD} -\end{equation*} -in which the rows are exact, but not the columns -(and $\alpha_k\circ\rho_k\neq 0$). -The top and bottom rows of the diagram split canonically: -$$Q_2=Q_1\oplus Q_3\quad\text{ and }\quad P_2=P_1\oplus P_3,$$ -and these splittings are compatible: -$$ -\alpha_2(\rho_2(0,q_3))= -\big(\,0,\,\alpha_3(\rho_3(q_3))\,\big)\tag{$*$} -$$ -for $q_3\in Q_3$. -I consider the Tate hypercohomology groups -$${\Bbb H}^0(\Gamma, Q_3\overset{\rho_3}\longrightarrow X _3)\quad\text{ and } - \quad{\Bbb H}^0(\Gamma,X _1\overset{\alpha_1}\longrightarrow P_1),$$ -where both short complexes are in degrees $(-1,0)$. -Below I construct "by hand" a canonical coboundary homomorphism -$$\delta\colon\, {\Bbb H}^0(\Gamma, Q_3\to X _3)\,\longrightarrow\, -{\Bbb H}^0(\Gamma,X _1\to P_1),$$ - -Question. How can I get this coboundary homomorphism from a kind of general theory? - -Remark. For a group $\Gamma$ of order 2 (and also for any cyclic group $\Gamma$) the Tate cohomology and hypercohomology are periodic with period 2. -Therefore, our $\delta$ is a map -$${\Bbb H}^1(\Gamma,\, Q_3\to X_3\to 0)\, \longrightarrow \, -{\Bbb H}^2(\Gamma,\, 0\to X_1\to P_1),$$ -where both complexes are in degrees $(-2,-1,0)$. -Construction. -We start with $[ q_3, x_3]\in {\Bbb H}^0(\Gamma, Q_3\overset{\rho_3}\longrightarrow X _3)$. -Here $( q_3, x_3)\in Z^0(\Gamma,Q_3\to X _3)$, that is, -\begin{equation} -q_3\in Q_3,\quad x_3\in X_3,\quad -\,^{\gamma\kern -0.8pt} q_3+ q_3=0,\qquad \,^{\gamma\kern -0.8pt} x_3- x_3=\rho_3( q_3).\tag{$**$} -\end{equation} -We lift canonically $ q_3$ to -$$ q_2=(0, q_3)\in Q_1\oplus Q_3= Q_2,$$ -and we lift $ x_3$ to some $ x_2\in X _2$. -We write -$$\alpha_2( x_2)=( p_1, p_3)\in P_1\oplus P_3=P_2,$$ -where $ p_3=\alpha_3( x_3)\in P_3$ and $ p_1\in P_1$. -We set -$$ x_1=\,^{\gamma\kern -0.8pt} x_2- x_2-\rho_2( q_2).$$ -Since by $(*)$ we have -$$\,^{\gamma\kern -0.8pt} x_3- x_3=\rho_3( q_3),$$ -we see that $ x_1\in X _1$. -We compute: -$$\,^{\gamma\kern -0.8pt} x_1+ x_1=\,^{\gamma\kern -0.8pt}(\,^{\gamma\kern -0.8pt} x_2- x_2)-{}^{\gamma\kern -0.8pt}\rho_2(0, q_3)+ - (\,^{\gamma\kern -0.8pt} x_2- x_2)-\rho_2(0, q_2)=-\rho_2(0,\,^{\gamma\kern -0.8pt} q_3+ q_3)=0$$ -by $(**)$. -Furthermore, -\begin{align*} -\alpha_1( x_1)&=\,^{\gamma\kern -0.8pt}\alpha_2(x_2)-\alpha_2(x_2)-\alpha_2(\rho_2(q_2))\\ -&=\,^{\gamma\kern -0.8pt}( p_1, p_3)-( p_1, p_3)-( 0,\alpha_3(\rho_3( q_3)))\\ -&=\big(\,^{\gamma\kern -0.8pt}p_1-p_1,\,^{\gamma\kern -0.8pt}p_3-p_3-\alpha_3(\rho_3(q_3))\big)\\ -&=\big(\,^{\gamma\kern -0.8pt}p_1-p_1,\,\alpha_3(\,^{\gamma\kern -0.8pt}x_3-x_3-\rho_3(q_3))\big)\\ -&=(\,^{\gamma\kern -0.8pt} p_1- p_1,0) -\end{align*} -by $(*)$ and $(**)$. -Thus -$$\alpha_1(x_1)=\,^{\gamma\kern -0.8pt} p_1-p_1.$$ -We see that $(x_1, p_1)\in Z^0(\Gamma, X _1\overset{\alpha_1}\longrightarrow P_1)$. -We set -$$\delta[ q_3, x_3]=[ x_1, p_1]\in {\Bbb H}^0(\Gamma,X _1\to P_1).$$ -A straightforward check shows that the map $\delta$ is a well-defined homomorphism. - -REPLY [3 votes]: I believe easiest way to handle this is in the formalism of triangulated categories. You can do it in various ways: either work with the unbounded derived category or (probably easier) replace each module $M$ with $\operatorname{Hom}_\Gamma(\mathcal R,M)$ where $\mathcal R$ is the complete resolution for $\Gamma$, i. e. the standard unbounded 2-periodic complex $$\cdots\xrightarrow{1-\gamma}\mathbb Z[\Gamma]\xrightarrow{1+\gamma}\mathbb Z[\Gamma]\xrightarrow{1-\gamma}\mathbb Z[\Gamma]\xrightarrow{1+\gamma}\cdots$$of $\Gamma$-modules. -Let then $X_1\to X_2\to X_3\to\Sigma X_1$ be an exact triangle in arbitrary triangulated category, and let $Q_3\to X_2\to P_1$ be arbitrary morphisms with zero composite. Let $P$ be the fiber of $X_1\to P_1$ and let $Q$ be the cofiber of $Q_3\to X_3$. Our aim is to construct from all that a canonical map $Q\to\Sigma P$. It turns out that there is such a map which is moreover an isomorphism if and only if $Q_3\to X_2\to P_1$ is exact. -Since the composite $Q_3\to X_2\to P_1$ is zero, the map $X_2\to P_1$ factors through cofiber of $Q_3\to X_2$, $X_2\to Q_0$, and the map $Q_3\to X_2$ factors through the fiber $P_0\to X_2$ of $X_2\to P_1$. Thus all in all $X_1\to P_1$ factors into the composite $X_1\to X_2\to Q_0\to P_1$, while $Q_3\to X_3$ factors into the composite $Q_3\to P_0\to X_2\to X_3$. -First note that in these circumstances the cofiber of $Q_3\to P_0$ is isomorphic to the fiber of $Q_0\to P_1$; denoting it by $H$, the composite $P_0\to H\to Q_0$ is the composite $P_0\to X_2\to Q_0$. -We get eight instances of the octahedron axiom, telling us that for various composites $f\circ g$ there are exact triangles $\operatorname{fibre}(f)\to\operatorname{cofibre}(g)\to\operatorname{cofibre}(f\circ g)\to\operatorname{cofibre}(f)=\Sigma\operatorname{fibre}(f)$ and $\operatorname{fibre}(g)\to\operatorname{fibre}(f\circ g)\to\operatorname{fibre}(f)\to\operatorname{cofibre}(g)=\Sigma\operatorname{fibre}(g)$. Strictly speaking, not all of them are needed, but for completeness let me list them all. - - - - -The composable pair -gives the exact triangle - - - - -$Q_3\to P_0\to X_2$ -$H\to Q_0\to P_1\to\Sigma H$ - - -$Q_3\to X_2\to X_3$ -$X_1\to Q_0\to Q\to \Sigma X_1$ - - -$Q_3\to P_0\to X_3$ -$\color{red}{P\to H\to Q\to\Sigma P}$ - - -$P_0\to X_2\to X_3$ -$P\to X_1\to P_1\to\Sigma P$ - - -$X_1\to X_2\to Q_0$ -$Q_3\to X_3\to Q\to\Sigma Q_3$ - - -$X_1\to X_2\to P_1$ -$P\to P_0\to X_3\to\Sigma P$ - - -$X_1\to Q_0\to P_1$ -$\color{red}{P\to H\to Q\to\Sigma P}$ - - -$X_2\to Q_0\to P_1$ -$Q_3\to P_0\to H\to\Sigma Q_3$ - - - - -To put it all in a single diagram - in what follows, lines with three objects on them represent exact triangles; everything commutes.<|endoftext|> -TITLE: Is every sequentially $\sigma(E',E)$-continuous linear functional on a dual Banach space $E'$ necessarily a point evaluation? -QUESTION [7 upvotes]: $\newcommand{\bf}[1]{\mathbb #1}\newcommand{\sc}[1]{\mathscr #1}$ -A duality between two vector spaces $E$ and $F$ over $\bf K$ ($= {\bf R}$ of ${\bf C}$) -is, by definition, a bilinear form -$$ - \langle \cdot, \cdot\rangle :E\times F\to \bf K, - $$ -such that, if $\langle x, y\rangle =0$ for every $x$ in $E$, then $y=0$. And vice-versa. -Given a duality as above, one defines the weak -topology on $F$, usually denoted $\sigma (F,E)$, to be the coarsest topology according to which the linear functionals -$$ - y\in F\mapsto \langle x, y\rangle \in \bf K - $$ -are continuous for every $x$ in $E$. -It is a classical fact that every $\sigma (F,E)$-continuous linear functional $\varphi :F\to \bf K$, may be represented by a vector in -$E$ in the sense that there exists a (necessarily unique) $x$ in -$E$ such that -$$ - \phi(y) = \langle x, y\rangle ,\quad\forall y\in E. - $$ -One could therefore ask: -Question. Does the above still hold if continuity is replaced by sequential continuity. In other words, -must every sequentially $\sigma (F, E)$-continuous linear functional on $F$ be represented by a vector in $E$. -Before the reader jumps to the task of proving or disproving it, let me say that unfortunately the answer is negative, a -counter example being presented below. -So let me specialize this a bit by restricting to the situation in which $E$ is a Banach space and $F$ is its -topological dual, with the canonical duality -$$ - \langle x, \varphi \rangle = \varphi (x), \quad \forall x\in E, \quad \forall \varphi \in E'. - $$ -To be precise: -Question. Let $E$ be a Banach space and let $\varphi $ be a linear functional on $E'$ which is sequentially -$\sigma (E',E)$-continuous. Is $\varphi $ necessarily represented by a vector in $E$? -This is obviously true if $E$ is reflexive and I think I can also prove it for $E=c_0$, as well as for $E=\ell ^1$. - -A COUNTER EXAMPLE -Let -$E=\sc F(H)$ be the set of all finite-rank operators on Hilbert's space, and -$F=\sc B(H)$, with duality defined by means of the trace, namely -$$ - \langle S, T\rangle = \text{tr}(ST), \quad\forall S\in \sc F(H), \quad\forall T\in \sc B(H). - $$ -In this case $\sigma \big (\sc B(H),\sc F(H)\big )$ turns out to be the weak operator topology (WOT), which coincides -with the sigma weak operator topology ($\sigma $-WOT) on bounded subsets of $\sc B(H)$. -Since WOT-convergent sequences are bounded by Banach-Steinhauss, we have that the WOT-convergent sequences are the same -as the $\sigma $-WOT convergent ones. It follows that every $\sigma $-WOT-continuous linear functional on $\sc B(H)$ is also -WOT-continuous. Making a long story short, for every trace class operator $S$ on $H$ of infinite rank, the linear functional -$$ - T\in \sc B(H) \mapsto \text{tr}(ST)\in {\bf C} - $$ -is sequentially WOT-continuous, but it is not represented by an operator in $\sc F(H)$. - -REPLY [7 votes]: Mikael de la Salle points out this is true when $E$ is separable, as shown in Corollary V.12.8 of Conway, A Course in Functional Analysis, 2e. -For a non-separable counterexample, consider the uncountable ordinal space $[0, \omega_1]$, which is compact Hausdorff, and $E = C([0, \omega_1])$. By the Riesz representation theorem, $E'$ is the space of signed Radon measures $\mu$ on $[0, \omega_1]$ with its total variation norm. Let $\varphi(\mu) = \mu(\{\omega_1\})$. This is clearly not represented by any vector in $E$ since the function $1_{\{\omega_1\}}$ is not continuous, but I claim $\varphi$ is sequentially $\sigma(E', E)$ continuous. -Let $\mu_n$ be a sequence converging to 0 in $\sigma(E', E)$ and fix $\epsilon > 0$. Since each $\mu_n$ is Radon, so is its total variation measure $|\mu_n|$, and thus we can approximate $\{\omega_1\}$ in $|\mu_n|$-measure from outside by open sets. So there exists $\alpha_n < \omega_1$ such that $|\mu_n|((\alpha_n, \omega_1)) < \epsilon$. Let $\alpha = \sup_n \alpha_n < \omega_1$; then $|\mu_n((\alpha, \omega_1))| \le |\mu_n|((\alpha, \omega_1)) < \epsilon$ for every $n$. -Define $f : [0, \omega_1] \to \mathbb{R}$ by $$f(x) = \begin{cases} 0, & x \le \alpha \\ 1, & x > \alpha \end{cases}$$ -and note that $f$ is continuous. Now -$$\varphi(\mu_n) = \mu_n(\{\omega_1\}) = \mu_n((\alpha, \omega_1]) - \mu_n((\alpha, \omega_1)) = \int f\,d\mu_n - \mu_n((\alpha, \omega_1)).$$ -But by assumption $\int f\,d\mu_n \to 0$, and $|\mu_n((\alpha, \omega_1))| < \epsilon$, so we conclude $\varphi(\mu_n) \to 0$.<|endoftext|> -TITLE: Idempotent completions in K-theory -QUESTION [6 upvotes]: I have a reference request on following comment I found in -nLab article on Karoubian categories & envelopes. It states: - -The Karoubian envelope is also used in the construction of the -category of pure motives, and in K-theory. - -Almost every introduction to motives containing the basic -constructions (eg Manin's "Correspondences, motifs and monoidal -transformations" as 'standard' reference) includes the -fundamental part where one passes from the category $(\mathsf{Sm}/k)$ of smooth curves over $k$ -to its Karoubian closure. -On the other hand I'm not sure to which construction in -K-theory where the Karoubian envelope is involved, the quoted -sentence refers. -The first naïve observation is that for a ring $R$ the $K_0(R)$ -in algebraic K-theory is obtained as a certain quotient group -("Grothendieck group") of the set of projective $R$-modules. -This set of projective $R$-modules can be reinterpreted as the -Karoubian envelope of the set of free $R$-modules. -Question: Is this the only explicit usage of Karoubian envelope -in constructions in K-theory or are there more general cases? -I'm quite not sure if the remark above only refers to this 'baby' case -with $K_0(R)$, or there are more general constructions in K-theory where -Karoubian envelopes are involved. -For example another nLab article on the more general development -of K-theory nowhere explains where Karoubian envelope is explicitly used -as a technical tool for certain constructions. Therefore I would like to know -if there are some recommendable papers on development of K-theory -where such constructions involving Karoubian envelopes are discussed. - -REPLY [3 votes]: In Schlichting's paper Negative K-theory of derived categories. Math. Z. 253, 97–134 (2006), a definition of negative $K$-theory of triangulated categories is given. These are abelian group valued functors $\mathbf{K}_{i}$ with $i\leq 0$. -These groups agree with the known negative $K$-theories, for example in the Thomason-Trobaugh paper mentioned by Prof. Grayson. However they use idempotents in an essential way. -In particular, if $E$ is an exact category then the bounded derived category $D^b(E)$ is triangulated and $\mathbf{K}_{0}(E)$ is defined to be the usual $K_0$ of $\widetilde{E}$, the idempotent completion of $E$. -If $E$ is idempotent complete, then the unbounded derived category $D(E)$ exists, and the $\mathbf{K}_{-1}(E)$ turns out to be the quotient of the abelian monoid (under direct sum) of isomorphism classes of idempotents in $D(E)$ by the submonoid of split idempotents. In particular $K_{-1}(E)=0$ if and only if $D(E)$ is idempotent complete. -More generally, if $A$ is an idempotent complete triangulated category then $\mathbf{K}_{-1}(A)$ is zero if and only if the Verdier quotient $B/A$ is idempotent complete for all full triangulated embeddings $A\to B$ with $B$ idempotent complete. -Incidentally, Schlichting conjectured that whenever $A$ is a small abelian category, then $\mathbf{K}_{i}(A)=0$ for all $i<0$. -This conjecture was generalised to stable $\infty$-categories by Antieau, Gepner, and Heller in K-theoretic obstructions to bounded t-structures. Invent. math. 216, 241–300 (2019), wherein they conjecture that $\mathbf{K}_{i}(A)=0$ for $i<0$ whenever $A$ is a stable $\infty$-category with a bounded $t$-structure. -Neeman gave a very simple and elegant counter-example to both these conjectures in a recent preprint https://arxiv.org/abs/2006.16536.<|endoftext|> -TITLE: How big is the least non-$\Sigma^1_1$-pointwise-definable ordinal? -QUESTION [7 upvotes]: There's a large countable ordinal which has cropped up (as a lower bound!) in a computable structure theory problem I'm playing with. At present I don't really understand how big it is, and I'm curious where it fits in amongst better-understood ordinals. (I do have a kind of upper bound, but it's weird and not very helpful to me.) -Say that $\alpha$ is $\Sigma^1_1$-pd iff for every $x\in L_\alpha$ there is some $\Sigma^1_1$ formula $\varphi$ in the language of set theory without parameters such that $\{x\}=\varphi^{L_\alpha}$. (Here "$\Sigma^1_1$" refers to quantification over subsets of $L_\alpha$, not $\omega$, so this is quite broad.) I'm interested in the following: - -How large is the smallest non-$\Sigma^1_1$-pd ordinal, $\eta$? - -I'm interested in either "theory-oriented" (e.g. "$\eta\ge$ the least $\alpha$ such that $L_\alpha\models\mathsf{KP\omega}$ + '$\omega_1$ exists'") or "stability-oriented" (e.g. "$\eta\le$ the least $\alpha$ such that $L_\alpha\prec_1L_{\omega_1}$") information about $\eta$. A genuine characterization would be great, but I suspect there isn't a snappy one; I'll settle for any interesting bounds (lower or upper) on $\eta$. -(To be fair there are a couple immediate projectum-flavored observations, namely that every $\Sigma^1_1$-pd ordinal is $\Pi^1_2$-projectible to $\omega$ and that if $\alpha$ is $\Sigma^1_1$-projectible to $\omega$ then $\alpha$ is $\Sigma^1_1$-pd, but that line of thought doesn't seem to give me much detailed information about $\eta$. So I'd be happy with projectum-flavored information as well, but I suspect that's the wrong way to go.) - -REPLY [8 votes]: Claim: Let $\kappa$ be least such that $L_\kappa$ is admissible and -$L_\kappa\models$``$\omega_1$ exists'' -and let $\alpha=\omega_1^{L_\kappa}$. Then $\alpha$ is the least -non-$\Sigma^1_1$-pd ordinal. Moreover, the 1st projectum of $L_\kappa$ is -$\rho_1^{L_\kappa}=\omega_1^{L_\kappa}=\alpha$, and the 1st standard parameter is $p_1^{L_\kappa}=\{\alpha\}$. -(Thus, $\kappa$ is also the least admissible such that -$\rho_1^{L_\kappa}>\omega$.) -The following lemma is probably standard, but I had not noticed it before: -Lemma: Let $\kappa$ be such that $L_\kappa$ is admissible and $\delta<\kappa$ -such that $L_\kappa\models$``$\delta$ is a cardinal''. Then -$\rho_1^{L_\kappa}\geq\delta$. -Proof: Suppose not. Then $L_\kappa=\mathrm{Hull}_1^{L_\kappa}(\rho\cup\{p\})$ -for some $\rho<\delta$ and $p\in L_\kappa$. (That is, the $\Sigma_1$-hull of -parameters in $\rho\cup\{p\}$.) Therefore $L_\kappa\models$``For every -$\beta<\delta$ there is an ordinal $\gamma$ such that -$\beta\in\mathrm{Hull}_1^{L_\gamma}(\rho\cup\{p\})$''. But then by -admissibility, -there is $\gamma<\kappa$ which works simultaneously for all $\beta<\delta$, -which yields a surjection $f:\rho\to\delta$ with $f\in L_\kappa$, contradicting -that $\delta$ was a cardinal there. -Proof of Claim: For the moment let $\alpha$ be any ordinal, and let -$\kappa_\alpha$ be the least $\kappa>\alpha$ such that $L_\kappa$ is admissible. -Let $\kappa=\kappa_\alpha$. Note that if there is -$\beta\in[\alpha,\kappa)$ such that $L_\beta$ projects to $\omega$, then one -can -identify the least such $\beta$ in a $\Sigma^1_1$-over-$L_\alpha$ way (using -the -usual relationship between illfounded models and admissibles), and therefore -$\alpha$ is $\Sigma^1_1$-pd. -So suppose that $\alpha=\omega_1^{L_\kappa}$, where $\kappa=\kappa_\alpha$. -Then -$L_\alpha\preceq_1 L_\kappa$, by condensation, and hence -$\mathrm{Hull}_1^{L_\kappa}(\alpha)=L_\alpha$. But -$L_\kappa=\mathrm{Hull}_1^{L_\kappa}(\alpha+1)$. For if this hull had ordinal -height -$\beta<\kappa$, just let $\psi$ be some $\Sigma_0$ formula and $p,d\in L_\beta$ such that -$L_\beta\models\forall x\in d\ \exists y\ \psi(p,x,y)$ but $\beta$ is least -such; then because $p,d$ are in the hull, so is $\beta$ actually, -contradiction. Combined with the lemma, this gives $\rho_1^{L_\kappa}=\alpha$ -and -$p_1^{L_\kappa}=\{\alpha\}$. -Now suppose for a contradiction that $\alpha$ is $\Sigma^1_1$-pd. I claim -that $\rho_1^{L_\kappa}=\omega$ and $p_1^{L_\kappa}=\{\alpha\}$, contradicting -the previous paragraph. For this, we show that -$L_\kappa=\mathrm{Hull}_1^{L_\kappa}(\{\alpha\})$. -We can convert $\Sigma^1_1$-over-$L_\alpha$ to -$\Pi_1^{L_\kappa}(\{\alpha\})$ formulas (i.e. $\Pi_1$ over $L_\kappa$, in -parameter $\alpha$). (This must be a standard generalization of the situation when $\alpha=\omega$: Given a $\Sigma^1_1$ formula -$\varphi(x)=\exists y\psi(x)$, consider the game where player I plays elements -of $L_\alpha$ and player II also plays such elements, and player II also decides -which of these elements go into a predicate $y$, and then player II wins iff -the -elements played produce a structure $(M,\in,y)$ such that -$(M,\in,y)\models\psi(x)$. -Then for $x\in L_\alpha$, $(\varphi(x))^{L_\alpha}$ iff there is a winning -strategy for player II in the game iff the standard analysis of the game (see below) does not -yield -a winning strategy for player I in $L_\kappa$, and the latter is a -$\Pi_1^{L_\kappa}(\{\alpha\})$ statement.) -(Edit:) The analysis of the game: The game, at least once one sets up the rules appropriately, has open payoff for player I. Let $E$ be the set of partial plays of the game of even length. Let $S$ be the set of all $p\in E$ from which player I has a winning strategy (for the game continuing from $p$). We rank $S$, writing $S_{\beta}$ for the set of partial plays of rank ${<\beta}$. Let $S_0=\emptyset$ and $S_1$ be the set of $p\in E$ where player I has already won. For limit $\beta$ let $S_\beta=\bigcup_{\gamma<\beta}S_\gamma$. And for $\beta\geq 1$ let $S_{\beta+1}$ be the set of $p\in E$ such that there is $x\in L_\alpha$ such that for all $y\in L_\alpha$, we have $(p,x,y)\in S_\beta$. Note that $\beta\leq\gamma\implies S_\beta\subseteq S_\gamma\subseteq S$. Let $\beta_\infty$ be the least $\beta$ such that $S_\beta=S_{\beta+1}$. Then $S=S_{\beta_\infty}$. Now (by setting up the game rules appropriately) $E$ is definable over $L_\alpha$, and note that it follows that $S_\beta\in L_\kappa$ for each $\beta<\kappa$, and that $\left_{\beta<\kappa}$ is $\Sigma_1^{L_\kappa}(\{\alpha\})$. Now $\beta_\infty\leq\kappa$. For otherwise we have some $p\in S_{\kappa+1}\backslash S_\kappa$, so for all $x\in L_\alpha$ there is $y\in L_\alpha$ such that $(p,x,y)\in\bigcup_{\beta<\kappa}S_\beta$, and note this contradicts admissibility. So player I wins iff $\emptyset\in S_\kappa$, and player II wins iff $\emptyset\notin S_\kappa$. -So, because $\alpha$ is -$\Sigma^1_1$-pd, for every $x\in L_\alpha$ there is a $\Pi_1$ formula $\varphi$ -such that $x$ is the unique $x'\in L_\alpha$ -such that $L_\kappa\models\varphi(x',\alpha)$. But then -$x\in\mathrm{Hull}_1^{L_\kappa}(\{\alpha\})$. For we have that for every -$x'\in L_\alpha$ with $x'\neq x$, $L_\kappa\models\neg\varphi(x',\alpha)$. -So by admissibility, there is $\beta<\kappa$ such that for every $x'\in -L_\alpha$ with $x'\neq x$, we have $L_\beta\models\neg\varphi(x',\alpha)$. But then -note that the least such $\beta$ can be detected in a -$\Sigma_1^{L_\kappa}(\{\alpha\})$ -manner, and from that we can compute $x$, so $x$ is in the hull, as desired. -This is a contradiction, showing that $\alpha$ is not $\Sigma^1_1$-pd. -The full Claim follows by putting the things above together above. -Remark: Let $\alpha$ be an ordinal and $\kappa=\kappa_\alpha$. Then conversely -to one point above, every $\Pi_1^{L_\kappa}(\{\alpha\})$ subset of $L_\kappa$ is -$\Sigma^1_1$-over-$L_\alpha$.<|endoftext|> -TITLE: How prove this Webb inequality? -QUESTION [7 upvotes]: I posted this question on Math StackExchange but did not get any answer. I am trying my luck here. - -Let $a_{1},a_{2},\dotsc,a_{n+1}$ be a sequence of distinct non-zero real numbers with -$$\sum_{j=1}^{n+1}a^2_{j}=1,\qquad\sum_{j=1}^{n+1}a_{j}=0.$$ -Show -\begin{equation} -\tag{1} -\label{1} -0<\sum_{k=1}^{n+1}\dfrac{1}{\lvert a_{k}\rvert}\prod_{\substack{j=1 \\ j\neq k}}^{n+1}\dfrac{a_{k}}{a_{k}-a_{j}}\le\sqrt{2}. -\end{equation} - -I found the equality on the right-hand side when $n=1$. But I can't prove this inequality \eqref{1}. First of all, this inequality is a bit like Lagrange's interpolation formula https://math.stackexchange.com/questions/500139/prove-1-sum-i-0n-frac1x-i-prod-j-neq-i1-frac1x-j-x-i-prod-i. I tried to prove it using Lagrange's interpolation formula but can't. - -REPLY [4 votes]: Here is the proof that the sum is non-negative. -Denote $f(x)=x^{n-2}|x|=x^n/|x|$. Then -$$ -A:=\sum_{k=1}^{n+1}\dfrac{1}{|a_{k}|}\prod_{j=1,j\neq k}^{n+1}\dfrac{a_{k}}{a_{k}-a_{j}}= -[x^n]\sum_{k=1}^{n+1}f(a_k)\prod_{j\ne k}\frac{x-a_j}{a_k-a_j}=:[x^n]h(x), -$$ -where the polynomial $h(x)=Ax^n+\ldots$, $\deg h\leqslant n$, interpolates $f$ in points $a_1,\ldots,a_{n+1}$. -The function $h(x)-f(x)$ is $(n-2)$ times continuously differentiable and has $n+1$ roots at $a_i$'s, thus by Rolle theorem its $(n-2)$-st derivative $\frac{n!}2Ax^2+Bx+C-(n-1)!|x|$ has three distinct roots. But if $A<0$, this function is strictly concave and can not have three roots. So $A\geqslant 0$. -Also this sum appears in the theory of splines and in the algebraic combinatorics / integrable probability. Probably everything is done and written somewhere.<|endoftext|> -TITLE: Great graduate courses that went online recently -QUESTION [186 upvotes]: In 09.2020 by pure chance I discovered the YouTube channel of Richard Borcherds where he gives graduate courses in Group Theory, Algebraic Geometry, Schemes, Commutative Algebra, Galois Theory, Lie Groups, and Modular forms! (and an undergraduate courses in Theory of numbers and Complex analysis). -I watched so far about 300 of his videos (about 90%) and they are really great. Borcherds is an amazingly good lecturer (for my taste). It is also clear, that these lectures were worked out/improved through years, since Borcherds was giving similar ones in Berkeley (one can find some lecture notes by students online). I would guess, that currently there are some other great lecturers that started to upload their courses on YouTube (or some other platforms). For this reason a question. -Question. If you watched recently an online graduate course (free for all), and found it brilliant, by a lecturer whom you find great, and believe that the course taught you something, could you please share the info about it. -The motivation for this question is to spread information about exciting things happening in mathematical life/education lately. - -REPLY [3 votes]: Here we go again. - -Integration Theory by Professor Marcus Carlssons, link: https://youtu.be/EzoaGgBHJEQ - -Graduate Real Analysis by Professor Emanuel Carneiro, link: https://youtube.com/playlist?list=PLp0hSY2uBeP8hajKOVGZ9oIPjG3HKMfoY - -Graduate course on “Groups and Galois Theory” by Professor Yuly Billig, link: https://youtube.com/playlist?list=PLu6jbin1VpDBGWvctA_vGFLXzJD8YsA3F - -Differential topology course at IMPA by Professor Vinicius Ramos, link: https://youtu.be/yOoWugX1AOI - -Introduction to Algebraic K-Theory, link: https://youtube.com/channel/UCNefIuLog1bTwR0rav3p_5A - -https://youtube.com/c/UndergraduateMathematics YouTube channel is crazy. Just check out entire playlist of that channel. That channel is a window to thousand different graduate level math courses. - -Introduction to Stacks and Moduli by Professor Jarod Alper, link: https://youtube.com/playlist?list=PLhFI5R_xInjdhtWuhgYlA8NZGXO-unnl4 - -Higher Algebra, link: https://youtube.com/playlist?list=PLsmqTkj4MGTDenpj574aSvIRBROwCugoB. Topological Cyclic Homology, link: https://youtube.com/playlist?list=PLsmqTkj4MGTB8pNGvW0iuKUFmBlOSke-C - -SubRiemannian geometry course by Professor Enricole Le Donne, link: https://youtube.com/playlist?list=PLleGBpoKCrJn21-tCrNRk6JAyh4gwtgks - -Area formula-Geometric Measure Theory, link: https://youtube.com/playlist?list=PL_c66HM3gCshuMPW7BgVdWg_I8OitI7mC - -Category Theory by Professor Ivo De Los Santos Vekemans, link: https://youtube.com/playlist?list=PLHYBb0FtzAK9TBFhZ8eZzJRc5gW1NNp2K - -Algebraic Geometry, link: https://youtube.com/c/AntonMosunov - -Stanford Algebraic Geometry Seminar, link: https://youtube.com/playlist?list=PLdBhxg1X10QTkMv9XKj7h0naLhwKY-qw_<|endoftext|> -TITLE: Intersection of two Jordan curves lying in the rectangle -QUESTION [7 upvotes]: Given a rectangle $ABCD$. -Let a Jordan curve $L_1$ joins the vertexes $A$ and $C$ -and all points of $L_1$ belong the rectangle. -Let a Jordan curve $L_2$ joins the vertexes $B$ and $D$ -and all points of $L_2$ belong the rectangle. -Then there exists a point of $L_1$ and $L_2$ intersection. -Has this statement a simple proof? -The statement may be connected with the Jordan curve theorem -but I think its proof must be more simple. -Where one can see any proof of the statement? -Thanks. - -REPLY [5 votes]: There's a very slick way to demonstrate this using the Hex theorem [which is known to be equivalent to the Brouwer fixed-point theorem]. In a game of Hex, two players compete to connect opposite sides of a rhombus-shaped board built out of regular hexagons. Hex cannot end in a draw: the only way for the blue player to block the red from making a connection is for blue to make a connection. -                                                  -Whether we prove OP's statement for a rectangle, or for a rhombus made by adjoining two equilateral triangles, makes no difference as the two are homeomorphic. So let $A, C$ be the far corners of a rhombus with angles $60°$ and $120°$, respectively, and let $B, D$ be the near corners. The Jordan arcs $L_i$, $i = 1, 2$, are images of continuous injections $\gamma_i : [0, 1] \to ABCD$ where each $L_i$ maps $0, 1$ to opposite corners of the rhombus, $L_1$ (red) connects $A$ to $C$, $L_2$ (blue) connects $B$ to $D$. - -As $L_1$ and $L_2$ are continuous images of the compact set $[0, 1]$, they are also compact subsets of the metric space $ABCD$. So to show that they intersect, it suffices to show that the distance $$d(L_1, L_2) = \inf \{ d(x, y): x \in L_1, y \in L_2 \}$$ is $0$. -This is because in any metric space, if $A$ is compact and $B$ is closed, there always exist points $x \in A$, $y \in B$ so that $d(x, y) = d(A, B)$. -But if $d(L_1, L_2) = 0$, that implies $d(x, y) = 0$ for some $x \in L_1, y \in L_2$, and it's an axiom of metric spaces that that's only possible if $x = y$, which means $x$ is also in $L_2$ and $L_1 \cap L_2$ is nonempty. -Now for the argument that $d(L_1, L_2) = 0$. For any $N \geq 2$, we may cover $ABCD$ with an $N \times N$ Hex board whose upper left corner is at $A$ and lower right corner is at $C$. For such a Hex board, the diameter of a single tile is upper bounded by $k/N$ for some absolute constant $k$. If we color all the tiles that intersect $L_1$ red, we get a red-tiled winning Hex path connecting the top to the bottom, while if we color all the tiles that intersect $L_2$ blue, we get a blue-tiled winning Hex path connecting the left to the right: - -Now superimpose these two boards. If the winning path for red and the winning path for blue somehow didn't interfere with each other, that would contradict the Hex theorem. So there must be at least one purple tile that belongs to both paths: - -But any purple tile must contain points from both the red Jordan arc $L_1$ and the blue Jordan arc $L_2$, which implies that $$d(L_1, L_2) \leq \text{ diameter of a single tile } \leq \frac{k}{N},$$ and since $N$ was arbitrary, that implies $d(L_1, L_2) = 0$ and the two Jordan arcs have at least one point in common. Done.<|endoftext|> -TITLE: Chiral homology for the Virasoro algebra and/or affine Lie algebra -QUESTION [11 upvotes]: I want to understand what concrete analytical objects are found in chiral homology of higher degree of a vertex algera (-module) $M$. More precisely: I can obtain conformal blocks on a surface $\Sigma$ for the Virasoro algebra as invariants of $M$ under the action of vector fields - for the torus one can then indeed work out a modular function depending on the structure of the torus and possibly the location of singularities. One can also compute elements in it by gluing the vertex operator on a 3-punctured sphere to a torus (a.k.a graded character of $M$) -But what about higher homologies: In the Lie algebra standard complex the chains are functions depending on an $n$-tuple of (here) vector fields up to...how does this relate for the torus to modular functions or something similar? -Same question for affine Lie algebra at negative level - I have found the great work of Gaitsgory, but I would like to know if there is any concrete analytic realization of the elements by...? -[But maybe it is simply the wrong question !?] -Also, I would already be very happy for reference like to as for references you might have on the chiral cohomology of the Virasoro algebra on surfaces (I know the computation for $M$ trivial, but for say $M$ a irrep of a minimal model?). -Thanks alot in advance, -Simon Lentner - -REPLY [9 votes]: There isn't much written up on how to compute the higher chiral homology of vertex algebras. Besides the original work of Beilinson and Drinfeld that covers in particular the universal cases of Virasoro and Affine algebras you won't find much more. There's an unwritten theorem by Gaitsgory that proves the vanishing for simple Affine algebras at integral level. -In the case of elliptic curves and in the limit of the nodal elliptic curve, together with Van Ekeren we related the first chiral homology group to the first Hochschild homology of the Zhu algebra (just as comformal blocks are related to the zeroth Hochschild homology) in -https://arxiv.org/abs/1804.00017 -Unfortunately that relies on a technical condition of vertex algebras being "classically free", that is that their classical limit is isomorphic to the arc algebra of the C2 quotient. This was shown to be the case for minimal models $Vir_{p,p'}$ if and only if $p=2$. And some simple affine algebras at integral level. -We can now relax this condition to vertex algebras to the case then the classical limit is a quotient of the arc algebra by a finitely generated differential ideal. This is still very hard to prove. The only known example of this to my knowledge is the Ising model $Vir_{3,4}$ in -https://arxiv.org/abs/2005.10769 -Using this we can now prove the vanishing of the first chiral homology group of an arbitrary elliptic curve (not necessarily the nodal limit) with coefficients in either the 2,p' minimal models or the Ising model. As well as recover some of Gaitsgory's results. That article should come out soon, it's being finalized for a while now.<|endoftext|> -TITLE: What sorts of extra axioms might we add to ZFC to compute higher Busy Beaver numbers? -QUESTION [27 upvotes]: First, some context. Ever since I was a high schooler, I have been fascinated with large numbers. As I have grown in mathematical maturity, I have become both disappointed and fascinated to see that the process of naming larger and larger numbers requires a sort of philosophical tradeoff. -For example, an ultrafinitist might reject $10^{10^{100}}$ as being a valid representation of a number. A finitist might reject numbers defined using ordinal collapsing functions with large cardinals. A person who believes only in the constructible universe might reject the same ordinal collapsing functions if cardinals incompatible with $V=L$ are used. In general, the more you are willing to accept philosophically, the "larger" a number you can name concisely. -The interesting thing about these tradeoffs is that some of these philosophical stances have concrete impacts on the "real" world. For example, there is a number $n$ for which new axioms in addition to those of ZFC would be needed to prove the value of $BB(n)$. So in a way, this "real world" value changes depending on our philosophical stance. -My question then is: what sorts of axioms might we accept to strengthen as much as possible the values of $BB(n)$ which we can prove? And more generally, is there a way we can evaluate whether certain statements independent of ZFC are "true" based on their implications for the "real world" value of $BB(n)$ (if there is even a "true" value for this number). -EDIT: useful reading for those not familiar with the phenomenon of independence of $BB(n)$ from ZFC: - -Adam Yedidia, Scott Aaronson, A Relatively Small Turing Machine Whose Behavior Is Independent of Set Theory, Complex Systems 25(4) (2016), journal, arXiv:1605.04343 - -EDIT: As requested, an example of a "real world" consequence of a set theoretical axiom would be how Con(ZFC) would prove the Turing machine in the above paper doesn’t halt, whereas the negation of Con(ZFC) proves it halts in finite time. - -REPLY [8 votes]: Knowing the values of the Busy Beaver function is the same as knowing the truth values of $\Pi^0_1$-statements (ie statements of the form $\forall n \in \mathbb{N} \ P(n)$ for decidable properties $P$). Knowing a particular $\mathrm{BB}(m)$ means knowing all $\Pi^0_1$-statements of complexity $m$ (for a suitable complexity measure). -For any c.e. theory $T$, $\mathrm{Con}(T)$ is a $\Pi^0_1$-statement. Moreover, if $T$ is of the form "$\mathrm{ZFC}$ + some short axiom", the complexity of $\mathrm{Con}(T)$ (in the sense above) will not be much more than the complexity of $\mathrm{Con}(\mathrm{ZFC})$. Hence if we want to settle values of the Busy-Beaver function on "medium-sized" inputs, we would need to add axioms that settle a lot of consistency statements. -Proving that adding an axiom actually increases the range of provable Busy-Beaver values will be non-trivial. It does not suffice to show that it settles more $\Pi^0_1$-statements, but one needs to show that actually increases the least complexity of an undetermined $\Pi^0_1$-statement. Unfortunately, the complexity notion here is essentially Kolmogorov complexity -- which means that we cannot compute the complexity of a logical formula from that formula.<|endoftext|> -TITLE: Measure the failure of colimit to commute with taking free loops (or Hochschild homology)? -QUESTION [12 upvotes]: For a space1 $X$, let $\mathcal{L}X = \mathrm{Maps}(S^1, X)$ be the free loop space. -Inclusion of constant loops gives a natural map $X \to \mathcal{L}X$. This is not a homotopy equivalence unless $X$ is contractible, because it splits the fibration $\Omega X \to \mathcal{L} X \to X$. -I would like to know that it is not an equivalence ``because'' loops do not commute with (homotopy) colimits. -That is: suppose $X$ is presented as a CW complex, i.e. as a colimit of contractible spaces: $X = \mathrm{Colim}\, D_\alpha$. Then the natural map -$$X = \mathrm{Colim} \, D_\alpha = \mathrm{Colim} \, \mathcal{L} D_\alpha \to -\mathcal{L} \, \mathrm{Colim} \, D_\alpha = \mathcal{L} X $$ -is (homotopic to) the inclusion of constant loops. - -It is possible to see that $X \mapsto \mathcal{L} X$ is not an equivalence by computing some invariant which, in general, measures the failure of loops to commute with colimits? - -Here I have in mind that if I wanted to know how a functor failed to be exact, I would study its derived functors. - -When I search the internet for loop spaces and colimits, invariably I find myself reading about ``calculus of functors''. - -Is the calculus of functors going to help me here? - - -In fact for my purposes I am ultimately interested in the analogous question for the functor ``Hochschild homology'' from dg categories to chain complexes, so would be perfectly happy with an answer to the above question after taking chains. -(As for what these purposes are, it has to do with a problem about dynamics in contact manifolds. An explanation of how that is related is a bit afield and perhaps too long to include here, but you can see this short note.) - -1 To avoid more complaints in the comments, let us say a connected CW complex. - -REPLY [2 votes]: If you consider Goodwillie calculus for functors ${\rm Top} \to Ch_{\mathbb Z}$, then a functor will be linear if and only if it preserves all small (homotopy) colimits (as in Maxime's comment). Indeed, in the stable setting, a square is a pushout if and only if it is a pullback, so excisive functors to a stable target preserve pushouts, and therefore all small colimits. So the functor $X \mapsto C_*(LX)$ will not be linear. -There may be issues when $X$ is not simply connected and I don't know a reference, but it seems to me that you should be able to reconstruct the Goodwillie tower of $X \mapsto C_*(LX)$, from the (co)-Hochschild model for $C_*(LX)$ by truncating the Hochschild complex. In particular it seems that the $n$th associated homogeneous functor is $C_*(X)^{\otimes n}$ (up to a shift).<|endoftext|> -TITLE: Asymptotics of ratios of polynomially recursive sequences -QUESTION [5 upvotes]: A sequence $a_n$ is said to be polynomially recursive (P-recursive) if it satisfies: -$$p^{[r]}(n)a_{n+r}+\cdots+p^{[1]}(n)a_{n+1}+\cdots + p^{[0]}(n)a_n=0$$ -where $p^{[i]}(t)\in \mathbb{Q}[t]$ are polynomials with rational coefficients, with $p^{[0]},p^{[r]}$ not identically zero. -For example, $a_n:=n!$ is one such sequence since: $2a_{n+2}-(n+2)a_{n+1}-(n+1)(n+2)a_n=0$, with initial conditions $a_0:=0, a_1:=1$. -Fix a set of polynomials $\{p^{[i]}(t)\}_{i=0}^r$, and suppose $a_n,b_n$ are a pair of sequences that satisfy the same above recurrence, with different initial conditions. Furthermore, suppose that both sequences aren’t ultimately periodic or constant. -Define $L :=\lim_{n\rightarrow\infty} \frac{a_n}{b_n}$. - -Main Question: Is it obvious when $L\in (0,\infty)$? In other words, when does $L$ exist, is non-zero and non-infinite? Aside from numerically evaluating the limit for large enough $n$, are there any algorithmic methods for deducing that $L \in (0,\infty)$? - -There are many known (non-trivial) results about the growth rates of such sequences, for example results due to Poincaré, Birkhoff and Trjitzinsky, Wimp and Zeilberger, and Mezzarobba and Salvy. However, I'm unable to find good references related to my question, especially as a function of initial conditions. The main difficulty I have is that I’m not sure how to find good lower bounds on the growth rates of such sequences. - -REPLY [2 votes]: This is a difficult question in general; see for instance - -https://people.mpi-sws.org/~joel/publications/positivity_and_minimality_holonomic21abs.html -https://people.mpi-sws.org/~joel/publications/holonomic-second-order21abs.html - -for recent work on special cases. (As you noted, the paper by Bruno Salvy and myself that you mention is purely about upper bounds and thus not terribly relevant.) -However, there are sufficient conditions for $L$ to be nonzero and finite that can be verified algorithmically and cover a fair number of (“easy”) cases. In particular, by passing to the differential equation on the generating series, one may be able to get “asymptotic expansions with error bounds” of the form, say, $|a_n - α f(n)| ≤ g(n)$, $|b_n - β f(n)| ≤ h(n)$ where $f$, $g$, $h$ are explicit functions with $g, h = o(g)$ and $α$, $β$ are constants that can be bounded from both sides. For more details on this kind of ideas, see for example - -https://arxiv.org/abs/2011.08155 -https://hal.archives-ouvertes.fr/hal-03291372/<|endoftext|> -TITLE: On the state of the art on closed $(n-1)$-connected $2n$ manifolds -QUESTION [8 upvotes]: In the paper "Classification of $(n - 1)$-Connected $2n$-Manifolds" by C.T.C.Wall (Annals of Mathematics , Jan., 1962, Second Series, Vol. 75, No. 1 (Jan., 1962), pp. 163-189), Wall studies $(n - 1)$-Connected $2n$-Manifolds with a small ball removed and proves a classification result for such manifolds in terms of algebro-topological invariants, namely the intersection form on the middle cohomology and a homotopy theoretic invariant which varies finitely. -In the introduction Wall makes the following remark (I quote): "In a subsequent paper, the author intends to study the diffeomorphisms of the manifolds here obtained; in particular, to give a complete set of invariants of isotopy of a diffeomorphism, and to consider more carefully the actual diffeomorphism classification of closed $(n-1)$-connected $2n$-manifolds (which is not settled in this paper, even when our results are complete.)" -There are two parts to my question: - -Did the mentioned paper appear? - -I should mention that I have looked through the titles of Wall's (100+) subsequent articles and not found a title directly related to this problem, hence the question. -I should also mention what is known in low dimensions. For $2n = 4$ this includes the smooth Poincare conjecture so is open to my knowledge, up to homeomorphism it has been settled by Freedman. In dimension $2n = 6$, I believe that it is known due to work of Wall, Jupp and Zubr that such a manifold should be diffeomorphic to a connect sum of $S^3 \times S^3$'s. -I ask a second part: - -Let $n$ be an even integer, $n \geq 4$. What is the state of the art of the homoemorphism classification of closed $(n-1)$-connected $2n$-manifolds? - -Note also in Question 2 I have intentionally slightly modified the problem, since manifolds with dimension $2 \mod 4$ and the issue of different smooth structures are not of primary interest to me. - -REPLY [11 votes]: The classification problem of smooth oriented closed $(n-1)$-connected $2n$-manifolds for $n\ge3$ splits into three parts. - -Classify smooth almost closed compact oriented $(n-1)$-connected $2n$-manifolds, where almost closed means that the boundary is a homotopy sphere. -Understand those homotopy spheres that arise as boundaries of almost closed compact oriented smooth $(n-1)$-connected $2n$-manifolds. -Understand inertia groups of smooth oriented closed $(n-1)$-connected $2n$-manifolds. - -In the work you mentioned, Wall achieved a classification of type 1. in terms of what he calls n-forms. Since then several authors (Wall, Kosinski, Schultz, Stolz, ...) have obtain partial results regarding 2. and 3. Most recently Burklund--Hahn--Senger and Burklund--Senger settled the last open cases and completed the classification. -Regarding your second question: In the topological category, 2. and 3. are vacuous since all high-dimensional homotopy spheres are topologically trivial. Wall's original approach to 1. only uses tools that have since been established in the topological category (mostly by Kirby--Siebenmann), so his approach goes through and reduces the classification to understanding $\pi_n(BSTop(n))$. To get at the latter, you can compare $BSTop(n)$ to $BSTop$ and use that its homotopy fibre receives a highly-connected map from $SO/SO(n)$.<|endoftext|> -TITLE: Product of two reflections lying in a parabolic subgroup of a Coxeter group -QUESTION [10 upvotes]: Let $(W,S)$ be a Coxeter group, $I\subseteq S$ a subset of simple reflections, and $W_I \subseteq W$ the corresponding parabolic subgroup (we could also assume $|W_I|<\infty$, if needed). - -Let also $t_1,t_2\in W$ be two reflections (i.e. elements in $W$ conjugated to some $s_1,s_2 \in S$ respectively) such that $t_1t_2\in W_I$ but $t_i\notin W_I$, $i=1,2$. -Is it then true that the only possibility is $t_1t_2=e$ (i.e. $t_1=t_2$)? -If I look at the geometric picture with alcoves and reflection hyperplanes this seems to me to be true, but I don't know how to prove it in full generality via geometric arguments. I rather tried with some combinatoric method (e.g. using the strong exchange condition or some result about parabolic double cosets) but I have succeeded only in the case $t_i \in S$ for at least one $i$. -Thanks a lot for any comment about that! - -REPLY [4 votes]: $\DeclareMathOperator\Fix{Fix}\DeclareMathOperator\Stab{Stab}$I want to share a possible solution that I found just developing the idea contained in Nathan's deleted answer. His argument just needed a representation $V$ with the following two properties: -(i) $\Fix(t_1t_2)=\Fix(t_1)\cap \Fix(t_2)$, if $t_1$, $t_2$ are distinct reflections in $W$. -(ii) $\exists v_0\in V$ such that $\Stab(v_0)=W_I$ -With such a representation in hand one can easily conclude as follows (assuming $t_1\neq t_2$): -$t_1t_2\in W_I \Rightarrow v_0\in \Fix(t_1t_2)=\Fix(t_1)\cap \Fix(t_2) \Rightarrow t_1,t_2 \in W_I$. -As discussed in previous comments, the geometric representation of $W$ satisfies (i) but fails (ii), while its dual satisfies (ii) but fails (i). -However, a certain enlargement of the geometric representation (having the dual of it as a quotient) does the job: -Claim: Let $V$ be a finite dimensional real vector space with given linearly independent vectors $\{e_s\}_{s\in S}\subseteq V$ and given linearly independent linear forms $\{e_s^\vee\}_{s\in S}\subseteq V^*$ such that $\langle e_t,e_s^\vee\rangle=-2\cos(\frac{\pi}{m_{st}})$ $\forall s,t\in S$. Assume also that $V$ has the smallest possible dimension to satisfy these requests. Then $s\cdot v\mathrel{:=}v-\langle v,e_s^\vee\rangle e_s$ turns $V$ into a representation of $W$ that satisfies (i) and (ii). -The fact that such a vector space $V$ always exists is straightforward, while the fact that the above formula actually defines a representation of $W$ can be found in Soergel's article Kazhdan–Lusztig polynomials and indecomposable bimodules over polynomial rings (Proposition 2.1) (actually the original article is in German, the linked one is just an English preprint). The same Proposition also shows that this representation is reflection faithful: this in particular means that $\Fix(w)$ is a hyperplane if and only if $w$ is a reflection in $W$ and that distinct reflections fix different hyperplanes, which implies at once (i). -For (ii), we use the fact (contained in the proof of the above mentioned Proposition) that, if $K\mathrel{:=}\{v\in V \mathrel\vert \langle v, e_s^\vee\rangle=0\ \forall s\in S\}$ (so $K$ is a subrepresentation of $V$, where $W$ acts trivially), then $V/K$ is isomorphic to the dual of the geometric representation of $W$, and so it contains a vector with stabilizer equal to $W_I$. As a consequece, for any $v_0\in V$ lying in the preimage of such a vector one has $\Stab(v_0)\subseteq W_I$. To show the other inclusion, let $s \in I$: one has then $s\cdot v_0\in v_0+K$, but by definition one also has $s\cdot v_0=v_0-\langle v_0, e_s^\vee\rangle e_s$. Since $e_s\notin K$ ($\langle e_s,e_s^\vee\rangle=2$) it has to be $\langle v_0,e_s^\vee\rangle=0$, i.e. $s\in \Stab(v_0)$, as desired.<|endoftext|> -TITLE: Is there a nonzero solution to this infinite system of congruences? -QUESTION [7 upvotes]: Is there a triple of nonzero even integers $(a,b,c)$ that satisfies the following infinite system of congruences? -$$ -a+b+c\equiv 0 \pmod{4} \\ -a+3b+3c\equiv 0 \pmod{8} \\ -3a+5b+9c\equiv 0 \pmod{16} \\ -9a+15b+19c\equiv 0 \pmod{32} \\ -\vdots \\ -s_na + t_nb + s_{n+1}c \equiv 0 \pmod{2^{n+1}} \\ -\vdots -$$ -where $(s_n)$ and $(t_n)$ are weighted tribonacci sequences defined by -$$ -s_1=s_2=1, \\ -s_3=3, \\ -s_n = s_{n-1} +2s_{n-2} + 4s_{n-3} \text{ for } n>3, -$$ -and -$$ -t_1=1, \\ -t_2=3, \\ -t_3=5, \\ -t_n = t_{n-1} +2t_{n-2} + 4t_{n-3} \text{ for } n>3. -$$ -I think there are no nonzero solutions, but I haven't been able to prove this. Computationally, I found there are no nonzero solutions for integers $a$, $b$, and $c$ up to $1000$. -Note the $s_n$ and $t_n$ are always odd, and that the ratios $\frac{s_n}{s_{n-1}}$ and $\frac{t_n}{t_{n-1}}$ approach $2.4675...$. - -REPLY [9 votes]: Let $u_n = a s_n + b t_n + c s_{n+1}$. The stronger claim is true: for large enough values of $n$, -the number $u_n$ will be exactly divisible -by a fixed power of $2$ that doesn't depend on $n$. -Let $u_n = a s_n + b t_n + c s_{n+1}$ then (by induction) -$$u_{n} = u_{n-1} + 2 u_{n-2} + 4 u_{n-3}.$$ -The polynomial $x^3 - x^2 - 2 x - 4$ is irreducible and has three roots $\alpha_1$, $\alpha_2$, and $\alpha_3$ in $\overline{\mathbf{Q}}$. -By the general theory of recurrence relations, -$$u_n = A_1 \alpha^n_1 + A_2 \alpha^n_2 + A_3 \alpha^n_3$$ -for constants $A_1$, $A_2$, $A_3$. Since $u_n \in \mathbf{Q}$, we may additionally deduce that $A_i$ lie in $\mathbf{Q}(\alpha_1,\alpha_2,\alpha_3)$. -That is because we can solve for $A_i$ using the equation -$$\left( \begin{matrix} \alpha_1 & \alpha_2 & \alpha_3 \\ \alpha^2_1 & \alpha^2_2 & \alpha^2_3 \\ \alpha^3_1 & \alpha^3_2 & \alpha^3_3 \end{matrix} \right) -\left( \begin{matrix} A_1 \\ A_2 \\ A_3 \end{matrix} \right) = \left( \begin{matrix} u_1 \\ u_2 \\ u_3 \end{matrix} \right)$$ -and the matrix on the left is invertible (Vandermonde). In fact we deduce the stronger claim that -any Galois automorphism sending $\alpha_i$ to $\alpha_j$ sends $A_i$ to $A_j$. -(Simply consider the action of the Galois group on both sides of this equation, -noting that the $A_i$ are determined uniquely from this equation.) In particular, if one of the $A_i = 0$, -then all of the $A_i = 0$. -But now fix an embedding of $\overline{\mathbf{Q}}$ into $\overline{\mathbf{Q}}_2$. From the Newton Polygon, -we see that there is one root (call it $\alpha_1$) of valuation $0$, and the other two roots have valuation $1$. Hence -$$\|A_1 \alpha^n_1 \|_2 = \|A_1\|_2, \quad \|A_2 \alpha^n_2 \|_2 = \|A_2\|_2 \cdot 2^{-n}, \quad \|A_3 \alpha^n_3 \|_2 = \|A_3\|_2 \cdot 2^{-n}.$$ -In particular, if $A_1 \ne 0$, then (by the ultrametric inequality) $\| u_n \|_2 = \|A_1\|$ for $n$ large enough. Hence we deduce that either the $2$-adic valuation -of $u_n$ is eventually constant (as claimed) or that $A_1 = 0$ and so $A_i = 0$ for all $i$, which implies that $u_n = 0$ for all $n$. But if $u_1 = u_2 = u_3 = 0$, -then -$$\left( \begin{matrix} 1 &1 & 1 \\ 1 &3 & 3 \\ 3 & 5 & 9 \end{matrix} \right) -\left( \begin{matrix} a \\ b \\ c \end{matrix} \right) = \left( \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right)$$ -The matrix on the left is invertible which implies that $a=b=c=0$. - -REPLY [5 votes]: $u_n=s_na + t_nb + s_{n+1}c$ satisfies the same recurrence relation as $s_n$ and $t_n$: $u_n = u_{n-1} +2u_{n-2} + 4u_{n-3}$. The question is whether $2^{n+1}\mid u_n$. -Since $v_n=u_n/2^{n+1}$ satisfies -$v_n = \displaystyle\frac{v_{n-1} +v_{n-2} + v_{n-3}}{2}$ -the answer is affirmative only if there are $v_0, v_1, v_2$ (not all 0) such that $v_n$ is always integral. -EDIT. -As sharply noticed by the OP, the attempt below was wrong, since a matrix I claimed to be invertible (mod $2$) is in fact singular. A similar, more computational argument does work (mod $5$). - -It's clear that (mod $2$) such a sequence $v_n$ must follow either one of the 3-periodic patterns $000$ and $110$ (up to shifts). In the $000$ case keep dividing entire sequence $(v_n)$ by $2$ until one term is odd, and then shift the sequence to start with that term, so it's back to the $110$ case. Therefore it must be that -$$\require{cancel}\cancel{\det\left (\begin{smallmatrix}v_1 & v_2 & v_3\\ v_2 & v_3 & v_4\\ v_3 & v_4 & v_5 \end{smallmatrix}\right )\equiv \det\left (\begin{smallmatrix}1 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 1 \end{smallmatrix}\right )\!\!\!\!\pmod{2} \ne0}$$ -CORRECTED ARGUMENT. -Notice that -$$D=\det\left (\begin{matrix}v_1 & v_2 & v_3\\ v_2 & v_3 & v_4\\ v_3 & v_4 & v_5 \end{matrix}\right )=\det\left (\begin{matrix}v_1 & v_2 & v_3\\ v_2 & v_3 & \frac{v_1+v_2+v_3}{2}\\ v_3 & \frac{v_1+v_2+v_3}{2} & \frac{v_1+3v_2+3v_3}{4} \end{matrix}\right )\\= -\frac{-4v_3^3+4v_2 v_3^2+2v_1 v_3^2+v_2^2 v_3+5v_1 v_2 v_3-v_1^2 v_3-3v_2^3-2v_1 v_2^2-2v_1^2 v_2-v_1^3}{4}$$ -is $\equiv 0 \pmod{5}$ if and only if $v_1\equiv v_2\equiv v_3\equiv 0 \pmod{5}$. This is proved by the following snippet of code: -awk -vp=5 'BEGIN { - for(a=0; a -TITLE: Reference request: A multidimensional generalization of the fundamental theorem of calculus -QUESTION [11 upvotes]: $\newcommand\R{\mathbb R}$Let $f\colon\R^p\to\R$ be a continuous function. For $u=(u_1,\dots,u_p)$ and $v=(v_1,\dots,v_p)$ in $\R^p$, let -$[u,v]:=\prod_{r=1}^p[u_r,v_r]$; -$u\wedge v:=\big(\min(u_1,v_1),\dots,\min(u_p,v_p)\big)$; $u\vee v:=\big(\max(u_1,v_1),\dots,\max(u_p,v_p)\big)$; -$$\int_u^v dx\, f(x):= -(-1)^{\sum_{r=1}^p\,1(u_r>v_r) }\int_{[u\wedge v,u\vee v]}dx\,f(x).$$ -Let $F\colon\R^p\to\R$ be any antiderivative of $f$, in the sense that -$$D_1\cdots D_p F=f,$$ -where $D_j$ is the operator of the partial differentiation with respect to the $j$th argument; it is assumed that the result of this repeated partial differentiation does not depend on the order of the arguments with respect to which the partial derivatives are taken. Let $[p]:=\{1,\dots,p\}$. For each set $J\subseteq[p]$, let $|J|$ denote the cardinality of $J$. -Then it is not hard to establish the following multidimensional generalization of the fundamental theorem of calculus (Lemma 5.1): -\begin{equation} - \int_u^v dx\, f(x)=\sum_{J\subseteq[p]}(-1)^{p-|J|}F(v_J), -\end{equation} -where $v_J:=\big(v_1\,1(1\in J)+u_1\,1(1\notin J),\dots,v_p\,1(p\in J)+u_p\,1(p\notin J)\big)$. -Has anyone seen this or similar statement elsewhere? (I am only asking about references, not proofs.) - -REPLY [6 votes]: For an elementary fact like this, which may have been reinvented a thousand times, it is hard to find the first paper where this appeared. However, let me give some missing context. There is a whole industry in constructive quantum field theory and statistical mechanics about related "smart" interpolation formulas or Taylor formulas with integral remainders. These are used to perform so-called cluster expansions. -For the OP's identity, there is no loss of generality in taking $u=(0,0,\ldots,0)$ and $v=(1,1,\ldots,1)$. In this case, via Möbius inversion in the Boolean lattice, the formula comes from the following identity. -Let $L$ be a finite set. Let $f:\mathbb{R}^L\rightarrow \mathbb{R}$, $\mathbf{x}=(x_{\ell})_{\ell\in L}\mapsto f(\mathbf{x})$ be a sufficiently smooth function, and let $\mathbf{1}=(1,\ldots,1)\in\mathbb{R}^L$, then -$$ -f(\mathbf{1})=\sum_{A\subseteq L}\int_{[0,1]^A}d\mathbf{h} -\left[\left(\prod_{\ell\in A}\frac{\partial}{\partial x_{\ell}}\right)f\right](\psi_A(\mathbf{h})) -$$ -where $\psi_A(\mathbf{h})$ is the element $\mathbf{x}=(x_{\ell})_{\ell\in L}$ of $\mathbb{R}^L$ defined from the element $\mathbf{h}=(h_{\ell})_{\ell\in A}$ in $[0,1]^A$ by the rule: -$x_{\ell}=0$ if $\ell\notin A$ and $x_{\ell}=h_{\ell}$ if $\ell\in A$. -Of course one needs to 1) apply this to all $L$'s which are subsets of $[p]$, 2) use Möbius inversion in the Boolean lattice, and 3) specialize to $L=[p]$, and this gives the OP's identity. -The above formula is the most naive one of its kind used to do a "pair of cubes" cluster expansion. See formula III.1 in the article -A. Abdesselam and V. Rivasseau, "Trees, forests and jungles: a botanical garden for cluster expansions". -It is also explained in words on page 115 of the book -V. Rivasseau, "From Perturbative to Constructive Renormalization". -Now the formula is a particular case of a much more powerful one, namely, Lemma 1 in -A. Abdesselam and V. Rivasseau, "An explicit large versus small field multiscale cluster expansion", -where one sums over "allowed" sequences $(\ell_1,\ldots,\ell_k)$ of arbitrary length of elements of $L$, instead of subsets of $L$. The notion of allowed is based on an arbitrary stopping rule. The above identity corresponds to "allowed"$=$"without repeats", or the stopping rule that one should not tack on an $\ell$ at the end of a sequence where it already appeared. By playing with this kind of choice of stopping rule one can use Lemma 1 of my article with Rivasseau, to prove the Hermite-Genocchi formula, the anisotropic Taylor formula by Hairer in Appendix A of "A theory of regularity structures" and many other things. When $f$ is the exponential of a linear form for instance, one can obtain various algebraic identities as in the MO posts -rational function identity -Identity involving sum over permutations -I forgot to mention, one can use Lemma 1 to derive the Taylor formula from calculus 1. This corresponds to $L$ having one element and defining allowed sequences as the ones of length at most $n$. See -https://math.stackexchange.com/questions/3753212/is-there-any-geometrical-intuition-for-the-factorials-in-taylor-expansions/3753600#3753600<|endoftext|> -TITLE: In the internal language of the topos of sheaves on a topological space, can we define locally constant real-valued functions? -QUESTION [17 upvotes]: For the purposes of this question, in a Grothendieck topos, we will call “definable” the objects and relations obtained from the terminal object, the natural numbers object and the subobject classifier, by taking finite products, finite coproducts, exponentials (internal homs) and taking subobjects defined by [edit 2021-02-16] finitary formulas in the internal language (using previously defined objects and relations). (I'm saying this a bit concisely in the hope that there are no major subtleties.) -In particular, if $X$ is a topological space and we consider the topos of sheaves on $X$, the sheaf of continuous functions with values in each one of the following is definable (along with its usual algebraic structure): - -$\mathbb{N}$ with the discrete topology (this is the natural numbers object), - -$\mathbb{Z}$ with the discrete topology (this is the Grothendieck group of the previous), - -$\mathbb{Q}$ with the discrete topology (this is the fraction field of the previous), - -$\mathbb{R}$ with the usual (Euclidean) topology (by Dedekind cuts). - - -Let us call $\mathbf{N}, \mathbf{Z}, \mathbf{Q}, \mathbf{R}$ the corresponding definable objects of the topos. -Now for a long time I thought one could not define the sheaf of continuous functions with values in - -$\mathbb{Q}$ with the usual (i.e. induced by $\mathbb{R}$) topology, - -but I serendipitously realized that you can, namely it is given by the following object: -$$\{x\in\mathbf{R} : \forall y\in\mathbf{R}. ((\forall z\in\mathbf{Q}.(y\mathrel{\#}z)) \Rightarrow (x\mathrel{\#}y))\}$$ -where $x\mathrel{\#}y$ stands for $(xy)$ or, equivalently, $\exists z\in\mathbf{R}.(z\cdot(x-y)=1)$. -(This is easy to see: first note that $\{x\in\mathbf{R} : \forall y\in\mathbf{Q}. (x\mathrel{\#}y)\}$ defines the sheaf of continuous functions with values in $\mathbb{R}\setminus\mathbb{Q}$ with the usual topology, then repeat the reasoning.) -So now I am curious to know whether the “converse” is possible: - -$\mathbb{R}$ with the discrete topology; - -in other words: -Question: is the sheaf of locally constant real-valued functions on $X$ definable, uniformly in $X$, as a subobject of $\mathbf{R}$ in the topos of sheaves on $X$? -I imagine there is little hope of finding a good answer to the very general question “for which topological spaces $Y$ is the sheaf of continuous $Y$-valued functions on $X$ definable as an object in the topos?”, but of course, if someone wants a crack at it rather than the particular case above, by all means do! - -REPLY [2 votes]: I can offer a helpful observation. Andrew Swan and I proved in Every metric space is separable in function realizability that in Kleene's function realizability topos every metric space is separable. One consequence of this is that in the internal language of a topos one cannot construct a non-separable metric space, or an uncountable set with decidable equality. Moreover, it is consistent to assume that every set with decidable equality is countable (Theorem 2.5, but see the comment following it.) -While this is not a complete answer to your question, it shows that if there is a definition of the discrete $\mathbb{R}$ in sheaves, we will not be able to prove intuitionistically that it is uncountable and has decidable equality.<|endoftext|> -TITLE: Happy ants never leave compact domain? -QUESTION [22 upvotes]: I am curious if the following seemingly simple question has an easy answer? -Consider an ant population of $N$ ants that lives in $\mathbb R^2$. Each ant can be labeled by some coordinate $x\in \mathbb R^2.$ -Ants like to be close to their peers but also not too close. The optimal distance between the center of two ants is $5^{2/3}$. So given two ants $x_i,x_j \in \mathbb R^2$. Their happiness $H$ is $$H(x_i,x_j):=\operatorname{max}\{-\vert \vert x_i-x_j \vert^{3/2} -5 \vert,-10\}.$$ -Distances $\vert x_i-x_j\vert \le 1$ are not allowed since ants do not like to get too close. -The purpose of the maximum, in the definition of $H$ above, reflects that there is no attraction between two ants anymore at a certain distance. -Now consider the total happiness $H_N:=\sum_{i0$ such that, for any configuration of $N$ ants, such that the graph with vertices ants and edges pairs of ants separated by a distance of at most $15^{2/3}$ is connected, the total happiness is at most $\lambda N - c R$ where $R$ is the minimal radius of a disc enclosing all the ants. - -The motivation for this conjecture is that, whatever configuration gives the maximum $\lambda$ happiness per ant, the boundary of the ant colony will form a flaw in that configuration, leading to a loss from the maximal $\lambda N$ happiness proportional to the size of the boundary. Under this mild connectedness hypothesis, $R$ is bounded by the size of the boundary, explaining the lost happiness proportional with $R$. -Combined with the statement that, for all $N$, there exists a configuration of happiness at least $\lambda N - O (\sqrt{N})$, this implies an upper bound as you desire. -However, I am not sure how to prove the conjecture without a more precise understanding of the optimal configuration - it is possible to imagine, say, packing ants in some highly efficient way which can't be continued past a certain curve, leading to very happy configurations with large boundary. - -I will now prove the existence statement. -Take a configuration of a very large number of ants that attains an average happiness of $\lambda-\epsilon$. Let $r$ be minimal such that a ball of radius $r$ can pack at most $n$ ants. Consider a random disc of radius $r$ in this large configuration. The expected number of ants in this disc is $ d \pi r^2$ where $d$ is the density. The expected total happiness of the ants in the disc is $(\lambda-\epsilon) d \pi r^2$. The expected number of ants in the disc within a distance $15^{2/3}$ of the boundary is $ O ( d \pi r)$ and so the expected loss to the happiness of the ants in the disc from removing all the ants outside the disc is $O( d \pi r)$. So the expected total happiness of the ants in the disc, once the outside ants are removed, is $$ \geq (\lambda - \epsilon ) d \pi r^2 - O( d \pi r).$$ -So there must exist some configuration of $\leq n$ ants whose average happiness is at least -$$ \frac{ (\lambda - \epsilon ) d \pi r^2 - O( d\pi r)}{ d \pi r^2} = \lambda - \epsilon - O \left( \frac{1}{r} \right)$$ -We have $r \approx \sqrt{n}$ so this is $$\lambda - \epsilon - O \left(\frac{1}{\sqrt{n} }\right).$$ Taking $\epsilon$ to $0$, we see that there is a configuration of $m \leq n$ ants with happiness $\geq m \lambda - O\left( \frac{m}{\sqrt{n}}\right)$. -I claim that there is a configuration of exactly $N$ ants with happiness $\lambda N - O ( \sqrt{N})$. To see this, take $n_1= N$ and find a configuration of $m_1$ ants with average happiness $\geq \lambda - O( \frac{1}{\sqrt{n_1}})$, then repeat it $ \lfloor \frac{n_1}{m_1} \rfloor$ times, leaving room for $n_2 = n_1 - m_1 \lfloor \frac{n_1}{m_1} \rfloor$ ant. Then apply the previous existence result again to find a configuration of at most $n_2$ ants with average happiness $\geq \lambda - O ( \frac{1}{ \sqrt{n_2} })$ , and iterate. The total happiness is $$\geq \lambda N - \sum_i \left\lfloor \frac{n_i}{m_i} \right\rfloor O \left( \frac{m_i}{ \sqrt{n_i}}\right) \geq \lambda N - \sum_i O \left(\sqrt{n_i} \right) = \lambda N - O (\sqrt{N})$$ since $n_{i+1} < n_i/2$ so $n_i < N/ 2^{i-1}$.<|endoftext|> -TITLE: Is Qcoh(X) locally presentable? -QUESTION [5 upvotes]: Let $X$ be a scheme. Is the category $QCoh(X)$ of quasi-coherent sheaves on $X$ locally presentable? If so, can we say anything about the $\kappa$ for which $QCoh(X)$ is locally $\kappa$-presentable? (e.g. is it always finitely presentable? Or related to the $\kappa$ of Gabber's result?) -I'm particularly interested in the case where $X$ is quasi-comact quasi-separated (qcqs). - -In my searching for references, I've come across answers ranging from "we don't know", "when qcqs", to "always", and would appreciate some clarity. - -REPLY [10 votes]: Zariski descent tells us that -$$\operatorname{QCoh}(X)=\lim_{U\subseteq X} \operatorname{QCoh}(U)$$ -where $U$ ranges through all open affines and the limit is taken in the $(2,1)$-categorical sense. Since small limits of presentable categories are presentable and $\operatorname{QCoh}(\operatorname{Spec}R)=\operatorname{Mod}_R$ is presentable, we have that that $\operatorname{QCoh}(X)$ is presentable. -I'm not sure how to get bounds on the accessibility degree in general, though when $X$ is qcqs it is compactly generated (i.e. $\omega$-presentable) by the argument in this answer by Denis-Charles Cisinski. -PS: following recent trends in homotopy theory, I'm dropping the "locally" from "locally presentable'.<|endoftext|> -TITLE: Metrics of non-negative sectional curvature on $S^7$-bundles over $S^8$ -QUESTION [6 upvotes]: In Curvature and symmetry of Milnor spheres, Grove and Ziller construct metrics of non-negative sectional curvature on $S^3$-bundles over $S^4$ (by using a cohomogeneity one action). In the same paper, they ask whether this can be done in other dimensions (see Problem 5.1). Does anybody know whether there has been progress on this matter? I would particularily like to know if the total space of linear $S^7$-bundles over $S^8$ carry metrics of non-negative sectional curvature. - -REPLY [8 votes]: My understanding is that this is generally unknown. Of course, a few of the total spaces (e.g., $S^7\times S^8$, $S^{15}$, and the unit tangent bundle of $S^8$) are homogeneous spaces, so admit a non-negatively curved metric. -For the most interesting class of $S^7$ bundles over $S^8$ (the exotic $15$-dimensional spheres), more is known. First, none of these exotic spheres (regardless of whether or not they have the bundle structure) is a biquotient or a homogeneous space. This is a result of Totaro and independently by Kapovitch and Ziller. -Second, the cohomogeneity one approach of Grove and Ziller will not equip exotic $15$-spheres with non-negatively curved metrics. More generally, the generalization of this approach by Goette, Kerin, and Shankar (which recently established the existence of non-negatively curved metrics on all exotic $7$-spheres) cannot work on exotic $15$-spheres. The issue is that both approaches require the two singular leaves to both have codimension $2$. But Galaz-García, Kerin, and myself recently proved that a simply connected rational cohomology sphere $\Sigma^n$ has such a foliation only when $n\in \{2,3,5,7\}$.<|endoftext|> -TITLE: What is equivariant chains on a representation sphere? -QUESTION [6 upvotes]: For a finite group $G$ and a finite-dimensional real representation $V$ of $G$, denote by $S^V$ the one-point compactification of $V$, with basepoint at infinity. - -What is the reduced chain complex $C_*(S^V,\infty)$ as an object of the derived category of $G$-representations? - -Admittedly this is a somewhat open-ended question. One could regard $C_*(S^V,\infty)$ itself already as an "explicit" chain complex of $G$-representations. One can also write down a "more explicit" presentation of it in terms of the lattice of subgroups of $G$ and the subrepresentations of $V$ that they fix. Is there a nice succinct answer here, perhaps identifying $C_*(S^V,\infty)$ with another known object in the derived category $G$-representations which would be natural from other (e.g. purely representation theoretic) points of view? - -REPLY [2 votes]: Exercise 10 of section 1 of Chapter II of tom Dieck's book Transformation Groups gives you one answer to your question. It reads: -Let $S(V)$ be the representation sphere of a finite group $G$. Show directly that $S(V)$ admits a $G$-equivariant triangulation by looking at the convex hull of $\{\pm ge_i \ | \ g \in G; e_1, \dots, e_m \in S(V) \text{ basis for } V \}$. -So he is describing an easy-to-define $G$-CW structure on $S(V)$. Then $S^V$ will be two cones on $S(V)$ attached together, so one can easily read off its $G$-CW structure as well. -In degree $i$, the associated cellular chain complex $C^{CW}_*(S^V)$ will have one copy of the induced representation $1_H^G$ for each $i$ cell of the form $G/H \times D^i$. Exploring a detailed answer to his exercise will reveal a more detailed answer to your question, as the stablizers of families of the elements $e_i$ will relate to the representation $V$. \ No newline at end of file