diff --git "a/stack-exchange/math_overflow/shard_19.txt" "b/stack-exchange/math_overflow/shard_19.txt" deleted file mode 100644--- "a/stack-exchange/math_overflow/shard_19.txt" +++ /dev/null @@ -1,26593 +0,0 @@ -TITLE: Milnor Conjecture on Lie groups for Morava K-theory -QUESTION [18 upvotes]: A conjecture by Milnor state that if $G$ is a Lie group, then the map $B(G^{disc})\to BG$ sending the classifying space of $G$ endowed with the discrete topology to the classifying space of the topological group $G$ induces an isomorphism on homology with $\mod p$ coefficients. -In chromatic homotopy theory, there are more "fields of characteristic p" than just finite fields, namely we have the Morava $K$-theories $K(p,n)$. -Question: Do we know "more of" the Milnor conjecture for those ring-spectra then for $\mathbb{F}_p$? (ultimately, but probably too ambitious, can it be proven for those ring spectra even though it is still open for $\mathbb{F}_p)$? -By "more of" I mean any progress that is special for this case and don't work for $\mathbb{F}_p$ coefficients. - -REPLY [7 votes]: Consider a map $f\colon X\to Y$ of spaces (such as $B(G^{\text{disc}})\to B(G)$). Say that $f$ is a $K(n)$-equivalence if $K(n)^*(f)\colon K(n)^*(Y)\to K(n)^*(X)$ is an isomorphism. We will allow the case $n=\infty$ (corresponding to $K(\infty)^*(X)=H^*(X;\mathbb{F}_p)$) but not the case $n=0$ (corresponding to $K(0)^*(X)=H^*(X;\mathbb{Q})$. -Using the Atiyah-Hirzebruch spectral sequence -$$ H^i(Cf;K(n)^j) \Longrightarrow K(n)^{i+j}(Cf), $$ -where $Cf$ is the cofibre of $f$, we see that if $f$ is a $K(\infty)$-equivalence then it is a $K(n)$-equivalence for all $n$. -Conversely, suppose that $f$ is a $K(n)$-equivalence for $N -TITLE: Minimum number of double points over all immersed disks -QUESTION [7 upvotes]: Let $K$ be a knot in the boundary of a compact smooth 4-manifold $X$, and suppose that $K$ is the the kernel of $\pi_1(\partial X) \to \pi_1(X)$. Then $K$ is the boundary of some immersed disk $D \to X$ that has some fine number of double points. I am interested in knowing how to compute minimum number of such double points, minimized over all such immersed disks. For now, I will call it the immersion number of $K$ - denoted $I(K)$, for fun. Has this quantity been studied in the literature? -One thing to note about $I(K)$ is that it provides a lower bound on the minimum genus of an orientable surface bounding $K$ (the "4-ball genus" if $X = B^4$). I am interested in knowing examples of knots (even/especially with $X = B^4$) where $I(K)$ and the minimum genus of an orientable surface properly embedded in $X$ bounding $K$ are not equal. - -REPLY [4 votes]: For knots in $S^3=\partial B^4$ this is called the four-ball crossing number or clasp number. -For knots in $S^3$ you have the following inequalities relating $I$ to the $4$-ball genus $g_4$ and the unknotting number $u$ -$$g_4\leq I \leq u.$$ -In general, these are no equalities. For counterexamples, you can have a look at the Table of Knot Invariants.<|endoftext|> -TITLE: Are there continua in $\infty$-topoi? -QUESTION [13 upvotes]: If topology were invented for algebraic geometry or logic, in ignorance of Euclidean space, we might reasonably regard connected compact Hausdorff spaces as pathological, or even doubt their existence. After all, Hausdorffness is a "separation" condition whereas connectedndess is a "nonseparation" condition -- it would seem strange that they should be compatible if you'd only seen spectral spaces or Stone spaces but not Euclidean spaces. -Luckily though, we know about Euclidean space and spaces built from it, and their central importance to mathematics. But they are very special, being as they are balanced between separatedness and nonseparatedness. For instance, every connected compact abelian group is a torus. Every number field has finitely many infinite places -- which are mysterious in a way that finite places are not. And so forth. -When it comes to $\infty$-topos theory as a generalization of topology, I wonder what the infinite places are. All the generalized concepts are there: - -A proper geometric morphism (HTT 7.3.1.4) $\mathcal Y \to \mathcal X$ is one which satisfies a Beck-Chevalley condition, stably under pullback. - -So a compact $\infty$-topos $\mathcal X$ is one such that the global sections geometric morphism $t_\ast: \mathcal X \to Spaces$ is proper. -And a Hausdorff $\infty$-topos $\mathcal X$ is one such that the diagonal $\mathcal X \to \mathcal X \otimes \mathcal X$ is proper (here $\mathcal X \otimes \mathcal X$ is the Lurie tensor product, which is the product in the category of toposes by HTT 7.3.3.9). - -Generalizing from ordinary topos theory, an $\infty$-connected geometric morphism is one such that the left adjoint $t^\ast: Spaces \to \mathcal X$ to global sections is fully faithful. -Generalizing again from ordinary topos theory, a locally $\infty$-connected topos is one such that $t^\ast$ has a further left adjoint $t_!: \mathcal X \to Spaces$. - - -Question: -What's an example of a compact Hausdorff, $\infty$-connected, locally $\infty$-connected $\infty$-topos other than $Spaces$ itself? - -There are various ways to relax these conditions: - -If we don't require $\infty$-connectedness, then any étale topos (i.e. a slice $Spaces/X$) satisfies the remaining conditions; I'd be interested in examples other than these. -If we don't require Hausdorffness, then we have a weakening of the notion of a cohesive topos (i.e. a topos which is local (a strengthening of compactness) and strongly connected (a strenghtening of connected + locally connected)). My sense is that there are lots of cohesive toposes, so I think that Hausdorffness may be what sets this question apart from questions about cohesion. -We could relax compactness to local compactness or exponentiability or even do away with it altogether. I'd be interested in such examples. -... - -REPLY [12 votes]: Every contractible finite CW complex $X$ satisfies these conditions. This follows from results in Section 7.3 of HTT and Appendix A of HA: we have $Shv(X) \otimes Shv(X)=Shv(X\times X)$ since $X$ is locally compact, proper morphisms between locally compact Hausdorff spaces give proper morphisms of $\infty$-topoi (so $Shv(X)$ is compact and Hausdorff), contractible topological spaces have trivial shape (so $t^*$ is fully faithful), and CW complexes are locally of constant shape (so $t_!$ exists). -The Hilbert cube is a non-CW example. -A non-$0$-localic example is presheaves on an $\infty$-category with finite sums, or more generally any weakly contractible $\infty$-category $C$ such that $C$ and $C_{(x,y)/}$ (for all $x,y\in C$) admit coinitial functors from finite $\infty$-categories (these guarantee compactness and Hausdorfness, respectively).<|endoftext|> -TITLE: How to write K-theory Conner-Floyd Chern classes in terms of Adams operations? -QUESTION [10 upvotes]: From Adams, we know that the algebra of (unstable, degree-zero) cohomology operations $K^0(BU)$ can be written as formal infinite linear combinations of canonical generators -$$\mu_n := \sum_{i=0}^{n} (-1)^{n-k}\binom{n}{k} \psi^k$$ -However, from the collapse of the Atiyah-Hirzebruch spectral sequence for BU, we also know that $K^0(BU) \cong \mathbb{Z}[[c_1^K,c_2^K,\ldots ]]$ where $c_i^K$ are the Conner-Floyd Chern classes (where I'm renormalizing them to degree zero by an appropriate power of $t\in \pi_2KU$). -Thus, it should be possible to write the Chern classes in terms of the Adams operations. How can I find these expressions? -Doing the reverse is not so bad, using Hirzebruch's theory of genera: I get that $\psi^k$ is $(1+c_1^K+c_2^K+\ldots)^k$. But unfortunately I'm completing lacking in the power series wizardry that would allow me to invert this. -This was wrong, since the multiplication of characteristic series of genera does not correspond to anything over on the cohomology operations side, coinciding neither with cup product nor composition (which are themselves distinct). - -REPLY [10 votes]: I am not sure about the facts you mention, and I don't think I'll quite answer your question, but here are some facts I do know. -First, it is not the case that all $KU$-operations can be written as (even infinite) sums of Adams operations; Adams operations are additive, and general $KU$-operations need not be. So I won't say how to write the Chern classes in terms of Adams operations, but I'll try to say something about how they relate; I apologize if I am just repeating facts you are already familiar with. -One technical remark is that $KU$-operations aren't governed by $KU(BU)$, but really by $KU(\mathbb{Z}\times BU)$. I'll talk about $ku({-}) = [{-},BU]$ instead, which is valued in nonunital rings and has operations governed by $ku(BU) = \mathbb{Z}[[c_1,c_2,\ldots]]^+$, the augmentation ideal of $KU(BU)$. From now on, all my rings will be nonunital, but you can feel free to add a unit and think of them as augmented rings instead. I think my $c_i$'s are the same as yours; they are so that if we write $ku(BU(1)) = \mathbb{Z}[[u]]^+$ with $u=l-1$ with $l$ the tautological line bundle, then pullback under summation $ ku(BU)\rightarrow ku(BU(1)^{\times n})$ sends $c_i$ to the $i$'th elementary symmetric polynomial in $u_1,\ldots,u_n$. -Adams operations can be defined in any $\lambda$-ring, and $ku({-})$ is valued in $\lambda$-rings in the usual way. So you can write $\psi^n$ as a polynomial in $\lambda^1,\ldots,\lambda^n$, but not conversely. I think the relation is -$$ -t\frac{d}{dt}\log(1+\sum_{n\geq 1}\lambda^n(x) t^n) = \sum_{n\geq 1}(-1)^{n+1}\psi^n(x)t^n, -$$ -but I would not swear on the signs. Moreover, you can identify sums of Adams operations as exactly the additive operations acting on all $\lambda$-rings. Since this relates $\lambda$-operations and Adams operations, I'll just say how $\lambda$-operations relate to Chern classes. -For $\lambda$-rings like $ku({-})$ that are comprised of things like degree zero virtual bundles, it's useful to introduce the $\gamma$-operations. If I set $\lambda_t(x) = 1+\sum_{n\geq 1}\lambda^n(x)t^n$ for a formal variable $t$, and similarly define $\gamma_t$, then these are uniquely determined by asking first that $\gamma_t(x+y) = \gamma_t(x)\gamma_t(y)$, and second that if $\lambda_t(l)=1+lt$ then $\gamma_t(l-1)=1+(l-1)t$. You can explicitly relate these with the $\lambda$-operations via $\gamma_t = \lambda_{t/(1-t)}$, and a $\lambda$-ring structure is equivalent to a $\gamma$-ring structure. -Finally, a splitting principle argument lets you show that the class $c_n\in ku(BU)$ corresponds to the operation $\gamma^n$.<|endoftext|> -TITLE: Equivalence of idempotents and projective modules over nonunital rings -QUESTION [5 upvotes]: For a nonunital ring $R$ (or "rng") one has to be a little bit careful when considering the category of (left or right) $R$-module and, furthermore, the standard equivalent definitions of projective modules are in general not equivalent anymore (see e.g.https://math.stackexchange.com/questions/120458/projective-modules-over-rings-without-unit). To fix terminology: by projective module we mean the standard categorical definition https://en.wikipedia.org/wiki/Projective_module. -Let $M_n(R)$ denote the set of $(n\times n)$-matrices with entries from $R$ and let $e\in M_n(R)$ and $f\in M_m(R)$ be idempotents. It is easy to show that $M=eR^n$ and $N=fR^m$ are projective modules. For a unital ring, the statement that $M$ is isomorphic to $N$ is equivalent to saying that there exists $u\in GL_k(R)$ such that $e=ufu^{-1}$ (tacitly assuming an appropriate $k$ and an embedding for $e$ and $f$ into $M_k(R)$). Now, what is the corresponding statement for nonunital rings? -The usual algebraic definition of equivalence of idempotents state that $e\sim f$ if there exists $x,y$ such that $e=xy$ and $f=yx$. It is easy to see that this implies that $M\simeq N$ (multiplication by $y$ gives the isomorphism $M\to N$). Now, is the converse true? That is, if $M\simeq N$ then there exists $x,y$ such that $e=xy$ and $f=yx$? -The standard proof for unital rings uses the fact that a projective module is a direct summand of a free module, which may no longer be true for nonunital rings. - -REPLY [4 votes]: If $R$ is a ring (not necessarily unital) and $e\in R$ is an idempotent, then it is still the case that $Hom_R(eR,M)\cong Me$ for any right $R$-module $M$ via $\phi\mapsto \phi(e)$. It immediately follows that, if $e,f\in R$ are two idempotents, then $eR\cong fR$ if and only if there exists $a\in eRf$ and $b\in fRe$ with $ab=e$ and $ba=f$. This is equivalent to your condition once you notice that if $e=xy$ and $f=yx$, then putting $a=exf$ and $b=fye$ you have that $ab=exffye=exyxye=e$ and $ba=fyeexf=fyxyxf=f$. -Nothing here uses that you have a ring. You could work with semigroups and right $S$-sets. Then you are reducing to standard facts about Green's relations.<|endoftext|> -TITLE: Closed formulas for the character of the symmetric group -QUESTION [10 upvotes]: I know the Murnaghan–Nakayama rule, but I am wondering if there is any closed formulas for the character of the symmetric group. I know the following: -$$\chi_{n}(\sigma) = 1$$ -$$\chi_{11...1}(\sigma) = sgn(\sigma)$$ -$$\chi_{n-1,1}(\sigma) = fix(\sigma)-1$$ -$$\chi_{21...1}(\sigma) = sgn(\sigma)(fix(\sigma) - 1)$$ -Are they any other simple formulas like these? I know that the answer is no for the general case, but maybe there is in simple cases, like for the other hook partitions or for rectangle partition? -Thanks in advance! -Étienne - -REPLY [7 votes]: For generalizing the formulas in your question, see http://www.combinatorics.org/ojs/index.php/eljc/article/view/v16i2r19 and Examples 1.7.13 and 1.7.14 in Macdonald's Symmetric Functions and Hall Polynomials, 2nd ed. -For a different formula, see https://www.researchgate.net/publication/227299451_Stanley%27s_Formula_for_Characters_of_the_Symmetric_Group.<|endoftext|> -TITLE: Commutativity up to homotopy implies strict commutativity, for lifting problems -QUESTION [7 upvotes]: Suppose we have a commutative diagram -$\require{AMScd}$ -\begin{CD} -A @>>> X \\ -@VVV & @VVV \\ -W @>>> Y\\ -\end{CD} -where the map $A\rightarrow W$ is a cofibration and the map $X\rightarrow Y$ is a fibration. Suppose also that there exists a map $W\rightarrow X$ that makes the diagram commute up to homotopy. -Is then true that we can find a map $W\rightarrow X$ that makes the diagram strictly commute? - -REPLY [7 votes]: I believe the answer is yes. The kind of lift you're asking about was studied extensively in the paper "On Fibrant objects in model categories" by Valery Isaev. Apply Proposition 3.4, with $I = \{A \to W\}$. Then, because $f:X\to Y$ is a fibration, it has RLP with respect to $J_I$, because $J_I$ consists of trivial cofibrations. Hence, Proposition 3.4 says that, having RLP up to relative homotopy with respect to $I$ implies $f$ has RLP with respect to $I$, as you desire. -There may be older or more elementary proofs of this fact, but Isaev's paper is what sprung to mind. Also, it works in much more general settings than Top, and you might find lots of useful facts in it, for whatever you're working on.<|endoftext|> -TITLE: Terminology: algebraic structure for "floating point" arithmetic -QUESTION [12 upvotes]: "floating point arithmetic" is a terminology that refer to the arithmetic perform over (finite) representation of real number. See the wikipedia article for more details. -In the formal specification of floating point arithmetic (that should be used by all the major programming languages), it is specified that a "Not A number" (NaN) value should be a number. -If we abstract the fact that floating point arithmetic care only of finitely many values and abstract a bit, we get a sort of arithmetic arithmetic over $\mathbb{R}\cup\{\textrm{NaN}\}$ with $\textrm{NaN}$ being a zero for any arithmetic operation. Formally, for all $x$, -$$x\times \textrm{NaN} = x + \textrm{NaN} = \frac{x}{0} = \frac{0}{0}= \cdots =\textrm{NaN}$$ -Is there any algebraic structures $(E,+,\times)$ having axioms allowing this or is it just to arbitrary to have been introduced even in weird part of algebra? - -REPLY [12 votes]: The abstract picture here is that of adjoining an undefined or bottom element to a (partial) algebra. Doing this to a total algebra is boring, but it is a useful trick to turn partial algebras into total ones. -One just picks an element $\bot$ not in the carrier set, and extends the operations as follows: Any operation applied to a tuple outside its original domain yields $\bot$, including any tuple containing a $\bot$ somewhere. -The structure you are describing is then just the $1$-point totalirization of the partial algebra $(\mathbb{R},+,\times,/)$. The algebras arising in such a manner will hardly ever have very familiar properties. In particular, I don't think people name them directly, and would rather speak about the underlying partial algebras.<|endoftext|> -TITLE: Bounded deformation vs bounded variation -QUESTION [6 upvotes]: Let $BV(\mathbb R^n; \mathbb R^n)$ be the space of (vector-valued) functions of bounded variation and let $BD(\mathbb R^n;\mathbb R^n)$ the space of functions with bounded deformation. They are made up respectively of functions $u$ for which the full distributional derivative -$$ -Du \in \mathcal M(\mathbb R^n) -$$ -is represented by a measure with finite total variation and of the functions for which the symmetric part of the distributional derivative -$$ -Eu := \frac{Du+(Du)^t}{2} \in \mathcal M(\mathbb R^n) -$$ -is represented by a measure with finite total variation. -If $n=1$ of course the two definitions coincide. For $n\ge 2$ they are different, but I do not find an explicit example. - -Q. Let $n\ge 2$. Find an element in $BD \setminus BV$. - -Is a characterization of such functions available somewhere in the literature? - -REPLY [7 votes]: Example 7.7 in -L. Ambrosio, A. Coscia, Alessandra, G. Dal Maso, -Fine properties of functions with bounded deformation. -Arch. Rational Mech. Anal. 139 (1997), no. 3, 201–238.<|endoftext|> -TITLE: Central extensions of loop groups -QUESTION [8 upvotes]: Let $LG=\operatorname{Maps}(S^1,G)$ be the loop group of a compact Lie group $G$. I should add some adjectives to $G$, but for sake of simplicity let's just take $G=SU(2)$. -There is a central extension -$$1\to S^1\to\widetilde{LG}\to LG\to 1$$ -(these are classified by "level" in $H^3(G)$, but we may as well restrict attention to the "universal" such extension corresponding to a generator of this group). The constructions of this central extension that I have found so far (e.g. the one in Pressley--Segal) all go via first defining a closed $2$-form on $LG$, arguing it defines a unique $S^1$-bundle, and then putting a group structure on this bundle. - -Is there a more intrinsic definition of $\widetilde{LG}$? - -In other words, given a loop $\gamma:S^1\to G$, I would like to have an intrinsically defined principal $S^1$ homogeneous space (or, equivalently, a $1$-dimensional complex vector space). -For example, here is an answer "up to homotopy". Since $\pi_1(G)=\pi_2(G)=0$ and $\pi_3(G)=\mathbb Z$, given any loop $\gamma:S^1\to G$, the space of extensions $\bar\gamma:D^2\to G$ (i.e. $\bar\gamma|_{\partial D^2=S^1}=\gamma$) is homotopy equivalent to $\Omega^2G$ which is connected with fundamental group $\mathbb Z$. If we take the $1$-truncation of this space (add cells to kill all higher homotopy groups), we get $S^1$ (up to homotopy). -This gives an "intrinsically defined" space homotopy equivalent to $S^1$ defined in terms of a given loop $\gamma:S^1\to G$ (although it's not very explicit, and has questionable meaning/use). What about an honest one-dimensional complex vector space? (with a natural meaning). Even better, can we define intrinsically a holomorphic line bundle over the complexified loop group $LG_{\mathbb C}$? - -REPLY [6 votes]: Fix a cocycle $\omega\in C^3(G, \mathbb R)$ such that $\omega(H_3(G, \mathbb Z)) = \mathbb Z$. (For $G = SU(2)$, we can take $\omega$ to be the standard volume form.) Fix a loop $L$ in $G$, and let $\partial^{-1}(L) \subset C_2(G, \mathbb Z)$ denote the set of 2-chains whose boundary is $L$. For $x,y \in \partial^{-1}(L)$, define an equivalence relation $x \sim y$ if $\omega(\partial^{-1}(x - y)) \in \mathbb Z$. Then $\partial^{-1}(L)/\sim$ is an $S^1$ torsor. -The above paragraph defines an $S^1$-bundle over $LG$, but it doesn't tell you how to multiply two elements of this bundle. For this, we use the fact that $\pi_2(G)$ is trivial and redefine $\partial^{-1}(L)$ to be maps of $D^2$ into $G$ which restrict to $L$ on the boundary. We can multiply the maps $D^2 \to G$ pointwise. This group structure will extend to the quotient if we choose $\omega$ to be invariant under left and right multiplication by elements of $G$. (I have not thought about this last claim as carefully as I should, so be skeptical here.) -Assuming I did not make a stupid error in the previous paragraph (time constraints!), we can summarize as follows: Elements of $\widetilde{LG}$ are represented by maps to $D^2$ to $G$, and two such maps are considered equivalent if the $\omega$-volume they cobound is an integer.<|endoftext|> -TITLE: Determining the conjugacy classes of a wreath product $G \wr S_n$ -QUESTION [5 upvotes]: If $G$ is a finite group and its conjugacy classes are known, can the conjugacy classes of the wreath product $G \wr S_n \cong G^n \rtimes S_n$ be determined? - -REPLY [5 votes]: This is indeed handled in Section 4.2 of James and Kerber's book on Representations of the Symmetric group. -We also considered this problem in Section 2 our paper describing a computer algorithm for computing conjugacy class representatives in permutation groups. -J. Cannon and D. Holt, Computing conjugacy class representatives in permutation groups, -J. Algebra 300 (2006), 213--222. -I can cut and paste the main bit of theory you need. The problem easily reduces to finding those classes that map onto a specific element of $S_n$. -Let $A$ be any group, let $P$ be a permutation group acting -on the set $\{ 1\ldots d\}$, and let $W := A \wr P$. -Then $W$ is a semidirect product of $A^d$ by $P$, and elements -of $W$ have the form $(g,x)$ with $g \in P$, $x \in A^d$, and -$x = (x_1,\ldots,x_d)$ with $x_i \in A$. In general, for $x \in A^d$, -we denote the $i$-th component of $x$ by $x_i$. -The action of $P$ on $A^d$ in $W$ is given by: -$$(g,1)^{-1}(1,(x_1,\ldots,x_d))(g,1) = (1,(x_{1^{g^{-1}}},\ldots, -x_{d^{g^{-1}}})).$$ -Hence, for $g \in P$, $x,z\in A^d$, we have -\begin{equation}\label{conjeqn} -(1,z)^{-1}(g,x)(1,z) = (g,y), \ \mathrm{where}\ -y_i = z^{-1}_{i^{g^{-1}}}x_iz_i \ \mathrm{for}\ 1 \le i \le d. -\end{equation} -$\textbf{Theorem}$ With the above notation, -let $x^{(1)},\ldots,x^{(k)}$ be representatives of the conjugacy classes of $A$. -Fix an element $g \in P$, and let $r_1,r_2,\ldots,r_s$ be representatives of -the cycles of $g$ in its action on $\{ 1\ldots d\}$. Then -$$ -R := \{\,(g,x) \mid x_i = 1 \ \mathrm{for}\ i \not\in \{r_1,\ldots,r_s\},\, - x_i \in \{x^{(1)},\ldots,x^{(k)}\}\ \mathrm{for}\ -i \in \{r_1,\ldots,r_s\}\,\} -$$ -is a set of representatives of the $A^d$-classes of elements of $W$ -of the form $(g,x)$ for $x \in A^d$. -The size of the $A^d$-class $R_{gx}$ containing $(g,x)$ is the product of the sizes of the classes of the $x_i$ in $A$ with $i \in \{r_1,\ldots,r_d\}$ with the lengths of the cycles of $g$ that contain $r_1,\ldots,r_d$. -The above describes the $A^d$-classes of $W$; i.e. the orbits under conjugation by the subgroup $A^d$ of $W$. To get the classes of $W$ itself, first of all we only consider those $R_{gx}$ as $g$ ranges over a set of class representatives of $P$. Then, for each such $g \in W$, the $A^d$-class representatives $R_{gx}$ are fused under the action of $C_P(g)$. This action is the same as the action of $C_P(g)$ on the cycles of $g$, so is easy to calculate. -Then the size of a $W$-class $R_{gx}$ is the product of the size of the class of $g$ in $P$ with the sum of the the sizes of the $A^d$-classes $R_{gx'}$ that lie in the orbit under the action of $C_P(g)$.<|endoftext|> -TITLE: Orientations of Planar Graphs -QUESTION [10 upvotes]: Let $G$ be a $2$-edge-connected graph drawn in the plane (such that the edges intersect only at the endpoints). I want to orient -the edges of $G$ such that for each vertex $v$, there are no -three consecutive edges (in the clockwise direction) such that all of them -are oriented towards $v$ or all of them are oriented outwards from $v$. -Does such orientation always exist? - -REPLY [10 votes]: Such an orientation always exists, here is a proof. -Take your 2-edge-connected graph $G$, and consider its dual graph $D$. $D$ has a proper 4-coloring in which each face of $D$ contains at most 3 different colors (add a vertex inside each face of $D$, connect it to all the vertices of the face, and apply the four color theorem to the resulting graph). Now, orient each edge of $D$ from the smaller color to the larger color. Note that there is no facial directed path on more that 2 edges in $D$ (otherwise, this would be a path with all 4 colors). Now, transfer the orientation of the edges of $D$ to the edges of $G$ in the natural way, and you get the desired result. -(in the first version of this post, the proof only gave that in 4-edge-connected plane graphs, you can find the desired orientation, and in 2-edge-connected plane graphs, you can find an orientation in which no four consecutive edges around a vertex have the same orientation)<|endoftext|> -TITLE: Products, coproducts and equalizers in category of lattices -QUESTION [7 upvotes]: Background: Let $\mathbf{Lat}$ be the 2-category of lattices which can be viewed as a subcategory of the 2-cateogry of posets $\mathbf{Pos}$, that is, objects in $\mathbf{Pos}$ that have all finite products and coproducts (a.k.a meets and joins in lattice-speak); we may or may not require 2-morphisms in $\mathbf{Lat}$ to preserve meets and joins (i.e continuous and cocontinuous). -I am considering (what I call) cellular sheaves valued in lattices which are just functors $F: X \rightarrow \mathbf{Lat}$ where $X$ is the face relation poset of a cell complex. In order to do "sheaf theory" with sheaves valued in $\mathbf{Lat}$, it would be nice to have a notion of a coproduct and product in this category. I think product is fairly clear: just use the product partial order; meets and joins are what you think they would be. -As far as a coproduct, I am not sure. -If anyone has any suggestions? I have heard of a "free product of lattices" but it is not defined in a language I can understand. Not even clear to me that the "free product" that Gratzer defines is unique. -By "sheaf theory", I mainly mean taking limits and colimits over all the stalks. Another property I would like (or like to know doesn't hold) is the existence of all equalizers and (maybe if coproducts exist) coequalizers which would guarantee the existence of all small (co)limits. That would be lovely. -In general, looking for references on lattice theory, "non-abelian" sheaf theory, or anything about categories and lattice theory. -Thanks in advance! - -REPLY [4 votes]: I'm not sure about sheaf theory, but limits and colimits in categories of lattices are routine to construct. You just have to be clear about your categorical setup. Consider the following categories: - -$sLat^{\vee}$, the category of posets with $(\vee,\bot)$ and morphisms which preserve these. -$Lat$, the category of posets with $(\vee,\wedge,\bot,\top)$ and morphisms which preserve these. -$DistLat$ the full subcategory of $Lat$ on the lattices where $\wedge$ distributes over $\vee$. -... - -and variants like - -$Lat^{unbbd}$, the category of posets with $(\vee,\wedge)$ and morphisms which preserve these. -$Lat^{\uparrow bdded}$, the category of posets with $(\vee,\wedge,\top)$ and morphisms which preserve these. -... - -In each case, we have a variety in the sense of universal algebra, i.e. a category where an object is a set with some $n$-ary functions on it satisfying some universal equations (note that the poset structure can be recovered from $\vee$ or $\wedge$) and morphisms being functions which commute with all the functions. Other examples are categories like groups, rings, etc. -In any such category $\mathcal C$, one has all limits and colimits. They can be constructed in the following way. There is a forgetful functor $U: \mathcal C \to Set$. This functor has a left adjoint $F: Set \to \mathcal C$, which just sends a set to the set of all words in the function symbols, quotiented by the universal relations that hold in objects of $\mathcal C$. For example in the case of groups, $F(X)$ is the free group on the set $X$. -This adjunction $F \dashv U$ yields a monad $UF: Set \to Set$ from which the category $\mathcal C$ can be recovered as the category of algebras. Limits and colimits can be constructed in a standard way. For limits, take the limit of the underlying sets and extend the functions in the obvious way. Colimits are a bit more complicated, but you can read about them here. Filtered colimits, though, are easy -- just take the colimit of the underlying sets and extend the operations in the obvious way (these are what you need to compute stalks). Reflexive coequalizers are likewise computed at the level of the underlying sets. The trickiest class of colimits are coproducts. The coproduct $\amalg_i C_i$ is constructed by taking $F(\amalg_i U(C_i))$ and quotienting by an equivalence relation (for instance, the coproduct of groups is the free amalgam). Alternative descriptions may be available depending on $C$ -- for example see here.<|endoftext|> -TITLE: Which 3-manifolds are known to admit exotic pairs of bounding 4-manifolds? -QUESTION [10 upvotes]: Let $M$ be a compact connected three manifold. By an exotic pair of bounding 4-manifolds, I mean two smooth 4-manifolds $X_1,X_2$ such that $X_1$ and $X_2$ are homeomorphic but not diffeomorphic, and $\partial X_1$ and $\partial X_2$ are homeomorphic to $M$. -I imagine that all 3-manifolds have exotic pairs of bounding 4-manifolds. In fact, I would imagine that we could just take some exotic pair of closed 4-manifolds $W_1, W_2$ and any bounding 4-manifold $X$ for $M$ and then just take $X_1 = X \sharp W_1$ and $X_2 = X \sharp W_2$. -Does this process always produce an exotic pair for $M$? -Do all 3-manifolds have an exotic pair that they bound? - -REPLY [11 votes]: Here is a list of 3-manifolds $Y$ that are boundaries of exotic 4-manifolds https://arxiv.org/pdf/1901.07964.pdf - -If either $Y$ or $-Y$ (i.e with reverse orientation) has a contact structure with non-trivial contact invariant. - -If $Y$ or $-Y$ has weak symplectic filling. - -If $Y$ bounds both positive and negative definite 4 manifolds. - - -In case of 1) and 2) those manifolds bounds simply-connected manifolds with infinitely many exotic structures and in case of 3rd case we cannot gurantee the simply-connected condition. -The above classes cover all Seifert fibered 3-manifolds, all 3 manifolds that admits Taut folitaion, all irreducible 3 manifolds with 1st Betti number strictly bigger than zero or $M\# -M$. -Conjecturally we covered all the irreducible 3-manifolds in the above 3 cases. One obstruction when dealing with reducible manifold is that most of the 4-manifolds invariant vanishes under connected sum. So it is still an open problem if all 3-manifolds bounds exotic 4-manifolds. Hope in some near future we (or someone else) will find some clever way to deal with all 3-manifolds. -In Theorem 1.13 above we gave a general construction which holds for every 3-manifolds (because all 3-manifolds admit contact structures). But we do not know how to prove that they all are not diffeomorphic in general.<|endoftext|> -TITLE: Young's natural representation of the symmetric group -QUESTION [18 upvotes]: The literature on the representation theory of the symmetric group contains some terminology that I find puzzling, and I am wondering if someone here knows the full story. -One of the standard ways to construct the irreducible representations of the symmetric group is to define Specht modules. This construction produces an explicit basis for the modules. If one now writes down the representing matrices with respect to this basis, the result is often referred to as Young's natural representation. -From a modern point of view at least, this terminology seems a little strange because it attributes "the same thing" to both Specht and Young. Now one possible explanation is that when Specht and Young were working on this stuff, it didn't seem like the same thing, and it's only today that they look the same. Indeed, in Specht's paper, he writes: - -Zwischen den Youngschen Arbeiten und der vorliegenden bestehen daher kaum - irgendwelche Zusammenhaenge ausser den rein ausserlichen, die darauf - beruhen, dass die hier verwendeten kombinatorischen Hilfsmittel haufig - auch von Herrn A. Young, freilich zu ganz anderen Zwecken herangezogen - werden. - -My German is poor but I think this translates to: - -Between Young's work and the present work there exist hardly any - connections except the purely superficial one that the combinatorial tools - used here are also used by Mr. A. Young, albeit for entirely different - purposes. - -I tried to look up Young's papers, but found them daunting, and in particular I could not immediately locate anything that looked like "Young's natural representation." Apparently I'm not the only one who is daunted by Young's papers, because here's a quote from some lecture notes of G. D. James: - -The representation theory of the symmetric groups was first studied - by Frobenius and Schur, and then developed in a long series of papers - by Young. Although a detailed study of Young's work would undoubtedly - pay dividends, anyone who has attempted this will realize just how - difficult it is to read his papers. The author, for one, has never - undertaken this task, and so no reference will be found here to any - of Young's proofs, although it is probable that some of the techniques - presented here are identical to his. - -So my question is, can someone point specifically to a place in Young's papers where he discussed what we would nowadays call "Young's natural representation"? And does anyone know the history of how the term "Young's natural representation" came to have its current meaning? - -REPLY [8 votes]: Thanks to Richard Stanley for the pointer to Garsia and McLarnan's paper, Relations between Young's natural and the Kazhdan–Lusztig representations of $S_n$, Advances in Math. 69 (1988), 32–92. -Young's fourth paper ("QSA IV") is: - -Alfred Young, On Quantitative Substitutional Analysis (Fourth Paper), - Proc. London Math. Soc. (2) 31 (1930), no. 4, 253–272. - -Note that the year of publication is slightly confusing because the running head of the paper itself says "Nov. 14, 1929" but the volume of the journal was actually published in 1930. Following MathSciNet, I have given the year as 1930, but I have also seen citations of the paper that give 1929 as the year. -I have stared at QSA IV for some time but have failed to fully decipher the notation, so I hesitate to personally vouch for the claim that it describes the same matrix representation that one gets by taking (the usual basis for) Specht modules. However, in addition to Garsia and McLarnan, the book Substitional Analysis by Daniel Rutherford—which by the way is a very useful guide to Young's work—also states that QSA IV describes a recipe for (what we now call) Young's natural representation, so I believe that this claim is true. -It is understandable to me that Specht regarded his work as different from Young's. What I can say from my (limited) understanding of QSA IV is that Young did not construct anything resembling Specht modules, and that Young's recipe for constructing representing matrices came from considering the action of the symmetric group on (what we would now call) primitive idempotents. -There is an interesting remark that Garsia and McLarnan make in their paper (writing in 1988): - -Very few authors today have much familiarity with Young's natural representation. The various presentations of Specht modules and the work of Garnir tend to hide the simplicity and beauty of Young's construction. … Young's natural can be constructed at once by a very simple combinatorial procedure which applies to all permutations. Moreover, the proof that the procedure is valid is actually quite short and elementary. - -One reason that Young's construction of his natural representation "fell off the radar" for a while may be that the exposition in Rutherford's aforementioned book does not follow Young's construction exactly. Young derives the natural representation first and only later derives the orthogonal representation, whereas Rutherford does it the other way around, making the natural representation seem like an afterthought. In his review of Rutherford's book in the Bulletin of the American Mathematical Society, G. de B. Robinson even goes so far as to say: - -[T]he natural representation appears as an anti-climax. Though reference to it had to be included, this reviewer would have preferred that it be in an appendix. The material of §§28–31 has historical and actual value, but it serves to obscure the magnitude of Young's real achievement, the orthogonal representation. - -I took a quick look at Robinson's own book on the representation theory of the symmetric group and I think he does not bother at all with Young's natural representation. Anyway, it seems that for the reader who wants to understand Young's natural representation without going through Specht modules, Garsia and McLarnan's account is the most readable one.<|endoftext|> -TITLE: Proof of ¬(¬1 ⊗ ¬1) in tensorial logic -QUESTION [6 upvotes]: I believe I once had a proof of this proposition, but it's been lost to the mists of time and old hard drives, so who knows if it was correct, and try as I might I can't seem to reproduce it. -Is it possible, in Melliès' tensorial logic, to give a proof of ¬(¬1 ⊗ ¬1) (a.k.a. 1 ⅋ 1)? Equivalently, is there an arrow in a dialogue category (with monoidal unit 1 and negation functor ¬) from ¬1 to 1? - -REPLY [6 votes]: Your notation is a bit confusing because usually in linear logic and related systems $\top$ denotes the unit of additive conjuction (i.e., categorically, the terminal object), but then when you formulate your question in categorical terms you clearly say that by $\top$ you mean the monoidal unit, which instead is traditionally denoted by $1$ (this is the notation used by Melliès, for instance in his habilitation thesis, see p.83-84). -So, the answer depends on what is $\top$. If it is the terminal object, then $\lnot(\lnot\top\otimes\lnot\top)$ is indeed provable; if it is the monoidal unit (which I am going to denote by $1$ here), then $\lnot(\lnot 1\otimes\lnot 1)$ is not provable. -In the former case, here's a proof: -$$ -\begin{array}{rcl} -\lnot\top & \vdash & \top \\ -\hline -\lnot\top,\lnot\top & \vdash & \bot \\ -\hline -\lnot\top\otimes\lnot\top & \vdash & \bot \\ -\hline -& \vdash & \lnot(\lnot\top\otimes\lnot\top) \\ -\end{array} -$$ -In the latter case, non-provability comes from the non-provability of $1⅋1$ in linear logic without mix. If you want to see it directly in tensorial logic, simply observe that, by cut-elimination (which holds in tensorial logic) and reversibility of the $\lnot$-right and $\otimes$-left rules, derivability of $\vdash\lnot(\lnot 1\otimes\lnot 1)$ is equivalent to derivability of $\lnot 1,\lnot 1\vdash\bot$, which may only come from a proof of $\lnot 1\vdash 1$, which is unprovable because no rule (except cut) admits such a sequent as its conclusion, as shown by a straightforward inspection of the various rules (cf. for instance p.84 of Melliès habilitation thesis).<|endoftext|> -TITLE: A binomial determinant formula: a new variant -QUESTION [9 upvotes]: In a previous MO question, the OP asks a proof for $\det_{1\leq i,j\leq n}\left(\binom{i}{2j}+\binom{-i}{2j}\right)=1$. Subsequently, Gjergji Zaimi generalized the problem to -$$\det_{1\le i,j\le n}\left( \binom{x_i}{2j}+ \binom{-x_i}{2j}\right)=\prod_{i=1}^n x_i^2 \prod_{i -TITLE: "Strange" proofs of existence theorems -QUESTION [29 upvotes]: This question isn't related to any specific research. I've been thinking a bit about how existence theorems are generally proven, and I've identified three broad categories: constructive proofs, proofs involving contradiction/contrapositive, and proofs involving the axiom of choice. -I'm convinced that there must be some existence theorem that can be proven without any of these techniques (and I'm fairly confident that I've probably encountered some myself in the past haha), but I can't come up with any examples at the moment. Can anyone else come up with one? I'd also like to stipulate the following conditions: - -The proof can't piggyback on another existence theorem whose proof involves one of the above-mentioned devices. -It has to be a theorem of ZF - no exotic and "high power" axioms allowed! - -Now for the interesting question: is there any existence theorem (again in ZF) such that every one of its proofs is of this type? Has anyone investigated something like this? If so, what results exist? -Edit: -This issue came up a few times in the comments: here I use "constructive" in its weaker sense (i.e. a constructive proof is merely one that constructs an object and shows that it satisfies the required properties). The stronger sense - that the proof may not use the law of excluded middle or involve any infinite objects - is not what I'm invoking. - -REPLY [2 votes]: C. De Lellis & L. Székelyhidi proved the existence of weird solution of the Euler solutions for an incompressible perfect fluid. These solutions violate the conservation of energy in an arbitrary way. The proof, based on so-called "convex integration" uses in a crucial way Baire's category theorem. Notice that the result does not depend upon the axiom of choice. Whether the proof is constructive depends on how much you consider Baire as a constructive argument.<|endoftext|> -TITLE: Non-existence of continuous extension of continuous linear operator defined on non-dense subspace -QUESTION [7 upvotes]: Bounded Extension from Dense Subspace Theorem. Suppose that $Μ$ is a dense subspace of a normed space $X$, that $Y$ is a Banach space, and that $T_0: Μ \to Y$ is a bounded linear operator. Then there is a unique continuous function $T: X \to Y$ that extends $T_0$. This function $Τ$ is a bounded linear operator, and $\|Τ\| = \|T_0\|$. -Megginson's Introduction to Banach Space Theory (and several other books) points out that if $M$ is not dense in $X$, then there might not exist a continuous extension of $T_0$ at all. The example given is this: - -If $X=\ell^{\infty}$, $M=Y=c_0$, and $T_0 = Id:c_0 \to c_0$, then $T_0$ cannot be continuously extended to an operator from $X \to Y$. - -However, verifying this example is not exactly trivial. Megginson obtains it as a corollary of non-trivial theorem of Philips which says: - -$c_0$ is an uncomplemented closed subspace of $\ell^{\infty}.$ - -Philips original proof is difficult. A shorter but still non-trivial proof was published by Whitley. Megginson's book gives Whitely's proof and a precise reference. -Here is an answer that gives two other examples of closed uncomplemented subspaces of Banach spaces: https://math.stackexchange.com/a/108289/570438 . -But again the proofs are not easy. -I wonder if these examples are a bit of overkill. -Question A Is there a simpler example of Banach spaces $X$ and $Y$, a non-dense subspace $M$ in $X$, and a bounded linear operator $T_0:M \to Y$ such that $T_0$ cannot be continuously extended to an operator from $X \to Y$? In particular, is there an example that doesn't require first showing that $M$ is uncomplemented in $X$? -Question B Suppose we have Banach spaces $X$ and $Y$, a non-dense subspace $M$ in $X$, and a bounded linear operator $T_0:M \to Y$ such that $T_0$ cannot be continuously extended to an operator from $X \to Y$. Does that imply $M$ is uncomplemented in $X$? -Note: This question is a modified cross-post of https://math.stackexchange.com/questions/2928488/simple-example-that-density-of-the-subspace-cannot-be-omitted-from-the-bounded-e - -REPLY [5 votes]: The examples are not overkill. There isn't easy way to show that a closed subspace is not complemented. For instance, $\ell_2$ isn't complemented in $L_1$ but again the proof isn't easy. The example you mention is probably the easiest one. There are also examples of Orlicz sequence spaces which contain some $\ell_p$ but not complemented. The proof is 'elementary' (but not easier than your $c_0$ example) if you are familiar with Orlicz spaces, see Example 4.c.6 in Lindenstrauss-Tzafriri's book -However, the only spaces all of whose subspaces are complemented are the ones isomorphic to a Hilbert space. This is a beautiful theorem of Lindenstrauss and Tzafriri. So every non-Hilbertian space has a subspace which is not complemented.<|endoftext|> -TITLE: Is the following weak version of second Hardy-Littlewood conjecture already known? -QUESTION [5 upvotes]: Very recently I was going through my previous MSE posts and I stumbled upon some of them regarding the Second Hardy-Littlewood Conjecture which states that, - -For all $x,y\ge 2$ we have, $$\pi(x)+\pi(y)\ge \pi(x+y)$$where $\pi$ is the Prime Counting Function. - -Observing that the Second Hardy-Littlewood Conjecture is equivalent to the following, - -For all $k\ge 1$ and $y\in \mathbb{R}$ the following holds, $$\pi(ky)+\pi(y)\ge \pi((k+1)y)$$where $\pi$ is the Prime Counting Function. - -I was wondering whether proving the following weak version of the conjecture is something significant or is it already known, - -Proposition. For all $k\ge 1$ there exists $M_{k}>0$ such that for all $y\ge M_{k}$ we have, $$π(ky)+π(y)>π((k+1)y)$$ - -I searched the internet for something similar to this but even though I found some results closely related to the above, I could't find this exact result. -So my questions are, - -Is the above proposition well-known? If so can anyone point me out to the paper/book that contains a proof of it or even better to some theorem from which this result follows? -If not then is the proof of this result considered a significant result? - -I agree that the second question may seem to be more opinionated but currently this is the best version that I can come up with. I will be glad to receive constructive suggestions on making this question more specific and answerable - -REPLY [8 votes]: I have not seen your proposition before, but it follows routinely from the prime number theorem. -Indeed, there exists an absolute constant $c>0$ such that -$$\pi(ky)+\pi(y)-\pi((k+1)y)=\mathrm{Li}(ky)+\mathrm{Li}(y)-\mathrm{Li}((k+1)y)+O_k\left(y e^{-c\sqrt{\log y}}\right),$$ -where -$$\mathrm{Li}(ky)+\mathrm{Li}(y)-\mathrm{Li}((k+1)y)=\int_0^y\left(\frac{k}{\log(kt)}+\frac{1}{\log t}-\frac{k+1}{\log(k+1)t}\right)dt.$$ -For fixed $k$ and $t\to\infty$, the integrand is -\begin{align*}&\frac{k}{\log t}\left(1-\frac{\log k+o(1)}{\log t}\right)+\frac{1}{\log t}-\frac{k+1}{\log t}\left(1-\frac{\log(k+1)+o(1)}{\log t}\right)\\[8pt]&=\frac{(k+1)\log(k+1)-k\log k+o(1)}{\log^2 t},\end{align*} -whence there exists $C_k$ such that -$$\mathrm{Li}(ky)+\mathrm{Li}(y)-\mathrm{Li}((k+1)y)\gg_k\frac{y}{\log^2 y},\qquad y\geq C_k.$$ -As $\log^2 y$ grows much slower than $e^{c\sqrt{\log y}}$, we conclude that there exists $M_k$ such that -$$\pi(ky)+\pi(y)-\pi((k+1)y)\gg_k\frac{y}{\log^2 y},\qquad y\geq M_k.$$<|endoftext|> -TITLE: Corollary for Casselman-Shalika formula -QUESTION [7 upvotes]: Assume $\pi$ is an unramified representation of $GL_n(F)$, where $F$ is a p-adic field. And $\phi$ is an unramified vector for $\pi$. Assume $W_{\phi}$ is a Whittaker function associated to $\phi$. Then whether $|W_{\phi}(1)|^2=\frac{1}{L(1,\pi,Ad)}$? -I will appreaciate if you can show me a detailed calculation of the equality in an specific example. Thank you all the time. - -REPLY [7 votes]: The answer depends on your normalisation of $W$. Perhaps the most standard normalisation of $W$ is such that $W(1_n) = 1$, where $1_n$ is the $n \times n$ identity matrix. -Let $\pi$ be a unitary spherical representation of $\mathrm{GL}_n(F)$ with Satake parameters $\alpha_1,\ldots,\alpha_n$. Let $\widetilde{W}$ be the spherical vector of the contragredient representation $\widetilde{\pi}$. Then it is true that $$\langle W, \widetilde{W} \rangle = \int_{N_{n - 1}(F) \backslash \mathrm{GL}_{n - 1}(F)} W\begin{pmatrix}g&0\\ 0&1\end{pmatrix} \widetilde{W}\begin{pmatrix}g&0\\ 0&1\end{pmatrix} \, dg$$ -is equal to -\[\frac{L(1,\pi \otimes \widetilde{\pi})}{\zeta_F(n)} = (1 - q^{-n}) \prod_{j = 1}^{n} \prod_{k = 1}^{n} \frac{1}{1 - \alpha_j \alpha_k^{-1} q^{-1}}.\] -Here $q$ is the cardinality of the residue field of $F$. -In fact, a more general statement is true, where $\pi$ is allowed to be ramified and $W$ is the newform. A proof appears in Section 7 of "Large sieve inequalities for $\mathrm{GL}(n)$-forms in the conductor aspect" by Akshay Venkatesh. -There is another nice method to prove this, which is to use the method of Michitaka Miyauchi in "Whittaker functions associated to newforms for $\mathrm{GL}(n)$ over $p$-adic fields", using Theorem 4.1, which is the Shintani-Casselman-Shalika formula for newforms, together with Cauchy-Schur identities, since the integral over $N_{n - 1}(F) \backslash \mathrm{GL}_{n - 1}(F)$ reduces to an integral over -\[\{\mathrm{diag}(\varpi^{f_1},\ldots,\varpi^{f_{n - 1}},1) : f_1 \geq \cdots \geq f_{n - 1} \geq 0\},\] -where $\varpi$ is a uniformiser for $F$. -Remarkably, such a result is also true, once suitably modified, for archimedean $F \in \{\mathbb{R},\mathbb{C}\}$, even if $\pi$ is ramified. For $\mathrm{GL}_2$, this is quite standard; for $\mathrm{GL}_n$ and $\pi$ unramified, this is due to Stade; for $\mathrm{GL}_n$ and $\pi$ ramified, this is Theorem 5.10 of my paper with Yeongseong Jo based on recent work of mine.) -All this is the adèlic version of the more classical statement that -\[\frac{|a_f(1)|^2}{\langle f,f\rangle} = \frac{1}{2\Lambda(1, \operatorname{ad} f)}\] -for an even Maass form $f$ on $\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}$ with first Fourier coefficient $a_f(1)$.<|endoftext|> -TITLE: Which knots are singularities of a hyperbolic cone-manifold structures on $S^3$? -QUESTION [14 upvotes]: Which knots $K\subseteq S^3$ are such that there is a hyperbolic cone-manifold structure on $S^3$ that has (exactly) $K$ as the singular locus? -What if in Question 1 we restrict the cone angles to be $\leq \pi$? -Is it true that, if $M$ is a cone-manifold as in question 2, its volume is at most the simplicial volume of $S^3-K$ times $v_3$? - -I would also appreciate partial answers: it is known of some knots that aren't singularities of hyperbolic cone-manifold structures? Is it known of some non-hyperbolic knots that are? What is known about their volume? - -REPLY [6 votes]: Let me add some remarks in addition to Roberto's answer. -Question 2. has the same answer as Question 1: a knot is the singular locus of a hyperbolic cone metric iff the knot has hyperbolic complement (complete finite-volume hyperbolic metric) iff there is a cone metric with angle $\leq \pi$. This follows from Thurston's hyperbolic Dehn filling theorem: the complete metric may be thought of as a metric with cone angle zero, and the angle may be perturbed to be $\epsilon < \pi$. -For 3., the volume relation holds whether or not the cone angles are $\leq \pi$. When the angles are $\leq \pi$, Kojima shows that the cone angle may be continuously deformed to $0$, and that the volume increases during this deformation by Schlafli's formula. As far as I know, an analogous deformation result is still unknown for cone angles $\geq \pi$. However, there is a global comparison result which implies that the volume is at most the simplicial volume of the knot complement. This follows from the "natural map" technique of Besson-Courtois-Gallot. One may also prove this using Gromov's approach of measurable cycles and simplicial volume (Ben Klaff did this in his thesis).<|endoftext|> -TITLE: Global regularity for Neumann problem -QUESTION [7 upvotes]: Let $\Omega\subset \mathbb{R}^d $ be a bounded open subset ($d\in \mathbb{N}$) and denote $\partial\Omega$ its boundary which we assume to be Lipschitz. The classical inhomogeneous Neumann problem for Laplace operator associate to data $f:\Omega\to\mathbb{R}$ and $g: \partial\Omega \to \mathbb{R}$ (measurable functions) consists on finding a function $u:\Omega\to \mathbb{R}$ satisfying -\begin{equation}\label{eqlocal-Neumann}\tag{$N_1$} --\Delta u = f \quad\text{in}~~~ \Omega \quad\quad\text{ and } \quad\quad \frac{\partial u}{\partial \nu}= g ~~~ \text{on}~~~ \partial \Omega. -\end{equation} - -In the standard setting one usually choose $ f$ in $L^2(\Omega)$ or in the dual space of $H^1(\Omega)$ and $g$ can be choose in the trace spaces of $H^1(\Omega)$ denote by $H^{\frac{1}{2}}(\partial\Omega)$ or in its dual $H^{-\frac{1}{2}}(\partial\Omega)$. -Assume $f\in L^{2}(\Omega)$ and $g \in H^{1/2}(\partial\Omega)$. We have the following Green-Gauss formula -$$\label{eqgreen-Gauss} -\int_{\Omega} (-\Delta) u v \, \mathrm{d}x = \int_{\Omega} \nabla u \cdot \nabla v \, \mathrm{d}x- \int_{\partial \Omega} \gamma_{1} u \gamma_{0}v \, \mathrm{d}\sigma(x), \quad u\in H^{2}(\Omega) ~\hbox{and}~v\in H^{1}(\Omega). -$$ -Henceforth, on $\partial \Omega$ we merely write $\gamma_0 v= v$ and $ \displaystyle\gamma_1 v=\displaystyle \frac{\partial v}{\partial \nu} $. -Clearly from this Green-Gauss formula, if $u\in H^{2}(\Omega)$ and solves \eqref{eqlocal-Neumann} then $u $ satisfies the variational problem -$$\label{eqlocalvar-Neumann}\tag{$V_1$} - \int_{\Omega} \nabla u \cdot \nabla v \, \mathrm{d}x= \int_{\partial \Omega} f v \, \mathrm{d}x + \int_{\partial \Omega}gv \, \mathrm{d}\sigma(x), \qquad \hbox{for all } ~~v\in H^{1}(\Omega). -$$ -In particular if we put $v=1$ the above formulation becomes -$$\label{eqlocalcompatible-Neumann}\tag{$C_1$} - \int_{\Omega}f\mathrm{d}x+ \int_{\partial \Omega}g\mathrm{d}\sigma(x)=0 -$$ -which is the compatibility condition. -Vice versa we have the following global regularity result. -Theorem -Assume $\Omega\subset \mathbb{R}^{d}$ is bounded open with $C^2$-boundary. If a function $ u \in H^{1}(\Omega)$ is solution to \eqref{eqlocal-Neumann} with $f\in L^{2}(\Omega)$ and $g \in H^{1/2}(\partial\Omega)$ then it belongs to $ H^{2}(\Omega).$ - -Question: In which book or recommendable reference can I find the proof of the above theorem? - I know that the proof of this Theorem for the corresponding Dirichlet problem has been done in the book by Brezis or by Evans. Patently, both references avoid the inhomogeneous Neumann problem. - -Remark -Moreover, observe that if $u$ solves \eqref{eqlocal-Neumann} or \eqref{eqlocalvar-Neumann} so does $\tilde{u} = u+c$ for every $c\in\mathbb{R} $ (that is invariant under additive constant). - -REPLY [5 votes]: Proposition 7.7 of chapter 5 in the first volume of "Partial Differential Equations" by M. E. Taylor (Springer 1996) proves the regularity result for $C^\infty$ boundaries. (Notice that any two solutions differ at most by an additive constant.) The argument of the proof should go through for boundaries with $C^{2,1}$ regularity, because then the Neumann trace $\gamma_1:H^2(\Omega)\to H^{1/2}(\partial \Omega)$ is known to be surjective, implying that it suffices to consider only the case $g=0$. I am not sure whether the assumption $\partial\Omega\in C^2$ is sufficient. -Among others the following sources contain detailed results about elliptic boundary problems which in particular apply to the Neumann boundary problem for the Laplace operator: J. Wloka, Partial Differential Equations, CUP, 1987; Chazarain and Piriou, Introduction to the theory of linear partial differential equations, North-Holland 1982, Chapter 5. The latter source uses the method of the Calderon projector to reduce an elliptic boundary value problem to a pseudo-differential equation on the boundary.<|endoftext|> -TITLE: $\pi$ in terms of polygamma -QUESTION [6 upvotes]: The computer found this, but couldn't prove it. -Let $\psi(n,x)$ denote the polygamma function. -With precision 500 decimal digits we have: -$$ \pi^2 = \frac{1}{4}(15 \psi(1, \frac13) - 3 \psi(1, \frac16)) $$ -Is it true? -In machine readable form: - pi^2 == 1/4*(15*psi(1, 1/3) - 3*psi(1, 1/6)) - -REPLY [14 votes]: Note that -$$ -\psi(m,x) =(-1)^{m+1} m! \sum_{k=0}^{\infty} \frac{1}{(x+k)^{m+1}}. -$$ -Therefore -$$ -\psi(m,1/6) = (-1)^{m+1} m! \sum_{k=0}^{\infty} \frac{1}{(k+1/6)^{m+1}} =(-1)^{m+1} m! 6^{m+1} \sum_{n\equiv 1 \mod 6} \frac{1}{n^{m+1}}. -$$ -Writing the condition $n\equiv 1 \mod 6$ as $n\equiv 1 \mod 3$ but not $4 \mod 6$, the above is -\begin{align*} -&(-1)^{m+1} m! 6^{m+1} \Big( \sum_{n\equiv 1 \mod 3} \frac{1}{n^{m+1}} - \frac{1}{2^{m+1}} \sum_{n\equiv 2 \mod 3} \frac{1}{n^{m+1}}\Big)\\ -&= 2^{m+1} \psi(m,1/3)-\psi(m,2/3). -\end{align*} -We also have -$$ -\psi(m,1/3) +\psi(m,2/3) = (-1)^{m+1} m! 3^{m+1} \sum_{n \not\equiv 0\mod 3} \frac{1}{n^{m+1}} = (-1)^{m+1} m! (3^{m+1} -1) \zeta(m+1). -$$ -From these two relations, clearly we have a linear relation connecting $\psi(m,1/6)$, $\psi(m,1/3)$ and $\zeta(m+1)$: namely, -$$ -\psi(m,1/6) = (2^{m+1}+1) \psi(m,1/3)+(-1)^m m! (3^{m+1}-1) \zeta(m+1). -$$<|endoftext|> -TITLE: How to visualize local complete intersection morphisms? -QUESTION [7 upvotes]: As the question title asks for, how do others visualize local complete intersection morphisms? My experiment in asking people in real life didn't pan out, so I'm consulting the MO algebraic geometry community. Bonus points for pictures! - -REPLY [4 votes]: As Dan Petersen said the comments, an lci morphism $f:X\to Y$ is precisely one that factors Zariski-locally as $X\to Y\times \mathbf{A}^n \to Y$, where $X\to Y\times \mathbf{A}^n$ is a regular embedding (on affines, this means that you are looking at a closed subscheme defining by the vanishing of an ideal generated by a regular sequence) and $Y\times \mathbf{A}^n \to Y$ is the projection map. A regular embedding should be thought of as a closed subscheme which has a tubular neighborhood (every closed immersion of smooth schemes is a regular embedding), so an lci morphism is precisely a morphism for which it is reasonable to talk about the normal bundle. -This can be made more precise. Namely, an lci morphism $f:X\to Y$ can be characterized as a morphism such that the relative cotangent complex $L_{Y/X}$ is perfect, and has Tor-amplitude in $[-1,0]$. The latter statement means that for every quasicoherent sheaf $\mathscr{F}$ on $Y$, the derived tensor product $L_{Y/X}\otimes^\mathbf{L} \mathscr{F}$ has cohomology concentrated in degrees $-1$ and $0$. (If your definition of an lci morphism is as a morphism which is locally given by quotienting by a regular sequence, then one direction of the above statement is proved as Tag 08SL.) In any case, the conormal bundle $N_{Y/X}$ can now be defined as $L_{Y/X}[-1]$, so that the normal bundle of $f$ is $N_{Y/X}^\vee = L_{Y/X}^\vee [-1]$. Since $f$ is an lci morphism, this is actually a vector bundle over $Y$, so it makes sense to call this the normal bundle. This should make sense: if $f:Y = \mathrm{Spec}(B) \to X = \mathrm{Spec}(A)$ is a surjective morphism and $I = \ker(A\to B)$, then $H^1(L_{Y/X}) = I/I^2$.<|endoftext|> -TITLE: What is the expected value of the submeasure of a random set? -QUESTION [5 upvotes]: Let $N \in \mathbb N$ and suppose that $\phi$ is a submeasure on $[1,N] = \{1,2,\dots,N\}$, by which I mean that $\phi$ is a function $\mathcal P ([1,N]) \rightarrow \mathbb R$ such that -i. $A \subseteq B$ implies $\phi(A) \leq \phi(B)$, -ii. $\phi(A \cup B) \leq \phi(A) + \phi(B)$ for any $A, B \subseteq [1,N]$, and -iii. $\phi(\emptyset) = 0$. -Suppose we choose a subset $X$ of $[1,N]$ at random. We do not know anything about the probability distribution used to select $X$, except that for every $i \in [1,N]$ the probability $P(i \in X)$ that $i$ is in the random set $X$ is at least $\varepsilon > 0$. -I would like to have a lower bound for the expected value of $\phi(X)$ in terms of $\varepsilon$ and $\phi([1,N])$. I conjecture that - -$$E(\phi(X)) \,\geq\, \frac{1}{2} \cdot \varepsilon \cdot \phi([1,N])$$ - -but cannot seem to prove it. I would be happy if there is any constant $c > 0$ (independent of $N$, but possibly dependent on $\varepsilon$) such that - -$$E(\phi(X)) \,\geq\, c \cdot \varepsilon \cdot \phi([1,N]).$$ - -(So my conjecture is that taking $c = \frac{1}{2}$ will do, but if you have a proof for some smaller value of $c$ then I would still love to see it.) My question is simply whether either of these bounds is true. -Comments: - -If $\phi$ is a measure instead of a submeasure, then it follows from the linearity of expectation that $E(\phi(X)) \geq \varepsilon \cdot \phi([1,N])$ (an even better lower bound than the conjectured one above). Indeed, if $\phi$ is a measure then -$$E(\phi(X)) = E\left( \sum_{i \in X}\phi(\{i\}) \right) = E\left( \sum_{i \in [1,N]}\phi(\{i\} \cap X) \right) = \sum_{i \in [1,N]}E(\phi(\{i\} \cap X)) = \sum_{i \in [1,N]}P(i \in X)\phi(\{i\}) \geq \varepsilon \sum_{i \in [1,N]}\phi(\{i\}) = \varepsilon \cdot \phi([1,N])$$ -Despite the previous comment, the factor of $\frac{1}{2}$ in my conjecture is necessary. For example, $\phi$ could be a submeasure on $[1,N]$ that assigns $\phi(\emptyset) = 0$, $\phi([1,n]) = 2$, and $\phi(A) = 1$ for all other $A \subseteq [1,N]$. If $X$ is selected uniformly at random from among the $N$ sets of the form $[1,N] \setminus \{i\}$, then we have $E(\phi(X)) = 1$ but $\varepsilon \cdot \phi([1,N]) = 2 - \frac{2}{N} \approx 2$. - -REPLY [5 votes]: Let $n$ be chosen arbitrarily and let $S_k=\{1,2,\ldots,n\}^k$ and let $\mathcal P$ be the collection of all Cartesian products of the form $A_1\times\ldots\times A_k$ where $|A_i|=n-1$. For $Z\subset S_k$, define $\phi(Z)=\min\{j\colon \exists P_1,\ldots,P_j\in \mathcal P:P_1\cup\ldots\cup P_k\supset Z\}$. -I claim that $\phi(S_k)>k$. The proof is by induction. In the case $k=1$, $\phi(S_1)=2$. Now suppose the result holds for $k$ and suppose we have a covering of $S_{k+1}$. Let the first element be $P_1$, which we may suppose by re-labelling is $\{2,\ldots,n\}^{k+1}$. Now the remaining elements are required to cover $\{1,\ldots,n\}^{k}\times \{1\}$. In particular, projecting them onto the first $k$ coordinates, they are required to cover $\{1,\ldots,n\}^k$. By the inductive hypothesis, the number of additional elements in the cover exceeds $k$, so that $\phi(S_{k+1})>k+1$ as required. -In particular, we have $\phi(S_n)>n$. Now if $X$ is a random variable taking values in $\mathcal P$, all with equal probability, then we see that for each element of $S_n$, the probability that it is contained in $X$ is $\epsilon=(\frac{n-1}n)^n\approx e^{-1}$. -So for any $c>0$, we have $\mathbb E\phi(X)=1\le c\epsilon \phi(S_n)$ for all sufficiently large $n$.<|endoftext|> -TITLE: Pages from a known textbook on Euclidean geometry? -QUESTION [6 upvotes]: Do you recall having seen the attached pages in a textbook once? If so, would you be so kind as to share its bibliographic record (or the main items in it) with me below? -A teacher provided us xerox copies of these exercises maaany years ago, but I did not have the presence of mind at the moment to ask him where he had gotten them from. -Let me thank you in advance for your attention and for allowing me to pose this question in spite of the fact that it may be a wee bit borderline for the site... - -REPLY [13 votes]: These are pages 4 and 5 from Paul Eigenmann, Geometrische Denkaufgaben, Klett+Balmer, Zug, 1982 (ISBN 3-264-72231-3).<|endoftext|> -TITLE: Dualizable presheaves with respect to Day convolution -QUESTION [6 upvotes]: This question was posted on MSE and got very little attention, so I'm also posting it here. -Let $\mathcal{C}$ be a closed symmetric monoidal category and let $PSh(\mathcal{C}):=Fun(\mathcal{C}^{op}, \mathbf{Set})$ be its category of presheaves regarded as a closed symmetric monoidal category via Day convolution of presheaves. -Is there a nice description of the dualizable objects of $PSh(\mathcal{C})$ -in terms of the dualizable objects of $\mathcal{C}$? For example, could it be that the dualizable presheaves $PSh(\mathcal{C})_{fd}$ consists of objects given as filtered (co)limits of dualizable objects in $\mathcal{C}$? - -REPLY [6 votes]: Lemma: In a closed symmetric monoidal category where the unit object $1$ is tiny (meaning $\text{Hom}(1, -)$ preserves colimits), every dualizable object is tiny. - -Proof. If $x$ is dualizable, then $\text{Hom}(x, -) \cong \text{Hom}(1, x^{\ast} \otimes (-))$. Since the monoidal structure is assumed to be closed, $x^{\ast} \otimes (-)$ preserves colimits, and by assumption so does $\text{Hom}(1, -)$. $\Box$ -The unit object in $\text{Psh}(C)$ is the presheaf represented by the unit object; since colimits in presheaf categories are computed pointwise, every representable presheaf is tiny, so the lemma applies and we conclude that every dualizable object in $\text{Psh}(C)$ is tiny. But it's standard that the tiny objects in $\text{Psh}(C)$ are precisely the retracts of the representable presheaves. Among these, the dualizable objects include the presheaves represented by dualizable objects in $C$, as well as (I think?) their retracts. I don't know if it's possible for a nontrivial retract of a presheaf represented by a non-dualizable object to be dualizable.<|endoftext|> -TITLE: Why are the medians of a triangle concurrent? In absolute geometry -QUESTION [17 upvotes]: This fact holds true in absolute geometry, and I would like to see an elementary synthetic proof not using the classification of absolute planes (Euclidean and hyperbolic planes) and specific models. Actually I know such a proof (from Skopets - Zharov book), but it uses the third dimension which has to be justified itself. -UPDATE. Here is the aforementioned proof using stereometry, maybe somebody may see how to get pure 2d-proof. -Let triangle $ABC$ lie in a horizontal plane $\pi$. Draw unit vertical segments $AA', BB'$ above $\pi$ and $CC'$ below $\pi$. The segments $AC$ and $A'C'$ pass through the midpoint $K$ of $AC$, $BC$ and $B'C'$ through the midpoint $M$ of $BC$. Let the medians $BK$ and $AM$ meet at $G$, the segments $BA'$ and $AB'$ at $P$. The planes $BA'C'$ and $AB'C'$ have three common points $C',P,G$ which are therefore concurrent. Thus their projections to $\pi$ are concurrent. But $C'$ projects to $C$, $P$ projects to the midpoint $N$ of $AB$ (by the symmetry of the quadrilateral $A'ABB'$). Hence $G$ belongs to the third median $CN$ and we are done. - -REPLY [8 votes]: Below is a summary of Hjelmslev's argument as it is explained in Bachmann's book, see the references to German and Russian editions in Misha's answer. The goal of this answer is at first to help people not speaking Russian or German, and at second to summarize quite a long theory (when you start to read this proof, you are referred to previous lemmata and definitions, then again and so on.) -The main idea is to identify any line or point with reflection in this line/point. So we multiply the lines and points as elements of the group of isometries. Also, let $[AB]$ denote a line through points $A$ and $B$. -This viewpoint allows, in particular, to define the pencil of lines: it is the set of lines for which the composition of any three reflections is again a reflection in some line. In particular, the set of lines which share a common point is a pencil: if $a,b,c$ are three lines through point $P$, then $\mathcal{R}:=bc$ is a rotation with centre $P$, there exists a line $d$ through $P$ for which $ad=\mathcal{R}$ too, thus $abc=a\mathcal{R}=aad=d$ is a reflection in $d$. -Another important example of a pencil is the set of lines orthogonal to a given line $p$. Indeed, let $a,b,c$ be three lines orthogonal to $p$, let them meet $p$ at points $A,B,C$ respectively. It is easy to find a point $D\in p$ such that $ABCD$ (it is a product of reflections) is identical on $p$. Let $d$ be a line through $D$ perpendicular to $p$. Then $abcd$ is identical on $p$ where it is the same as $ABCD$ and preserves the half-plane bounded by $p$, thus $abcd=\rm{id}$ and $abc=d$. Analogous argument shows that $AbC=d$ in this example. -To show how this stuff works, define an isogonal conjugation. Let $ABC$ be a triangle with side lines $a,b,c$ (respective as usual). Let $a_1,b_1,c_1$ be lines through $A,B,C$ respectively belonging to some pencil (for example, sharing a common point). Then we may find the lines $a_2,b_2,c_2$ through $A,B,C$ respectively such that $a_2b=ca_1$ (that implies $a_2c=a_2bbc=ca_1bc=(ca_1b)^2ba_1=ba_1$, we used that $ca_1b$ is a reflection in some line), $b_2c=ab_1$, $c_2a=bc_1$. It yields $a_2c_2b_2=(ca_1b)(bc_1a)(ab_1c)=c(a_1c_1b_1)c$ is a reflection in a line, since so is $a_1c_1b_1$. Thus $a_2,b_2,c_2$ belong to a pencil, which is called isogonally conjugate to the pencil $(a_1,b_1,c_1)$. If both pencils correspond to points, these two points are isogonally conjugate with respect to the triangle $ABC$. -A more general result which is proved in the same spirit is the following -9 Lines Theorem. If $x_1,x_2,x_3$ are distinct isometries and so are $y_1,y_2,y_3$ and $x_iy_j$ is a reflection in a line for 8 pairs $(i,j)\in \{1,2,3\}\times \{1,2,3\}$, then all these lines are from the same pencil and the ninth product is also a reflection in a line from this pencil. -It implies the following -Coupling theorem. Let $a_1,b_1,c_1,\ell,a_2,b_2,c_2$ be seven lines from the same pencil such that $a_2=\ell a_1\ell,b_2=\ell b_1\ell, c_2=\ell c_1 \ell$ are symmetric to $a_1,b_1,c_1$ with respect to $\ell$ and $a_1,b_1,c_1$ pass through the vertices $A,B,C$ respectively of $\triangle ABC$. Then the pencils $(a,a_2)$, $(b,b_2)$, $(c,c_2)$ share a common line. In particular, if there exist the points $a_2\cap a, b_2\cap b, c_2\cap c$, they are collinear. -This is a general stuff, and now consider the midpoints $A_1,B_1,C_1$ of sides of a triangle $ABC$. Let $c_1\ni C_1$ be the perpendicular bisector to $AB$. We start with -Midline lemma. The midline $q$ through $A_1$ and $B_1$ is perpendicular to $c_1$. -Proof. Let $h$ be a perpendicular from $C$ to $q$. The composition of reflections $A_1qB_1$ is a reflection via some line $r\perp q$ (proved above), on the other hand it maps $B$ to $A$, so $r=c_1$. -Now we prove that the medians of $ABC$ are concurrent. Let the points $P,Q$ be symmetric to $A,C_1$ with respect to $A_1$, respectively, $k\ni A_1$ be the perpendicular bisector to $AP$. By Midline Lemma applied to $\triangle AA_1C$ we have $[QB_1]\perp k$. Consider the pencil $\Phi$ of lines orthogonal to $k$. It contains the line $\ell=[AP]$, $r_C\ni C, r_B\ni B, [QB_1]=r_Q\ni Q$, $r_1\ni C_1$. Note that the projections of $B$ and $C$ onto $k$ are symmetric with respect to $A_1$. It implies that $r_B$ is the image of $r_C$ under reflection in $\ell$ (in other words, $r_B=\ell r_C \ell$), analogously $r_Q=\ell r_1\ell$. The lines $r_C,\ell, r_1$ of the pencil $\Phi$ pass through the vertices of $\triangle AC_1C$. Thus there images under reflection in $\ell$ meet the opposite sides in concurrent points. These images are $r_B,\ell,r_Q$ respectively, they meet corresponding opposite sides of $\triangle AC_1C$ in the points $B,[CC_1]\cap [AA_1],B_1$ respectively. Thus by Coupling Theorem these three points are collinear as we need.<|endoftext|> -TITLE: How is Chern-Simons theory related to Floer homology? -QUESTION [18 upvotes]: Chern-Simons theory (say, with gauge group $G$) is the quantum theory of the Chern-Simons functional -$$CS(A)=\frac{k}{8\pi^2}\int_M \text{Tr}\left(A\wedge dA + \frac{2}{3}A\wedge A \wedge A\right)$$ -while instanton Floer homology is roughly the Morse homology of the moduli space of $G$-connections on $M$ with the Chern-Simons functional $CS(A)$ as Morse function. -Other than the superficial fact that they both use the Chern-Simons functional $CS(A)$ in some way, is there any deeper connection between the Chern-Simons theory and the Floer homology? - -Added 10/11, just to clarify what I am asking : -There must be many different aspects of the connection between Chern-Simons theory and the Floer homology. For instance, let me list a few of what I've heard : - -(Kronheimer & Mrowka, 2010) For a link $K$, there is a spectral sequence whose $E_2$ page is the Khovanov homology of $K$ that abuts to the orbifold instanton homology $I^\natural (K)$. -(Gukov, Putrov, Vafa) Both Heegaard Floer homology and Chern-Simons gauge theory arise naturally in the context of $6\text{d }\mathcal{N}=(0,2)$ SCFT. - -I am interested in learning more about these connections, and also want to know other aspects of connection between CS theory and Floer theory; So I think my question is twofold : -for one, I want a conceptual explanation of why the two theories should be related, and -for two, I want to expand the above list of known facts revealing some deeper connection between the two theories. -Any comment is more than welcome. - -REPLY [5 votes]: I'm far from an expert, and I apologize if this is too basic / philosophical / vague. -In instanton Floer homology, the functional $CS(A)$ plays the role of the potential energy function for a $4$d field theory. In Chern-Simons theory, $CS(A)$ plays the role of the action for a $3$d field theory. -Let's consider the analogous situation in the original setting for Floer homology (as in Supersymmetry + Morse theory). We have a Riemannian manifold $(M,g)$ and a function $f: M \to \mathbb R$ , analogous to the Chern Simons functional. We have two options: -First, we can use $f$ as a potential. This corresponds to a $1$d field theory (or classical mechanical system) whose fields are $\gamma(t) \in {\rm Map}([a,b],M)$, and whose action is $S(\gamma) = \int |\dot \gamma(t)|^2 + f(\gamma(t))dt$. This theory is dependent on the metric $g$, but the "vacuum states" of the quantum mechanical system stay the same as we deform the metric-- the vector space of vacuum states is the analog of the Floer groups. -Second we can use $f$ as the action of a $0$d field theory. The fields are ${\rm Map}(*,M) = M$, and the action is just $f(m)$. Physically, this describes a static system. The integral of $\exp(if)$ is the analog of the Chern-Simons invariants. -These two theories are related in the sense that if we let kinetic energy term of the $1$d field theory $|\dot \gamma(t)|^2$ tend to zero (for instance by making the metric $g$ small), the "limit" should be related to the $0$d field theory. This is the limit where the potential energy is very large relative to the kinetic energy, and the dynamical system approaches a static system. -This is to say, the $0$d field theory is a "dimensional reduction" of the $1$d field theory. -Analogously, we expect the $3$d Chern Simons theory to be a dimensional reduction of the $4$d Donaldson/Floer theory. Dimensional reduction is closely related to decategorification, for instance by taking Euler characteristics. The Kronheimer-Mrowka theorem you mention implies that Khovanov homology has the same Euler characteristic as instanton Floer homology. So they both categorify the Jones Polynomial / Chern simons invariant, as this (very) heuristic picture would suggest.<|endoftext|> -TITLE: How should I think about the module of coinvariants of a $G$-module? -QUESTION [15 upvotes]: Let $G$ be a group, $M$ a $G$-module, then the group of coinvariants is the module $M_G := M/I_GM$, where $I_G$ is the kernel of the augmentation map $\epsilon : \mathbb{Z}G\rightarrow \mathbb{Z}$. -The dual notion of $G$-invariants is very simple, and for most $G$-modules, it's easy to tell whether or not for example $M^G = 0$. -On the other hand, I don't have a good intuition for $M_G$. -I'd appreciate a list of results that say something about this module of coinvariants, and criteria which might help recognize when it vanishes. Even facts in the case where $G$ is a finite, abelian, or even finite abelian group would be welcome. -References would also be helpful. - -REPLY [12 votes]: I will say how to compute the module of coinvariants---this will give a criterion to say if it vanishes. We work over $\mathbb{Z}$, since this is the most general setting, but the following analysis works over any commutative ring. -Given a $G$-module $M$, pick generators $x_1, \ldots, x_k \in M$. These are elements so that every other $m \in M$ is a $\mathbb{Z}$-linear combination of the elements $gx_i$ where $i \in \{1, \ldots, k\}$ and $g \in G$. This is the same as saying that the $x_i$ generate $M$ as a left $\mathbb{Z}G$-module. -If $M \cong \bigoplus_{i=1}^k \mathbb{Z}Gx_i$, then $M$ is free as a left $\mathbb{Z}G$-module, and so the coinvariant module is just $\bigoplus_{i=1}^k \mathbb{Z}x_i$, the free $\mathbb{Z}$-module on the same generators. If $M$ is not free on the $x_i$, then there must be some relations. In other words, the surjection $\bigoplus_{i=1}^k \mathbb{Z}Gx_i \twoheadrightarrow M$ must have some kernel. -Let's suppose that $G$ is finite, or otherwise ensure that the kernel of this map is finitely generated. Pick generators for the kernel $y_1, \ldots, y_l$. We obtain an exact sequence -$$ -\bigoplus_{j=1}^l \mathbb{Z}Gy_j \xrightarrow{\varphi} \bigoplus_{i=1}^k \mathbb{Z}Gx_i \to M \to 0. -$$ -Since the map $\varphi$ is a map from one free module to another, it is given by a $l \times k$ matrix with entries in the ring $\mathbb{Z}G$. Let $\epsilon\varphi$ be the same matrix, but where we have replaced every group element $g \in G$ with the integer $1 \in \mathbb{Z}$. So $\epsilon\varphi$ is an integer matrix. Writing $\mathbb{Z}$ for the integers with the trivial right action of $G$, we have -\begin{align} -M_G &\cong \mathbb{Z} \otimes_G M \\ -&\cong \mathbb{Z} \otimes_G \mathrm{coker} \, \varphi \\ -&\cong \mathrm{coker} \left( \mathbb{Z} \otimes_G \varphi \right) \\ -&\cong \mathrm{coker} \left( \epsilon \varphi \right). -\end{align} -In other words, the coinvariants are still generated by the same generating set; however, any group elements appearing in the relations are ignored. For example, any relation $gx_1 = x_2$ holding in $M$ just becomes $x_1=x_2$ in $M_G$. Similarly, the relation $x_1 + gx_2 = hx_3$ becomes $x_1 + x_2 = x_3$, $x_1-gx_1=0$ becomes $x_1-x_1=0$, etc. This gives a systematic way of obtaining a presentation for the $\mathbb{Z}$-module $M_G$ from a presentation for the $\mathbb{Z}G$-module $M$.<|endoftext|> -TITLE: To prove the fiber above a codimension 1 point contains a geometrically integral open subscheme -QUESTION [6 upvotes]: Suppose $f:X\rightarrow \mathbb{P}_k^n$ is a proper smooth morphism, where $k$ is an algebraically closed field. If $f$ admits a rational section, can we prove that the fiber of $f$ above any codimension 1 point contains a geometrically integral open subscheme? - -REPLY [7 votes]: This is true: -Because $f$ is smooth, any fibre $X_y$ is smooth. If $y$ has codimension $1$, then the rational section $\sigma \colon \mathbb P^n_k \dashrightarrow X$ is defined at $y$ by the valuative criterion of properness. Thus, $X_y$ has a rational point $\sigma(y)$. If $U \subseteq X_y$ is the connected component of $\sigma(y)$, then $U$ is a smooth connected $\kappa(y)$-scheme with a rational point, hence geometrically connected by Tag 04KV.<|endoftext|> -TITLE: Are there logical systems where formal proofs are not computer verifiable? -QUESTION [7 upvotes]: In a set-theoretic system using first-order logic, every proof could be written as a goal followed by a finite sequence of sentence where each one is justified by an axiom or previously established sentence and the last line is the goal. Computers can easily verify proofs like these using the rules of first-order logic. -In dependent type theory, proofs come in the form of programs, and you know your proof is correct if it can be compiled (or type-checked). -Do there exist logics/foundations where rigorous proofs can not be checked by computers? Or where a checking program might never terminate for a correct proof? - -REPLY [8 votes]: If you take "proof" as a convincing argument (to error probability smaller than arbitrary epsilon) that a proposition is true, then two other possibilities come to mind: - -Probabilistically checkable proofs, where you have a conventional first-order proof (like your secret unpublished proof of RH) and you can convince me through a cryptographic-like protocol that you really do have a proof, without revealing how the proof works. Zero-knowledge proofs are related to this; and -(Due to L. Levin) if you believe there are physically realizable ways of generating algorithmically random numbers (e.g. by rolling dice), you can make almost-certainly-true but unprovable statements, like that the string produced by 1000 bits of coin flips can't be compressed to less than half its original length (this is an assertion about the Kolmogorov complexity, which is uncomputable). - -It follows from the second example above that the physics assertion that you can generate random numbers through quantum processes is scientifically unverifiable, if you believe the Church-Turing thesis.<|endoftext|> -TITLE: Hyperbolic manifolds containing totally geodesic hypersurfaces which themselves contain totally geodesic hypersurfaces -QUESTION [5 upvotes]: I will preface this by saying that while I am familiar with the general theory of (semi)-Riemannian manifolds, I am a complete novice when it comes to the specifics of hyperbolic manifolds (I am using them for examples when computing cohomology with local coefficients). I was originally trained as a physicist, so I am also not very familiar with field theory (which comes up in the construction). So apologies in advance if there are some "dumb" questions here. -The purpose of my question is to understand the following construction referred to in the paper Local rigidity of hyperbolic manifolds with geodesic boundary by Kerckhoff and Storm, where they state: - -"For each dimension $n>2$ [Gromov and Thurston] construct an infinite number of closed hyperbolic n-manifolds $V$ with the following properties: $V$ has a codimension 1 embedded, totally geodesic submanifold $M$ which itself has a codimension 1 embedded, totally geodesic submanifold $P$ (so $P$ is codimension 2 in $V$)." - -The Gromov & Thurston paper being referred to is Pinching Constants for Hyperbolic Manifolds. It appears that the brief Section 1 contains the relevant construction, which I have snipped from the linked pdf: -I am unfortunately confused about "why" this construction gives us a hyperbolic manifold containing a codimension 1 totally geodesic hypersurface itself containing a totally geodesic codimension 1 hypersurface. To be more specific, why use this $\Phi_n$ instead of the usual Lorentzian quadratic form, and why finite index subgroups of this particular $\Gamma_n$? - -REPLY [3 votes]: The fixed point set of the involution in the universal cover is a totally geodesic copy of hyperbolic space of codimension 1. So that gives you the first totally geodesic hypersurface in some finite cover of the given thing. And since the hypersurface is the same type of lattice, there is a finite cover of it that has a totally geodesic hypersurface by the same argument. There is a single finite cover of the original thing doing both at once.<|endoftext|> -TITLE: Derived equivalences of Artin algebras with finitistic dimension zero -QUESTION [5 upvotes]: Let $A$ be an Artin algebra of finitistic dimension zero and $B$ an algebra derived equivalent to $A$. Does $B$ also have finitistic dimension zero? -In case this is true, this might generalise the result that selfinjective algebras are closed under derived equivalence. (Since selfinjective algebras have finitistic dimension zero and they are exactly the Gorenstein algebras with finitistic dimension zero (and being Gorenstein is preserved under derived equivalence)). -edit: It might be a better question to ask for the two-sided condition. That is: -Let $A$ be an Artin algebra of left and right finitistic dimension zero and $B$ an algebra derived equivalent to $A$. Does $B$ also have left and right finitistic dimension zero? -I think in this form it might be true. Non-selfinjective examples with left and right finitistic dimension zero are quiver algebras where at each vertex there is at least one loop. - -REPLY [3 votes]: Let $A$ be the radical square zero algebra whose quiver has two vertices $1$ and $2$, with a loop at each vertex and an arrow from vertex $1$ to vertex $2$. Then the projective (right) modules have structure -$$P_1=\matrix{&1&\\1&&2}\mbox{$\quad$ and $\quad$}P_2=\matrix{2\\2},$$ -and the left and right finitistic dimensions are both zero. -There is a tilting complex that is the direct sum of complexes $T_a:=\dots\to0\to P_2\to P_1\to0\to\dots$ and $T_b:=\dots\to0\to0\to P_1\to0\to\dots$, with endomorphism algebra having projective right modules (according to my hand calculations) with structure -$$Q_a=\matrix{&a&\\a&&b\\&&a\\&&b}\mbox{$\quad$ and $\quad$}Q_b=\matrix{b\\a\\b},$$ -and since there is an injective map $Q_b\to Q_a$, the right finitistic dimension is greater than zero.<|endoftext|> -TITLE: Is there any paper which summarizes the mathematical foundation of deep learning? -QUESTION [26 upvotes]: Is there any paper which summarizes the mathematical foundation of deep learning? -Now, I am studying about the mathematical background of deep learning. -However, unfortunately I cannot know to what extent theory of neural network is mathematically proved. -Therefore, I want some paper which review the historical stream of neural network theory based on mathematical foundation, especially in terms of learning algorithms (convergence), and NN’s generalization ability and the NN’s architecture (why deep is good?) -If you know, please let me know the name of the paper. -For your reference, let me write down some papers I read. - -Cybenko, G. (1989). Approximation by superpositions of a sigmoidal function. Mathematics of control, signals and systems, 2(4), 303-314. -Hornik, K., Stinchcombe, M., \& White, H. (1989). Multilayer feedforward networks are universal approximators. Neural networks, 2(5), 359-366. -Funahashi, K. I. (1989). On the approximate realization of continuous mappings by neural networks. Neural networks, 2(3), 183-192. -Leshno, M., Lin, V. Y., Pinkus, A., \& Schocken, S. (1993). Multilayer feedforward networks with a nonpolynomial activation function can approximate any function. Neural networks, 6(6), 861-867. -Mhaskar, H. N., \& Micchelli, C. A. (1992). Approximation by superposition of sigmoidal and radial basis functions. Advances in Applied mathematics, 13(3), 350-373. -Delalleau, O., \& Bengio, Y. (2011). Shallow vs. deep sum-product networks. In Advances in Neural Information Processing Systems (pp. 666-674). -Telgarsky, M. (2016). Benefits of depth in neural networks. arXiv preprint arXiv:1602.04485. -Barron, A. R. (1993). Universal approximation bounds for superpositions of a sigmoidal function. IEEE Transactions on Information theory, 39(3), 930-945. -Mhaskar, H. N. (1996). Neural networks for optimal approximation of smooth and analytic functions. Neural computation, 8(1), 164-177. -Lee, H., Ge, R., Ma, T., Risteski, A., \& Arora, S. (2017). On the ability of neural nets to express distributions. arXiv preprint arXiv:1702.07028. -Bartlett, P. L., \& Maass, W. (2003). Vapnik-Chervonenkis dimension of neural nets. The handbook of brain theory and neural networks, 1188-1192. -Kawaguchi, K. (2016). Deep learning without poor local minima. In Advances in Neural Information Processing Systems (pp. 586-594). -Kingma, D. P., \& Ba, J. (2014). Adam: A method for stochastic optimization. arXiv preprint arXiv:1412.6980. -Duchi, J., Hazan, E., \& Singer, Y. (2011). Adaptive subgradient methods for online learning and stochastic optimization. Journal of Machine Learning Research, 12(Jul), 2121-2159. -Tieleman, T., \& Hinton, G. (2012). Lecture 6.5-RMSProp, COURSERA: Neural networks for machine learning. University of Toronto, Technical Report. -Zeiler, M. D. (2012). ADADELTA: an adaptive learning rate method. arXiv preprint arXiv:1212.5701. -Yun, C., Sra, S., \& Jadbabaie, A. (2017). Global optimality conditions for deep neural networks. arXiv preprint arXiv:1707.02444. -Zeng, J., Lau, T. T. K., Lin, S., \& Yao, Y. (2018). Block Coordinate Descent for Deep Learning: Unified Convergence Guarantees. arXiv preprint arXiv:1803.00225. -Weinan, E. (2017). A proposal on machine learning via dynamical systems. Communications in Mathematics and Statistics, 5(1), 1-11. -Li, Q., Chen, L., Tai, C., \& Weinan, E. (2017). Maximum principle based algorithms for deep learning. The Journal of Machine Learning Research, 18(1), 5998-6026. -Zhang, C., Bengio, S., Hardt, M., Recht, B., \& Vinyals, O. (2016). Understanding deep learning requires rethinking generalization. arXiv preprint arXiv:1611.03530. -Kawaguchi, K., Kaelbling, L. P., \& Bengio, Y. (2017). Generalization in deep learning. arXiv preprint arXiv:1710.05468. - -REPLY [10 votes]: Mathematics of Deep Learning (2017) - -This tutorial will review recent work that aims to provide a - mathematical justification for several properties of deep networks, - such as global optimality, geometric stability, and invariance of the - learned representations. - -Deep Learning: An Introduction for Applied Mathematicians (2018) - -This article provides a very brief introduction to the basic ideas - that underlie deep learning from an applied mathematics perspective. - Our target audience includes postgraduate and final year undergraduate - students in mathematics who are keen to learn about the area.<|endoftext|> -TITLE: show this nice and hard inequality with $ \prod_{i=1}^{n}|x_{i}-y_{i}| -TITLE: Gerhard Frey, "Links between stable elliptic curves and certain diophantine equations" -QUESTION [12 upvotes]: I am searching for the article by Gerhard Frey, which has indicated a connection between Fermat's Last Theorem and the Taniyama-Shimura Conjecture. The reference is give as - -Gerhard Frey, Links between stable elliptic curves and certain diophantine equations, Annales Universitatis Saraviensis 1, 1-40 (1986). - -Unfortunately, I am not able to find this publication. Is there an online-version of it available somewhere, or is contacting University Saarbruecken library the only option? - -REPLY [8 votes]: Please find here the scanned version of the manuscript Links between stable elliptic curves and certain diophantine equations by Gerhard Frey.<|endoftext|> -TITLE: Reference request: an elementary result on characters of finite abelian groups -QUESTION [6 upvotes]: The referee of a paper I submitted to a journal asked me to include a reference of the following elementary result on characters of finite abelian groups: -Let $A$ be a finite abelian group of order $N$ and let -$\hat A$ be its dual group. Let $a\in A$ have order $h$. Then -$$\prod_{\chi\in\hat A}(1-\chi(a)T)=(1-T^h)^{N/h}.$$ -I don't want to include a proof because one of the good things about this paper (I hope not the only one) is that is short. -I have searched in books about abelian groups, finite groups, representations, and number theory, but I could not find it. As usual, the only place I could find it is in one of the (magnificent) "blurbs" by Keith Conrad. -Does anyone knows a book where I can actually find this result? - -REPLY [3 votes]: This is essentially proved in Rosen's "Number Theory in Function Fields", page 109, Lemma 8.14. - - - -This is also proved in Lang's "Algebraic Number Theory" (2nd edition), page 230. It is the equation with (*) in its beginning. The context is abelian extensions, so some of the notation make it seem number-theoretic, but the argument is general.<|endoftext|> -TITLE: Lelong numbers and integrability of psh functions -QUESTION [5 upvotes]: Let $\varphi$ be a plurisubharmonic function in the unit ball $B_1\subset \mathbb{C}^n$ with $\varphi\le 0$. Suppose that the Lelong number $\nu(\varphi,0)0$. Does it follow that there exists $\alpha>0$, possibly depending on $k$, such that $\int_{B_{\frac{1}{2}}}e^{-\alpha\varphi}dvol(z)\le C$? -Note that this does not immediately follow from the definition of Lelong number. Since $\nu(\varphi,0)=\lim\inf_{z\rightarrow0}\frac{\varphi(z)}{\log|z|}$, so $\nu(\varphi,0) -TITLE: A property of an ultrafilter -QUESTION [9 upvotes]: Let $\mathcal U$ be a free ultrafilter on a set $X$. For $n\in\mathbb N$ let $\mathcal F$ be a family of $n$-element subsets of $X$ such that $\bigcup\mathcal F\in\mathcal U$. - -Question. Is there a set $U\in\mathcal U$ and a subfamily $\mathcal E\subset\mathcal F$ such that $\bigcup\mathcal E\in\mathcal U$ and $|U\cap E|=1$ for every $E\in\mathcal E$? - -I do not know the answer even for $n=2$ and countable set $X$. - -REPLY [2 votes]: Consider this with assuming that $\mathcal{F}$ consists of finite subsets of bounded cardinal; let $n$ be the max of these cardinals and let us argue by induction on $n$. The goal is to find $\mathcal{E}\subset\mathcal{F}$ and $U\in\mathcal{U}$ such that $U\subset\bigcup\mathcal{E}$ and and $|E\cap U|\le 1$ for every $E\in\mathcal{E}$. -First, consider a maximal subset $\mathcal{E}$ of the cover such that no element is in the union of others (we call this ``minimal"). So $V:=\bigcup\mathcal{F}=\bigcup\mathcal{E}$. Write $V=V_1\cup V_2$, where $V_2$ is the set of elements of $V$ that belong to at least two elements of $\mathcal{E}$. The minimality of $\mathcal{E}$ implies that no $E\in\mathcal{E}$ is contained in $V_2$ (i.e., has nonempty intersection with $V_1$). -Since the intersection of elements of $\mathcal{E}$ with $V_1$ are pairwise equal or disjoint, we can partition $V_1$ into $n$ subsets each intersecting, each element of $\mathcal{E}$ in at most a singleton. Hence we can conclude if $V_1\in\mathcal{U}$ (with $U$ being one of those $n$ subsets of $V_1$). -Otherwise, $V_2\in\mathcal{U}$. Since $|E\cap V_2|\le n-1$ for all $E\in\mathcal{E}$, we can conclude by induction (and find $U\subset V_2$). -PS Nik Weaver posted an answer while I was writing this one, but I still post, although I guess it's the same idea.<|endoftext|> -TITLE: How to handle results from an unpublished paper? -QUESTION [14 upvotes]: I am writing a paper right now, and part of the paper makes use of a (trivial) generalization of a number of really nice theorems and constructions from a paper that was never made public. The author has left pure mathematics and has no intention of publishing the paper, but I received a copy directly from him several years ago. -The results and constructions are crucial to my paper, and since I am working with a minor generalization, I think I do need to include the proofs, especially since they aren't available. I don't want it to look like I'm taking credit for the results or plagiarizing, but some of the proofs are basically copies. I have a bunch of disclaimers at the top of the section and continually remind the reader that all of these results are in the original paper. -Do I need to not only give credit at the top of the section, but also give credit for every observation, statement, lemma, and diagram? -PS It seems like the community wiki checkbox is gone. I'd appreciate if a moderator could do that for me. -Edit: I contacted the author, and he agreed to put it on the arXiv himself. Going to leave the question up as a community wiki for anyone else who runs into this problem. - -REPLY [3 votes]: Why not you approach the author and ask to join forces to write as a co-author if the other person is willing to do so? That could be a fine resolution and fair to both parties.<|endoftext|> -TITLE: SYT and contents of a partition -QUESTION [7 upvotes]: Let $\lambda$ be an integer partition, denote the number of Standard Young Tableaux of shape $\lambda$ by $f_{\lambda}$. This number is computed by the formula -$$f_{\lambda}=\frac{n!}{\prod_{u\in\lambda}h_u}$$ -where $h_u$ is a hook length. It is also well-known that -$$\sum_{\lambda\vdash n}f_{\lambda}^2=n!\tag1$$ -Recall the notation for the content of a cell $u=(i,j)$ in a partition is $c_u=j-i$. - -Question. Is this true? - $$\sum_{\lambda\vdash n}f_{\lambda}^2\prod_{u\in\lambda}(t+c_u)=n!\,t^n.$$ - -NOTE. An affirmative answer would imply (1): divide both sides by $t^n$ and take the limit $t\rightarrow\infty$. - -REPLY [6 votes]: Notice that $f_{\lambda}$ is the number of standard Young tableaux of shape $\lambda$, whereas $f_{\lambda}\frac{\prod_{u\in \lambda} (t+c_u)}{n!}$ is the number of semistandard Young tableaux of shape $\lambda$ and content $t$. It was mentioned in a previous question that the identity -$$\sum_{|\lambda|=n}\left|\text{SSYT}(\lambda)\right|\left|\text{SYT}(\lambda)\right|=t^n.$$ -can either be proved bijectively from RSK (as Darij explains in the comments) or by appealing to Schur-Weyl duality (as David explains in the link), which gives an isomorphism of $S_n\times GL(V)$ representations -$$\sum_{|\lambda|=n} Sp_{\lambda} \boxtimes S_{\lambda}(V) \cong V^{\otimes n}$$ -where $V$ is a $t$ dimensional vector space, $S_{\lambda}$ is the Schur functor, and $Sp_{\lambda}$ the corresponding Specht module. Our identity follows by taking the dimension of both sides.<|endoftext|> -TITLE: Distribution of 'square classes' of binary quadratic forms -QUESTION [6 upvotes]: Let $f$ be a binary quadratic form with integer coefficients and non-zero discriminant $D$. Suppose for simplicity that $D$ is a fundamental discriminant (which in particular implies that $f$ is primitive). We say that $f$ represents a square mod $D$ if for any $a$ representable by $f$ we have -$$a \equiv \square \pmod{p}$$ -for each prime $p | D$. Note that this condition only depends on the $\text{GL}_2(\mathbb{Z})$-equivalence class of $f$. We can thus say that the $\text{GL}_2(\mathbb{Z})$-class of $f$, $[f]$, is a square class if for a prime $q$ representable by $f$ and co-prime to $D$ satisfies $q \equiv \square \pmod{p}$ for $p | D$, and if $D \equiv 0 \pmod{4}$ (respectively $8$), then $q \equiv 1 \pmod{4}$ (respectively $8$). By our earlier discussion, the choice of $q$ representable by $f$ does not matter. -How are 'square classes' distributed in the ideal class group of $\mathcal{O} = \mathcal{O}_{\mathbb{Q}(\sqrt{D})}$? Heuristically, for each (odd) prime $p | D$, half of the classes of the ideal class group should be square mod $p$ and the other half should not, so the number of square classes ought to be $h_2(D) 2^{-\omega(D)}$ where $h_2(D)$ is the class number of $\mathcal{O}$ and $\omega(n)$ denotes the number of prime divisors of $n$. - -REPLY [3 votes]: The standard definition for a form to belong to the principal genus of forms with fundamental discriminant $d$ is that the primes $p$ coprime to $d$ that -the form $Q$ represents satisfy $(d_1/p) = \ldots = (d_t/p)$, where -$d =d_1 \cdots d_t$ is the factorization of $d$ into prime discriminants. In particular, $Q$ is allowed to represent primes $p \equiv \pm 1 \bmod 8$ if -$d_1 = 8$ occurs in the factorization, and primes $p \equiv 1, 3 \bmod 8$ if $d_1 = -8$ occurs (for example, $40 = 5 \cdot 8$ and $-40 = 5 \cdot (-8)$ are -factorizations into prime discriminants, $40 = -5 \cdot (-8)$ is not since $-5$ is not a discriminant). -Gauss's principal genus theorem states that a form is in the principal genus if and only if the equivalence classof $Q$ is a square in the class group. The genus of a form is the sign vector $((d_1/p), \ldots, (d_t/p))$, -and there are forms for every sign vectors whose entries have product $+1$.<|endoftext|> -TITLE: Simple Lie algebras: making subspaces 'very transversal' -QUESTION [8 upvotes]: Let $G$ be a Lie group or group of Lie type whose Lie algebra $\mathfrak{g}$ is simple. Because the Lie algebra is simple, for any proper subspace $V\subset \mathfrak{g}$, -there is a $g\in G$ such that $g V g^{-1} \ne V$. Is it the case that there is a $g\in G$ such that $V$ and $g V g^{-1}$ are "as transversal as possible", meaning that $\dim(V + g V g^{-1}) = \min(2 \dim(V), \dim(G))$? Is this the case, at least, for $G$ a classical group? - -REPLY [9 votes]: This is not true. Namely, in the 8-dimensional $\mathfrak{sl}_3$, consider the 4-dimensional Lie subalgebra $\mathfrak{v}$ consisting of matrices of the form -$$\begin{pmatrix} a & x & z\\ 0 & -2a & y\\ 0 & 0 & a\end{pmatrix}.$$ -(This is the centralizer of $E_{13}$ in $\mathfrak{sl}_3$.) I claim that $g\mathfrak{v}g^{-1}\cap \mathfrak{v}\neq 0$ for every $g\in \mathrm{GL}_3$. -Indeed, let $\mathfrak{b}$ be the normalizer of $\mathfrak{v}$, namely the Lie subalgebra of upper triangular matrices of trace zero. Let $\mathfrak{d}$ be the subalgebra of diagonal matrices: it is a Cartan subalgebra in both $\mathfrak{b}$ and $\mathfrak{sl}_3$. It is known that the intersection of any two Borel subalgebras contains a Cartan subalgebra $\mathfrak{d}'$. This applies to the intersection $\mathfrak{b}\cap g\mathfrak{b}g^{-1}$. Conjugating by some upper triangular matrix conjugating $\mathfrak{d}'$ into $\mathfrak{d}$, we can suppose that $\mathfrak{d}'=\mathfrak{d}$. So $g\mathfrak{b}g^{-1}$ is one of the 6 Borel subalgebras containing $\mathfrak{d}$ (images of $\mathfrak{b}$ by the Weyl group $\mathfrak{S}_3$). The condition $[\mathfrak{b},\mathfrak{b}]\cap g[\mathfrak{b},\mathfrak{b}]g^{-1}=0$ forces $g\mathfrak{b}g^{-1}$ to be the opposite Borel subalgebra $\mathfrak{b}_-$, that is, the Lie subalgebra of lower triangular matrices. So $g$ maps the unique flag preserved by $\mathfrak{b}$ to the unique flag preserved by $\mathfrak{b}_-$. So $g$ is "south-east"-triangular. Right-multiplying $g$ by a suitable element of $[\mathfrak{b},\mathfrak{b}]$, we can suppose that $g$ is an anti-diagonal matrix. Then we see that $g$ centralizes $\mathfrak{d}\cap \mathfrak{v}$, hence conjugates $\mathfrak{d}\cap \mathfrak{v}$ into itself and this contradicts $\mathfrak{v}\cap g\mathfrak{v}g^{-1}=0$.<|endoftext|> -TITLE: Sphere spectrum, Character dual and Anderson dual -QUESTION [10 upvotes]: The homotopy groups of the sphere spectrum are the stable homotopy groups of spheres. -However, could you help me to appreciate the mathematical meanings of the following: - -What is the significance of the meanings of "Anderson dual" to the sphere spectrum? (For example, I understand the significant meanings of Pontryagin dual and torsion subgroup. Can "Anderson dual" be explained as simple as that of Pontryagin dual?) -What is the significance of the meanings of (a shift of) $I\mathbb{C}^{\times}$, a “character dual” to the sphere spectrum? -(A shift of) the Anderson dual $I\mathbb{Z}_1$ to the sphere spectrum? - -Please feel free to correct my statements and to manifest the meaning behind my question. Many thanks! - -REPLY [16 votes]: The Anderson dualizing spectrum $I_\mathbf{Z}$ can be defined as follows. Consider the functor $X\mapsto \mathrm{Hom}(\pi_{-\ast} X,\mathbf{Q/Z})$ from the homotopy category of spectra to graded abelian groups. Since $\mathbf{Q/Z}$ is an injective $\mathbf{Z}$-module, this functor is representable by a spectrum $I_\mathbf{Q/Z}$, called the Brown-Comenetz dualizing spectrum. After $p$-completion, this spectrum is equivalent to the spectrum you denoted $I\mathbf{C}^\times$. If you do the same construction with $\mathbf{Q/Z}$ replaced by $\mathbf{Q}$, you obtain the Eilenberg-Maclane spectrum $\mathbf{Q}$. There is a canonical map $\mathbf{Q}\to I_\mathbf{Q/Z}$, and the fiber of this map is defined to be $I_\mathbf{Z}$. This has the property that it is $(-1)$-coconnective, $\pi_0 I_\mathbf{Z} \cong \mathbf{Z}$, $\pi_{-1} I_\mathbf{Z} = 0$, and $\pi_{-n} I_\mathbf{Z} = \mathrm{Hom}(\pi_{1-n} S, \mathbf{Q/Z})$. -There are several reasons why the Anderson dualizing spectrum is significant. I'll try to explain some of them, in the order in which I understand them the best. - -It is the dualizing sheaf of the affine derived scheme $\mathrm{Spec}(S)$. In order for this to make sense, I have to tell you what "dualizing sheaf" means. Let $A$ be a connective $\mathbf{E}_\infty$-ring (this definition fails drastically in the nonconnective setting). Then, an $A$-module $\omega_A$ is said to be a dualizing sheaf if: - -$\omega_A$ is coconnective, -$\pi_n \omega_A$ is a finitely generated $\pi_0 A$-module for every $n$, -the map $A\to \mathrm{Map}_A(\omega_A, \omega_A)$ is an equivalence (this is the dualizing property), and -$\omega_A$ has finite injective dimension as an $A$-module, i.e., there is some integer $n$ such that for every discrete $A$-module $M$, the groups $\pi_k\mathrm{Map}_A(M,\omega_A)$ vanish for $k>n$. - -You can check that $I_\mathbf{Z}$ satisfies all of these properties when $A = S$. Moreover, one can prove (see Proposition 6.6.2.1 of Lurie's SAG) that dualizing sheaves are unique up to elements of $\mathrm{Pic}(A)$; when $A = S$, this group is just $\mathbf{Z}$, generated by $S^1$. It follows that any other dualizing sheaf for $S$ is equivalent to a suspension of $I_\mathbf{Z}$. -There is an analogue of Serre duality in this spectral setting. Let me first explain the even periodic case, and then discuss what happens in the $K(n)$-local setting. In the even periodic setting, one might consider objects like $KU$, $KO$, and $\mathrm{TMF}$ and its variants. All of these can be obtained as the global sections of sheaves of $\mathbf{E}_\infty$-rings on Deligne-Mumford stacks ($\mathrm{Spec}(\mathbf{Z})$, $\mathrm{Spec}/\!\!/C_2$, and the moduli stack $\mathcal{M}_{\mathrm{ell}}$ of elliptic curves, respectively). Each of these Deligne-Mumford stacks are --- in the classical world --- smooth. In particular, they are Gorenstein. -One can actually show that the associated derived stacks are Gorenstein in the spectral world, too. (The cases mentioned above are due to Heard-Stojanoska, and Stojanoska, respectively.) One finds the extremely interesting statement, e.g., that $I_\mathbf{Z} \mathrm{TMF} := \mathrm{Map}(\mathrm{TMF}, I_\mathbf{Z})$ is in the Picard group of $\mathrm{TMF}$ (which is known to be cyclic, by work of Mathew-Stojanoska). This can be generalized: one can prove that most reasonable derived locally even periodic Deligne-Mumford stacks $\mathfrak{X}$ satisfy the property that $\mathrm{Map}(\Gamma(\mathfrak{X}, \mathcal{O}_\mathfrak{X}), I_\mathbf{Z})$ is in the Picard group of $\Gamma(\mathfrak{X}, \mathcal{O}_\mathfrak{X})$. I have a proof of the latter fact here, as Theorem 3.14; I'm sure experts already know of this result. -One practical consequence of this is the following. For a spectrum $A$, define $I_\mathbf{Z} A = \mathrm{Map}(A,I_\mathbf{Z})$. Then there is a short exact sequence -$$0\to \mathrm{Ext}^1_\mathbf{Z}(\pi_{-d-1} X, \mathbf{Z}) \to \pi_d I_\mathbf{Z} X \to \mathrm{Hom}_\mathbf{Z}(\pi_{-d} X, \mathbf{Z}) \to 0.$$ -Now, suppose that $E$ is a spectrum such that $I_\mathbf{Z} E \simeq \Sigma^k E$ for some $k$ (so that $I_\mathbf{Z} E \in \mathrm{Pic}(E)$). Suppose we want to understand the $E$-theory of a spectrum $X$. Apply the short exact sequence above to the spectrum $E\wedge X$; this gives a short exact sequence -$$0\to \mathrm{Ext}^1_\mathbf{Z}(E_{-d-1} X, \mathbf{Z}) \to (I_\mathbf{Z} E)^d X \to \mathrm{Hom}_\mathbf{Z}(E_{-d} X, \mathbf{Z}) \to 0.$$ -But the middle term is, by the "Anderson self-duality" of $E$, equivalent to $E^{d+k} X$. This gives an interesting and useful "universal coefficients" exact sequence. -Another reason one might care about the spectrum $I_\mathbf{Z}$ is because of its appearance in the Picard group of the $K(n)$-local category, which goes by the name Gross-Hopkins duality. Let $n>0$. After $K(n)$-localizing, the difference between $I_\mathbf{Q/Z}$ and $I_\mathbf{Z}$ vanishes: $L_{K(n)} I_\mathbf{Q/Z} \simeq \Sigma L_{K(n)} I_\mathbf{Z}$. Suppose $X$ is an $E_n$-local spectrum. We define another spectrum $I_n X = \Sigma L_{K(n)} I_\mathbf{Z} X \simeq L_{K(n)} \mathrm{Map}(L_n X, I_{\mathbf{Q/Z}})$, and let $I_n = I_n L_{K(n)} S$. Then, Gross and Hopkins proved two remarkable statements: first, the spectrum $I_n$ is invertible in the $K(n)$-local category (this is not very surprising), and second, there is an isomorphism $(E_n)^\vee_\ast I_n \simeq \Sigma^{n^2-n} (E_n)_\ast[\det]$ of $(E_n)_\ast[\![\Gamma_n]\!]$-modules, where $\det:\Gamma_n \to \mathbf{Z}_p^\times$ is the determinant map, and $\Gamma_n$ is the Morava stabilizer group at height $n$. (I've implicitly been fixing the Honda formal group over $\mathbf{F}_{p^n}$.) -The proof of this result is really cool; it passes through an important object in rigid analytic geometry (the Gross-Hopkins period map). In any case, if $p\gg n$, then one can actually identify $I_n$ itself with a $K(n)$-local object $S[\det]$, called the determinantal sphere. This object is incredibly interesting, and plays an important role in the chromatic story at height $n$. -The Anderson dualizing spectrum also plays an important (but unpublished) role in orientation theory. The following simple question is an open problem: let $\kappa$ be a perfect field of characteristic $p>0$, and let $H$ be a formal group of finite height over $\kappa$. Denote by $E(\kappa, H)$ the associated Morava $E$-theory. Then, is there an $\mathbf{E}_\infty$-orientation $MU \to E(\kappa,H)$? One can reduce this to understanding the spectrum of units $gl_1 E(\kappa,H)$. In unpublished work, Hopkins and Lurie have shown that the fiber of the map $gl_1 E(\kappa, H) \to L_n gl_1 E(\kappa, H)$ (which is known as the discrepancy spectrum) is equivalent to $\Sigma^n I_\mathbf{Q/Z}$ in nonnegative degrees (they also did more). This allows one to construct a map $\Sigma^{n+1} H\mathbf{Z} \to gl_1 E(\kappa, H)$, the very existence of which is enough to obstruct the existence of certain $\mathbf{E}_\infty$-orientations of $E(\kappa, H)$. -I have heard that there is some relationship with TQFTs, but I don't know how that goes. There are people here more qualified than me who can say more about this. - -REPLY [9 votes]: One way to think of these spectra is in terms of the cohomology theories they define. In other words, if -$E$ is a spectrum, what is $[E, I\mathbb Z]$? This is less of a description of what they are and more of a -description of what they do; there are probably other, more conceptual answers to your question. -1. $I\mathbb Z$ and $\Sigma^n I\mathbb Z$ -The universal coefficient theorem describes how to compute cohomology groups from homology groups: there is a short -exact sequence -$$ 0\longrightarrow \mathrm{Ext}^1(H_{n-1}(X), \mathbb Z)\longrightarrow H^n(X)\longrightarrow \mathrm{Hom}(H_n(X), \mathbb -Z)\longrightarrow 0,$$ -and it splits noncanonically. -If you try to do this for generalized -cohomology, nothing so nice -is true, and the whole story is more complicated. -Nonetheless, part of the story can be salvaged: if you try this with stable homotopy groups (the homology theory -represented by the sphere spectrum), you obtain the cohomology theory represented by the Anderson dual of the -sphere. That is, for any spectrum $X$ there is a short exact sequence -$$ 0\longrightarrow \mathrm{Ext}^1(\pi_{n-1}(X), \mathbb Z)\longrightarrow [X, \Sigma^n I\mathbb Z]\longrightarrow -\mathrm{Hom}(\pi_n(X), \mathbb Z)\longrightarrow 0,$$ -and it splits noncanonically. (See Freed-Hopkins, §5.3.) -There's a more general version of this in skd's answer. -2. $I\mathbb C^\times$ and $\Sigma^n I\mathbb C^\times$ -These spectra provide an analogue of Pontrjagin duality. If $A$ is an abelian group, the set of maps $A\to\mathbb -C^\times$ is an abelian group under pointwise multiplication, and this is called the Pontrjagin dual of $A$. An -analogue for spectra might be the assignment $X\mapsto \mathrm{Hom}(\pi_nX, \mathbb C^\times)$, the group of -characters of the $n$th homotopy group of $X$. This is precisely what $[\Sigma^n X, I\mathbb C^\times]$ -is; more broadly, one could describe $I\mathbb C^\times$ as the spectrum whose cohomology theory is calculated by -$$(I\mathbb C^\times)^n(X) = \mathrm{Hom}(\pi_{-n}(X), \mathbb C^\times).$$ - -As an addendum, I'm guessing this question arose because of the appearance of these spectra in physics, -specifically in the classification of invertible topological field theories. Following Freed-Hopkins, the -classification of invertible TQFTs $\mathsf{Bord}_n\to \mathsf C$, where $\mathsf C$ is some target symmetric -monoidal $(\infty, n)$-category, is equivalent to the abelian group of homotopy classes of maps between the -classifying spectrum of the groupoid completion of $\mathsf{Bord}_n$ and the classifying spectrum of the groupoid -of units of $\mathsf C$. The classifying spectrum of $\mathsf{Bord}_n$ is determined by -Schommer-Pries, and for certain reasonable choices of $\mathsf C$, we get -$\Sigma^nI\mathbb C^\times$ and $\Sigma^{n+1} I\mathbb Z$ (for small $n$, and conjecturally for all $n$). -For small $n$, we know some good chocies for $\mathsf C$: for example, if we let $n = 1$, we can take $\mathsf C$ -to be the category of super-vector spaces, and for $n = 2$ we can take the Morita 2-category of superalgebras. In -both cases, the classifying spectrum is the connective cover of $\Sigma^n I\mathbb C^\times$. This suggests that in -higher dimensions $n$, we might find symmetric monoidal $n$-categories $\mathsf C$ whose classifying spectra -continue this pattern, seeing more and more of $I\mathbb C^\times$, and the calculations of SPT phases coming out -of physics provide heuristic evidence for this conjecture. -This used the discrete topology on $\mathbb C$. If you instead give $\mathbb C$ the usual topology when defining the -categories of super-vector spaces or superalgebras, you get different classifying spectra: the connective covers of $\Sigma^{n+1} I\mathbb Z$ for $n = -1$, resp. $2$. Conjecturally, this pattern also continues further. Using the usual topology on $\mathbb C$ corresponds to classifying -deformation classes of invertible TQFTs rather than isomorphism, and again physics calculations provide some evidence for the conjecture.<|endoftext|> -TITLE: Do an unlinked trefoil and figure-eight cobound an annulus in $B^4$? -QUESTION [6 upvotes]: Let $K_1$ the trefoil (left or right hopefully does not matter?) and let $K_2$ be the figure-eight knot in $S^3 = \partial B^4$. Are there any smooth properly embedded annulus $A$ in $B^4$ with $\partial A = K_1 \coprod K_2$? -I'm really interested in knowing some general obstructions for this sort of thing for more general knots $K_1,K_2$ so any ideas are welcome. I figured that I would ask the question with two specific knots just for fun. - -REPLY [11 votes]: The relationship you're asking for is called concordance. Determining if knots are concordant is quite difficult: there are many concordance invariants, but no kind of global picture of what it means. -One concordance invariant is the knot signature. The trefoil has signature $-2$ and the figure-8 knot has signature $0$, so they are not concordant. I got this data from the wonderful website Table of Knot Invariants. -As far as where to read about knot concordance, a good place to start is Livingston and Naik's book-in-progress here. - -EDIT: Knots $K_0$ and $K_1$ are usually said to be (smoothly) concordant if there is a smoothly embedded annulus in $S^3 \times [0,1]$ connecting $K_0 \times \{0\}$ and $K_1 \times \{1\}$. In the comments, I was asked to sketch why this is the same as having an annulus connecting $K_0$ and $K_1$ in $B^4$ (assuming that $K_0$ and $K_1$ are unlinked). -There are a lot of ways to prove this. Here's a brief sketch of one of the easiest. If $D$ is a small open round ball in $S^3$, then -$$\left(S^3 \times [0,1]\right) \setminus \left(D \times [0,1]\right) \cong B^4.$$ -There is a tiny issue in that the right hand side is a manifold with corners, but as always corners can be smoothed. Using this diffeomorphism, we immediately see that if there is a smoothly embedded annulus in $B^4$ connecting $K_0$ and $K_1$, then after possibly moving $K_0$ and $K_1$ around we can obtain a smoothly embedded annulus in $S^3 \times [0,1]$ connecting $K_0 \times \{0\}$ and $K_1 \times \{1\}$; this annulus avoids $D \times [0,1]$. Conversely, if we can find an annulus in $S^3 \times [0,1]$ connecting $K_0 \times \{0\}$ and $K_1 \times \{1\}$, then homotoping everything we can assume that this annulus avoids some $D \times [0,1]$, and then transport it to an annulus in $B^4$ connecting $K_0$ and $K_1$. -The above sketch elides various identification we are making, but once you understand the geometry (draw a picture!) you'll see that they are all pretty clear.<|endoftext|> -TITLE: For which primes does this iterated function act transitively? (Sort of a finite analogue of Collatz conjecture.) -QUESTION [8 upvotes]: Background: I was trying to prove something having to do with cyclic group actions on matroids and was able to show that what I want holds if a particular elementary-looking number-theoretic property holds, but upon closer inspection it holds for some primes and not others and now I am wondering if this "elementary" problem is known/studied somewhere or perhaps it is an open problem--I have no idea since I only stumbled into it entirely accidentally. Even learning what area it should be considered, what potential tools there are for studying it, and what else it might be related to (even just keywords to search for!) would be helpful. -The problem statement: Let $p$ be an odd prime and consider the permutation of the set $\{1,\ldots,\frac{p-1}{2}\}$ defined by sending $n$ to $\frac{n}{2}$ if $n$ is even and sending it to $\frac{p-n}{2}$ if $n$ is odd. For which primes $p$ does this permutation have a single orbit? -Here are some examples of iterating this function: -$p=7$) $1 \mapsto 3 \mapsto 2 \mapsto 1$, so it is transitive. -$p=11$) $1 \mapsto 5 \mapsto 3 \mapsto 4 \mapsto 2 \mapsto 1$, so it is transitive. -$p=13$) $1 \mapsto 6 \mapsto 3 \mapsto 5 \mapsto 4 \mapsto 2 \mapsto 1$, so it is transitive. -$p=17$) $1 \mapsto 8 \mapsto 4 \mapsto 2 \mapsto 1$: here there are two orbits of size 4. -This reminds at least loosely in spirit of the Collatz conjecture but I presume (and hope!) it is much simpler since for each $p$ there are only finitely many numbers to consider. We tried some computer experiments and found lots of primes for which there is a single orbit and also lots for which there is not; we put the sequence of "good" primes in the OEIS and did not find any matches, which makes me suspect there is not something simple like a congruence condition on the prime, but I don't know any number theory so I really have no idea. - -REPLY [10 votes]: It's easier to work with the inverse permutation. Starting from $1$ and working backwards we end up with a sequence $1=2^0\to \pm 2^1\to \pm 2^2\to \cdots \to\pm 2^{\frac{p-1}{2}}=1$. Where the signs are uniquely determined so that $\pm 2^r\in \{1,2,\dots,\frac{p-1}{2}\}$. This cycle covers all numbers in $\{1,2,\dots,\frac{p-1}{2}\}$ exactly once if and only if $2^r\neq \pm 1\pmod{p}$ for any $1\le r\le \frac{p-3}{2}$. -Proposition: This happens precisely when either $2$ is a primitive root for $p$, or if $p=7\pmod 8$ and $2$ has order $\frac{p-1}{2}$. -Proof: In the situation that $2^{\frac{p-1}{2}}=-1\pmod{p}$ we have that $2^r$ for $r=1,2,\cdots, p-1$ forms a complete system of residues, therefore $2$ is a primitive root for $p$. We are left with the case where $2^{\frac{p-1}{2}}=1\pmod{p}$. This means that $2$ has order $\frac{p-1}{2}$ in $\mathbb Z/p\mathbb Z$ and moreover $-1$ is not of the form $2^r\pmod{p}$. In turn, this implies $2$ is the square of a primitive root, and so $2^r$ covers all quadratic residues which means $-1$ is not a quadratic residue. Since $\left(\frac{2}{p}\right)=1$ and $\left(\frac{-1}{p}\right)=-1$ we have $p=7\pmod 8$.<|endoftext|> -TITLE: Do finite simplicial sets jointly detect isomorphisms in the homotopy category? -QUESTION [9 upvotes]: Let $\mathcal{H}$ denote the homotopy category associated with the Kan-Quillen model structure on $\mathbf{sSet}$. Suppose we have a map $f\colon X \to Y$ between Kan complexes, such that for every finite simplicial set $K$ we have an isomorphism of the form: $$\mathcal{H}(K,X) \cong \mathcal{H}(K,Y)$$ induced by postcomposition with $[f]$. Is it true that $f$ is then a weak equivalence? - -REPLY [7 votes]: The answer is no. Otherwise, it would follow from Brown's representability theorem (and here I mean very specifically Theorem 2.8 from Brown's 1965 paper Abstract Homotopy Theory) that every "half-exact" functor on the homotopy category of unbased simplicial sets is representable. That is however false by my answer here. See also this post and the discussion below for some context.<|endoftext|> -TITLE: The discrete Hardy-Littlewood-Sobolev inequality -QUESTION [9 upvotes]: Let $p>1$, $q>1$, $0<\lambda<1$ be such that -$\frac{1}{p}+\frac{1}{q}+\lambda=2$. Suppose that -$(a_{k})\in \ell^{p}(\mathbb{Z})$ and $(b_{k})\in \ell^{q}(\mathbb{Z})$. -It is known ([1,2,3]) that -$$\sum_{j\neq k}\frac{a_{j}b_{k}}{|j-k|^{\lambda}}\leq C_{p,q} \parallel a \parallel_{p}\parallel b \parallel_{q}.$$ -If $\frac{1}{p}+\frac{1}{q}=1$ and $\lambda=1$, then the estimate fails. -Namely we have -$$\sum_{j\neq k,\, j,k=1,...,N }\frac{a_{j}b_{k}}{|j-k|}\geq C\log{N} \parallel a \parallel_{p}\parallel b \parallel_{q}.$$ -Question: What estimates do we still have when $\frac{1}{p}+\frac{1}{q}= 1$ -and $\lambda>1$ ? I expect the inequality to hold. Does it? -Observe that when $\frac{1}{p}+\frac{1}{q}<1$ and $\lambda>1$ the inequality fails. -A counterexample is $a_{k}=b_{k}=1$ for which we have -$\parallel a \parallel_{p}\parallel b \parallel_{q}=N^{\frac{1}{p}+\frac{1}{q}}$, while $\sum_{j\neq k,\, j,k=1,...,N }\frac{1}{|j-k|^{\lambda}}\geq C N$. -[1] G. H. Hardy, J. E. Littlewood, and G. Polya. Inequalities, volume 2. Cambridge at the University Press, 1952. -[2] Congming Li, John Villavert, An extension of the Hardy-Littlewood-Pólya inequality, Acta Mathematica Scientia, 31 (6), (2011), 2285-2288. -[3] Ze Cheng,Congming Li, An Extended Discrete Hardy-Littlewood-Sobolev Inequality, Discrete Contin. Dyn. Syst. 34 (5), (2014), 1951-1959 (arXiv:1306.1649, doi: 10.3934/dcds.2014.34.1951). - -REPLY [3 votes]: Write $$\sum_{j\ne k,\,j,k=1,\dots,N} \frac{|a_kb_j|}{|j-k|^{\alpha}}=\sum_{d=1}^{N}\frac{1}{d^{\alpha}} \Big(\sum_{l=1}^{N-d} |a_lb_{l+d}| + \sum_{l=1}^{N-d} |a_{l+d}b_{l}|\Big).$$ -By Holder's inequality one has -$$\sum_{l=1}^{N-d} |a_lb_{l+d}|\le \Big(\sum_{l=1}^{N-d} |a_l|^p \Big)^{1/p}\Big(\sum_{l=1}^{N-d} |b_{l+d}|^q\Big)^{1/q}\le \Big(\sum_{k=1}^{N} |a_k|^p \Big)^{1/p}\Big(\sum_{k=1}^{N} |b_{k}|^q\Big)^{1/q}$$ -for every $d\in \{1,\dots,N\}$. -Similarly, -$$\sum_{l=1}^{N-d} |a_{l+d} b_{l}|\le \Big(\sum_{k=1}^{N} |a_k|^p \Big)^{1/p}\Big(\sum_{k=1}^{N} |b_{k}|^q\Big)^{1/q}.$$ -Therefore, if $\alpha>1$, then you get the estimate -$$\sum_{j\ne k,\,j,k=1,\dots,N} \frac{|a_kb_j|}{|j-k|^{\alpha}} \le 2\zeta(\alpha)\Big(\sum_{k=1}^{N} |a_k|^p \Big)^{1/p}\Big(\sum_{k=1}^{N} |b_{k}|^q\Big)^{1/q}.$$ -If $\alpha=1$, then you get -$$\sum_{j\ne k,\,j,k=1,\dots,N} \frac{|a_kb_j|}{|j-k|^{\alpha}} \le C (\log N)\Big(\sum_{k=1}^{N} |a_k|^p \Big)^{1/p}\Big(\sum_{k=1}^{N} |b_{k}|^q\Big)^{1/q}$$ -for some constant $C>0$.<|endoftext|> -TITLE: Probability of commutation in a compact group -QUESTION [13 upvotes]: It is well known that if $G$ is a finite group, then the probability that two elements commutte is either $1$ (if $G$ is abelian) or less than or equal to $\frac58$. -If instead $K$ is a compact group, there exists a unique probability over $K$ that is left-invariant (Haar measure). The same problem as above still makes sense: What is the probability that two elements commutte ? In general, this probability can be non-trivial, because $K$ may have several connected components, being a (semi-)direct product of its neutral component $K_0$ with a finite group. If $K_0$ is abelian (a torus), then we are led back to the finite-case result. -So let me assume that $K=K_0$, that is $K$ is connected. Is it possible that the probability that two elements commutte be non-trivial, namely $0 0$. Then the set $A^{-1} A$ contains an open neighborhood of the identity. In particular, if $A$ is a subgroup of $G$ then $A$ is open. - -You can find it as Exercise 17.13 (ii) in Kechris, Classical Descriptive Set Theory. You begin by showing that the function $x \mapsto \mu(xA \triangle A)$ is continuous, which uses the regularity of the measure or else Lusin's theorem. (The proof I know uses dominated convergence and hence only shows that this function is sequentially continuous, which is why I am unsure about the non-metrizable case.) -So let $E = \{(x,y) : xy = yx\} \subset K \times K$. This set is closed, and in particular, measurable. The probability of two elements commuting is $p = (\mu \times \mu)(E)$ where $\mu$ is the left Haar probability measure on $K$. Let $C_x$ denote the centralizer of $x$, which is closed. We have $(x,y) \in E$ iff $y \in C_x$, and thus by Fubini's theorem -$$p = \iint 1_E(x,y)\,\mu(dy)\,\mu(dx) = \iint 1_{C_x}(y)\,\mu(dy)\,\mu(dx) = \int \mu(C_x) \,\mu(dx).$$ -If we suppose $p >0$, then the set $B = \{x : \mu(C_x) > 0\}$ must have positive measure. But if $\mu(C_x) > 0$, then by our lemma, $C_x$ is an open subgroup of $K$. It was already closed, and $K$ is connected, so $C_x = K$, i.e. $x$ is in the center $Z$ of $K$. Thus $B \subset Z$, so $Z$ is a closed subgroup with positive measure. Applying our lemma again, $Z$ is open and so $Z=K$, i.e. $K$ is abelian. -There is also be a "Baire category" analogue of this statement in the Polish case: if $E$ is nonmeager in $K \times K$, then $K$ is abelian. You could prove it by a nearly identical argument: use the Pettis lemma (Kechris 9.9) in place of Steinhaus above, and the Kuratowski-Ulam Theorem (Kechris 8.41) in place of Fubini. (Though maybe there's a simpler argument, since a nonmeager closed set has to have nonempty interior.)<|endoftext|> -TITLE: Is there a set $S\subseteq [0,1]$ with $|S|=2^{\aleph_0}$ and distinct pairwise distances? -QUESTION [6 upvotes]: Short version of question. Is there a set $S\subseteq [0,1]$ with $|S|=2^{\aleph_0}$ such that all points of $S$ have distinct pairwise distances? -Formal version of question. If $X$ is a set, let $[X]^2=\big\{\{x,y\}:x\neq y\in X\big\}$. Let $(X,d)$ be a metric space. Let $d_{\text{set}}:[X]^2\to \mathbb{R}$ be defined by $$\{x,y\}\in [X]^2\mapsto d(x,y)=d(y,x).$$ We say $S\subseteq X$ has the distinct pairwise distance property (dpdp) if the restriction of $d_{\text{set}}$ to $S$ is injective. Is there a set $S\subseteq [0,1]$ with (dpdp) and $|S|=2^{\aleph_0}$, where $[0,1]$ is endowed with the Euclidean metric? - -REPLY [13 votes]: Consider $\Bbb R$ as a vector space over $\Bbb Q$, choose (Zorn) a basis $S$ consisting (after rescaling with rational scalars) of elements in $(0,1)$. Then we have no relation of the shape $\pm(s_1-s_2)=(s_3-s_4)$ for $s_1,s_2,s_3,s_4\in S$. The cardinality of the basis is also the required one. - -REPLY [8 votes]: You can actually make $S$ a (copy of the) Cantor set. -First a claim: if $[a_1,b_1]$, $[a_2,b_2]$, ..., $[a_n,b_n]$ is a finite set of intervals, ordered left-to-right ($b_1 -TITLE: Minimal, uniquely ergodic but not Lebesgue-ergodic? -QUESTION [11 upvotes]: So here's my question: -Does there exist a minimal diffeomorphism of class at least $\mathcal{C^2}$ of a compact manifold X which is - -minimal -uniquely ergodic with unique probability measure $\mu$ -not ergodic with respect to the Lebesgue measure ? - -I don't really see why these requirements should contradict each other but I haven't been able to find an example. Note that the regularity hypothesis is necessary as (see R.W.'s answer below): there are $\mathcal{C}^1$ circle diffeomorphisms that satisfy those conditions, but one could argue that they are a bit artificial since as soon as the derivative is required to have bounded variation this can no longer be true. -I would also be happy with any example that is just a piecewise diffeomorphism! - -REPLY [2 votes]: I couldn't manage to find an online version, but this paper by Yoccoz provides an example of a diffeomorphism of the $2$-dimensional torus which is the product of two analytic circle diffeomorphisms, which is minimal, uniquely ergodic, and totally dissipative for Lebesgue measure (hence it cannot be ergodic).<|endoftext|> -TITLE: Counting configurations on a 2xn board under restrictions -QUESTION [7 upvotes]: Find the number of ways of selecting k cells from a $(2\times n)$-board such that no two selected cells share a side (non-adjacent). -For $n=3$ and $k=2$, the answer is $8$; for $n=5$ and $k=3$, the answer is $38$. -Counting manually works for small numbers. - -Question. Is there any formula that can be derived for large numbers? - -REPLY [3 votes]: Index the chosen cells from left to right, and let $c_i$ and $r_i$ be the column and row of cell $i$. Consider how many times your chosen cells switch from being in the bottom row to being in the top as you go from left to right, and call the number of switches $s$. The configuration is uniquely specified by: - -A choice of whether the first cell is in the top or bottom row. -Which cells are in the opposite row from the previous one. -The "distance" between subsequent cells, where distance between cell $i$ and $i+1$ is $c_{i+1}-c_{i}$ if they are in opposite rows and $c_{i+1}-c_{i}-1$ if they are in the same row. This definition ensures that distance 1 corresponds to the closest the two cells can be in either case. We must also specify the column of the first cell $c_1$ so we have a starting point. - -Example: - -The switches happen between cells 1 & 2 and between cells 3 & 4. The distances between successive cells are 1 (between cells 1 & 2), 1 (between 2 & 3), and 3 (between cells 3 & 4). -For fixed $s$, we can enumerate the number of possibilities for each of the above bullet points: - -There are two choices for the first row. -There are $\binom{k-1}{s}$ ways to choose where the switches occur. -We can evaluate the sum of the distances plus $c_{1}$ plus $n-c_{k}$: If we switched rows between every two chosen cells, it would just be $n$, but every time we don't switch we lose $1$ because the cells can't be adjacent. So the sum is $n-\left(\text{number of non-switches}\right) = n-(k-1-s)$. This sum is split into $k+1$ boxes, with at least 1 in each of the first $k$ boxes. Hence there are $\binom{n+1+s-k}{k}$ choices here. - -So the total number of combinations for fixed $s$ is $2\binom{k-1}{s}\binom{n+1+s-k}{k}$, and the total number of arrangements is the sum of this over $s=0,\dots,k-1$, $$\sum_{s=0}^{k-1} 2\binom{k-1}{s}\binom{n+1+s-k}{k}$$ -Happily, this agrees with your calculations for $n=3,k=2$ and $n=5,k=3$.<|endoftext|> -TITLE: Hodge theory (after Deligne) -QUESTION [23 upvotes]: In an interview with Deligne on the Simons Foundation website, I heard Robert MacPherson say that at the time Deligne's papers on Hodge theory were being published, the results seemed absolutely miraculous to specialists in Algebraic Geometry. -Can anyone explain why this is ? - -REPLY [21 votes]: A brief answer. -First of all, the results are miraculous. Deligne's Hodge II and Hodge III give just a few example applications of the kind of results you can prove using mixed Hodge theory; these are great theorems which just fall out of the general theory. People quickly figured out more applications, like the Hodge-Deligne polynomial, which is itself a complete miracle - and its existence just falls out of the general theory. -But there is also the fact that the results seemingly came out of nowhere. In fact Deligne was motivated by the philosophy of motives, which was not very well known/understood at the time outside the initiated few around Grothendieck. To my knowledge the only accounts of motives in the literature at the time were Kleiman's paper in the Oslo proceedings and Manin's paper on the blow-up formula for Chow motives, and it would be a long time until anyone talked about "mixed motives". The weight filtration appeared naturally in the $\ell$-adic cohomology when you considered a variety which was not necessarily smooth and projective, and the Frobenius eigenvalues on $\mathrm{Gr}^W_i H^k$ would be of absolute value $q^{i/2}$, so they would look like part of the degree $i$ cohomology of a smooth proper variety. And "by motivic philosophy" that phenomenon should correspond to something on the Hodge theory side. But without this motivic philosophy as guidance the weight filtration looks completely ad hoc and unmotivated.<|endoftext|> -TITLE: Exotic $R^4$ as the universal covering space -QUESTION [44 upvotes]: Is there a smooth compact 4-manifold whose universal covering is an exotic $R^4$, i.e. is homeomorphic but not diffeomorphic to $R^4$? -Remark. I am aware of examples (due to Mike Davis) of compact $n$-manifolds whose universal covering spaces are fake $R^n$'s i.e. are contractile but not homeomorphic to $R^n$. - -REPLY [37 votes]: This is problem 4.79(A) of Kirby's 1995 problem list, contributed by Gompf. It is my impression that it is still open. -There is some small progress: Remark 7.2 in this article observes that their constructions imply that a specific countable set of examples of exotic $\Bbb R^4$s cannot possibly cover a closed manifold. This is not a huge reduction, as only countably many exotic $\Bbb R^4$s could possibly be covers of the countable set of closed smooth 4-manifolds, anyway. But at least the examples are somewhat explicit. -I couldn't find any other references to this problem in the literature, but that doesn't mean there aren't any. -UPDATE: I emailed Bob Gompf; he is not aware of any recent progress.<|endoftext|> -TITLE: Generalizing Polar Decomposition of Matrices -QUESTION [5 upvotes]: I am trying to find a certain proof of polar decomposition of complex matrices which I think should exist more generally for a certain class of Lie groups. Recall that the polar decomposition of a nonsingular complex matrix $A \in \text{GL}_n(\mathbb{C})$ is an expression $A = UP$, where $U \in \text{GL}_n(\mathbb{C})$ is unitary and $P \in \text{GL}_n(\mathbb{C})$ is positive definite. -My attempt arises with a well known analogy between four classes of matrices and four sets of complex numbers: -\begin{array}{|c|c|} -\hline -\text{Subset of }\text{Mat}_{n \times n } (\mathbb{C}) & \text{Corresponding set in } \mathbb{C} \text{ } \\ -\hline -\text{Hermitian matrices} & \text{Real numbers}\\ \hline -\text{Skew-Hermitian matrices} & \text{Imaginary numbers} \\ \hline -\text{Hermitian positive definite matrices} & \text{Positive real numbers} \\ \hline -\text{Unitary matrices} & \text{The unit circle in } \mathbb{C}\\ \hline -\end{array} -The analogy is justified through the observation that each set of matrices in the left column is precisely the set of unitarily diagonalizable matrices with eigenvalues in the corresponding set on the right. Moreover, the exponential map of matrices sends hermitian matrices to positive definite matrices and skew-hermitian to unitary matrices, a further similarity. -In showing polar decomposition, we take from the case for $\mathbb{C}$; $\text{exp} : \mathbb{C} \rightarrow \mathbb{C}^*$ sends the reals to the positive reals, and imaginary numbers to the unit circle. Taking $\text{exp}$ of $z = \text{Re}(z) + i \ \text{Im}(z)$ gives us the polar decomposition $\text{exp}(z) = e^{\text{Re}(z)} e^{i\ \text{Im}(z)}$ into a positive real number and a complex number on the unit circle. This analogy is presumably where polar decomposition gets its name. -Put $\mathfrak{g} = \text{Mat}_{n \times n } (\mathbb{C})$, the Lie algebra of $G = \text{GL}_n (\mathbb{C})$. I would like to use the decomposition $\mathfrak{g} = i \mathfrak{h} \oplus \mathfrak{h}$, where $\mathfrak{h}$ is the sub-$\mathbb{C}$-vector space of hermitian complex matrices and $i \mathfrak{h}$ is the sub-Lie algebra of skew-Hermitian matrices to achieve a decomposition $A = UP$ of any $A \in \text{GL}_n (\mathbb{C})$ into a unitary matrix $U$ and a positive definite matrix $P$, making use of the exponential map. However, there is an obvious obstruction: $\text{exp}(A + B)$ is not $\text{exp}(A) \text{exp}(B)$ necessarily. So the question is, how might we work around this problem? -Now, the subgroup $U \subset G$ of unitary matrices acts on the sub-$\mathbb{C}$-vector space $P \subset G$ of hermitian positive definite matrices by $(U, P) \mapsto U^* P U$, so this theorem will show that $G$ has a semi-direct product decomposition $G = U \times P$ with product $(U, P)(U', P') \mapsto (UU' , U'^* P U' P)$. -It is possibly useful to note that the map $i \mathfrak{h} \rightarrow \text{End}_{\mathbb{C}}(\mathfrak{h})$ sending $s$ to $\text{ad}_s$, where $\text{ad}_s (h) = sh - hs$ is a Lie-algebra representation. - -REPLY [4 votes]: The basic answer to the question here is "Yes, there is a strong analogy via the Iwasawa decomposition for a semisimple Lie group". If I were trying to study this kind of question, I'd probably start with a MathSciNet search, which of course does require access. For example, searching for "lie group" and "polar decomposition" (written this way in two Anywhere boxes) returns about 38 items with authors such as Kostant along with many Russians. Browsing through such a list gives some insight into the historical development. -Since this is pretty far from my actual specialization, I can't comment in more detail on what you might find out. But the main point is that Math Overflow might not be the place to start. - -REPLY [3 votes]: I have always understood the polar decomposition as the matrix analogue of $z=|z|e^{i\,{\rm arg}z}$ -so a decomposition into modulus and argument; the OP want the matrix analogue of a decomposition into real and imaginary parts, $z=e^{{\rm Re}\,z}e^{i\,{\rm Im}\,z}$, which I have never encountered and may not exist for the reason mentioned by the OP ($e^{X+iY}\neq e^Xe^{iY}$). -In any case, if we do follow the first approach, then $P=(A^∗A)^{1/2}$ -and $U=AP^+$ (with $P^+$ the pseudo-inverse) gives the unique polar decomposition $A=UP$, with $P$ Hermitian positive semidefinite and $U^+=U^\ast$. -Generalizations of this polar decomposition exist (notably for non-square matrices, see -The Canonical Generalized Polar Decomposition, 2009), but these generalizations also follow the modulus-argument decomposition (and not the real-imaginary decomposition).<|endoftext|> -TITLE: Modules over Hopf Algebras and $E_2$-algebras -QUESTION [7 upvotes]: Justin Young has a paper on the brace bar-cobar duality between hopf algebras and $E_2$-algebras: https://arxiv.org/pdf/1309.2820.pdf -I was wondering if anybody knows of a nice relationship between the categories of left modules over a Hopf Algebra and left modules over the underlying e1 structure of an $E_2$ algebra. -In my mind, one should be able to use the $E_2$ structure to turn left modules into bi-modules in a systematic way and then use that the category of bi-modules is monoidal. -We know that the category of left modules over a Hopf algebra is monoidal. How are these monoidal categories related? -Also, a big related question I have is: can we pick an $E_3$-algebra related to the Hopf/$E_2$-algebra such that the category of left modules is braided and gives knot invariants similar to the Jones polynomial and other braided category invariants (from tangles)? -(Note: if any clarification is needed in my question please let me know and I will edit it.) - -REPLY [6 votes]: Let $A$ be a brace algebra and $B$ the Koszul dual bialgebra. There is a natural adjunction -$$ -\Omega\colon \mathrm{CoMod}_B\rightleftarrows \mathrm{LMod}_A -$$ -where, for instance, the functor $\mathrm{LMod}_A\rightarrow\mathrm{CoMod}_B$ sends $M\mapsto M\otimes B$ equipped with some standard differential. -For $\mathrm{LMod}_A$ the natural notion of weak equivalence is that of quasi-isomorphism and in $\mathrm{CoMod}_B$ a map $C_1\rightarrow C_2$ is a weak equivalence if $\Omega C_1\rightarrow \Omega C_2$ is a quasi-isomorphism. The localization of $\mathrm{LMod}_A$ with respect to quasi-isomorphisms is the derived category of $A$-modules and the localization of $\mathrm{CoMod}_B$ with respect to $\Omega$-quasi-isomorphisms is the coderived category of $B$-comodules. The above adjunction induces an adjoint equivalence between the corresponding localizations (see https://arxiv.org/abs/0905.2621 for details on all of these constructions). -$\mathrm{CoMod}_B$ indeed has a natural monoidal structure. However, $\mathrm{LMod}_A$ does not have any natural monoidal structure (there is no strict way to turn left modules over a brace algebra into right modules). There is, however, a monoidal structure on the derived category of $A$-modules. It is obtained using the Dunn--Lurie additivity equivalence $\mathcal{A}\mathrm{lg}_{\mathbb{E}_2}\cong \mathcal{A}\mathrm{lg}(\mathcal{A}\mathrm{lg})$. -Brace bar-cobar duality gives an adjunction $\mathrm{Alg}(\mathrm{CoAlg})\rightleftarrows \mathrm{Alg}_{\mathrm{Brace}}$. After inverting corresponding weak equivalences you obtain another equivalence of $\infty$-categories $\mathcal{A}\mathrm{lg}(\mathcal{A}\mathrm{lg})\cong \mathcal{A}\mathrm{lg}_{\mathbb{E}_2}$. As far as I know, nobody has compared Dunn--Lurie additivity for $\mathbb{E}_2$ with brace bar-cobar duality. Assuming such a comparison, the monoidal structure on the coderived category of $B$-comodules will indeed coincide with the monoidal structure on the derived category of $A$-modules.<|endoftext|> -TITLE: Lattice points in a square pairwise-separated by integer distances -QUESTION [5 upvotes]: Let $S_n$ be an $n \times n$ square of lattice points in $\mathbb{Z}^2$. - -Q1. What is the largest subset $A(n)$ of lattice points in $S_n$ that have the - property that every pair of points in $A(n)$ are separated by - an integer Euclidean distance? - -Is it simply that $|A(n)| = n$? -And similarly in $\mathbb{Z}^d$ for $d>2$? - -          - - -          - -$5 \times 5$ lattice square, $5$ collinear points. - - - -Q2. What is the largest subset $B(n)$ of lattice points in $S_n$, - not all collinear, that have the - property that every pair of points in $B(n)$ are separated by - an integer Euclidean distance? - - -          - - -          - -$5 \times 5$ lattice square, $4$ noncollinear points. - - -A $9 \times 9$ example with $5$ noncollinear points, -also based on $3{-}4{-}5$ right triangles, -is illustrated in the Wikipedia article -Erdős–Diophantine graph. - -REPLY [3 votes]: @FedorPetrov's idea leads in fact to the solution of Q1. It seems that it may also provide some progress for Q2, but this is beyond this answer. -I denote $[n]=\{1,2,\dots,n\}$; with standard conventions, we have $2[n]=\{2,4,\dots,2n\}$ and $2[n]-1=\{1,3,5,\dots,2n-1\}$. Let $A$ be a subset in $[n]^2$ with integer pairwise distances. We show by the induction on $n$ that - -(1) $|A|\leq n$, and (2) if $|A|=n$, then $A$ is collinear in a line parallel to a coordinate axis. - -The base cases $n=1,2$ are clear. For the step, we use Fedor's observation that no two points in $A$ have opposite parities of abscissas and oppositve parities of ordinates. So we may assume all ordinates are odd. Now $A=A_0\sqcup A_1$, where the abscissas of the points in $A_0$ are even, and those for $A_1$ are odd. -Set $k=\lceil n/2\rceil\geq 2$. If $|A_0|, |A_1|\leq k-1$, then $|A| -TITLE: A link between hooks, contents and parts of a partition -QUESTION [8 upvotes]: Let $\lambda$ be an integer partition: $\lambda=(\lambda_1\geq\lambda_2\geq\dots\geq0)$. Denote its conjugate partition by $\lambda'$. For example, if $\lambda=(4,3,1)$ then $\lambda'=(3,2,2,1)$. -Recall also the notation for the content of a cell $u=(i,j)$ in a partition is $c_u=j−i$. -I have made the following observation which seems interesting enough to ask here. - -QUESTION. Is this true? It might even be known. Is it? Any reference? - $$\sum_{i\geq1}\lambda_i^2=\sum_{u\in\lambda}(h_u+c_u).$$ - -REMARK 1. The above identity implies -$$\sum_{i\geq1}(\lambda_i^2+(\lambda_i')^2)=2\sum_{u\in\lambda}h_u.$$ -REMARK 2. It also implies that -$$\sum_{\lambda\vdash n}\sum_{i\geq1}\lambda_i^2=\sum_{\lambda\vdash n}\sum_{u\in\lambda}h_u.$$ - -REPLY [8 votes]: Another approach is to notice that $$\sum_{u\in\lambda}h_u=\sum_{u}d_u$$ -where $d_u=i+j-1$ for $u=(i,j)$. -Proof: The easiest way to see this is that both sides count the number of pairs $\{(i_1,j_1),(i_2,j_2)\}\in\lambda$ such that either $i_1=i_2$ and $j_1\le j_2$ or $j_1=j_2$ and $i_1\le i_2$. - -Using this we have -$$\sum_{u\in \lambda}(h_u+c_u)=\sum_{u\in \lambda} (c_u+d_u)=\sum_{(i,j)\in \lambda} (2j-1)=\sum_{i\geq 1}\sum_{j=1}^{\lambda_i}(2j-1)=\sum_{i\geq1} \lambda_i^2$$ -which gives us the equality we wanted.<|endoftext|> -TITLE: Elliptic curves and $GL(2)$ Iwasawa theory -QUESTION [5 upvotes]: Let $E$ be a elliptic curve without complex multiplication over a number field $F$. Let $F_n=F[E_{p^n}]$ and $F_{\infty}=F[E_{p^\infty}]$. So by a well know theorem of Serre, the Galois group $Gal(F_\infty/F)$ is an open subgroup of $GL_2(\mathbb{Z}_p)$. -Consider the natural norm maps on $E_{p^\infty}(F_m)$. I read somewhere that for $m$ sufficiently large, these maps are essentially just multiplication by $p^4$ and so the projective limit of the inverse system $E_{p^\infty}(F_m)$ is zero. -I don't know the proof of this. Can anyone provide me with a proof? That will be very helpful. - -REPLY [4 votes]: For $n$ large enough the Galois group of $F_n/F_{n-1}$ identifies with the kernel $G$ of $\operatorname{GL}_2(\mathbb{Z}/p^{n+1}\mathbb{Z}) \to \operatorname{GL}_2(\mathbb{Z}/p^{n}\mathbb{Z})$. Let $P$ and $Q$ be a basis of $E_{p^{n+1}}$. The norm of $P$ is written as a sum over $a$, $b$, $c$, $d$ modulo $p$ as -\begin{align*} N_G(P) &= \sum_{a,b,c,d} \begin{pmatrix} 1+ap^n & bp^n \\ cp^n & 1+dp^n\end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix}\\ -&= \sum_{a,b,c,d}\bigl( (1+ap^n)P+cp^nQ\bigr)\\ -&= p^4P+p^{n+3}(\sum_{a=0}^{p-1} a)\, P + p^{n+3} (\sum_{c=0}^{p-1} c)\, Q \\ -&= p^4 P\end{align*} -and similar for $Q$ so it is multiplication by $p^4$ on all of $E_{p^{n+1}}$.<|endoftext|> -TITLE: Is there a good name for the operation that turns $A\operatorname{-mod}$ and $B\operatorname{-mod}$ into $A\otimes B\operatorname{-mod}$? -QUESTION [12 upvotes]: The name pretty much says it all: there's a well-defined operation on categories equivalent to modules over some ring: if $\mathcal{C}_1=A\operatorname{-mod}$ and $\mathcal{C}_2=B\operatorname{-mod}$, then $\mathcal{C}_1\boxtimes\mathcal{C}_2 =A\otimes B\operatorname{-mod}$ is independent of the choice of $A$ and $B$ (since Morita equivalence of one of the factors will induce Morita equivalence of the tensor product). - -Is there some clever name for this operation? - -Apologies if I should know this; it's not such an easy thing to Google for. - -REPLY [8 votes]: I say “Deligne-Kelly tensor product” for this. Technically Deligne tensor product is for certain abelian categories and right exact functors, and Kelly is for finitely cocomplete categories and right exact functors, while here you want locally presentable categories and cocontinuous functors. But they’re all similar enough, see Section 3.1-3.2 of Ben-Zvi-Brochier-Jordan for a quick summary and some references. -(Aside: If you want understand the technical differences between Deligne and Kelly's versions the must-read paper is Lopez Franco.)<|endoftext|> -TITLE: Is there an explanation of analogies between the cross-ratio and the Riemann curvature tensor? -QUESTION [29 upvotes]: Define the cross-ratio of four real or complex numbers as follows: -$$[a,b,c,d] = \frac{(a-c)(b-d)}{(a-d)(b-c)}.$$ -Then its logarithm has the same symmetries as the curvature tensor: -$$\log[a,b,c,d] = -\log[b,a,c,d] = -\log[a,b,d,c] = \log[c,d,a,b].$$ -Moreover, if $[a,b,c,d] = \lambda$, then $[b,c,a,d] = 1 - \lambda^{-1}$ and $[c,a,b,d] = (1-\lambda)^{-1}$, which implies an analog of the algebraic Bianchi identity: -$$\log[a,b,c,d] + \log[b,c,a,d] + \log[c,a,b,d] = \pi i.$$ -Is there something behind these coincidences? - -REPLY [19 votes]: There is an indirect connection which goes via the representation theory of the symmetric group. The symmetries of the Riemann tensor are equivalent to saying that $R$ transforms according to the two-dimensional irreducible representation of $\mathbb S_4$ corresponding to the partition $[2,2]$. On the other hand let us consider the moduli space $M_{0,4}$ parametrizing four distinct ordered points on the Riemann sphere up to Möbius transformations. The cohomology group $H^1(M_{0,4},\mathbf C)$ is $2$-dimensional, and transforms according to the representation $[2,2]$ under its natural action of $\mathbb S_4$. But we may compute this cohomology group as the space of holomorphic $1$-forms on $M_{0,4}$ with at most logarithmic poles at infinity. Moreover, the cross-ratio $\chi$ may be considered as a holomorphic function on $M_{0,4}$ and its logarithmic derivative $d \log(\chi)$ is such a 1-form with log poles. (Note that differentiating your analogue of the Bianchi identity gets rid of the $\pi i$, so we really get something exactly like the usual Bianchi identity!)<|endoftext|> -TITLE: Improvement of Chernoff bound in Binomial case -QUESTION [6 upvotes]: We know from Chernoff bound -$P\bigg(X \leq (\frac{1}{2}-\epsilon)N\bigg)\leq e^{-2\epsilon^2 N}$ where -$X$ follows Binomial($N, \frac{1}{2}$). -If I take $N=1000, \epsilon=0.01$, the upper bound is 0.82. -However, the actual value is 0.27. Can we improve this Chernoff bound? - -REPLY [7 votes]: An explicit high-accuracy bound on the Kolmogorov distance between a properly standardized binomial distribution $B(N,p)$ and the standard normal distribution is provided by the theorem on page 129 of Uspensky's book "Introduction to Mathematical Probability" (1937). In your case, of $p=1/2$, this theorem implies that for any integer $x$ and any natural $N\ge100$ -\begin{equation} - |P(X\le x)-\Phi(z)|<\delta:=\frac{0.52}N+\exp(-\tfrac34\,\sqrt N), -\end{equation} -where $\Phi$ is the standard normal cumulative distribution function and $z:=(x-Np+1/2)/\sqrt{N/4}$. In particular, for your $N=1000$, we have $\delta=0.000520000050\ldots<\frac1{1923}$. So, for $N=1000, \epsilon=0.01$, Uspensky's result implies that the true probability (which is $0.27398\ldots$) is in the interval $[0.27397\ldots-\frac1{1923},0.27397\ldots+\frac1{1923}]\subset[0.27345, 0.27450]$, which is much more informative than -the Chernoff and entropy upper bounds on $0.27398\ldots$, which both are $\approx0.82$. -As was commented by user RaphaelB4, the Chernoff and entropy upper bounds are bounds on probabilities of large deviations. Those probabilities are small, in contrast with your probability $0.27398\ldots$, which makes large deviation bounds unsuitable in your example. Also, there are large deviation bounds much better than Chernoff's. E.g., in a more general situation one has an upper bound on large deviation probabilities that is asymptotic to the corresponding Gaussian tail.<|endoftext|> -TITLE: A variant of Cauchy-type functional equation conjecture -QUESTION [5 upvotes]: Let $f:\mathbb{C}\to \mathbb{C}$ be a complex function such that -$$|f(x-y)|=|f(x)-f(y)|,\qquad x,y\in\mathbb{C}.$$ -Is it true that $$f(x+y)=f(x)+f(y),\qquad x,y\in\mathbb{C}?$$ -The answer is affirmative when $f:\mathbb{R}\to\mathbb{R}$ is a real function, and the proof is not difficult. The above generalization seems harder: I inferred it from a conjecture that I saw in a paper. - -REPLY [13 votes]: counterexample -$$f(x)=1-e^{\Re(x) i}$$<|endoftext|> -TITLE: Maps from 2-Torus to SO(3) -QUESTION [6 upvotes]: Could someone please point me to a reference for topologically nontrivial maps from 2-Torus to SO(3), and how they are classified? [I'm a physicist, so a simple explanation would be useful] - -REPLY [16 votes]: Another way to see that there are four homotopy classes of maps $S^1\times S^1 \to SO(3)$ is to use the fact that $SO(3) = \mathbb{RP}^3$. So by cellular approximation, -\begin{align*} -[S^1\times S^1, SO(3)] &= [S^1\times S^1, \mathbb{RP}^3]\\ -&= [S^1\times S^1, \mathbb{RP}^{\infty}]\\ -&= [S^1\times S^1, K(\mathbb{Z}_2, 1)]\\ -&= H^1(S^1\times S^1; \mathbb{Z}_2)\\ -&\cong \mathbb{Z}_2^2. -\end{align*} -More generally, the same argument shows that the set of homotopy classes of maps $\Sigma_g \to SO(3)$ is in bijective correspondence with $H^1(\Sigma_g; \mathbb{Z}_2)$. In particular, there are $2^{2g}$ such classes, which is consistent with the statement in the final paragraph of Dan Ramras' answer. -As $\pi_2(G) = 0$ for a path-connected, finite-dimensional Lie group $G$, we can attach cells of dimension at least four to $G$ to obtain a $K(\pi_1(G), 1)$ which has the same three-skeleton as $G$. Hence, for any surface $\Sigma$ (orientable or otherwise) we have -\begin{align*} -[\Sigma, G] &= [\Sigma, K(\pi_1(G), 1)]\\ -&= H^1(\Sigma; \pi_1(G))\\ -&\cong \operatorname{Hom}(\pi_1(\Sigma), \pi_1(G))\\ -&\cong \operatorname{Hom}(H_1(\Sigma; \mathbb{Z}), \pi_1(G)) -\end{align*} -where the last step uses the fact that $\pi_1(G)$ is abelian. In particular, if $\Sigma$ is an orientable surface of genus $g$, then $H_1(\Sigma; \mathbb{Z}) \cong \mathbb{Z}^{2g}$ and therefore there are $|\pi_1(G)|^{2g}$ such maps. Again, this is consistent with Dan Ramras' answer.<|endoftext|> -TITLE: Probability of satisfying a word in a compact group -QUESTION [22 upvotes]: This question is inspired by Probability of commutation in a compact group, which asked whether $P(xyx^{-1}y^{-1} = 1)$ could take values strictly between $0$ and $1$ on a compact connected group. That question turned out to have a rather easy answer, but one that was quite particular to the word $xyx^{-1}y^{-1}$. - -Question: Let $w$ be an arbitrary word in $k$ letters, and let $G$ be a compact connected group. Let $x_1, \dots, x_k$ be drawn uniformly from $G$. Can $P(w(x_1, \dots, x_k)=1)$ be strictly between $0$ and $1$? - -By the Peter--Weyl theorem we may assume that $G$ is a compact connected Lie group. -There is some low-hanging fruit, like $w = [[x,y],z]$, but that's still very specialized. Admittedly I am not even clear on the words $w = x^n$. - -REPLY [20 votes]: The negative answer follows easily from a very useful fact that should be better known, so I am writing it explicitly. -Fact (Tannaka, Chevalley): -Every (connected) compact Lie group is isomorphic as a topological group to the group of real points of a (connected) reductive affine real algebraic group. -Moreover, the Haar measure on such a group is given by a volume form on the corrseponding variety. -We thus may view (for a given word $w$ in the rank $k$ free group $F_k$) the word map $w:G^k\to G$ as a morphism of real varieties and its solution space (the fiber over the identity element) as a the real points of a closed real subvariety. Note that $G^k$ is an irreducible variety, as $G$ is by its connectedness, thus this subvariety will be either $G^k$ itself, or a lower dimensional one. Its measure, accordingly, will be either 0 or 1. - -We could be more precise: - -If $G$ is trivial then the measure is 1. -If $G$ is non-trivial abelian then it is isomprphic to a torus $\text{SO}(2)^n$, and it is easy to see that the measure is 1 iff $w$ is in the commutator group of $F_k$, otherwise the measure is 0. -If $G$ is non-abelian then it contains a free group on $k$ generators (by Tits alternative, if you want to hit it with a hammer), thus if $w\neq 1$ the corresponding subvariety is proper and its measure is necessarily 0. If $w=1$ then the measure of 1.<|endoftext|> -TITLE: Laplacian of an infinite graph and connected components -QUESTION [5 upvotes]: For a finite graph with undirected, unweighted edges, a well-known result is that the dimension of the null space of the Laplacian matrix gives the number of connected components. Does this result apply to infinite graphs as well? -The infinite graphs I'm interested are locally finite. That is, the degree of each node is finite. In my case, the number of nodes is countably infinite, there are no self-edges, and the edges are undirected. -At least according to the PDF from this course: -http://www.maths.nuigalway.ie/~rquinlan/linearalgebra/section3-1.pdf -the connected components theorem does not assume anything about finite graphs. Can someone provide a reference (a paper or text) where this is explicitly discussed? -Thank You! - -REPLY [6 votes]: As Uri Bader points out, the infinite tree has an infinitely dimensional space of harmonic functions, so this is an answer to the philosophical part of the question: The way you prove that all harmonic functions on a finite graph are constant is by using a maximum principle (the value of a function at a vertex is the average over the neighbors, so if there is a vertex where the value of the function is maximal, the function must assume the same value on all the neighbors, etc. However, a function on an infinite set need not assume its supremum, which is where the argument breaks down.<|endoftext|> -TITLE: Pontryagin square and $\frac{1}{2}(\mathcal{P}(x) -x^2) =x \cup_1 Sq^1 x$ -QUESTION [7 upvotes]: The Pontryagin square, maps $x \in H^2({B}^2\mathbb{Z}_2,\mathbb{Z}_2)$ to $ \mathcal{P}(x) \in H^4({B}^2\mathbb{Z}_2,\mathbb{Z}_4)$. Precisely, -$$ - \mathcal{P}(x)= x \cup x+ x \cup_1 2 Sq^1 x. -$$ -The $\cup_1$ is a higher cup product. The $Sq^1 x= x \cup_1 x$. -It shall be true that -$$ -\mathcal{P}(x) \mod 2= x \cup x. -$$ - -Question 1: - If $ x \cup x =0 \mod 2$, and if $ x \cup_1 Sq^1 x =0 \mod 2$, is it true that - $$ -\mathcal{P}(x) =0 \mod 4? -$$ - -If not, please provide some counter examples. - - --- - -Question 2: - $\mathcal{P}(x)$ is a well-defined invariant for the cobordism $\Omega^4_{SO}(B^2 \mathbb{Z}_2)=\mathbb{Z}_4$. - Is - $$ -\frac{1}{2}(\mathcal{P}(x) -x^2) \mod 2 = x \cup_1 Sq^1 x \mod 2 -$$ - a well-defined invariant - of the cobordism $\Omega^4_{Spin}(B^2 \mathbb{Z}_2)=\mathbb{Z}_2$? -Question 3: Please provide the correct way to write the cobordism generator of $$\Omega^4_{Spin}(B^2 \mathbb{Z}_2)=\mathbb{Z}_2.$$ - -REPLY [5 votes]: (1) No, not for any reasonable interpretation of your condition "$x \cup_1 Sq^1 x = 0$". Consider $M= S^2 \times S^2$, let $y \in H^2(M; \mathbb{Z})$ be the sum of the two obvious generators, and let $x \in H^2(M; \mathbb{Z}_2)$ be the mod 2 reduction of $y$. Then $P(x)$ is the mod 4 reduction of $y^2 = 2$, but certainly $x^2 = 0$ mod 2 and $Sq^1 x = 0$. -(2) No, I don't think either side of your equation is a well-defined element of $H^4(M; \mathbb{Z}_2)$ in general. Rather, Wu's theorem implies that if $M$ a closed spin 4-manifold and $x \in H^2(M;\mathbb{Z}_2)$ then -$x^2 = Sq^2x = w_2 x = 0$ (mod 2), so since the mod 2 reduction of $P(x)$ equals $x^2$ we get that $P(x) \in 2\mathbb{Z}_4$. -Hence $\frac12 P(x)$ can be interpreted as a well-defined homomorphism $\Omega_4^{Spin}(B^2\mathbb{Z}_2) \to \mathbb{Z}_2$. -(3) Taking $M = S^2 \times S^2$ and $x$ as in (1) shows that $\frac12 P(x)$ is non-trivial.<|endoftext|> -TITLE: A link between hooks and contents: Part II -QUESTION [8 upvotes]: This is a question in the spirit of an earlier problem. -Let $\lambda$ be an integer partition: $\lambda=(\lambda_1\geq\lambda_2\geq\dots\geq0)$. -Recall also the notation for the content of a cell $u=(i,j)$ in a partition: $c_u=j−i$. Further, let $h_u$ denote the hook-length of the cell $u$. -The below identity is rather cute for which I don't remember a reference or a proof, so here I ask. - -Identity. Given a partition $\lambda\vdash n$, it holds that - $$\sum_{u\in\lambda}h_u^2=n^2+\sum_{u\in\lambda}c_u^2.$$ - -For example, if $\lambda=(4,3,1)\vdash 8$ then the hooks are $\{6,4,3,1,4,2,1,1\}$ and the contents are $\{0,1,2,3,-1,0,1,-2\}$. Hence -\begin{align} -LHS=6^2+4^2+3^2+1^2+4^2+2^2+1^2+1^2&=84 \\ -RHS=\mathbf{8^2}+0^2+1^2+2^2+3^2+(-1)^2+0^2+1^2+(-2)^2&=84. -\end{align} - -REPLY [4 votes]: Similarly to the other question, one can give a direct proof, but it requires a few more identities. Given a cell $u\in \lambda$ with coordinates $(i,j)$ we can check that $h_u+c_u=\lambda_i+\lambda'_j-2i+1$ and $h_u-c_u=\lambda_i+\lambda'_j-2j+1$ -so we can compute -$$\sum_{u\in \lambda} \left(h_u^2-c_u^2\right)=\sum_{(i,j)\in \lambda}\left(\lambda_i+\lambda'_j-2i+1\right)\left(\lambda_i+\lambda'_j-2j+1\right)$$ -$$=\sum_{(i,j)\in \lambda}\left(\lambda_i^2+(2-2i-2j)\lambda_i+\lambda_j'^2+(2-2i-2j)\lambda_j'\right)+2\sum_{(i,j)\in \lambda}\lambda_i\lambda_j'+\sum_{(i,j)\in\lambda}(2i-1)(2j-1).$$ -In order to simplify this we can compute the summands individually as follows: -$$\sum_{(i,j)\in \lambda}(\lambda_i^2-2i\lambda_i)=\sum_i (\lambda_i^2-2i\lambda_i)\sum_{j=1}^{\lambda_i}1=\sum_i (\lambda_i^3-2i\lambda_i^2)\tag{1}$$ -$$\sum_{(i,j)\in \lambda}(\lambda_j'^2-2j\lambda_j')=\sum_{j}(\lambda_j'^3-2j\lambda_j'^2)\tag{1'}$$ -$$\sum_{(i,j)\in \lambda}(2j-2)\lambda_i=\sum_i\lambda_i\sum_{j=1}^{\lambda_i}(2j-2)=\sum_{i}(\lambda_i^3-\lambda_i^2)\tag{2}$$ -$$\sum_{(i,j)\in \lambda}(2i-2)\lambda_j'=\sum_j (\lambda_j'^3-\lambda_j'^2)\tag{2'}$$ -$$\sum_{(i,j)}\lambda_i\lambda_j'=\sum_i \lambda_i\sum_{j=1}^{\lambda_i}\lambda_j'=\sum_i i\lambda_i^2+\sum_{i_1 -TITLE: A curious process with positive integers -QUESTION [42 upvotes]: Let $k > 1$ be an integer, and $A$ be a multiset initially containing all positive integers. We perform the following operation repeatedly: extract the $k$ smallest elements of $A$ and add their sum back to $A$. Let $x_i$ be the element added on $i$-th iteration of the process. The question is: is there a simple formula describing $x_i$, or can they be computed faster than simulating the process? One can easily see that for $k = 2$ we have $x_i = 3i$, but no simple pattern is evident for $k > 2$. -UPD: thought I would add some actual numbers and observations. -Here's what happens for $k = 3$ (bold numbers are those not initially present in the set): -$x_1 = 1 + 2 + 3 = \mathbf{6}$ -$x_2 = 4 + 5 + 6 = \mathbf{15}$ -$x_3 = \mathbf{6} + 7 + 8 = \mathbf{21}$ -$x_4 = 9 + 10 + 11 = \mathbf{30}$ -$x_5 = 12 + 13 + 14 = \mathbf{39}$ -$x_6 = 15 + \mathbf{15} + 16 = \mathbf{46}$ -The sequence continues with $54, 62, 69, 78, 87, 93, 102, 111, 118, 126, 135, \ldots$ -One observation is that extra numbers are far apart from each other, enough so that no two extra numbers end up in the same batch, hence each batch is either a run of $k$ consecutive numbers, or a run of $k - 1$ numbers with one extra. -If we look at consecutive differences $\Delta_i = x_{i + 1} - x_i$, we get a sequence $9, 6, 9, 9, 7, 8, 8, 7, 9, 9, 6, 9, 9, 7, 8, 9, 6, \ldots$ It looks like it can be split into triples with sum $24$ (implying $x_{3i + 1} = 6 + 24i$ for whatever reason?). Further update: a similar pattern seems to persist for any $k$: empirically $x_{ki + 1} = k(k + 1) / 2 + i(k^3 - k)$ for any integer $i \geq 0$. -Looking further down the sequence, there's a hint of periodicity which never seems to amount to much. (Since this answer was becoming cluttered I've removed the huge table of differences, but one can probably find them in the edit history.) -UPD: Bullet51 discovered what seems to be a complete solution for the case $k = 3$. Understanding how and why it works may be the key to cracking the general case as well. -UPD: Following in Bullet51's steps I decided to try my hand at constructing some finite state machines for larger $k$ (see their answer below for the legend). This resulted in pictures I feel painfully obliged to share. -$k = 4$: - -$k = 5$: - -$k = 6$: - -$k = 7$: - -I've verified all of these FSMs for the first $10^7$ differences in each case. Hopefully someone can make sense of what's going on here. - -REPLY [2 votes]: Here is another way of studying these processes. In the sequel $k=3$ but it is clear how to generalize. -The idea is to look at the matrices formed by three consecutive triples used in the insertion process. -To this end we write the triples produced by the process into an infinite matrix $S$ with $3$ columns (indexed by 0,1,2), and rows -indexed $1,2,,\ldots$. -The $i$-th row of $S$ consists of the triple (in order) which sums to $x_i$. -So the top of $S$ is -\begin{align*} - \begin{matrix} - c0&c1&c2&\;\;x\\ - 1&2&3&\;\;6\\ - 4&5&6&\;\;15\\ - 6&7&8&\;\;21\\ - 9&10&11&\;\;30\\ - 12&13&14&\;\;39\\ - \ldots -\end{matrix}\end{align*} -We let $M(n)$ be the $3\times 3$ submatrix of $S$ which consists of the rows $3n-1,3n,3n+1$ of $S$, -and let $T_0(n),T_1(n),T_2(n)$ be the $3 \times 3$ matrices -\begin{align*} - T_0(n):=\begin{pmatrix} - 8n-4 & 8n-3 & 8n-2\\ - 8n-2 & 8n-1 & 8n\\ - 8n+1 & 8n+2 & 8n+3\\ - \end{pmatrix},\;\; - T_2(n):=\begin{pmatrix} - 8n-4 & 8n-3 & 8n-3\\ - 8n-2 & 8n-1 & 8n\\ - 8n+1 & 8n+2 & 8n+3\\ - \end{pmatrix}\,\;\end{align*} -\begin{align*} - T_1(n):=\begin{pmatrix} - 8n-4 & 8n-3 & 8n-2\\ - 8n-1 & 8n-1 & 8n\\ - 8n+1 & 8n+2 & 8n+3\\ -\end{pmatrix}\;\;. -\end{align*} -Our aim is to show: -Lemma: for all $n$ $$M(n)\in\{T_0(n),T_1(n),T_2(n)\}$$ -Auxiliary considerations: -(1) Insertion of $x_n$. -$x_n$ is inserted after the first appearance of its value. Since $n-1+x_n$ elements of $S$ are less or equal than $x_n$ -$x_n$ will be inserted in row $\lceil{n+x_n \over 3}\rceil$ and in column $(n+x_n-1)\bmod 3$. -(2) Succession rules. Assume that $M(j)=T_i(n)$. Where will the produced $x$s be inserted? -Application of the insertion rule gives: -for $i=0$ : -$x_{3j-1}$ goes to row $8n-3+j$ , colummn $1$. -$x_{3j}$ goes to row $8n-1+j$ , colummn $2$. -$x_{3j+1}$ goes to row $8n+3+j$ , colummn $0$. -for $i=1$ -$x_{3j-1}$ goes to row $8n-3+j$ , colummn $1$. -$x_{3j}$ goes to row $8n-1+j$ , colummn $0$. -$x_{3j+1}$ goes to row $8n+3+j$ , colummn $0$. -for $i=2$ -$x_{3j-1}$ goes to row $8n-3+j$ , colummn $0$. -$x_{3j}$ goes to row $8n-1+j$ , colummn $2$. -$x_{3j+1}$ goes to row $8n+3+j$ , colummn $0$. -In other words: if $S(j)$ is of type $T_0$ it generates a type sequence $120$ later, type $T_1$ generates $100$, type $T_2$ generates $020$. -. -Proof of the lemma:(sketch) comparison shows that $M(1)=T_0(1)$. Iterated use of the succession rules -now shows that $(M(2),M(3),M(4))=(T_1(2),T_2(3),T_0(4))$, the next block of nine matrices has -indices $100, 020, 120$, and in general the indices of a following $3^{k+1}$ block are generated by substituting -$120$ for $0$, $100$ for $1$ and and $020$ for $2$ in the previous $3^k$ block. End of proof. -Some corollaries: -(1) for all n $$x_n\in \{8n-1,8n-2,8n-3\}$$ -Thus $$1+\lfloor{n+x_n \over 3}\rfloor=3n\;\;.$$ -(2) the row $3n+1$ is $8n+1,8n+2,8n+3$ and $x_{3n+1}=24n+6$ -Open question: has the replicative sequence above a simple ternary description? - -ADDED: the general case. -Slightly reformulating, we found above for $k=3$: -(1) the behaviour of $(x_n)$ is governed by the type sequence $(t_n)$, $x_n=8n-2$ if $t_n=0$, $x_n=8n-1$ if $t_n=1$, and $x_n=8n-3$ if $t_n=2$ -(in the description above $t_n$ can be visualized as the column index of $x_n$) -(2) the sequence $(t_n)$ is the fixed point starting from $0$ of the $3$-uniform morphism $\phi$ of the free monoid $\{0,1,2\}^*$ -given by $0\rightarrow 012,\;1\rightarrow 010,\;2\rightarrow 002$ -The general case can be treated in the same way, giving: -(1) the behaviour of $(x_n)$ is governed by its type sequence $(t_n)$. -in case $k$ is odd: -$t_n\in\{-\tfrac{(k-1)}{2},\ldots,\tfrac{k-1}{2}\}$ (the representatives of the residues $\bmod \,k$ in $\{-\tfrac{(k-1)}{2},\ldots, \tfrac{k-1}{2}\}$) -and for all $n$ -$$x_n=(k^2-1)n-\tfrac{k(k-1)}{2} + 1 +t_n$$ -(2a) let $b(0)=(0,1,2\ldots,k-1)$. For $p=1,..,\tfrac{k-1}{2}$ construct the word $b(p)$ from $b(0)$ by adding $p$ ($\bmod\,k$) to the value $\tfrac{k+1}{2}$ (leaving the rest unchanged). For $p=-\tfrac{(k-1)}{2},\ldots,-1$ -construct the word $b(p)$ by adding $p$ ($\bmod k$) to the value ${k-1 \over 2}$ (leaving the rest unchanged). -Then the sequence $(t_n)$ (with representatives $0,1,\ldots,k-1$ $\bmod k$) is the fixed point starting from $0$ of the $k$-uniform morphism $\phi$ of the free monoid $\{0,1,\ldots,k-1\}^*$ -given by $p\rightarrow b(p)$. -in case $k$ is even: -$t_n\in\{1,\ldots,k-1\}$ (i.e. $t_n=0$ does not happen), and for all $n$ -$$x_n=(k^2-1)n-\tfrac{k^2}{2} + 1 +t_n$$ -(2b) -for $p\in \{0,\ldots k-1\}$ let $b(p)= (\tfrac{k}{2},\tfrac{k}{2}+1,\ldots k-1,p,1\ldots,\tfrac{k}{2}-1)$ (with $p$ at position $\tfrac{k}{2}+1$). -Then the sequence $(t_n)$ is the fixed point starting from $\tfrac{k}{2}$ of the $k$-uniform morphism $\phi$ of the free monoid $\{0,1,\ldots,k-1\}^*$ -given by $p\rightarrow b(p)$. -By Cobham's theorem ("a sequence arising as (image under a coding of) a fixed point of a $k$-uniform morphism is $k$-automatic") the sequence $(t_n)$ is therefore $k$-automatic. -By the closure properties of the set of $k-$automatic sequences (under shifts, parallel generation and taking functions of output elements) then also the sequence $(\Delta_n)$ with -$$\Delta_n=(k^2-1)+t_{n+1}-t_n=x_{n+1}-x_n$$ -is $k$-automatic. (That is, $\Delta_n$ can be computed from the base-$k$ -digits of $n$ with a finite state machine of the type desribed above.)<|endoftext|> -TITLE: What are the monomorphisms of ($\infty$-)toposes? -QUESTION [17 upvotes]: There are standard notions of "surjections" and "embeddings" of toposes. However, not every surjection is an epimorphism, and not every regular monomorphism is an embedding. (EDIT: as Alexander Campbell points out in the comments, in this higher context, regular mono does not imply mono. So perhaps embeddings are not as strange as I make them out to be here. I still find it strange that most surjections are not epimorphisms, though.) -Let $f: \mathcal Y \to \mathcal X$ be a geometric morphism. Recall that $f$ is said to be - -surjective if $f^\ast$ is conservative (or equivalently, $f^\ast$ is comonadic.) -an embedding if $f_\ast$ is fully faithful (equivalently, $f^\ast$ is a localization). - -I'd say the correct notion of monomorphism / epimorphism is the $(\infty,1)$-categorical one: $f$ is a monomorphism iff the canonical square $f \circ 1 = f \circ 1$ is a pullback, and dually for epimorphisms. Since $(\infty,1)$-colimits of topoi are computed by taking $(\infty,1)$-limits of the inverse image functors between the underlying categories, $f$ is an epimorphism iff $f^\ast$ is a monomorphism of $(\infty,1)$-categories. That is, - -$f$ is an epimorphism iff $f^\ast$ is a replete subcategory inclusion, i.e. $f^\ast$ reflects the property of being isomorphic and is an inclusion of path components on hom-spaces. - -(Note that the coalgebras for any accessible left exact comonad on $\mathcal Y$ is an $\infty$-topos $\mathcal X$ which admits a canonical surjection from $\mathcal Y$ which will typically not be an epimorphism.) -As for monomorphisms, clearly if $f$ is an embedding, then it is a monomorphism. But not even every "regular monomorphism" is an embedding (EDIT: which is not to say that not every monomorphism is an embedding -- see Alexander Campbell's comment below!). For example, if $F,G: C \to D$ are functors, then the $(\infty,1)$-equalizer of the induced geometric morphisms $Psh(C) \to Psh(D)$ is presheaves on the iso-inserter of $F$ and $G$. The canonical map $Psh(IsoIns(F,G)) \to Psh(C)$ typically fails to be an embedding. Anyway, this leaves me with the question: -Question: What are the monomorphisms of topoi, or of $\infty$-topoi? -I expect this may be complicated, given how complicated monomorphisms of affine schemes are (a category which behaves in some ways similarly to the category of toposes). -Note that because every embedding is a monomorphism, by the surjection / embedding factorization system it suffices to determine which surjections are monomorphisms. - -REPLY [6 votes]: Edit: My original answer contained a big mistake, that I can't fix. A long time I had thought ago about monomorphisms of locales, and I wrongly convince myself that everything would generalizes to toposes, but I now realize things are more complicated than this... All my apologies about this. -I still can use what I said before to give an example of monomorphisms that are not embeddings, which is what I will explain now. -Given a locale $L$, there is an other locale $DL$, called the dissolution of $L$, endowed with a geometric morphism $ DL \rightarrow L$, which is universal for geometric morphism $p:K \rightarrow L$ such that for each open subspace $u \subset L$ (I.e. elements $u \in \mathcal{O}(L)$ of the frame corresponding to $L$), $p^* u$ is a complemented open subspace in $K$. $DL$ is explicitly constructed as the frames of nuclei of $L$, see for example Sketches of an elephant section C.1 for this construction. -Claim: For any non-boolean locale $L$, the geometric morphism $Sh(DL) \rightarrow Sh(L)$ is a monomorphisms of topos (or $\infty$-topos) that is not an embedding. -Indeed, $DL \rightarrow L$ is always a surjection, so if it is an embeddings it is an isomorphism, which only happen when $L$ is boolean. -Moreover, the map $DL \rightarrow L$ is also clearly a monomorphism in the $1$-category of locale: maps to $DL$ are a subset of maps to $L$ (those that send every open to a complemented element) -But as the functor $L \mapsto Sh(L)$ from the category of locales to the category of topos/$\infty$-topos is a right adjoint, it preserves finite limits and monomorphism. So $Sh(DL) \rightarrow Sh(L)$ is a monomorphism in the category of topos/ $\infty$-topos as well.<|endoftext|> -TITLE: Does $\mathrm{SL}_{n}(\mathbb{Z}/p^{2})$ have the same number of conjugacy classes as $\mathrm{SL}_{n}(\mathbb{F}_{p}[t]/t^{2})$? -QUESTION [21 upvotes]: Let $p$ be a prime; $\mathbb{F}_{p}$ is the field with $p$ elements -and $\mathbb{F}_{p}[t]$ the ring of polynomials in $t$ over $\mathbb{F}_{p}$. - -Does $\mathrm{SL}_{n}(\mathbb{Z}/p^{2})$ have the same number of conjugacy classes as $\mathrm{SL}_{n}(\mathbb{F}_{p}[t]/t^{2})$? - -When $p$ does not divide $n$ this follows from a theorem of P. Singla (see this paper). Note that the case when $p$ divides $n$ in this paper has a gap (see Section 5 here). In fact, when $p$ does not divide $n$, we have the stronger statement that the number of irreducible characters of degree $d$ is the same for both groups, for every $d$. However, we do not know the answer to the question in the title in general when $p$ divides $n$. -One can check that the answer is yes when $p=n=2$ (10 conjugacy classes) -and for $p=n=3$ (127 conjugacy classes), using GAP (the $n=2$ case -can also be done by hand), but for $n=4$, $p=2$ I don't know the -answer, mainly because the only way I know to create the group over -$\mathbb{F}_{p}[t]/t^{2}$ in GAP is via generators, and this seems -to be very computationally inefficient. - -REPLY [7 votes]: I would have preferred to not answer my own question, but here it goes. Yes, the two groups have the same number of conjugacy classes and in fact, the groups $\mathrm{SL}_{n}(W_{2}(\mathbb{F}_{q}))$ and $\mathrm{SL}_{n}(\mathbb{F}_{q}[t]/t^{2})$, for $q$ a power of a prime $p$ dividing $n$, have the same number of irreducible characters of dimension $d$ for every integer $d>1$. -This is proved here (arXiv link here).<|endoftext|> -TITLE: Is there another quantum deformation of sl(2)? -QUESTION [5 upvotes]: By looking at defining relations of standard deformation of $\mathfrak{sl}_2$, which are: -$$ -[E,F] = \frac{q^{H}-q^{-H}}{q-q^{-1}}, \quad [H,E] = 2E, \quad \text{ and } \quad [H,F] = -2F, -$$ -some questions come around. -For example, one can check that Jacobi identity is satisfied, but it would be also satisfied if one considers an arbitrary function $Fun(D)$ instead of the original one $\frac{q^{H}-q^{-H}}{q-q^{-1}}$. -I know that for the algebra to be quantum, the conditions of Hopf algebra have to be enjoyed. -But still, can we imagine another deformation for $\mathfrak{sl}_2$? Or there is a theorem which says that it is the only deformation? -Thanks in advance! - -REPLY [4 votes]: Regarding your second question, on other possible deformations of $sl(2)$: -There have been various studies on (multi-parametric) deformations of Lie algebras -as has already been mentioned in the comments to the OP- during the last decades: - An example of a $2$-parameter deformation $sl_{pq}(2)$ which leads to a quantum group is given by: -$$ -[H,E_{\pm}]=\pm E_{\pm}, \ \ \ [E_+,E_-]=\frac{q^{2H}-p^{-2H}}{q-p^{-1}} -$$ -where $E_-$ stands for $F$,up to a suitable rescaling and $p$, $q$ are complex parameters. - (setting $p=q$ and rescaling the generators produces the $q$-deformation described in the OP as a special case). You can find more details at arXiv:math/0506539, where this deformation is studied and it is proved that it admits a class of infinite dimensional representations with no analogue in the undeformed or in the $q$-deformed case. -Another -similar- example can be found in the article: A two-parameter deformation of the universal enveloping algebra of $sl(3,C)$, by J.F. Cornwell. A detailed discussion on the hopf structure of the deformed algebra and its implications on the usual hopf structure(s) of the undeformed algebra is also included. -In Introduction to quantum algebras, by M.R. Kibler, two parameter deformations such as $u_{pq}(2)$ and $u_{pq}(1,1)$ are studied: their hopf algebraic structures are investigated and their realizations (that is: homomorphisms or isomorphisms) with two parameter deformations of the Weyl algebras and the angular momentum algebras are used as a tool of investigating their representations.<|endoftext|> -TITLE: equidistributed parameters on graphs -QUESTION [7 upvotes]: Let $\mathcal G_n$ be the set of (isomorphism classes of unlabelled) simple graphs on $n$ vertices. -I wonder whether there are any 'interesting' combinatorial parameters $a,b: \mathcal G_n\to \mathbb N$, which are conjecturally equidistributed, that is, -$$ -\sum_{G\in\mathcal G_n} q^{a(G)} = \sum_{G\in\mathcal G_n} q^{b(G)}. -$$ -I would also be interested in such parameters where the proof is not entirely straightforward. - -REPLY [4 votes]: As a spin-off of https://mathoverflow.net/a/321171/3032, we have -equidistribution of the two parameters -$ -a(G) = \begin{cases} -1 & \text{if $G$ has no vertices of degree $1$}\\ -0 & \text{otherwise} -\end{cases} -$ -and -$ -b(G) = \begin{cases} -1 & \text{if $G$ has no two vertices with the same set of neighbours}\\ -0 & \text{otherwise} -\end{cases} -$ -This was shown bijectively by Kilibarda. His bijection preserves connectedness (but I have not yet understood it). -Kilibarda, Goran, Enumeration of unlabelled mating graphs, Graphs Comb. 23, No. 2, 183-199 (2007). ZBL1116.05038.<|endoftext|> -TITLE: Lower density of numbers not summable by consecutive integers -QUESTION [9 upvotes]: Let us call a positive integer $n\in\mathbb{N}$ consecutively summable if there are positive integers $m, k < n$ such that $$n=\sum_{i=0}^k (m+i).$$For $A\subseteq \mathbb{N}$ we set the lower density of $A$ to be $$\text{ld}(A)=\text{lim inf}_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}.$$ -If $N$ is the set of positive integers that are not consecutively summable, do we have $\text{ld}(N)=0$? - -REPLY [12 votes]: It is known that the only numbers not "consecutive summable" are the powers of $2$. -This is easy to prove: If $m+m+1+\ldots m+k=2^a$ then $2^a=\frac{(k+1)(2m+k)}{2}$. -This means that $2^{a+1}=(k+1)(2m+k)$ which is impossible since the differences form the two parentheses is an odd number. (And they should be both powers of two). -It is not that difficult to show that every number not of the form $2^a, a\in \mathbb{N}$ can be "consecutive summable". -I think the shortest way to prove both is here. - -This shows that $ld(A)=0$. - -REPLY [3 votes]: Let us denote $B=\mathbb N\setminus N$, i.e., the set of consecutive-summable numbers. -There already is an answer precisely characterizing the sets $N$ and $B$. (Which is much better result than just estimating the density.) If we are only interested in the density, another way to get that it is equal to zero is to notice the following. -If we fix a $k\in\mathbb N$, then we get the numbers of the form -$$n=km+\sum_{i=0}^k k = km + T_k$$ -where $T_k$ is the $k$-th triangular number. So we see that with finitely many exceptions $B$ contains the arithmetic progression $k\mathbb N+a$, where $a=T_k \bmod k$. -That means that $N$ is almost (i.e., with finitely many exceptions) contained in the union of the remaining $(k-1)$ congruence classes modulo $k$. -If we take any coprime $k_1,\ldots,k_n$, then we have that $N$ is almost contained in the union of corresponding congruence classes modulo $k_1\cdots k_n$. (Here we are using the Chinese remainder theorem which gives us the correspondence between the system of remainder modulo $k_1,\dots,k_n$ and a single congruence class modulo $k_1\cdots k_n$.) This gives us an estimate for the upper density -$$\operatorname{ud}(N) \le \frac{(k_1-1)\cdot (k_2-1) \cdot (k_n-1)}{k_1\cdot k_2 \cdot k_n} = \prod_{j=1}^n \left(1-\frac1{k_j}\right).$$ -In particular, if we take $k_n=p_n$ to be the $n$-th prime number, we get that -$$\operatorname{ud}(N) \le \prod_{j=1}^n \left(1-\frac1{p_j}\right)$$ -and since this is true for every $n$ we have -$$\operatorname{ud}(N) \le \prod_{j=1}^\infty \left(1-\frac1{p_j}\right)=0.$$ -You may notice that this argument would work for any set which can be related in a similar way to the union of distinct congruence classes if we have $\sum \frac1{k_j}=\infty$.<|endoftext|> -TITLE: Generating function for $3$-core partitions -QUESTION [8 upvotes]: Let $\lambda$ be an integer partition: $\lambda=(\lambda_1\geq\lambda_2\geq\dots\geq0)$. Further, let $h_u$ denote the hook-length of the cell $u$. -We call $\lambda$ a $t$-core partition if none of its hooks $h_u$ equals $t$. Define $c_t(n)$ to be the number of partitions of $n$ that are $t$-core partitions. It's well-known that -$$\sum_{n\geq0}c_t(n)\,q^n=\prod_{k=1}^{\infty}\frac{(1-q^{tk})^t}{1-q^k}.$$ -For example, $\sum_{n=0}^{\infty}c_2(n)\,q^n=\sum_{k=0}^{\infty}q^{\binom{k+1}2}$. -Now, consider only those partitions of $n$ with distinct parts and let $d_t(n)$ be the number of such partitions that are $t$-cores. Then it is easy to see $d_2(n)=c_2(n)$. - -QUESTION. Is this true? -$$\sum_{n\geq0}d_3(n)\,q^n=\sum_{k\geq0}q^{k^2} -+\sum_{k\geq1}q^{2\binom{k+1}2}.$$ - -Note that I have simplified the generating function from -$$\frac12\prod_{n\geq1}(1-q^{2n})(1+q^{2n-1})^2+\prod_{n\geq1}(1-q^{2n})(1+q^{2n})^2-\frac12.$$ - -REPLY [13 votes]: The set of $3$-core partitions can be described explicitly. -Theorem The partition $\lambda=\{\lambda_1,\lambda_2,\dots\}$ of length $k$ (that is, $\lambda_k > 0$ but $\lambda_{k+1} = \lambda_{k+2} = \cdots = 0$) is a $3$-core if and only if the sequence of differences $\{\lambda_1-\lambda_2,\lambda_2-\lambda_3,\dots,\lambda_k - \lambda_{k+1}\}$ is of the form $\{2,2,\dots,2,1,0,1,0,\dots ,1\}$ or $\{2,2,\dots,2,0,1,0,1,\dots ,1\}$. -Proof: It is easy to check by hand that a $3$-hook appears in the situations where - -a) some member of the sequence is $\geq 3$, -b) two members of the sequence in a row are $0$'s, -c) there is a $1$ in the sequence that is not followed by a $0$, -d) there is a $0$ in the sequence that is not followed by a $1$. - -These correspond to the four possible shapes of a $3$ hook-strip in the boundary. If all of these patterns are avoided, then the partition has no hooks of length $3$. -Therefore the partitions with distinct parts that are $3$-cores have a difference sequence $\{2,2,\dots,2\}$ or $\{2,2,\dots,2,1\}$. The size of the partitions in the first case are given by $2\binom{k+1}{2}$ and the sizes in the second case are given by $k^2$, where $k\geq 1$, and this implies your generating function.<|endoftext|> -TITLE: Reference for the algebro-geometric proof of Matsumoto theorem -QUESTION [13 upvotes]: Matsumoto proved in his PhD thesis that if $F$ is a field then $$K_2(F)=(F^*\otimes F^*)/(x\otimes (1-x)).$$ -The original Matsumoto proof as it is written in Milnor's book on algebraic K-theory looks not really nice to me and one guy told me that there is another proof of this fact that uses "sheaves of groups on Severi-Brauer varieties" which seems nicer to me. This guy told me that it is due to Vaserstein (Васерштейн) but it seems that Vaserstein was interested in other questions and couldn't give this proof. Perhaps you know whom this proof belongs to and I'd be very grateful if you could give me the reference to the article. - -REPLY [10 votes]: Apparently, what this one guy meant is Merkurjev's proof of Merkurjev-Suslin theorem, which was writen down by A. R. Wadsworth and later by W. van der Kallen. The latter paper starts with the statement of Matsumoto theorem and the proof does indeed use cohomology of sheaves of K-groups on Severi-Brauer varieties. -As for the nicer proofs of Matsumoto theorem, one can take a look at A new approach to Matsumoto's theorem by K. Hutchinson or $(t^2−t)$-reciprocities on the affine line and Matsumoto's theorem by F. Keune.<|endoftext|> -TITLE: Definition of unitary representation of $\mathbf G(\mathbb A_k)$ -QUESTION [6 upvotes]: Let $k$ be a global field, and let $G = \mathbf G(\mathbb A_k)$ for a connected, reductive group $\mathbf G$ over $k$. In these notes by Jayce Getz and Heekyoung Hahn, a unitary representation of $G$ is a Hilbert space $V$ together with a continuous homomorphism $\pi: G \rightarrow \operatorname{GL}(V)$ whose image is contained in the group $U(V)$ of unitary operators on $V$. -What is the topology on $\operatorname{GL}(V)$ (which I assume is the group of bounded linear operators on $V$) being considered here? Is it the induced topology coming from the norm topology? -I am trying to compare this definition with one given by Gerald Folland in A Course in Abstract Harmonic Analysis, which requires that for each $v \in V$ the map $g \mapsto \pi(g)v$ be continuous $G \rightarrow V$, where $V$ is taken in the norm topology. Are these two definitions of unitary representations different? -This matters because one later defines the Fell topology on the unitary dual $\hat{G}$ of $G$, and I want to know which representations are actually in $\hat{G}$. - -REPLY [4 votes]: Considering a topological group $G$, a Hilbert space $V$ and a corresponding unitary representation, that is a homomorphism $\pi:G\to U(V)$, the following are equivalent: - -$\pi$ is continuous when $U(V)$ is taken with the weak operator topology. -$\pi$ is continuous when $U(V)$ is taken with the strong operator topology. -For every $v\in V$, the orbit map $G\to V$ given by $g\mapsto gv$ is continuous. -The action map $G\times V\to V$ given by $(g,v)\mapsto \pi(g)(v)$ is continuous. - -In fact, the implications $4 \Rightarrow 3\Rightarrow 2\Rightarrow 1$ are trivially true, while $1\Rightarrow 4$ follows from the uniform convexity of $V$ (thus an analogue is valid for any isometric representation on a uniformly convex space). -The standard terminology is to refer to $\pi$ as a continuous unitary representation if it satisfies the properties above. - -It should be mentioned that for non-discrete locally compact groups (eg for your $\mathbf{G}(\mathbb{A}_k)$) unitary representations are almost never continuous when $U(V)$ is endowed with the norm topology, so a careful writer is unlikely to make such an assumption without mentioning it explicitly.<|endoftext|> -TITLE: Pontryagin square, Postnikov square and their consistency formulas -QUESTION [5 upvotes]: $\mathcal{P}_2$ is Pontryagin square -$$H^{2i}(M,\mathbb Z_{2^k})\to H^{4i}(M,\mathbb{Z}_{2^{k+1}}).$$ - -$\mathfrak{P}$ is the Postnikov square $$H^2(M,\mathbb Z_3)\to H^5(M,\mathbb Z_9).$$ - - - - -question (i) Can Pontryagin square and Postnikov square coincide with the Steenrod square, or the Generalized "Bockstein homomorphism" $\beta_p$, $\beta_p'$, $\beta_{2^n}$ or other homomorphisms? - -question (ii) Are there useful consistency formulas for these above "Pontryagin square" and "Postnikov square"? - - - - -Comments about question (i) -We know for $\mathbb Z_2$-valued cocycles $z_n$, -$$ -Sq^{n-k}(z_n) \equiv z_n\cup_{k} z_n -$$ -is always a cocycle. Here $Sq$ is called the Steenrod square. -More generally -$h_n \cup_{k} h_n$ is a cocycle if $n+k =$ odd and $h_n$ is a cocycle. -If we define a generalized Steenrod square for -cochains $c_n$: -$$ -\tilde Sq^{n-k} c_n \equiv c_n\cup_{k} c_n + c_n\cup_{k+1} d c_n . -$$ -We can check -$$ - d \tilde Sq^{k} c_n = d( -c_n\cup_{n-k} c_n + c_n\cup_{n-k+1} d c_n ) -$$ -$$ -= \tilde Sq^k d c_n, \;\;\;\; k=\text{odd} -$$ -$$ -=\tilde Sq^k d c_n +(-)^{n} 2 \tilde Sq^{k+1} c_n , \;\;\;\; k=\text{odd}. -$$ -This $$\tilde Sq^{2} c_2 \equiv c_2\cup_{0} c_2 + c_2\cup_{1} d c_2$$ almost is the same as the Pontryagin square $\mathcal{P}_2$ above, for ($i=1,k=2$ above) -$$H^{2}(M,\mathbb Z_{2})\to H^{4}(M,\mathbb{Z}_{4}),$$ -for -$$\mathcal{P}_2 (x_2) \equiv x_2\cup_{0} x_2 + x_2\cup_{1} d x_2.$$ - -Are these generalized Steenrod squares known? ($\tilde Sq^{n-k} c_n$) Where can I find more discussions along this? - - -Comments about question (ii) -For example, for Steenrod square, the total Stiefel-Whitney class $w=1+w_1+w_2+\cdots$ is related to the -total Wu class $u=1+u_1+u_2+\cdots$ through the total Steenrod square -$$ w=Sq(u),\ \ \ Sq=1+Sq^1+Sq^2+ \cdots . -$$ -Therefore, -$w_n=\sum_{i=0}^n Sq^i (u_{n-i})$. -The Steenrod squares have the following properties: -$$ -Sq^i(x_j) =0, \ i>j, \ \ -Sq^j(x_j) =x_jx_j, \ \ Sq^0=1, -$$ - -Do we have something similar for thse "Bockstein homomorphism?" $\beta_p$, $\beta_p'$, $\beta_{2^n}$? - -REPLY [7 votes]: You may be interested in the following paper: -Massey, W. S., Pontryagin squares in the Thom space of a bundle, Pac. J. Math. 31, 133-142 (1969). ZBL0188.28504. -Massey proves an analogue for the Pontryagin square of Thom's formula $w_k=\Phi^{-1}Sq^k(u)$ for the Stiefel-Whitney classes, which seems relevant to your question (ii).<|endoftext|> -TITLE: Equivalence of surjections from a surface group to a free group -QUESTION [14 upvotes]: Let $g \geq 2$. Let $S = \langle a_1,b_2,...,a_g,b_g | [a_1,b_1] \cdots [a_g,b_g] \rangle$ be the fundamental group of a genus $g$ surface and let $F_g$ be a free group with $g$ generators. Given two surjections $f_1,f_2 : S \to F_g$ is there a way to determine if there are automophisms $\phi: S \to S$ and $\psi: F_g \to F_g$ so that $f_1 = \phi \circ f_2 \circ \psi$? -Is there an example of two surjections $f_1,f_2$ that are not equivalent in the above way? -I asked the question on MSE before but didn't get much. - -REPLY [9 votes]: This is true, and it is written up in lemma 2.2 of "The co-rank conjecture for 3--manifold groups" by C. Leininger and A. Reid https://arxiv.org/abs/math/0202261. They state the result in slightly different language, that is they prove that any such epimorphism is induced by choosing a genus $g$ handlebody.<|endoftext|> -TITLE: Relation between Fourier coefficients and Satake parameters -QUESTION [11 upvotes]: Let $L(s)$ be an automorphic L-function (attached to a self contragredient automorphic representation on $GL(3)$), according to the following notations for $s$ of sufficiently large real part: -$$L(s) = \sum_{k=0}^\infty \frac{a_n}{n^s} = \prod_{p} \left( 1 - \alpha(p)p^{-s} \right)^{-1} \left( 1 - \beta(p)p^{-s} \right)^{-1} \left( 1 - \gamma(p)p^{-s} \right)^{-1}$$ -Straightforwardly developing the Euler product provides expressions of the Fourier coefficients $a_n$'s in terms of the Satake parameters $\alpha(p)$, $\beta(p)$ and $\gamma(p)$. I am not particularly aware of others standard useful relations between them. I bumped into the following one: -$$a_{p^k} = \frac{ -\left| \begin{array}{ccc} -\alpha(p)^{k+2} & \beta(p)^{k+2} & \gamma(p)^{k+2} \\ -\alpha(p) & \beta(p) & \gamma(p) \\ 1 & 1 & 1 -\end{array} \right| -}{ -\left| \begin{array}{ccc} -\alpha(p)^{2} & \beta(p)^{2} & \gamma(p)^{2} \\ -\alpha(p) & \beta(p) & \gamma(p) \\ 1 & 1 & 1 -\end{array} \right| -}$$ -I guess this can be verified, but even the case $k=1$ seems obscure to me. I do not want to believe such a formula to be a (verifiable) accident. Despite it works computationally, am I missing something lying behind? How strongly is the self-contragredience assumption necessary? -Any insight is welcome, as well as other ways to embrace the relations between spectral parameters and coefficients. - -REPLY [3 votes]: Let $s_{k}(\alpha_1(p),\ldots,\alpha_n(p))$ be the complete homogeneous symmetric polynomial of degree $k$ in variables $\{\alpha_1(p),\ldots,\alpha_n(p)\}$. If $\mathrm{Re}(s)$ is sufficiently large, then -$(*)~\sum_{k=0}^{\infty}\cfrac{a_{p^k}}{p^{ks}}=\prod_{j=1}^n (1-\alpha_j(p) p^{-s})^{-1} = \sum_{k=0}^{\infty}s_{k}(\alpha_1(p),\ldots,\alpha_n(p))p^{-ks}$. -This can be easily verified since the LHS is a product of geometric sums. To recover your second identity, we realize that if we let $\lambda(1)=k$ and $\lambda(j)=0$ for all $j\geq 2$, then -$s_k(\alpha_1(p),\ldots,\alpha_n(p))=\cfrac{\det[(\alpha_i(p)^{\lambda(j)+n-j})_{ij}]}{\det[(\alpha_i(p)^{n-j})_{ij}]}$, -where $i,j\in\{1,\ldots,n\}$. Note that the numerator is an alternating symmetric polynomial in $\{\alpha_1(p),\ldots,\alpha_n(p)\}$ (this follows from standard determinant properties), and hence is divisible by the Vandermonde determinant in the denominator. None of this relies on whether the underlying representation is self-contragredient. -The polynomial $s_k(\alpha_1,\ldots,\alpha_n)$ is a special case of the more general Schur polynomials. These are the characters of polynomial irreducible representations of $\mathrm{GL}(n)$. Moreover, $(*)$ is a special case of Cauchy's identity, which serves as a sort of orthogonality statement for Schur polynomials.<|endoftext|> -TITLE: Making the Fourier transform quantitative -QUESTION [7 upvotes]: I am undergraduate Physics student and understand that this is a professional mathematics forum. But due to perhaps broader interest, I hope this question is suitable for this website. -I understand the Fourier transform is a Hilbert space isometry -$F:L^2(\mathbb R^d) \rightarrow L^2(\mathbb R^d).$ -In Quantum Mechanics we are told that intuitively the Fourier transform transforms narrow signals into extended ones, I would like to make this precise. -The best example which illustrate this, is when one considers the Fourier transform on Schwartz distributions, $F(\delta_x)(k)=e^{ikx},$ but this is very unquantitative. -A good decay measure in "Fourier space" seem to be Sobolev spaces since -$$\left\lVert f \right\rVert_{H^s(\mathbb R^d)} = \left( \int_{\mathbb R^d} \left\lvert F(f) \right\rvert^2 (1+\vert x \vert^2)^{s} \ dx \right)^{1/2}$$ -Let us fix a signal $f \in L^2(\mathbb R^d)$ of unit norm. -I call the signal $f$ to be $\varepsilon,\delta-$localized if there is a ball $B(x,\delta)$ such that $\left\lVert f1_{B(x,\delta)} \right\rVert_{L^2} \ge 1-\varepsilon$. -I would say that the intuitive explanation of the Fourier transform should then imply that if $f$ is in the $H^s$ Sobolev space, with $s>0$ and $\left\lVert f \right\rVert_{H^s} \le k$ then for any $\delta(k)>0$ there is $\varepsilon'(\delta(k))>0$ such that $f$ cannot be $\varepsilon'(\delta(k)),\delta(k)-$localized. -Is this true? -Can one find this $\varepsilon'$ explicitly? -The above question is motivated by the statement that if a function is in $H^s$ with bounded norm, then it decays sufficiently rapidly in Fourier space that it cannot be too localized in real space. - -REPLY [4 votes]: Notice that $\|f1_{B(0,\delta)}\|_{L^2}\to 0$ as $\delta\to 0$. So the answer to the question is no because it should reasonably include the requirement $\varepsilon'<1$. -- Rapid decrease of the Fourier transform $F(f)$ implies smoothness of $f$ but, without further assumptions, it does not imply localization.<|endoftext|> -TITLE: Product of arithmetic progressions -QUESTION [7 upvotes]: Let $(a_1,a_2\ldots,a_n)$ and $(b_1,b_2,\ldots,b_n)$ be two permutations of arithmetic progressions of natural numbers. For which $n$ is it possible that $(a_1b_1,a_2b_2,\dots,a_nb_n)$ is an arithmetic progression? -The sequence is (trivially) an arithmetic progression when $n=1$ or $2$, and there are examples for $n=3,4,5,6$: -$n=3$: $(11,5,8), (2,3,1)$ -$n=4$: $(1,11,6,16), (10,4,13,7)$ -$n=5$: $(8,6,4,7,5), (4,9,19,14,24)$ -$n=6$: $(7,31,19,13,37,25), (35,11,23,41,17,29)$ - -REPLY [2 votes]: From Yaakov's answer, take the second difference $(u_i-2u_{i+1}+u_{i+2})$ to remove (1,2,...,n) and the constant term. Then you have (n-2)x3 matrix $W=[D^2(u),D^2(v),D^2(uv)]$ that we want to have a null vector. So calculate $det(W^tW)$, and you need the determinant to be zero for a solution. I ran the calculations and got no solutions for 8 or 9. Case 9 took 5 hours on Matlab.<|endoftext|> -TITLE: What is the current status of the Arnold conjecture? -QUESTION [10 upvotes]: Let $(M, \omega)$ be a symplectic manifold. V.Arnold conjectured that the number of fixed points of a Hamiltonian symplectomorphism is bounded below by the number of critical points of a smooth function on $M$. -The conjecture was proved in a weaker (homological) version using Floer homology, with two additional conditions: - -Compactness of $M$; -Non-degeneracy of the fixed points of the considered Hamiltonian symplectomorphism. - -In this setting, it is now fully understood that the number of fixed points is at least the sum of Betti number of $M$. - -I am wondering where the general statement stands today. More precisely: - -is compactness really necessary to Floer's framework ? -what is known for degenerate Hamiltonians ? -what is known for the critical points lower bound, rather than the Betti sum ? - -Thanks a lot - -REPLY [8 votes]: A recent (2017) overview of the status of Arnold's conjecture is given in The number of Hamiltonian fixed points on symplectically aspherical manifolds.<|endoftext|> -TITLE: Explicit proof that $c_0$-module $\ell_\infty$ is not projective -QUESTION [9 upvotes]: It is well known in narrow circles that the homological dimension (in the sense of relative Banach homology) of $c_0$-module $\ell_\infty$ is 2. As the corollary, this module is not projective. This proof is rather involved, its main ingridient is a lack of a right inverse for the mapping: -$$ -\Delta:c_0\;\hat{\otimes}\;c_0\to(c\;\hat{\otimes}\;c_0)\oplus(c_0\;\hat{\otimes}\;\ell_\infty): x\;\hat{\otimes}\;y\mapsto (x\;\hat{\otimes}\;y)\oplus(x\;\hat{\otimes}\;y) -$$ -in the category of left Banach $c_0$-modules. -I would like to see a more direct proof of non-projectivtity. The standard route would be to show that there is no right inverse $c_0$-morphism for the mapping $\pi:c \;\hat{\otimes}\; \ell_\infty\to \ell_\infty \colon a\; \hat{\otimes}\; x\mapsto a\cdot x$, where $c$ is the Banach space of convergent sequences. -Does anyone have an idea how to prove non-projectivity more or less directly? - -REPLY [3 votes]: If $\ell^\infty$ was projective then any Banach limit $L:\ell^\infty \to \mathbb{C}$ would extend to a $c_0$-module morphism $\bar{L}:\ell^\infty \to c$ -such that $L=\lim\circ \bar{L}$, but such does not exist. -Assume such $\bar{L}$ does exist. -Consider $r=\bar{L}(1)-1\in c$ -and observe that $\lim r=\lim\bar{L}(1)-1=L(1)-1=0$, thus $r\in c_0$. -Let $I\subset \mathbb{N}$ be the set of indices $n$ for which $r_n=-1$ and let $V<\ell^\infty$ be the vector space of sequences supported on $I$. Note that $I$ is finite, thus $V$ is finite dimensional. -For $s\in c_0$, $\bar{L}(s)=s\bar{L}(1)=sr+s$. -It follows that $c_0\cap \ker\bar{L}=V$. -But, as $V$ and $c$ are separable and $\ell^\infty$ is not, we can find $x\in \ker \bar{L}$ which is not in $V$. We consider the element $s=(1/n)\in c_0$ and observe that $sx$ is in $c_0\cap\ker\bar{L}$ but not in $V$. This is a contradiction.<|endoftext|> -TITLE: The outer automorphism of the dihedral group $D_4$ and quartic polynomials -QUESTION [6 upvotes]: Let $L$ be a dihedral quartic field. That is, we have $[L : \mathbb{Q}] = 4$ and the Galois closure $M$ of $L$ is a degree 8 extension of $\mathbb{Q}$ with Galois group isomorphic to the dihedral group $D_4$ of order 8. It is well-known (see the diagram on page 6 of this paper: https://arxiv.org/abs/1704.01729) that $M$ contains five subfields which are degree 4 extensions of $\mathbb{Q}$; say $L = L_1, L_2, L_0, L_3, L_4$. In particular, $L_1, L_2$ are Galois conjugates and contain a common quadratic field $K_1$, $L_0$ is a Galois extension of $\mathbb{Q}$ and has Galois group isomorphic to $C_2 \times C_2$, while $(L_3, L_4)$ is related to $(L_1, L_2)$ via an outer automorphism which comes from the group theory of $D_4$, which we denote by $\phi$. In this language, we identify $L_3 = \phi(L_1), L_4 = \phi(L_2)$ and $L_3, L_4$ are Galois conjugates. -Given $L$, it is of course possible (say via the primitive element theorem) to find a quartic polynomial $f$ such that for some root $\alpha$ of $f$ we have $L = \mathbb{Q}(\alpha)$. -My question is, given a quartic polynomial $f$ such that for some root $\alpha$ of $f$ we have $L = \mathbb{Q}(\alpha)$ is a dihedral quartic field, how to find a polynomial $g$ whose coefficients are algebraic functions of the coefficients of $f$ such that for some root $\beta$ of $g$ we have $\phi(L) = \mathbb{Q}(\beta)$? Moreover, can one take advantage of the invariant theory of the homogenized binary quartic form $F(x,y) = y^4 f(x/y)$? That is, is $g$ naturally given by some quartic covariant of $F$? - -REPLY [4 votes]: Let $\alpha=\alpha_1, \alpha_2, \alpha_3, \alpha_4$ be the conjugates of $\alpha$ in $M$, numbered so that one of the 4-cycles in the Galois group permutes them in the order $\alpha_1 \mapsto \alpha_2 \mapsto \alpha_3 \mapsto \alpha_4 \mapsto \alpha_1$. -Thus $L_1=\mathbb{Q}(\alpha_1) = \mathbb{Q}(\alpha_3)$ is fixed by that automorphism which interchanges $\alpha_2$ and $\alpha_4$. (Think of a square with corners labeled $1,2,3,4$ being reflected through the diagonal passing through corner number $1$. And $L_0$ corresponds to the central inversion symmetry of the square.) -The outer automorphism of $D_4$ amounts to looking at how rigid motions act on the set of unoriented edges of the square instead of on the corners: what fixes an edge is reflecting the square across the bisector. We're thus looking for an element fixed e.g. under simultaneously exchanging $\alpha_1$ with $\alpha_4$ and $\alpha_2$ with $\alpha_3$, and not fixed under the other non-identity permutations coming from the Galois group. In particular, it must not remain fixed when we reflect the chosen edge onto its opposite edge. -My first suggestion had been to set $\beta_1=(\alpha_4-\alpha_1)^2 - (\alpha_3-\alpha_2)^2$, but that turns out not to work at all - it always ends up in one of the quadratic subfields (and as has been pointed out in the comments, it sometimes vanishes outright!). -Revised approach: By adding a rational number if necessary, we can assume from the start that $\alpha_1 = -\alpha_3$ is traceless over the quadratic subfield of $L_1$, and thus of the form $\pm\sqrt{m\pm\sqrt{n}}$ with rational $m$ and $n\ne 0$, and its conjugates correspond to all the sign combinations. -As a warm-up exercise, $\gamma_1=\alpha_1/\alpha_4$ is then conjugate to its inverse (under $1\leftrightarrow 4, 2\leftrightarrow 3$) and also to its negative inverse $\alpha_2/\alpha_1$ (under a 4-cycle), and not fixed by either operation; it is fixed however by the central involution $1\leftrightarrow 3, 2\leftrightarrow 4$, and thus it generates the abelian quartic field $L_0$. -Shifting things away from the traceless subspace destroys these relationships: $(1+\alpha_1)/(1+\alpha_4)$ generically generates the normal closure $M$. (When it doesn't, replace $1$ with another rational number.) And adding its inverse to it produces an element invariant under the desired involution: -$$\beta_1=\frac{1+\alpha_1}{1+\alpha_4} + \frac{1+\alpha_4}{1+\alpha_1}$$ -It remains to compute the symmetric functions of $\beta_1$ and its conjugates and to express them in terms of those of the $\alpha_i$.<|endoftext|> -TITLE: Are there enough meromorphic functions on a compact analytic manifold? -QUESTION [5 upvotes]: Let $X$ be a compact complex analytic manifold, $D\subset X$ an irreducible smooth divisor, given as zeroes of a global meromorphic function $f\in {\mathfrak M} (X)$. Are there enough other meromorphic functions defining $D$? -Here is a precise question: can one find, for each $x\in X$, a meromorphic function $g\in {\mathfrak M} (X)$ such that $g$ is defined (i.e., not $\infty$) at $x$ and $D={\mathrm{zeroes}}(g)$? - -REPLY [10 votes]: First, let's clarify some possible confusion. A compact complex manifold does not admit non-constant holomorphic functions, so, assuming that $X$ admits non-constant meromorphic functions, you actually want the divisor of such a function to include poles, not just zeros. Infinity is thus a legitimate value. The points at which the function is (truly) undefined are called indeterminacy points. Following pretty much the notation and approach of Encyclopedia of Mathematics, \url{https://www.encyclopediaofmath.org/index.php/Meromorphic_function} this is the distinction: -Let $\Omega$ be a complex manifold (at this stage, we do not require compactness, algebraicity or anything more, and I will stick to the notation $\Omega$ to emphasize that the description which follows is local). Let $\mathcal{O}$ be the sheaf of germs of holomorphic functions on $\Omega$, and for each point $x \in \Omega$ let $\mathcal{M}_x$ denote the field of fractions of the ring $\mathcal{O}_x$ (the stalk of the sheaf $\mathcal{O}$ over $x$). Then $\mathcal{M}=\bigcup \mathcal{M}_x$is naturally endowed with the structure of a sheaf of fields, called the sheaf of germs of meromorphic functions in $\Omega$. A meromorphic function in $\Omega$ is defined as a global section of $\mathcal{M}$, i.e., a continuous mapping $f: x \to f_x$ such that for all $x \in \Omega$, $f_x \in \mathcal{M}_x$. The polar set $P_f$ (of codimension $1$) and the set of indeterminacy $N_f \subset P_f$ (of codimension at least $2$) are defined as follows: Let $f_x=\varphi_x/\psi_x, \quad \varphi_x, \psi_x \in \mathcal{O}_x$, with $\psi_x$ not identically $0$. Then $x \in P_f$ if $\psi_x(x)=0$ and $x \in N_f$ if $\varphi_x(x)=\psi_x(x)=0$. So at each point $x \in P_f\setminus N_f$ (a pole) one can define the value of $f$ to be $\lim_{y \to x}f(y)=\infty \in \mathbb{P}^1$. I cannot think of any general condition that would imply $N_f = \emptyset$. Even on an algebraic manifold this is not guaranteed. See the comment below. -Now to answer the question: in a compact complex manifold a meromorphic function (if admissible) is uniquely determined by its divisor, up to multiplication. If you are asking for ``another" meromorphic function with the same divisor, all you get will be a scalar multiple of the original one. -To broaden the context a bit, given a divisor $D$ in $\Omega$, finding a meromorphic function $f$ in $\Omega$ with prescribed divisor $D$ is one of Cousin problems (the multiplicative one). Its solvability depends on some cohomological conditions on the manifold. -Finally, a question ``how many meromorphic functions are there" (not what you are asking) can be also approached from the point of view of Siegel's theorem: if $X$ is a compact, connected, complex manifold of dimension $n$ and $\mathcal{M}(X)$ denotes the field of (globally defined) meromorphic functions on it, then the transcendence degree of $\mathcal{M}(X)$ over $\mathbb{C}$ does not exceed $n$. -Edit: If you are going to change $f$ into $g$ while keeping the zero part $Z_f$ of the divisor of $f$, remember that on a manifold $Z_f=P_{1/f}$. So you want to solve the (additive) Cousin problem of finding a meromorphic function $h$ on $X$ with prescribed polar part (that of $1/f$). This will be solvable in general if $H^1(X,\mathcal{O})=\emptyset$. If $H^1(X,\mathcal{O})\neq \emptyset$, a solution might be still possible in specific cases. Then $Z_{1/h}=Z_f$. If you want the polar set $P_{1/h}=Z_h$ to avoid a specific point $x \in X\setminus Z_f$, solve your problem on $X\setminus \{x\}$.<|endoftext|> -TITLE: Classification of smooth algebraic surfaces with a smooth morphism to $\Bbb P^1$ -QUESTION [7 upvotes]: Let $k$ be an algebraically closed field, it is well known that $\mathbb P^1$ is simply connected, but how about smooth projective surfaces $X$ with a smooth morphism to $\Bbb P^1$? -Except the case $X$ is a product of curves or a projective bundle like Hirzebruch surfaces, could we classify all such $X \rightarrow \Bbb P^1$ ? What about higher dimensional cases? -Motivation: Shafarevich conjecture over function field. - -REPLY [14 votes]: Let $k$ be an algebraically closed field. -Let $f:X\to \mathbb{P}^1$ be a smooth proper morphism with fibres of dimension one. Note that the fibres of $f$ are geometrically connected by Stein factorization and the fact that $\mathbb{P}^1$ is simply connected (Riemann-Hurwitz). - - -Theorem. The morphism $f$ is isotrivial. - - -Proof. Let $g$ be the genus of the fibres. Clearly, if $g=0$, then all geometric fibres are isomorphic. Thus, if $g=0$, then $f$ is isotrivial. Next, assume that $g=1$. Then, as the moduli space of elliptic curves is affine, the moduli map associated to the Jacobian of $f$ is constant, so that $Jac(f)$ has constant $j$-invariant. It follows that $f$ is isotrivial. Finally, it is a theorem of Moret-Bailly that any genus $g>1$ curve over $\mathbb{P}^1$ is isotrivial; see Lemme 5 in [1]. QED -[1] Laurent Moret-Bailly. Un théorème de pureté pour les familles de courbes lisses. C. R. Acad. Sc. Paris, t. 300, Serie I, n. 14, 1985. -Remark. If $k$ has characteristic zero, then Moret-Bailly's theorem is due to Parshin. Parshin's theorem follows from the "hyperbolicity" of the moduli stack of genus $g$ ($g>1$) curves. There are many notions of "hyperbolicity" lurking around there, and all of them imply the statement you want (in characteristic zero). (More generally, if $k$ is of characteristic zero, any smooth proper morphism $X\to \mathbb{P}^1$ whose fibres are smooth proper connected varieties with ample canonical bundle is isotrivial by work of Kovács, Migliorini, Viehweg-Zuo.) -Let $F$ be a fibre of $f$ over a closed point of $\mathbb{P}^1_k$. -If $g>1$, then the Isom-scheme $Isom(F\times \mathbb{P}^1, X)\to \mathbb{P}^1$ is finite etale and thus trivial. Thus, $f$ is trivial. -If $g=1$, then probably the family $X\to \mathbb{P}^1$ is trivial Will Sawin's argument below. -If $g=0$, as Daniel Loughran explains in the comments below, the (isotrivial) morphism $f$ has a section by Tsen's theorem, so that $X$ is a Hirzebruch surface.<|endoftext|> -TITLE: Matrix determinant inequality proof without using information theory -QUESTION [9 upvotes]: Let $A$ be a $k \times n$ orthogonal matrix; i.e., $AA^T = I_{k \times k}$. For $1 \leq j \leq n$, let the squared norm of the $j$-th column of $A$ be denoted by $\alpha_j^2$; i.e., -$$\sum_{i=1}^k a_{ij}^2 = \alpha_j^2.$$ -Let $\Lambda = \operatorname{Diag}(\lambda_1, \lambda_2, \dotsc, \lambda_n)$ be a positive definite matrix. Define a function $F$ from the space of p.d. diagonal matrices to $\mathbb R$ as follows: -$$ F(\Lambda) = \log \lvert A \Lambda A^T\rvert-\sum_{j=1}^n \alpha_j^2 \log \lambda_j.$$ -To show: $F(\Lambda) \geq 0$, with equality only when $\Lambda$ is a multiple of identity. -Background: This result is true and follows from an inequality known in information theory as Zamir and Feder's entropy power inequality. To obtain the result, we may apply Zamir and Feder's inequality to Gaussian random variables with covariances $\lambda_i$. (A statement of the inequality may be found here: Rioul - Information-theoretic proofs of entropy power inequalities, Proposition 9, (65c).) -I am interested in knowing if there is an alternate proof of this result that does not rely on entropy inequalities, and uses linear algebraic tools instead. - -REPLY [8 votes]: Here's a more general result. -From Choi's inequality, we know that for a positive linear map $\Phi$ and an operator convex function $f$, we have $\Phi(f(\Lambda)) \ge f (\Phi(\Lambda))$. Let $\Phi(\Lambda)=A\Lambda A^T$, and let $f(t)=-\log t$, then we obtain -\begin{align*} --A\log(\Lambda)A^T \ge -\log(A\Lambda A^T)\\ -\Leftrightarrow\qquad\log(A\Lambda A^T) \ge A\log(\Lambda)A^T. -\end{align*} -This operator inequality is stronger than the what we need. Simply take the trace of both sides to conclude $F(\Lambda) \ge 0$ (using the observation that $\log\det(Z)=\text{tr}\log Z$, and since $\Lambda$ is diagonal).<|endoftext|> -TITLE: Find an integer coprime to sequence sum -QUESTION [9 upvotes]: Given two positive integers $a -TITLE: Borel-Écalle re-summation and resurgence: criteria and results -QUESTION [13 upvotes]: This is about the theory of Borel-Écalle re-summation and resurgence, see Refs below. -This states that the perturbative series (say of the vacuum expectation value of an operator $\mathcal{O}$ in quantum field theory) should be thought of as part of the so-called trans-series expansion, containing both perturbative and non-perturbative corrections: -$$\langle \mathcal{O} \rangle -= \sum_{k=0}^{\infty} c_{0,k} g^k+ \sum_{I} e^{-\frac{S_I}{g^2}} \left( \sum_{k=0}^{\infty} c_{I,k} g^k \right)+ \dots \;, -$$ -where $I$ is a label for the saddle points of the action $S$, and $S_I$ denotes the value of the action at the $I$-th saddle point; -the coefficients $c_{I,k}$ represent the perturbative expansions around the $I$-th saddle point. -Assuming we know all the saddle points contributing to the path integral, -all the coefficients $c_{0,k}$ and $c_{I,k}$ can be computed by perturbative methods, -and we obtain a trans-series. -Now we hope to turn the trans-series into a well-defined function -$\langle \mathcal{O} \rangle(g)$ as a function -of the coupling constant with the help of the resurgence theory. -If that function is resurgent, -we may adiabatically continue back to -the large value of the coupling constant along a suitable choice of path in the complex plane. - -Questions: - - -(1). What are the precise mathematical conditions/criteria to turn the trans-series above into a well-defined function as a function -of the coupling constant with the help of the resurgence theory? - - -When we say that the function is resurgent, does it require infinite differentiable or an infinite continuable? - -p.s. The resurgence in the physics literature may refer to a stronger statement. A large-order asymptotic growth (as $k$ large) of the perturbative coefficients $c_{0,k}$ around the trivial saddle point contains the information of the non-perturbative saddle points within the same topological sector. - -(2). What well-controlled mathematical properties can we state precisely for the above function $\langle \mathcal{O} \rangle$? What are the mathematical rigorous results for the re-summation and resurgence? - -The context is partially inspired by this post. - - -J. Écalle, Les fonctions résurgentes. Tome I, vol. 5 of Publications Mathématiques d’Orsay 81 [Mathematical Publications of Orsay 81]. Université de Paris-Sud, Département de Mathématique, Orsay, 1981. Les algèbres de fonctions résurgentes. [The algebras of resurgent functions], With an English foreword. - -J. Écalle, Les fonctions résurgentes. Tome II, vol. 6 of Publications Mathématiques d’Orsay 81 [Mathematical Publications of Orsay 81]. Université de Paris-Sud, Département de Mathématique, Orsay, 1981. Les fonctions résurgentes appliquées à l’itération. [Resurgent functions applied to iteration]. - -O. Costin, Asymptotics and Borel summability, vol. 141 of Chapman & Hall/CRC Monographs and Surveys in Pure and Applied Mathematics. CRC Press, Boca Raton, FL, 2009. - -C. Mitschi and D. Sauzin, Divergent series, summability and resurgence. I, vol. 2153 of Lecture Notes in Mathematics. Springer, [Cham], 2016. Monodromy and resurgence, With a foreword by Jean-Pierre Ramis and a preface by Éric Delabaere, Michèle Loday-Richaud, Claude Mitschi and David Sauzin. - -REPLY [11 votes]: I can only offer a partial answer here. I don't really know how resurgence is used in QFT, so I am going to talk about the general principles. -First thing first. The definition of an asymptotic series does not need complex analysis. However an asyptotic series in a real setting is much less useful. Lets say that a function $f$ admits $x$ as an asymptotic series to $+\infty$. Now define the function $g$ to be $1$ on $[n,n+1)$ if $n$ is even and $0$ if $n$ is odd. Then the function $f\cdot g$ also admits $x$ as an asymptotic serries. So in the real setting, we do not have any form of uniquness and we cannot extract any information (well maybe the word any is too strong here). -So when we talk about resurgent functions we are talking necessarily about complex analytic functions. Of course in the complex setting, having one derivative is equivalent to having all of them and a bit more. I will come back to endlessly continuability later. -So now let's look at a somewhat trivial example: the ODE $-x''(t)+x(t)=1/t$. -If we assume that the solution is an analytic function in a neighbourhood of infinity we can substitute $\tilde x(t)=\sum_{n\ge0}\frac{c_n}{t^n}$ in the ODE and collect terms to find eventually that -$$\tilde x(t)=\sum_{n\ge0}\frac{(2n)!}{t^{2n+1}}.$$ -And we realise that our assumption was wrong because this series has $0$ radius of convergence. So let's resum it. Let's us $s$ as the dual variable of $t$. The Borel transform of this is -$$\hat x(s)=\sum_{n\ge0}s^{2n}.$$ -This series converge in the unit disc. -Then because we were "lucky", we can easily extent it beyond the unit disc: -$$ \hat x(s) = \frac{1}{1-s^2}. $$ -Then the Laplace transform of this is a solution of the ODE. But which Laplace transform? -We can define 2 here. One by integrating on the positive real axis and the other by integrating on the negative real axis. In practice we can move the contour of integration freely as long as we don't hit a singularity of the function. The singularities are of course $\pm i$, this is the reason that we have 2 Laplace transforms. -So the solutions of the ODE are -$$ x^\pm(t) = \int_0^{\pm\infty}\frac{e^{-ts}}{1-s^2}ds. $$ -These are two particular solutions, i.e. $x^\pm(t)$ tends to $0$ as $t$ tends to $\pm\infty$. -OK, let's talk about the resurgence here. Unfortunately, the term resurgent function is used liberally to describe 3 different objects. In the above example we can call a resurgent function $\tilde x$, $\hat x$ or $x^\pm$. I think that it is more appropriate to call $\hat x$ a resurgent function and nothing else, so let's go with that. -So ok, what do we get from what we did in the example? We know that we have 2 solution and we know that these 2 solutions are not the same. But how different are they? First of all what is the domain of definition of these solutions? I will not go into details here but for $x^+$ it is $\mathbb C$ minus the negative real axis with the origin. If this is not clear to you why, you should look into the literature for the Laplace transform. The domain of definition for $x^-$ is the reflection with respect to the imaginary axis. -Then we can take the difference $x^+(t)-x^-(t)$ for non-real $t$ and this is well defined. In order to calculate this difference we notice that since -$$ x^\pm(t) = \int_0^{\pm\infty}\frac{e^{-ts}}{1-s^2}ds, $$ -we have -$$ x^+(t)-x^-(t) = \int_{-\infty}^\infty\frac{e^{-ts}}{1-s^2}ds. $$ -However as it is written this integral does not converge, we have to deform the contour of integration. So let's choose $t$ with negative imaginary part. This means that we have to tilt the contour of integration "upwards on both sides". Then because the singularities of the function are just poles we get -$$ x^+(t)-x^-(t) = \oint_i\frac{e^{-ts}}{1-s^2}ds, $$ -which means the closed integral around the complex number $i$. So not only we know that the 2 solutions are different, but also we can explicitely compute their difference. -OK, so what is the big deal with resurgence? In general the situation is as follows. We want an analytic solution, but the solution fails to be analytic and we get an asymptotic series. Then we want to construct a solution and we need to use resummation. The problem we find there is that typically the solution we construct is different in different sections of the complex plane and the way it is different depends on the singularities of the Borel transform, $\hat x$ in our case. The theory of resurgent functions gives a toolset that allows us to analyse the singularities of such functions. Before this theory this was largely science fiction. -So let's go back to the endlessly continuability business. Initially we have $\hat x$ as a convergent series, i.e. it is a function defined in a disc, but we need to take its Laplace transform. The first thing we do is to extend it to (hopefully) its maximal domain, which means that it need to extend to infinity in a sector. If moreover the function is of exponential growth in this sector, then we can define the Laplace transfor. No other conditions are needed. -However, if we want to use the resurgence's toolkit we need to have a functions with isolated singularities. With meromorphic functions what isolated singularities are is straightforward. If the function is defined in a cover of $\mathbb C$, in other words "it has branching singularities", this becomes tricky. Let's say that this means that if we project the singularities of the funcition on $\mathbb C$, they are isolated points. This condition is stricter than what the theory needs, but I hope your case is covered by it. -So what do you need in order to rigorously prove your statement? All the above. You need to have a well defined asymptotic series. You need to prove that it's Borel transform converges. You need to prove that this functions is endlessly continuable. And finally you need to have exponential bounds for the growth of the functions in "most" directions. -I hope this helps. Unfortunately I only touched the surface of this amazing theory. In reality the rabbit hole is too deep. I hope this helps for now. Ask if something is unclear. -ADDED LATER: -I forgot to comment on this transseries business. So going back to the example we had -$$ x^+(t)-x^-(t) = \oint_i\frac{e^{-ts}}{1-s^2}ds. $$ -Let's move this our the origin: -$$ x^+(t)-x^-(t) = \oint_0\frac{e^{-t(s+i)}}{1-(s+i)^2}ds = e^{-it}\oint_0\frac{e^{-ts}}{1-(s+i)^2}ds. $$ -So you see that we get exponential in front of a "series" (in this case it is not really a series). -In the general case we get a function that has many singularities that are ramification points, let's look at another simple example: the finite difference equation $x(t+1)-x(t)=1/t^2$. -The Borel transform of this equation is $e^{-s} \hat x(s)-\hat x(s) = s$, so we get $$ \hat x(s) = \frac{s}{1-e^{-s}}. $$ -Now this equation has a pole on $2\pi i n$ for all integers $n$ but the origin. -We can do basically the same as above, i.e. define 2 solutions (that are of course not general) and take their difference. But now the difference of the solutions is: -$$ x^+(t)-x^-(t) = \sum_{n=1}^\infty\oint_{2\pi i n}\frac{s\,e^{-ts}}{1-e^{-s}}ds $$ -and if we move all the integrals to the origin we get -$$ x^+(t)-x^-(t) = \sum_{n=1}^\infty e^{-2\pi i n t} \oint_{0}\frac{(s+2\pi i n)\,e^{-ts}}{1-e^{-s-2\pi i n}}ds. $$ -Of course these integrals do not give us series because the function is simple. However in the general case, if for example add a non-linear term in the example, we will get ramification points instead of ajust a pole and then we will get an actual transseries. -There is one last thing that I was wondering if I should write. I decided to do so, but it may be confusing at least in the beginning. OK here we go: -Now that you read all the previous, I have to tell you that I lied. A lot. So the classical Borel transform is defined as the formal inverse of the Laplace transform and it devides by a factorial. This is the reason that we can get a convergent series. However, in resurgence we need a generalization of this. -I will just give you some references because this is too big to be written here. -So the proper way to define a resurgent function is through hyperfunction, which are a generalisation distributions. I think the easiest textbook on the subject is "Introduction to Hyperfunctions and Their Integral Transforms" by Urs Graf. -Resurgent functions can be thought as a very special class of hyperfunctions. The paper Introduction to 1-summability and resurgence gives an introduction but I think it is not so easy to follow if you don't know already this. I looked around a bit and I found this PhD thesis that explains it a bit. It is rather dense, but potentially easier. Read Chapter 2. -Also I don't know if you know them already but Sternin and Shatalov are probably more relevant to you. They have a book titled "Borel-Laplace Transform and Asymptotic Theory: Introduction to Resurgent Analysis" which for some reason is very expensive and not all libraries have it. Take a look also at their papers. -A word of caution about the book though. Their definition of resurgent functions is not the same as Ecalle's. They define a smaller class that has some extra properties, which means that probably they are the only ones using this definition of resurgence. Nevertheless it can help you understand the theory.<|endoftext|> -TITLE: Can we inductively define Wadge-well-foundedness? -QUESTION [7 upvotes]: For a topological space $X$ (which I'll identify with its underlying set of points), we define the Wadge preorder $Wadge(X)$: elements of the preorder are subsets of $X$, and the ordering is given by $A\le_{W(X)}B$ iff there is a continuous map $f:X\rightarrow X$ with $f^{-1}(B)=A$. -Actually, it is sometimes better to tweak the definition; if this affects the answer, feel free to use it instead, or indeed any other improvement on the naive Wadge hierarchy. -There is a natural substructure of $Wadge(X)$, namely its wellfounded part: $$WWF(X)=\{A\subseteq X: \neg\exists (B_i)_{i\in\omega}(B_0=A, B_i>_{W(X)}B_{i+1})\}.$$ (Here "$<_{W(X)}$" means "$\le_{W(X)}$ and not $\ge_{W(X)}$," as expected.) -For example, Borel determinacy implies that $WWF(\omega^\omega)$ contains the class of Borel sets, and under AD we in fact get $WWF(\omega^\omega)=\mathcal{P}(\omega^\omega)$. -Now, faced with a natural well-founded preorder, my instinct is that it should consist of the objects which can be "built up from below" by some natural procedure. But I don't see that this is the case here. So I want to ask: - -Main question: Is there a way to think of $WWF(X)$ this way? Phrased a bit more abstractly, is $WWF(X)$ the least fixed point of some reasonable operator on $\mathcal{P}(\mathcal{P}(X))$, at least for "reasonable" $X$? - -Even for $X=\omega^\omega$, this isn't clear to me. Indeed, it's not even clear to me that the usual "tame part" of $Wadge(\omega^\omega)$ is all of $WWF(\omega^\omega)$. So maybe the following is worth resolving on its own: - -Secondary question: Is there an $A\in WWF(\omega^\omega)$ which is Wadge incomparable with (say) both the set of reals coding well-orderings and the complement of that set? (That is, the Wadge degrees $\Pi^1_1$ and $\Sigma^1_1$.) - -REPLY [3 votes]: Long comment: -It is difficult to answer "no" to the main question, given the unlimited potential interpretations of "reasonable". Still, I would be very surprised by a positive answer. -The reason is an observation by Peter Hertling that already on $\mathbb{R}$ we can build a descending chain of sets $(A_m)_{m < \omega}$ where each $A_m$ is a finite union of half-open intervals and open. Precisely, let -$$A_n := [0,1) \cup (2,3) \cup \ldots \cup (2n,2n+1) \cup [2n+2,2n+3)$$ -Since I believe that $\mathbb{R}$ ought to be reasonable space, this would leave us with needing a reasonable construction of sets that on $\mathbb{R}$ avoids already finite unions of (half)-open intervals, while creating at least all Borel sets on $\omega^\omega$. -Of course, the main question seems to be already very interesting on $\omega^\omega$ alone; so this example only serves to show that the scope should probably be limited a bit. -Going for Pequignot's tweaked definition would remove this obstacle, so that might be another approach..<|endoftext|> -TITLE: Can $\Delta^{1}_{2}$ separate degrees of constructibility? -QUESTION [7 upvotes]: Suppose that $\phi(x)$ is a $\Delta^{1}_{2}$-formula (without parameters) and let $A:=\{x\subseteq\omega:\phi(x)\}$. It is clear that, e.g. if there are Cohen-generics over $L$, then $A$ cannot be the set of constructible reals, roughly because $\Delta^{1}_{2}$ cannot distinguish between sufficiently high Cohen-generics in $L$ and Cohen-generics over $L$. I wonder whether this can be strengthenend in the following way: -Given appropriate largeness assumptions (existence of generic filters, large cardinals...), at least one of $A$ and $\mathfrak{P}(\omega)\setminus A$ contains real numbers of all degrees of constructibility. In other words, can $\Delta^{1}_{2}$ "separate" degrees of constructibility in a nontrivial way? - -REPLY [5 votes]: Expanding on Douglas Ulrich's answer: -Note that this question is only meaningful when there are nonconstructible reals since otherwise there is only one constructibility degree. -If there is a nonconstructible real and $A$ is $\Delta_2^1$ set, then either $A$ or $\mathbb{R} \setminus A$ is a $\Sigma_2^1$ set of reals with a nonconstructible element. Without loss of generality, suppose $A$ has the nonconstructible real. By the Mansfield-Solovay theorem (using the $\omega_1$-Suslin representation via the Shoenfield Tree $S$ for $A$ which belongs to $L$), there is a constructible tree $T$ so that $[T] \subseteq A$. -Every constructible tree is an $L$-pointed tree. So $[T]$ has every $L$-degree.<|endoftext|> -TITLE: Adding constraints as penalty with $\| \cdot \|_0$ norm -QUESTION [6 upvotes]: In the paper Image Denoising Via Sparse and Redundant Representations Over Learned Dictionaries (page 2), the authors rewrite the minimization problem -\begin{align} -\min_{\alpha \in \mathbb R^k} \| \alpha \|_0 && s.t. && \|D \alpha - y \|_2^2 \leq T, -\end{align} -where $D \in \mathbb R^{n \times k}$ and $y \in \mathbb R^n$, into -\begin{align} -\min_{\alpha \in \mathbb R^k} \| D \alpha - y \|_2^2 + \mu \| \alpha \|_0 -\end{align} -and state that - -for a proper choice of $\mu$, the two problems are equivalent. - -There is no reference in the paper that explains why this would be true and in fact, I don't believe it is true, since $\| \cdot \|_0$ is not even convex. Does such a $\mu$ really exist? -I don't know if this is important, but the following assumptions were made: $y$ is a noisy version of $x$ with zero mean white noise of variance $\sigma^2$, and that there exists $x$ so that $\| D \alpha - x \|_2 \leq \epsilon$ with $\| \alpha \|_0 \leq L \ll n$. The $T$ from the first equation above is dictated by $\varepsilon$ and $\sigma$. -I already asked two professors who also don't believe the statement is true. However, it has to come from somewhere. Does it maybe work with $\| \cdot \|_1$? Or is this just an "analytic application" of a penalty method? - -REPLY [3 votes]: The claim in the paper is false. -Since the problem is not convex, the claim does not follow from general results. However, there are some results in this direction in quite general cases: -If $x^*$ is the unique minimizer of $\min_x F(x) + G(x)$, then it is a solution of the constrained problem -$$ -\min_x F(x)\qquad\text{s.t.}\qquad G(x) \leq T -$$ -for $T = G(x^*)$. The converse direction (even with uniqueness) is false in general (I guess that this is folklore, but I typed up a result and counterexample in "Necessary conditions for variational regularization schemes, D Lorenz, N Worliczek, Inverse Problems 29 (7), 075016", Theorem 2.3 and Example 2.4). -In this special case you can consider the 2d problem -$$ -\min_x \|x\|_0\qquad\text{s.t.}\qquad \| x - \begin{bmatrix}2\\2\end{bmatrix}\|_2^2\leq 1 -$$ -which gives you the optimal value $2$ and the solutions are the whole feasible set. As far as I see, you can't realize this optimal set for -$$ -\min_x \mu\|x\|_0 + \| x - \begin{bmatrix}2\\2\end{bmatrix}\|_2^2 -$$ -for any $\mu$.<|endoftext|> -TITLE: Cohomology of neighborhood of $\mathbb{C}\mathbb{P}^1$ in $\mathbb{C}\mathbb{P}^n$ -QUESTION [6 upvotes]: Let $\mathbb{C}\mathbb{P}^1$ be embedded linearly to $\mathbb{C}\mathbb{P}^n$ with $n>1$. (Such an embedding is given in coordinates by $[x:y]\mapsto [x:y:0:\dots: 0]$.) - -Is it true that for any open neighborhood $U$ (in the analytic topology) of $\mathbb{C}\mathbb{P}^1$ there exists a smaller neighborhood $V$ of the latter such that - $$H^i(V, \mathcal{O})=0 \mbox{ for any } i>0,$$ - where $\mathcal{O}$ is the structure sheaf? - -REPLY [6 votes]: It is not hard to see that the -first cohomology is infinite-dimensional. -Take a complement $M:=CP^3 \ CP^1$ -and consider a projection -$\pi:\; M \mapsto CP^1$. It is not -hard to see that $M$ is -isomorphic to the total space of -the bundle $O(1)^2$. The fibers -of $\pi$ are Stein, hence -$R^i\pi_*F=0$ for $i>0$ and any coherent sheaf -$F$, and cohomology $H^i(O_M)$ are the -same as $H^i(\pi_* O_M)$. -However, $\pi_* O_M= Sym^*(O(-1)^2)$, -because the regular functions on the -total space of $O(1)^2$ are $Sym^*(O(-1)^2)$. -However, $H^1(Sym^*(O(-1)^2))$ is -infinite-dimensional. -Same argument works for smaller -neighbourhoods $U\supset CP^1$, as -long as the fibers of $\pi:\; U \mapsto C P^1$ -remain Stein. -Same is true for a neighbourhood of a -rational line $C$ in a complex manifold -if the normal bundle $NC$ is ample, -but the proof is more complicated.<|endoftext|> -TITLE: Cardinality of families of subsets of $\mathbb{N}$ whose intersections are finite -QUESTION [11 upvotes]: Does there exist an uncountable $P \subset \mathcal{P}(\mathbb{N}) $ with the property that for any distinct $x,y \in P$, $|x \cap y|$ is prime? -A more general, but likely harder, question: is it possible to characterize the set $\mathcal{A}$ of all subsets $S \subset \mathbb{N}$ with the following property: there is some uncountable $P \subset \mathcal{P}(\mathbb{N})$ such that for all distinct $x,y \in P$, $|x \cap y| \in S$. -It's easy to show no $S$ finite belongs to $\mathcal{A}$. It's also not hard to show every $\text{mod} \ m$ equivalence class, belongs to $\mathcal{A}$. -One reasonable sounding idea is that this is related to the asymptotic density of $S$ in $\mathbb{N}$. - -REPLY [2 votes]: The following argument uses less set-theory than the earlier ones! -For each positive real number $\alpha$ we can write the binary expansion $\alpha=\sum_{i=0}^{\infty} a_i 2^{N-i}$ with $a_0=1$ and $a_i\in\{0,1\}$ for $i\geq 1$. To avoid ambiguity (from an endless sequence of 1's) we also assume that for all $j$ there is a $k>j$ so that $a_k=0$. -We take $S(\alpha)=\{ \alpha_n : n>0 \}$ where $\alpha_n=\sum_{i=0}^n a_i 2^{N-i}$. Then $S(\alpha)\subset\mathbb{Q}$, so we have an uncountable collection of subsets of $\mathbb{Q}$ which is in bijection with $\mathbb{N}$. Moreover, $S(\alpha)\cap S(\beta)$ is finite when $\alpha\neq\beta$. (It consists of those truncated binary expansions of $\alpha$ and $\beta$ which are equal.) -Now take the subset of reals which consists of those $\alpha$ for which $a_i=1$ unless $i$ is prime. It is not too difficult to see that these $\alpha$ are also uncountable. For example you can map reals to such numbers by "filling in 1's" in the binary expansion as required so that the old $a_i$'s become the new $a_{\pi(i)}$ where $\pi(n)$ is the $n$-th prime. -Moreover, it follows that $S(\alpha)\cap S(\beta)$ is a prime for every such $\alpha$ and $\beta$.<|endoftext|> -TITLE: Is there any significance to Bousfield localization in the non-derived context? -QUESTION [12 upvotes]: The term "Bousfield localization" of a category $C$ is used in roughly two different ways: - -There is a general usage (as in model categories or triangulated categories), which $\infty$-categorically just means a reflective subcategory of $C$. -There is also a more restrictive usage (as when talking about spectra), which requires $C$ to be monoidal, and means a reflective subcategory where the class of maps being localized at is of the form $\{X \mid E \otimes X = 0\}$ for some fixed $E \in C$. - -In this question, I'm interested in the more restrictive usage (2). -In this sense, Bousfield localization makes sense in either an ordinary monoidal category or in a monoidal $\infty$-category (for that matter, the more general usage makes sense in either an ordinary category or in an $\infty$-category). But it's typically only discussed in an $\infty$-categorical setting (e.g. in model categories or triangulated categories). - -Question 0: Is there a good reason why Bousfield localization for ordinary categories (in sense (2)) is rarely discussed? - -I think the answer may be "yes" because the behavior of Bousfield localization may be quite different in the two settings, and it seems somehow "better" in the $\infty$-categorical setting. But I'm not sure how to articulate this. -Here are two examples of what I mean: - -$E = \mathbb Z/p$: - -When $C = Ab$ is the (ordinary) category of abelian groups and $E = \mathbb Z / p$, the Bousfield localization consists of the abelian groups which have no nonzero infinitely $p$-divisible elements. -But when $C = D(Ab)$ is the $\infty$-category of chain complexes of abelian groups (localized at the quasi-isomorphisms) and $E = \mathbb Z/p$, the Bousfield localization consists of chain complexes whose homology is $p$-complete. - -$E = \mathbb Z_{(p)}$: - -When $C = Ab$ and $E = \mathbb Z_{(p)}$, the Bousfield localization consists of abelian groups which are $\ell$-torsionfree for $\ell\neq p$. -When $C = D(Ab)$ and $E = \mathbb Z_{(p)}$, the Bousfield localization consists of chain complexes whose homology is a $\mathbb Z_{(p)}$-module. - - -By "different behavior", I mean, to a first approximation, that even though $D(Ab)$ is "the natural $\infty$-categorical counterpart to $Ab$", in these cases it's not the case that the restriction of the $E$-Bousfield localization in $D(Ab)$ to $Ab$ coincides with the $E$-Bousfield localization in $Ab$ itself. -Part of the problem is that I'm not exactly sure what qualifies as "being in the $\infty$-categorical setting". After all, an ordinary category is in particular an $\infty$-category. But maybe for concreteness, I'll ask a slightly less vague version of the question: - -Question 1: If $T$ is a tensor triangulated category with a $t$-structure, and $E \in T^{heart}$, is there any reason to think about the Bousfield localization of $T^{heart}$ at $E$ rather than the Bousfield localization of $T$ at $E$? - -REPLY [6 votes]: I think that one of the reasons why these localizations are rarely discussed in ordinary categories is that their application in a derived context often relies on stability. -Generally Bousfield localization starts with a collection of maps $S$ that you want to make into equivalences, defines an object $Y$ to be local if $Hom(B,Y) -\to Hom(A,Y)$ is an isomorphism for any $f:A \to B$ in $S$, and then asks for a reflective ocalization onto the class of local objects. -Given $E$, you can either - -localize at the maps $A \to B$ such that $E \otimes A \to E \otimes B$ is an equivalence, or -localize at the maps $X \to 0$ such that $E \otimes X$ is trivial. - -In a stable setting, these are equivalent because $A \to B$ is an $E$-equivalence if and only if $E \otimes (B/A)$ is trivial. However, in an unstable setting the localizations are quite different. My suspicion is that people are often implicitly interested in applications of type (1) and that's why there's less of an emphasis on the types of localizations that you're discussing. -As a remark, in the unstable (but still homotopical) setting we often do construct nullifications by Bousfield localizing with respect to maps $W -\to *$. However, the class of $W$ we are nullifying often has no description using $E \otimes W$ for any $E$ because they are often described using certain generators or a homology theory.<|endoftext|> -TITLE: Reflection-invariant monomial ideals and Alexander duality -QUESTION [6 upvotes]: First we give some definitions from Section 3 of the paper Monomials, Binomials, and Riemann-Roch by Manjunath and Sturmfels and then we restate a claim from that paper offered without proof. Finally we provide an example that seems to contradict that claim. The question is -Question: Is the example given below a counterexample to the claim? And if not, why not? -Definitions -Fix an Artinian monomial ideal $I$ of a polynomial ring $K[\mathbf{x}] = K[x_1, \dots, x_n]$. A monomial $\mathbf{x}^{\mathbf{b}}$ is a socle monomial of $I$ if $\mathbf{x}^{\mathbf{b}} \notin I$ and $x_i\mathbf{x}^{\mathbf{b}} \in I$ for all $i$. Let $\mathrm{MonSoc}(I)$ be the set of all socle monomials of $I$. -Def: $I$ is reflection invariant if there is a canonical monomial $\mathbf{x}^{\mathbf{K}}$ such that the map that sends a monomial $\mathbf{x}^{\mathbf{b}} \mapsto \mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{b}}$ is an involution on $\mathrm{MonSoc}(I)$. -Following these definitions the authors note the following. -The Claim -Claim: $I$ is reflection invariant with canonical monomial $\mathbf{x}^{\mathbf{K}}$ if and only if the monomial ideal generated by $\mathrm{MonSoc}(I)$ equals the Alexander dual $I^{[\mathbf{K} + \mathbf{e}]}$ where $\mathbf{e} = (1,1,\dots, 1)$. -The (Counter?) Example -Let $I = \langle a^4,~ab^2,~b^3,~a^3c,~abc,~c^3 \rangle \subset K[a,b,c]$ and let $\mathbf{K} = (3,2,2)$. Then -$$\mathrm{MonSoc}(I) = \left\{a^{3}b,~a^{2}c^{2},~b^{2}c^{2}\right\}.$$ -By (the constructive proof of) Proposition 5.2 in this paper - the ideal $J = \langle a^4, a^2b, b^3, ac, b^2c, c^3 \rangle$ is the unique Artinian ideal with -$$\mathrm{MonSoc}(J) = \left\{\mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{b}} \mid \mathbf{x}^{\mathbf{b}} \in \mathrm{MonSoc}(I)\right\}.$$ -Moreover, the same algorithm can be used to show that $I$ is the unique Artinian ideal with -$$\mathrm{MonSoc}(I) = \left\{\mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{c}} \mid \mathbf{x}^{\mathbf{c}} \in \mathrm{MonSoc}(J)\right\}.$$ -In particular, the map that sends a monomial $\mathbf{x}^{\mathbf{b}} \mapsto \mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{b}}$ is an involution on $\mathrm{MonSoc}(I)$, so $I$ is reflection-invariant with canonical monomial $\mathbf{x}^{\mathbf{K}}$. We now get a contradiction to the claim above by computing the Alexander dual (in Macaulay2, for example) and noting that the minimal generators of $I^{[(4,3,3)]}$ are $\{a^4bc,~a^2b^3c,~ab^2c^3\} \neq \mathrm{MonSoc}(I)$. -Again, the question is -Question: Is the example just given a counterexample to the above claim? And if not, why not? - -REPLY [3 votes]: Note: The following is the result of an email exchange with one of the coauthors of the paper cited in the OP. -In the definition of a reflection-invariant monomial ideal the requirement that the map $\phi: \mathbf{x}^{\mathbf{c}} \mapsto \mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{c}}$ be an involution on $\mathrm{MonSoc}(I)$ is not the usual definition of involution, i.e., that the map is its own inverse. Instead, the map $\phi$ is required to be both an involution on and a permutation of $\mathrm{MonSoc}(I)$. This additional requirement is not mentioned in the paper, though upon closer reading, it is used in a number of the proofs. -As for the example in the OP, the map $\phi$ is clearly not a permutation of $\mathrm{MonSoc}(I)$. Moreover, one can show that $I$ is not reflection-invariant for any choice of $\mathbf{K}$. So the ideal $I$ is not reflection-invariant. Also, with this new restriction on permissible involutions it is easy to prove the note following the definition of reflection invariance in the paper in question. -Edit: Thanks to @benblumsmith's comment the heart of the matter is now clear. The map that sends $\mathbf{x}^{\mathbf{b}} \mapsto \mathbf{x}^{\mathbf{K}-\mathbf{b}}$ has domain and codomain equal to $\mathrm{MonSoc}(I)$ (as opposed to the set of all monomials in the quotient ring $K[\mathbf{x}]/\left\langle x_1^{K_1+1}, \dots, x_n^{K_n+1}\right\rangle$). So it's clear that the example in the OP is not a counterexample.<|endoftext|> -TITLE: Cokernel of map of étale sheaves -QUESTION [6 upvotes]: Let $p:\mathbb{G}_m\to \operatorname{Spec} k$ be the structure map, and let $T$ be an algebraic $k$-torus viewed as an étale sheaf over $k$. Why is the cokernel of the canonical map $T\to p_*p^*T$ canonically isomorphic to the cocharacter lattice $L$ of $T$? -If $\operatorname{Spec}A$ is affine scheme, it seems to me that $p^*T=T\times \mathbb{G}_m$, so $$p_*p^*T(A)=p_*((T\times \mathbb{G}_m))(A)=T(A[t^{\pm 1}])\times \mathbb{G}_m(A[t^{\pm 1}]).$$ Say that $A=K$ is a field. Then $$p_*p^*T(K)=L\oplus\mathbb{Z},$$ but why is the image of $T(K)$ equal to $\mathbb{Z}$? - -REPLY [5 votes]: There is a small issue with your computation related to $\mathbb G_m$-linearity. By the $\operatorname{Gal}(k^{\text{sep}}/k)$-module view on étale sheaves on $\operatorname{Spec} k$, it suffices to understand the $k^{\text{sep}}$-points. We compute -\begin{align*} -p_*p^*T(k^{\text{sep}}) &= p^*T(\mathbb G_{m,k^{\text{sep}}}) = \operatorname{Hom}_{\mathbb G_m}(\mathbb G_{m,k^{\text{sep}}}, T \times \mathbb G_m) \\ -&= \operatorname{Hom}_k(\mathbb G_{m,k^{\text{sep}}}, T) = \operatorname{Hom}_{k^{\text{sep}}}(\mathbb G_{m,k^{\text{sep}}}, T_{k^{\text{sep}}})\\ -&=\operatorname{Hom}_{k^{\text{sep}}}(k^{\text{sep}}[T], k^{\text{sep}}[x,x^{-1}]). -\end{align*} -The image of the unit $T \to p_*p^*T$ will be the subgroup $\operatorname{Hom}_{k^{\text{sep}}}(k^{\text{sep}}[T],k^{\text{sep}})$. If¹ $T_{k^{\text{sep}}} \cong \mathbb G_m^n$, then we get -\begin{align*} -\operatorname{Hom}_k(k[T],k^{\text{sep}}[x,x^{-1}]) &= \operatorname{Hom}_{k^{\text{sep}}}(k^{\text{sep}}[t_1^{\pm 1}, \ldots, t_n^{\pm 1}],k^{\text{sep}}[x,x^{-1}])\\ -&= \Big((k^{\text{sep}})^\times \times \mathbb Z\Big)^n,\\ -\operatorname{Hom}_k(k[T],k^{\text{sep}}) &= \operatorname{Hom}_{k^{\text{sep}}}(k^{\text{sep}}[t_1^{\pm 1}, \ldots, t_n^{\pm 1}],k^{\text{sep}}) \\ -&= \Big((k^{\text{sep}})^\times\Big)^n, -\end{align*} -so the quotient is just $\mathbb Z^n$. The Galois structure is that of the cocharacter lattice of $T$. $\square$ -(In the above, all $\operatorname{Hom}$ sets indicate morphisms of rings or schemes, not of group schemes.) - -¹Every torus splits over $k^{\text{sep}}$. See for example Lemma B.1.5 of Conrad's notes.<|endoftext|> -TITLE: Convergence of the intertwining operator as a vector valued integral -QUESTION [5 upvotes]: Let $G$ be a connected, reductive group over a $p$-adic field with parabolic $P = MN$ defined by a set of simple roots $\theta \subset \Delta$. For $(\pi,V)$ a representation of $M$, and $\nu \in \mathfrak a_{M,\mathbb C}^{\ast}$, we have the induced representation -$$I(\nu,\pi) = \operatorname{Ind}_P^G \pi q^{\langle \nu+\rho,H_M(-)\rangle}$$ -of $G$. For $w$ in the Weyl group sending $\theta$ to $\theta' \subset \Delta$, and $P' = M'N'$ corresponding to $\theta'$, we have the intertwining operator $A = A(\nu,\sigma,w): I(\nu,\pi) \rightarrow I(w(\nu),w(\pi))$ defined by -$$A(f)(g) = \int\limits_{N_w} f(w^{-1}ng)dn$$ -where $N_w$ is generated by the root subgroups of positive roots made negative by $w^{-1}$. The given integration takes place in the vector space $V$, and I am trying to understand: - -What is the meaning of this vector valued integral? -Why does the integral converge (whatever that means, depending on the answer to my first question) for $\nu$ in a suitable cone? - -I had asked a question about the meaning of the integral before, but I am sorry to say that after all this time I still do not understand what is going on. Paul Garrett provided an answer in which he suggested that we should not think of $V$ as having the discrete topology, but having a locally convex, quasi-complete topological vector space structure (coming as a colimit of its f.d. subspaces) in which one could make sense of the integral as a Pettis integral. That is, we should show that there exists a vector $v = A(f)(g)$ in $V$ with the property that for all $v^{\ast}$ in the algebraic dual of $V$, -$$\langle v^{\ast},v \rangle \rangle = \int\limits_N \langle v^{\ast}, f(w^{-1}ng)\rangle dn$$ -He also suggested that taking a good maximal compact subgroup $K$ of $G$, so that we have $G = PK = P'K$, we could use the fact that elements of the induced representation are determined by their effect on $K$ to reduce to the case where the vector valued integrals are just finite sums. I still have not figured out how to do this, and wanted to ask math overflow again for help. -These intertwining operators are unfortunately still very much a mystery to me, and I have not seen any reference explain them rigorously. - -REPLY [2 votes]: EDIT: This doesn't work, because $n \mapsto f(k_n)$ is not well defined. -There is another way I was just thinking of this vector valued integral, and several people (in particular, Paul Garrett) have already explained things to me in this way, but I was not able to understand what they were saying at the time. I will write what I have, and if it's wrong, hopefully someone will point out my error. -So we choose a maximal compact open subgroup $K$ of $G$, in good position relative to $P$, so that we have $G = PK$. For each $n \in N_w$, we choose $p_n \in P$ and $k_n \in K$ such that $w^{-1}n = p_nk_n$. Define $f_{\nu}(n) = q^{\langle \nu + \rho, H_M(p_n)\rangle}$. This is a well defined locally constant function on $N_w$. Also, for a given $f \in I(\nu,\pi)$, the map $n \mapsto f(k_n)$ is a well defined locally constant function $N_w \rightarrow V$, and we have -$$f(w^{-1}n) = f_{\nu}(n)f(k_n) \in V$$ -Since $f|_K:K \rightarrow V$ is locally constant, $\{f(k_n) : n \in N_w \}$ is a finite set. -Now to make sense out of $\int\limits_{N_w} f(w^{-1}ng)dn$, we may replace $f$ by $R_g(f)$ and assume $g = 1$. Thus we want to make sense out of $\int\limits_{N_w} f(w^{-1}n)dn$. For $v \in V$, define -$$ A_v = \{ n \in N_w : f(k_n) = v\}$$ -which is an open set in $N_w$, and is empty for almost all $v$. Naively, I'm going to write -$$\int\limits_{N_w} f(w^{-1}n)dn = \sum\limits_{v \in V} \int\limits_{A_v} f(w^{-1}n)dn = \sum\limits_{v \in V} \int\limits_{A_v}f_{\nu}(n)f(k_n)dn$$ -$$ = \sum\limits_{v \in V} \bigg(\int\limits_{A_v}f_{\nu}(n)dn \bigg)v$$ -where the sum over $V$ is really just a finite sum. So the vector valued integral over $N_w$ can be defined to be $ \sum\limits_{v \in V} \bigg(\int\limits_{A_v}f_{\nu}(n)dn \bigg)v$, provided each Lebesgue integral -$$\int\limits_{A_v}f_{\nu}(n)dn$$ -converges. But each such Lebesgue integral converges if and only if -$$\sum\limits_{v \in V} \int\limits_{A_v}f_{\nu}(n)dn = \int\limits_{N_w} f_{\nu}(n)dn$$ -converges. So the intertwining operator makes sense provided that $\int\limits_{N_w} |f_{\nu}(n)|dn < \infty$.<|endoftext|> -TITLE: mixing theorem with definition (definition with proof) -QUESTION [5 upvotes]: I often find myself writing a definition which requires a proof. You are defining a term and, contextually, need to prove that the definition makes sense. -How can you express that? What about a definition with a proof? -Sometime one can write the definition and then the theorem. But often happens that many definition which should stay together need to be split -because a theorem is required in between. -A tentative example: -Definition (rational numbers) -Let $\sim$ be the equivalence relation on $\mathbb Z^*\times \mathbb Z$ given -by -$$ - (q,p) \sim (q',p') \iff pq' = p'q. -$$ -We define $\mathbb Q= (\mathbb Z^*\times \mathbb Z)/\sim$. -On $\mathbb Q$ we define addition and multiplication as follows -$$ - [(q,p)] + [(q',p')] = [(qq',pq'+p'q)] \\ - [(q,p)] \cdot[(q',p')] = [(qq',pp')] -$$ -With these operations and choosing -$0_\mathbb Q=[(1,0)]$, and $1_\mathbb Q=[(1,1)]$ -turns out that $\mathbb Q$ is a field. -Proof. -We are going to prove that $\sim$ is indeed an equivalence relation, -that addition and multiplication are well defined and that the resulting -set is a field. [...] - -REPLY [2 votes]: I disagree with the implication that it's always necessary, when trying to write something to be understood, that every sentence has to be a logical consequence of previous sentences or assumed background knowledge. -If you're careful about it, and you say this is what you're doing, I think it can be pedagogical to state a definition first and then show that it makes sense. Especially if the verification is routine and you don't want the definition to be buried among a mass of trivialities. I can give examples of this from my own writing.<|endoftext|> -TITLE: A class number estimate -QUESTION [7 upvotes]: Let $\mathcal{D} = \{D \in \mathbb{Z} : D \equiv 0, 1 \pmod{4}\}$ be the set of discriminants. It is well-known that each element in $\mathcal{D}$ is the discriminant of a primitive binary quadratic form, namely $x^2 - (D/4)y^2$ when $D \equiv 0 \pmod{4}$ and $x^2 + xy - (D-1)y^2/4$ when $D \equiv 1 \pmod{4}$. -For a primitive binary quadratic form $f$ the group $\text{SL}_2(\mathbb{Z})$ acts on it via substitution. Denote by $[f]$ the $\text{SL}_2(\mathbb{Z})$-equivalence class of $f$. Note that the discriminant is a $\text{SL}_2(\mathbb{Z})$-invariant, so that for each $g, h \in [f]$ we have $\Delta(g) = \Delta(h)$, so we may write $\Delta([f])$, the common value for $\Delta(g), g \in [f]$, to be the discriminant of the equivalence class. For each $D \in \mathcal{D}$, put -$$\displaystyle h(D) = \#\{[f] : f \text{ primitive}, \Delta([f]) = D\}$$ -for the class number of primitive binary quadratic forms of discriminant $D$. -Observe that the set of squares is contained in $\mathcal{D}$, and it is well-known that $h(n^2) = \phi(n)$, and representatives of the distinct $\text{SL}_2(\mathbb{Z})$-classes of discriminant $n^2$ are given by -$$\displaystyle ax^2 + nxy : \gcd(a,n) = 1, 1 \leq a \leq n-1.$$ -My question is, what about $h(-4n^2)$? Is there a nice characterization of them like the case of positive squares? Equivalently, can one find all reduced forms, i.e., a form $f(x,y) = ax^2 + bxy + cy^2$ with $|b| \leq a \leq c, a,c > 0$ of discriminant $-4n^2$? -Certainly, factoring $n^2 = u^2 v^2$ with $\gcd(u,v) = 1$ gives a primitive form $(ux)^2 + (vy)^2$ of discriminant $-4n^2$... but there are other forms such as $4x^2 + 4xy + 5y^2$ with discriminant $-64$ that do not arise this way. - -REPLY [5 votes]: This is from Buell, Binary Quadratic Forms, pages 109-119. -Note that $h(-16) = h(-4) = 1.$ -When $p$ is an odd prime, we use the Legndre symbol in -$$ h(-4p^2) = \frac{p - (-1|p)}{2} $$ -After that, if $u > 1,$ while $u$ is allowed odd or even as needed; if $p$ does not divide $u,$ then -$$ h(-4u^2p^2) = h(-4 u^2) \cdot \left(p - (-1|p) \right). $$ -If $p$ divides $u,$ -$$ h(-4u^2p^2) = h(-4 u^2) \cdot p. $$ -$$ h(-16u^2) = 2 h(-4 u^2) $$ -Let's see; note that $p - (-1|p)$ for an odd prime $p$ is always divisble by $4,$ so these class numbers are even, and the max power of $2$ grows. This is fair, the number of genera is growing with each new prime and is always a power of two, while each discriminant has a constant number of classes per genus. -For example, with odd $n > 1,$ if $$ n = \prod_p p^{e_p} $$ we get -$$ h(-4n^2) = \frac{1}{2} \prod_p \left( p - (-1|p)\right) p^{e_p -1} $$ -The conclusion is that $h(-4n^2)$ is mostly pretty similar in size to $\frac{n}{2} .$ In order to deviate much from that, it is most efficient to have $n$ squarefree odd, with either $n = 5 \cdot 13 \cdot 17 ...$ the product of the consecutive primes $1 \pmod 4$ up to some bound $P,$ or $n = 3 \cdot 7 \cdot 11 \cdot 19 ...$ the product of the consecutive primes $3 \pmod 4$ up to some bound. In these cases, the growth or shrinkage of $h$ is still bounded by Merten's Theorem, in one version (H+W_thm427) -$\sum_{p \leq x} \frac{1}{p} = \log \log x + B + o(1).$ I guess if we are taking half the primes according to mod 4, we would expect the estimate to be halved. -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= - n -4 n^2 h n factored - 1 -4 1 factor n = 1 - 2 -16 1 factor n = 2 - 3 -36 2 factor n = 3 - 4 -64 2 factor n = 2^2 - 5 -100 2 factor n = 5 - 6 -144 4 factor n = 2 * 3 - 7 -196 4 factor n = 7 - 8 -256 4 factor n = 2^3 - 9 -324 6 factor n = 3^2 - 10 -400 4 factor n = 2 * 5 - 11 -484 6 factor n = 11 - 12 -576 8 factor n = 2^2 * 3 - 13 -676 6 factor n = 13 - 14 -784 8 factor n = 2 * 7 - 15 -900 8 factor n = 3 * 5 - 16 -1024 8 factor n = 2^4 - 17 -1156 8 factor n = 17 - 18 -1296 12 factor n = 2 * 3^2 - 19 -1444 10 factor n = 19 - 20 -1600 8 factor n = 2^2 * 5 - 21 -1764 16 factor n = 3 * 7 - 22 -1936 12 factor n = 2 * 11 - 23 -2116 12 factor n = 23 - 24 -2304 16 factor n = 2^3 * 3 - 25 -2500 10 factor n = 5^2 - 26 -2704 12 factor n = 2 * 13 - 27 -2916 18 factor n = 3^3 - 28 -3136 16 factor n = 2^2 * 7 - 29 -3364 14 factor n = 29 - 30 -3600 16 factor n = 2 * 3 * 5 - 31 -3844 16 factor n = 31 - 32 -4096 16 factor n = 2^5 - 33 -4356 24 factor n = 3 * 11 - 34 -4624 16 factor n = 2 * 17 - 35 -4900 16 factor n = 5 * 7 - 36 -5184 24 factor n = 2^2 * 3^2 - 37 -5476 18 factor n = 37 - 38 -5776 20 factor n = 2 * 19 - 39 -6084 24 factor n = 3 * 13 - 40 -6400 16 factor n = 2^3 * 5 - 41 -6724 20 factor n = 41 - 42 -7056 32 factor n = 2 * 3 * 7 - 43 -7396 22 factor n = 43 - 44 -7744 24 factor n = 2^2 * 11 - 45 -8100 24 factor n = 3^2 * 5 - 46 -8464 24 factor n = 2 * 23 - 47 -8836 24 factor n = 47 - 48 -9216 32 factor n = 2^4 * 3 - 49 -9604 28 factor n = 7^2 - 50 -10000 20 factor n = 2 * 5^2 - 51 -10404 32 factor n = 3 * 17 - 52 -10816 24 factor n = 2^2 * 13 - 53 -11236 26 factor n = 53 - 54 -11664 36 factor n = 2 * 3^3 - 55 -12100 24 factor n = 5 * 11 - 56 -12544 32 factor n = 2^3 * 7 - 57 -12996 40 factor n = 3 * 19 - 58 -13456 28 factor n = 2 * 29 - 59 -13924 30 factor n = 59 - 60 -14400 32 factor n = 2^2 * 3 * 5 - 61 -14884 30 factor n = 61 - 62 -15376 32 factor n = 2 * 31 - 63 -15876 48 factor n = 3^2 * 7 - 64 -16384 32 factor n = 2^6 - 65 -16900 24 factor n = 5 * 13 - 66 -17424 48 factor n = 2 * 3 * 11 - 67 -17956 34 factor n = 67 - 68 -18496 32 factor n = 2^2 * 17 - 69 -19044 48 factor n = 3 * 23 - 70 -19600 32 factor n = 2 * 5 * 7 - 71 -20164 36 factor n = 71 - 72 -20736 48 factor n = 2^3 * 3^2 - 73 -21316 36 factor n = 73 - 74 -21904 36 factor n = 2 * 37 - 75 -22500 40 factor n = 3 * 5^2 - 76 -23104 40 factor n = 2^2 * 19 - 77 -23716 48 factor n = 7 * 11 - 78 -24336 48 factor n = 2 * 3 * 13 - 79 -24964 40 factor n = 79 - 80 -25600 32 factor n = 2^4 * 5 - 81 -26244 54 factor n = 3^4 - 82 -26896 40 factor n = 2 * 41 - 83 -27556 42 factor n = 83 - 84 -28224 64 factor n = 2^2 * 3 * 7 - 85 -28900 32 factor n = 5 * 17 - 86 -29584 44 factor n = 2 * 43 - 87 -30276 56 factor n = 3 * 29 - 88 -30976 48 factor n = 2^3 * 11 - 89 -31684 44 factor n = 89 - 90 -32400 48 factor n = 2 * 3^2 * 5 - 91 -33124 48 factor n = 7 * 13 - 92 -33856 48 factor n = 2^2 * 23 - 93 -34596 64 factor n = 3 * 31 - 94 -35344 48 factor n = 2 * 47 - 95 -36100 40 factor n = 5 * 19 - 96 -36864 64 factor n = 2^5 * 3 - 97 -37636 48 factor n = 97 - 98 -38416 56 factor n = 2 * 7^2 - 99 -39204 72 factor n = 3^2 * 11 -100 -40000 40 factor n = 2^2 * 5^2 - n -4 n^2 h n factored - -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=<|endoftext|> -TITLE: To what extent can we characterise the image of the topological Chern character? -QUESTION [21 upvotes]: For a finite CW complex $X$, the Chern character gives an isomorphism -of finite-dimensional vector spaces: -$$ - ch : K^*(X)\otimes \mathbb{Q} \to H^*(X, \mathbb{Q}). -$$ -The vector space $V = H^*(X, \mathbb{Q})$ thus comes equipped with two natural maximal-rank lattices: - -$L_H = H^*(X, \mathbb{Z}) / T$ (where $T$ is the torsion), -$L_K = ch(K^*(X) / T')$ (where $T'$ is the torsion). - - -What we can say about the relationship between $L_H$ and $L_K$? - -Here is simple necessary relationship. Let $n=[\dim X/2]$; using -denominators of size at most $n!$, the Chern character can be -expressed as a linear combination of Chern classes. Chern classes -are integral, so we must have: -$$ - n! \cdot L_K \subseteq L_H. -$$ - -What other necessary conditions are known? Is there a known set of -necessary and sufficient conditions? I.e., any $(V, L_H, L_K)$ -satisfying them can be realised by some $X$. - -For what it's worth, my motivation comes from spheres. Suppose we define a topological invariant $k$ to be the smallest natural number such that $k\cdot L_K \subseteq L_H$ (which presumably captures only a tiny bit of the relationship). Then the fact that $k=1$ for spheres allows a slick proof that the spheres admit no almost complex structure above dimension six. (In fact $L_K = L_H$ for spheres.) -Incidentally, the above ignores the $\mathbb{Z}/(2)$-grading. I haven't thought about this but I'd even be interested in the case when $X$ is even-dimensional with no odd-dimensional cohomology (e.g., a non-singular complex quadric). - -REPLY [2 votes]: I remember, dimly, work of J.Frank Adams,published in "Topology" (journal), in the early 1960's, on this kind of problem, -Bruno Harris<|endoftext|> -TITLE: Enhancing Grothendieck's universes and Grothendieck's axiom: Feferman's universe -QUESTION [7 upvotes]: A Grothendieck's universe is such a set $U$ so that - - -$\forall x \in U, x \subseteq U$, - -$\forall x,y \in U, \{x,y\} \in U$, - -$\forall x \in U, \mathcal{P}(x) \in U$, - -given a family $(X_i)_{i \in I}$ such that $I \in U$ and any $X_i \in U$ we have $\bigcup_{i \in I} X_i$, - -$\mathbb{N} \in U$. - - - -We can "do set theory" inside a single Grothendieck's universe. However, the existence of a single universe wasn't enough for Grothendieck, he introduced the axiom of universes: - -Let $X$ be a set. Then there is a universe $U$ such that $X \in U$. - -In particular, this guarantees that given a universe $U$, there is always a universe $V$ so that $U \in V$, hence we can enlarge a given universe. -Now, I'm personally interested in not the universe itself but it's enhancement based on a model-theoretic concept of "elementary substructure". Here Michael Shulman introduces a theory which he calls $ZMC/S$ (which is based on S.Feferman's theory $ZFC/S$) which takes $ZFC$ and adds a constant symbol $S$ which is a set which is both a universe and an which satsfies the following reflection principle: - -Let $\phi(x_1,...,x_n)$ be a formula in the language of set theory. Then -$$\forall x_1...\forall x_n, (\phi(x_1,...,x_n) \Longleftrightarrow \phi^S(x_1,...,x_n)),$$ - -where $\phi^S(x_1,...,x_n)$ denoted the formula obtained from $\phi(x_1,...,x_n)$ by restricting all quantifiers to the set $S$. -Assume this is the approach we wish to take for the foundations of category theory (I do not wish to discuss whether it is useful with respect to ordinary universes). Is a single such set $S$ enough for all purposes of category theory or do we instead need a proper class of similar universes, such as introduced there? Note that both Michael Shulman in his notes and Joel David Hamkins in his aforementioned answer claim that these theories are both equiconsistent with the assumption that "$Ord$ is Mahlo", which would imply that that they are equiconsistent with each other. However, consistency is not what interests me right now, what I want to understand is there there a matter of enlarging a universe when working with a single Feferman-Shulman universe $S$ as it is when working with a single Gorthendieck universe? - -REPLY [11 votes]: I haven't thought deeply about this question, but shooting from the hip my reaction is that one might sometimes need two or three such universes, or maybe even countably many, but proper-class many seems unlikely to be needed. -The point of Feferman's ZFC/S (and its slight modification ZMC/S) for category theory is that when we want to prove general category-theoretic theorems and then apply them to an arbitrary structure $K$, we no longer have to choose a universe containing $K$ to define the categories to which we apply our theorems. Instead, we prove our theorems about "the" reflective universe $S$, so that a priori they only make statements about "small" structures, and then apply the reflection property of $S$ to deduce that the same statements we proved about small structures using category theory are true about "large" structures as well. -Thus, if we proceed in this way, it seems to me that the only situation in which we might need more than one $S$-like universe is if while doing the abstract category theory we need more than one universe at the same time. This might happen: for instance, we might want to talk about a 2-category of categories which contains as an object a category of sets. I've wanted to do that myself. Or a 3-category of 2-categories containing such a 2-category; I think maybe once I wanted to do that too. Maybe nowadays it would be an $(\infty,1)$-category containing as an object an $(\infty,1)$-category of $(\infty,1)$-categories, etc. etc. While I certainly can't rule out that we might want proper-class-much such nesting, right now I'm not thinking of any situation in which we would need to go beyond finitely or at most countably many such categories. -(But, this is the sort of statement that tends to get proven wrong as soon as it is made. All readers are invited to take it as a challenge: find an argument in abstract category theory that uses proper-class-many universes at the same time.)<|endoftext|> -TITLE: Naturality of PD model of a CDGA -QUESTION [5 upvotes]: In the paper "Poincaré duality and commutative differential graded algebras", Lambrechts and Stanley constructed PD model for cdga with simply connected cohomology. My question is: if $A$ and $B$ are two quasi-isomoprhic CDGA, and $\hat{A}$ and $\hat{B}$ be their PD models respectively, are $\hat{A}$ and $\hat{B}$ quasi-isomorphic as PD algebras? - -REPLY [3 votes]: By the work of Hamilton and Lazarev, cyclic $C_{\infty}$-algebras are uniquely determined up to cyclic $C_{\infty}$-quasi-isomorphism by the cyclic (PD) algebra structure on their cohomology. It is a beautiful result. More precisely: -Suppose $A$ and $B$ are simply connected cyclic $C_{\infty}$-algebras which are $C_{\infty}$-quasi-isomorphic and the induced isomorphism in cohomology is one of PD algebras. Then there exists a cyclic $C_{\infty}$-quasi-isomorphism between $A$ and $B$. The statement does not hold for $A_{\infty}$ and $L_{\infty}$-algebras. -Actually, something a bit stronger is true. It is the following version of the transfer theorem: -Suppose A is a cyclic $C_{\infty}$-algebra. Then there exists a cyclic $C_{\infty}$-algebra structure on the cohomology $H(A)$ such that: -1) the pairing in $H(A)$ is induced by the pairing in $A$ -2) the $C_{\infty}$-algebra structure on $H(A)$ in particular has structure maps $m_1=0$, $m_2=$ the product induced by the product of $A$. -3) there is a cyclic $C_{\infty}$-quasi-isomorphism between $A$ and $H(A)$ -4) the cyclic $C_{\infty}$-algebra structure on $H(A)$ is unique up to cyclic $C_{\infty}$-ISOMORPHISM. -PD CDGA's are particular examples of cyclic $C_{\infty}$-algebras. Note that this does not give a positive answer to the question since having a cyclyc $C_{\infty}$-quasi-isomorphism between two PD CDGA's is weaker than having a zig zag with intermediate PD CDGAs. However, this does provide certain uniqueness for the PD models which, for example, is useful for studying the invariance of certain string topology operations at the level of cohomology. -Hamilton and Lazarev wrote several papers about this. They called cyclic $C_{\infty}$-algebras "symplectic" $C_{\infty}$-algebras. A relevant paper is "Symplectic $C_{\infty}$-algebras" which appeared in Moscow Mathematical Journal in 2008. -@Najib Idrissi, I wonder if this implies your result?<|endoftext|> -TITLE: Do commutative rings without unity have the IBN property? -QUESTION [10 upvotes]: Let $R$ be a commutative rng, i.e. a commutative ring without an identity element. - -Does $R$ still have the Invariant Basis Number (IBN) property? - -Recall that a ring is said to have the IBN property if $R^m \cong R^n \Rightarrow m=n$. -All commutative rings have the IBN property, but the standard proof I know makes an essential use of the existence of a maximal ideal by Zorn's Lemma and passing to the residue field, which depends on the presence of a unit. - -REPLY [16 votes]: (I will write my comment as an answer.) -The answer is "not necessarily" for the way IBN is defined in the problem. For a counterexample, let $R=\oplus^{\omega} \mathbb Z$ -with zero multiplication. -But the definition of IBN in the problem is not the right one. It IS true that when $R$ is a commutative nonunital ring, the f.g. free $R$-modules have uniquely determined rank. The reason is that, when $R_1$ is the ring obtained from $R$ be formally adding a unit element, then the forgetful functor from the category of $R_1$-modules to the category of $R$-modules is an equivalence. The free $n$-generated $R$-module is $R_1^n$, not $R^n$.<|endoftext|> -TITLE: Intuitive proof of Golden-Thompson inequality -QUESTION [5 upvotes]: Sutter et al. [1] in their paper "Multivariate Trace Inequalities" give an intuitive proof of the following Golden-Thompson inequality: -For any hermitian matrices $A,B$: -$$ -\text{tr}(\exp{(A+B)}) \le \text{tr} \exp{(A)}\exp{(B)}. -$$ -Lemmas involve the spectral pinching method which uses the eigendecomposition $A=\sum_{\lambda}\lambda P_\lambda$ where the $\lambda$ are eigenvalues and $P_\lambda$ corresponding projectors which are mutually orthogonal. -The spectral pinching map with respect to $A$ is defined as -$$ -\mathcal{P}_A: X \mapsto \sum_{\lambda} P_\lambda XP_\lambda -$$ -and then come these properties for any $X \geq 0$: -1) $\mathcal{P}_A[X]$ commutes with $A$ -2) $\text{tr} \mathcal{P}_A[X]A = \text{tr} X A$ -3) \begin{align} -\mathcal{P}_A[X] &= \sum_{\lambda \in \text{spec}(A)} P_\lambda XP_\lambda\\ -&= \frac{1}{|\text{spec}(A)|} \sum_{y=1}^{|\text{spec}(A)|}U_yXU_y^*\\ -&\geq \frac{1}{|\text{spec}(A)|} X -\end{align} -where $\text{spec}(A) = \{\lambda_1, \lambda_2, \dots, \lambda_{|\text{spec}(A)|}\}$ -and $U_y = \sum_{z=1}^{|\text{spec}(A)|} \exp{\frac{i2\pi yz}{|\text{spec}(A)|}}P_{\lambda_z}$ satisfies $UU^T=I$ -1) and 2) are straightforward to follow. But I cannot understand how the second equality and the first inequality hold true for 3). -[1] https://arxiv.org/abs/1604.03023 - -REPLY [4 votes]: $\newcommand{\al}{\alpha} -\newcommand{\la}{\lambda}$ -Let $\la_1,\dots,\la_n$ be the distinct eigenvalues of $A$, so that $|\text{spec}(A)|=n$. Then -\begin{multline} -\sum_{y=1}^n U_yXU_y^*=\sum_{y=1}^n\sum_{u,v=1}^n e^{i2\pi yu/n}P_{\la_u}XP_{\la_v}e^{-i2\pi yv/n}\\ -= -\sum_{u,v=1}^n P_{\la_u}XP_{\la_v}\sum_{y=1}^n e^{i2\pi y(u-v)/n} -=\sum_{u,v=1}^n P_{\la_u}XP_{\la_v}n1_{\{u=v\}} - =n\sum_{u=1}^nP_{\la_u}XP_{\la_u}, -\end{multline} -so that the 2nd equality in 3) holds. -Now, as pointed out in the cited paper, $U_yXU_y^*\ge0$ for all $y$, whereas $U_n=I$. So, the inequality in 3) follows as well.<|endoftext|> -TITLE: $BG$ the stack, $BG$ the simplicial presheaf -QUESTION [5 upvotes]: I have a theoretical question about comparing two objects that I have recently come across. -For concreteness, let us work over the category $C$ of schemes over $k$. Let $G$ be an algebraic group over $k$. One can construct the stack $BG$ as a fibered category in groupoids and as a simplicial scheme $(BG)_n=G^n$ (with certain face and degeneracy maps which I don't specify). What is the precise relation between these two versions of $BG$? Can I obtain one from the other? -I think I might have heard that "the two constructions are equivalent because the simplicial $BG$ has no higher homotopy groups". Does this make sense? How does this implication work? -Any answer that could help me better understand the relation between these two $BG$s is very welcome. - -REPLY [16 votes]: The two constructions are not quite equivalent. Let me write $\mathbf BG$ for the stack and $B_\bullet G$ for the simplicial scheme to better distinguish between them. There is a third relevant player, $BG$, which is the presheaf of ∞-groupoids on $C$ presented by $B_\bullet G$. -The precise relation between these three objects is the following: - -$\mathbf BG$ is the fppf sheafification of $BG$ -$BG$ is the colimit of the simplicial object $B_\bullet G$ -$B_\bullet G$ is the Čech nerve (= $0$-coskeleton) of either $\mathrm{Spec}(k) \to \mathbf BG$ or $\mathrm{Spec}(k) \to BG$ - -In more detail, algebraic stacks over $k$ (in the sense of Artin, say) form a full subcategory of the $2$-category of fppf sheaves of groupoids (classically, "stacks in groupoids") on $C$, which is itself a full subcategory of the ∞-category of presheaves of ∞-groupoids on $C$. Thus, both $\mathbf BG$ and $BG$ live in this ∞-category (in fact they both belong to the subcategory of presheaves of groupoids), and one is the fppf sheafification of the other. The étale sheafification suffices if $G$ is smooth. -The reason $\mathbf BG$ and $BG$ are not the same is that there is a unique homotopy class of map $X\to BG$ from any scheme $X$, but homotopy classes of maps $X\to \mathbf BG$ are in bijection with isomorphism classes of $G$-torsors on $X$. In fact $BG$ is the full subpresheaf of $\mathbf BG$ spanned by the trivial $G$-torsors. -(Added details about 2.) -$BG$ being the colimit of $B_\bullet G$ is the manner in which simplicial sets give rise to ∞-groupoids. Of course, one must define ∞-groupoids at some point and one way to do that is to start with simplicial sets and invert weak equivalences, so that we have a localization functor {simplicial sets} → {∞-groupoids}. Once higher category theory is set up however, it turns out that this functor is the composition of the inclusion of simplicial sets into simplicial ∞-groupoids and of the colimit over $\Delta^{op}$ functor (this is the standard fact that a simplicial set is canonically the homotopy colimit of itself). In practice this is often a more useful way to think about it.<|endoftext|> -TITLE: A combinatorial property of uncountable groups -QUESTION [6 upvotes]: Let $A,B$ be two uncountable sets in a group $G$ such that for any elements $x,y\in G$ the intersection $xA\cap yB$ is finite. Let $\Phi:G\to 2^G$ be a function assigning to each element $x\in G$ some finite set $\Phi(x)\subset G$. - -Question. Is it true that there exist elements $x,y\in G$ and $a\in A\setminus\Phi(x)$ and $b\in B\setminus\Phi(y)$ such that $xa=yb$? - -Comment. The answer is affirmative if $ab=ba$ for any $a\in A$ and $b\in B$. In this case we can choose two elements $a\in A\setminus\Phi(b)$ and $b\in B\setminus \Phi(a)$ and put $x=b$ and $y=a$. - -REPLY [2 votes]: Unfortunately (for my further plans) this question has negative answer. Just take any two disjoint uncountable sets $A,B$ and consider the free group $G$ over the union $A\cup B$. Let $\Phi:G\to 2^G$ be the function assigning to each $g\in G$ the set of letters in the irreducible representation of $g$. -Now assume that there exist elements $x,y\in G$ and $a\in G\setminus\Phi(x)$ and $b\in G\setminus \Phi(y)$ such that $xa=yb$. Since $a\notin\Phi(x)$ the letter $a$ does not appear in the irreducible representation of $x$ and hence $\Phi(y)=\Phi(xab^{-1})=\Phi(x)\cup\{a,b\}$, which contradicts the choice of $b\notin \Phi(y)$. - -By the way, my purpose was to resolve Question 2.2 from this survey of Protasov. This question asks if the countability of a group $G$ is equivalent to the normality of the finitary ballean on $G$. It is known that a commutative or free group is countable if and only if its finitary ballean is normal.<|endoftext|> -TITLE: English translation of "Les aspects probabilistes du contrôle stochastique" -QUESTION [5 upvotes]: I am looking for an English translation of "Les aspects probabilistes du contrôle stochastique" written by Nicole El Karoui, or knowledge whether it exists. -Other references with similar content on Snell envelopes are also appreciated. -Reference info: Les aspects probabilistes du contrôle stochastique. (French) [The probabilistic aspects of stochastic control] Ninth Saint Flour Probability Summer School—1979 (Saint Flour, 1979), pp. 73–238, -Lecture Notes in Math., 876, Springer, Berlin-New York, 1981. Springerlink. - -REPLY [6 votes]: There is no English translation of El Karoui's lecture notes, however her work on Snell envelopes is described in Reflected Solutions of Backward SDE'S, and Related Obstacle Problems for PDE's. For a text book treatment of Snell envelopes see Methods of Mathematical Finance by Karatzas and Shreve.<|endoftext|> -TITLE: A combinatorial property of uncountable groups, II -QUESTION [8 upvotes]: Problem 1. Is it true that each uncountable group $G$ contains two subsets $A,B\subset G$ such that -1) for any $x,y\in G$ the intersection $xA\cap yB$ is finite and -2) for any function $\Phi:G\to 2^G$ assigning to each element $g\in G$ a finite subset $\Phi(g)\subset G$ there are two elements $x,y\in G$ and points $a\in A\setminus\Phi(x)$ and $b\in B\setminus \Phi(y)$ such that $xa=yb$. - -Remark. Such sets $A,B$ do exist if $G$ contains a subgroup $H$ that admits a homomorphism onto a group $\Gamma$ that contains an uncountable subset $U$ with infinite centralizer $C(U)=\bigcap_{u\in U}\{x\in\Gamma:xu=ux\}$. This means that a counterexample if exists, should be very non-commutative, like a Jonsson group, constructed by Shelah. -We recall that a group $G$ is Jonsson if it is uncountable but all proper subgroups of $G$ are countable. - -Problem 2. What is the answer to the Problem for simple Jonsson groups (constructed by Shelah)? - -Comment. Problem 1 is a combinatorial reformulation of Question 2.2 from this survey of Protasov. Question 2.2 asks if the countability of a group is equivalent to the normality of its finitary ballean. This question was also repeated (as Problem 12.6) in the paper "The normality and bounded growth of balleans" of Banakh and Protasov. - -REPLY [3 votes]: Problems 1 and 2 both have affirmative answers (implying that the finitary ballean of any uncountable group is normal). -Two cases are possible: -I. There exists a countable subgroup $A\subset G$ and an uncountable subset $B\subset G$ such that $bAb^{-1}\cap A$ is infinite for all $b\in B$. Replacing $B$ by a smaller uncountable set, we can assume that the family $(bA)_{b\in B}$ is disjoint. The latter condition can be used to show that the sets $A,B$ satisfy the condition 1 of the Problem. -We claim that for any function $\Phi:G\to [G]^{<\omega}$ there are elements $x,y\in G$ and $a\in A\setminus\Phi(x)$ and $b\in B\setminus \Phi(y)$ such that $xa=yb$. -Since $B$ is uncountable, there exists an element $b\in B\setminus\bigcup_{a\in H}\Phi(a)$. The set $bAb^{-1}\cap A$ is infinite and hence contains some element $a\notin\Phi(b)$. Put $x=b$ and $y=bab^{-1}\in A$. Observe that $xa=ba=yb$, $a\notin\Phi(b)=\Phi(x)$ and $b\notin\Phi(y)$. -II. There exists a countable infinite subgroup $A\subset G$ and an uncountable set $B'\subset G$ such that $bAb^{-1}\cap A$ is finite for every $b\in B'$. Replacing $B'$ by a smaller uncountable subset, we can aditionally assume that the family $(AbA)_{b\in B'}$ is disjoint. -We claim that the sets $A$ and $B=\{aba^{-1}:a\in A,\;b\in B'\}$ satisfy the conditions 1 and 2. The condition 1 will follow as soon as we check that for every $x\in G$ the intersection $xA\cap B$ is finite. Assuming that this intersection is not empty, we can find elemente $b\in B'$ and $a,\alpha\in A$ such that $x\alpha=aba^{-1}$. Taking into account that the family $(AvA)_{v\in B}$ is disjoint, we conclude that $xA\cap B=abA\cap B\subset abA\cap b^A$ where $b^A=\{gbg:g\in A\}$. -Given any element $z\in abA\cap b^A$, we can find elements $\alpha,g\in A$ with $ab\alpha=z=gbg^{-1}$ and conclude that $a^{-1}g=b\alpha gb^{-1}\in A\cap bAb^{-1}$. So, the element $g$ belongs to the finite set $F=a(A\cap gAg^{-1})$ and then $z=gbg^{-1}\in b^F:=\{fbf^{-1}:f\in F\}$. Therefore, $xA\cap B=abA\cap b^A$ is contained in the finite set $b^F$, which means that the sets $A,B$ satisfy the condition 1. -Now given any function $\Phi:G\to [G]^{<\omega}$, we shall find elements $x,y\in G$ and $a\in A\setminus\Phi(x)$ and $b\in B\setminus \Phi(y)$ such that $xa=yb$. -Using the uncountability of the set $B'$, choose an element $u\in B'\setminus \bigcup_{a\in A}a\Phi(a)a^{-1}$. -Find $a\in A\setminus \Phi(u)$. Put $y=a\in A$, $x=u$ and $b=a^{-1}ua\in B$. It follows that $xa=ua=aa^{-1}ua=yb$. -Also $a\notin\Phi(u)=\Phi(x)$ and $b=a^{-1}ua\notin \Phi(a)$ (as $u\notin a\Phi(a)a^{-1}$).<|endoftext|> -TITLE: To what extent can a von Neumann algebra be determined by its projection lattice structure? -QUESTION [6 upvotes]: Let $ M, N $ be von Neumann algebras, $ P $ (resp. $Q$) the projection lattice of $M$ (resp. $N$). Any isomorphism $ \varphi : M \to N $ on the level of involutive algebras induces an isomorphism $ \varphi_L : P \to Q $ of lattices. My question is, what can we say about the relation between $ M $ and $ N $, knowing only that their projection lattice structures are isomorphic? -For example, - -Suppose $ M $ is a factor, and $ P $ is isomorphic to $ Q $ as lattices, is $ N $ a factor as well? If so, do they have the same type? -Are there any examples where $ P $ and $ Q $ are isomorphic as lattices, but $ M $ and $ N $ are not isomorphic as involutive algebras? - -Here is some of my thoughts about question 1. -I believe that we can conclude $ N $ is also a factor in this case. If I remember correctly, we can identify factors as von Neumann algebras which can not be further decomposed as the direct sum of two (smaller) von Neumann algebras (can anyone give a reference for a proof of this fact, or in the case I am wrong, a counterexample?). Such a direct sum decomposition of a von Neumann algebra would be reflected on its projection lattice structure. -As for the types of $M$ and $N$. By the fundamental theorem of projective geometry, it is easy to see that if $ M $ is a type $ I_n $ factor, $ N $ a type $ I_m $ factor, and $ P $ is isomorphic to $ Q $ as lattices, then $ m = n $. This naturally raises the question of whether the same can be said for factors of other types. -Question 2 seems more intractable to me. But it sure is interesting in its own right. I think maybe some easy examples can be constructed, as it seems too wild to conjecture that $M$ is isomorphic to $N$ if $P$ is isomorphic to $Q$. Yet, I can not produce such an example after some effort. -Perhaps answers to these questions are well-known among experts, as they seem rather basic to me. In this case, please kindly point out the relevant references. - -REPLY [2 votes]: Related to Which complete orthomodular lattices arise from von Neumann algebras? but I post only one answer here. -A improved version of the question is: extend von Neumann equivalence for finite factors -to an equivalence (again in absence of type I$_1$ and I$_2$ components) between a -real vNa $A$ (as ring or $*$-ring), its ($*$-)ring of classical quotients $Q(A)$, given by locally measurable possibly unbounded operators, and its (ortho)lattice of projections $L(A)$. -This is answered here explicitly in presence of orthocomplementation and only implicitly in absence. -Here I make the involutionless case explicit, after a recall of some things from that post. - - [Usual disclaimer: in 1992 - 1994 my brain was still half working, and I am confident about things that I knew at that time. In 2014 much less so, and today nothing at all. Luckily these are things that I mostly studied back then so unless I completely mess up my rememberings the great picture is correct] - - - [Disclaimer$^2$. My point of view wants to maximize interaction between lattice and rings, with language translations of equivalent conditions. It searches to minimize computations (leaved to luckily already known results). Not everyone likes this (see Kadison's memoir of a Kaplansky - Segal interaction about operator algebras). Moreover, a complete picture of equivalences was liked by von Neumann much more than by others, and is usually considered bad didactics.] - -Recall from the type III post, for a real -or complex, here better seen as real polarizable, vNa $A$ with no type I$_1$ or type I$_2$ components: - -The lattice $L(A)$ (without involution) is equivalent to the classical ring $Q(A)$ of (right, and left by involution) quotients of $A$ realized as -locally measurable operators. -following Saito (1971) - - - [When $A$ is finite i.e. $L$ is modular, then the all affiliated operators are (locally) measurable and $Q$ is also the classical and maximal ring of (right) quotients of $A$, with unique extension of the involution. If not, then the involution does not extend to the maximal ring of maximal right quotients: it is always regular and right self-injective, so a involution can be there only when its coordinatized lattice is a continuous geometry.] - -The equivalence of $*$-rings between $A$ and $Q$ is given by the explicit Saito - Berberian construction of $Q$ from $A$ (Saito, theorem 4.2), -with the reconstruction of $A$ from $Q$ given by the "subring of bounded elements". -As seen by Saito and Berberian, $Q$ is a Baer$^*$-ring with the same projections as $A$, and the classical ring of quotients (every $q\in Q$ has the form $a/b$ with $a,b\in A$ and $b$ invertible in $Q$). -This implies that $Q$ as ring depends only on the ring $A$, and as Baer / Rickart ring its lattice is the same as that of $A$: -even if $Q$ has more idempotents than $A$, the order associated to the -divisibility preorder (is the associated lattice $L$ and) is the same thanks to projections giving a unique representative in each equivalence class. -The idempotents for one side generate the ring and for the other side they are identified with particular complementary pairs in $L$. In detail: -when matrix units of order at least 3 exist -(and the decomposition and dimension theory, -which depends only on $L$, gives a reduction to such a case), -the ring is generated by its idempotents $e,f,...$ with the relations $e\oplus f=e+f-fe$ when $ef=0$, and these relations depends only upon the lattice: -a $e$ "is" a complementary modular pair in $L$ (thanks to O-symmetry, all usual modularity conditions on a pair are equivalent; "independent" pairs) "kernel and image of $e$"; -$ef=0$ is "image of the first in the kernel of the second" and then the partial operation $\oplus$ on idempotents is "independent join of the images, dually for kernels". -Here the $Q$ is the largest possible ring over $L$ to be realized by right ideals generated by a idempotent -i.e. with algebraic complement: all "independent" complementary pairs in $L$ give a idempotent in $Q$ (and conversely only "independent" complements in $L$ can come from a Rickart ring). -The algebraic sum of corresponding subspaces of $H$ (kernel and image of the idempotent as linear operator) could be only dense -(and so one has a densely defined linear operator); the important point is the validity of the four modularity conditions for a pair in $L(A)$ (not in the larger $L(H)$); -this permits the construction of the ring from the generators and relations, realized by concrete linear algebra computations with closed subspaces of $H$. - - The purely algebraic setting of this is common in the general theory of rings of quotients (see for example Rowen's book about ring theory); one ends up with only densely defined operators, and it is decisive the identification of maps which are defined and coincide on a "dense" common part. Here one has endomaps and a modular lattice setting (abelian subobjects); the "inspiring" cases are instead with distributive lattices and functions towards a fixed external structure, like the real numbers: maps which are continuous and coincident on a dense open subset (a dense G$_\delta$ set in a Baire setting; outside a set in a ideal of zero measure set is another common setting; rational maps and Zariski topology in algebraic geometry). - - - While $A$ can be reconstructed from $L$ following the "linear algebra proof" of von Neumann coordinatization using linear endomorphisms of $H$, for $Q$ one must use also tricks (identify operators equal on a "large/dense" common domain, as usual in the theory of ring of quotients); however, for any countable subring of $Q$ a common dense domain of definition exists (like intersection of dense G$_\delta$ sets in Baire spaces, or co-negligible sets in measure spaces) that permits to realize the subring with linear operators in a dense subspace of $H$. - -Note that what matters here is not what kind of ring of quotients form the locally measurable operators, or the other specific properties above -(like the fact that the idempotents of $Q$ are exactly the "independent" complementary pairs in $L$, and not other kinds of complementary pairs). -Even if having an equivalence also with $Q$ is a nice complement to the global picture, the above is only used as model for the reconstruction of $A$ from $L$, using another kind of complementary pairs. - -The ring $A$ is instead equivalent to the lattice $L$ with the additional structure of S. Maeda semiorthogonality: "ideals generated by idempotents with product 0". -The reconstruction of $A$ from $L$ is as above, except that one now uses only the semiorthogonal pairs (which are more restricted than the independent ones). - -[As seen below, these are exactly the complementary pairs such that the associated algebraic direct sum decomposition of the (Hilbert) vector space $H$ -on which $A$ acts as "ring of operators" is also a topological decomposition, i.e. the skew-projections are continuous (and open, thanks to the Banach setting). -$A$ is realized as the ring of bounded linear operators associated to $L$ when $L$ is realized in a $H$. $Q$ instead includes also unbounded ones.] -(The semiorthogonality relation is also studied by L. Herman, -who relates it with Topping nonasymptoticity, and implicitly by -Bures -"modular separation" in 1984, with definitive improvements by S. Maeda in 1986 with the paper "Modular pairs in the lattice of projections of a von Neumann algebra"). - -So to complete the equivalence of $L$ with $A$ it is enough to see that the semiorthogonality can be recovered (in the class of vNa) from the lattice alone $L$. -This recovering (the only point not explicit in the Type III post) is given below. Note that once $L$ and $A$ are known to be equivalent one immediately has -(by general categorical arguments already used by von Neumann, even before categories were defined), that $L$ equipped with a antiautomorphism, involution, orthocomplementation -is equivalent to $A$ equipped with a antiautomorphism, involution, proper (a.k.a. positive definite: $xx^*=0\Rightarrow x=0$) involution. - -Lattice reconstruction of semiorthogonality: -first for type I factors, then in general. -For $A$ a type I factor i.e. for the lattice $L$ of closed subspaces of a Hilbert space $H$ one has: -a disjoint pair $X,Y$ is dually modular iff the algebraic sum $X+Y$ is again closed iff the natural map from $X\oplus Y$ [as external direct sum of topological vector spaces] -to $X+Y$ (as topological vector subspace of $H$) is a linear and topological isomorphism (Mackey, 1943 - 1945, in Banach spaces). -Dually modular is then equivalent to all other modularity conditions (in Hilbert spaces, one has O-symmerty of the lattice, -hence cross-symmetry, M-symmetry and dually. See F. Maeda, S. Maeda, theory of symmetric lattices). -Then following L. Herman and S. Maeda one identifies semiorthogonal pairs in type I factors with these "independent" pairs. -Another equivalent condition: there exists a new orthocomplementation in $L$ i.e. (see again Maeda Maeda for the equivalence) a scalar product in $H$ with the same topology as the original one -that makes the pair $X,Y$ orthogonal. (A orthogonal pair is semiorthogonal. Conversely, one can take $X,Y$ as Hilbert subspaces of $H$ and use the Hilbert space direct sum structure $X\oplus Y\oplus (X+Y)^\perp$ -which is linearly and topologically isomorphic to $H$ with the natural map. Note: the new scalar product is given by a linear change of variables, hence the new involution in $A$ is obtained from the old with a internal automorphism). -This last equivalent condition shows how the "compatibility" (simultaneous experimentability) relation for quantum logic propositions (defined in orthomodular structures as "to be contained in a boolean subalgebra") -can be defined in the lattice $L(H)$ without ortocomplementation: $X,Y$ "commute" (another name for that relation, since for projections in $A$ it coincides with commutativity in the ring) iff -there are $X',Z,Y'$ in $L$ which are independent (pairwise disjoint modular pairs, and also the sum of two independent from the third) such that $X'\oplus Z=X$, $Y'\oplus Z=Y$. -[The usual definition of compatibility uses existence of such orthogonal decompositions, but the above shows that existence of semiorthogonal ones is equivalent]. -Once one has defined "commutativity" in purely lattice terms in $L(H)$, one can define "faithful vNa representations" for a general $L$ (without involution) in a type I factor: -poset (hence complete lattice) isomorphism between $L$ and a subset of $L(H)$ (poset of closed subspaces, topology is fixed but not the scalar product) which is its own double-commutant. -[One obtains exactly the $L(A)$ for a spatial vNa in $H$, and one sees that all scalar products that give the same complete topology give the same $L(A)$ as lattices, even if the induced -orthocomplementation is not the same]. Then one has, following S. Maeda 1986: -a disjoint pair is semi-orthogonal iff -in (one hence) all such representations of $L$ it goes to a independent (modular, semiorthogonal, nonasymptotic) pair in $L(H)$. -This is a purely lattice reconstruction (without a chosen orthogonality). - - Note that this can only work for vNa. For representing JBW algebras, one must use also octonionic orthoplanes together with Hilbert lattices $L(H)$. - - - Note that a "finite" $A$ gives a modular $L(A)$ but not all among its pairs are modular pairs in the larger $L(H)$; semiorthogonality is stronger than modular disjoint. (It is weaker than orthogonality, being "orthogonality for a compatible orthocomplementation" which can be different from the originally given one) - -Another equivalent definition of semiorthogonality of a disjoint pair is "absolute modularity": -for all faithful representations of $L$ as complete sub ortholattice in a vNa, the pair goes to a (dually, cross, etc.) modular pair. - - For AW$^*$ (and its Jordan analogue) the last version (absolute modularity, using all AW$^*$-embeddings) might work or not: since the basic relations among idempotents which are used here survive enlarging the ring, semiorthogonality is "absolute for (complete otholattice / AW$^*$) embeddings", but here we lack a class of cases (like the type I factors for spatial embeddings of vNa) with two properties together: (a) semiorthogonality is characterized as modular disjoint in these cases; (b) every AW$^*$ embeds as double commutant in one such case. Perhaps (probably?) type III AW$^*$ factors (including the wild ones) might work for this class. They should work using the more general AW$^*$-embeddings, but this is unpleasant: to be independent from orhocomplementation, one must say "for each orthocomplementation ..." - -Complementary notes: - -The present case is analogous to the case of "elementary geometry". Really it is a generalization, since purely real $A$ are included here together with those with a $i=-i^*$, $ix=xi$, $i^2=-1$. -The real type I$_n$ case is "elementary geometry". The affine structure (over a euclidean hence uniquely orderable field) can be defined by pure incidence -(or equivalently by betweenness); it does not uniquely define a orthogonality structure -(polar reciprocity on the hyperplane at infinity, or metric on the vector space of translations), but all such additional structures are isomorphic (classification of quadratic forms, only one case is anisotropic). - -In a sense, von Neumann calling his algebras "rings of operators" is almost justified also for modern readers: the ring structure uniquely defines the real algebra structure, but not -the complex one (which exists iff a $i$ exists, and existence of $i$ can be important, for example for some applications to physics, but the specific choice of $i$ is not, even for physics; -unnaturality of the complex structure is emphasized also by the existence of factors with a $i$ where all ring i.e. real algebra anti-automorphisms must change sign to $i$). On the other side, -a "compatible" involution structure is also not unique, and again it is unique up to isomorphisms; but in this case it is important not only the existence, but also the specific choice. - - -It is almost like Veblen that in 1904 pretended that (real) euclidean geometry can be axiomatized as affine (betweenness without orthogonality) because a compatible metric structure is unique up to (non-unique) isomorphism. -It took Tarski to clearly spell out that no, unique up to a unique isomorphism is needed, and so Tarski axiomatized the various elementary geometries with more primitive concepts, -even a redundant system of primitive concepts so that axioms are easier to state (directly using primitive concepts) and understand. And the same did von Neumann with quantum logic; -even if the name he choose for the "analytic" equivalent of his "synthetic" axioms would suggest a Veblen-like oversight ("$*$-ring of operators" would have been the name), -he was conceptually unacceptable like the two best logicians (of his time, and not only). - -Notable non-counterexample: for complex type I$_n$ factors, there exist orthocomplementations in $L$ (the Grassmannian of a finite dimensional projective geometry) i.e. proper involutions in $A$ -which do not come from a Hilbert space structure on the vector space $H$: one takes a discontinuous involution of the complex numbers (and any vector basis of $H$, then declared orthonormal -for a hermitian form for the discontinuous involution. By Birkhoff and von Neumann, 1936, all orthocomplementations on $L$ i.e. proper involutions on $A$ come in this way). -These are not counter examples; here one asks for unicity of the $A$ in the special class of vNa that corresponds to $L$, -and one has only shown coordinatization by another $A$ outside that class: the ring is the same, but the involution makes $A$ not $*$-ring isomorphic to any vNa -unless the complex involution gives a fixed subfield isomorphic to the reals; this is possible ($z^*=f^{-1}(\bar{f(z)})$ with $f$ discontinuous automorphism), but when it happens the -lattice with involution is again isomorphic to the standard one, so that unicity (up to non-unique isomorphism) in the special class is not disproved. - -Already Mackey and Kakutany in 1944 and 1946 -knew and proved that this complex type I$_n$ case is the only "almost counterexample" possible for type I factors; in all other cases, all possible orthocomplementations of the lattice and proper involutions -of the ring give a vNa (and always a isomorphic one, over a given lattice). The delicate point is to show that in the complex infinite dimensional case (the other cases being easier by absence of exotic involutions on -the skew field) the fact that one uses only closed linear subspaces for a Banach topology excludes discontinuous involutions. - -There is a way to represent a generic ring isomorphism between $A$ and $A'$ (equivalently, rings automorphisms of $A$) in terms of $*$-rings isomorphisms (i.e. real vNa isomorphism) -and spatial, internal automorphisms of $B(H)$ for a suitable Hilbert space representation. - -First a theorem of Kaplansky 1953 -for complex semisimple Banach algebras: - -if $\phi$ is a ring isomorphism from one semi-simple Banach algebra $A$ onto another, then $A$ is a direct sum $A_1\oplus A_2\oplus A_3$ with $A_1$ finite dimensional, $\phi$ linear on $A_2$, and $\phi$ conjugate linear on $A_3$. - -In our case the part $A_1$ is uninteresting (see the non-counterexample above), so one can assume a real linear $\phi$. -Yood 1958 -explicitly covers the real case (among other generalizations). -Still in 2010 -generalizations were produced. -Then one has a result of Gardner, 1964, -where C$^*$-algebras are complex and isomorphisms $\psi$ are linear, not only ring ones; -preservation of $*$ is not requested and continuity is part of the thesis: "In the proof, we make use of the fact that $\psi$ is necessarily bounded [((Dixmier 1957)), p. 15, Exercise 5]"): - -Even for W$^*$-algebras, the question has remained open: if $A$ and $A'$ are algebraically isomorphic, are they necessarily $*$-isomorphic? -See, e.g. [((Sakai 1962)), p. 1.53, Problem (i)]. -In this note, the above question is answered affirmatively for the more inclusive class of C$^*$-algebras [Theorem 3]. -Theorem 2 gives the structure of isomorphisms of C$^*$-algebras, showing that each is, in a certain canonical sense, spatial in nature. - -Spatial: there is a Banach space isomorphism between the direct sum of all GNS representations relative to pure states of $A$ and $A'$ which extends the given isomorphism between $A,A'$. -So, when both "atomic universal" representations are given in the same Hilbert space $H$, a inner automorphism of $B(H)$ implements the given isomorphism between $A,A'$. -Another description is the "polar decomposition" of such a isomorphism: Okayasu, 1974: -"any isomorphism of von Neumann algebras is decomposed uniquely as the product of a $*$-isomorphism and an automorphism implemented by an invertible positive element." -"Any isomorphism of C$^*$-algebras is decomposed uniquely as the product of a $*$-isomorphism and a positive automorphism, and this decomposition is norm-continuous." -[One can use the complexification $A+iA$ of a real or complex $A$ to obtain a complex-linear ring isomorphism between the complexifications and apply the theorem]. -There are also modern proofs pag. 1761. -One can also use unicity of the topology on a real or complex semisimple Banach algebra, Johnson 1967, -to have continuity of real-linear ring isomorphism among (real or complex) C$^*$-algebras. -Another proof, Aupetit 1982 -using the spectral radius and linearity. -"The main idea in the proof is to remark that the spectral radius is independent of the Banach algebra norm." - -Recent related papers (the authors might or might be not fully aware of the relevant old literature on the subject, I cannot say): - -https://arxiv.org/abs/2006.08959 -https://arxiv.org/abs/2010.01627 -https://arxiv.org/abs/2010.03176 -(Note also https://arxiv.org/abs/2107.05806 which considers $*$-regular subrings of a vNa i.e. of $B(H)$; the author might be unaware of the important work of C. Herrmann and coauthors around 2000 - 2015 in strictly related areas). -The different points of view give not only different proofs, but also theorems with nonzero boolean difference in applicability, for example: -ortholattice complete hom. among AW$^*$-algebras 2014. - -Note older results concerning Jordan maps, example -but here are relevant really much older results about Jordan maps. - -This is related to the present question, as shown by the equivalence of the Jordan structure on self-adjoint elements (or also the structure of "effects" with spectrum in $[0,1]$) with $A$ and $L$ including the involution. - -Is the "general linear" group (invertible elements of $A$) or its unitary subgroup ($uu^*=1=u^*u$) another equivalent structure? This is considered (also for other rings known to be equivalent to their associated lattice) in -old papers by Ehrlich, [[with modern followes and old ones like -Whitesitt who considered, before Maeda, the generalization of regular rings where idempotent generated ideals form a sublattice of all right ideals; -then Maeda considered the corresponding generalization of Rickart rings using annihilators of pairs of idempotents; equivalently, of the products of pairs of idempotents (in place of arbitrary elements)]]. -Above all the (Feldman -) Dye theorem [Ann. of Math. 61 and 63, (1956)] shows (sometimes only in the finite $A$ i.e. modular $L$ case) -a "almost equivalence" of the above structures ($L$ or $A$ or $Q$) with or without involution -with their associated group (invertible element of the ring, or unitary elements of the $*$-ring). -However, note "almost" (up to a finite, order two, quotient of the group of automorphisms: -like the relation between a algebraically closed field with involution and its polarized version, i.e. the field of pairs in a real closed field): -the linear group (with its group of automorphism and anti-automorphisms) -can only determine $L$ i.e. $A$ with their group of automorphisms and antiautomorphisms -(an "unoriented", slightly weaker structure than $L$ i.e. $A$, like betweenness on a totally ordered set is weaker than the total order, and separation of pairs of points is weaker than cyclic order on a circle / projective line). -(For the structures with involution and unitary group the case is even more delicate, Dye notes I$_{2n}$ complex exceptions.) -This is a kind of "Klein Erlanger program" but with the group alone, without a externally given set for its action; -in this, it is like Tits buildings. After this one expects also a study of the Lie structure on $A$, and how much it can determine $A$ i.e. $L$; also this exists but I never looked at it. - -More details for the case with involution excluded might appear in future here; some other complements are already there.<|endoftext|> -TITLE: Are Gray codes in $\{0,1\}^n$ isomorphic? -QUESTION [6 upvotes]: Let $n\in\mathbb{N}$ be a positive integer. Two elements of $\{0,1\}^n$ form an edge if and only if their Hamming distance equals $1$. It is known that $\{0,1\}^n$ endowed with this graph structure possesses Hamiltonian cycles, also known as Gray codes. -If $c_1, c_2$ are Hamiltonian cycles in $\{0,1\}^n$, is there a graph isomorphism $\varphi:\{0,1\}^n\to \{0,1\}^n$ such that $c_1$ gets mapped onto $c_2$? - -REPLY [16 votes]: If you consider the Boolean addresses of points on the hypercube as base $2$ expansions of integers, then the vertices of the $4$-dimensional cube may be labeled $0, 1, \ldots, 15$. With this labeling, no two of the following Hamiltonian paths differ by a cube automorphism: -$(0,2,6,7,15,14,10,8,12,13,9,11,3,1,5,4)$, -$(0,2,6,14,10,8,12,13,9,11,15,7,3,1,5,4)$, -and -$(0,1,3,2,6,4,5,7,15,13,12,14,10,11,9,8)$. -This is how I reasoned: -(1) The vertices of the $n$-cube may be identified with the points in $\mathbb R^n$ whose coordinates are in $\{0,1\}^n$. (Call this a geometric realization of the $n$-cube.) It is from this representation that points obtain Boolean addresses. For example, the vertex $(0,1,0,1)\in\mathbb R^4$ of the geometric realization of the $4$-cube has address 0101, hence will be labeled $5$. -(2) Any automorphism of the $n$-cube is determined by what it does on a single point and its immediate neighbors. -(3) It follows from (2) that each automorphism of the geometric realization from (1) is induced by a rigid motion of $\mathbb R^n$, since there are enough rigid motions to permute vertices transitively and then to permute the neighbors arbitrarily. The main point derived from this is that the automorphisms of the cube preserve its hyperfaces. -(4) From (3) we derive an invariant of Hamiltonian cycles in the $n$-cube: the $n$-term sequence of integers counting how many times the cycle is cut by hyperplanes which divide the cube into two parallel hyperfaces. -(5) The cut sequence for my first Hamiltonian cycle is $(2,4,4,6)$. The cut sequence for the second cycle is $(2,2,4,8)$. -The cut sequence for the third cycle is $(2,2,6,6)$. - - -Here is an alternative way to view what is written above. -My first Hamiltonian path, written in terms of Boolean addresses, is: -${\small (0000,0010,0110,0111,1111,1110,1010,1000,1100,1101,1001,1011,0011,0001,0101,0100)}$. -You can check that it is a Hamiltonian cycle by observing that exactly one bit changes each step of the cycle. The invariant of this cycle, which I called the cut sequence, is just the unordered sequence of the number of bit-changes per digit through the cycle. For example, the sequence of least significant digits of addresses in this Hamiltonian cycle is -$(0,0,0,1,1,0,0,0,0,1,1,1,1,1,1,0)$, -and there are four places in the cycle where the least significant bit changes. One can show that there are four places where the $2$nd-least significant bit changes, six places where the next least significant bit changes, and two places where the most significant bit changes. Thus, the (unordered) cut sequence is $(4, 4, 2, 6)$. -The second Hamiltonian cycle, written in terms of Boolean addresses is: -${\small (0000,0010,0110,1110,1010,1000,1100,1101,1001,1011,1111,0111,0011,0001,0101,0100)}$. -It is easy to read off that the $3$rd least significant digit goes through $8$ bit changes, which is the important distinction between the two Hamiltonian cycles. -The third Hamiltonian cycle is: -${\small (0000,0001,0011,0010,0110,0100,0101,0111,1111,1101,1100,1110,1010,1011,1001,1000)}$. - -I did not locate any Hamiltonian cycle in $\{0,1\}^4$ with cut sequence $(4,4,4,4)$, but also I do not see a reason why such a Hamiltonian cycle cannot exist. - -Edit. -To follow up on comments of Gerry Myerson and verret, it is shown in -Gilbert, E. N. -Gray codes and paths on the n-cube. -Bell System Tech. J 37 1958 815-826. -That there are 9 equivalence classes of Hamiltonian cycles in the 4-cube modulo the symmetry group of the cube. Also, it is shown in -Abbott, H. L. -Hamiltonian circuits and paths on the n-cube. -Canad. Math. Bull. 9 1966 557-562. -that the total number of Hamiltonian cycles in the $n$-cube is at least -$$\left(7\sqrt{6}\right)^{2^n-4},$$ -which grows doubly exponentially, while the size of the automorphism group of the $n$-cube is $2^n\cdot n!$, which grows singly exponentially $\left(2^{O(n\log(n)}\right)$. -These results provide a quick solution to this problem.<|endoftext|> -TITLE: Is taking the positive part of a measure a continuous operation? -QUESTION [8 upvotes]: Here is question I tried to answer for some time - it seems to be straightforward, but I have trouble figuring it out. -Let $\Omega$ be a compact domain in $\mathbb{R}^n$. For any signed Borel measure $\mu$ on $\Omega$ let $\mu_+$ denote its positive part (obtained by Hahn-Jordan decomposition). My question is: - -Is taking the positive part a continuous operation, i.e. does $\mu^n\to \mu$ in the space signed Borel measures imply $\mu^n_+\to \mu_+$ in the same space? - -Of course, the answer would be positive if -$$\|w^n_+-w_+\|_{\mathfrak{M}}\leq \|w_n-w\|_{\mathfrak{M}}$$ -where $\|\cdot\|_{\mathfrak{M}}$ denotes the variation norm, but I could not see this. - -REPLY [5 votes]: The Borel measures on $\Omega$ can be identified with continuous linear functionals $\newcommand{\bR}{\mathbb{R}}$ $\mu:C(\Omega)\to\bR$. Assume that $\Omega$ is compact so $C(\Omega)$ is Banach space. Denote by $\Vert-\Vert$ the sup norm on $C(\Omega)$ and by $\newcommand{\eM}{\mathscr{M}}$ $\eM(\Omega)$ the dual of $C(\Omega)$ equipped with the dual norm -$$ -\Vert \mu\Vert_*:=\sup_{\Vert f\Vert\leq 1}|\mu(f)|. -$$ -Set -$$ -C(\Omega)_+:=\big\{ f\in C(\Omega):\;\; f(x)\geq 0,\;\;\forall x\in\Omega)\big\}. -$$Let $\mu\in \eM(\Omega)$ and $f\in C(\Omega)$. Then, for any $f\in C(\Omega)_+$ we have (see Theorem 4.3.2. of R.E. Edwards: Functional Analysis) -$$ -\mu_+(f)=\sup_{0\leq g\leq f} \mu(g). -$$ -Let $\mu,\nu\in \eM(\Omega)$, and $f\in C(\Omega)_+$. Then for any $0\leq g\leq f$ we have -$$ -\mu(g)-\nu(g)\leq \Vert\mu-\nu\Vert_*\Vert g\Vert \leq \Vert\mu-\nu\Vert_*\Vert f\Vert. -$$ -Hence, for any $0\leq g\leq f$ -$$ -\mu(g)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert+\nu(g), -$$ -and, symmetrically, -$$ -\nu(g)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert+\mu(g). -$$ -Taking the sup on both sides of the above inequalities we deduce -$$ -\mu_+(f)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert+\nu_+(f)\implies \mu_+(f)- \nu_+(f)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert, -$$ -$$ -\nu_+(f)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert+\mu_+(f)\implies \nu_+(f)- \mu_+(f)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert. -$$ -Hence -$$ -\big\vert (\mu_+-\nu_+)f\big\vert\leq \Vert\mu-\nu\Vert_*\Vert f\Vert. -$$ -This implies -$$ -\Vert \mu_+-\nu_+\Vert_*\leq \Vert\mu-\nu\Vert_*. -$$<|endoftext|> -TITLE: Does this multiplicative function have a name? If so, what is known about it? -QUESTION [6 upvotes]: It is well-known that the Euler $\phi$-function is multiplicative: that is, for co-prime positive integers $m,n$ we have $\phi(mn) = \phi(m)\phi(n)$. Thus it is defined by its values on prime powers. We know that $\phi(p^k) = p^{k-1} (p-1)$ for all primes $p$. -What about the multiplicative function $\psi$ defined on the primes by $\psi(p^k) = p^{k-1} (p+1)$? Does it have a name? If so, what's known about it? -For example, can one evaluate $\sum_{n \leq X} \psi(n)$? - -REPLY [8 votes]: This is also called the Dedekind $\psi$ function: -$$ -\psi(n):=n\prod_{p|n}(1+p^{-1}) -$$ -See also A001615 and A158523.<|endoftext|> -TITLE: Is there a clear-cut analogue of the strong form of Serre's Conjecture for residually reducible Galois Representations? -QUESTION [9 upvotes]: Let $p$ be a prime and $\mathbb{F}$ a finite field of characteristic $p$. The theorem of Khare and Wintenberger roughly states that an irreducible, odd Galois representation $\bar{\rho}:G_{\mathbb{Q}}\rightarrow GL_2(\mathbb{F})$ which is ramified at finitely many primes lifts to a modular Galois representation associated to an eigenform of optimal weight and level. The optimal level of this form is the prime to $p$ part of the Artin conductor of $\bar{\rho}$. This was the strong form of Serre's conjecture. -Hamblen and Ramakrishna showed that the hypothesis of irreducibility may be relaxed by showing that under some conditions, a reducible and indecomposable $\bar{\rho}$ lifts to a Galois representation associated to an eigenform. However, they are not able to optimize the level of the eigenform. -My question is the following: is it expected that the optimal level is the prime to $p$ part of the Artin conductor of $\bar{\rho}$? It may be the case that it is not exactly this, and if it is not, is there an explicit counterexample to this? - -REPLY [3 votes]: In addition to François' answer, here is what can be said if you only seek for an isomorphism between the semi-simplification of your residual representation $\overline{\rho}$ and the semi-simplification of the reduction of a $p$-adic representation attached a Hecke eigenform. -Consider an odd mod $p$ Galois representation $\overline{\rho}=\nu_1\oplus\nu_2$ of Serre weight $k$ and level $N$ (coprime to $p$) and assume $p>k+1$. Then, there exist $\epsilon_1,\epsilon_2$ two Galois characters unramified at $p$ such that $\overline{\rho}\simeq\epsilon_1\oplus\epsilon_2\chi_p^{k-1}$. Set $\eta=\epsilon_1^{-1}\epsilon_2$. Then, there is a newform $f$ of (optimal) weight $k$ and level $N$ and a prime ideal $\mathfrak{p}$ over $p$ in $\overline{\mathbf{Q}}$ such that we have $\overline{\rho}\simeq\overline{\rho}_{f,\mathfrak{p}}^{ss}$ if and only if we have $B_{k,\eta}=0$ or $\eta(\ell)\ell^k=1$ for some prime $\ell$ dividing $N$. -Here, $\eta(\ell)=\eta(\mathrm{Frob}_\ell)$ if $\eta$ is unramified at $\ell$ and $\eta(\ell)=0$ otherwise. (Roughly speaking $B_{k,\eta}$ is the mod $p$ reduction of the $k$-th Bernoulli number associated with a lift of $\eta$.)<|endoftext|> -TITLE: What does it mean for a category to admit direct integrals? -QUESTION [30 upvotes]: Given an infinite countable group $G$, the category of unitary representations of $G$ admits direct integrals. -Namely, given a measure space $(X,\mu)$ and a measurable family of unitary $G$-reps $(H_x)_{x\in X}$, one can form a new representation -$$ -H=\int^\oplus H_x = \left\{\text{$L^2$ functions $f:X \to \coprod_{x\in X} H_x$}\right\}, -$$ -called the direct integral of the $H_x$'s. -Conversely, given $H\in \mathrm{Rep}(G)$ and an abelian sub-von Neumann algabra $A\subset \mathrm{End}_G(H)$ of the von Neumann algebra of endomorphisms of $H$, one can find a measure space $X$ s.t. $A=L^\infty X$, and a measurable family of unitary representations $(H_x)_{x\in X}$ such that $H=\int^\oplus H_x$. -The category of unitary representation of $G$ is said to admit direct integral decompositions. - -What does it mean for a $W^*$-category to admit direct integrals? -What does it mean for a $W^*$-category to admit direct integral decompositions? - -Presumably, "admitting direct integrals" is not a property of a category, but is extra structure instead. What is this extra structure? What axioms does this extra structure satisfy? -Is "admitting direct integral decompositions" a property of a category that admits direct integrals? - -Here are some examples of categories that admit direct integral decompositions. -Example1: The category of representations of a separable C* algebra. -Example2: Let $A$ be a separable C* algebra, let $(p_n)_{n\in\mathbb N}$ be an increasing sequence of projections in $A$, and let $p\in A$ be yet another projection which is bigger than all the $p_n$. -Then the category of all representations $(H,\pi)$ of $A$ such that $\pi(p_n)\to \pi(p)$ in the strong topology is another example of a category that admits direct integral decompositions. -Example3 (very slight generalisation of Example2): Let $A$ be a separable C* algebra. Let $S$ be a countable set. For every $\alpha\in S$, let $(p^\alpha_n)_{n\in\mathbb N}$ be an increasing sequence of projections, and let $p^\alpha\in A$ be another projection which is bigger than all the $p^\alpha_n$. -Then we can consider the category of all representations $(H,\pi)$ of $A$ such that $\forall \alpha\in S,\pi(p^\alpha_n)\to \pi^\alpha(p)$ in the strong topology. - -REPLY [6 votes]: The following is an argument for showing that "having direct integral" is definitely not a property nor a "property-like-structure" of $W^*$-categegories, but a real, non-trivial additional structure. I.e. that in examples where they exists they are not intrinsic to the $W^*$ structure. That does not answer the question, but I felt like it was worth mentioning as it shows that lots of approach to this question are bound to fail, and hopefully give some hints for what kind of additional structure a solution should have. -Let $C=C([0,1])$, and I'm considering the category of representations of $C$, it admit direct integral, but the $W^*$-category have a lot of automorphisms that do not preserves these representations. -Indeed the category of representation of $C$ is equivalent to the category of normal representation of its enveloping Von Neuman algebra $C''$. As any commutative von Neuman algebra $C''$ can be split into an atomic part and a diffuse part. -One has one atom for each element of $[0,1]$, and they corresponds to the irreducible representations $\chi_x$ of $C$. -So for any bijection of $[0,1]$ there is an automorphism of $C''$ that justs permute the atom and leave the diffuse part unchanged. On the $W^*$-category it permutes the irreducible representation (and their orthogonal sums) while leaving the diffuse part of each representation unchanged. -As any representation is a direct integral of irreducible, these direct integral are clearly not going to be preserved by the automorphisms: -for example if I simply chose the automorphism $1-x$ of $[0,1]$ then the representation $L^2([0,1/2]) = \int_0^{1/2} \chi_x$ should become (if integral were preserved) under the automorphism $L^2[1/2,1] = \int_{1/2}^{1} \chi_x$ and these representation are not even isomorphic. -Even worse (but I have not checked this last part in details), in the case (as above) where the automorphism of $[0,1]$ is chosen measurable, it seems to me that the isomorphisms of the $W^*$-category preserve the natural notions of measurable fields of representations and morphisms. Indeed unless I'm mistaken the splitting of representations in "atomic+diffuse" can be applied to the measurable fields and if the permutation of the atomic part is measurable it can be applied to fields as well. So this type of structure will be insuficient. (this seem to for example completely prevent what is suggested at the end of Tobias Fritz answer).<|endoftext|> -TITLE: Eigenvalues of Laplace-Beltrami on half sphere -QUESTION [7 upvotes]: Let $ \Delta_\theta$ denote the Laplace-Beltrami operator on $S^{N-1}$. The eigenvalues of this are well known. I assume the same is the case of this operator on the upperhalf sphere; say $ S^{N-1}_+$ with zero Dirichlet boundary conditions. Does anyhow know where I can find a reference for these? -thanks -Craig - -REPLY [14 votes]: Using symmetry, you can extend any Dirichlet eigenfunction on the upper half-sphere to the entire sphere $\mathbb{S}^{N-1}$. Therefore, the spectrum of the upper hemisphere is a subset of the spectrum of the full sphere. You are searching for the spherical harmonics which vanish on the great circle $x_N \equiv 0$. The reference that I've seen that explicitly constructs the $N-1$ dimensional spherical harmonics is the following paper of Frye and Efthimiou: https://arxiv.org/pdf/1205.3548.pdf -In theory, this reduces your question to a combinatorial problem involving Legendre polynomials, though I haven't solved out the combinatorics explicitly. For the 2-sphere, it seems like the eigenfunctions (and their eigenvalues) you are looking for are the $Y_l^m$ where $m+l$ is odd. From this, you can see the that spectrum is $l(l+1)$ but with less degeneracy than with the full sphere. - -REPLY [4 votes]: Tools for computing eigenvalues on disks in constant-curvature space forms are worked out in Chapter II, section 5 of Chavel's book Eigenvalues in Riemannian Geometry although skimming it I do not see the spectrum itself explicitly written out. Basic idea is separation of variables in polar coordinates.<|endoftext|> -TITLE: Seeking a more symmetric realization of a configuration of 10 planes, 25 lines and 15 points in projective space -QUESTION [15 upvotes]: I've got ten (projective) planes in projective 3-space: -\begin{align} -&x=0\\ -&z=0\\ -&t=0\\ -&x+y=0\\ -&x-y=0\\ -&z+t=0\\ -&x-y-z=0\\ -&x+y+z=0\\ -&x-y+t=0\\ -&x+y-t=0 -\end{align} -If I did not make a mistake somewhere, their intersections produce $25$ lines and $15$ points, each line containing $3$ points, each plane containing $6$ lines and $7$ points. Ten of the points belong to $4$ lines and $4$ planes each and five of them belong to $7$ lines and $6$ planes each. $15$ lines belong to $2$ of the planes and $10$ of them belong to $3$ of the planes. -The above indicates that this configuration is highly symmetric, is it known? How to compute its automorphism group? Where to look? My goal is to find another more symmetric realization. - -REPLY [13 votes]: With the aid of the answer I now managed to find a better realization. At the expense of some of the symmetries it can be nicely drawn in 3d space - it is just the barycentric subdivision of a tetrahedron: - -The planes are faces (4 of them) and the planes through an edge and the barycenter of the tetrahedron (+6); the lines are edges (6), the lines joining a vertex with the barycenter of its opposite face (+4) or with the midpoint of one of its non-incident edges (+12), as well as lines through midpoints of opposite edges (+3); and the points are vertices, edge midpoints, face barycenters, and the barycenter of the tetrahedron (4+6+4+1). Additional symmetry that is not well apparent in this realization is that there is a full S$_5$. In the projective 3-space it acts transitively on all five yellow points (i. e. vertices of the tetrahedron and its barycenter), on all ten blue points, all ten orange lines, all fifteen green lines, and all ten planes.<|endoftext|> -TITLE: Direct proof of "Nuclear implies $C_{red}^*(G) \cong C^*(G)$" -QUESTION [6 upvotes]: It is well-known that for a discrete group $G$ the following statements are equivalent: - -$C_{red}^*(G)$ nuclear -$C_{red}^*(G) \cong C^*(G)$ canonically i.e. there exists an *-isomorphism between the full and the reduced group $C^*$-algebras extending the identity on the corresponding conolution algebra. - -All proves of this I have seen so far show a chain of equivalent statements usually starting with assuming amenability of the group $G$ (which is also equivalent). -For me the two statements above seem to be of the same flavour therefore I'm wondering if there is a (short) direct way to see the equivalence (or at least the implication of (i) to (ii)). - -REPLY [5 votes]: The following direct proof uses Fell's absorption principle, as well as its (easy) corollary: the "diagonal" map $C^\ast(G) \hookrightarrow C^\ast_r(G) \otimes_{\max{}} C^\ast_r(G)$ is injective (see Theorem 8.2 in Pisier's book). -$(i) \Rightarrow (ii)$: If $C^\ast_r(G)$ is nuclear, then the composition -\begin{equation} -C^\ast(G) \to C^\ast_r(G) \otimes_{\max{}} C^\ast_r(G) = C^\ast_r(G) \otimes_{\min{}} C^\ast_r(G) \subseteq \mathbb B(\ell^2(G) \otimes \ell^2(G)) -\end{equation} -is faithful, and this representation is $\lambda \times \lambda$ where $\lambda$ is the left regular representation. By Fell's absorption, $\lambda \times \lambda$ and $\lambda \times I$ are equivalent, where $I$ is the trivial representation, so $\lambda \colon C^\ast(G) \to \mathbb B(\ell^2(G))$ is faithful. Hence $C^\ast(G) = C^\ast_r(G)$. -$(ii)\Rightarrow (i)$: Assume $C^\ast(G) = C^\ast_r(G)$, or even weaker, that the trivial representation $\tau \colon C^\ast_r(G) \to \mathbb C$ is well-defined. Let $D$ be an arbitrary $C^\ast$-algebra. Let $\pi_G \colon C^\ast_r(G) \to \mathbb B(H)$ and $\pi_D \colon D \to \mathbb B(H)$ be representations with commuting images such that $\pi := \pi_G \times \pi_D \colon C^\ast_r(G) \otimes_{\max{}} D \to \mathbb B(H)$ is faithful. Note that we have a faithful representation -\begin{equation} -\lambda \otimes \pi \colon C^\ast_r(G) \otimes_{\min{}} (C^\ast_r(G) \otimes_{\max{}} D) \to \mathbb B(\ell^2(G) \otimes H). -\end{equation} -Hence we get -\begin{eqnarray*} -\| \sum_{i=1}^n u_{g_i} \otimes d_i\|_{C^\ast_r(G)\otimes_{\max{}} D} &=& \| \sum_{i=1}^n \tau(u_{g_i}) \otimes (u_{g_i} \otimes d_i)\|_{\mathbb C \otimes_{\min{}} (C^\ast_r(G)\otimes_{\max{}} D)} \\ -&\leq& \| \sum_{i=1}^n u_{g_i} \otimes (u_{g_i} \otimes d_i)\|_{C^\ast_r(G) \otimes_{\min{}} (C^\ast_r(G)\otimes_{\max{}} D)} \\ -&=& \|\sum_{i=1}^n \lambda(g_i)\otimes \pi_G(u_{g_i}) \pi_D(d_i) \|_{\mathbb B(\ell^2(G) \otimes H)}. -\end{eqnarray*} -Let $U\in \mathbb B(\ell^2(G) \otimes H)$ be the unitary given by -\begin{equation} -U\delta_g \otimes \xi = \delta_g \otimes \pi_G(g) \xi. -\end{equation} -Note that this unitary implements a unitary equivalence $\lambda \times \pi_G \sim \lambda \times I$. -As $\pi_G$ and $\pi_D$ have commuting images, it easily follows that $U$ and $1\otimes \pi_D$ commute. Hence -\begin{eqnarray*} -&& \|U^\ast \sum_{i=1}^n \lambda(g_i)\otimes \pi_G(u_{g_i}) \pi_D(d_i) U \|_{\mathbb B(\ell^2(G) \otimes H)}\\ -&=& \| \sum_{i=1}^n \lambda(g_i) \otimes \pi_D(d_i)\|_{\mathbb B(\ell^2(G) \otimes H)} \\ -&=& \| \sum_{i=1}^n u_{g_i} \otimes d_i\|_{C^\ast_r(G) \otimes_{\min{}} D}. -\end{eqnarray*} -Thus $C^\ast_r(G) \otimes_{\max{}} D = C^\ast_r(G) \otimes_{\min{}} D$, so $C^\ast_r(G)$ is nuclear.<|endoftext|> -TITLE: Quantum homology of $(S^2 \times S^2,\omega_{FS}\oplus \omega_{FS})$ and Poincare duality -QUESTION [7 upvotes]: I am having some issues computing Poincare duality for the quantum homology $QH(M)$ when $(M,\omega)=(S^2 \times S^2,\omega_{FS}\oplus \omega_{FS})$. I am using the simple novikov ring $\Lambda$ consisting of polynomials in variables $t$ and $t^{-1}$. I know that a basis for $QH(M)$ (over $\Lambda$) is given by the elements -$$ -[pt], \ [M], \ A:=[\{pt\}\times S^2], \ B:=[S^2 \times \{pt\}]. -$$ -Does someone know (or know a reference) how to determine which classes are Poincare dual to each other in this scenario? I imagine that $[pt]$ would be Poincare dual to $[M]$, but I am not sure. - -REPLY [3 votes]: A bit late for this one, but I'll still post the answer for future visitors. -Poincaré duality on the quantum homology is just the same as Poincaré duality on normal homology, see for example the famous PSS paper [1, Section 2]. This means that for an element $\alpha= \sum_{A\in \Gamma} \alpha_A e^A$ with $\alpha_A\in H_{k+2c_1(A)}(M)$ and $\Gamma$ the image of $\pi_2(M)$ under the Hurewicz homomorphism, the Poincaré dual is given by -$$ -PD(\alpha)=\sum_{A\in\Gamma}PD(\alpha_A)e^A. -$$ -To get to your case with polynomials on $t,t^{-1}$ you just have to choose a basis for $H_2(M)$. -So $PD$ on your basis is just $PD$ on normal homology, which is uniquely characterised by the homology classes. -[1] "Symplectic Floer-Donaldson theory and quantum cohomology"; Piunikhin, Sergey, Dietmar Salamon, and Matthias Schwarz.<|endoftext|> -TITLE: References for Riemann surfaces -QUESTION [31 upvotes]: I know this question has been asked before on MO and MSE (here, here, here, here) but the answers that were given were only partially helpful to me, and I suspect that I am not the only one. -I am about to teach a first course on Riemann surfaces, and I am trying to get a fairly comprehensive view of the main references, as a support for both myself and students. -I compiled a list, here goes in alphabetical order. Of course, it is necessarily subjective. For more detailed entries, I made a bibliography using the bibtex entries from MathSciNet: click here. - -Bobenko. Introduction to compact Riemann surfaces. -Bost. Introduction to compact Riemann surfaces, Jacobians, and abelian varieties. -de Saint-Gervais. Uniformisation des surfaces de Riemann: retour sur un théorème centenaire. -Donaldson. Riemann surfaces. -Farkas and Kra. Riemann surfaces. -Forster. Lectures on Riemann surfaces. -Griffiths. Introduction to algebraic curves. -Gunning. Lectures on Riemann surfaces. -Jost. Compact Riemann surfaces. -Kirwan. Complex algebraic curves. -McMullen. Complex analysis on Riemann surfaces. -McMullen. Riemann surfaces, dynamics and geometry. -Miranda. Algebraic curves and Riemann surfaces. -Narasimhan. Compact Riemann surfaces. -Narasimhan and Nievergelt. Complex analysis in one variable. -Reyssat. Quelques aspects des surfaces de Riemann. -Springer. Introduction to Riemann surfaces. -Varolin. Riemann surfaces by way of complex analytic geometry. -Weyl. The concept of a Riemann surface. - -Having a good sense of what each of these books does, beyond a superficial first impression, is quite a colossal task (at least for me). -What I'm hoping is that if you know very well such or such reference in the list, you can give a short description of it: where it stands in the existing literature, what approach/viewpoint is adopted, what are its benefits and pitfalls. Of course, I am also happy to update the list with new references, especially if I missed some major ones. -As an example, for Forster's book (5.) I can just use the accepted answer there: According to Ted Shifrin: - -It is extremely well-written, but definitely more analytic in flavor. - In particular, it includes pretty much all the analysis to prove - finite-dimensionality of sheaf cohomology on a compact Riemann - surface. It also deals quite a bit with non-compact Riemann surfaces, - but does include standard material on Abel's Theorem, the Abel-Jacobi - map, etc. - -REPLY [5 votes]: For a hyperbolic perspective: Buser's "geometry and spectrum of compact riemann surfaces" is a nice book.<|endoftext|> -TITLE: Are there knots that can be distinguished by the Alexander-Conway polynomial, but not the Alexander polynomial? -QUESTION [9 upvotes]: On page 9 of Kauffman's Formal Knot theory, Kauffman claims - -The Alexander-Conway Polynomial is a true refinement of the Alexander Polynomial. Because it is defined absolutely (rather than up to sign and powers of variables) it is capable of distinguishing many links from their mirror images - a capability not available to the Alexander polynomial - -Does anybody know of any examples of this happening? Or when two knots have the same Alexander polynomial but different Alexander-Conway polynomials? - -REPLY [8 votes]: Let $\overline{L}$ denote the mirror image of a link $L$. The (reduced) Alexander-Conway polynomial $D_L(t)$ of an $n$-component link $L$ satisfies $D_{\overline{L}}(t)=(-1)^{n+1}D_L(t)$. More generally, the multivariable potential function $\nabla_L(t_1,\ldots,t_n)$ also satisfies $\nabla_{\overline{L}}(t_1,\ldots,t_n)=(-1)^{n+1}\nabla_L(t_1,\ldots,t_n)$. So you can use these sign-refined versions to distinguish links from their mirror images for links with an even number of components.<|endoftext|> -TITLE: How does multiplication affect degrees? -QUESTION [5 upvotes]: Let $M(n) \sim \mathbb{A}^{n^2}$ be the space of $n$-by-$n$ matrices, seen as an affine space over a field $K$, and endowed with the usual matrix multiplication. Let $V$ and $W$ be subvarieties of $M(n)$. The product $V\cdot W$ is a constructible set (by Chevalley's theorem); write $\overline{V\cdot W}$ for its Zariski closure. -Is it the case that $\deg(\overline{V\cdot W}) \leq \deg(V) \cdot \deg(W)$? - -Note: this is a less ambitious version of a question I previously asked (Bézout and products in algebraic groups). - -REPLY [5 votes]: This is false. Let $G$ be the open subset $\textbf{GL}_2$ in the affine space $\textbf{Mat}_{2\times 2} = \mathbb{A}^4$ with affine coordinates $x_{1,1}$, $x_{2,1}$, $x_{1,2}$ and $x_{2,2}$. Let $V$ be the one-dimensional subvariety, -$$ -V = \text{Zero}(x_{2,1},x_{2,2}-1,x_{1,1}-x_{1,2}) = \left\{\left[ \begin{array}{rr} s & s \\ 0 & 1 \end{array} \right] : s\in \mathbb{A}^1 \right\}. -$$ -Similarly, let $W$ be the following one-dimensional subvariety, -$$ -V = \text{Zero}(x_{2,1},x_{1,1}-1,x_{2,2}-x_{1,2}) = \left\{\left[ \begin{array}{rr} 1 & t \\ 0 & t \end{array} \right] : t\in \mathbb{A}^1 \right\}. -$$ -Each of $V$ and $W$ is a one-dimensional, affine linear variety. In particular, each has degree $1$. Yet the product set $\overline{V\cdot W}$ is the following two-dimensional variety, -$$ -W = \text{Zero}(x_{2,1},2x_{1,1}x_{2,2}-x_{1,2}) = \left\{\left[ \begin{array}{rr} s & 2st \\ 0 & t \end{array} \right] : (s,t)\in \mathbb{A}^2 \right\}. -$$ -This has degree $2$.<|endoftext|> -TITLE: HKR generalized character theory question regarding a certain construction -QUESTION [6 upvotes]: In that paper https://web.math.rochester.edu/people/faculty/doug/mypapers/hkr.pdf Hopkins-Kuhn-Ravenel introduced the idea of generalized character corresponding to a complex-oriented cohomology theory $E^*$. They construct an $E^*$-algebra $L(E^*)$, which is defined to be the colimit: ${\rm colim} E^*(B\mathbf{Z}^n_p)$ (More details regarding the whole construction can be found in the link.) -Later on, they use this construction to define the invariant ring $L(E^*)^{{\rm Aut}(\mathbf{Z}^n_p)}$, where ${\rm Aut}(\mathbf{Z}^n_p)$ acts as $E^*$-algebra homomorphisms ($\mathbf{Z}_p$ denotes the additive group of $p$-adic integers). To prove this they define $L_r(E^*)=E^*(B\mathbf{Z}_{p^r})$, and the natural ${\rm Aut}(\mathbf{Z}_{p^r})$ action on gives $L_r(E^*)^{{\rm Aut}(\mathbf{Z}_{p^r})}=p^{-1}E^*$. -My question has to do with the proof of the above: What I understand is that the invariant rings $L_r(E^*)^{{\rm Aut}(\mathbf{Z}_{p^r})}=p^{-1}E^*$, induce a direct system of $E^*$-algebras and the colimit must be $L(E^*)^{{\rm Aut}(\mathbf{Z}^n_p)}$. However, they don't give any proof hence should be somehow straightforward why the colimit is the invariant ring $L(E^*)^{{\rm Aut}(\mathbf{Z}^n_p)}$ (not at all to me). Can you explain me please if my understanding makes sense? If yes, probably an explanation why the above colimit converges on the invariant ring $L(E^*)^{{\rm Aut}(\mathbf{Z}^n_p)}$ would be really helpful. If not, a sort of insight would be very appreciable! - -REPLY [4 votes]: I think what you understand is correct. Write $A_r= L_r(E^*)$, which fit into a direct system $A_r\to A_{r+1}\to \cdots$. Let $G=\mathrm{Aut}(\mathbb{Z}_p^n)$, which acts compatibly on every $A_r$ (it actually acts on $A_r$ through the quotient group $\mathrm{Aut}((\mathbb{Z}/p^r)^n)$). -If $\cdots \to A_r\to A_{r+1}\to \cdots$ is a sequence of injective maps between $G$-sets, then it is straightforward to show that $\mathrm{colim} (A_r^G) \to (\mathrm{colim} A_r)^G$ is an isomoprhism. In fact, you only need them to be injective for all $r\geq R$ for some $R$. -It remains to show that the $A_r\to A_{r+1}$ are injective. I don't see a proof of this in the paper, but it is surely true. -Here's a proof that $A_r\to A_{r+1}$ is injective if $r\geq1$. There's probably a better proof, but it's what comes to mind now. I use the description in the proof of [HKR, 6.5]: -$$ -A_r = S_r^{-1} E^*B\Lambda_r, -$$ -where $\Lambda_n=(\mathbb Z/p^r)^n$, and $S_r\subseteq E^*B\Lambda_r$ is a certain multiplicatively closed subset, which can be defined as follows: -$$ -S_r = \{ c(\alpha):=(B\alpha)^*(x)\; | \; \alpha\in \Lambda_r^*\smallsetminus\{0\} \},\qquad \Lambda_r^*:=\mathrm{Hom}(\Lambda_r, U(1)), -$$ -where $x\in E^*BU(1)$ is a chosen coordinate of the formal group. Although the set $S_r$ depends on the choice of $x$, the fraction ring $A_r$ doesn't, because any two coordinates differ by a unit in $E^0BU(1)$. Note that the direct system of $A_r$ comes from the inverse system $\cdots\to\Lambda_{r+1}\to \Lambda_r\to\cdots$ -Note that $c(\alpha^p)=[p](c(\alpha))= c(\alpha)f(c(\alpha))$ where $[p](x)$ is the $p$-series of the formal group and $f(x)=[p](x)/x$ is a power series. This implies that if we invert $c(\alpha^p)$ then we automatically invert $c(\alpha)$ as well. For every $\alpha\in \Lambda_r^*\smallsetminus\{0\}$ with $r\geq1$, some $\alpha^{p^k}$ is in the image of $\Lambda_1^*\to \Lambda_r^*$, so in fact -$$ -A_r = S_1^{-1}E^*B\Lambda_r, -$$ -where $S_1$ really means the image of $S_1\subset E^*B\Lambda_1$ under the map induced by $\Lambda_r\to \Lambda_1$. Since $E^*B\Lambda_r\to E^*B\Lambda_{r+1}$ is injective and $S_1^{-1}E^*B\Lambda_1$ is flat over $E^*B\Lambda_1$, the claim follows.<|endoftext|> -TITLE: Reconstructing a curve in $S^2$ from intersections with great circles -QUESTION [5 upvotes]: Take $S^2$ with its standard metric. The space of great circles in $S^2$ can be identified with the real projective plane $\mathbb{R}P^2$. Let $X$ be an embedded circle in $S^2$; associate to it a function $f_X:\mathbb{R}P^2\rightarrow \mathbb{Z}\cup\{\infty\}$ which counts the number of intersection points (with multiplicity) of $X$ with given great circle. Can we reconstruct $X$ from $f_X$? -Remark: I think from some version of Crofton formula, it should be possible to determine the length of $X$ from $f_X$. - -REPLY [2 votes]: The answer to this question, as stated, is "no". If infinitely many intersections are allowed, it is easy to construct plenty of different curves, each of them having infinitely many intersections with every great circle.<|endoftext|> -TITLE: Weak parabolic maximum principle on Riemannian manifolds -QUESTION [7 upvotes]: I'm studying by myself Mean Curvature Flow and I'm reading the paper "Interior estimates for hypersurfaces moving by mean curvature" by Klaus Ecker and Gerhard Huisken, specifically, I'm reading the following theorem: - -$\textbf{Theorem 2.1}$ Let $R > 0$ and $\textbf{x}_0 \in R^{n+1}$ be arbitrary and define $\varphi(\textbf{x},t)= R^2 - |\textbf{x} - \textbf{x}_0| - 2nt$. If $\varphi_+$ denotes the positive part of $\varphi$ we have the estimate -$$v(\textbf{x},t) \varphi_+(\textbf{x},t) \leq \sup_{M_0} v \varphi_+$$ -as long as $v(x,t)$ is defined everywhere on the support $\varphi_+$. - -The authors denote $\textbf{x} := F(p,t)$ for a solution of the MCF, $\textbf{x}_0 := F(p,0) = M_0$ and define a gradient function $v := \left( \langle \nu, \omega \rangle \right)^{-1}$, where $\nu$ is the unit normal vector of the hypersurface and $\omega$ is some fixed vector such that $\langle \nu, \omega \rangle > 0$ and $|\omega| = 1$. They defined in the proof a function $\eta := \left( R^2 - r \right)^2$, where $r := |\textbf{x}| - 2nt$ (they assumed w.l.o.g. that $\textbf{x}_0 = 0$) and they got - -$$\left( \frac{d}{dt} - \triangle \right) v^2 \eta \leq -12 v \nabla v \cdot \nabla \eta + \eta^{-1} \nabla \eta \cdot \nabla (v^2 \eta)$$ - -They stated - -If we replace $\eta$ by $(\varphi_+)^2$ this computation remains valid on the support of $\varphi_+$ as long as $v$ is defined. The weak parabolic maximum principle then implies the result. - -I would like to know what is this weak parabolic maximum principle. -This is what I thought about my question: -Initially, I thought that could be the weak parabolic maximum principle on $\mathbb{R}^n$ which is common to see in PDE courses, but the problem is that I'm working with a manifold which receive a local treatment, then I thought that the weak parabolic maximum principle on $\mathbb{R}^n$ could be extended to a manifold, but I couldn't extend the result. -I found a maximum principle applied on MCF in this lecture notes (it's the theorem $2.2.1$ on page $17$), but I couldn't see how this helps me to conclude the inequality of the theorem $2.1$ of the paper. - -REPLY [7 votes]: Firstly, you have the wrong inequality (there is a small typo in the paper). Young's inequality is typically written for nonnegative numbers, but for any $a,b\in\Bbb R$ we have -\begin{align*} --ab&\le |ab|\\ -&\le \frac{1}{2}a^2+\frac{1}{2}b^2. -\end{align*} -In the context of the Ecker--Huisken, -\begin{align*} --6v \nabla v\cdot\nabla\eta&\le |6v\nabla v\cdot\nabla\eta|\\ -&\le 6|\nabla v|^2\eta+6|\nabla |\mathbf x|^2|^2v^2. -\end{align*} -When rearranged, -$$ -6|\nabla v|^2\eta-6|\nabla |\mathbf x|^2|^2v^2 \le 6v \nabla v\cdot\nabla\eta.$$ -We therefore have -$$(\partial_t-\Delta)v^2\eta\le \eta^{-1}\nabla\eta\cdot\nabla(v^2\eta).$$ -Here is the kind of theorem you need now. - -Weak Maximum Principle. Let $M$ be a smooth manifold and consider a linear parabolic operator $$Lu=\partial_tu-a^{ij}(X)\nabla_{ij}u+b^i(X)\nabla_iu$$ - on a bounded domain $\Omega\subset M\times[0,\infty)$. If $u\in C^{1,2}(\Omega)\cap C^0(\overline\Omega)$, $Lu\le 0$ in $\Omega$ and if $u\le 0$ on $\mathcal P\Omega$ (parabolic boundary), then $u\le 0$ in $\Omega$. -Proof. The same proof as Lieberman Lemma 2.3 works here. Here is the idea: Let $w=e^{-t}u$. Then $w$ and $u$ have the same sign and - $$\partial_tw=e^{-t}(\partial_tu-u).$$ Substituting - $$\partial_tu\le a\cdot\nabla^2u+b\cdot \nabla u$$ - gives - $$\partial_tw\le a\cdot\nabla^2 w+b\cdot\nabla w-w.$$ Let $X=(x,t)$ be a point where $w$ assumes a positive maximum. By hypothesis, $X\in \overline\Omega\setminus\mathcal P\Omega$. Then $a\cdot\nabla^2w(X)\le 0$ (here we use the fact that $a$ is a symmetric, positive definite matrix), $\nabla w(X)=0$, and $\partial_tw(X)\ge 0$. This leads to a contradiction if $X\in \Omega$. Therefore $X\in \partial\Omega\setminus\mathcal P\Omega$. This is a bit more technical, but morally if $w$ is positive somewhere on this set then for some time slice of $\Omega$ close to the "top," there will be a point $X^*$ in the slice which is a spatial maximum and such that $\partial_t w(X^*)\ge 0$. In this case we get a contradiction as well. See Lieberman for details. - -Now we define $u=(v\varphi_+)^2-(v\varphi_+)^2(0).$ We consider $\Omega$ to be the set of spacetime points $X$ satisfying $\varphi(X)>0$. Then $\Omega$ is actually a cone and $\mathcal P\Omega$ is just $\{\varphi(X)=0\}\cup \{\Omega\cap \{t=0\}\}$. Clearly $u\le 0$ here, so the WMP gives -$$u\le 0\quad\text{on }\Omega.$$ -Therefore, -$$v\varphi_+(X)\le \sup_{x\in \Omega\cap \{t=0\}}v\varphi_+(x,0)\quad \text{for }X\in\Omega.$$ -Using the fact that $\varphi_+=0$ outside of $\Omega$, this gives the desired inequality.<|endoftext|> -TITLE: Further Developments of Lieb-Schultz-Mattis theorem in Mathematics -QUESTION [6 upvotes]: The Lieb-Schultz-Mattis theorem [1] and its higher-dimensional generalizations [2] says that a translation-invariant lattice model of spin-1/2's cannot allow a non-degenerate ground state preserving both spin rotational and translation symmetries. -Another way to state Lieb-Schultz-Mattis theorem is -that an insulator with half-odd-integer spin per unit cell cannot have a trivial gapped ground state: In 1+1 spacetime dimension, the ground state must either break the translational symmetry (say along the $X$-direction as the lattice translational symmetry group of integer $\mathbb{Z}$) or be gapless (many low energy states in the large/infinite size volume limit of the system), while in higher dimensions the system may also spontaneously break the SO(3) spin rotational symmetry or support Topological quantum field theory (TQFT) at low energy. -There are many later developments in physics. - -Question: I wonder whether there are also some developments in mathematics for rigorous proofs or other extensions of Lieb-Schultz-Mattis theorem [1]? (In particular, since Elliott H. Lieb is a mathematical physicist and professor of mathematics.) - - -[1] Two soluble models of an antiferromagnetic chain, Elliott Lieb, Theodore Schultz, Daniel Mattis, Annals of Physics -Volume 16, Issue 3, December 1961, Pages 407-466 -[2] Lieb-Schultz-Mattis in higher dimensions, -M. B. Hastings, -Phys. Rev. B 69, 104431 – Published 29 March 2004 - -REPLY [3 votes]: A "rigorous proof" of [2] in the OP has been published by Nachtergaele and Sims, Commun. Math. Phys. 276 (2007) 437-472. - -For a large class of finite-range quantum spin models with - half-integer spins, we prove that uniqueness of the ground state - implies the existence of a low-lying excited state. For systems of - linear size $L$, of arbitrary finite dimension, we obtain an upper - bound on the excitation energy (i.e., the gap above the ground state) - of the form $(C\log L)/L$. This result can be regarded as a - multi-dimensional Lieb-Schultz-Mattis theorem and provides a rigorous - proof of a recent result by Hastings.<|endoftext|> -TITLE: Embedding Turing machine -QUESTION [6 upvotes]: I have some questions about Turing machines. Is there an embedding method where you embed Turing machines, finite automata into continuous space or graphs? Or are there geometrical approaches to analyze Turing machines or automata? ( embed them into some spaces and apply geometrical tools) Thank you - -REPLY [3 votes]: Let $\mathcal{M}$ be a class of binary functions acting on strings in $\Sigma^*$, -along with a "size" function $|\cdot|:\mathcal{M}\to\{1,2,\ldots\}$ with the property that there are only finitely many $M\in\mathcal{M}$ of a given size. -Turing machines and finite automata are of this form, where size can be taken as the number of states. Then you can define the following symmetric positive definite kernel on $\Sigma^*$: -$$ K_n(x,x') = \sum_{M\in\mathcal{M}, |M|\le n} -M(x)M(x'). -$$ -In words, this counts the number of $M$ on at most $n$ states which accept both $x$ and $x'$. This kernel then embeds $\Sigma^*$ into a corresponding Reproducing Kernel Hilbert Space. One can ask what languages can be defined by linear separators under this kernel. This topic was explored in -https://www.sciencedirect.com/science/article/pii/S0304397508004581 -and -http://www.jmlr.org/papers/volume10/kontorovich09a/kontorovich09a.pdf -.<|endoftext|> -TITLE: Intuition for pseudo-points and the inductive step in Johnstone's proof of Deligne's completeness theorem -QUESTION [10 upvotes]: In Johnstone's Topos Theory appears the following lemma. -7.41 Lemma. Let $P$ be a pseudo-point of $\mathsf C$, $X$ a $J$-sheaf on $\mathsf C$, and $x,y$ two distinct element of $P(X)$. Let $(V_j\to V)$ be a finite $J$-covering family in $\mathsf C$, and $v$ and element of $P(h_V)$. There there exists a refinement $Q$ of $P$ such that (a) the images of $x,y$ under the natural map $P(X)\to Q(X)$ are distinct, and (b) the image of $v$ in $Q(h_V)$ is in $\bigcup _j\; \operatorname{im}(Q(h_{V_j})\to Q(h_V))$. -So the refinement continues to "separate" $x,y$, which have to do with the sheaf $X$ (but not the covering family $(V_j\to V)$), but has the additional property (b), which seems to be only about representables associated to the covering family $(V_j\to V)$ (and not the sheaf $X$). -Questions. - -What is a nice conceptual intuition for the notion of a pseudo-point? Is there a geometric intuition? -What's the idea behind property (b)? - -This lemma seems to be topological in nature, but I just can't seem to unpack the content of property (b) of the refinement. - -REPLY [13 votes]: Johnstone's proof based on this lemma follows in fact the original proof of Deligne which can be found in the appendix "Criteria for the existence of points" to SGA 4, vol. VI. Both the original and a TeX reedition from 2015 can be found here. -The idea behind the notion of pseudo-point is very simple; it is just a description of a point using only the site. Since a point of a topos is by Diaconescu's theorem the same as a continuous flat functor from the site to $\mathcal{S}et$, when the site has finite limits this is the same as a filtered colimit of representable functors that in addition sends covering families to jointly surjective arrows in $\mathcal{S}et$. Therefore, one can identify the opint with a filtered diagram $(U_i)_{i \in I}$ in the site, which is precisely what Johnstone calls a pseudo-point. -As for property (b), it is just the inductive condition that ensures that in the end covering families are sent to jointly surjective arrows in $\mathcal{S}et$. Indeed, the continuous flat functor corresponding to the pseudo-point, evaluated in $C$, is precisely the filtered colimit $\lim [U_i, C]$, and condition (b) translates to a refinement of the filtered diagram to ensure the property above. -This way of proving Deligne's theorem follows closely Henkin's proof of Gödel's completeness theorem (which is why Deligne's theorem is referred to as a completeness theorem), since the inductive process of refining the pseudo-point is analogous to that of adding constants that witness existential statements. In both cases, one has to perform this addition countably many times since the introduction of new constants increases the set of formulas of the theory, which in Deligne's case boils down to applying lemma 9.4 repeatedly to produce a countable sequence of refinements of the pseudo-point. The analogy became evident with Joyal's proof of the completeness theorem for coherent logic, which is inspired by Deligne's proof and follows precisely this pattern.<|endoftext|> -TITLE: Lagrange four squares theorem -QUESTION [35 upvotes]: Lagrange's four square theorem states that every non-negative integer is a sum of squares of four non-negative integers. Suppose $X$ is a subset of non-negative integers with the same property, that is, every non-negative integer is a sum of squares of four elements of $X$. -$\bullet$ Is $X=\{0,1,2,\ldots\}$? -$\bullet$ If not what is a minimal set $X$ with the given property? - -REPLY [3 votes]: Back to my original suggestion, it appears that taking $X$ to be numbers with at most four prime factors is (basically) enough. Indeed, the following paper due to Tsang and Zhao show the following: every sufficiently large integer $N \equiv 4 \pmod{24}$ can be written as $x_1^2 + x_2^2 + x_3^2 + x_4^2 = N$, with $x_i \in P_4$ for $i = 1,2,3,4$. Here $P_r$ is the set of numbers with at most $r$ prime factors. I am guessing one only needs to take $r$ slightly larger, and allowing for possibly non-primitive representations, to cover the remaining congruence classes.<|endoftext|> -TITLE: Weak*-convergence of signed measures -QUESTION [7 upvotes]: Let $X$ be a compact Hausdorff space and let $M(X)$ denote the space of signed measures that is naturally dual to $C(X)$, the space of continuous functions on $X$. I am interested whether the following condition: -$$\nu_n(O) \to \nu(O)$$ -for every open set $O\subset K$ is sufficient for weak* convergence of $\nu_n$ to $\nu$, at least when $X$ is zero-dimensional but not necessarily metrizable. Thank you. - -REPLY [2 votes]: The answer is positive if the sequence $(\nu_n)$ is bounded; please see Theorem IV.9.15 in Dunford/Schwartz, vol. 1. PS: I just notice that this was already observed by Christian a couple of minutes ago.<|endoftext|> -TITLE: Principal curvatures of $\mathbb{R}^{n^2}$-embedded SO(n) -QUESTION [5 upvotes]: It's well known that the sectional curvatures of a Lie group, endowed with a left-invariant metric have a nice closed-form formula $k(X,Y) = \frac{1}{4} \|[X Y]\|^2$. -I'm wondering if the following (extrinsic) question has a simple answer: let me consider the natural embedding of $\mathrm{SO}(n)$ into $\mathbb{R}^{n^2}$, such that the induced Euclidean metric agrees with the left-invariant metric. -Can one derive closed-form expressions for the principal curvatures of the Weingarten map (for a normal direction, as the co-dimension is $> 1$) ? - -REPLY [12 votes]: What you are asking about is the second fundamental form of the embedding. Since this is a Lie group, it's enough to know what the second fundamental form is at the identity matrix $I_n=e$. Since the tangent space of $\mathrm{SO}(n)$ at the identity is the space of $n$-by-$n$ skew-symmetric matrices and the normal space is the space of $n$-by-$n$ symmetric matrices, the second fundamental form, which is a quadratic map from the tangent space to the normal space, is simply given by the order-2 term in the exponential series that gives the embedding. Thus, the quadratic map $\mathrm{I\!I}_e:T_e\mathrm{SO}(n)\to N_e$ (i.e., from the tangent space to the normal space), is given by -$$ -\mathrm{I\!I}_e(A) = \tfrac12 A^2. -$$ -Added remark [28 Oct 2018] I was asked how to prove this. Here is a little extra detail on this computation: When one writes the vector space $M_n$ of $n$-by-$n$ symmetric matrices with real entries as the direct sum of the subspaces $A_n$ (the $n$-by-$n$ anti-symmetric matrices) and the subspace $S_n$ (the $n$-by-$n$ symmetric matrices), the exponential map $\exp: A_n\to \mathrm{SO}(n)\subset M_n$ splits into two parts as -$$ -\exp(a) = I_n + a + \tfrac12a^2 + \cdots = (a + \tfrac16a^3+\cdots)+(I_n+\tfrac12a^2+\cdots) = \sinh(a) + \cosh(a), -$$ -where $\sinh:A_n\to A_n$ is a local diffeomorphism near $a = 0$, and $\cosh(a):A_n\to S_n$ is smooth. Thus, writing $b = \sinh(a)$, we have, for small $b\in A_n$, -$$ -\exp(a) = b + \cosh(\sinh^{-1}b) = b + \sqrt{I_n+b^2}. -$$ -Hence, near the identity $I_n\in\mathrm{SO}(n)$, the submanifold $\mathrm{SO}(n)$ can be regarded as a graph in $M_n = A_n\oplus S_n$ of the form -$$ -\bigl(b, \sqrt{I_n+b^2}\,\bigr). -$$ -Since one has the convergent Taylor series expansion $\sqrt{I_n+b^2} = I_n + \tfrac12b^2 + \cdots$, the result follows. -To compute the principal curvatures in a given direction $S\in N_e$, where $S$ is a symmetric $n$-by-$n$ matrix, you just need to take the inner product of $S$ with the above map and find the eigenvalues of the quadratic form -$$ -Q_S(A) = \tfrac12 \mathrm{tr}(SA^2) = \tfrac12 \mathrm{tr}(ASA). -$$ -with respect to the natural quadratic form $Q(A) = -\mathrm{tr}(A^2)$. By equivariance, it suffices to consider the case in which $S$ is diagonal, with eigenvalues (i.e., diagonal entries) $s_1,\ldots,s_n$. Then one finds that the eigenvalues of $Q_S$ with respect to $Q(A)$ are of the form $\tfrac12(s_i{+}s_j)$ for $1\le i -TITLE: Is this bound uniform in $N$? -QUESTION [6 upvotes]: I encountered this small combinatorial problem and do not quite know how to solve it: -Consider a set $\mathbf N:=\left\{1,2,....,N \right\}.$ This set has $\binom{N}{2}$ many subsets of cardinality $2.$ Thus, we can introduce variables $x_1,...,x_{\binom{N}{2}}$ taking a value in $\mathbb R$ where each of the variables is associated to precisely one subset of $\mathbf N$ of cardinality $2.$ -In other words there is a unique correspondence: -$x_i \leftrightarrow M_i$ where $M_i \subset \mathbf N$ and $\left\lvert M_i \right\rvert =2.$ -I would like to know whether the following estimate is true: -$$ \frac{1}{N} \sum_{i,j=1}^{\binom{N}{2}} x_i x_j \le C \sum_{i,j=1}^{\binom{N}{2}} \left\lvert M_i \cap M_j \right\rvert x_i x_j ?$$ -for some $C$ independent of $N$ and all $x_i,x_j$? -EDIT: -It holds true if $x_i=1$, since then the left hand side is $\binom{N}{2}^2/N=\frac{1}{4}(N-1)^2N$ and the right hand side is, using the hypergeometric distribution, $$2\binom{N}{2}+ \binom{N}{2}^2\frac{4}{N}=N(N-1)+N(N-1)^2.$$ -So an obvious scaling argument does not disprove the estimate. However, it is not clear to me whether this estimate is true in general? -I emphasize that the left-hand side $\frac{1}{N} \sum_{i,j=1}^{\binom{N}{2}} x_i x_j$ can be interpreted as $$\frac{1}{N} \langle (x_1,..,x_{\binom{N}{2}}),\mathbb 1 (x_1,..,x_{\binom{N}{2}})\rangle $$ -where $\mathbb 1$ is the matrix with all entries equal to one. This one has one non-zero eigenvalue with eigenvector $(1,...1)$ and all other eigenvalues are zero. That's why I was particularly curious to study that case separately. -So does this inequality hold true in general? - -REPLY [7 votes]: Let $u_i\in \mathbb{R}^N$ be a vector with two coordinates equal to 1 and other equal to 0, corresponding to the characteristic vector of the set $M_i$. Then your inequality rewrites as $N^{-1}(\sum x_i)^2\leqslant C(\sum x_i u_i)^2$. Denote $v=(1,1,\dots,1)\in \mathbb{R}^N$. Then $(u_i,v)=2$ (here $(u,v)$ stands for the inner product of vectors $u,v$) and we have $2\sum x_i=(\sum x_i u_i,v)$. Thus $$\left(\sum x_i u_i\right)^2\cdot N=\left(\sum x_i u_i\right)^2\cdot v^2\geqslant \left(\sum x_i u_i,v\right)^2=4\left(\sum x_i\right)^2$$ -as desired.<|endoftext|> -TITLE: Residually finite group surjective to nonresidually finite group with finitely generated kernel -QUESTION [10 upvotes]: As is described in the title, is there a known example such that there is a surjective homomorphism of groups $$f: G\rightarrow H,$$ with $G$ and $H$ finitely presented, $G$ is residually finite, and $H$ is non-residually finite, such that $\ker f$ is finitely generated? - -REPLY [10 votes]: You can obtain such a surjection for any finitely presented non-residually finite group $H$ using Daniel Wise's residually finite Rips construction, which is the main result of this paper: A Residually Finite Version of Rips's Construction.<|endoftext|> -TITLE: What is a spectrum object in $\infty$-topoi? -QUESTION [8 upvotes]: For any spectrum $E$, there is a "discrete" topos spectrum $(Spaces / E_n)_n$. And I believe any topos spectrum is a localization of a "discrete" one. Are there any "non-discrete" topos spectra? -To be precise, let $Topoi$ be the $\infty$-category of $\infty$-topoi and geometric morphisms (pointing in the direction of the right adjoint). Note that $Spaces$ is terminal in $Topoi$. Let $Sp(Topoi)$ be the stabilization of this category: an object is a sequence $(E_n)_{n \in \mathbb Z}$ of pointed $\infty$-toposes equipped with equivalences $E_n = \Omega E_{n+1}$ where $\Omega$ is the loops functor on the category $Topoi_\ast$ of pointed toposes. The question is -Question: Are there examples of objects of $Sp(Topoi)$ other than the "discrete" ones referred to above? -As alluded to above, I think there is a major restriction on the possibilities: the $\Omega$ functor on pointed $\infty$-categories lands in spaces. Because the "presheaves" functor preserves limits, the $\Omega$ functor on presheaf toposes lands in slices of $Spaces$. Since every $\infty$-topos embeds in a presheaf topos and embeddings are stable under finite limits, this implies that if $(E_n)_n$ is a topos spectrum, then each $E_n$ embeds in (i.e. is a localization of) a slice of $Spaces$. I think this sounds pretty restrictive -- I don't know an example of a topos that embeds into a slice of $Spaces$ which is not itself a slice of $Spaces$! - -REPLY [7 votes]: Following up on the answer of Simon Henry, let us prove the following statement. For a pro-space $\hat{X} = \{X_i\}_{i \in I}$, we let $Spaces_{/\hat{X}}$ denote the $\infty$-topos defined as the (cofiltered) limit in $Topoi$ of the $I$-family of étale topoi $Spaces_{/X_i}$. We will refer to such $\infty$-topoi as pro-étale $\infty$-topoi. -Claim: Suppose that ${\cal X}$ is an $\infty$-topos which is a loop object in $Topoi$. Then ${\cal X}$ is a left exact localization of a pro-étale $\infty$-topos $Spaces_{/\hat{X}}$ for $\hat{X} \in Pro(Spaces)$. If ${\cal X}$ is a double loop object then ${\cal X}$ itself is a pro-étale $\infty$-topos. In particular, every spectrum object in $Topoi$ consists of pro-étale $\infty$-topoi. -Proof: -Let ${\cal Y}$ be an $\infty$-topos equipped with two points $x_*,y_*: Spaces \to {\cal Y}$. Before considering the associated limit $Spaces \times_{\cal Y} Spaces$ we can consider the corresponding lax limit (or comma object) $Spaces \times^{\rm lax}_{\cal Y} Spaces$. We claim that this comma object exists and is furthermore a pro-étale $\infty$-topos. Indeed, let ${\cal Z}$ be an $\infty$-topos and let $p_*: {\cal Z} \to Spaces$ denote the terminal map. Then the data of a natural transformation $x_*p_* \Rightarrow y_*p_*$ is equivalent, by adjunction, to the data of a natural transformation $y^*x_* \Rightarrow p_*p^*$ of functors from $Spaces$ to $Spaces$. We note that both $y^*x_*$ and $p_*p^*$ are left exact functors and are hence corepresentable by pro-spaces, where the pro-space $Shp({\cal Z})$ corepresnting $p_*p^*$ is also known as the shape of ${\cal Z}$. Let $\hat{P}_{x,y} \in Pro(Spaces)$ denote the pro-space corepresenting $y^*x_*$. We then get that the data of a natural transformation $x_*p_* \Rightarrow y_*p_*$ is equivalent to the data of a map of pro-spaces $Shp({\cal Z}) \to \hat{P}_{x,y}$. We now recall that the formation of shapes ${\cal Z} \mapsto Shp({\cal Z})$ is left adjoint to the functor $\hat{X} \mapsto Spaces_{/\hat{X}}$ from pro-spaces to $\infty$-topoi. We may hence conclude that the data of a natural transformation $x_*p_* \Rightarrow y_*p_*$ is equivalent to the data of a geometric morphism ${\cal Z} \to Spaces_{/\hat{P}_{x,y}}$. We may then conclude that, if we let $q_*: Spaces_{/\hat{P}_{x,y}} \to Spaces$ be the terminal geometric morphism, then we have a canonical natural transformation $\tau:x_*q_* \Rightarrow y_*q_*$ which exhibits $Spaces_{/\hat{P}_{x,y}}$ as the desired comma object $Spaces \times^{\rm lax}_{\cal Y} Spaces$. Now let ${\cal P}_{x,y} \subseteq Spaces_{/\hat{P}_{x,y}}$ be the maximal left exact localization (see HTT 6.2.1.2) of $Spaces_{/\hat{P}_{x,y}}$ contained in the reflexive accessible subcategory -$$\{X \in Spaces_{/\hat{P}_{x,y}} | \tau_X:x_*q_*X \to y_*q_*X \text{ is an equivalence}\} \subseteq Spaces_{/\hat{P}_{x,y}}.$$ -Comparing universal properties we see that ${\cal P}_{x,y} \simeq Spaces \times_{\cal Y} Spaces$ represents the corresponding limit. In particular, for every points $x_*: Spaces \to {\cal Y}$ the loop $\infty$-topos ${\cal P}_{x,x} \simeq \Omega_x{\cal Y}$ is a left exact localization of a pro-étale $\infty$-topos. -Now suppose that ${\cal X}$ is an $\infty$-topos which is a double loop object, i.e., ${\cal X} \simeq \Omega_x{\cal Y}$ where ${\cal Y}$ itself is a loop object in $Topoi$. By the above we then have that ${\cal Y}$ is a left exact localization of pro-étale $\infty$-topos $Spaces_{/\hat{Y}}$, for some pro-space $\hat{Y} = \{Y_i\}_{i \in I} \in Pro(Spaces)$. Then $Spaces_{/\hat{Y}} = \lim_i Spaces_{/Y_i}$ and hence the space of points $y_*: Spaces \to Spaces_{/\hat{Y}}$ is naturally equivalent to the space $\lim_i Y_i = {\rm Map}_{Pro(Spaces)}(\ast,\hat{Y}) \in Spaces$. In this case, if $y_*: Spaces \to Spaces_{/\hat{Y}}$ corresponds to a compatible collection of points $y_i \in Y_i$ then -$$ \Omega_{y}Spaces_{/\hat{Y}} = \Omega_{y}\lim_i Spaces_{/Y_i} \simeq \lim_i \Omega_{y_i} Spaces_{/Y_i} \simeq \lim_i Spaces_{/\Omega_{y_i} Y_i} = Spaces_{/\Omega_{y}\hat{Y}} .$$ -Furthermore, if $y_*: Spaces \to Spaces_{/\hat{Y}}$ is a point which factors as $Spaces \stackrel{x_*}{\to} {\cal Y} \hookrightarrow Spaces_{/\hat{Y}}$ then $\Omega_y (Spaces_{/\hat{Y}}) \simeq \Omega_x {\cal Y}$. It then follows that ${\cal X} \simeq \Omega_x {\cal Y} \simeq Spaces_{/\Omega_y\hat{Y}}$ is a pro-étale $\infty$-topos, as desired. -$\Box$ -Remarks: -1) At this point one may be tempted to conclude that every spectrum object in $Topoi$ is the image of a spectrum object in $Pro(Spaces)$. This is very possibly the case, but a-priori it does not follow from the above claim. All that can be deduced is that if we denote by $\widehat{Etale} \subseteq Topoi$ the full subcategory spanned by pro-étale topoi, i.e., the essential image of $Pro(Spaces) \to Topoi$, then every spectrum object in $Topoi$ comes from a spectrum object in $\widehat{Etale}$. However, since the functor $Pro(Spaces) \to \widehat{Etale}$ is not fully-faithful it is not a-priori clear if the map $Sp(Pro(Spaces)) \to Sp(\widehat{Etale})$ is essentially surjective. In other words, there could, in principle, be a spectrum object ${\cal X}_0,{\cal X}_1,...$ in $Topoi$ in which every ${\cal X}_i$ is a pro-étale spectrum ${\cal X}_1 \simeq Spaces_{/\hat{X}_i}$ but the structure equivalences $\varphi_i:{\cal X}_i \stackrel{\simeq}{\to} \Omega{\cal X}_{i+1}$ do not come from equivalences of pro-spaces $f_i:\hat{X}_i \stackrel{\simeq}{\to} \Omega\hat{X}_{i+1}$ (up to finitely many $\varphi_i$'s we can always arange it up to equivalence, but it's not clear if we can arrange all the $\varphi_i$'s at once; there is an obstruction to this which lies in a suitable $\lim^1$ set). -2) The claim that $Sh(S^1)$ is not an infinite loop object in $Topoi$ can be deduced from the above claim, at least if we assume that the truncation functor $\tau_{\leq 0}: Topoi \to Topoi_0$ from $\infty$-topoi to $0$-topoi (i.e., locales) preserves cofiltered limits (it seems to me that this claim should be deducible from the fact that cofiltered limits on both cases are computed in ${\rm Cat}_\infty$, see HTT 6.3.3.1). Assuming this, suppose that we had a CW complex $X$ such that $Sh(X)$ was an infinite loop object in $Topoi$. By the claim we have that $Sh(X) \simeq Spaces_{/\hat{X}} = \lim_i Spaces_{/X_i}$ for some pro-space $\hat{X} = \{X_i\}_{i \in I}$. By the commutativity of cofiltered limits and truncations we then have that the local $O(X)$ is equivalent to the local $\lim_i \tau_{\leq 0}(Spaces_{/X_i}) = \lim_i{\rm Sub}(\pi_0(X_i))$, where ${\rm Sub}(\pi_0(X_i))$ is the locale of subsets of $\pi_0(X_i)$. Since $X$ is Hausdorff it is sober and hence we may deduce that $X$ is homeomorphic to the limit $\lim_i \pi_0(X_i)$ (computed in topological spaces). But $X$ is a CW-complex and hence locally connected, and so $X \cong \lim_i \pi_0(X_i)$ would have to be a discrete set. In particular, $Sh(S^1)$ is not an infinite loop object (or even a double loop object).<|endoftext|> -TITLE: Are conformal maps between Riemannian manifolds real-analytic? -QUESTION [11 upvotes]: This is a cross-post. - -Let $M,N$ be oriented smooth ($C^{\infty}$) $n$-dimensional Riemannian manifolds, and let $f:M \to N$ be a smooth orientation-preserving weakly* conformal map. -Do there exist real-analytic structures on $M,N$ that make $f$ real-analytic? - -I only assume the metrics are $C^{\infty}$. -Every smooth manifold has a unique real-analytic structure (up to diffeomorphism) compatible with its smooth structure. -A reasonable starting point would be to know whether every $C^{\infty}$ conformal map between real-analytic manifolds with real-analytic metrics is real-analytic. (I don't know a reference for that; anyway, what I am asking seems harder). -*A weakly conformal map is a map whose differential at every point is either conformal or zero. (This is equivalent to $df^Tdf =(\det df)^{\frac{2}{n}} \, \text{Id}_{TM}$). -Motivation: -I am trying to understand if smooth weakly conformal maps whose differential vanishes at a point are constant (for dimensions $n \ge 3$). This seems to be the case for analytic maps, hence my interest in the possible analyticity of such maps. - -For the Euclidean case, this follows directly by Liouville's theorem: -For $n=2$, every such map is complex-analytic. Let $\Omega \subseteq \mathbb{R}^n$ be an open subset, $n \ge 3$, and let $f:\Omega \to \mathbb{R}^n$ be a smooth conformal map. By Liouville's theorem, $f$ is of the form -$$ f(x)=b+\alpha\frac{1}{|x-a|^\epsilon}A(x-a),$$ -where $A$ is an orthogonal matrix, and $\epsilon \in \{0,2\}, b \in \mathbb{R}^n,\alpha \in \mathbb{R},a \in \mathbb{R}^n \setminus \Omega$. -So, up to translations and dilations, -$ f(x)=\frac{A x}{|x|^2}$ (where $ 0 \notin \Omega$) which is real-analytic as a multiplication of two analytic maps. ($1/x^2$ is analytic on $\mathbb{R} \setminus \{0\}$). - -REPLY [4 votes]: This is not an answer. Just a possible vector of attack for this problem. -According to the Theorem 11.4.6 of Harmonic Morphisms Between Riemannian Manifolds by Paul Baird and John C. Wood, every non-constant weakly conformal map between manifolds of the same dimension $n\geq 3$ is a conformal local diffeomorphism. The automorphism group of $(M, [g])$ is a Lie group whose Lie algebra is the algebra of conformal Killing vector fields. -According to the Theorem 3.8 of Normal BGG solutions and polynomials every conformal Killing vector field is given by a polynomial expression (of degree at most 2) in normal coordinates and normal frame. Given a Cartan geometry $(\pi: \mathcal{G} \to M, \omega \in \Omega^1(\mathcal{G}, \mathfrak{g})$, these normal coordinates and frames are given by flow flow of $\omega^{-1}(X)$ and by projection of that flow through $\pi.$<|endoftext|> -TITLE: Relation between the Axiom of Choice and a the existence of a hyperplane not containing a vector -QUESTION [10 upvotes]: In a lot of problems in linear algebra one uses the existence, for each $E$ vector space over a field $k$, and each $x\in E$, of a Hyperplane $H$ such that $E=k\cdot x \oplus H$ (Let us denote $\mathcal{P}$ this property). With Zorn's Lemma, the existence of a such $H$ is trivial. However, as this seems weaker than the existence of a basis of $E$, maybe $\mathcal{P}$ does not imply the Axiom of choice. Thus my question is: is $\mathcal{P}$ equivalent to the Axiom of Choice and if not, is there a weaker form of the Axiom of Choice equivalent to $\mathcal{P}$? - -REPLY [20 votes]: It is not hard to see that this statement is equivalent to "In every vector space, for every vector $v$ there is a functional $f$ such that $f(v)=1$". -If $\cal P$ holds, then the projection onto $k\cdot x$ is a linear functional which is nontrivial; if there is a nontrivial functional then choose $x$ which is mapped to $1_k$, and consider $H$ as the kernel of the functional. Some $\cal P$ is equivalent to "If $V$ is nontrivial, then $V^*$ is nontrivial". -As of October 2018, it is still unknown whether or not this is equivalent to the axiom of choice in full. But there are some intermediate results, for example it is consistent that there is a vector space over a field $k$ whose dual is trivial, for any fixed $k$. Moreover, this is consistent with $\sf DC_\kappa$ for any prescribed $\kappa$. -One should remark that restricting to Banach spaces and continuous functionals, the nontriviality of the [topological] dual is equivalent to the Hahn–Banach theorem. -Let me also add that Marianne Morillon showed that in $\sf ZFA$, $\cal P(\Bbb Q)$, namely restricting our attention to vector spaces over $\Bbb Q$, is not enough to deduce that every vector space over $\Bbb Q$ admits a basis, and that a weaker form of choice called "Axiom of Multiple Choice" implies $\cal P$ for fields of characteristics $0$. - -Morillon, Marianne, Linear forms and axioms of choice., Commentat. Math. Univ. Carol. 50, No. 3, 421-431 (2009). ZBL1212.03034. - -These two facts are not telling us a whole lot about what happens in $\sf ZF$, since in $\sf ZF$ the axiom of multiple choice implies full choice.<|endoftext|> -TITLE: Matrix rescaling increases lowest eigenvalue? -QUESTION [8 upvotes]: Consider the set $\mathbf{N}:=\left\{1,2,....,N \right\}$ and let $$\mathbf M:=\left\{ M_i; M_i \subset \mathbf N \text{ such that } \left\lvert M_i \right\rvert=2 \text{ or }\left\lvert M_i \right\rvert=1 \right\}$$ -be the set of all subsets of $\mathbf{N}$ that are of cardinality $1$ or $2.$ -The cardinality of the set $\mathbf M$ itself is $\binom{n}{1}+\binom{n}{2}=:K$ -We can then study for $y \in (0,1)$ the $K \times K$ matrix -$$A_N = \left( \frac{\left\lvert M_i \cap M_j \right\rvert}{\left\lvert M_i \right\rvert\left\lvert M_j \right\rvert}y^{-\left\lvert M_i \cap M_j \right\rvert} \right)_{i,j}$$ -and -$$B_N = \left( \left\lvert M_i \cap M_j \right\rvert y^{-\left\lvert M_i \cap M_j \right\rvert} \right)_{i,j}.$$ -Question -I conjecture that $\lambda_{\text{min}}(A_N)\le \lambda_{\text{min}}(B_N)$ for any $N$ and would like to know if one can actually show this? -As a first step, I would like to know if one can show that $$\lambda_{\text{min}}(A_N)\le C\lambda_{\text{min}}(B_N)$$ for some $C$ independent of $N$? -In fact, I am not claiming that $A_N \le B_N$ in the sense of matrices. But it seems as if the eigenvalues of $B_N$ are shifted up when compared with $A_N.$ -Numerical evidence: -For $N=2$ we can explicitly write down the matrices -$$A_2 =\left( -\begin{array}{ccc} - \frac{1}{y} & 0 & \frac{1}{2 y} \\ - 0 & \frac{1}{y} & \frac{1}{2 y} \\ - \frac{1}{2 y} & \frac{1}{2 y} & \frac{1}{2 y^2} \\ -\end{array} -\right) \text{ and }B_2 = \left( -\begin{array}{ccc} - \frac{1}{y} & 0 & \frac{1}{y} \\ - 0 & \frac{1}{y} & \frac{1}{y} \\ - \frac{1}{y} & \frac{1}{y} & \frac{2}{y^2} \\ -\end{array} -\right)$$ -We obtain for the lowest eigenvalue of $A_2$ (orange) and $B_2$(blue) as a function of $y$ - -For $N=3$ we get qualitatively the same picture, i.e. the lowest eigenvalue of $A_3$ remains below the lowest one of $B_3$: -In this case: -$$A_3=\left( -\begin{array}{cccccc} - \frac{1}{y} & 0 & 0 & \frac{1}{2 y} & 0 & \frac{1}{2 y} \\ - 0 & \frac{1}{y} & 0 & \frac{1}{2 y} & \frac{1}{2 y} & 0 \\ - 0 & 0 & \frac{1}{y} & 0 & \frac{1}{2 y} & \frac{1}{2 y} \\ - \frac{1}{2 y} & \frac{1}{2 y} & 0 & \frac{1}{2 y^2} & \frac{1}{4 y} & \frac{1}{4 y} \\ - 0 & \frac{1}{2 y} & \frac{1}{2 y} & \frac{1}{4 y} & \frac{1}{2 y^2} & \frac{1}{4 y} \\ - \frac{1}{2 y} & 0 & \frac{1}{2 y} & \frac{1}{4 y} & \frac{1}{4 y} & \frac{1}{2 y^2} \\ -\end{array} -\right)\text{ and } B_3=\left( -\begin{array}{cccccc} - \frac{1}{y} & 0 & 0 & \frac{1}{y} & 0 & \frac{1}{y} \\ - 0 & \frac{1}{y} & 0 & \frac{1}{y} & \frac{1}{y} & 0 \\ - 0 & 0 & \frac{1}{y} & 0 & \frac{1}{y} & \frac{1}{y} \\ - \frac{1}{y} & \frac{1}{y} & 0 & \frac{2}{y^2} & \frac{1}{y} & \frac{1}{y} \\ - 0 & \frac{1}{y} & \frac{1}{y} & \frac{1}{y} & \frac{2}{y^2} & \frac{1}{y} \\ - \frac{1}{y} & 0 & \frac{1}{y} & \frac{1}{y} & \frac{1}{y} & \frac{2}{y^2} \\ -\end{array} -\right)$$ - -REPLY [3 votes]: The matrices are of the form -$$ -A=\begin{pmatrix} 1 & C \\ C^* & D \end{pmatrix}, \quad\quad -B= \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} A \begin{pmatrix} 1 & 0\\ 0&2\end{pmatrix} , -$$ -with the blocks corresponding to the sizes of the sets $M_j$ involved. -Let $v=(x,y)^t$ be a normalized eigenvector for the minimum eigenvalue $\lambda $ of $B$. Then -$$ -(x,2y)A\begin{pmatrix} x \\ 2y \end{pmatrix} = \lambda -$$ -also, but this modified vector has larger norm. -So the desired inequality $\lambda_j(A)\le\lambda_j(B)$ (for all eigenvalues, not just the first one) will follow if we can show that $B\ge 0$. This is true for $y=1$ because in this case we can interpret -$$ -|M_j\cap M_k|=\sum_n \chi_j(n)\chi_k(n) -$$ -as the scalar product in $\ell^2$ of the characteristic functions, and this makes -$v^*Bv$ equal to $\|f\|_2^2\ge 0$, with $f=\sum v_j\chi_j$. -For general $y>0$, we have $B(y)=(1/y)B(1) + D$, for a diagonal matrix $D$ with non-negative entries.<|endoftext|> -TITLE: Asymptotic behavior of sum linked with Lagrange interpolation -QUESTION [15 upvotes]: I already asked this a few weeks ago with no answer, so let me formulate differently. -In performing Lagrange interpolation with nodes 1/n, one encounters the sum -$$S(f)=\dfrac{1}{N!}\sum_{n=0}^N(-1)^{N-n}\binom{N}{n}n^Nf(n)\;.$$ -1) If $f(n)=n^{-a}$ for any real (or complex ?) $a$ it seems that -$S(f)$ is asymptotic as $N\to\infty$ to -$$\dfrac{2^a}{\Gamma(1-a)}N^{-2a}$$ -(including when $a$ is a positive or $0$ integer). This is probably easy but I haven't found the result, certainly in the literature. -2) My real question is what is the asymptotic when $f(n)=\psi'(n)$, where $\psi$ is the logarithmic derivative of the gamma function. I believe that up to smaller order terms it is $C^{-N}$, but what is $C$ ? I also believe that we have the same asymptotic with the same $C$ for higher derivatives of $\psi$. -ADDED for Q 2): I had a heuristic which gave $C=2\pi$, but the correct value seems to be $C=8.077...$. -UPDATE!!!: Don Zagier has completely answered Q 1), and almost completely Q 2: the constant $C$ is in fact a complex number $x_0$ solution of $\exp(x_0-1)=x_0$, whose modulus is indeed approximately $8.076$ - -REPLY [5 votes]: For a finite non-empty set $\Omega=\{\omega_1,\dots,\omega_{N+1}\}\subset \mathbb{R}$, denote by $\Phi_{\Omega}$ the linear functional which maps a function $g:\Omega\mapsto \mathbb{R}$ to $$\Phi_{\Omega}(g)=\sum_{\omega\in \Omega} \frac{g(\omega)}{\prod_{\tau\in \Omega\setminus \omega} (\omega-\tau)}.$$ -This expression may be viewed as a coefficient of $x^{N}$ in the polynomial $P_{\Omega}(g)$ which interpolates $g$ on the set $\Omega$. Note that for any polynomial $p(x)\in \mathbb{R}[x]$ we have $\Phi_{\Omega}(p)=[x^N]P_{\Omega}(p)$ (as usual, $[x^N]$ stands for a coefficient of $x^N$) and $P_{\Omega}(p)$ is a remainder of $p(x)$ modulo the polynomial $\prod_{\omega\in \Omega}(x-\omega)$. Therefore $\Phi_{\Omega}(h)=0$ whenever $\deg h<|\Omega|-1$, and $\Phi_{\Omega}(x^{N+k})=h_k(\Omega)$, -where $h_k(\Omega)$ stands for $$h_k(\Omega)=h_k(\omega_1,\dots,\omega_{N+1})=\sum_{i_1\leqslant i_2\dots\leqslant i_k} \omega_{i_1} \omega_{i_2}\dots \omega_{i_k},$$ -the complete symmetric polynomial. This is a basic fact from the theory of symmetric functions, the short way to see it is to observe that $h_k(\Omega)$ is a coefficient of $x^{k}$ in the product $\prod_{\omega\in \Omega}(1+\omega x+\omega^2x^2+\dots)=\prod_{\omega\in \Omega} (1-\omega x)^{-1}$ and expanding this rational function as a linear combination of elementary fractions $1/(1-\omega x)$. -You are interested in the case when $\Omega=\{0,1,\dots,N\}$ and $g(x)=x^Nf(x)$. It is less or more natural to expect that $\Phi_{\Omega}(g)$ behaves like the coefficient of $[x^N]$ in the Taylor polynomial -$p_N(x)=\frac1{N!}g^{(N)}(N/2)(x-N/2)^N+\dots$ of the function $g$ expanded in the point $N/2$. Let us justify this for $g(x)=x^{N-a}$, this gives your asymptotics $\frac1{N!}g^{(N)}(N/2)=(N!)^{-1}(N/2)^{-a}(1-a)(2-a)\dots(N-a)\sim 2^aN^{-2a}/\Gamma(1-a)$ for large $N$. -We need the following -Lemma. Assume that $\Omega$ is a symmetric w.r.t. 0 set. Then $h_k(\Omega)=0$ for odd $k$ and $h_{2m}(\Omega)=h_m(\Omega^2\setminus 0)$, where $\Omega^2=\{\omega^2:\omega\in \Omega\}$. In particular, this gives the estimate $|h_{2m}(\Omega)|\leqslant \max(\Omega)^{2m}\cdot \binom{[N/2]+m-1}m$. -Proof. The formula is seen from the generating function $\prod 1/(1-\omega x)$. The estimate is obtained by counting the number of terms and estimating each of them as $\max(\Omega)^{2m}$. -Now write $g(x):=x^Nf(x)=p_N(x)+r_N(x)$, we have $\Phi_{\Omega}(g)=\Phi_{\Omega}(p_N)+\phi_{\Omega}(r_N)$, where -$p_N(x)=\frac1{N!}g^{(N)}(N/2)(x-N/2)^N+\dots$ is the Taylor polynomial of the function $g(x)$ at the point $N/2$ and $r_N(x)$ is the remainder. As said beforуб the first summand $P:=\Phi_{\Omega}(p_N)$ behaves as expected, and it remains to estimate the second summand. It is more convenient to do it for any $\Omega\subset (0,N)$ symmetric w.r.t. $N/2$, the estimate would be uniform in $\Omega$, so it continues to $\Omega\subset [0,N]$. -You may expand $r_N(x)$ as a sum $\sum_{k\geqslant 1} \frac1{(N+k)!}g^{(N+k)}(N/2)(x-N/2)^{N+k}$. Let $\Omega=\{N/2+\rho_1,N/2+\rho_2,\dots,N/2+\rho_N\}$, we have $\rho_i\in (-N/2,N/2)$ and the set $\Omega_s=\Omega-N/2$ is symmetric w.r.t. 0. Now the series in $k$ and the interpolating operator $\Phi_{\Omega}$ may be permuted (simply because the sum in this operator is finite for any fixed $N$). -We get $$\Phi_{\Omega}(r_N)=\sum_{k\geqslant 1} \frac1{(N+k)!}g^{(N+k)}(N/2) h_k(\Omega_s).$$ Comparing $\frac1{(N+k)!}g^{(N+k)}(N/2)$ and $P=\frac1{N!}g^{(N)}(N/2)$ we see (remember that $g^{(N)}(x)=Cx^{-a}$ for certain $C$) that -$$ -\left|\frac{P^{-1}}{(N+k)!}g^{(N+k)}(N/2)\right|\leqslant \frac{(2/N)^k k! k^{A}}{(N+1)(N+2)\dots (N+k)} -$$ -where a constant $A$ depends on $a$ (so that $|a|(|a|+1)\dots (|a|+k)\leqslant k! k^A$ for all $k\geqslant 2$). -Multiplying this by the estimate provided by Lemma we totally get (denote $k=2m$) at most -$$ -\sum_{m=1}^\infty \frac{(2/N)^{2m} (2m)! (2m)^{A}}{(N+1)(N+2)\dots (N+2m)}\cdot \frac{(N/2+1)\dots (N/2+m)}{m!}\cdot (N/2)^{2m} -$$ -It is easy that if we denote $m$-th term by $x_m$, than $x_m=O_a(m^{-2}N^{-1})$, where a constant in $O(\cdot)$ depends only on $A$ (which in turn depends only on $a$): look at the ratio $x_{m+1}(m+1)^2/(x_m m^2)$, it eventually decreases. This gives $O(1/N)$ after summation by $m$. -(to be continued for other functions $f$). -For $\psi'$ you may use the formula $\psi'(x)x^N=\int_0^\infty \frac{t}{1-e^{-t}}e^{-xt}x^Ndt$ and interchange the integral and interpolating functional, since interpolating the function $x^Ne^{-xt}$ in the points $0,1,\dots,N$ is sort of pleasant: -$$ -\frac1{N!}\sum_{n=0}^N(-1)^{N-n}\binom{N}ne^{-nt}n^{N}=\frac1{N!}(d/dt)^N (1-e^{-t})^N -$$ -We integrate by parts to get the answer in the form -$$ -\int_0^\infty -dt\cdot \frac{d}{dt}\frac{t}{1-e^{-t}}\cdot \frac1{N!}(d/dt)^{N-1} (1-e^{-t})^N. -$$ -Next, we write -$$ -\frac1{N!}(d/dt)^{N-1} (1-e^{-t})^N=\frac1N[z^{N-1}](1-e^{-t-z})^N -$$ -for using Lagrange inversion formula in the following form: -$$ -\frac1N[z^{N-1}]\varphi(z)^N=[w^N] g(w) -$$ -where $f(z)=z/\varphi(z)$ [ah, of course, it is not at all your $f$ from the post] and $g$ is functional inverse of $f$. In our situation $f(z)=f_t(z)=z/(1-e^{-t-z})$ and so $g(w)=g_t(w)$ is the solution of the equation $g/(1-e^{-t-g})=w$. The growth of coefficients of the analytic function $g$ is determined by its first singularity. The first singularity of the inverse function $g$ is the first critical value of the initial function $f$ (that is, in the point closest to the origin where $f'=0$). Consider $t=0$, I guess for $t>0$ the singularities are further from 0 (need to check of course). Then $f(z)=z/(1-e^{-z})$, the equation $f'(z)=0$ rewrites as $e^{z_0}=z_0+1$, then $f(z_0)=z_0+1$ is a solution of $x_0=e^{x_0-1}$ as you conjecture. -(to be continued)<|endoftext|> -TITLE: Finally dense implies dense -QUESTION [5 upvotes]: I am reading the article "A convenient category for directed homotopy" by Fajstrup and Rosicky and I have a doubt about the proof of Proposition 3.5. The setting is the following: -let $\cal{C}$ be a concrete category, with forgetful functor $U\colon \cal{C}\to \mathsf{Set}$. -Definition 1. A full subcategory $\cal{I}$ of $\cal{C}$ is said to be finally dense in $\cal{C}$ if, for every object $X$ in $\cal{C}$ there exists a family of maps $(f_\lambda\colon C_{\lambda}\to X)_{\lambda\in \Lambda}$ such that a map $g\colon UX\to UY$ lifts to a morphism $\bar{g}\colon X\to Y$ iff $g\circ Uf_\lambda$ lifts to a morphism $\bar{g_\lambda}$ for every $\lambda$. -Definition 2. A full subcategory $\cal{I}$ of $\cal{C}$ is said to be dense in $\cal{C}$ if every object $X$ in $\cal{C}$ is isomorphic to the colimit of the canonical functor $\mathcal{I}/X\to \cal{C}$. -In Proposition 3.5 of the aforementioned paper, $\cal{C}$ is actually the category $\cal{K}_{\cal{I}}$ of $\cal{I}$-generated objects, for a topological construct $\cal{K}$ and a full subcategory $\cal{I}$ of $\cal{K}$. In particular, this means that we have a forgetful functor $U\colon\cal{K}\to \mathsf{Set}$ which has a right and a left adjoint, that associate to every set $S$ the discrete and indiscrete object on $S$, respectively. -The statement I'm a bit puzzled about is that, "since $\cal{I}$ is finally dense in $\cal{C}$, then it is dense." -I can see why this should be true if we could show that, for any $X\in \cal{K}$ the natural map -$$\mathrm{colim}_{C\to X}UC\to UX$$ -is a bijection. -If either $\cal{I}$ contains the discrete object on the terminal object, or the discrete object and the indiscrete object on the terminal object conincide, then I know how to conclude, since the aformentioned cocone would contain all the points of $UX$ (In the second case it is enough to consider all constant maps $C\to *_I=*_D\to X$. However, in a more general setting I don't know why such a statement should hold. -Here comes the question: -Am I missing something? Is there a reason why the above statement should hold in a more general setting? -If the answer is no, do you have any example of a topological category where UX and the canonical colimit displayed above do not coincide? - -REPLY [9 votes]: You are right, in general $\mathcal I$ should contain the discrete object $D_1$ on the singleton. A general result is in my paper Codensity and binding categories (Theorem 1.3). In Proposition 3.5 of the aforementioned paper, $D_1$ is $\mathcal I$-generated and thus it can be added to $\mathcal I$ without changing $\mathcal I$-generated objects. Then Proposition 3.5 follows.<|endoftext|> -TITLE: How to explain to an engineer what algebraic geometry is? -QUESTION [42 upvotes]: This question is similar to this one in that I'm asking about how to introduce a mathematical research topic or activity to a non-mathematician: in this case algebraic geometry, intended as the most classical complex algebraic geometry for simplicity. -Of course some of the difficulties of the present question are a subset of those of the linked question. But I think I want to be more precise here, about what's the pedagogical/heuristic obstacle I want to bypass/remove/etc, which is, after all, a detail. -So, say the engineer is happy with the start - -"Algebraic geometry is the study of solutions of systems of polynomial equations in several variables..." - -An engineer certainly understands this. - -"...with complex coefficients..." - -Here she's starting to feel a bit perplexed: why complex numbers and not just reals? But she can feel comfortable again once you tell her it's because you want to have available all the geometry there is, without hiding anything - she can think of roots of one-variable polynomials: in $(x-1)(x^2+1)$ the real solutions are not all there is etcetera. -Happy? Not happy. Because the engineer will inevitably be lead to think that what algebraic geometry consists of is fiddling around with huge systems of polynomial equations trying to actually find its solutions by hand (or by a computer), using maybe tricks that are essentially a sophisticated version of high school concepts like Ruffini's theorem, polynomial division, and various other tricks to explicitly solve systems that are explicitly solvable as you were taught in high school. - -Question. How to properly convey that algebraic geometry mostly (yeah, I know, there are also computational aspects but I would contend that the bulk of the area is not about them) doesn't care at all of actually finding the solutions, and that algebraic geometers rarely find themselves busy with manipulating huge polynomial systems, let alone solving them? In other words, how to explain that AG is the study of intrinsic properties of objects described by polynomial systems, without seeming too abstract and far away? -Also, how would you convey that AG's objects are only locally described (or rather, in the light of the previous point, describable) by those polynomial systems in several variables? - -REPLY [4 votes]: I would just say that very roughly speaking it's a subject where you are doing geometry and thinking about geometry, but you write about it formally like it is algebra and you use algebra (which can be fairly esoteric). I have an example from page 119 of 'Introduction to Homological Algebra' by Weibel (the application is originally due to Hartshorne). -Let $R = \mathbb{C}[x_1, x_2, y_1, y_2]$, $P=(x_1,x_2)R$, $Q=(y_1,y_2)R$, and $I = P \: \cap \: Q $. -Since $P$, $Q$, and $m = P+Q = (x_1, x_2, y_1, y_2)R$ are generated by regular sequences, it can be shown that the outside terms in the Mayer-Vietoris sequence -$H^3_P(R) \bigoplus H^3_Q (R) \rightarrow H_I^3(R) \rightarrow H^4_m(R) \rightarrow H^4_P(R) \bigoplus H^4_Q (R) $ -vanish, which tells us that $H^3_I(R) \simeq H^4_m(R) \neq 0$. (The cohomology is local cohomology here). -Esoteric and abstract to the typical engineer? Perhaps, but leaving aside the words and jargon, this implies that the union of two planes in $\mathbb{C}^4$ which meet at a point cannot be described as the solutions of only two equations $f_1 = f_2 =0$, which is a concrete geometric fact.<|endoftext|> -TITLE: The roots of unity in a tensor product of commutative rings -QUESTION [36 upvotes]: For $i\in\{1,2\}$ let $A_i$ be a commutative ring with unity whose additive group is free and finitely-generated. Assume that $A_i$ is connected in the sense that $0$ and $1$ are unique solutions of the equation $x^2=x$ in $A_i$. Denote by $\mu(A_i)$ the group of roots of unity of $A_i$, i.e., the elements of finite multiplicative order in $A_i$. - -Problem. Is the map $$\mu(A_1)\times\mu(A_2)\ \longrightarrow\ \mu(A_1\otimes_{\mathbb Z}A_2),\;\;(u,v)\ \longmapsto\ u\otimes v,$$ - surjective? - -(The problem is posed 9.07.2018 by Tristan Tilly from Leiden University on page 25 of Volume 2 of the Lviv Scottish Book). -Prize for solution: An eternal welcome in Netherlands, and a bottle of liquor of your choice. - -REPLY [2 votes]: $\newcommand\F{\mathbf{F}}$ -$\newcommand\Z{\mathbf{Z}}$ -$\newcommand\Q{\mathbf{Q}}$ -$\newcommand\Gal{\mathrm{Gal}}$ -$\newcommand\GL{\mathrm{GL}}$ -$\newcommand\SL{\mathrm{SL}}$ -Some more examples of groups with corresponding maps $w$: -Let $G = \GL_2(\F_p)$ and $V = \SL_2(\F_p)$, and consider $(\Z/p)^{\times} \subset G$ via the map -$$c \rightarrow \left( \begin{matrix} c & 0 \\ 0 & 1 \end{matrix} \right).$$ -Then one can take $w: G \rightarrow \F_p$ to be (say) the $[1,1]$ entry of $G$, because then -$$w \left( \left( \begin{matrix} w & x \\ y & z \end{matrix} \right) -\left( \begin{matrix} c & 0 \\ 0 & 1 \end{matrix} \right) -\left( \begin{matrix} r & s \\ t & u \end{matrix} \right)\right) -= c p w + r x$$ -is affine linear in $c$. Hence this gives an admissible $w$. Moreover, for $G$ to occur as a Galois group, we would want a representation: -$$\rho: \Gal(\overline{\Q}/\Q) \rightarrow \GL_2(\F_p)$$ -which is unramified outside $p$ with cyclotomic determinant and so that the restriction to inertia was -$$\rho |_{I_p} = \left( \begin{matrix} \varepsilon & 0 \\ 0 & 1 \end{matrix} \right)$$ -where $\varepsilon$ is the mod-$p$ cyclotomic character (which gives the canonical identification of $\Gal(\Q_p(\zeta_p)/\Q_p)$ with $(\Z/p)^{\times}$). -Speyer asked in another question whether one could prove that $V_{p-2} = 0$ using anything simpler than Herbrand --- perhaps with the idea that any simple direct negative answer to this question would give a new proof. This example is even worse --- to rule out the existence of such a representation $\rho$, I can't see any simpler argument than using the proof of Serre's conjecture by Khare-Wintenberger (the case $N(\rho) = 1$ and $k(\rho) = 2$). Even worse, if one replaces $\GL_2$ with $\GL_3$ and maps $(\Z/p)^{\times}$ to $G$ via $\mathrm{diag}(c,1,1)$, it becomes an open problem to show that a corresponding extension with Galois group $\GL_3(\F_p)$ and representation $\rho$ does not exist --- although various standard conjectures certainly imply that it does not. This strongly suggests that proving the answer to the original question is "no" will be very hard. So either one should - -Try to prove the answer is "conditionally no" by using conjectures in number theory, after more fully understanding the group theoretic condition. -Try to prove the answer is "yes". - - -earlier: This answer is basically a response to Speyer (although he omitted this part of the argument, so hopefully it is correct). edit: This doesn't seem to work. -Suppose that $G = \Gal(K/\Q)$, where $K$ is unramified over $\Q(\zeta_p)$. Let $V = \Gal(\Q(\zeta_p)/\Q)$. As noted by Speyer, byconsidering the inertia group, we have that that $G$ is a semi-direct product, so $G = (\Z/p \Z)^{\times} \ltimes V$, and elements of $G$ have the form $(c,g):=(c,0)(0,g)$. Since the genus field of $\Q(\zeta_p)$ is trivial, we certainly have $V = [G,G]$. Let us now assume that -$$H = [G,[G,G]] = [G,V] \ne V.$$ -Then we can define a function $w: G \rightarrow \Z/p$ on $G$ as follows: -$$\psi(c,g) = \begin{cases} c, & g \in H, \\ 0, & g \notin H. \end{cases}$$ -For fixed $x$ and $y$ in $V$, -$$\psi((0,x)(c,0)(0,y^{-1})) = \psi(c,c^{-1} xc y^{-1}) = \psi(c,[c^{-1},x] x y^{-1}).$$ -Now $[c^{-1},x] \in [G,V] = H$, and hence -$$\psi((0,x)(c,0)(0,y^{-1})) = \begin{cases} c, & xH = yH, \\ 0, & \text{otherwise}.\end{cases}$$ -If I understand correctly, this satisfies the properties of $w$ desired by Speyer. -edit:I guess this doesn't work --- or rather not in an interesting way --- the conditions are never satisfied! They force $G/H$ to be abelian and then $G/H = G/V$.<|endoftext|> -TITLE: Extending colouring of graphs using small number of colours -QUESTION [5 upvotes]: Conjecture (Csóka-Lippner-Pikhurko). If $G$ is a graph with each vertex of degree $\le d$ with at most $d-1$ pendant edges properly coloured, then this pre-colouring can be extended to all edges of $G$, using $d+1$ colours in total. -If proved, this will directly give new bounds on questions of Albert (2010) & Marks (2016) on measurable Vizing's theorem. - -(This problem was written 23.08.2018 by Oleg Pikhurko on page 51 of Volume 2 of the Lviv Scottish Book). - -REPLY [3 votes]: The main motivation for us stating this conjecture was that, if the conjecture is true, then Vizing's theorem holds for every graphing. Since the latter result was recently proved by Jan Grebik and me (in https://arxiv.org/abs/1905.01716) via a different route, the conjecture is not so interesting now.<|endoftext|> -TITLE: Relating bordism groups of different dimensions -QUESTION [5 upvotes]: Let - -$M_d$ - -be a $d$-manifold generator of a subgroup of bordism group -$$ -\Omega_d^{G}, -$$ -or further generalization - -$$ -\Omega_d^{G}(K(\mathcal{G},n+1)), -$$ - -which $G$ is the given structure including the tangent bundle structure (such as the SO or Spin) and the internal gauge bundle structure (such as an additional compact Lie group). The $K(\mathcal{G},n)$ is the Eilenberg–MacLane space of $\mathcal{G}$; if $n$ > 1, then $\mathcal{G}$ must be abelian. -Generally, $M_d$ cannot be written as $M'_{d-1} \times S^1$, nor $\tilde M_{d-n} \times T^n$, where $T^n$ is the $n$-torus. - -Here are my questions: - - -(1) However, are there certain cases that - $$M_d \overset{?}{=}M_{d-n} \times T^n,$$ - such that - $M_{d-n}$ is also a $(d-n)$-manifold generator of a subgroup of bordism group - $$ -\Omega_{d-n}^{G}(K(\mathcal{G},1))? -$$ - Now $K(\mathcal{G},1)=B\mathcal{G}$ is the classifying space of $\mathcal{G}$. -(2) Are there actually mathematical proofs or theorems stating the similar structures given above relation the bordism group generators of $\Omega_d^{G}(K(\mathcal{G},n+1))$ to the bordism group generators of $\Omega_{d-n}^{G}(K(\mathcal{G},1))$, for whatever integer $n$? If so, please state the results and please provide the Refs? - -Many thanks. - -REPLY [6 votes]: Just a few remarks: - -Bordism of Eilenberg--Mac Lane spaces can be identified with bordism of pairs consisting of a closed manifold with a cohomology class. More precisely, $\Omega^G_d(K(\mathcal{G},n+1))$ can be described as bordism classes of pairs $(M^d,x)$ where $M^d$ is a closed, $d$-dimensional $G$-manifold and $x\in H^{n+1}(M^d;\mathcal{G})$ is a cohomology class. -For any (reduced) generalized homology theory $E_*$ there is a loop-suspension homomorphism -$$ -E_k(\Omega X)\to E_{k+1}(\Sigma\Omega X)\to E_{k+1}(X). -$$ -In particular, since $K(\mathcal{G},i)=\Omega K(\mathcal{G},i+1)$ for $i\ge1$ there are homomorphisms -$$ -\Omega^G_{d-n}(K(\mathcal{G},1))\to \Omega^G_{d-n+1}(K(\mathcal{G},2))\to \cdots \to \Omega^G_{d}(K(\mathcal{G},n+1)). -$$ -One might expect that the homomorphisms $$\Omega^G_{d-n+i-1}(K(\mathcal{G},i))\to \Omega^G_{d-n+i}(K(\mathcal{G},i+1))$$ appearing in 2. are described geometrically by taking the bordism class of a pair $(M,x)$ to that of the pair $(M\times S^1,x\times \sigma)$, where $\sigma\in H^1(S^1;\mathbb{Z})$ denotes a generator and $\times$ denotes cross product. Something like this must be true, but care needs to be taken with reduced vs. unreduced homology theories.<|endoftext|> -TITLE: A Shelah group in ZFC? -QUESTION [12 upvotes]: In his famous paper "On a problem of Kurosh, Jonsson groups, and applications" of 1980, Shelah constructed a CH-example of an uncountable group $G$ equal to $A^{6641}$ for any uncountable subset $A\subset G$. -Let us call a group $G$ -$\bullet$ $n$-Shelah if $G=A^n$ for each subset $A\subset G$ of cardinality $|A|=|G|$; -$\bullet$ Shelah if $G$ is $n$-Shelah for some $n\in\mathbb N$; -$\bullet$ almost Shelah if for each subset $A\subset G$ of cardinality $|A|=|G|$ there exists $n\in\mathbb N$ such that $A^n=G$; -$\bullet$ Jonsson if each subsemigroup $A\subset G$ of cardinality $|A|=|G|$ coincides with $G$. -$\bullet$ Kurosh if each subgroup $A\subset G$ of cardinality $|A|=|G|$ coincides with $G$. -It is clear that for any group $G$ the following implications hold: - -finite $\Leftrightarrow$ 1-Shelah $\Rightarrow$ $n$-Shelah $\Rightarrow$ Shelah $\Rightarrow$ almost Shelah $\Rightarrow$ Jonsson $\Rightarrow$ Kurosh. - -In the mentioned paper, Shelah constructed a ZFC-example of an uncountable Jonsson group and also a CH-example of an uncountable 6641-Shelah group. - -Problem 1. Can an infinite (almost) Shelah group be constructed in ZFC? -Problem 2. Find the largest possible $n$ (which will be smaller than 6640) such that each $n$-Shelah group is finite. - -This result of Protasov implies - -Theorem (Protasov). Each countable Shelah group is finite. - -It is easy to show that each 2-Shelah group is finite. - -Problem 3. Is each 3-Shelah group finite? - -REPLY [11 votes]: I'm not sure what makes an answer to a question with several problems of very variable difficulty a good answer :) -Anyway: - -there's no "3-Shelah" group. That is, every infinite group admits a subset $W$ such that $W^3\neq G$ and $|W|=G$. (Actually one can arrange $W\cup W^2\cup W^3\neq G$.) - -Let $G$ be an infinite group. Let $A$, by Zorn, be a maximal subset such that $1\notin A\cup A^2\cup A^3$. Denote by $\langle A\rangle$ the subgroup generated by $A$, and $G^{(6)}$ the subgroup of $G$ generated by $\{g^6:g\in G\}$; clearly $G^{(6)}$ is normal in $G$. -For every $g\in G\smallsetminus A$, the maximality implies that $$1\in (A\cup\{g\})\cup(A\cup\{g\})^2\cup (A\cup\{g\})^3.$$ Since $1\notin A\cup A^2\cup A^3$, this means that -$$1\in \{g\}\cup Ag\cup gA\cup\{g^2\}\cup A^2g\cup AgA\cup gA^2\cup g^2A\cup gAg\cup Ag^2\cup\{g^3\}.$$ -Hence one of the following holds: $g=1$ or $g^2=1$ $g^3=1$ or $g\in A^{-1}$, or $g^2\in A^{-1}$ or $g\in (A^2)^{-1}$. -Hence, $g^6\in \langle A\rangle$ for all $g\in G$; equivalently, $G^{(6)}\subset \langle A\rangle$. -In $G/G^{(6)}$, every element satisfies $x^6=1$. Since groups of exponent 6 are solvable, it follows that either $G=G^{(6)}$, or $G$ has a normal subgroup of index 2 or 3. In the last two cases, we define this subgroup as $W$. Otherwise, $\langle A\rangle=G$, that is, $A$ generates $G$. In particular, since $G$ is infinite, $|A|=|G|$, so we put $A=W$.<|endoftext|> -TITLE: Is it expected that the mod $p$ representation determines a normalized Hecke newform of fixed weight for p large enough? -QUESTION [6 upvotes]: Mazur's conjecture on the image of Galois representations of Elliptic curves states that for $N$ large enough there is a unique elliptic curve $E$ over $\mathbb{Q}$ giving rise to a fixed mod $N$ Galois representation $\bar{\rho}: G_{\mathbb{Q}}\rightarrow GL_2(\mathbb{Z}/N\mathbb{Z})$. -Is there a similar expectation for normalized Hecke newforms of a fixed weight? In greater detail, let $p$ be a prime, $\bar{\rho}: G_{\mathbb{Q}}\rightarrow GL_2(\mathbb{Z}/p\mathbb{Z})$ be a fixed irreducible Galois representation and $k\geq 2$ a fixed integer, is the set of normalized Hecke newforms $f$ with weight $k$ and rational coefficients whose associated residual Galois representation coincides with $\bar{\rho}$ expected to be a finite set when $p$ is large enough? If not, is there a heuristic why it isn't the case? - -REPLY [4 votes]: This is not true. Take $k=2$, and $p \geq 5$. Let $\ell$ be a prime such that $p$ divides $\ell-1$. Then we know (by Mazur) that there exists a newform of weight $2$ and level $\Gamma_0(\ell)$ whose residual semi-simple representation is $\overline{\rho} = 1 \oplus \overline{\chi}_p$ where $\overline{\chi}_p$ is the modulo $p$ cyclotomic character. The set of such $\ell$ is infinite. Note however that the newform we get almost never have rational coefficients (I think except if $p=5$ and $\ell=11$ in which case $X_0(11)$ is an elliptic curve).<|endoftext|> -TITLE: Künneth formulas/theorem for bordism groups and cobordisms? -QUESTION [9 upvotes]: We are familiar with Künneth theorem: -The Kunneth formula is given by $R$ as a ring, $M,M'$ as the R-modules, $X,X'$ are some chain complex. The Kunneth formula shows the cohomology of a chain complex $X \times X'$ in terms of the cohomology of a chain complex $X$ and another chain complex $X'$. For topological cohomology $H^d$, we have -$$ - H^d(X\times X', M\otimes_R M') -\simeq \Big[\oplus_{k=0}^d H^k(X,M)\otimes_R H^{d-k}(X',M')\Big]\oplus -$$ -$$\Big[\oplus_{k=0}^{d+1} -\text{Tor}_1^R(H^k(X,M),H^{d-k+1}(X',M'))\Big] . -$$ -$$ -H^d(X\times X',M) -\simeq \Big[\oplus_{k=0}^d H^k(X,M)\otimes_{\mathbb Z} H^{d-k}(X',\mathbb{Z})\Big]\oplus -$$ -$$ -\Big[\oplus_{k=0}^{d+1} -\text{Tor}_1^{\mathbb Z}(H^k(X,{M}),H^{d-k+1}(X',\mathbb Z))\Big]. -$$ -The above is valid for both topological cohomology $H^d$ and group cohomology $\mathcal{H}^d$ (for $G'$ is a finite group): -$$ -\mathcal{H}^d(G\times G',M) -\simeq \Big[\oplus_{k=0}^d \mathcal{H}^k(G,M)\otimes_{\mathbb Z} \mathcal{H}^{d-k}(G',\mathbb Z)\Big]\oplus -$$ -$$ -\Big[\oplus_{k=0}^{d+1} -\text{Tor}_1^{\mathbb Z}(\mathcal H^k(G,M),\mathcal H^{d-k+1}(G',\mathbb Z))\Big]. -$$ - - -Questions: - -[1]. Do we have similar results of Künneth theorem for bordism groups $\Omega_d^{...}(...)$? Schematically, maybe something like - $$ -\Omega_d^{...}(...) \simeq \oplus_n \Omega_n^{...(1)}(...) \otimes \Omega_{d-n}^{...(2)}(...)? -$$ -[2]. Can we, and, how can we interpret the decompositions of bordism group generators as manifolds $\Sigma_d$: - $$ -\Sigma_d \sim \Sigma_{d-n}^{(1)} \times \Sigma_{n}^{(2)}? -$$ - where $d$-manifold generators are related to the $d-n$-manifold generator and - $n$-manifold generator. Are these general, or are they only special cases? - -Refs are welcome. Thanks. - -REPLY [8 votes]: If you work with the unoriented bordism groups $\Omega^O_*(X)=MO_*(X)$ then there is a Künneth isomorphism. This is just because there is a natural isomorphism $MO_*(X)=H_*(X;\mathbb{Z}/2)\otimes MO_*$, where $MO_*$ is a graded polynomial ring over $\mathbb{Z}/2$ with one generator $x_i$ in each degree $i>0$ not of the form $2^j-1$. I think that all of this was already known to Thom. The key point is that $H_*(MO;\mathbb{Z}/2)$ is a cofree comodule over the dual Steenrod algebra, by a fairly concrete and straightforward calculation. -There are also some interesting things to say if you are willing to work with the complex bordism groups $MU_*(X)=\Omega^U_*(X)$ instead of the real ones. Here it is known that $MU_*=\mathbb{Z}[a_1,a_2,a_3,\dotsc]$,so in particular this is a free abelian group. Consider the following conditions: - -$H_*(X)$ is free abelian -$MU_*(X)$ is a free $MU_*$-module -$MU_*(X)$ is a flat $MU_*$-module -$MU_*(X)$ is Landweber exact -$MU_*(X)$ is torsion-free -The Künneth map $MU_*(X)\otimes_{MU_*}MU_*(Y)\to MU_*(X\wedge Y)$ is an isomorphism for all $Y$. - -First, 1 implies 2. Indeed, there is an Atiyah-Hirzebruch spectral sequence $H_*(X)\otimes MU_*\Rightarrow MU_*(X)$, and the differentials are controlled by homotopy groups of spheres in nonzero dimensions so they are torsion-valued, so they must be zero. The spectral sequence therefore collapses and the claim follows easily. Once you have recalled the definition of Landweber exactness it is easy to see that 2 implies 3 implies 4 implies 5. The Landweber exact functor theorem says that 4 implies 6. By taking $Y$ to be a generalised Moore spectrum $S/(v_0^{i_0},\dotsc,v_n^{i_n})$ one can also check that 6 implies 4. -Because $MU_*(X)$ is actually an $MU_*MU$-comodule rather than just an $MU_*$-module, I think it works out that 5 implies 4. However, I am surprised to find that this very natural question has never occurred to me before. -Now suppose we want to consider the oriented real bordism groups $\Omega^{SO}_*(X)=MSO_*(X)$. The coefficient ring $MSO_*$ is somewhat complicated: it is the direct sum of a polynomial algebra over $\mathbb{Z}$ with a module over $\mathbb{Z}/2$, and does not have any very simple presentation. However, there is a map $MU\to MSO$ of ring spectra, and this gives a map $MU_*\to MSO_*$ of coefficient rings, and thus a natural map -$$ MSO_* \otimes_{MU_*} MU_*(X) \to MSO_*(X). $$ -If $H_*(X)$ is free abelian then we have seen that $MU_*(X)$ is a free $MU_*$-module. It is then not hard to see that the above map is an isomorphism and so $MSO_*(X)$ is free over $MSO_*$. Assuming that, it is not hard to deduce that the Künneth map -$$ MSO_*(X)\otimes_{MSO_*}MSO_*(Y)\to MSO_*(X\wedge Y) $$ -is again an isomorphism for all $Y$.<|endoftext|> -TITLE: Branching Rule for alternating groups -QUESTION [9 upvotes]: Let $A_n$ be the alternating group of degree $n$. What is the branching rule for the subgroup $A_{n-1}\subset A_n$, i.e., the structure of the restriction of ordinary irreducible representations of $A_n$ to $A_{n-1}$? Are there some nice books or references which provide detailed answer to this question? - -REPLY [9 votes]: This is answered in Theorem 4 of my paper Comparison of Gelfand-Tsetlin Bases for Alternating and Symmetric Groups, with Geetha Thangavelu, which is published in Algebras and Representation Theory, and is also available on the arXiv. See also Ruff, O.: Weight theory for alternating groups. Algebra Colloq. 15(03), 391–404 (2008). -The alternating groups have two types of representations, those corresponding to mutually conjugate non-self-conjugate pairs $(\lambda,\lambda')$ which are simply the restriction of the irreducible representation $V_\lambda$ or $V_{\lambda'}$ of $S_n$ to $A_n$, and two corresponding to each self-conjugate partition $\lambda$, denoted $V_\lambda^\pm$, which are the two irreducible summands of the irreducible representation $V_\lambda$ of $S_n$, when resticted to $A_n$. -If $\lambda$ is a partition of $n$ and $\mu$ is a partition of $n-1$, write $\mu\in \lambda^-$ if the representation $V_\lambda$ of $S_n$ contains the representation $V_\mu$ of $S_{n-1}$ upon restriction, then we have: - -If $\lambda$ and $\mu$ are non-self-conjugate then the representation $V_\mu$ of $A_{n-1}$ is contained in the restriction of $V_\lambda$ from $A_n$ to $A_{n-1}$ if either $\mu\in \lambda^-$, or $\mu'\in \lambda^-$. -If $\lambda$ is non-self-conjugate and $\mu\in \lambda^-$ is self-conjugate, then $V_\mu^\pm$ are both contained in the restriction of $V_\lambda$ from $A_n$ to $A_{n-1}$. -If $\lambda$ is self-conjugate and $\mu\in \lambda^-$ is non-self-conjugate, then $V_\mu$ is contained in the restriction of both $V_\lambda^\pm$ from $A_n$ to $A_{n-1}$. -Finally, if $\lambda$ and $\mu\in \lambda^-$ are both self-conjugate, then $V_\mu^+$ is contained in $V_\lambda^+$ and $V_\mu^-$ is contained in $V_\lambda^-$. This result is based on a careful choice of sign in defining the representations $V_\lambda^\pm$ (deciding which gets the $+$ sign, and which gets the $-$ minus sing among the irreducible $A_n$ representations contained in a self-conjugate representation of $S_n$). - -See the figure below. Please write to me if you would like an e-print of the the published version of the article.<|endoftext|> -TITLE: Progress towards a computational interpretation of the univalence axiom? -QUESTION [8 upvotes]: I will preface this by saying that I am not an expert on type theory. I am just a curious outsider slowly making my way through the HoTT book when I (rarely) have some spare time. -I am just curious if there has been any progress towards a computational interpretation of univalence? If not, is the general feeling that univalence is not compatible with computation? - -REPLY [2 votes]: The Arend programming language provides an alternative approach. It does not actually provide a computational interpretation. Instead, it provides a function converting isomorphism into a path, and it built-in a computation rule that transportation along a path constructed from univalence will reduce to the application to the function in the isomorphism. In other words, it's type theory with the following primitives (simplified for reading ease -- the actual definitions are slightly more general): - -Isomorphism-to-path: iso : (Iso A B) -> A = B -Transport: coe : A = B -> A -> B -Computation rule coe (iso f) ==> f - -If you look at the library, you'll see that even with these simple rules we can do a lot of HoTT already (from a type theoretical perspective, imagine a constructive version of MLTT with various extensionalities -- the community has done a lot with MLTT, and we can do everything there in Arend as well. Also, it has higher inductive types, so we can have some HoTT things there). -This style is similar to the HoTT-Agda library, but unlike the library, Arend doesn't introduce notions of REWRITING and stuffs -- because Agda wants to have a general notation of custom reduction rule, and Arend focus on HoTT. The underlying type theory is called HoTT-I by the author.<|endoftext|> -TITLE: The Stone-Čech compactification of the fixed point set -QUESTION [6 upvotes]: Let $G$ be a discrete group and $X$ be a Tychonoff $G$-space. Then there -exists a $G$-action on Stone-Čech compactification $\beta X$. If the -fixed point set $X^{G}\neq \emptyset $, then the Stone-Čech -compactification of the fixed point set is the fixed point set of the Stone-Čech compactification, i.e., $\beta \left( X^{G}\right) =\left( \beta -X\right) ^{G}$? - -REPLY [5 votes]: Not without further assumptions. -First create an ordered space $X$ by identifying $\langle0,\omega_1\rangle$ and $\langle1,\omega_1\rangle$ in the product $2\times(\omega_1+1)$ to a point, $\Omega$ say. Then make the product $X\times X$ and take out $\langle\Omega,\Omega\rangle$. You can rotate the resulting space $Y$ around that hole over $\pi$ by mapping $\bigl<\langle i,\alpha\rangle,\langle j,\beta\rangle\bigr>$ to $\bigl<\langle 1-i,\alpha\rangle,\langle 1-j,\beta\rangle\bigr>$ and similarly for points with $\Omega$ as a coordinate. This map $f:Y\to Y$ has no fixed points but $\beta Y=X$ and $\beta f$ does have a fixed point: $\langle\Omega,\Omega\rangle$. -Now take the sum $Y\cup\{0\}$ and extend $f$ by $f(0)=0$. Then $f$ has one fixed point and $\beta f$ has two. As $f$ is its own inverse this gives us an action of $\mathbb{Z}/2\mathbb{Z}$ that is a counterexample.<|endoftext|> -TITLE: A divisibility of q-binomial coefficients combinatorially -QUESTION [11 upvotes]: Let a and b be coprime positive integers. Then the number a+b divides the binomial coefficient ${a+b \choose a}$. I know how to prove this combinatorially - for example after choosing an ordered set of a+b elements, there is a free action of a cyclic group of order a+b on the set of a-element subsets (and you can even choose a distinguished representative of each orbit if you so wish). -Passing to q-binomial coefficients, the quantum number $[a+b]_q$ divides the q-binomial coefficient $\begin{bmatrix}{a+b}\\ {a}\end{bmatrix}_q$. -My question today is whether this divisibility fact has a combinatorial proof or interpretation. -I remark that one can also ask whether or not -$$\frac{1}{[a+b]_q}\begin{bmatrix}{a+b}\\ {a}\end{bmatrix}_q\in \mathbb{N}[q],$$ -which would likely follow from any such combinatorial argument. When a+b=p is prime, I know this is the case by constructing algebraically a free $\mathbb{F}_p[d]/(d^p)$-module structure on the cohomology of the Grassmannian. I haven't done enough numerical experiments for non-prime values of a+b to be confident that this polynomial has non-negartive coefficients in general. - -REPLY [4 votes]: Supposing $\gcd(a,b)=1$ throughout this answer. Haiman was able to prove that -$$\frac{1}{[a+b]_q}\begin{bmatrix}{a+b}\\ {a}\end{bmatrix}_q\in \mathbb{N}[q]$$ -by interpreting it as the Hilbert series of some quotient of a polynomial ring. See propositions 2.5.(2-4) in his paper "Conjectures On The Quotient Ring By Diagonal Invariants". -Regarding combinatorial proofs: It is known in rational Catalan land that the quantity $\frac{1}{a+b}\binom{a+b}{a}$ counts paths from $(0,0)$ to $(a,b)$ with steps $(1,0),(0,1)$ that don't go below the $y=bx/a$ line -simultaneous $a,b$ core partitions -and there are explicit statistics on both of these sets which conjecturally give $\frac{1}{[a+b]_q}\begin{bmatrix}{a+b}\\ {a}\end{bmatrix}_q$ (see conjectures 2 and 3 in this paper) but to my knowledge they are still open.<|endoftext|> -TITLE: Symplectic Lefschetz fibrations in terms of factorization in symplectic mapping class group -QUESTION [6 upvotes]: There is a well-known theorem stating that there is a bijection between diffeomorphism classes of Lefschetz fibrations over $S^2$ whose general fiber is a closed orientable surface $\Sigma_g$ of genus $g\geq 2$ and factorizations of identity in the mapping class group $\Gamma(\Sigma_g)$ up to Hurwitz moves and global conjugation. It is explained here, for example. -Is there an analogue of this theorem for symplectic Lefschetz fibrations in higher dimensions? In other words, are symplectomorphism classes of symplectic Lefschetz fibrations in bijective correspondence with factorizations of identity in the symplectic mapping class group of the general fiber? - -REPLY [3 votes]: Up to homotopy, giving a Lefschetz fibration over $S^2=\mathbb C\cup\{\infty\}$ with singular fibers over $\{e^{2\pi ik/N}\}_{0\leq k -TITLE: Explicit Left Adjoint to Forgetful Functor from Cartesian to Symmetric Monoidal Categories -QUESTION [22 upvotes]: There is a forgetful functor from the category of (small) Cartesian monoidal categories (a symmetric monoidal category in which the tensor product is given by the categorical product) to the category of symmetric monoidal categories. This functor is known to have a left adjoint and right adjoint. The right adjoint is easy to describe: it takes a symmetric monoidal category to its category of cocommutative coalgebras. I am looking for an explicit description of the left adjoint though. Does anyone know what it is? -There is a MO question about this, but it does not give an explicit description: Cartesian envelope of a symmetric monoidal category -I am also interested in the left adjoint of the forgetful functor with domain semi-Cartesian monoidal categories, which are symmetric monoidal categories in which the tensor unit is terminal. -Another way to look at this is from the point of view of the "commutative algebra of categories," as described (in the quasicategorical case) here: https://arxiv.org/abs/1606.05606. Using that machinery, i.e. knowing that Cartesian monoidal categories are precisely the symmetric monoidal categories which are modules over the solid semi-ring category $Fin^{op}$, it would suffice to have an explicit description of the tensor product on (presentable) symmetric monoidal categories (that recovers the structure described in the above reference). So maybe this is known? - -REPLY [4 votes]: I don't know the answer, although I've thought about this quite a bit. I think the best that can be said in general is that the free cartesian monoidal category on $\mathcal{C}^\otimes$ is a full subcategory of $\text{Fun}^\otimes(\mathcal{C}^\otimes,\text{Set}^\times)^\text{op}$, which is the category of symmetric monoidal functors. That is Theorem 3.10 of https://arxiv.org/abs/1606.05606. -There is some hope of giving a more explicit description if we know something about maps into tensor products. Here are three examples. (I would love to understand what they have in common.) -Baby example: If $\mathcal{C}^\times$ is already cartesian monoidal, then $\mathcal{C}_{/Y\times Z}\cong\mathcal{C}_{/X}\times_\mathcal{C}\mathcal{C}_{/Y}$. That is, a map $X\rightarrow Y\times Z$ corresponds to two maps $X\rightarrow Y$ and $X\rightarrow Z$. In this case, the free cartesian monoidal category on $\mathcal{C}$ is $\mathcal{C}$ itself. -Example 2: If $\mathcal{C}^\amalg$ is cocartesian monoidal, we may ask whether the coproduct behaves like a disjoint union (that is, $\mathcal{C}$ is disjunctive) in the following sense: any map $X\rightarrow Y\amalg Z$ decomposes canonically as the coproduct of two maps $X_Y\rightarrow Y$ and $X_Z\rightarrow Z$. That is, $\mathcal{C}_{/Y\amalg Z}\cong\mathcal{C}_{/Y}\times\mathcal{C}_{/Z}$. In this case, I believe that the free cartesian monoidal category can be described as a category of spans, something like the following: -Objects are just objects of $\mathcal{C}$. Morphisms from $X$ to $Y$ are spans $X\leftarrow T\rightarrow Y$, such that $T$ admits a decomposition $T_1\amalg\cdots\amalg T_n$ with each $T_i\rightarrow X$ an inclusion of a direct summand; that is, $T_i\amalg T_i^\prime\cong X$. -For example, when $\mathcal{C}=\text{Fin}$, the category of finite sets, the free cartesian monoidal category is spans of finite sets. The $\mathcal{C}=\text{Fin}$ case is a main result of the paper cited above (at the level of $\infty$-categories). The general case is not written down anywhere, as far as I know. -Example 3: Given an operad $\mathcal{O}$, there is a category $[\mathcal{O}^\otimes]$, whose objects are finite sets. A map from $S$ to $T$ is a function $f:S\rightarrow T$ and an $|f^{-1}(t)|$-ary operation for each $t\in T$. Concatenation gives $[\mathcal{O}^\otimes]$ a symmetric monoidal structure. -Note that $[\mathcal{O}^\otimes]$ is disjunctive (in the sense of Example 2) but not cocartesian monoidal. In this case, the free cartesian monoidal category is the full subcategory of $$\text{Fun}^\otimes([\mathcal{O}^\otimes],\text{Set}^\times)^\text{op}=\text{Alg}_\mathcal{O}(\text{Set})^\text{op}$$ spanned by the free $\mathcal{O}$-algebras. Classically, the free $\mathcal{O}$-algebra on $n$ generators is $$\coprod_{k\geq 0}\mathcal{O}(k)\times_{\Sigma_k}n^k.$$ That effectively computes the free cartesian monoidal category on $[\mathcal{O}^\otimes]$ as a type of span construction, similar to Example 2. See the paper above, Remark 5.1 (also Example 3.16). The span construction can be made precise without much difficulty for ordinary categories, but it is an open problem for $\infty$-categories.<|endoftext|> -TITLE: Low dimensional representations of $SL_n(\mathbb{Z}/p^\ell \mathbb{Z})$ -QUESTION [10 upvotes]: When $\ell = 1$ I know that the smallest non-trivial irreducible complex representations of $SL_n(\mathbb{Z}/p\mathbb{Z})$ has dimension $\frac{p^n - 1}{p-1} - 1$ (with maybe some exceptions for small values of $n$ or $p$). -For $\ell > 1$ we have the map $SL_n(\mathbb{Z}/p^\ell \mathbb{Z}) \to SL_n(\mathbb{Z}/p\mathbb{Z})$ so we still get a representation of this size. I was wondering if this is still the smallest possible dimension (again maybe with a finite number of exceptions for small $p$ and $n$), and if not are there known lower bounds for how small such a representation can be? -I found some calculations of characters for small values of $n$ that suggest this could be true, but otherwise couldn't find much. Is there anywhere in the literature that addresses this sort of question? I'm also interested in similar results for $Sp_{2n}(\mathbb{Z}/p^\ell \mathbb{Z})$. - -REPLY [2 votes]: For $n=2$ and $p$ odd: -The minimal possible dimension of a non-trivial irrep of $G_l=\mathrm{SL}_2(\mathbb{Z}/p^l)$ is always $(p-1)/2$, and the minimal possible dimension of a non-trivial primitive irrep of $G_l$ is $p^{l-2}(p^2-1)/2$. -This can be found in Shalika's paper Representation of the two by two unimodular group over local fields.<|endoftext|> -TITLE: Is the set of Lorentzian metrics metrizable? -QUESTION [9 upvotes]: Fix a differentiable non-compact manifold $M$. Denote by $\mathrm{Lor}(M) := \{\text{Lorentzian metrics on $M$}\}.$ One can define a topology on this set via: fix any open covering $\mathcal{A}$ on $M$. For each $g \in \mathrm{Lor}(M)$ and for each positive continuous function $r : M\to ]0,\infty[,$ one defines: -$$\mathcal V(g,r) := \{h \in \mathrm{Lor}(M) : \forall p \in M, |\nabla^kg_{ij}(p) - \nabla^kh_{ij}(p)| < r(p)\},$$ -where $g_{ij}, h_{ij}$ are the coordinates on some local chart with open domain contained on some open of $\mathcal{A}$. (I am sorry being a bit informal at this point, but I think it is clear what I mean). -Also, $\nabla^k$ is supposed to denote any $k$-derivative of the metric. This is a sub-basis for the $C^k$-topology on $\mathrm{Lor}(M)$. -My first question is: is this topology somehow metrizable? -Further, if $M$ was compact (and $\chi(M) = 0)$, then $\mathrm{Lor}(M)$ is not empty, by imposing the $L^2$ metric (for some Riemannian metric) on $\mathrm{Lor}(M)$ induces the same topology as that I have defined? I mean, if we are interested on statements like: - -(M,g) is locally causal if $g$ is close (on the sense of $C^r$-topology) to a causal metric $h$ on $M$, - -Does this can be interchanged by - -(M,g) is locally causal if $g$ is close to a causal metric $h$ on $M$ on the $L^2$-norm. - -I am also sorry it these questions don't make sense at all, at the end, my question is: in general, introducing a Riemannian metric for comparing Lorentzian metric is somehow inappropriate, in the sense it leads to lost of some information? - -REPLY [14 votes]: First of all, there is a bunch of basic things that you need to write in a slightly clearer way. If you try to topologize the set of Lorentzian metrics as you did, you need first: - -Restrict to the subset of Lorentz metrics of a given (say, $C^k$, $0\leq k\leq\infty$) regularity, otherwise your definition for the basic neighborhoods $\mathcal{V}(g,r)$ makes no sense. -Once you did the above (denote the resulting set by, say, $\mathrm{Lor}_k(M)$), replace $|\nabla^k g_{ij}(p)-\nabla^k h_{ij}(p)|$ by the sum $\sum_{0\leq j\leq k}|\nabla^j g_{ij}(p)-\nabla^j h_{ij}(p)|$ in the definition of $\mathcal{V}(g,r)$ for $k$ finite. If $k=\infty$, you must include such $\mathcal{V}(g,r)\doteq\mathcal{V}_k(g,r)$ for all $k\geq 0$ (or at least for all $k$ in an infinite subset of $\mathbb{N}\cup\{0\}$). - -This is the bare minimum. Ideally, it would be better to do things in a coordinate-free way: denote by $\nabla^k g$ the iterated covariant derivative of order $k$ of $g$ w.r.t. some (say, torsion-free) covariant derivative operator $\nabla$ on $M$ and define the pointwise Euclidean norm $|T|$ of a tensor field $T$ by lifting some reference Riemannian metric $e$ on $M$ to the corresponding tensor bundle. This provides a way to write a fiberwise scalar product on the jet bundle of order $k$ of the fiber bundle of Lorentzian metrics over $M$ (more generally, on the jet bundle of order $k$ of the vector bundle of covariant tensors of rank 2 over $M$). One can write these things in more detail, but this is more or less standard. Anyway, I think you get the gist. -With the trivialities out of the way (at this point, if something about them is not yet clear, please do let me know), we can begin to address your question proper. What you wrote above (with the tacit understanding that the amendments 1. and 2. above are included) are the basic neighborhoods for the $C^k$ Whitney topology of $\mathrm{Lor}_k(M)$. In fact, this topology does not depend on the choice of a reference Riemannian metric $e$ as above (the reason will be explained below). -If $M$ is compact, this topology is even normable if $k$ is finite (as a subset of the normable vector space of $C^k$ covariant tensor fields of rank two) and still metrizable if $k=\infty$ for then it coincides with the compact-open $C^k$ topology. However if $M$ is non-compact (as you assumed, since you seem to be ultimately interested in causality theory for Lorentzian metrics and this theory is nontrivial only for non-compact $M$) this topology is non-metrizable for all $k$ (even $k=0$). In fact, this topology is not even first-countable in this case. -This is easier to visualize in the case of $C^k$ scalar fields (i.e. $C^k$ real-valued functions) on $M$ instead of Lorentzian metrics: the connected component of $f\equiv 0$ in the $C^k$ Whitney topology of $C^k(M)$ is the space $C^k_c(M)$ of $C^k$ functions with compact support on $M$ with the usual inductive limit (locally convex vector space) topology. This topology is not first-countable, hence non-metrizable. More generally, the connected component of any $C^k$ function $f$ in the $C^k$ Whitney topology of $C^k(M)$ is precisely $f+C^k_c(M)$. One sees from this remark that the Whitney topologies get so fine when $M$ is non-compact, they become extremely disconnected. A similar fact holds for the $C^k$ Whitney topology in $\mathrm{Lor}_k(M)$ - the metrics $h$ in the connected component of $g\in\mathrm{Lor}_k(M)$ in this topology differ from $g$ only inside some compact subset of $M$ (depending on $h$). This remark also makes it clear why the choice of the Riemannian metric $e$ is not relevant to the definition of the $C^k$ Whitney topology, despite the fact that $M$ is not compact. -For (many!) details on the Whitney topologies, you may want to check The Convenient Setting of Global Analysis by Andreas Kriegl and Peter W. Michor (AMS, 1997), specially Chapter IX (Manifolds of Mappings). -The paper by Lerner quoted by Igor in his comment relates the Whitney topologies to structures which are natural to Lorentz metrics - e.g. conformal classes of Lorentzian metrics with representatives being $C^0$ Whitney-near to each other amounts to their light cones being close to each other, metrics which are $C^1$ Whitney-near to each other have their geodesics near to each other in some sense, and so on.<|endoftext|> -TITLE: Weinstein neighborhood theorem for Lagrangians with Legendrian boundary -QUESTION [12 upvotes]: $\require{AMScd}$ -Weinstein's neighborhood theorem says that every Lagrangian has a standard neighborhood. The more precise statement goes like this. - -Theorem 1: (Lagrangian Neighborhood Theorem) Let $(X,\omega)$ be a symplectic manifold and $L \subset X$ be a closed Lagrangian. Then there exists a neighborhood $U$ of $L$ in $X$ and a symplectomorphism $\varphi:U \simeq V \subset T^*L$ taking $L$ identically to the zero-section $L \subset T^*L$. - -Now let $(W,\lambda)$ be a Liouville domain. That is, $W$ is a compact manifold with boundary, and $\lambda$ is a $1$-form on $W$ such that $d\lambda$ is symplectic and $\lambda|_{\partial W}$ is a contact form. Furthermore, let $L \subset W$ be a compact Lagrangian sub-manifold with Legendrian boundary $\partial L \subset \partial W$. -My question is whether the following version of the neighborhood theorem holds in this setting. It seems to me that if it is true, then it should be standard, but I can't find a reference. - -Theorem 2 (Maybe?): There exists a neighborhood $U$ of $L$ in $W$ and a symplectomorphism of manifolds with boundary $\varphi:U \simeq V \subset T^*L$ taking $L$ identically to the zero-section $L \subset T^*L$. - -Remark On Proof Of Theorem 1: The basic result that the usual Lagrangian neighborhood theorem depends on is the following lemma (see [1] or McDuff-Salamon). - -Lemma: Let $X$ be a manifold with closed sub-manifold $S \subset X$, and let $\omega_0, \omega_1$ be two symplectic forms on $X$. Suppose that $\omega_0 = \omega_1$ on the fiber $T_sX$ for any $s \in S$. -Then there exists neighborhoods $N_0$ and $N_1$ of $S$ and a symplectomorphism $\varphi:N_0 \to N_1$ with $\varphi|_S = \text{Id}$ and $\varphi^*\omega_1 = \omega_0$. - -The proof is a version of the usual Moser trick. You find a $1$-form $\sigma$ in a neighborhood with $d\sigma = \omega_1 - \omega_0$ and then you integrate the vector-field $Z_t$ satisfying: -$$\iota_{Z_t}\omega_t = -\sigma\quad\text{where}\quad\omega_t = (1-t)\omega_0 + t\omega_1$$ -This gives you a family of diffeomorphisms with $\varphi^*_t\omega_t = \omega_0$ and you're done. If you try to run this proof on a sub-manifold $S \subset X$ with $\partial S \subset \partial X$, you run into the issue that $Z_t$ needs to be parallel to the boundary $\partial X$ in order for the flow to be well-defined. If I'm not mistaken, the criterion for this to be the case is: -$$ T(\partial X)^{\omega_t} \subset \ker(\sigma) \text{ on }\partial X $$ -Here $T(\partial X)^{\omega_t}$ is the characteristic foliation on $\partial X$ with respect to $\omega_t$, i.e. the symplectic perp to the tangent space to $\partial X$. It isn't clear to me that you can even accomplish the above inclusion for $\sigma$, or that you can upgrade $\sigma$ to a family $\sigma_t$ with this property. -Speculation On Validity Of Theorem 2: On a conceptual level, I can't decide whether or not Theorem 2 is too optimistic. Here is what makes me skeptical about it. -Theorem 2 would imply not only that the boundaries $\partial U \simeq \partial V$ of $U$ and $V$ were contactomorphic, but also that the characteristic foliations $T(\partial U)^{d\lambda}$ and $T(\partial V)^{d\lambda}$ on $U$ and $V$ near $\partial L$ were the same. The characteristic foliation of a contact hypersurface is generally very sensitive to the embedding of said hypersurface, and from that perspective a standard neighborhood in the vein of Theorem 2 would be a bit surprising. -I haven't pursued this idea enough to produce a counter-example unfortunately. - -REPLY [8 votes]: Theorem 2 is true, verbatim. I will give an outline of the proof here, since the details make it kind of long. If you would like a detailed write-up and you don't want to do it yourself, DM or email me. -The actual statement that is true is more general: you do not need the boundary $\partial X$ to be contact or for the boundary $\partial L$ to be Legendrian. - -Theorem: (Weinstein Neighborhood With Boundary) Let $(X,\omega)$ be a symplectic manifold with boundary $\partial X$ and let $L \subset X$ be a properly embedded, Lagrangain sub-manifold with boundary $\partial L \subset \partial X$ transverse to $T(\partial X)^\omega$. -Then there exists a neighborhood $U \subset T^*L$ of $L$ (as the zero section), a neighborhood $V \subset X$ of $L$ and a diffeomorphism $f:U \simeq V$ such that $\varphi^*(\omega|_V) = \omega_{\text{std}}|_U$. - -Proof: The proof has two steps. First, we construct neighborhoods $U \subset T^*L$ and $V \subset X$ of $L$, and a diffeomorphism $\varphi:U \simeq V$ such that: -\begin{equation} -\varphi|_L = \text{Id} \qquad \varphi^*(\omega|_V)|_L = \omega_{\text{std}}|_L \qquad T(\partial U)^{\omega_{\text{std}}} = T(\partial U)^{\varphi^*\omega} -\end{equation} -Here $T(\partial U)^{\omega_{\text{std}}} \subset T(\partial U)$ is the symplectic perpendicular to $T(\partial U)$ with respect to $\omega_{\text{std}}$ (and similarly for $T(\partial U)^{\varphi^*\omega}$. Second, we apply Lemma 1 (below) and a Moser-type argument to conclude the result. -For the first part, the proof proceeds like this. First you pick a metric on $L$ and use the exponential map in the usual way, to get a diffeomorphism $\varphi:U \simeq V$ with $U \subset T^*L$, $V \subset X$ and $\varphi^*\omega_{\text{std}} = \omega$ along $L$. Then we use Lemma 2 below to modify $\varphi$ in a collar neighborhood of $\partial U$ to satisdy $T(\partial U)^{\omega_{\text{std}}} = T(\partial U)^{\varphi^*\omega}$. -The second part is basically identical to the usual Moser argument. - -Lemma 1: (Fiber Integration With Boundary) Let $X$ be a compact manifold with boundary, $\pi:E \to X$ be a rank $k$ vector-bundle with metric and $\pi:U \to X$ be the closed disk bundle of $E$. Let $\kappa \subset T(\partial U)$ be a distribution on $\partial U$ invariant under fiber-wise scaling. Finally, suppose that $\tau \in \Omega^{k+1}(U)$ is a $(k+1)$-form such that: - \begin{equation} \label{eqn:fiber_integration_sigma} d\tau = 0 \qquad \tau|_X = 0 \qquad (\iota^*_{\partial X}\tau)|_\kappa = 0\end{equation} -Then there exists a $k$-form $\sigma \in \Omega^k(U)$ with the following properties. - \begin{equation} \label{eqn:fiber_integration_tau} d\sigma = \tau \qquad \sigma|_X = 0 \qquad (\iota^*_{\partial X}\sigma)|_\kappa = 0 \end{equation} - -The proof of Lemma 1 just involves examining the proof of the version of the Poincare Lemma in McDuff-Salamon, and checking that the primitive constructed there satisfies the 3rd property. Note that to apply this lemma, you need to show that the characteristic foliation of $\partial(T^*L) \subset T^*L$ is invariant under fiber-scaling, but this is a quite easy Lemma. - -Lemma 2: Let $U$ be a manifold and $L \subset U$ be a closed sub-manifold. Let $\kappa_0,\kappa_1$ be rank $1$ orientable distributions in $TU$ such that $\kappa_i|_L \cap TL = \{0\}$ and $\kappa_0|_L = \kappa_1|_L$. -Then there exists a neighborhood $U' \subset U$ of $L$ and a family of smooth embeddings $\psi:U' s\times I \to U$ with the following four properties. - $$ -\psi_t|_{\partial L} = \text{Id} \qquad d(\psi_t)_u = \text{Id} \text{ for }u \in L \qquad \psi_0 = \text{Id} \qquad [\psi_1]_*(\kappa_0) = \kappa_1 -$$ - Furthermore, we can take $\psi_t$ to be $t$-independent for $t$ near $0$ and $1$. - -The proof of Lemma 2 is straight-forward.<|endoftext|> -TITLE: Zero divisors in complex group algebras of residually finite groups -QUESTION [6 upvotes]: Conjecture. There exists a function $f:\mathbb{N} \rightarrow \mathbb{N}$ such that if $\alpha$ and $\beta$ are non-zero elements of the complex group algebra $\mathbb{C}[G]$ of a finite group $G$ such that $1\in \text{supp}(\alpha) \cap \text{supp}(\beta)$, $|\text{supp}(\alpha)|\leq |\text{supp}(\beta)|$ and $\alpha \cdot \beta =0$, then $\text{exp}\big (\langle \text{supp}(\alpha) \rangle \big) \leq f(|\text{supp}(\beta)|)$, where $\text{exp}(H)$ denotes the exponent of a finite group $H$ and $\text{supp}(\gamma)$ for an element $\gamma=\sum_{g\in G} \gamma_g \; g \in \mathbb{C}[G]$ denotes the set $\{ g\in G \;|\; \gamma_g \not=0\}$. -Motivation. If the conjecture is true, then the support of any zero divisor of the complex group algebra of any residually finite group generates a finite subgroup. -Proof. Let $\alpha$ be a non-zero element of $\mathbb{C}[G]$ for some residually finite group $G$ such that $\alpha \cdot \beta =0$ for -some non-zero $\beta \in \mathbb{C}[G]$. Let $H=\langle \text{supp}(\alpha) \rangle$. By Zelmanov's celebrated result on restricted Burnside problem, it is enough to show that the exponent of $H$ is finite. Since $H$ is finitely generated residually finite, there exists a descending series $H=N_1\geq N_2 \geq \cdots$ of normal subgroups $H$ of finite index such that $\cap_{i\in\mathbb{N}} N_i=1$. Note that there exists $k\in\mathbb{N}$ such that $\bar{\alpha} \cdot \bar{\beta}=0$ in $\mathbb{C}[H/N_i]$ for all $i\geq k$ and $|\text{supp}(\bar{\beta})|=|\text{supp}(\beta)|=:t$, where $\bar{}$ is the natural ring epimorphism from $\mathbb{C}[H]$ onto $\mathbb{C}[H/N_i]$. Now if the above conjecture is true then $\text{exp}(H/N_i)\leq f(t)$ and so $\text{exp}(H)$ is finite. This completes the proof. -If the above proof and conjecture are true then the complex group algebras of torsion-free residually finite groups have no zero divisor. - -REPLY [8 votes]: The conjecture does not hold. -Let $m \geq 1$ be an arbitrary integer. -Let $G = \langle x,y | x^m, y^5 , [x,y]\rangle$ be the finite abelian group isomorphic to a product of two cyclic groups $C_m \times C_5$. -Let $\alpha = xy +y -x -1$ and let $\beta = y^4 +y^3 +y^2 +y +1$ in $\mathbb{Z}[G]$. -Then $\alpha \beta = (x+1)(y-1)\beta= 0$. -The support of $\beta$ has $5$ elements and the support of $\alpha$ has $4$ elements. Moreover, the support of $\alpha$ generates the group $G$, which has exponent $\mathrm{exp}(G) \geq m$.<|endoftext|> -TITLE: What is Kontsevich's Hodge theory of path integrals? -QUESTION [11 upvotes]: I was reading about the appearance of Calabi-Yau manifolds in Feynman integrals, and I thought to wonder if there is such a thing as "infinite-dimensional Hodge theory". Googling the phrase turned up references to a 2010 lecture by Maxim Kontsevich called "Infinite-dimensional Hodge theory of path integrals". Does anyone know what its content was? - -REPLY [13 votes]: The most usual Hodge theory is related to integrals, periods of algebraic differential forms on algebraic varieties, whereas the physics path integral usually involves integration of an exponential quantity. So the first thing to understand is that there is a finite dimensional Hodge theory for "exponential integrals", i.e. integrals of exponential of algebraic objects. It is sometimes called "irregular Hodge theory" (because unlike the usual Gauss-Manin connection of Hodge theory which admits regular singularities, exponential periods satisfy in general differential equations with irregular singularities). One of the numerous motivations for this theory is mirror symmetry for Fano varieties. -Some relevant names are probably: Kontsevich, Barannikov, Sabbah, Esnault, Yu, Mochizuki... -So the thing which should be more related to path integrals should be an infinite dimensional version of irregular Hodge theory. Unlike the finite dimensional case, I don't think there is yet a rigorous mathematical theory. But one can take some finite dimensional phenomenons, like the Stokes phenomenon, and see if they have some infinite dimensional counterparts (e.g. for Chern-Simons theory). -On the website of the Simons Center, you can find two video lectures by Kontsevich, with the title "Exponential integrals": -http://scgp.stonybrook.edu/video_portal/video.php?id=1798 -and on the youtube channel of the IHES, you can find four video lectures with the same title: -https://www.youtube.com/watch?v=tM25X6AI5dY&index=1&list=PLx5f8IelFRgE098k6hVHriQyfMkoq--T0 -In both cases, these lectures start with some finite dimensional theory and go to some infinite dimensional examples towards the end.<|endoftext|> -TITLE: Injective indecomposable modules over Laurent polynomial rings -QUESTION [7 upvotes]: What does the injective envelope of $\mathbb C[x,x^{-1}]/(p(x,x^{-1}))$ as a $\mathbb C[x,x^{-1}]$-module look like where $p(x,x^{-1})$ is an irreducible element? I’m sure this is well known, but when Googling I mostly found stuff for ordinary polynomial rings. -I am particularly interested in the possibile dimensions of the injective indecomposables over $\mathbb C$, i.e., can they be countable. - -REPLY [4 votes]: If $R$ is a PID and $P$ is a nonzero prime ideal, then $E(R/P)=K/P_P$, where $K$ is the fraction field of $R$ and $R/P$ is viewed as a submodule of $K/P_P$ via the evident map. Indeed, one readily checks that $R/P\cong R_P/P_P$ is an essential submodule of $K/P_P$ and that $K/P_P$ is divisible, hence injective (injective and divisible are the same since $R$ is a PID). -This fact can be generalized to arbitrary ideals in Dedekind domains: - -Theorem. Let $R$ be a Dedekind domain with fraction field $K$ and let $I$ be an ideal of $R$ different from $0$ and $R$. Let $P_1,\dots,P_t$ denote the primes containing $I$ and let $S$ be the multiplicative set $R\setminus (P_1\cup\dots\cup P_t)$. Then the natural map $R/I\to K/S^{-1}I$ is an injective envelope of $R/I$. - -Proof. Observe that the image of any $s\in S$ in $R/I$ is invertible. Indeed, it is not contained in any maximal ideal of $R/I$. Thus, $R/I$ is $S$-divisble, and it follows that the natural map $R/I\to S^{-1}(R/I)=S^{-1}R/S^{-1}I$ is an isomorphism of $R$-modules. -Next, I claim that $K/S^{-1}I$ is the injective envelope of $S^{-1}R/S^{-1}I$ viewed as an $S^{-1}R$-module. This is a generalization of the fact mentioned above: The ring $S^{-1}R$ is a PID with finitely many primes, namely, $S^{-1}P_1,\dots,S^{-1}P_t$ and $S^{-1}I$ is a nonzero ideal contained in their product. Using this, it is routine to check that $S^{-1}R/S^{-1}I$ is an essential $S^{-1}R$-submodule of $K/S^{-1}I$, and the latter is injective over $S^{-1}R$ because it is divisible. -To finish the proof it is enough to show that if $M$ is an $R$-module such that $S^{-1}M$ is injective as an $S^{-1}R$-module, then $S^{-1}M$ is injective as an $R$-module. This follows by writing down the diagram definition of injectivity, noting that the diagram maps into its localization relative to $S$, and applying the injectivity of $S^{-1}M=S^{-1}(S^{-1}M)$ over $S^{-1}R$ to the latter diagram. - -Edit. Concerning your question about the dimension of $E(R/P)$, if $R$ is a PID containing a field $k$ and $P$ is a nonzero prime ideal such that $\dim_k (R/P)=1$ (in your question $k=\mathbb{C}$), then $\dim_k E(R/P)$ is countable. -Proof. -Suppose $P=pR$. As explained above, $E(R/P)=K/P_P$, where $K$ is the fraction field of $R$. -I claim that the set $\{1,p^{-1},p^{-2},\dots\}$ spans $K/P_P$ as a $k$-vector space. (In fact, it is a $k$-basis.) -Given $r\in K$, there is some positive integer $n$ such that $r\in p^{-n} R_P$. -Write $r=p^{-n}a$ with $a\in R_P$. -Since the natural map $k\to R/P\to R_P/P_P$ is an isomorphism, there is $\alpha_{-n}\in k$ such that $\alpha_n$ and $a$ have the same image in $R_P/P_P$. Thus, $a-\alpha_{-n}\in P_P=pR_P$ and we can write $r=\alpha_{-n} p^{-n}+r'$ with $r'=p^n(a-\alpha_{-n})\in p^{-n+1} R_P$. Applying the same argument to $r'$, we see that there is $\alpha_{-n+1}\in k$ such that $r=\alpha_{-n} p^{-n}+\alpha_{-n+1}p^{-n+1} +r''$ with $r''\in p^{-n+2}R_P$. Proceeding by induction, we eventually find that -$$r=\alpha_{-n} p^{-n}+\alpha_{-n+1}p^{-n+1} +\dots+ \alpha_0p^0+r_1$$ with $r_1\in P_P$. Thus, $r\equiv \alpha_{-n} p^{-n}+\alpha_{-n+1}p^{-n+1} +\dots+ \alpha_0p^0\bmod P_P$, which is what we want. -Remark. -One can elaborate this argument further to show that $E(R/P)=K/P_P$ can be described as the set of formal power series -$$ -\alpha_{-\ell} p^{-\ell}+\dots+\alpha_0 p^0 -$$ -with $\alpha_0,\dots,\alpha_{-\ell}\in k$ and $r\in\mathbb{N}$. -The action of $R$ is given as follows: Given $r\in R$, write it as $r=\beta_0+\beta_1p+\dots+\beta_t p^t+r_{t+1}$ with $r_{t+1}\in p^{t+1}R$ and $\beta_0,\dots,\beta_t\in k$ for $t$ sufficiently large (i.e. $t>\ell$). Compute the formal product $(\alpha_{-\ell} p^{-\ell}+\dots+\alpha_0 p^0)(\beta_0+\beta_1p+\dots+\beta_t p^t)$ and truncate the positive $p$-powers.<|endoftext|> -TITLE: What is the automorphic interpretation of the Weil conjectures over finite fields -QUESTION [15 upvotes]: I am very much a beginner in the theory of automorphic forms and I might (will?) make mistakes in what follows. Please correct me. -A loose interpretation of the Langland's philosophy is that to any variety $X/\mathbb Z$, we should be able to find an automorphic form $f$ so that $\zeta(X) = \zeta(f)$ where $\zeta(f)$ is the Hasse-Weil zeta function and $\zeta(f)$ is the L-function associated to an automorphic form. -Question 1: What automorphic objects do the zeta functions of varieties over finite fields correspond to? -Tentative answer: It seems to me that these should correspond to euler factors at a prime of an automorphic form. (Lift the variety to characteristic 0, find the zeta function of that and take the corresponding automorphic form). -Is this on the right track? -Question 2: If so, what property of the automorphic forms does the Riemann hypothesis for the Weil conjectures over finite fields (proved by Deligne) correspond to? More generally, what do the Weil Conjectures correspond to on the automorphic side. -I can see two possibilities: -1) They correspond to some conjectural property of automorphic forms and the proof of the Weil conjectures actually tells us something about (a subclass of) automorphic forms. -2) They correspond to some known (easy to prove?) property of automorphic forms and the "reason" Deligne/Grothendieck had to work so hard is because we don't know how to associate automorphic forms to motivic L functions. -Question 3: Which one is right? - -REPLY [7 votes]: This is a brief answer; possibly others have different opinions about this. -Question 1: The Langlands conjectures gives a correspondence between Galois representations and automorphic forms. So a very naive way to view the zeta function of a variety over a finite fields from this philosophy is to look for a Galois representation which it comes from. However, this is exactly the point of the Weil conjectures; the zeta function has a description in terms of the action of the absolute Galois group of the finite field on the etale cohomology of the variety. These indeed give the Euler factors. -Question 2: The Riemann hypothesis for the Weil conjectures over finite fields corresponds to the Ramanjuan conjecture for automorphic forms. In fact, this was one of Deligne's original applications of the Weil conjectures, to proving the Ramanjuan conjecture for the Ramanjuan tau function. -Question 3: The answer is that it is a mixture of 1) and 2). For 1), as I said above, the Weil conjectures give you results towards the Ramanjuan conjecture for automorphic forms. For 2), it is a standard result that each Euler factor of an automorphic $L$-function is a rational function in $p^{s}$, but proving the rationality of the zeta functions of varieties over finite fields was one of the first difficult steps in the proof of the Weil conjectures.<|endoftext|> -TITLE: What is the minimum worst-case length of an element removal game? -QUESTION [6 upvotes]: A game is played as follows. There is a set $X = \{1, \ldots, n\}$. Player 1 is trying to find a "locally minimal subset" $M \subseteq X$ - that is, player 2 has said that $M$ is good, and also that every subset $M - \{x\}$ for $x \in M$ is bad. -Formally, play proceeds as follows: A move from player $1$ is any subset $S \subseteq X$ that is has not previously played. Its first move is always to play $X$. -Player 2 then moves by providing an answer $A_S \in \{0, 1\}$. $A_X = 1$ but may otherwise be chosen arbitrarily. -Play alternates between players one and two, terminating when there is some set $M$ such that player 1 has played all of $M$ and $M - \{x\}$ for all $x \in M$, and in response player 2 has played $A_M = 1$ and $A_{M - \{x\}} = 0$. -Player 1 is trying to finish in as few moves as possible. Clearly it can terminate play (just enumerate every subset in size-increasing order until player 2 says yes - the first such value will be a suitable $M$). -A naive greedy algorithm gives player 1 a winning strategy in $O(n^2)$ moves: At each stage, try removing one element at a time from your current best answer. If player 2 says yes, use that as your new current best answer. If it says no for all subsets with an element removed, the game is over. -Given any such greedy strategy that tries removing only one element at a time, player 2 can clearly force player 1 to play $O(n^2)$ moves by always saying yes to the last subset tried of the current set. -Is it true that any strategy for player 1 can similarly be forced to play $O(n^2)$ moves to win? -I conjecture (and am pretty confident in this conjecture) that it is, but the details keep escaping me. Essentially the idea should be to just force the algorithm to decay to the greedy case by blocking off all attempts, but I keep getting lost in the details - for example, you can't do it naively because player 2 could have already blocked off some paths with its previous moves. -Some Things I've Tried -(Updated to elaborate on what doesn't and might work) -The strategy that blocks the greedy algorithm from completing in fewer than $O(n^2)$ steps is to simply return $A_S = 0$ whenever $|S| \leq 1$ or when $S$ is not the last set of size $|S|$ that is a subset of the current smallest good set. This does not work for non-greedy algorithms because you can play the following moves: - -$X - \{i\}$ for $i \in \{1, \ldots, n - 2\}$ -$X - \{n - 1, i\}$ for $i \in \{1, \ldots, n - 2\}$ -$X - \{n\}$ -$X - \{n - 1\}$ - -When player 1 plays the greedy-blocking strategy and player 2 plays as above, the set $X - \{n - 1\}$ is a suitable minimal set. -I believe something like the following might work: -Let $\pi$ be a permutation of $X$. The $\pi$ strategy for player 1 is that $A_S = 1$ if and only if $|S| > 1$ and $S = \{\pi_1, \ldots, \pi_{|S|}\}$. -When player 1 plays the $\pi$ strategy the final answer must be $\{\pi_{n - 1}, \pi_n\}$. -I believe the following strategy works for player 1: Maintain a set of permutations $P$. Initially $P$ is the set of all permutations. Whenever player 2 plays $S$, if the $A_S = 1$ if and only if it would be $1$ when playing any $\pi \in P$. Otherwise $A_S = 0$ and we update $P$ to remove all $\pi$ which would not give that answer. -The thing I haven't been able to prove is that player 1 can continue this strategy for long enough. Intuitively the argument that it should be is as follows: - -If you pick a $\pi$ uniformly at random, then any greedy algorithm will play expected $O(n^2)$ moves. -Any attempt to reduce the size by more than one step at a time succeeds with very low probability. In particular if you have $U_1, \ldots, U_{k}$ with $k < \frac{n(n - 1)}{2}$ and $|U_i| < n - 1$ then the expected number of $U_i$ with $A_{U_i} = 1$ for the randomly chosen $\pi$ is $< 1$ and in particular the probability that none of them succeed is $> 0$. This suggests that the non-greedy algorithm is dominated for this permutation by some "greedy core". - -But at this point I'm handwaving wildly and again haven't been able to make the details of this intuition work out. - -REPLY [3 votes]: I think that there's a simple adversary strategy: -Answer 0 unless it would block all paths from $X$ to the second level from below, i.e., to the $2$-element sets. (So for $\le 1$ element sets we always answer 0.) -Such a game can end only by receiving a 1 answer on the second level for some $(a,b)$. But since this secures the path from $X$ to the second level, there is a path from $X$ to $(a,b)$. Without loss of generality, we can suppose that $a=1$, $b=1$ and the elements of the path are $(1,\ldots, i)$. Since $(1,2)$ was the last chance to reach the second level, every other path from each of $(1,\ldots, i)$ to the second level must be blocked. For $(1,\ldots, n)$, there are $n-1$ disjoint paths that end, respectively, in $(j,n)$; we need at least $n-1$ zeros to block these and they cannot block any other path from an $(1,\ldots, i)$ to the second level. Similarly, we need further $n-2$ zeros to block the paths from $(1,\ldots, n-1)$ to the second level etc, in total $n^2/2+\Theta(n)$, which matches the trivial upper bound. (Possibly with a little more care even the exact value can be determined.)<|endoftext|> -TITLE: Extension of Dickson's theorem on integers of the form $a^2+b^2+2c^2$ -QUESTION [11 upvotes]: Theorems V in this paper of L.E. Dickson states that the following two sets are equal. $$E=\{a^2+b^2+2c^2 \ | \ a,b,c \in \mathbb{Z}\} \ \text{ and } \ F=\mathbb{N} \setminus \{4^k(16n+14) \ | \ k,n \in \mathbb{N}\}.$$ -Let $\mathbb{A}$ denote $\mathbb{N}$ or $\mathbb{Z}$. Consider the following set: $$E(\mathbb{A}) = \left\{\frac{1}{2}\|u+v \|^2 \text{ with } u,v \in \mathbb{A}^3 \text{ and } \|u \| = \|v \| \right\}.$$ -Let $u= (a,b,c)$, $v = (b,-a,c) \in \mathbb{Z}^3$. Then $\|u \| = \|v \|$ and $\frac{1}{2}\|u+v \|^2 = a^2+b^2+2c^2.$ -It follows that $F= E \subseteq E(\mathbb{Z})$. Now, by Legendre's three-square theorem, $E(\mathbb{Z}) \subset F$ also. -Then, we have an extension of Dickson's theorem as $E(\mathbb{Z}) = F$. Now, what about $E(\mathbb{N})$? -Take $u=v \in \mathbb{N}^3$, then $\frac{1}{2}\|u+v \|^2 = 2 \|u \|^2$, so by Legendre's three-square theorem, $E(\mathbb{N})$ contains the even part $F$. The computation below shows that $E(\mathbb{N})$ contains every odd number less than $95362$, except those in $I=\{ 5, 23, 29, 65, 167 \}$, suggesting that $E(\mathbb{N}) = F \setminus I$. - -Question: Is it true that, for $u,v \in \mathbb{N}^3$ with $\|u \| = \|v \|$, the form $\frac{1}{2} \|u+v \|^2$ covers every odd - number, except those in $\{ 5, 23, 29, 65, 167 \}$? - -Application: this answer proves that the form $\| A\|^2$ covers every natural number for $A \in M_3(\mathbb{Z})$. - A positive answer to the above question would prove this result for $A \in M_3(\mathbb{N})$. - -For the convenience of the reader, the answer of Philipp Lampe (of what was Question 1 in a previous version) was incorporated in the post. - -Computation -sage: L=cover(135) -sage: set([2*i+1 for i in range(47681)])-set(L) -{5, 23, 29, 65, 167} - - -Code -# %attach SAGE/3by3.spyx - -from sage.all import * - -cpdef cover(int r): - cdef int a1,a2,a3,b1,b2,b3,x,n - cdef list L - L=[] - for a1 in range(r): - for a2 in range(a1+1): - for a3 in range(a2+1): - x=a1**2+a2**2+a3**2 - for b1 in range(isqrt(x)+1): - for b2 in range(isqrt(x-b1**2)+1): - for b3 in range(isqrt(x-b1**2-b2**2)+1): - if a1**2+a2**2+a3**2==b1**2+b2**2+b3**2: - n=((a1+b1)**2+(a2+b2)**2+(a3+b3)**2)/2 - if is_odd(n) and not n in L: - L.append(n) - return L - -REPLY [9 votes]: given your interest: the list of all $A x^2 + B y^2 + C z^2$ with ordered positive coefficients, such that the represented numbers can be described by congruences -Ummm. Allowing mixed terms, all 913 (probably) regular positive forms<|endoftext|> -TITLE: Questions on group and Nakayama algebras from a book -QUESTION [5 upvotes]: Recall that a Nakayama algebra (also called serial algebras in the literature) is an algebra such that every indecomposable module has a unique composition series. -In the book "Classical artinian rings and related topics" from 2009 by Yoshitomo Baba und Kiyoichi Oshiro one can find the following two questions at the end of the book: - - -Let G be a finite group and K be an algebraic closure of - a field k. If kG is a Nakayama algebra, is KG a Nakayama algebra? And - how is the converse? -For which algebraically closed fields $K$ and for which $n$ are the group algebras $KS_n$ and $KA_n$ Nakayama algebras, where $S_n$ is the symmetric and $A_n$ the alternating group? - - -(the original formulation for 2. is: "Let K be an algebraically closed field. Are there infinitely -many $KS_n$ or $KA_n$ which are non-semisimple Nakayama algebras?") -Is an answer to those problems known now? -edit: It would be interesting to know whether there are quick computer programs available to test whether $KS_n$ or $KA_n$ are Nakayama algebras for $K$ being the algebraic closure of a prime field and given $n$. - -REPLY [2 votes]: For question 2 and $S_n$, suppose $\text{char}(K)=p$ and $KS_n$ is a Nakayama algebra. Then $KS_n$ must have finite representation type, and so $S_n$ must have cyclic Sylow $p$-subgroups, which means $n<2p$ (and for $KS_n$ to be non-semisimple we must have $n\geq p$). -Then every non-simple block has cyclic defect, and it is well-known that cyclic defect blocks of symmetric groups have Brauer trees that are lines with $p-1$ edges and multiplicity one. But such a Brauer tree algebra is only a Nakayama algebra if the number of edges is at most two. -So $KS_n$ is a non-semisimple Nakayama algebra if and only if either - -$p=2$ and $2\leq n<4$, or -$p=3$ and $3\leq n<6$. - -(Probably there’s a similar answer for $KA_n$.)<|endoftext|> -TITLE: Counting spanning trees of a planar graph -QUESTION [7 upvotes]: I know through Kirchoff's Theorem, one can calculate the number of spanning trees via the determinant of a Laplacian. This has complexity $O(N^{2.373}$). I was wondering if anyone was aware of a method which computes this faster, at least for planar graphs. - -REPLY [9 votes]: The determinant of matrices whose support corresponds to incidence matrices of planar graphs (this includes the Laplacian matrix of a planar graph, or more precisely its cofactors) can be calculated in $O(n^{1.5})$ with the algorithm given in - -R.J. Lipton, D. Rose, R.E. Tarjan - Generalized nested dissection - SIAM J. Numer. Anal., 16 (1979), pp. 346-358 - -which uses the planar separator theorem (see also nested dissection). -Another algorithm that runs in $O(n^2)$ but sometimes more efficient to implement is the one based on delta-wye graph reduction (a set of combinatorial substitution rules on the graph) given in - -C.J. Colbourn, J.S. Provan, D. Vertigan A new approach to solving three combinatorial enumeration problems on planar graphs Discrete Appl. Math., 60 (1995), pp. 119-129<|endoftext|> -TITLE: Does every positive continuous function have a non-negative interpolating polynomial of every degree? -QUESTION [22 upvotes]: Let $f:[a,b] \to (0,\infty)$ be a continuous function. Then is it necessarily true that for every $n\ge 1$, we can find $n+1$ distinct points $\{x_0,x_1,...,x_n\}$ in $[a,b]$ such that the interpolating polynomial $p_n(x)$ of $f$ at those points is non-negative on $[a,b]$ i.e. $p_n(x) \ge 0,\forall x \in [a,b]$ ? -NOTE: $p_n(x)$ is the unique, degree at most $n$, polynomial such that $f(x_i)=p_n(x_i),\forall i=0,1,...,n$. - -REPLY [3 votes]: This question was recently studied in that paper: -F. Charles, M. Campos-Pinto, B. Després, Algorithms for positive polynomial approximation, hal-01527763, -assuming that the function $f$ is Lipschitz on the interval. Let $[a,b]=[0,1]$. The interpolating polynomials $p_{n}$ are seeked in the form -\begin{align*} -p_{n}(x) & =a_{p}^{2}(x)+x(1-x)b_{p-1}^{2}(x),\quad n=2p+1,\\ -p_{n}(x) & =xa_{p}^{2}(x)+(1-x)b_{p}^{2}(x),\quad n=2p, -\end{align*} -with $a_{p}$ and $b_{p}$ polynomials of degree $p$, and $b_{p-1}$ a polynomial of degree $p-1$, which are classical representations for non-negative polynomials in $[0,1]$. -The interpolation points are chosen to be $0$, $1$, and $n-1$ points in $(0,1)$ which are obtained through a converging fixed point algorithm for a map defined on $(0,1)^{n-1}$. -Some details when $n=2$ or $n=3$ : first, let $f(y)=g^{2}(y)$ with $g(y)>0$. - -when $n=2$, $p_{2}(x)$ is seeked in the form $a_{1}^{2}(x)+x(1-x)b_{0}^{2}$. One sets -$a_{1}(0)=g(0)>0$ and $a_{1}(1)=-g(1)<0$ so that there exists $\alpha\in(0,1)$ with $a_{1}(\alpha)=0$. It then suffices to choose $b_{0}$ such that $f(\alpha)=\alpha(1-\alpha)b_{0}^{2}$. -when $n=3$, $p_{3}(x)$ is seeked in the form $xa_{1}^{2}(x)+(1-x)b_{1}^{2}(x)$. One shows that there exists $0<\alpha<\beta<1$ such that -\begin{align} -a_{1}(1) & =g(1), \quad a_{1}(\alpha) =-g(\alpha)/\sqrt{\alpha}, \quad a_{1}(\beta) =0,\\ -b_{1}(0) & =-g(0), \quad b_{1}(\alpha) =0, \quad b_{1}(\beta) =g(\beta)/\sqrt{1-\beta}, -\end{align} -which is checked through direct calculations. - -The general case $n\geq4$ consists in generalizing the above conditions and showing that there exists $n-1$ interpolation points in $(0,1)$ that satisfy these conditions.<|endoftext|> -TITLE: The Precise Meaning of the Moduli Space of Flat Connections? -QUESTION [10 upvotes]: Questions: I would like to have a precise description of the meanings of the Moduli Space of Flat Connections, such that it is understandable by mathematical physicists and physicists. -For 3d Chern-Simons (CS) theory, I suppose that the following is an interpretation. - -the Moduli Space of Flat Connections of CS theory = the phase space of the classical Chern-Simons field theory $\equiv$ the classical phase space -the quantization of the classical phase space = the Hilbert space of ground states and zero modes of quantum Chern-Simons theory. -By quantization, we mean that replacing the Poisson bracket in the classical phase space $\{x, p\}$; by the commutator of matrix operators $[X, P]$. - -For 4d Yang-Mills (YM) theory, what would it be the Moduli Space of Flat Connections? - -YM flat connections are in the classical phase space? -YM non-flat connections are also in the classical phase space? -the Moduli Space of YM theory = the phase space of the classical YM field theory $\equiv$ the classical phase space of both flat connections and non-flat connections? -the Moduli Space of Flat Connections of YM theory = the classical phase space of only the flat connections part? -What will be the quantization of the Moduli Space of YM theory? -What will be the quantization of the Moduli Space of Flat Connections of YM theory? - -REPLY [6 votes]: Let $P \to M$ be a principal $G$-bundle. The moduli space of flat connections on $P$ is, by definition, the space $\mathcal{M} = \mathcal{C}_0 / \mathcal{G}$, where $\mathcal{C}_0$ denotes the subspace of flat connections on $P$ and $\mathcal{G}$ is the group of (local) gauge transformations. Whether $M$ is $3$ or $4$-dimensional does not make a difference (for the definition, the properties of $\mathcal{M}$ of course depend on the topology of $M$). -If you want to speak of a configuration or phase space, you need to split your equations into space and time direction (at least in the naive interpretation you need an evolution to have a meaningful notion of a phase space). So, for example, for 4-dimensional Yang-Mills you choose a splitting $M = \mathbb{R} \times \Sigma$ and decompose the YM-equations according to get the non-abelian analog of the Maxwell equations. The configuration space of the theory is then the space of $G$-connections over $\Sigma$ and phase space is the cotangent bundle.<|endoftext|> -TITLE: A set of prime numbers -QUESTION [11 upvotes]: Consider a non-empty set $S$ of primes, with the property that, for every finite subset $S'\subset S$, all the primes dividing $\left(\prod_{k\in S'}k\right)+1$ are in $S$. -For instance, it can easily be proven that $2\in S$ (if not, then the smallest member $q$ of $S$ is odd, hence, $q+1$ is even, and thus $2\in S$). In fact, with this way, $2+1=3\in S$, $2\cdot 3+1=7\in S$, $2\cdot 7+1=5\in S$, and so on. -It seems this set must contain all primes, but I could not prove it. -I tried to use that if $p_n$ is the smallest prime that is not contained in $S$, then $p_1,p_2,\dots,p_{n-1}\in S$. $p_n\sim n\log n$ for $n$ large, and if I could somehow show that there is enough residues that can be constructed using subset-products of $p_1,\dots,p_{n-1}$ this should have been handled, but how? Can anybody see a way to do it? - -REPLY [7 votes]: If I'm not mistaken, it is true that $S$ must contain all primes. -First of all it is obvious that $S$ is infinite -- indeed, as Euclid teach us, if $S$ is finite then $\left(\prod\limits_{p\in S} p\right) + 1$ is coprime to all primes in $S$ and therefore has at least one prime factor not in $S$. -Now let $p$ be a prime number. We will show that $p\in S$. Assume by contradiction that $p\notin S$. Let $R = \{r_1, \ldots , r_k\}$ be residues modulo $p$ that appear infinitely often as residues of primes from $S$ modulo $p$. Since $S$ is infinite $k \ge 1$. Now let $A = \{a_1, \ldots, a_n\}$ be all elements of subgroup of $(\mathbb{Z}/p\mathbb{Z})^*$, spanned by all $r_i$. Also denote by $b$ product of all other primes from $S$ whose residues modulo $p$ are not from $R$. -Note that we can form each $a_i$ as product of different primes from $S$ (and moreover we can use only primes congruent to numbers from $R$ modulo $p$). Then we can also have $ba_i$. -If some of $ba_i + 1$ is congruent to $0$ mod $p$ then we found $p$ and we are done. Thus we can assume that $ba_i + 1$ is never congruent to $0$. Also note that $ba_i$ never congruent to $0$ (since $p\notin S$ by assumption) and all $ba_i$ are pairwise distinct(since $p$ is prime). Therefore $ba_i + 1$ is never congruent to $1$. Finaly note that $1\in A$. From all this we can deduce that there is some $i$ such that $ba_i + 1\notin A$. Let $C$ be corresponding number. Let $q$ be any of its prime divisors. Note that $q$ can't be from exceptional set since we added all of them through $b$. Thus $q$ is congruent to some $r_i$. But then $ba_i + 1\in A$, as product of primes from $R$ -- a contradiction.<|endoftext|> -TITLE: Is there a version of Fischer-Riesz theorem for Banach space? -QUESTION [8 upvotes]: $( \Omega,F, P )$: a measurable space equipped with a finite measure -$(B , \Vert \cdot \Vert) $ : a Banach space with $\mathcal{B}$ as its borelian $\sigma$-algebra -$p$ : a constant bigger than $1$ -Define $L^p(\Omega, B )$ the vector space that contain all $( F, \mathcal{B})$-measurable function $f$ such that : -$ \vert \Vert f \Vert \vert = \sqrt[p]{ \int_{\Omega} \Vert f \Vert ^p \cdot dP } < \infty$ -I'm looking for a version of Riesz-Fischer theorem which affirms that: -Proposition: -$\left( L^p(\Omega, B ) , \vert \Vert \cdot \Vert \vert \right)$ is a Banach space -With some quick calculations, I have the feeling that this proposition is quite easy to be proved. But as we all know, it's always better to have a reliable reference. -So my question is: "Is the above proposition true? And does anyone have references to this matter?" - -REPLY [11 votes]: With the definitions in the OP, this is false. It is OK if the Banach space $B$ is separable and $(\Omega,\mathcal F, P)$ is an arbitrary probability space. It is OK if the Banach space $B$ is arbitrary and $(\Omega,\mathcal F,P)$ is a perfect measure space. But for arbitrary $B$ and $(\Omega,\mathcal F, P)$, it can fail. It can fail in many different ways. -(A theorem of Charles Stegall: if $(\Omega,\mathcal F,P)$ is a perfect probability space, $B$ is a metric space, and $f : \Omega \to B$ is $(\mathcal F, \mathcal B)$-measurable, then there is a set $\Omega_1 \subseteq \Omega$ of measure $1$ such that $f(\Omega_1)$ is separable.) -Here is the simplest way in which it may fail. Write $\mathcal B = \mathrm{Borel}(B)$. Let $L^p(\Omega,B)$ be the set of all functions $f : \Omega \to B$ such that $f$ is $(\mathcal F, \mathcal B)$-measurable, and -$$ -\int_\Omega \|f(\omega)\|^p\;dP(\omega) < \infty . -$$ -It is possible that there are $f,g \in L^p(\Omega,B)$ such that $f+g \notin L^p(\Omega,B)$ because $f+g$ is not even $(\mathcal F , \mathcal B)$-measurable. -Example I -Let $T$ be a discrete space with cardinal $\frak{a} > 2^{\aleph_0}$. Let $B = l^2(T)$, that is, a Hilbert space with orthonormal basis of cardinal $\frak{a}$. For each $t \in T$ let $e_t \in l^2(T)$ be defined by: $e_t(t) = 1$ and $e_t(s) = 0$ if $t\ne s$. This system of "unit vectors" is an orthonormal basis of the space $B$. -Let $\Omega = T \times T$ be the Cartesian square. Let $\mathrm{Borel}(T)$ be the Borel sigma-algebra on $T$, which is of course the power set of $T$. -Let the sigma-algebra $\mathcal{F} = \mathrm{Borel}(T) \otimes \mathrm{Borel}(T)$, the product sigma-algebra. The reason for requiring that $\mathrm{card}(T) > 2^{\aleph_0}$ is so that the diagonal -$$ -\Delta := \{(t,t) \in \Omega : t \in T\}, -$$ -although closed, is not in $\mathcal F$. See HERE. -We do not care what the probability measure $P$ is. (In an extreme case it could even be the point mass at a single point.) -Finally we are ready. Define $f : \Omega \to B$ by -$$ -f\big((u,v)\big) = e_u, -$$ -That is: Given $\omega = (u,v)$ in $\Omega$, we take its first component, and use the corresponding unit vector. Similarly, define $g : \Omega \to B$ by -$$ -g\big((u,v)\big) = -e_v, -$$ -using the second component and a minus sign. -I claim that $f, g \in L^p(\Omega,B)$ but $f+g$ is not. -First: $f$ is $(\mathcal F, \mathcal B )$-measurable. Indeed, if -$Q \in B$ is Borel, then $f^{-1}(Q) \in \mathcal F$ because -$f^{-1}(Q) = \widetilde{Q} \times T \in \mathcal F$ where -$\widetilde{Q} = \{t \in T : e_t \in Q\}$. -So $f$ is $(\mathcal F, \mathcal B )$-measurable. Similarly -$g$ is $(\mathcal F, \mathcal B )$-measurable. -Next, -$$ -\int_\Omega \|f(\omega)\|^p\,dP(\omega) = 1 < \infty. -$$ -(Regardless of what the probability measure $P$ is, the integral of the constant $1$ is $1$.) -So $f \in L_p(\Omega,B)$. Similarly, $g \in L_p(\Omega,B)$. -Now we claim the sum $f+g$ is not measurable. Indeed, even more, we claim that $\{\omega\in \Omega : f(\omega)+g(\omega) = 0\} \notin\mathcal F$. (Since $\{0\}$ is closed, this shows $f+g$ is not measurable.) Indeed, -$$ -\{\omega : f(\omega) + g(\omega) = 0\} = -\{(u,v) : e_u-e_v = 0\} = -\{(u,v) : u=v\} = \Delta. -$$ -As noted above, $\Delta \notin \mathcal F$ -End of Example I<|endoftext|> -TITLE: Do we need the Weber function to generate ray class fields of imaginary quadratic fields of class number one? -QUESTION [8 upvotes]: I'm a bit confused by the role of the Weber function in generating ray class fields of imaginary quadratic fields of class number one. More specifically, let $K$ be such a field and $E$ an elliptic curve defined over $\mathbf{Q}$ with CM by $\mathscr{O}_K$. Let $\mathfrak{m}$ be a modulus for $K$. To get the ray class field for $\mathfrak{m}$, we consider the torsion group $E[\mathfrak{m}]$. This is a free $\mathscr{O}_K/\mathfrak{m}$-module of rank $1$, and $G_K$ acts on it by $\mathscr{O}_K$-linear automorphisms. Thus we have a character $\alpha \colon \mathrm{Gal}(K(E[\mathfrak{m}])/K) \hookrightarrow \mathscr{O}_K^\times$. Moreover, the Main Theorem of CM tells us that $\alpha(\mathrm{rec}(s))$ acts by $\lambda(s) s_\mathfrak{m}^{-1}$ for $\lambda \colon \mathbf{A}_{K, f}^\times \rightarrow K^\times$ the CM Großencharacter. Note that if $s \equiv 1 \pmod {\mathfrak{m}}$, this tells us that $\alpha(\mathrm{rec}(s)) = \lambda(s) \pmod {\mathfrak{m}}$, and that if $s = (\gamma)$ is a principal idele, that $\lambda(s) = s_\mathfrak{m} \pmod{\mathfrak{m}}$. -Now, at least if $\mathfrak{m}$ is divisible by the conductor of $\lambda$ (i.e. $\mathfrak{m}$ is divisible by the primes of bad reduction of $E$), we see that $\alpha(\mathrm{rec}(s))$ only depends on the image of $s$ in $\mathrm{Cl}_\mathfrak{m}(K)$. Thus, we have a map $\alpha \circ \mathrm{rec} \colon \mathrm{Cl}_{\mathfrak{m}}(K) \rightarrow \mathrm{Gal}(K(E[\mathfrak{m}])/K) \hookrightarrow (\mathscr{O}_K/\mathfrak{m})^\times$. -If we compose $\alpha \circ \mathrm{rec}$ with the map $(\mathscr{O}_K/\mathfrak{m})^\times \rightarrow (\mathscr{O}_K/\mathfrak{m})^\times/\mathscr{O}^\times = \mathrm{Cl}_\mathfrak{m}(K)$, we certainly get an isomorphism, since we can see that $\alpha \circ \mathrm{rec}(\pi_v) = \lambda(\pi_v)$ is a generator of $v$ for all but finitely many primes $v$ of $K$ which are split over $\mathbf{Q}$. -In the sources I've read on this, it seems that one replaces $\alpha \colon \mathrm{Gal}(K(E[\mathfrak{m}])/K) \hookrightarrow \mathscr{O}_K^\times$ with $\mathrm{Gal}(K(h(E[\mathfrak{m}]))/K) \hookrightarrow (\mathscr{O}_K/\mathfrak{m})^\times/\mathscr{O}_K^\times$ where $h$ is a Weber function. This shows that $K(h(E[\mathfrak{m}]))$ is the ray class field of $K$ with conductor $\mathfrak{m}$. One sometimes comments that the field $K(E[\mathfrak{m}])$ might not be abelian over $K$ in general, since $E$ is only defined over the Hilbert class field of $K$. But in the class number one case, this problem goes away. -So what happens to $\alpha$? We've shown that when $\mathfrak{m}$ is divisible by the conductor of $\lambda$, $K(E[\mathfrak{m}])$ is an abelian extension of $K$ such that the reciprocity map kills ideals with generators which are congruent to $1$ mod $\mathfrak{m}$, so it must be contained in the ray class field with conductor $\mathfrak{m}$, i.e. the subfield $K(h(E[\mathfrak{m}])$. Thus, these are the same field. Is $\alpha$ surjective, or is the image some index $2$ (or $4$ or $6$ for the cases of extra automorphisms, I suppose) subgroup of $(\mathscr{O}_K/\mathfrak{m})^\times$ mapping isomorphically onto the ray class group? -If $\mathfrak{m}$ is not divisible by the conductor of $\lambda$, $K(h(E[\mathfrak{m}]))$ is still the ray class field of conductor $\mathfrak{m}$, but I don't think my proof shows the reciprocity map into $\mathrm{Gal}(K(E[\mathfrak{m}])/K)$ kills principal ideals with a generator which is congruent to $1$ mod $\mathfrak{m}$. Are these still the same field? If not, $K(E[\mathfrak{m}])$ is some abelian extension of $K$, so it must sit inside $K(h(E[\mathfrak{n}]))$ for some $\mathfrak{n}$ - which one? - -REPLY [2 votes]: A quick comment (which I have to post as an answer, since I was locked out of my old account): The construction of $K(j(E), E[\mathfrak{m}])$ you describe in the general gives abelian extensions of the Hilbert class field $K(j(E))$ which might not be abelian over $K$. The role of the Weber function $h$ is to select out the subfields which are abelian over $K$. Geometrically, fixing a model for $E$, this $h$ can be defined as any finite map $h: E \longrightarrow E/\operatorname{Aut}(E)\approx E/\mathcal{O}_K^{\times}$, and does not depend on the choice of model. (Analytically, it is defined as $z \in {\bf{C}}/\Lambda \approx E({\bf{C}}) \longrightarrow (\wp(z, \Lambda), \wp'(z, \Lambda))$, and does not depend on the choice of lattice $\Lambda$). Hence, the way to construct the ray class field of conductor $\mathfrak{m}$ of $K$ in general is to adjoin to the coordinates of the images in $X := E/\operatorname{Aut}(E)$ of the torsion points $E[\mathfrak{m}]$. If $O_K^{\times}$ has order 2, then the homomorphism $E \rightarrow X$ is given by the coordinate $x$; if the order is 4 it is given by $x^2,$ and if the order is 6 then it is given $x^3$. In other words, the ray class extension of conductor $\mathfrak{m}$ of $K$ is generated by $j(E)$ and the coordinates $x$, $x^2$, or $x^3$ respectively, which to be explicit is $K(j(E), h(E[\mathfrak{m}]))$ rather than $K(j(E), E[\mathfrak{m}])$. However, when the class number of $K$ is one, then it is usually $K(E[\mathfrak{m}])$ as you suggest, but could be a larger abelian extension of $K$. Sorry that this comment does not answer the original question about the exact location of $K(E[\mathfrak{m}])$. (This should come down to keeping track of action by elements of $\operatorname{Aut}(E) = \mathcal{O}_K^{\times}$ on torsion points in the proof of the main theorem). If I have time, then I will try to post an answer later.<|endoftext|> -TITLE: Convergence of Newton's method -QUESTION [14 upvotes]: For a polynomial $P$ of degree $n$ with real coefficients and with $n$ distinct real roots, the Newton's method $z_{n+1} = z_n - {P(z_n) \over P'(z_n)}$ converges for almost all initial values $z_0$ in $\mathbb R$ (or almost all $z_0$ in $\bf C$ with respect to the area measure) to a root of $P$. This is a result due to M. Lyubich (~ 1984). -I think I remember that for a polynomial with complex coefficients, almost all initial values $z_0$ has an orbit that converges to a periodic orbit in ${\mathbb C} \cup \{\infty\}$, but there are examples where that orbit is not a root of $P$. -Unfortunately, I can't remember who is the author of that result and I would like to find a reference. -EDIT: the result is actually false. There are polynomials whose Newton's method has a periodic Siegel disk, see e.g. this answer. In that case, there is an open set of points whose orbit's $\omega$-limit set is a circle. - -REPLY [7 votes]: Your statement that iterates of the Newton method converge to a cycle almost everywhere is equivalent to the statement that for every polynomial $f$ -the Julia set of the rational function $z-f(z)/f'(z)$ has zero area. -This is unlikely to be true, but I do not know a published counterexample. -For the state of the art on Newton Method for polynomials, I recommend these papers: -MR1859017 -J. Hubbard, D. Schleicher, S. Sutherland, -How to find all roots of complex polynomials by Newton's method. -Invent. Math. 146 (2001), no. 1, 1–33. -MR3659421 D. Schleicher, R. Stoll, Newton's method in practice: Finding all roots of polynomials of degree one million efficiently. Theoret. Comput. Sci. 681 (2017), 146–166.<|endoftext|> -TITLE: Reference request: Oldest calculus, real analysis books with exercises? -QUESTION [13 upvotes]: Per the title, what are some of the oldest calculus, real analysis books out there with exercises? Maybe there are some hidden gems from before the 20th century out there. -Edit. Unsolved exercises are fine. Same with solved. - -REPLY [4 votes]: Maria Agnesi is perhaps best (only?) known for the "witch of Agnesi" today, but her calculus textbook has much more than that to recommend it. I'm not sure if it contains "exercises" in your sense of the word, though.<|endoftext|> -TITLE: Randomly picking $k$ members of $\{1,\ldots,n\}$ -QUESTION [6 upvotes]: Every day, I randomly pick a sample consisting of $k$ members of $\{1,\ldots,n\}$ where $k\leq n$. I stop as soon as every number of $\{1,\ldots,n\}$ has been picked at least once. Let $S$ be the number of days needed to reach my goal. What is the expected value of $S$? - -REPLY [4 votes]: I think you should have already a good estimate just comparing with the classical coupon colector's problem (CCP). -Let us consider the CCP and pick the numbers one by one. We define the following stopping times -$T_i$ = The first $t\geq T_{i-1}$ such that there are $k$ different numbers picked between $T_{i-1}$ and $t$ -Then the set of number picked at $T_i$ is the same as the set picked in your process after $i$ days. Moreover, the random variable $X_i=T_{i+1}-T_i$ are iid random variables (of same law $X$). And we have $$ t=\sum_{i:T_i\leq t} X_i+(t-T_i)$$ -If we call $T$ the time the classical coupon colector has picked all its numbers, we have $$T=\sum_{i:T_i\leq T} X_i+(T-T_i)$$ In your process, $S$ is the first $i$ such that $T_i\geq T$. Therefore -$$\mathbb{E}(T)=\mathbb{E}(\sum_{i -TITLE: Do eigenfunctions determine the geometry of a manifold? If so, do finitely many suffice? -QUESTION [19 upvotes]: Let $X$ be a smooth, Riemannian manifold. It is known that the geometry of $X$ can be recovered from its heat kernel $k_{t}(x,y)$, using Varadhan's Lemma: $\displaystyle\lim_{t \to 0} t \log k_{t}(x,y) = -\frac{1}{4}d^{2}(x,y)$. Since the heat kernel $k_{t}(x,y)$ can be expressed in terms of eigenfunctions and eigenvalues, -$$k_{t}(x,y) = \sum_{i=0}^{\infty}e^{-\lambda_{i}t}\phi_{i}(x)\phi_{i}(y)$$ -we can say that the knowledge of the eigenvalues and eigenfunctions of $X$ determines its geometry. -Now, the following result of Bates: https://arxiv.org/pdf/1605.01643.pdf tells us that there is a constant $d(X)$, depending on the dimension and geometry of $X$, such that the map $X \to \mathbb{R}^{d(X)}$ sending $x \in X$ to $\langle \phi_{0}(x), \cdots, \phi_{d(X)}(x)\rangle$ is injective. -My question therefore is: we see that a map to Euclidean space using finitely many eigenfunctions is enough to recover the topological type of $X$. Can we also recover its metric (up to some simple transformation)? What if we used infinitely many eigenfunctions? In either case we do not have access to the eigenvalues. - -REPLY [3 votes]: Here is a sketch of an idea of how to show that the set $\mathcal{E}(g)\subset C^\infty(M)$ of all the eigenfunctions of the metric $g$ on a compact manifold $M$ determines $g$ up to a constant multiple. (Note that I do not assume that the corresponding eigenvalues are given, which would be easier.) Clearly, this is the best one can hope for, since for any constant $c>0$, $\mathcal{E}(cg) = \mathcal{E}(g)$. In fact, I think it's very likely that knowing a sufficiently 'large' finite subset of $\mathcal{E}(g)$ should be sufficient, but that remains to be seen. -The basic idea is this: Let $J^k(M)\to M$ denote the vector bundle consisting of the $k$-jets of smooth functions on $M$. When $M$ has dimension $n$, the bundle $J^k$ has rank ${n+k}\choose n$ over $M$. Given $g$, it is easy to show that there is a closed quadratic cone bundle $Q(g)\subset J^3(M)$ of codimension $n$ such that the $3$-jets of all the local eigenfunctions of $g$ lie in $Q(g)$. In fact, the $3$-jets of local eigenfunctions fill out $Q(g)$. -Note that $Q(g)$ is not a linear subbundle of $J^3(M)$ precisely because we are not specifying the eigenvalues of the eigenfunctions. Of course, the $3$-jets of local eigenfunctions with eigenvalue $\lambda$, fill out a linear subbundle $Q(g,\lambda)\subset J^3$, but the union of the $Q(g,\lambda)$ as $\lambda$ varies is not a linear subbundle. However, it is easy to show that, for any $x\in M$, the subset $Q_x(g)\subset J^3_x$ is the zero locus of an ideal generated by $n$ polynomials homogeneous of degree $2$ on the vector space $J^3_x$. ($Q_x(g)$ is not a smooth variety in $J^3_x$ but the singular locus is quite small.) -I claim that the subbundle $Q(g)\subset J^3$ determines $g$ up to a constant factor (assuming that $M$ is connected). Here is why: Let $Q^0(g)\subset Q(g)$ denote the subset consisting of those $3$-jets in $Q(g)$ whose $0$-jet vanishes. The projection $\pi^2_3(Q^0(g)_x)$ of $Q^0(g)_x$ into $J^2_x$ has codimension $n{+}1$ in $J^2_x$, cut out by the linear equation that says that the $0$-jet vanishes and the $n$ quadratic equations that then turn out to all be multiples of a single linear equation on the space of $2$-jets whose $0$-jet vanishes, as is easy to verify in local coordinates. -Consequently, it follows that there exist second order, elliptic differential operator of the form -$$ -L u = a^{ij}\,\frac{\partial^2u}{\partial x^i\,\partial x^j} + b^i\,\frac{\partial u}{\partial x^i} + c\,u -$$ -(with $(a^{ij})$ positive definite) such that every local eigenfunction of $g$ satisfies an equation of the form $Lu = \phi(u)u$ where $\phi(u)$ is a smooth function (that could depend on $u$), and, moreover, $L$ is unique up to scalar multiplication and the addition of a $0$-th order term. -Finally, the requirement that there exist a function $f>0$ such that $fL$ can be expressed in the divergence form -$$ -(fL)(u) = |h|^{-1/2}\frac{\partial}{\partial x^i}\left(|h|^{1/2}h^{ij}\,\frac{\partial u}{\partial x^j}\right) -$$ -is easily seen to imply that $h = cg$ for some constant $c\not=0$. Thus, $g$ can be recovered, up to a constant multiple, from knowledge of the subbundle $Q(g)\subset J^3$, as claimed. -Finally, what one would expect is that, when $M$ is compact, if we now look at the $3$-jets of the elements of $\mathcal{E}(g)$, i.e., the global eigenfunctions of $g$, that a sufficiently large subset will determine sufficiently many points in each $J^3_x$ that they will determine $Q(g)_x$, which, after all, is known to be cut out by $n$ homogeneous quadratic polynomials for each $x$. (Of course, the number of elements of $\mathcal{E}(g)$ needed to do this at each point could be large, even for $n=2$, but it will be finite.) Assuming such a 'density' result, $\mathcal{E}(g)$ will determine $Q(g)\subset J^3$, which, as we have seen, will determine $g$ up to a constant multiple.<|endoftext|> -TITLE: Definable functions in $o$-minimal structures -QUESTION [8 upvotes]: Consider the field of real numbers $(\mathbb R,+,\cdot)$. An expansion of $(\mathbb R,+,\cdot)$ is a tuple $(\mathbb R,+,\cdot,S)$, where $S$ is a collection of subsets of $\mathbb R^n$ for $n \in \mathbb N$. A subset of $\mathbb R^n$ is said to be definable in $(\mathbb R,+,\cdot,S)$, when it is definable by a first order formula in $(\mathbb R,+,\cdot)$ that is allowed to refer to sets in $S$. -A field expansion is said to be $o$-minimal if it allows to define exactly the same subsets of $\mathbb R$. It is a famous theorem of Wilkie (A. J. Wilkie, -Model completeness results for expansions of the ordered field of real numbers by restricted Pfaffian functions and the exponential function. J. Amer. Math. Soc. 9 (1996), no. 4, 1051–1094.) that $(\mathbb R,+,\cdot,G(\exp))$, where $$G(\exp) = \{(x,\exp(x)) \in \mathbb R^2 \mid x \in \mathbb R\},$$ is an $o$-minimal expansion of $(\mathbb R,+,\cdot)$. -A function $f \colon \mathbb R \to \mathbb R$ is definable if and only if its graph is a definable subset of $\mathbb R^2$. Clearly, $\exp \colon \mathbb R \to \mathbb R$ is definable in $(\mathbb R,+,\cdot,G(\exp))$ but not in $(\mathbb R,+,\cdot)$. -There is a remarkable dichotomy proved by Miller (Chris Miller, Exponentiation is hard to avoid. Proc. Amer. Math. Soc. 122 (1994), no. 1, 257–259.): - - -Theorem: Let $(\mathbb R,+,\cdot,S)$ be an $o$-minimal expansion of $(\mathbb R,+,\cdot)$. Then either - -the function $\exp \colon \mathbb R \to \mathbb R$ is definable, or -all definable functions are polynomially bounded. - - - -Now, consider $(\mathbb R,+,\cdot,G(\exp\circ \exp))$. This is clearly $o$-minimal, since $G(\exp\circ\exp)$ is definable in $(\mathbb R,+,\cdot,G(\exp))$. According to the theorem, $\exp \colon \mathbb R \to \mathbb R$ should now be definable! - - -Question: How do you define $\exp \colon \mathbb R \to \mathbb R$ in $(\mathbb R,+,\cdot,G(\exp\circ \exp))$? - - -There must be some first order formula in $(\mathbb R,+,\cdot)$ making reference to the set $G(\exp \circ \exp)$ that defines the function $\exp \colon \mathbb R \to \mathbb R$. Is it possible to write it down? -More generally, one might then also wonder if some (maybe unique) function $f \colon \mathbb R \to \mathbb R$ is definable in $(\mathbb R,+,\cdot,G(\exp))$ such that $f \circ f = \exp$. Maybe an answer to that is hidden in Kneser's construction of an analytic solution to this problem in (Hellmuth Kneser, Reelle analytische Lösungen der Gleichung $f(f(x))=e^x$ und verwandter Funktionalgleichungen. Journal für die reine und angewandte Mathematik (1950) Volume: 187, page 56-67.) - -REPLY [8 votes]: The derivative of $e^{e^x}$ is $e^x e^{e^x}$. So -$$y=e^x\ \leftrightarrow\ y = \lim_{h\rightarrow 0} \dfrac{e^{e^{x+h}} - e^{e^x}}{h e^{e^x}}$$ -which is a first-order statement.<|endoftext|> -TITLE: Gauss - Dirichlet class number formula -QUESTION [8 upvotes]: Let $p=8k+3$ be a prime. Then the class number of the imaginary quadratic field $\mathbb Q(\sqrt{-p})$ is given by -$$h(-p)=\frac 13\sum_{k=1}^{\frac{p-1}{2}}\left(\frac kp \right).$$ While this is certainly a very elegant and compact formula, does it have any nontrivial applications or consequences? (Note that I am not asking about the Dirichlet class number formula relating the value of the corresponding $L$-series at $1$ with the class number, but rather about its particular corollary). - -REPLY [20 votes]: Let $p=2n+1$ be a prime with $p\equiv3\pmod4$. By Wilson's theorem, $ (n!)^2\equiv1\pmod p$ and hence $n!\equiv\pm1\pmod p$. L. J. Mordell [Amer. Math. Monthly 68(1961), 145-146] used Dirichlet's class number formula $$\left(2-\left(\frac 2p\right)\right) h(-p)=\sum_{k=1}^n\left(\frac kp\right)$$ to deduce in few lines that $n!\equiv(-1)^{(h(-p)+1)/2}\pmod p$ if $p>3$. This is a nice application of Dirichlet's class number formula.<|endoftext|> -TITLE: The distribution of fractional parts $\Big\{ \frac{N}{n} \Big\}$ -QUESTION [6 upvotes]: Let $N$ be a large integer. What is known about distribution of fractional parts $\Big\{ \frac{N}{n} \Big\} \in [0,1)$ after division of $N$ by all odd numbers $n$ in the range $3 \leq n < \sqrt{N}$? - -REPLY [3 votes]: As $N \to \infty$, the set of fractional parts -$$ -\Big\{ \frac{N}{n}\Big\}, \quad 1 \leq n < \sqrt{N} -$$ -becomes asymptotically equidistributed in the Lebesgue measure of $[0,1]$. The same persists with the congruence restriction $n \equiv 1 \mod{2}$ asked by the OP, or indeed with $n$ running through any fixed arithmetic progression. The same equidistribution answer holds moreover in some other natural variants, such as taking $n$ to run through the primes of a fixed primitive arithmetic progression. Regarding this last point I happen to recall de la Vallee Poussin being amused in an 1898 paper by his observation (propriete assez curieuse; un rapprochement tres remarquable) that the mean value $1-\gamma$ of fractional parts $\Big\{\frac{N}{an+b} \Big\}$, $an + b \leq N$ was the same for all arithmetic progressions $an+b$, as well as for the prime values $an+b$ when $(a,b) = 1$. (Of course, $ \sum_{n \leq N} \frac{\Lambda(n)}{n} - N^{-1}\log{N!} = \frac{1}{N}\sum_{n \leq N} \Big\{\frac{N}{n}\Big\} \Lambda(n) \sim 1-\gamma$ is just the logarithmic form $\sum_{n \leq N} \frac{\Lambda(n)}{n} = \log{N} - \gamma +o(1)$ of the prime number theorem, easily equivalent to the usual forms $\psi(N) \sim N$ and $\pi(N) \sim N / \log{N}$.) -Why equidistribution? For simplicity of notation, let me omit the congruence condition on $n$; in practice these are handled straightforwardly as in de la Vallee Poussin's paper, in every situation where we can solve the problem without a congruence condition. It is enough to prove the moment asymptotic -$$ -\frac{1}{\sqrt{N}} \sum_{n < \sqrt{N}} \Big\{ \frac{N}{n} \Big\}^k \sim \frac{1}{k+1} = \int_0^1 t^k \, dt, \quad k = 1,2, \ldots, -$$ -or what amounts to the same, the estimate -$$ -\sum_{n < \sqrt{N}} B_k\Big(\Big\{ \frac{N}{n} \Big\} \Big) = o(\sqrt{N}) -$$ -for the sum of values of the periodized Bernoulli function $B_k(\{t\})$, for each $k = 1, 2, \ldots$. This is now a more natural form of posing the problem, for there is a conjecture of S. Chowla and H. Walum ("On the divisor problem," Proc. Symp. Pure Math., vol. VIII, 1965, pp. 138-143) that asserts the much stronger and best-possible ("square root cancellation") bound -$$ -\sum_{n < \sqrt{N}} B_k\Big(\Big\{ \frac{N}{n} \Big\} \Big) \ll_{k,\epsilon} N^{\frac{1}{4} + \epsilon}, -$$ -for any fixed $k \in \mathbb{N}$ and $\epsilon > 0$. This is one way to generalize the celebrated Dirichlet divisor conjecture, which is the case $k = 1$ with $B_1(\{t\}) = \{t\} - 1/2$. The Chowla-Walum conjecture thus implies, moreover, a quantitative estimate on the discrepancy (rate of equidistribution) of our fractional parts. It is discussed, for example, in this 1982 paper of R.A. MacLeod. -I have not given you the proof of the equidistribution, but I hope this can convince you of the answer to your question and how you may approach it. I would guess that Voronoi's method for the Dirichlet divisor problem could be adapted to prove the crude $o(\sqrt{N})$ bound needed for the qualitative equidistribution. A more precise study of the distribution of fractional parts $\{N / n\}$, $n \leq y$ is done in these papers of Saffari and Vaughan, but the proved results are very far from the discrepancy prediction that the Chowla-Walum conjecture implies on your question.<|endoftext|> -TITLE: Nielsen-Schreier with operations -QUESTION [5 upvotes]: The Nielsen-Schreier theorem states that subgroups of a free subgroup are free. -Is this hold also for groups with operations? -Explicitly, let $G$ be a fixed group. Let $F$ be a group with $G$-action which is free (as a group with $G$-action). Let $F'\subset F$ be a subgroup closed by the $G$-action. Then, must $F$ be free as a group with $G$-action? - -REPLY [9 votes]: Let $G$ be a two element group; the free group on two generators $x,y$ with the action of $G$ interchanging them is a free $G$-group (on one generator). Its subgroup generated by $xy^{-1}$ is closed under the $G$-action but is not a free $G$-group.<|endoftext|> -TITLE: Moduli space of flat connections over a Riemann surface -QUESTION [7 upvotes]: If I understand correctly, in the Refs below: -We can see that the moduli space of SU($N$) flat connections over a torus, is equivalent to a complex projective space $\mathbb{P}^{N-1}$ -Namely, -$$ -M_{\rm flat}=\mathbb{E} / {S}_N = \mathbb{P}^{N-1} -$$ -where $\mathbb{E}$ is given by -$$ -\mathbb E := \left\{ (\phi_1,\cdots, \phi_N) \equiv {(\mathbb T^2)}^N; \text{ subject to a constrain }\sum_i \phi_i=0 \right\} . -$$ -while the ${S}_N$ is the symmetric group usually denoted as $S_N$ (the Weyl group of SU$(N)$). -My questions - - -what is the moduli space of U(1) flat connections over a genus $g$-Riemann surface? - -what is the moduli space of SU(N) flat connections over a genus $g$-Riemann surface? - - - -Stable and Unitary Vector Bundles on a Compact Riemann Surface, M. S. Narasimhan and C. S. Seshadri, Annals of Mathematics, Second Series, Vol. 82, No. 3 (Nov., 1965), pp. 540-567 -Thank you for the kind comments and helps! - -REPLY [12 votes]: Both of these moduli spaces are discussed in the survey "Flat connections on oriented 2-manifolds" by Lisa Jeffrey. The theme of the first section of the paper is roughly as follows. An oriented 2-manifold can be viewed topologically (i.e. up to homeomorphism), as a smooth surface (up to diffeomorphism), or holomorphically (as a Riemann surface with a complex structure). As it happens, moduli spaces of flat connections can also be given topological, smooth, and holomorphic descriptions as well. For instance, the moduli space of flat U(1)-connections (defined for smooth surfaces) on a surface of genus $g$ can be viewed topologically as $U(1)^{2g}$ or in the holomorphic setting as the Jacobian of a Riemann surface (viewed as an algebraic curve). This is described in Section 1.2 of Jeffrey's paper with some more discussion in Section 2. Similar descriptions can be given for SU(N) (Section 1.3). -"What is such-and-such a space" is a rather vague question; the answer could be kind of tautological (e.g. I could just give you another name for the space without telling you anything new), unless you specify what exactly you want to know. If you asked "how can we study the topology of these moduli spaces", then sections 3 and 4 of Jeffrey's paper give a brief introduction to some techniques for extracting topological information (computing intersection numbers of cohomlogy classes). There's a lot more information out there on this question in the literature on representation varieties. See e.g. this MO question "Cohomology of representation varieties".<|endoftext|> -TITLE: Random pairs of commuting permutations -QUESTION [5 upvotes]: Let $\Omega_n \subseteq \mathrm{Sym}(n)^4$ be the set of all $4$-tuples $(\sigma_1,\sigma_2,\tau_1,\tau_2)$ of permutations of $\{1,\ldots,n\}$ such that $\sigma_j \tau_k = \tau_k \sigma_j$ for each pair $(j,k) \in \{1,2\}^2$. -Any four permutations determine a $8$-regular edge-labeled directed graph on $\{1,\ldots,n\}$. (Of course, the graph might not be simple.) I am interested in understanding the local structure of this graph when the four permutations are chosen uniformly at random from $\Omega_n$. In particular, I would like to know whether the commutativity constraint enforces any other kind of constraint on the structure of the graph, in the sense that a significant number of short cycles other than commutators need to occur. -A more precise formulation of the problem is as follows. Let $\mathbb{F}_r$ be the free group on $r$ generators and let $G = \mathbb{F}_2 \times \mathbb{F}_2$. Fixing an indexation for the generators of $G$, each element of $\Omega_n$ determines a unique homomorphism from $G$ to $\mathrm{Sym}(n)$. We identify a uniform random element of $\Omega_n$ with the associated random homomorphism, to be denoted $\phi$. Consider the following questions. -(1) Is it the case that for every nontrivial element $g \in G$ and every $\epsilon > 0$ the probability that $\frac{1}{n}|\mathrm{Fix}(\phi(g))| \leq \epsilon$ tends to $1$ as $n \to \infty$? I emphasize that $g$ and $\epsilon$ are fixed before the limit. -(2) Is it the case that for every nontrivial element $g \in G$ the expectation of the random variable $\frac{1}{n}|\mathrm{Fix}(\phi(g))|$ tends to $0$ as $n \to \infty$? -Clearly a positive answer to Question (1) implies a positive answer to Question (2). A positive answer to Question (1) is exactly the assertion that the uniform measures on $\Omega_n$ form a random sofic approximation to $G$. Intuitively, this would mean that a uniform random element of $\Omega_n$ has no 'unnecessary' structure. -Note that residual finiteness of $G$ implies that there exists a sequence $(\phi_n:G \to \mathrm{Sym}(n))_{n=1}^\infty$ of homomorphisms such that for each $g \in G$ the set $\mathrm{Fix}(\phi_n(g))$ is empty for sufficiently large $n$. -A final (vague) question is whether this model can be generated in any way other than its definition. The model given by a uniformly chosen $4$-tuple of permutations with no additional constraints is contiguous with a uniform random $8$-regular graph, and there are multiple ways of generating this. In this case the elementary theory shows that the analog of Question (1) with $G$ replaced by $\mathbb{F}_4$ has a positive answer. - -REPLY [10 votes]: This is not the case, in fact in both cases the limit is 1/2. This is because with probability going to 1 as $n \to +\infty$ a uniformly random morphism $\mathbb F_2 \times \mathbb F_2 \to \mathrm{Sym}(n)$ is trivial on one of the factors. The latter claim follows from Dixon's theorem that a random pair of independently chosen permutations almost surely generates a primitive group, in fact the full alternating or symmetric group so it has a trivial commutant, so for large $n$ a typical representation of $\mathbb F_2 \times F_2$ to $\mathrm{Sym}(n)$ has one factor acting with primitive image and the other trivially. -In terms of graphs this means that the graphs you defined converge to the mean of the Schreier coset graphs of $1 \times \mathbb F_2$ and $\mathbb F_2 \times 1$. -EDIT As noted in the comments the above does not contain an actual proof of its statements. To provide such a proof one needs to look onto Dixon's arguments a bit. The reference for Dixon's paper, which I will use below, is : The Probability of Generating the Symmetric Group, Math. Zeitschrift 110 (1969). -The first claim is that there are at most $(n!)^2(1 + O(C^{-n}))$ representations of $\mathbb F_2\times \mathbb F_2$ such that the first factor acts transitively. To prove this note that if it acts primitively then the image of the second factor has to act trivially. On the other hand even if it acts nonprimitively (but still transitively) its centraliser in $\mathrm{Sym}(n)$ is of cardinality at most $n$ (see eg. Lemma A.2 in https://arxiv.org/pdf/1805.03893.pdf). In the proof of his Lemma 2 Dixon shows that the number of such imprimitive representations is at most $n2^{-n/4}(n!)^2$. So we get in fine that the number of representations of $\mathbb F_2 \times \mathbb F_2$ where $\mathbb F_2 \times 1$ acts transitively is at most equal to -$$ -(1 + n^3 2^{-n/4})(n!)^2 -$$ -which is what we wanted. -The second claim is that the number of representations where $\mathbb F_2 \times \mathbb F_2$ acts transitively, but not $\mathbb F_2 \times 1$ is $O(C^{-n}(n!)^2)$. This is in fact simpler : the number of such representations where the orbits of the first factor are of size $2 \le d \le n/2$ is at most equal to -$$ -\frac{n!}{(n/d)!(d!)^{n/d}} \cdot (d!)^{2n/d} \cdot \left((n/d)!\right)^2 \cdot d^{n/d} = n!\cdot (d!)^{n/d} \cdot (n/d)! \cdot d^{n/d} -$$ -and by Stirling's approximation this is $O((n!)^{3/2}C^n)$. -It remains to check that nontransitive representations of $\mathbb F_2 \times \mathbb F_2$ are also rare. I'm going to give a formula which should be estimable by elementary means but leave the estimate for now. Using the two facts proven above we can estimate the number of such representations by: -$$ -\sum_\pi \frac{n!}{\prod_{1\le i \le n} \pi_i! (i!)^{\pi_i}} \prod_{1\le i\le n} \left(1 + O(C^{-i})\right) (i!)^{2\pi_i} \\= (n!)^2 \sum_\pi \prod_{1\le i \le n}\frac{1 + O(C^{-i})}{\pi_i!} \cdot\frac{\prod_{1\le i\le n} (i!)^{\pi_i}}{n!} -$$ -where the sum goes over all partitions of $n$ ($\pi = (\pi_1, \ldots, \pi_n$ with $\sum_i i\pi_i = n$) with $\pi_1 = \pi_n = 0$ (these correspond to actions where the first factor acts trivially or transitively). I think in the form on the second line it should be possible to evaluate the sum to prove that it is $o(1)$.<|endoftext|> -TITLE: Second Betti number of lattices in $\mathrm{SL}_3(\mathbf{R})$ -QUESTION [32 upvotes]: We fix $G=\mathrm{SL}_3(\mathbf{R})$. - -Let $\Gamma$ be a torsion-free cocompact lattice in $G$. Is $b_2(\Gamma)=0$? - -Here the second Betti number $b_2(\Gamma)$ is both the dimension of the cohomology group $H^2(\Gamma,\mathbf{Q})$ and the dimension of the de Rham cohomology in degree 2 of the locally symmetric space $\Gamma\backslash G/K$, $K=\mathrm{SO}(3)$. -From Kazhdan's Property T, I now that - -$b_1(\Gamma)=0$, and, -for every finite index subgroup $\Lambda$ of $\Gamma$, the restriction map $H^2(\Gamma,\mathbf{Q})\to H^2(\Lambda,\mathbf{Q})$ is injective. In particular, $b_2(\Lambda)\ge b_2(\Gamma)$. - -I do not know the answer to the question for a single example of $\Gamma$, so I would accept the answer in a single case. -Actually, for a fixed $\Gamma$, I would consider both a positive or negative answer as remarkable, because: - -If $b_2(\Gamma)=0$, then the (5-dimensional) locally symmetric space $\Gamma\backslash G/K$ is a rational homology sphere (see this MO question); -If $b_2(\Gamma)\neq 0$, then we have a central extension $1\to Z\to\widetilde{\Gamma}\to\Gamma\to 1$, with $Z\simeq\mathbf{Z}$, which does not split (and does not split in restriction to any finite index subgroup, since $\widetilde{\Gamma}$ inherits Property T). Since the fundamental group of $G$ is cyclic of order 2, one can deduce, using superrigidity see that any homomorphism of $\widetilde{\Gamma}$ into any connected Lie group is trivial on $2Z$ (and in particular $\widetilde{\Gamma}$ is linear). If so I'd be very curious about this exotic central extension. For instance, how is $Z$ distorted in $\widetilde{\Gamma}$? It cannot be more than quadratically distorted, because the Dehn function of $\Gamma$ is quadratic. - -[Edit, 2018 Dec 4] In addition, in the latter case case we have another pair of possibilities in which both alternatives appear as surprising. Indeed $H^2_\mathrm{b}(\Gamma,\mathbf{R})=0$ (vanishing of bounded cohomology: Theorem 1.4 of Monod-Shalom 2004). So either (a) in the above central extension, $Z$ is distorted (in contrast to central extensions coming from connected coverings of ambient Lie groups), or (b) $Z$ is undistorted, and this would be a central extension by $Z$ that not represented by a bounded cohomology class. I'm not sure this is known to exist. [/end edit] - - -In principle my question should be computer-answerable, if in a single case, one can implement a triangulation of the locally symmetric space, of reasonable size. - -Additional contextual notes: - -as far as I understand, the vanishing results (Matsushima, Zuckerman, Borel-Wallach...) for $b_2$ apply when $G$ is replaced by a simple Lie group of real rank $\ge 3$, hence don't apply here. -by Abert-Bergeron-Biringer-Gelander-Nikolov-Raimbault-Samet (Annals of Math 2017), we have, in $G$ arbitrary simple Lie group of rank $2$ and finite center, and $\Gamma_n$ strictly decreasing sequence of cocompact lattices, $b_2(\Gamma_n)=o([\Gamma:\Gamma_n])$. -for non-cocompact lattices in $G=\mathrm{SL}_3(\mathbf{R})$, the picture is a bit different since the (rational) cohomological dimension is 3 or 4 (rather than 5) and there is no Poincaré duality. For instance, for a finite index subgroup of $\mathrm{SL}_3(\mathbf{Z})$, the rational cohomological dimension is 3, the Euler characteristic is 0, and hence we have $(b_0,b_1,b_2,b_3,b_4,b_5)=(1,0,b_2,1+b_2,0,0)$. We can indeed (typically) have $b_2>0$: it is proved by Ash (Bull AMS, 1977), for $\Gamma=\mathrm{Ker}(\mathrm{SL}_3(\mathbf{Z})\to \mathrm{SL}_3(\mathbf{Z}/7\mathbf{Z}))$, that $b_2\ge 5814$. Is this evidence that cocompact lattices should also have nonzero $b_2$, I don't know. - -REPLY [14 votes]: The arithmetic cocompact lattices constructed in (6.7.1) of Witte-Morris' book all have torsion-free finite index subgroups with arbitrarily large second Betti number. -I will briefly recall the construction because this is necessary for the answer. Let $F$ be a totally real number field with elements $a,b,t \in F$ such that $\sigma(a), \sigma(b), \sigma(t) < 0$ for all but one real $\sigma$ embedding of $F$. Let $L = F(\sqrt{t})$ be the degree $2$ extension with Galois group $\mathrm{Gal}(L/F) = \{\mathrm{id}, \tau\}$ and -let $\mathcal{O}_L$ denote the ring of integers of $L$. Define -$$h = \left(\begin{smallmatrix}a & & \\ -& b & \\ - & & -1\end{smallmatrix}\right)$$ -The arithmetic group -$$\Gamma = \{ g \in \mathrm{SL}_3(\mathcal{O}_L) \mid \tau(g)^T h g = h\} \subseteq \mathrm{SU}(h,L/F) $$ -embeds as a cocompact lattice in $\mathrm{SL}_3(\mathbb{R})$ (via any one of the two embeddings $L \to \mathbb{R}$). -The non-trivial Galois automorphism $\tau$ of $L/F$ induces an automorphism of the algebraic group $\mathrm{SU}(h,L/F)$ which restricts to an automorphism $\tau$ of $\Gamma$. In particular, we get automorphisms $$\tau^j\colon H^j(\Gamma,\mathbb{C}) \to H^j(\Gamma, \mathbb{C})$$ in the cohomology. It is possible to calculate (or to bound) the Lefschetz number -$$ L(\tau) = \sum_{j=0}^5 (-1)^j\mathrm{Tr}(\tau^j) $$ in the cohomology of $\Gamma$. The methods for this have been worked out by Jürgen Rohlfs (and others) in the 80's and 90's. -The trick is to use the Lefschetz fixed point theorem on the associated locally symmetric space, which says that the Lefschetz number is the Euler characteristic of the set of fixed points. -In the specific example the fixed point set should consist of a bunch of surfaces (and a couple of isolated points). Indeed, -on $\mathrm{SL_3}(\mathbb{R})$ the automorphism is $g \mapsto h^{-1}(g^{-1})^Th$ and the group of fixed points is isomorphic to $SO(2,1)$. -Since surfaces have non-zero Euler characteristic the method yields a lower bound for the cohomology. Here one can find a decreasing sequence of finite index subgroups $\Gamma_n \leq \Gamma$ such that -$$ \sum_{j=0}^5 b_j(\Gamma_n) \gg [\Gamma:\Gamma_n]^{3/8} $$ -as $n \to \infty$; see Theorem 4 in my article. -Asymptotically we have the same lower bound for $b_2(\Gamma_n)$. We have Poincare duality and by property (T) we know that $b_1(\Gamma_n) = 0 =b_4(\Gamma_n)$. Since $b_0(\Gamma_n) = b_5(\Gamma_n) = 1$ and we can only have interesting cohomology in degrees $2$ and $3$, and moreover $b_2(\Gamma_n) = b_3(\Gamma_n)$. -(In the case of non-cocompact lattices a similar argument has been carried out by Lee and Schwermer. I did not check whether the other examples of cocompact arithmetic lattices in $\mathrm{SL}_3(\mathbb{R})$ have a useful algebraic finite order automorphism.)<|endoftext|> -TITLE: Why might the Lawson minimal surface $\xi_{1,2}$ have a Morse index 9? -QUESTION [9 upvotes]: In page 15 of the article New Applications of Min-max Theory, Andre Neves said that: a wishful thinking would suggest the Lawson minimal surface $\xi_{1,2}$ in the 3-sphere to have a Morse index 9, but there is no real evidence. -My question is, what suggests the index to be 9? Say, can we find a 9-parameter deformation of $\xi_{1,2}$ which decreases its area? - -REPLY [2 votes]: This is an old thread, but I figured I would add a short comment that may explain some of the reasoning behind the 'wishful thinking' that Neves refers to. -Let us, for all integer $p \geq 1$, write $\omega_p$ for the $p$-width of $\mathbf{S}^3$, that is the min-max value associated to a $p$-parameter Almgren-Pitts construction. Then -\begin{equation} -\omega_1 \leq \cdots \leq \omega_p \leq \cdots -\end{equation} -is an increasing, though not strictly increasing sequence. -Moreover, for each $p$ there is an embedded minimal surface $\Sigma_p \subset \mathbf{S}^3$ with area and Morse index satisfying -\begin{equation} -\mathrm{area} \, \Sigma_p = \omega_p -\quad \text{and} \quad \mathrm{index} \, \Sigma_p \leq p. -\end{equation} -Note that the index bound is not sharp in general. For example $\omega_1 = \cdots = \omega_4 = 4 \pi$ are all realised by an equatorial sphere, and therefore -\begin{equation} -\mathrm{index} \, \Sigma_4 = 1. -\end{equation} -The specific values of the widths are only known for relatively small $p$, but a PhD student of Neves, Charles Nurser, proved some estimates for the ninth width, namely: -\begin{equation} -2 \pi^2 < \omega_9 < 8 \pi. -\end{equation} -The lower bound means that $\Sigma_9$ is not a Clifford torus, and the upper bound that it is not a multiplicity-two copy of a minimal sphere. It must therefore be a 'new' surface, and it is not unreasonable to conjecture that it might be the Lawson surface $\xi_{1,2}$. -This is just a guess, but as far as the Morse index is concerned, perhaps the thinking was that $\omega_p > \omega_{p-1}$ might mean that $\mathrm{index} \, \Sigma_p = p$. For example $\omega_5 = 2 \pi^2 > 4 \pi = \omega_4$, and is realised by a Clifford torus, with Morse index five. However I must emphasise this is purely speculative, and I'd be very curious to hear from someone more knowledgeable.<|endoftext|> -TITLE: Curve-counting with fixed source -QUESTION [5 upvotes]: Suppose I fix a smooth projective curve $C$ of positive genus, and I have a smooth projective variety $X$. Do standard tools from GW theory (or any curve-counting theory for that matter) allow me to compute the (virtual) number of maps $C\to X$ with incidence conditions? I have only ever seen GW invariants defined to be zero when the expected homological dimension is positive. But do the usual techniques (e.g. localization if $X$ is toric) allow me to do something like take the forgetful map $\overline{M_{g,n}}(X,\beta)\to \overline{M_{g}}$ and intersect a positive (virtual-) dimensional locus with a fiber and hope to get a number out? -Apologies if this is a very naive question; I am not an expert in curve-counting by any means. - -REPLY [3 votes]: You can compute the GW invariants associated to maps with fixed domain curve $C$ of genus $g$ in terms of genus 0 invariants. The conceptual idea is that the invariants are independent of the choice of the fixed curve $C$, and so one can choose $C$ to be a rational curve with $g$ nodes, and then use the usual gluing axioms to rewrite the invariant as a genus 0 invariant with $2g$ additional insertions. -Let $X$ be the target, let $C$ be a fixed curve of genus $g$, $\beta\in H_2(X)$, and let $\gamma_i\in H^*(X)$. Then the fixed domain Gromov-Witten invariant would be defined as -$$\langle \gamma_1,\ldots,\gamma_n\rangle_{\beta,C}^X = \int _{[\overline{M}_{g,n} \, (X,\, \beta)]^{vir}} ev_1^*(\gamma_1)\cdots ev_n^* (\gamma_n) \cdot ct^*([C]^\vee)$$ -where $ct: \overline{M}_{g,n} \, (X,\, \beta)\to \overline{M}_g$ takes a stable map to its domain curve, with marked points forgotten and unstable components contracted. Here $[C]^\vee \in H^*(\overline{M}_g)$ is the class of the point associated to $C$. Then we can rewrite this in terms of usual genus 0 invariants by the following: -$$ -\langle \gamma_1,\ldots,\gamma_n\rangle_{\beta,C}^X = \frac{1}{g!2^g}\sum_{k_1,\ldots,k_g} \langle \eta_{k_1},\eta^{k_1},\ldots,\eta_{k_g},\eta^{k_g},\,\gamma_1,\ldots,\gamma_n\rangle_{\beta,0}^X -$$ -where $\eta_k$ is a basis for $H^*(X)$ and $\eta^k$ is the dual basis so that in particular the class of the diagonal in $H^*(X\times X)$ is given by $\sum_k \eta_k \otimes \eta^k$. -My combinatorial factor out front might not be right. Your mileage may vary.<|endoftext|> -TITLE: Maximum of a quantity for two normal orthogonal vectors in $\mathbb{R}^n$ -QUESTION [9 upvotes]: Let's define for every pair of vectors $u,v\in\mathbb{R}^n$, a quantity as follows: -$$f(u,v) = \sum_{1\leq i,j\leq n}|u_iu_j-v_iv_j|.$$ -I want to find: -$$M(n)= \max \{f(u,v): u,v\in \mathbb{R}^n, |u|=|v|=1, u\perp v\}.$$ -An easy estimate using the triangle inequality gives $M(n) <2n$, but it seems there should be better upper bounds. -Remark. 1) Let $a = \cos(\pi/8), b = \sin(\pi/8)$, then if for even numbers $n$ we define the vectors $u = \sqrt{\frac{2}{n}}(a,b,a,b,\dots)$ and $ v = \sqrt{\frac{2}{n}}(b,-a,b,-a,\dots)$, we have $f(u,v)=\sqrt{2}n\leq M(n)$. -2)A similar construction shows that for every positive integers $n,k$, $\frac{M(n)}{n}\leq \frac{M(nk)}{nk}.$ - -REPLY [3 votes]: Numerical approximations suggest $M(n)=\sqrt{2n^2-1}$ when $n$ is odd, with the max reached for some $u=(a,b,a,b,\dots,a)$, $v=(c,d,c,d,\dots,c)$. Solving exactly for such $u,v$ is easy and only implies a quadratic equation in one variable, $x=b^2$ for instance. Assuming $b>d>0$ without loss of generality, one finds $b^2 = \frac{1}{n-1}(1+\frac{n}{\sqrt{2n^2-1}})$ and deduce $a,c,d$ from the constraints $|u|=|v|=1$ and $u\cdot v=0$, yielding $M(n)\ge \sqrt{2n^2-1}$ when $n$ is odd. Now we still must show that more $a$ in $u$ or more than two distinct components in $u$ cannot improve the bound. -EDIT: Here are elementary details on an analysis for $u$ and $v$ restricted to -$$u=(a,a,\dots,a,b,b,\dots,b)\in\{a\}^{k}\times\{b\}^{\ell}$$ -$$v=(c,c,\dots,c,d,d,\dots,d)\in\{c\}^{k}\times\{d\}^{\ell}$$ -with $k+\ell=n$ odd or even and $0\lt d\le b$ and $0\lt a\le -c$. -$\langle u|u\rangle = \langle v|v\rangle = 1$ and $\langle u|v\rangle = 0$ translate into -$$ka^2+\ell b^2 = 1$$ -$$kc^2+\ell d^2 = 1$$ -$$kac+\ell bd = 0$$ -hence -$$(kac)^2 = (\ell bd)^2 = (\ell b^2)(\ell d^2) = (1-ka^2)(1-kc^2)$$ -simplifies into -$$ka^2+kc^2 = 1$$ -and likewise -$$\ell b^2+\ell d^2 = 1$$ -Furthermore -$$\begin{matrix} -k\ell(ab)^2 &=& (ka^2)(\ell b^2) &=& (1-\ell b^2)(\ell b^2) & &\\ -k\ell(cd)^2 &=& (kc^2)(\ell d^2) &=& (1-\ell d^2)(\ell d^2) &=& (\ell b^2)(1-\ell b^2)\\ -\end{matrix}$$ -hence $0\lt 1-\ell b^2$ and -$$ab = -cd = \sqrt{\frac{b^2-\ell b^4}{k}}$$ -This all enables to write -$$f(u,v) = k^2(c^2-a^2)+\ell^2(b^2-d^2)+2k\ell(ab-cd)$$ -$$= k(1-2ka^2)+\ell(2\ell b^2-1)+4k\ell ab$$ -$$= k(2\ell b^2-1)+\ell(2\ell b^2-1)+4k\ell ab$$ -and finally -$$f(u,v) = g(x) := 2n\ell x -n + 4\ell\sqrt{k}\sqrt{x-\ell x^2}$$ -where $0\lt x=b^2\lt \frac{1}{\ell}$. Actually $b^2$ cannot be arbitrarily close to $0$ because from our assumptions -$$1 = \ell b^2+\ell d^2 \le 2\ell b^2$$ -thus -$$\frac{1}{2\ell}\le x=b^2\lt \frac{1}{\ell}$$ -Now -$$g'(x)=2n\ell + 2\ell\sqrt{k}\frac{1-2\ell x}{\sqrt{x-\ell x^2}}$$ -starts positive and goes toward $-\infty$ on the mentioned interval for $x$, therefore $g(x)$ and $f(u,v)$ will be maximal when $g'(x)=0$, which happens when -$$ 2\ell x-1 = \frac{2n\ell}{2\ell\sqrt{k}}\sqrt{x-\ell x^2}$$ -or equivalently on the prescribed interval -$$ (2\ell x-1)^2 = \frac{n^2}{k}(x-\ell x^2)$$ -which is the quadratic equation -$$x^2-\frac{1}{\ell}x+\frac{k}{\ell(4k\ell+n^2)} = 0$$ -whose solutions are -$$x_\pm = \frac{1}{2\ell} \pm \frac{1}{2}\sqrt{\frac{1}{\ell^2}-\frac{4k}{\ell(4k\ell+n^2)}}$$ -$$= \frac{1}{2\ell} \pm \frac{1}{2}\sqrt{\frac{4k\ell+n^2-4k\ell}{\ell^2(4k\ell+n^2)}}$$ -$$= \frac{1}{2\ell} \left(1 \pm \frac{n}{\sqrt{4k\ell+n^2}}\right)$$ -where only $x_+$ is in $[\frac{1}{2\ell},\frac{1}{\ell})$. The maximum is then -$$g(x_+) = \frac{n^2}{\sqrt{4k\ell+n^2}} + 4\ell\sqrt{k}\sqrt{x_+-\ell x_+^2}$$ -The constant coefficient in the quadratic equation is also $x_+x_-$ thus -$$x_+-\ell x_+^2=x_+(1-\ell x_+)=x_+(\ell x_-)=\ell x_+x_- = \frac{k}{4k\ell+n^2}$$ -yields -$$g(x_+) = \frac{n^2+4\ell k}{\sqrt{4k\ell+n^2}} = \sqrt{4k\ell+n^2}$$ -In the end, when $u$ and $v$ are restricted to the shapes above, $f(u,v)$ is maximal for -$$a = +\sqrt{\frac{1}{2k} \left(1 - \frac{n}{\sqrt{4k\ell+n^2}}\right)}\;, \quad b = \sqrt{\frac{1}{2\ell} \left(1 + \frac{n}{\sqrt{4k\ell+n^2}}\right)}\\ -c = -\sqrt{\frac{1}{2k} \left(1 + \frac{n}{\sqrt{4k\ell+n^2}}\right)}\;, \quad d = \sqrt{\frac{1}{2\ell} \left(1 - \frac{n}{\sqrt{4k\ell+n^2}}\right)}\\ -$$ -and the maximum is -$$f(u,v)=\sqrt{4k\ell+n^2}$$ -Now allowing $k$ and $\ell=n-k$ to vary for $n$ fixed, one gets that $f(u,v)=\sqrt{4k\ell+n^2}$ is maximum over $k$ for $k=n/2$ when $n$ is even, in which case $f(u,v)=\sqrt{2n^2}$, and $k=(n\pm1)/2$ when $n$ is odd, in which case $f(u,v)=\sqrt{2n^2-1}$. We still have to show that other shapes for $u$ and $v$ cannot improve $f(u,v)$.<|endoftext|> -TITLE: Publishing a Simple Paper as an Undergraduate -QUESTION [41 upvotes]: First off I apologize if this question does not belong here, I would be happy to hear about any better locations to post this on. -I am a (first year) undergraduate mathematics student, and I recently discovered some interesting properties hidden in certain families of sequences. I ran these ideas past my math professors, and they said these are interesting ideas and the kind of thing I may want to consider publishing. I did pretty thorough research through my school's library and I am highly confident at this point that these ideas have not been written on before. -To give some context into the situation, I am attending a top 15 university, and my professors are all Harvard, Cambridge, Princeton, etc. educated, so I know they know what they are talking about. With that said, the ideas I discovered -- while interesting -- are really pretty trivial as they can be completely understood with background in Calculus III and some linear algebra, and all my proofs fit onto about 3 pages of LaTeX. -I would love to go ahead with trying to write up a paper on these ideas and get them published as I think that would be a great experience, however I do not want to run the risk of coming off as an over-confident "crank" type early on and be stained by that. Would you recommend I keep my cards close to my chest so to speak, or do you think I should put my ideas out there and see what happens? If I was to pursue publication, I would just submit to undergraduate journals, or maybe something else a little more specialized. - -REPLY [11 votes]: To those journals already mentioned, I can add College Mathematical Journal published by MAA. But my advise is to ask those professors who approved your result. They must be able to give a good recommendation, after you show them the finished paper. In any case, I recommend you to show your finished paper to a professor. -I have similar experience: I wrote my first paper on my second undergraduate year. -I showed it to my professor, he recommended a (mainstream) journal, and the paper was accepted. Several of my friends mathematicians had a similar experience.<|endoftext|> -TITLE: How do fractional tensor products work? -QUESTION [7 upvotes]: [I asked and bountied this question on Math SE, where it got several upvotes and a comment suggesting it was research-level, but no answers. So I'm reposting here with slight edits, but please feel free to close it if it's inappropriate. Also, I'm a physicist rather than a mathematician, so fancy answers might go over my head.] -In this blog post, Terry Tao discusses the $n$-fold tensor product of a one-dimensional vector space $V^L$ ($L$ is just a non-numeric label, not an exponent). He claims that - -With a bit of additional effort (and taking full advantage of the one-dimensionality of the vector spaces), one can also define spaces with fractional exponents; for instance, one can define $V^{L^{1/2}}$ as the space of formal signed square roots $\pm l^{1/2}$ of non-negative elements $l$ in $V^L$, with a rather complicated but explicitly definable rule for addition and scalar multiplication. ... However, when working with vector-valued quantities in two and higher dimensions, there are representation-theoretic obstructions to taking arbitrary fractional powers of [vectors]. - - -What is the "rather complicated but explicitly definable rule for addition and scalar multiplication"? -Is it easy to see why this construction doesn't work in higher than one dimension? (Not necessarily a rigorous proof, just intuition for what goes wrong.) -Could one extend the construction to include irrational exponents? -(a) Tao claims earlier in the blog post that the vector space needs to be totally ordered. Considering that the vector space is 1D, is this equivalent to the requirement that the underlying field be totally ordered? (b) What properties does the vector space (or underlying field) need to satisfy in order for this construction to work? Presumably it doesn't work for arbitrary ordered fields, because you certainly can't define a square root function $\mathbb{Q} \to \mathbb{Q}$. Does it only work for real vector spaces? - -REPLY [6 votes]: I'm not going to try to match Tao's notation. -I will use the word line to mean a one-dimensional real vector space. -Suppose that $L$ is a line. Consider the line $L^{\otimes 2}$. It has the following property: it has a well-defined notion of "positive" element. Indeed, for each $\ell \in L$, we declare that $\ell^{\otimes 2} \in L^{\otimes 2}$ is positive, and $-\ell^{\otimes 2}$ is negative. It is not too hard to show that every nonzero element $L^{\otimes 2}$ is either positive or negative and not both, and that if you multiply a nonzero element of $L^{\otimes 2}$ by a positive real number, then you do not change its sign, whereas multiplication by a negative real number changes the sign. -A line equipped with a notion of "positive" satisfying the above properties will henceforth be called a positive line. Given a positive line $L$, the subspace of nonzero positive elements will be denoted $L_{>0}$. -Suppose now that $L$ is a positive line. For each $r\in \mathbb R$, I will define a positive line $L^{\otimes r}$. The definition is the following. Given $\ell \in L_{>0}$, we declare that there is an element $\ell^{\otimes r} \in L^{\otimes r}_{>0}$. Given $x \in \mathbb R_{>0}$, we declare that $(x\ell)^{\otimes r} = x^r \ell^{\otimes r}$. Note that this makes sense because $x \in \mathbb R_{>0}$ and so $x^r$ is defined. -These declarations are enough to define the line $L^{\otimes r}$ up to isomorphism. Indeed, to define line, it suffices to declare one nonzero element in it — then all other elements are the real multiples of the declared element. So the only other thing needed is to declare how different elements are related, which is what we did. -Example. $L^{\otimes 0}$ is canonically isomorphic to $\mathbb R$, because it has a distinguished element in it. Namely, the element $\ell^{\otimes 0} \in L^{\otimes 0}$ doesn't depend on the choice of element $\ell \in L_{>0}$. -Example. If $L$ is an arbitrary line, then we can define its absolute value to be $|L| := (L^{\otimes 2})^{\otimes 1/2} = \sqrt{L^{\otimes 2}}$. Then we can define things like $|L|^{\otimes r}$ for $r \in \mathbb R$. -Example. When $r \in \mathbb Z$, this definition of $L^{\otimes r}$ agrees with the usual definition (meaning there is a canonical isomorphism). Note that the usual definition of integral powers of a line does not require that the line be positive. This is because, when $r \in \mathbb Z$, the function $x \mapsto x^r$ makes sense on all of $\mathbb R_{\neq 0}$ and not just on $\mathbb R_{>0}$. -There is a representation-theoretic way to say all this. The automorphism group of a positive line (meaning the linear automorphisms that take positive elements to positive elements) is the group $\mathbb R_{>0}$ of positive real numbers, acting by multiplication. It is isomorphic, via the logarithm map, to the group $\mathbb R$ of real numbers under addition. Its irreducible representations are indexed by the Pontryagin dual group, which is also isomorphic to $\mathbb R$. The positive line $L^{\otimes r}$ is "$r$ times the line $L$" in the space of representations of $\mathrm{Aut}(L) \cong \mathbb R_{>0}$. -The operation $L \mapsto |L|$ is important in calculus and manifold theory: it is necessary in order to define volume forms and integration theory on unoriented manifolds. For example, if $M$ is an $n$-dimensional manifold, then a "volume form" on $M$ is not a section of the line bundle $\bigwedge^n T^*_M$ (the top exterior power of the cotangent bundle), but is rather a section of the absolute value of that line bundle. Compare the formulas appearing in "$u$-substitution" in multivariable calculus.<|endoftext|> -TITLE: Random reflections unexpectedly produce banded distributions -QUESTION [8 upvotes]: Start with $p_1$ a random point on the origin-centered unit circle $C$. -At step $i$, select a random point $q_i$ on $C$, and a random mirror line -$M_i$ through $q_i$, and reflect $p_i$ in $M_i$ to point $p_{i+1}$. -(Note that if every $M_i$ passed through the origin rather than $q_i \in C$, -all $p_i$ would lie on $C$.) -In the example below, $p_1$ on $C$ reflects to $p_2$ near the origin, -$p_2$ reflects out to $p_3$ near $(1,1.5)$, etc. - -          - - -          - -Green connects $p_1,\ldots,p_5$. Dark lines are mirrors $M_i$. - - -I expected that repeating this process would produce some smooth -distribution around the origin, as the points reflect further and further out, -when the mirrors effectively pass through the origin at that expanded scale. -But I find not infrequently bands of density, as depicted below. - -    - - -    - -Each distribution includes $n=5000$ reflected points. The unit circle is red. -Approx. radii: $107,67,47,148$. - - -Extending $n$ to larger values seems to continue banding effects. -Here is the 3rd example above, the most uniform of those four, extended. -Its radius increases from $47$ to $225$: - -          - - -          - -Continuation of 3rd example above, to $n=25000$. Approx. radius $225$. - - - -Q. What explains this radial clustering/banding behavior? - -REPLY [2 votes]: As said by Sangchui Lee, we are interested in $r_i = |p_i|$ and we assume that the reflections induce enougth random rotation such that the "bands" of radius $r$ just reveal the number of time $r_i$ was closed to $r$. And we have -$$| p_{i+1}| =|p_i -2\langle q_i, n_i \rangle n_i|$$ -We define then a random march $X_i$ defined by $X_0=0$ and $$X_{i+1}=X_i-2\langle q_i, n_i \rangle n_i$$ -We immediately have that $|p_i|$ has the same law as $|X_i|$ and that for large $i$, $X_i$ behave like the brownian motion on $\mathbb{R}^2$. -One can calculate that the variance of $2\langle q_i, n_i \rangle n_i$ is $\sigma^2 = 2$, the points should be then in a disk of randius $\sim \sqrt{n \sigma^2 }=100$ for $n=5000$ and $224$ for $n=25000$ . -Conclusion for large $i$, $|p_i|$ is given by the Bessel process. And as Sangchui Lee said we are interested in the local time. -https://en.wikipedia.org/wiki/Local_time_(mathematics). -The Ray-Knigh theorem should be usefull here even if it is state for the Brownian motion and not for the Bessel process. -(original paper of Knigh https://www.jstor.org/stable/pdf/1993647.pdf)<|endoftext|> -TITLE: Subtori of groups of type E6 -QUESTION [7 upvotes]: Let $G$ be a semisimple algebraic group of type $E_6$, defined over a perfect field $k$ (so $G$ is a group scheme over $k$ and $G_{\bar{k}}$ is a semisimple algebraic group in the usual sense), and let $T$ be a $k$-subtorus of $G$ of rank $6$, so a (not necessarily split) maximal subtorus of $G$. -Does there exist a strictly smaller semisimple subgroup $H$ of $G$ such that $T$ is a subtorus of $H$? What is the type of such $H$? -For example, can you find $H$ of type $A_2\times A_2\times A_2$ (probably too optimistic)? - -REPLY [10 votes]: This question is precisely answered by Borel–de Siebenthal theory. Ignoring rationality issues (i.e., base changing to an algebraic closure), and fundamental groups, the possible types of $H$ are $E_6$, $A_1 + A_5$, and $A_2 + A_2 + A_2$ (as you hoped, corresponding to removing the root $\alpha_4$ in Bourbaki's notation).<|endoftext|> -TITLE: Is every element of $Mod(S_{g,1})$ a composition of right handed Dehn twists? -QUESTION [7 upvotes]: Let $S_{g,1}$ be the surface of genus $g \geq 1$ and $1$ boundary component. Let $Mod(S_{g,1})$ be the mapping class group in which we allow isotopies to rotate the action on the boundary (equivalently think of it as the mapping class group of the once-punctured surface of genus $g$). -Is every element of $Mod(S_{g,1})$ a composition of right-handed Dehn twists? -Note that this is true for $S_{g,0}$ as stated in page 124 of A primer on Mapping Class Groups by Farb and Margalit under the name of "a strange fact". -Edit: I am going to comment ThiKu's answer to avoid further confusion. -In A. Wand: Factorisation of Surface Diffeomorphisms and in Baker, Etnyre and Van Horn-Morris: Cabling, Contact Structures and and Mapping Class Monoids the authors, independently, provide with examples of diffeomorphisms in $Veer(\Sigma_{2,1},\partial \Sigma_{2,1})$ which are not in $Dehn^+(\Sigma_{2,1}, \partial \Sigma_{2,1})$. That is, right-veering diffeomorphisms which are not a product of right-handed Dehn twists. However, the mapping class group in which these results hold is $Mod( \Sigma_{2,1}, \partial \Sigma_{2,1})$, that is, the mapping class group of automorphisms fixing the boundary and isotopies fixing the boundary as well. This, a priori, does not yield counter-examples to my question (unless it does together with some other result that I do not know). -Observe that for all $g \geq 1$ there is a central extension -$$1 \to \mathbb{Z} \to Mod(\Sigma_{g,1}, \partial \Sigma_{g,1}) \to Mod( \Sigma_{g,1}) \to 1 $$ -which is not split in general. - -REPLY [4 votes]: Yes. As pointed out by Ian Agol, I actually answered my question. -I observed that the identity is a non-empty composition of right handed Dehn twists in $Mod(S_{g,1})$. A priori this is not trivial. I was thinking about monodromies on Brieskorn-Pham singularities $(x^p+y^q)$ which are freely periodic and a composition of right-handed Dehn twists (by morsifying the singularity). One can easily see that this solves the problem in the cases that I was asking originally since all surfaces $S_{g,1}$ appear as Milnor fibers of such singularities for all $g$. -EDIT: I removed the last part of the answer since it was not true in that generality and this answers completely what I wanted.<|endoftext|> -TITLE: Hochschild homology with coefficients in a certain bimodule -QUESTION [9 upvotes]: Let $A$ be a finite-dimensional $k$-algebra and $U$ and $V$ two finite-dimensional projective $A$-modules (maybe neither the finiteness nor projectivity has to play a role, but these requirements are satisfied in my problem). -Now consider the $A$-bimodule -\begin{align} -M := \mathrm{Hom}_A (U,A) \otimes \mathrm{Hom}_A (A,V) \ , -\end{align} where $\mathrm{Hom}_A (U,A)$ is a right $A$-module via right multiplication applied to $A$, - and $\mathrm{Hom}_A (A,V)$ a left $A$-module by right action on $A$ in the source (again by right multiplication); this ensures $\mathrm{Hom}_A (A,V) =V$ as left $A$-modules. -I would like to compute the Hochschild homology $HH_*(A;M)$. -If you write down the corresponding Hochschild complex, there is an augmentation by $\mathrm{Hom}_A (U,V)$. Does that induce a quasi-isomorphism? -Maybe such a result is known to experts? -Is there, more generally, a coefficient theorem I could use? -As usual, thank you for any hints. - -REPLY [8 votes]: (I'm assuming in the following that the base ring $k$ was a field.) -First, we note that for any $A$-bimodule of the form $M \otimes N$, where $M$ is a left $A$-module and $N$ is a right $A$-module, has an isomorphism -$$ -HH_*(A;M \otimes N) \cong Tor^A_*(N,M). -$$ -To see this, we note that there is an explicit simplicial isomorphism between the cyclic bar construction computing Hochschild homology and the two-sided bar construction computing Tor: -$$ -(M \otimes N) \otimes A^{\otimes p} \to N \otimes A^{\otimes p} \otimes M -$$ -In particular, the last map which moves the last factor of $A$ around and multiplies it on the left is carried simply to its left action on $M$. -Therefore, we find -$$ -HH_*(A;Hom_A(U,A) \otimes Hom_A(A,V)) \cong Tor^A_*(Hom_A(A,V), Hom_A(U,A)) \cong Tor^A_*(V, Hom_A(U,A)). -$$ -If $V$ is projective as a (right) $A$-module, then we find that the Tor-groups vanish for $* > 0$ and that the zero'th group is -$$ -V \otimes_A Hom_A(U,A). -$$ -The natural map augmentation that you are describing is then the natural map $V \otimes_A Hom_A(U,A) \to Hom_A(U,V)$, and (because $V$ is projective) this map is an isomorphism whenever $U$ is finitely presented. In particular, we don't need $U$ to be projective. (We could instead ask that $U$ is finitely generated projective and $V$ is arbitrary and get this result; this was stated in the comments already by მამუკა ჯიბლაძე.)<|endoftext|> -TITLE: Intuition about L^p spaces -QUESTION [24 upvotes]: I have read somewhere the following very nice intuition about $L^p(\mathbb{R})$ spaces. - -This graphic shows a lot of nice relations: -1) There is no inclusion between $L^p$ and $L^q$ -2) $L^p$ is the dual of $L^q$ for $\frac{1}{p} + \frac{1}{q} = 1$, since the boxes have the same shape, but just rotated -3) $L^2$ is self-dual -4) If $p -TITLE: $\Delta^{1}_{2}$ and degrees of constructibility $\textbf{on sets}$ -QUESTION [5 upvotes]: This is a follow-up to my question on $\Delta^{1}_{2}$ and degrees of constructibility of real numbers that was answered by the user "William", see here: Can $\Delta^{1}_{2}$ separate degrees of constructibility? -Let us say that the real number $r$ encodes the heriditarily countable set $x$ if and only if there is a bijection $f:\omega\rightarrow\text{tc}(x)\cup\{x\}$ such that $f(0)=x$ and $r=\{p(i,j):f(i)\in f(j)\}$, where $p$ is Cantor's pairing function and $\text{tc}(x)$ denotes the transitive closure of $x$. -Let us say that a $\Delta^{1}_{2}$-formula $\phi$ is "coding invariant" if and only if, for any two codes of the same set $x$, $\phi$ either holds of both or of neither of them; thus, $\phi$ expresses a classification of the heriditarily countable sets. If $x$ is heriditarily countable, we will say that $\phi$ holds of $x$ and write $\phi(x)$ if and only if $\phi$ holds of every real code of $x$. -My question now is whether the following holds: When $\phi$ is a coding invariant $\Delta^{1}_{2}$-formula, $A$ is the set of heriditarily countable sets of which $\phi$ holds and $\bar{A}$ is the set of heriditarily countable sets of which $\phi$ does not hold, does one of $A$ and $\bar{A}$ contain elements of all degrees of constructibility of heriditarily countable sets? -In other words, can $\Delta^{1}_{2}$ separate degrees of constructibility "$\textbf{on the set level}$"? -Note that the answer to Can $\Delta^{1}_{2}$ separate degrees of constructibility? does not immediately yield an answer here, for at least two reasons: (1) not every real number codes a set and (2) codes for the same set can come from very different degrees of constructibility. (3) [added after Douglas Ulrich's comment to Liang Yu's answer]: Not every heriditarily countable set is $L$-equivalent to a real number. -[Here was a wrong example of a heriditarily countable set not $L$-equivalent to a real number, which I deleted after Liang Yu's comment below.] - -REPLY [3 votes]: Here is a partial positive answer: i.e. either $A$ or $\bar{A}$ contain elements of all degrees of constructibility of reals. -Given an infinite set of numbers $x$, let $f:\omega\to \omega\cup \{x\} $ so that $f(n+1)=n$ and $f(0)=x$. Then the corresponded $r_x\equiv_T x$. Now if $A$ is a coding invariant $\Delta^1_2$-set, then either $A$ or the complement of $A$ contains a nonconstructible $r_x$. W.l.o.g, we assume that $A$ contains a nonconstructible $r_x$. Let $B=\{s\in A\mid s \mbox{ is a coding like }r_x\}=\{s \in A\mid \{p(i,j)\mid 1\leq i\leq j\}\subseteq s \subseteq \{p(i,j)\mid 1\leq i\leq j\}\cup\{p(i,0)\mid i\in \omega\}\}$ -Then $B$ is a $\Delta^1_2$ set and $r_x\in B$. So $B$ has a perfect subset $T\in L$. But for each real $s\in [T]$, there is a real $y$ such that $s=r_y\equiv_T y$. So the coded heriditarily countable sets range over all the $L$-degrees. -The partial answer is not quite far away the full one. Since for any heriditarily countable set x, there are two sets of numbers $y$ and $z$ so that $L[z]\cap L[y]=L[\{x\}\cup \mathrm{tc}(x)]$. - -Here is a partial negative answer to your question: I.e. there is a $\Pi^1_1$-formula as you required for which $A$ contain elements of all degrees of constructibility of reals but not all degrees of constructibility of heriditarily countable sets. -Given the function as above. Note that for any $x$ set of numbers and $s$ coding $x$, $r_x\leq_h s$. Now let $B=\{s\mid \exists r\leq_h s(r\cong s\wedge (r \mbox{ is }r_x \mbox{ for some }x))\}$ be a set defined by a $\Pi^1_1$-formual $\phi$. Then $\phi$ satisfies your requirements. Let $A$ be the corresponded collection of heriditarily countable sets of which $\phi$ holds. Then $A$ exactly contains all the constructible degrees of reals. If $\omega_1^L$ is countable, the set $\{(\alpha,g_{\alpha})\mid \alpha<\omega_1^L\}$, the sequence produced by a finite supported Cohen forcing of length $\omega_1^L$ over $L$, does not belong to $A$. Actually it has no a constructible degree of reals.<|endoftext|> -TITLE: Completeness number of ultrafilters -QUESTION [13 upvotes]: In what I write below, by "ultrafilter" I mean a non-principal ultrafilter. -Given an ultrafilter $U$ on some set $S$, let $\mu$ be the least cardinal such that $U$ is $\mu$-complete but not $\mu^+$-complete. Call this number the completeness number of $U$. It is easy to check that this must always be an infinite regular cardinal. -The question is which cardinals can appear as completeness numbers for ultrafilters. Stated otherwise: for which cardinals $\mu$ we can find examples of ultrafilters that have completeness number $\mu$. -For instance, if an ultrafilter is not countably complete, then its completeness number is $\aleph_0$ (and if there are no measurable cardinals then in fact every ultrafilter has completeness number $\aleph_0$). This means that $\aleph_0$ can be completeness number. -What about $\aleph_1$? or $\aleph_n$? or any other regular or even large cardinal $\mu$? -Perhaps this is well-known but I cannot find a reference. -Thanks. - -REPLY [20 votes]: Any countably incomplete ultrafilter has completeness number $\aleph_0$, and if $\kappa$ is measurable then any $\kappa$-complete non-principal ultrafilter on $\kappa$ has completeness number $\kappa$. I claim that these are the only possible completeness numbers. -To prove it, suppose $U$ is a non-principal ultrafilter on some set $A$ and that its completeness number is an uncountable cardinal $\kappa$. I'll prove that $\kappa$ is measurable by producing a $\kappa$-complete non-principal ultrafilter on $\kappa$. -Since $U$ is not $\kappa^+$-complete, we can fix $\kappa$ sets $X_\alpha\in U$ (where $\alpha$ ranges over ordinals $<\kappa$) such that then intersection $\bigcap_{\alpha<\kappa}X_\alpha\notin U$. By subtracting this intersection from each $X_\alpha$, we can assume, without loss of generality, that $\bigcap_{\alpha<\kappa}X_\alpha=\varnothing$. So we can define a function $f:A\to\kappa$ by letting $f(p)$ (for any $p\in A$) be the smallest $\alpha$ such that $p\notin X_\alpha$. Then let $V$ be the image of $U$ under this map $f$; that is, $V=\{Y\subseteq\kappa:f^{-1}[Y]\in U\}$. Since $f^{-1}$ preserves all Boolean operations, including infinitary ones, it is immediate that $V$ is, like $U$, a $\kappa$-complete ultrafilter. -It remains to check that $V$ is non-principal, but this is easy. For any $\alpha<\kappa$, the definition of $f$ implies that $f^{-1}[\{\alpha\}]$ is disjoint from $X_\alpha$ and is therefore not in $U$. So $\{\alpha\}$ is not in $V$.<|endoftext|> -TITLE: Subsets of a group with special property -QUESTION [6 upvotes]: Let $G$ be a finite group. We say a subset $A$ of $G$, $|A|=m$, is $(m,i)$-good, $m\geq 1$ and $0\leq i\leq m$, if there exist $g_A\in G$ such that we have $|gA\cap A|=m-i$. -I need some groups such that, for fixed $m$ and $i$, preferably $i$ is much smaller than $m$, the number of $(m, i)$-good subsets is large. -Is this property well known in group theory? Or, is it related to some well-known properties of groups for special values $m$ and $i$? -$\textbf{The motivation for this question:}$ These type of subsets are used to define some special type of participants in secret sharing schemes on groups. Participants are the members of group and subsets with this property can take some special values of share to construct the key. -I do not know how can I upload a large file here. You can download the text file from this address. If it is helpful, I can do this computation for each order of groups. -$\textbf{Addendum 1}:$ http://s8.picofile.com/file/8341377742/Order_8.txt.html -$\textbf{Addendum 2}:$ http://s9.picofile.com/file/8341380550/Order_9.txt.html -$\textbf{Addendum 3}:$ http://s9.picofile.com/file/8341379792/Order_12.txt.html -$\textbf{New question (10/2018)}:$ -Is it true that if the $(m,i)$-good numbers of two groups $G$ and $H$ are equal, then $G$ is isomorphic to $H$? -Thanks for helpful answers, references, and comments. - -REPLY [2 votes]: Any $m$-set is an $(m,0)$-set with $g$ the identity. It is easier to count the number of pairs $(g,S)$ where $S$ is $(m,i)-$good for that particular $g.$ If $g$ has order $2$ and $|G|=n=2t$ then, using that $g$, there are $\binom{t}k$ sets that are $(2k,0)$ and $\binom{t}k\binom{t-k}i2^i$ which are $(2k+i,i). $ An elementary abelian $2$ group has all non-identity elements of order $2$. Of course the same set could be $(m,i)$ for more than one $g$ but that might not be that big a consideration. -It doesn't really matter if the group is abelian. For $g$ of order $j,$ the exact counts using that $g$, are sums of terms each the product of a multinomial coefficient and some powers of constants which are counts for a cyclic group of order $j$. -For $C_s$ and a given generator $g$ consider the $2^s-1$ non-empty subsets and let $c_{m',i'}$ be the number that are $(m',i')-$good for that $g.$ -Then if $G$ is a group of order $n=st$ and $g$ is an element of order $s$ then from the $t$ right cosets of $$ $n_{(m',i')}$ of each type. The result will be $(m,i)-$ good (for that $g$) for $m=\sum n_{(m',i')}m'$ and $i=\sum n_{(m',i')}i'.$ Sum over all the possibilities to get the total count of $(m,i)$ sets for that $g$. The number of ways to do this is a multinomial coefficient. -To count $(m,i)-$good sets with a distinguished $g$ do the above for every combination giving the desired $(m,i)$ -In theory, do the above for each element of $G$ then somehow account for sets with multiple possible $g$.<|endoftext|> -TITLE: Example of nonvanishing Waldhausen Nil group -QUESTION [6 upvotes]: In a remarkable series of papers, both anticipating development in geometric topology and algebraic K-theory, specifically what we call now the Farrell-Jones conjecture, Waldhausen introduced obstructions to the Whitehead groups to satisfy a mayer-vietoris sequence. -See https://mathscinet.ams.org/mathscinet-getitem?mr=498807 -These obstructions are related to splitting of homotopy equivalences across submanifolds of high dimension ( >5) due to Sylvain Cappell: https://mathscinet.ams.org/mathscinet-getitem?mr=285010. -Since then , lots of effort has been put into showing vanishing results for Waldhausen Nil groups. I would like to know an easy example of non-vanishing of these groups. It would be interesting to know the minimal cohomological dimension of the groups involved in an amalgam of such example. Waldhausen sorts out in his original paper fundamental groups of surfaces, free groups, fundamental groups of submanifolds of the three dimensional spheres... - -REPLY [4 votes]: There are examples of non-vanishing nil groups due to Daniel Juan-Pineda. See Juan-Pineda, Daniel(MEX-NAMMO-IM) On higher nil groups of group rings. Homology Homotopy Appl. 9 (2007), no. 2, 95–100 (MSN).<|endoftext|> -TITLE: Algebraic exponential values -QUESTION [5 upvotes]: Is there a non-zero real number $t$ for which there exist infinitely - many prime numbers $p$ with $p^{it}$ an algebraic integer? - -I would even be surprised to find a real $t \neq 0$ with both $2^{it}$ and $3^{it}$ being algebraic integers. - -REPLY [8 votes]: The six exponentials theorem (look it up) states that if $x_1$ and $x_2$ are two complex numbers which are linearly independent over $\mathbb{Q}$ and if $y_1,y_2$ and $y_3$ are three complex numbers linearly independent over $\mathbb{Q}$ then at least one of the numbers $e^{x_i y_j}$ will be transcendental. -Therefore, for any real number $t$ it is not even possible for three primes $p_1,p_2$ and $p_3$, let alone infinitely many. Take $x_1=1$, $x_2=it$, then take $y_i=\text{log} p_i$ for $i=1,2,3$. Clearly, $y_1,y_2,y_3$ are independent over $\mathbb{Q}$ since $p_i$ are prime numbers and since $t$ is real, $x_1$ and $x_2$ are independt over $\mathbb{Q}$ as well. -The numbers $e^{x_1 y_j}$ are the primes $p_j$ so are algebraic. Therefore one of the numbers $e^{x_2 y_j}=p_j^{it}$ is not algebraic for $j=1,2,3$.<|endoftext|> -TITLE: Log-concavity of areas of level sets -QUESTION [7 upvotes]: Suppose $f: \mathbb{R}^d \to \mathbb{R}$ is a smooth convex function. -Consider the level sets of the function, namely $M_s = \{x: f(x) = s\}$. -Is it true/known that the surface areas of $M_s$ are log-concave as a function of $s$? -(This feels awfully like a Brunn-Minkowski style inequality, but I'm unsure if it follows from known results. If so, references are highly appreciated!) - -REPLY [9 votes]: Yes, this is true, and you are right, this follows from a generalization of the Brunn-Minkowski inequality. -Let $K_s = \{x \mid f(x) \le s\}$, so that $M_s = \partial K_s$. We have $K_s \supseteq (1-s)K_0 + sK_1$, thus the surface area of the former is $\ge$ the surface area of the latter. -The surface area of a convex body can be written as a mixed volume: -$$ -\mathrm{vol}_{n-1}(\partial K) = n \mathrm{vol}(\underbrace{K, \ldots, K}_{n-1}, B). -$$ -A general version of the Brunn-Minkowski inequality says that the function -$$ -\mathrm{vol}(\underbrace{(1-t)K_0 + tK_1, \ldots, (1-t)K_0 + tK_1}_{n-i}, L_1, \ldots, L_i)^{\frac1{n-i}} -$$ -is concave for any convex bodies $L_1, \ldots, L_i$. -It follows that the $(n-1)$-st root of the surface area of $K_t$ is concave. Concavity of $f^{\frac{1}{n-1}}$ implies concavity of $\log f$, and we are done. -References: -Gardner, R. J., The Brunn-Minkowski inequality, Bull. Am. Math. Soc., New Ser. 39, No. 3, 355-405 (2002). ZBL1019.26008, Section 17. -Burago, Yu. D.; Zalgaller, V. A., Geometric inequalities. Transl. from the Russian by A. B. Sossinsky, Grundlehren der Mathematischen Wissenschaften, 285. Berlin etc.: Springer-Verlag. XIV, 331 p.; DM 184,- (1988). ZBL0633.53002, p. 146. -Schneider, Rolf, Convex bodies: the Brunn-Minkowski theory, Encyclopedia of Mathematics and Its Applications. 44. Cambridge: Cambridge University Press. xiii, 490 p. (1993). ZBL0798.52001, Theorem 6.4.3.<|endoftext|> -TITLE: When $\int_M \exp(-d_M(x,y)^2/t) dvol(y)$ becomes constant for a Riemannian manifold $M$? -QUESTION [8 upvotes]: Let $(M,g)$ be a closed and connected Riemannian manifold. $d_M$ is its geodesic metric and $dvol_M$ is its standard volume measure. For each $t>0$, define a map $f:M\rightarrow\mathbb{R}_{>0}$ in the following way: -$$f_t(x):=\int_M e^{-\frac{d_M(x,y)^2}{t}} dvol_M(y).$$ -Then, when this $f_t$ becomes constant map? Obviously it becomes constant if $M$ is symmetric, like $S^n$. But do we have more rich characterization? -This seems a simple question, so it might be a classical one. But I'm not expert in differential geometry. Is there any related reference about this question? - -REPLY [2 votes]: Here are a few comments: -1) I like the example of a cylinder, slightly more complicated than $S^n$. This satisfies the constancy in the problem. It is symmetric in having a transitive group of isometries, but the group of isometries is not doubly transitive. -2) We can solve the problem if for any points $p$ and $q$, we can exhibit an isometry $f$ taking $p$ to $q$. And there is an obvious candidate for such an isometry: Choose a geodesic $\gamma$ from $p$ to $q$. Then parallel transport along $\gamma$ takes the tangent vectors at $p$ to tangent vectors at $q$, and takes small geodesics starting at $p$ to small geodesics starting at $q$. This creates a map $f:B(p,r)\rightarrow B(q,r)$, where $r$ is the minimum of the radii of injectivity at $p$ and $q$. We need to prove first that $f$ is an isometry on its domain, and then that $f$ can be extended to the whole space. -3) To prove that $f$ is an isometry, it suffices to show that the minimal ball $S$ containing $p_1$ and $p_2$ has the same volume as the minimal ball containing $f(p_1)$ and $f(p_2)$. Then, by David Speyer's reformulation of the hypothesis, the two balls have the same radii, and the same diameters, and the same distance from $p_1$ to $p_2$ as from $f(p_1)$ to $f(p_2)$. This may be useful because if $S$ has (e.g.) twice the volume of $f(S)$, then we can identify a decreasing sequence of balls $S_i$ where $S_i$ has twice the volume of $f(S_i)$, so we can identify a particular point where the isometry fails.<|endoftext|> -TITLE: Nonlinear sigma models with non-compact groups / target spaces -QUESTION [5 upvotes]: A nonlinear σ model (NLSM) describes a scalar field Σ which takes on values in a nonlinear manifold called the target manifold T. -The target manifold T is equipped with a Riemannian metric g. Σ is a differentiable map from Minkowski space M (or some other space) to T. -The Lagrangian density in chiral form is given by -$$ -{\displaystyle {\mathcal {L}}={1 \over 2}g(\partial ^{\mu }\Sigma ,\partial _{\mu }\Sigma )-V(\Sigma )}, -$$ -One can also add the Wess–Zumino–Witten term into this NLSM. -My question is that - - -Are there any mathematical studies and mathematical/physics uses of nonlinear sigma model (NLSM) with non-compact groups $G$ or non-compact target space T? - - -Are these theories "unitary"? -In all the context that I am familiar, I always deal with a NLSM of compact (Lie) groups $G$ or compact target space T. So any comments and lectures on non-compact cases are welcome. Thanks! - -REPLY [5 votes]: Of course there are. A nice early example is the Wess--Zumino--Witten model based on a non-semisimple group admitting a bi-invariant lorentzian metric: -@article{Nappi:1993ie, - author = "Nappi, Chiara R. and Witten, Edward", - title = "{A WZW model based on a nonsemisimple group}", - journal = "Phys. Rev. Lett.", - volume = "71", - year = "1993", - pages = "3751-3753", - doi = "10.1103/PhysRevLett.71.3751", - eprint = "hep-th/9310112", - archivePrefix = "arXiv", - primaryClass = "hep-th", - reportNumber = "IASSNS-HEP-93-61", - SLACcitation = "%%CITATION = HEP-TH/9310112;%%" -}<|endoftext|> -TITLE: Asymptotic behavior of a certain trigonometric partial sum -QUESTION [6 upvotes]: Let $a<0$ and $b>0$ be real numbers such that $a<-2b$. Let $n>1$ be a positive integer and consider the following partial sum: -$$ -f(n) = \frac{1}{(n+1)^2}\sum_{i=1}^{n}\sum_{j=1}^{n} (-1)^{i+j}\frac{(1-\cos(2x_i))(1-\cos(2x_j))}{-2a-2b(\cos(x_i)+\cos(x_j))}, \ \ \text{with }\ x_k:=\frac{k\pi}{n+1}. -$$ - -Problem. Numerical simulations seem to suggest that - $$\lim_{n\to\infty}\frac{f(n)}{c^{n}}= \infty$$ - for small enough $c>0$. Is there a way to prove this claim? If so, is it possible to characterize the values of $c$ that satisfy the above condition? - -I would like to stress that this is not an homework question, even if it may look so at a first glance. I encountered this fact in my research, and, after several unsuccessful proof attempts, I decided to post the problem here. If this is a trivial fact (it doesn't look so to me), please feel free to close this OP. - -REPLY [8 votes]: The desired inequality should be true iff -$$ -c < c_0 := (r - \sqrt{r^2-1})^2 -\quad\ \text{where} \quad\ -r = \frac{|a|}{2b} -$$ -(NB the hypotheses $b>0$ and $a < -2b$ imply $r>1$, so $0 < c_0 < 1$). -Numerical computation suggests that $f(n) \sim A c_0^n \left/ \sqrt{n} \right.$ -for some $A>0$. It should be possible to prove this by completing -the following analysis. -It will be convenient to set $N = n+1$, and sum over $j,k$ rather than -$i,j$ because we'll need $i = \sqrt{-1}$. -For $x,y \in ({\bf R}/{\bf Z})^2$ define -$$ -F(x,y) = \frac1{2b}\frac - {(1 - \cos(4\pi x)) (1 - \cos(4\pi y))} - {2r - (\cos 2\pi x + \cos 2 \pi y)}. -$$ -Then -$$ -f(n) = \frac1{N^2} - \sum_{j=1}^{N-1} \sum_{k=1}^{N-1} (-1)^{j+k} - F\Bigl(\frac{j}{2N}, \frac{k}{2N}\Bigr) \, . -$$ -Now $F(x,y) = F(-x,y) = F(x,-y)$, and -$$ -F(0,y) = F(1/2,y) = F(x,0) = F(x,1/2) = 0 -$$ -for all $x,y$ thanks to the factors -$1 - \cos(4\pi x)$ and $1 - \cos(4\pi y)$ in the numerator of $F$. -So we can write $f(n)$ as an alternating sum over $\frac1{2N}$-lattice points: -$$ -f(n) = \frac1{(2N)^2} - \sum_{j=1}^{2N} \sum_{k=1}^{2N} (-1)^{j+k} - F\Bigl(\frac{j}{2N}, \frac{k}{2N}\Bigr) -$$ -(each term in the original sum for $f$ appears four times here, -and the added terms with $j,k \in \{N, 2N\}$ all vanish). -But such a sum can be expressed in terms of the Fourier expansion -$$ -F(x,y) = \sum_{r\in\bf Z}\sum_{s\in\bf Z} \phi(r,s) \exp 2\pi i (rx+sy) -$$ -of $F$, which converges absolutely because $F$ is smooth. -The alternating sum of $\exp 2 \pi i (rx+sy)$ is $(2N)^2$ if -$(r,s) \equiv (N,N) \bmod 2N$, and zero otherwise. Therefore -$f(n) = \sum\!\sum_{r,s} \phi(r,s)$ with the sum extending over all $(r,s)$ -such that $r \equiv s \equiv N \bmod 2N$. The simplest terms -in this sum are the four coefficients $\phi(\pm N, \pm N)$, -which are all equal (since they satisfy $\phi(r,s) = \phi(-r,s) = \phi(r,-s)$ for all $r,s$). -Thus we expect that -$$ -f(n) \sim 4 \phi(-N,-N) = 4 \int_{-1/2}^{1/2} \int_{-1/2}^{1/2} - \exp (2\pi i N (x+y)) \, F(x,y) \, dx \, dy -$$ -unless $\phi(N,N)$ is unusually small. -We can now explain and quantify the observed exponential cancellation -in the alternating sum that defines $f$. For large $r,s$ -the Fourier coefficients $\phi(r,s)$ decay exponentially but no faster, -because $F$ is not just smooth but analytic for -$x,y \in ({\bf C}/{\bf Z})^2$ in some neighborhood of $({\bf R}/{\bf Z})^2$, -but does have singularities for some complex $(x,y)$. -We can shift the integral in the complex direction: -$$ -\phi(N,N) = e^{-4\pi N w} \int_{-1/2}^{1/2} \int_{-1/2}^{1/2} - \exp (2\pi i N (x+y)) \, F(x+iw,y+iw) \, dx \, dy -$$ -as long as $F$ is analytic for - $\left|\mathop{\rm Im}(x)\right|, \left|\mathop{\rm Im}(y)\right| \leq w$. -This is the case for all -$$ -w < w_0 := \frac1{2\pi} \cosh^{-1} r - = \frac1{2\pi} \log \bigl(r + \sqrt{r^2-1}\bigr) - = -\frac1{4\pi} \log c_0. -$$ -In fact we can take $w = w_0$: the integrand -$F(x+iw_0,y+iw_0)$ blows up at $(x,y)=(0,0)$, but the integral -still converges absolutely because if $\left| F(x+iw_0,y+iw_0) \right| > M$ -then $|x+y| \ll M^{-1}$ and $|x-y| \ll M^{-1/2}$. Thus -$$ -\phi(N,N) = c_0^N \int_{-1/2}^{1/2} \int_{-1/2}^{1/2} - \exp (2\pi i N (x+y)) \, F(x+iw_0,y+iw_0) \, dx \, dy. -$$ -This shows that $f(n) \ll c_0^N$, and thus also $f(n) \ll c_0^n$; -some more analysis of the double integral near the singularity at $(0,0)$ -should show that $f(n) / c^n \to \infty$ for all $c -TITLE: Representability of matroids over finite fields -QUESTION [6 upvotes]: I have several questions regarding representability of matroids. -Question 1. Does there exist a finite matroid that is representable over an infinite field, but is not representable over any finite field? -Question 2. Does there exist a finite matroid that is representable over a field of characteristic $0$, but is not representable over any field of positive characteristic? -Question 3. Does there exist a finite matroid that is representable over a field of characteristic $0$, but $\{\mathrm{char}(F): M \text{ is representable over }F\}$ is a finite set? -I care about Question 1 most, and find other quetions also interesting. I briefly checked Oxley, Matroid theory but did not find an answer. - -REPLY [7 votes]: The main results of Rado's Note on Independence Functions settle all three questions. The first few lines of Effective Versions of Two Theorems of Rado give a perfect recap of those results, so here they are verbatim: - -Our starting point is given by the following two theorems of Rado [5]. -Theorem 1 (Rado, 1957). Let $M$ be a matroid representable over a field $K$. Then $M$ is representable over a simple algebraic extension of the prime field of $K$. -Theorem 2 (Rado, 1957). Let $K$ be an extension field of $\mathbb{Q}$ of degree $N$ , and let $M$ be a matroid representable over $K$. Then there is a positive integer $c$ such that given any prime $p > c$ there is a positive integer $k = k(p) ≤ N$ such that $M$ is representable over $GF(p^k)$. For infinitely many $p$, $k(p) = 1$. -Together, these two theorems say that if a matroid is linearly representable, then it is representable over a finite field.<|endoftext|> -TITLE: Greedy simplices in an ultrametric space (generalized Bhargava $p$-orderings) -QUESTION [5 upvotes]: Let $\left(U, d\right)$ be a finite ultrametric space -- that is, $U$ is a finite set, and $d : U \times U \to \mathbb{R}_{\geq 0}$ is a metric on $U$ such that every $x, y, z \in U$ satisfy $d\left(x, z\right) \leq \max\left\{d\left(x,y\right), d\left(y,z\right)\right\}$ (in words: in any triangle, the longest two sides have the same length). -Fix $n \in \mathbb{N}$. An $n$-simplex shall mean an $\left(n+1\right)$-tuple $\left(u_0, u_1, \ldots, u_n\right)$ of elements of $U$. The sum-sequence of an $n$-simplex $\left(u_0, u_1, \ldots, u_n\right)$ shall mean the $n$-tuple $\left(s_1, s_2, \ldots, s_n\right) \in \mathbb{R}^n$, where $s_i = d\left(u_0, u_i\right) + d\left(u_1, u_i\right) + \cdots + d\left(u_{i-1}, u_i\right)$ for each $i\in \left\{1,2,\ldots,n\right\}$. -An $n$-simplex $\left(u_0, u_1, \ldots, u_n\right)$ is said to be greedy if and only if every $i \in \left\{1,2,\ldots,n\right\}$ and $v \in U$ satisfy -\begin{equation} -d\left(u_0, u_i\right) + d\left(u_1, u_i\right) + \cdots + d\left(u_{i-1}, u_i\right) \geq d\left(u_0, v\right) + d\left(u_1, v\right) + \cdots + d\left(u_{i-1}, v\right) . -\end{equation} -Intuitively speaking, a greedy $n$-simplex is what you get if you try to construct an $n$-simplex with the lexicographically largest possible sum-sequence by first picking any $u_0 \in U$, then picking any $u_1 \in U$ maximizing $d\left(u_0, u_1\right)$ (with $u_0$ already fixed), then picking any $u_2 \in U$ maximizing $d\left(u_0, u_2\right) + d\left(u_1, u_2\right)$, etc.. Note that there are many choices in this construction. Nevertheless, I suspect the following: - -Conjecture 1. Any two greedy $n$-simplices have the same sum-sequence. - -This would generalize Theorem 5 in Manjul Bhargava, The Factorial Function and Generalizations, The American Mathematical Monthly, Vol. 107, 2000, pp.783-799. Indeed, the set $S$ in that paper can be regarded as an ultrametric space (with the metric defined by $d\left(s, t\right) = p^{- v_p\left(s-t\right)}$, as usual), and then the $p$-orderings would be exactly the greedy $\infty$-simplices. (For the pedants: Of course, $S$ is not finite, and $\infty \notin \mathbb{N}$; but it is easy to reduce the claims to the case of finite $U$ and finite $n$.) -If Conjecture 1 holds, then the sum-sequence of any $n$-simplex is lexicographically $\leq$ to the sum-sequence of any greedy $n$-simplex. This makes me wonder if we can say something stronger: - -Conjecture 2. Let $\left(t_1, t_2, \ldots, t_n\right)$ be the sum-sequence of any $n$-simplex, and let $\left(s_1, s_2, \ldots, s_n\right)$ be the sum-sequence of any greedy $n$-simplex. Then, $t_1 + t_2 + \cdots + t_i \leq s_1 + s_2 + \cdots + s_i$ for any $i \in \left\{0,1,\ldots,n\right\}$. - -Note that proving this for $i = n$ is enough, because the first $i+1$ entries of a greedy $n$-simplex always form a greedy $i$-simplex. So Conjecture 2 is tantamount to saying that the perimeter of a greedy $n$-simplex (i.e., the sum of its edge-lengths) is $\geq$ to the perimeter of any $n$-simplex. In other words, the greedy algorithm succeeds in maximizing the perimeter. -I regret to say I have not thought much about this, as I am currently completely swamped between teaching, job hunting and some risible pretense of research. But the most obvious things don't work: Bhargava's proof for his Theorem 5 is number-theoretical and doesn't generalize. I know that we can define an equivalence relation $\sim$ on $U$ by letting $x \sim y$ if $x = y$ or $d\left(x, y\right)$ is the smallest nonzero value of $d$, and I know that any permutation of $U$ that preserves $\sim$ is an automorphism of the metric space $\left(U, d\right)$. I thought of replacing $U$ by $U / \sim$, but so far I don't really see how to translate things to the quotient. I am tagging this "matroid-theory" just in case, but I don't myself see how to reduce the above greedy algorithm to that of a matroid. -I wouldn't be surprised if this has practical applications: An ultrametric space $\left(U, d\right)$ can be viewed as a nested sequence of set partitions of $U$ (i.e., a set partition, then another that refines it, then another that refines it further, and so on), so some sort of classification at several levels of fine-grainedness. An $n$-simplex of maximum perimeter would then be a way to sample $n$ data points that are "as representative as possible of the whole population", in the sense of not being closer to each other than necessary. -EDIT: Note: Some of what I wrote in this post is false. The metric on the set $S$ in Bhargava's paper has to be defined by $d\left(s,t\right) = -v_p\left(s,t\right)$ (not by $d\left(s,t\right) = p^{-v_p\left(s,t\right)}$) in order for the $p$-orderings to be the greedy $\infty$-simplices. This requires slightly generalizing the definition of a metric: Instead of being a map to $\mathbb{R}_{\geq 0}$, it now has to be a map to $\mathbb{R} \cup \left\{-\infty\right\}$ which takes the value $-\infty$ only at pairs of the form $\left(a,a\right)$. This is no longer a metric in the proper sense of this word, but just a symmetric map on $U \times U$ that satisfies $d\left(x, z\right) \leq \max\left\{d\left(x,y\right), d\left(y,z\right)\right\}$. Soon, a preprint by Fedor Petrov and myself will come out which discusses these issues in greater detail. -EDIT2: The preprint is out: - -Darij Grinberg, Fedor Petrov, A greedoid and a matroid inspired by Bhargava's p-orderings, arXiv:1909.01965v3. - -It incorporates both Fedor's answer to this question and the conversation we had in the chat afterwards, and more. A followup paper is in the process of being written: - -Darij Grinberg, The Bhargava greedoid as a Gaussian elimination greedoid. - -REPLY [2 votes]: It looks that both claims are true. -For $V\subset U$ denote by $f(V)$ the sum of mutual distances between elements of $V$. -Exchange lemma. Let $V\subset U$ and $v\in U$, and let $w$ be a projection of $v$ on $V$ (that is, a point in $V$ that is closest to $v$). Then, $f(V\cup v\setminus w)\geqslant f(V)$ and $d(v,x)\geqslant d(w,x)$ for any $x\in V$. -Proof. If the latter inequality fails for some $x\in V$, we have $d(w,x)>d(v,x)\geqslant d(v,w)$, a contradiction to the ultra-triangle inequality for $\triangle vxw$. Summing up $d(v,x)\geqslant d(w,x)$ by $x\in V\setminus w$ we get $f(V\cup v\setminus w)\geqslant f(V)$. -Not we prove that for a greedy simplex $S=\{v_0,\dots,v_m\}$ we have $$f(S)\geqslant f(T)\,\,\,\,(\star)$$ for any $T\subset V$, $|T|=m+1$. This implies both of your conjectures. -We do this by choosing consecutively elements $u_0,u_1,\dots\in T$ such that $u_i$ is a projection of $v_i$ on $T\setminus \{u_0,\dots,u_{i-1}\}$. Then by the exchange lemma we have $d(v_i,u_j)\geqslant d(u_i,u_j)$ whenever $j>i$ (since $u_j \in T\setminus \{u_0,\dots,u_{i-1}\}$). Summing this by all pairs $j>i$ we get $f(T)\leqslant \sum_{i -TITLE: Embedding open connected Riemann Surfaces in $\mathbb{C}^2$ -QUESTION [7 upvotes]: This question arises in the context of a question asked on MSE: Are concrete Riemann surfaces Riemann domains over $\mathbb{C}$. Part of the answer to that question is the question above which is being asked here since it appears to be at a "research level". -Is there an example (or even an obstruction) to embedding an open connected Riemann surface into $\mathbb{C}^2$? -A theorem of Narasimhan (Narasimhan, R. "Imbedding of open Riemann surfaces", Göttingen Nachrichten, No. 7 (1960), pp. 159-165; also see American Journal of Mathematics Vol. 82, No. 4 (Oct., 1960), pp. 917-934) proves that -any open connected Riemann surface has a non-singular embedding in $\mathbb{C}^3$. -Extending this to $\mathbb{C}^2$ seems to be difficult as a linear projection would introduce nodal singularities in general. -There are examples of affine algebraic curves that do not embed algebraically into $\mathbb{A}^2$. One obstruction in this case is that for a smooth curve $X$ in $\mathbb{A}^2$, we have $\Omega^1_{X}$ is a trivial line bundle, whereas there are affine algebraic curves with non-trivial $\Omega^1_{X}$. However, this is not an obstruction in the case of an open Riemann surface since line bundles on it are holomorphically trivial! - -REPLY [5 votes]: One of the first example of proper embedding of open Riemann surface is, -Proper embedding of open unit disk in $\mathbb C^2$ which is one of corollary of Fatou-Bieberbach domains (holomorphic dynamics in $\mathbb C^2$). There are results about proper embedding of annulus as well. -But general question is still open which is known as Bell-Narasimman conjecture. -Bell-Narasimman conjecture has two parts (both are open) - -1) Every open Riemann surface can be embedded in $\mathbb C^2$ -2) Every embedded open Riemann surface can be properly embedded in $\mathbb C^2$.<|endoftext|> -TITLE: Algebraic independence of $P,Q,R$ or $E_2,E_4,E_6$ over $\mathbb C(z)$ -QUESTION [7 upvotes]: Let $P,Q,R$ be the Fourier series of the Eisenstein series $E_2,E_4,E_6$, that is, -$$ -P(q)=1-24\sum_{n=1}^{\infty}\sigma_1(n)q^n, -$$ -$$ -Q(q)=1+240\sum_{n=1}^{\infty}\sigma_3(n)q^n, -$$ -$$ -R(q)=1-504\sum_{n=1}^{\infty}\sigma_5(n)q^n. -$$ -I heard that $P,Q,R$ are algebraically independent over $\mathbb C(q)$, due to Mahler, but I cannot find a reference. Could you help me if you know the reference? -Meanwhile, I found 'On Algebraic Differential Equations Satisfied by Automorphic Functions', by Mahler, 1969. I do not think that this paper addresses this topic, but if it does, why does it imply the algebraic independence of $P,Q,R$? - -REPLY [10 votes]: I think the follwing is the answer of this question. It is not depend on the above comments because I didn't understand them. This is just my approach. -(But not fully my idea because it depends on a strong proposition which is already known, and the remaining part is just a corollary.) - -Proposition. Let $f$ be a modular form of weight $k$, defined over $\mathbb Q^{\mathrm{alg}}$. If $f$ is non constant, then the functions $f, Df$ and $D^2 f$ are algebraically independent over the function field $\mathbb C(q)$. - -The proof is in the book, 'Introduction to Algebraic Independence Theory', Y.V.Nesterenko and P.Philippon, (Eds), Ch.1 prop 1.1. Here $q=e^{2\pi i \tau}$ and $D=\frac{1}{2\pi i}{\tau}=q\frac{d}{dq}$. -Let $\Delta = E_4^3-E_6^2$, cusp form of weight 12. Due to Ramanujan, we have the following formulas : -$$ -DP=\frac{1}{12}(P^2-Q), \quad DQ=\frac{1}{3}(PQ-R), \quad DR=\frac{1}{2}(PR-Q^2). -$$ -Thus we have $D\Delta = P\Delta$. Take $D$ in both side and do some elementary algebra, we have -$$ -Q=13\frac{(D\Delta)^2}{\Delta^2}-12\frac{D^2 \Delta}{\Delta}. -$$ -By the proposition, $\{\Delta, D\Delta, D^2\Delta\}$ is algebraically independent over $\mathbb C(q)$, so $\{Q^3-R^2,P,Q\}$ is algebraically independent over $\mathbb C(q)$, so $\{P,Q,R^2\}$ is algebraically independent over $\mathbb C(q)$. This is equivalent to say that $R^2$ is transcendental over $\mathbb C(q)[P,Q]$, and this implies that $R$ is transcendental over $\mathbb C(q)[P,Q]$, i.e. $\{P,Q,R\}$ is algebraically independent over $\mathbb C(q)$.<|endoftext|> -TITLE: Can a Shelah semigroup be commutative? -QUESTION [7 upvotes]: A semigroup $S$ is called -$\bullet$ $n$-Shelah for a positive integer $n$ if $S=A^n$ for any subset $A\subset S$ of cardinality $|A|=|S|$; -$\bullet$ Shelah if $S$ is $n$-Shelah for some $n\in\mathbb N$. - -Question. Can an infinite Shelah semigroup be commutative? - -This problem was motivated by the following results: -Theorem (Shelah, 1980). For any infinite cardinal $\lambda$ with $\lambda^+=2^\lambda$ there exists a group $G$ of cardinality $|G|=\lambda^+$, which is a $6640$-Shelah semigroup. -Corollary. Under CH there exists a Shelah semigroup of cardinality $\aleph_1$. -Theorem (Protasov, 2010). Each countable Shelah semigroup is finite. -Theorem (folklore?). A commutative group is finite iff it is a Shelah semigroup. -Proposition (@YCor, 2018). A group is finite iff it is a 3-Shelah semigroup. -Theorem (Todorcevic, 1987). There is a commutative binary operation $\cdot:X\times X\to X$ of a set $X$ of cardinality $|X|=\aleph_1$ such that $X=A^2:=\{ab:a,b\in A\}$ for any uncountable subset $A\subset X$. - -Added in Edit, after reading the answer of Keith Kearnes who referred to the paper of Ralph McKenzie who studied Jonsson semigroups. -Let us recall that a semigroup $S$ is Jonsson if $S=\bigcup_{n\in\mathbb N}A^n$ for any subset $A\subset S$ of cardinality $|A|=|S|$. It is clear that each Shelah semigroup is Jonsson. - -Theorem (McKenzie, 1971). A Jonsson semigroup $S$ of infinite cardinality $\kappa$ is a non-commutative group if $\mathrm{cf}(\kappa)>\omega$ or $2^{<\kappa}\le\kappa$. - -McKenzie asked in his paper if this theorem remains true without set-theoretic assumptions. - -Problem (McKenzie, 1971). Is each infinite Jonsson semigroup a group? - -Is this problem of McKenzie still open? - -REPLY [12 votes]: An infinite Shelah semigroup must be a Jonsson semigroup (meaning that it is an infinite semigroup whose proper subsemigroups have lesser power). Therefore the following paper answers the question asked on this page: -McKenzie, Ralph -On semigroups whose proper subsemigroups have lesser power. -Algebra Universalis 1 (1971), no. 1, 21-25. -McKenzie proves that -(1) the only commutative Jonsson semigroups are the generalized cyclic groups, and -(2) under GCH every Jonsson semigroup is the underlying semigroup of a group. -To reiterate and overexplain: infinite + Shelah + commutative implies infinite + Jonsson + commutative implies generalized cyclic. But the generalized cyclic groups are not Shelah. Thus infinite + Shelah + commutative semigroups do not exist.<|endoftext|> -TITLE: Density-$c_0$ in $\ell^\infty$ -QUESTION [9 upvotes]: Let $A \subseteq \mathbb{N}$, define the upper density of $A$ as, -$$ -\overline{\delta}(A) := \limsup_{N\to\infty}\frac{|A\cap\{1,2,3,\cdots,N\}|}{N}. -$$ -This naturally leads to a weaker form of convergence of a sequence $(x_n)$ in $\mathbb{R}$: -A sequence $(x_n)$ converges in density to $x\in \mathbb{R}$ if for every $\epsilon>0$, the set, $$ -\{n\in \mathbb{N}: |x_n - x| \geq \epsilon\} -$$ has upper density zero. Denote this as $D\!-\! \lim_{n\to\infty}x_n =x$. This form of convergence is important in applications of Ergodic Theory to Ramsey Theory. -One can then consider the closed subspace of $\ell^\infty$, -$$ - \overline{\delta}c_0 := \{(x_n)\in \ell^\infty:D\!-\! \lim_{n\to\infty}x_n = 0\} -$$ -I would like to know if there are any references where the $\overline{\delta}c_0$ is studied in $\ell^\infty$. -Thanks in advance! - -REPLY [10 votes]: This type of convergence is often called statistical convergence. -The paper Constantin P. Niculescu, Gabriel T. Prajitura: Some open problems concerning the convergence of positive series -(arXiv:1201.5156) mentioned in connection with the history of this notion that: "The monograph of H. Furstenberg [13] outlines the importance of convergence in density in ergodic theory. In connection to series summation, the concept of convergence in density was rediscovered (under the name of statistical convergence) by Steinhaus [28] and Fast [12] (who mentioned also the first edition of Zygmund’s monograph [31], published in Warsaw in 1935)." - -Statistical convergence can be viewed as a special case of ideal convergence for the ideal of density zero sets -$$\mathcal I=\{A\subseteq\mathbb N; \delta(A)=0\}.$$ -(Or, if you prefer filters, you can take filter convergence w.r.t. the filter of sets with the full density $\delta(A)=1$.) -Of course, since the function $x\mapsto\operatorname{\mathcal I-lim} x_n$ is continuous w.r.t. the sup-norm, the set $c_0(\mathcal I)$ of all bounded $\mathcal I$-null sequences is a closed linear subspace of $\ell_\infty$. -If you are interested also in this generalization, i.e., the space of -$$c_0(\mathcal I)=\{x\in\ell_\infty; \operatorname{\mathcal I-lim} x_n=0\},$$ -then some related papers could be, for example: - -Paolo Leonetti: Continuous Projections onto Ideal Convergent Sequences, Results in Mathematics (2018), 73:114; doi: 10.1007/s00025-018-0876-8 (see also follow-up by paper by Tomasz Kania, A letter concerning Leonetti's paper `Continuous Projections onto Ideal Convergent Sequences', arXiv:1810.09383) -T. Šalát, B.C. Tripathy, M. Ziman: On $\mathcal I$-convergence field, Italian J. of Pure and Appl. Math, 17:45-54, 2005. -Remarks on ideal boundedness, convergence and variation of sequences, J. Math. Anal. Appl. 375 (2011) 431-435; doi: 10.1016/j.jmaa.2010.09.023, author's homepage (Here it is studied in $\ell_\infty(\mathcal I)$ rather than in $\ell_\infty$.)<|endoftext|> -TITLE: Quantum groups at $q=-1$ -QUESTION [5 upvotes]: For a Drinfeld--Jimbo quantized enveloping algebra $U_q(\frak{g})$, it is standard knowledge that the categories of modules are very different in the $q$ a root of unity, and $q$ not a root of unity case. The later being semi-simple, the former not. I wonder how the $q=-1$ case look like. Is it in any sense "less badly behaved" than the complex root of unity case? As a concrete question, for $q=-1$, how far is the category of modules from being semi-simple. - -REPLY [7 votes]: It depends exactly on your conventions, but in a form of the quantum group with a relation along the lines of -$$ -[E, F] = \frac{K - K^{-1}}{q - q^{-1}} -$$ -won't be well-defined at $ q = \pm 1$. To get phenomena associated with second roots of unity, one sets $q$ to be a fourth root of unity. -As an example of what I'm referring to, if $q$ is a primitive $l$th root of unity for $l$ odd, you get cyclic $l$-dimensional representations, while if $q$ is a primitive $2r$th root of unity, you get cyclic $r$-dimensional representations. -For this reason, a lot of literature excludes the $q = \pm 1$ case. -I'm not well-versed in the more combinatorial aspects of quantum groups, and I know that there are other presentations with different behaviors over different coefficient rings. You may want to find someone who knows more to expand on this answer.<|endoftext|> -TITLE: Cohomology of toric blowup -QUESTION [6 upvotes]: Let $n\geq2$. Let $G$ be a linear automorphisms group of prime order on $\mathbb{C}^n$. We assume that $0$ is the unique fixed point of $G$. -I consider the quotient $\mathbb{C}^n/G$. It is a toric variety, so I can consider the toric blowup: $\widetilde{\mathbb{C}^{n}/G}$. -I am interesting in the integral cohomology (singular cohomology) of $\widetilde{\mathbb{C}^{n}/G}$. Especially, I would like to prove that $H^{2k}(\widetilde{\mathbb{C}^{n}/G},\mathbb{Z})$ is torsion free for $k\leq n-1$. Do someone know an efficient method to deal with this kind of problems? Has this cohomology been studied somewhere? -I managed to prove that $H^2(\widetilde{\mathbb{C}^{n}/G},\mathbb{Z})$ and $H^{2n-2}(\widetilde{\mathbb{C}^{n}/G},\mathbb{Z})$ are torsion free using Danilov combinatorial result. However for the other cohomology groups, it becomes too technical to be done by hand. - -REPLY [2 votes]: You can find an answer here (Proposition 3.14).<|endoftext|> -TITLE: K-theory of finite diagram categories -QUESTION [5 upvotes]: Suppose $I$ is a finite $\infty$-category and $F:I\rightarrow\text{fCW}$ is a functor that takes values in finite CW complexes. For each $X\in I$, let $[F(X)]$ denote the class of $F(X)$ in $K_0(\text{fCW})\cong\mathbb{Z}$. -Since evaluation at $X$ is a right exact functor, $[F(-)]$ is a homomorphism $K_0(\text{fCW}^I)\rightarrow\mathbb{Z}$. As we let $X$ vary over equivalence classes of objects in $I$, these assemble into a homomorphism $$K_0(\text{fCW}^I)\xrightarrow{ev}\bigoplus_{X\in I/\cong}\mathbb{Z}.$$ Question: Is this an isomorphism? -Remark: There is a splitting like this one when fCW is replaced by abelian/stable categories. See for example https://arxiv.org/abs/0908.3417 -Edit: Tom Goodwillie points out this is not the isomorphism one would expect, but I am still curious whether $K_0(\text{fCW}^I)$ splits according to the objects of $I$ (and maybe their automorphism groups, as in the paper linked above), or whether it can be computed at all in general. -K-theory can mean so many things, I want to be clear: When I write $K_0(\mathcal{C})$, I mean the abelian group supporting a universal function $[-]:\mathcal{C}\rightarrow K_0(\mathcal{C})$ such that: - -if $X\cong Y$, then $[X]=[Y]$, -$[X\amalg Y]=[X]+[Y]$, -if $A\rightarrow B\rightarrow C$ is a cofiber sequence, $[B]=[A]+[C]$. - -REPLY [2 votes]: If $I = \mathbb Z/2\mathbb Z$ counts as a finite $\infty$ category (Edit: it doesn't), then the answer is no. The space $* \sqcup *$ admits both a free and a trivial $I$ action. We have an additive map given by taking homology and pairing with the alternating representation $$X \mapsto \chi(H^*(X, \mathbb C), {\rm alt}) = \sum_{i} (-1)^i \dim (H^i(X, \mathbb C) \otimes{ \rm alt})^{\mathbb Z/2}$$ which distinguishes between the two actions on $* \sqcup *$.<|endoftext|> -TITLE: Manifold generators of O-bordism invariants -QUESTION [5 upvotes]: If I understand correctly, I can obtain the $O$-cobordism group of -$$ -\Omega^{O}_3(BO(3))=(\mathbb{Z}/2\mathbb{Z})^4, -$$ -The 3d cobordism invariants have 4 generators of mod 2 classes, are generated by -$$g^3,$$ -$$g w_2'(V_{SO(3)}),$$ -$$w_3'(V_{SO(3)}),$$ -$$g w_1(T)^2.$$ -Denote: -The $T$ for the spacetime tangent bundle. -The $O(3)=\mathbb{Z}_2 \times SO(3)$. -The $V_{SO(3)}$ for the vector bundle of $SO(3)$. -Here $g$ is related to the $\mathbb{Z}_2$ generator of $g=H^1(B\mathbb{Z}_2,\mathbb{Z}_2)$. - -Questions: What are the corresponding 3-manifold generators of O-co/bordism invariants $\Omega_{O}^3(BO(3))$, for $g^3,$ $g w_2'(V_{SO(3)}),$ -$w_3'(V_{SO(3)}),$ -$g w_1(T)^2.$? - -Hint: I think the manifold generator for $g^3$ is $\mathbb{RP}^3$. -The manifold generator for $g w_1(T)^2$ is $S^1 \times \mathbb{RP}^2$? - -What are manifold generators of $g w_2'(V_{SO(3)}),$ and $w_3'(V_{SO(3)})$? Can each manifold generator of 4 generators of mod 2 classes be unique and distinct? - -REPLY [7 votes]: $\newcommand{\RP}{\mathbb{RP}}\newcommand{\ang}[1]{\langle #1\rangle}\newcommand{\R}{\mathbb R}$A representative of -a class in $\Omega_3^O(BO_3)$ is a 3-manifold $M$ together with a principal $O_3$-bundle $P\to M$. Principal -$O_3$-bundles are equivalent to rank-3 real vector bundles, so I will use 3-manifolds $M$ together with rank-3 -real vector bundles $E\to M$. In this setting, your invariants correspond to products of Stiefel-Whitney numbers of -$M$ and $E$, evaluated on the mod 2 fundamental class of $M$. -Specifically, I will use the more standard notation $g^3 = \ang{w_1(E)^3, [M]}$, $gw_2'(V_{SO(3)}) = -\ang{w_1(E)w_2(E), [M]}$, $w_3'(V_{SO(3)}) = \ang{w_3(E), [M]}$, and $gw_1(T)^2 = \ang{w_1(E)w_1(M)^2, [M]}$. For -brevity I'll write $w_1(E)^3$ for $\ang{w_1(E)^3, [M]}$, etc., and let $S(M,E)$ denote the tuple $(w_1(E)^3, -w_1(E)w_2(E), w_3(E), w_1(E)w_1(M)^2)$. -Let $\ell_{\RP^n}\to\RP^n$ denote the tautological line bundle (I'll also do this for $\RP^1 = S^1$); if $x\in H^1(\RP^n;\mathbb Z/2)$ denotes the -generator, then $w(\ell_{\RP^n}) = 1+x$. Then, a generating set for -$\Omega_3^O(BO_3)$ is -$$\{ (\RP^3, \ell_{\RP^3}^{\oplus 3}), (\RP^3, \ell_{\RP^3} \oplus\underline\R^2), (S^1\times\RP^2, -\ell_{S^1}\oplus\underline\R^2), (S^1\times\RP^2, \ell_{S^1}\oplus\ell_{\RP^2}\oplus\underline\R)\}. $$ -Specifically, using the Whitney sum formula one can calculate that - -$S(\RP^3, \ell_{\RP^3}^{\oplus 3}) = (1, 1, 1, 0)$, -$S(\RP^3, \ell_{\RP^3}\oplus\underline\R^2) = (1, 0, 0, 0)$, -$S(S^1\times\RP^2, \ell_{S^1}\oplus\underline\R^2) = (0, 0, 0, 1)$, and -$S(S^1\times\RP^2, \ell_{S^1}\oplus\ell_{\RP^2}\oplus\underline\R) = (1, 1, 0, 1)$. - -These four vectors are linearly independent in $\mathbb F_2^4$, so these manifolds generate the bordism group. -If you want manifolds equal to $1$ on one invariant and $0$ on the others, you can take disjoint unions of these -manifolds: for example, $(0, 1, 0, 0)$ is represented by the disjoint union of the last three generators. It may be -possible to find simpler representatives.<|endoftext|> -TITLE: How to find a conformal map of the unit disk on a given simply-connected domain -QUESTION [19 upvotes]: By the classical Riemann Theorem, each bounded simply-connected domain in the complex plane is the image of the unit disk under a conformal transformation, which can be illustrated drawing images of circles and radii around the center of the disk, like on this image taken from this site (Wayback Machine): - -I am interested in finding such transformations for the simply-connected domains having natural origin: oak and maple leaves: - -Is it possible to find and draw corresponding conformal maps? -Maybe there are some online instruments (like Wolframalpha or Maple) for doing such tasks. -The purpose of this activity is to obtain an attractive image for the cover of a textbook on univalent maps of the unit disk. - -REPLY [20 votes]: The geometry processing group at Carnegie Mellon University recently developed an algorithm called Boundary First Flattening that allows you to efficiently and interactively compute conformal parameterizations of triangle meshes. You can download the software here: https://geometrycollective.github.io/boundary-first-flattening/ -It is incredibly powerful and easy to use. Unlike almost all previous algorithms for conformal parameterization it allows for significant control over the boundary shape. - -REPLY [14 votes]: You may want to look at Don Marshall's Zipper algorithm: -https://sites.math.washington.edu/~marshall/zipper.html - -Added in Edit by T. Banakh. This Zipper algorithm yields the following image of the conformal map of the unit disk to an oak leaf. - -Many thanks to Prof. Donald E. Marshall for producing this image (which I post here with his permission).<|endoftext|> -TITLE: How does a statistical divergence change under a Lipschitz push-forward map? -QUESTION [5 upvotes]: Let $\mu, \nu$ be two probability measures on a space $X$ (assume Polish space). -$T: X \rightarrow Y$ is a Lipschitz-map that acts as a push-forward on these measures; let $\mu^\prime = T_{\#\mu}$ and $\nu^\prime = T_{\#\nu}$ be the resulting push-forward measures on the space $Y$. -Let $d(\mu, \nu)$ denote some distance function between measures. -What can be said about how $d(\mu, \nu)$ and $d(\mu^\prime, \nu^\prime)$ relate to each other? Especially, how is the relationship a function of the Lipschitz constant of $T$? -Of course the relationship is likely to be a function of the exact nature of $d()$ (e.g., KL-divergence, or Total-variation or Wassterstein .. ). -I suspect a lot must be known about this type of questions; what would be the right place to look at? - -REPLY [3 votes]: If $d(\mu,\nu)$ is taken to be the total variation metric, then Lipschitz and metric properties don't matter. This is due to a "data processing inequality" of sorts: applying a transformation can only make two distributions closer in TV. I'll illustrate this for discrete sets $X,Y$: -$$ d(\mu',\nu') =\sum_{y\in Y}|\mu'(y)-\nu'(y)| -=\sum_{y\in Y}|\sum_{x\in T^{-1}(y)}\mu(x)-\nu(x)| -\le\sum_{x\in X}|\mu(x)-\nu(x)|=d(\mu,\nu). -$$ -For other metrics, the Lipschitz constant will matter. For example, take two densities $\mu,\nu$ on the real line and put $T(x)=a x$ for some $a>0$. Then -$$ d_p(\mu',\nu'):=(\int |\mu'-\nu'|^p)^{1/p} = a^{(1-p)/p}d_p(\mu,\nu). -$$ -The last example is taken from Ch. 1 of https://www.szit.bme.hu/~gyorfi/nonpar_dens_en.html -; you may find the discussion there useful.<|endoftext|> -TITLE: Collecting proofs of the birth of the giant component -QUESTION [7 upvotes]: I want to collect different proofs of Erdös-Rényi result on the double jump of the largest connected component on $G(n,p)$ (or in $G(n,M)$. -I know the original proof of Erdös-Rényi, the proof that uses Galton-Watson processes and the short proof of Sudakov and Krivelevich. -Are there other (essentially different) proofs of the result? - -REPLY [2 votes]: Nachmias, Asaf, and Yuval Peres. "The critical random graph, with martingales." Israel Journal of Mathematics 176, no. 1 (2010): 29-41. -https://arxiv.org/abs/math/0512201<|endoftext|> -TITLE: Is $\text{PSL}_2(\mathbb{F}_{p^m})$ known to be a Galois group over $\mathbb{Q}$ for $m>1$? -QUESTION [13 upvotes]: Let $\mathbb{F}$ be a finite field of characteristic $p$, is it known that $\text{PSL}_2(\mathbb{F})$ can be realized as a Galois extension of $\mathbb{Q}$ for any/all cases when $\mathbb{F}$ is not $\mathbb{F}_p$? - -REPLY [15 votes]: It is known that ${\rm PSL}_{2}(\mathbb{F})$ can be realized for some $\mathbb{F}$ but definitely not all at present. According to David Zywina's note here, ${\rm PSL}_{2}(\mathbb{F}_{27})$ is the smallest non-abelian finite simple group for which it's not yet known if it occurs as a Galois group over $\mathbb{Q}$. -EDIT: As noted by the OP, David Zywina's paper gives realizations of ${\rm PSL}_{2}(\mathbb{F}_{\ell^{3}})$ for $\ell \equiv \pm 2, \pm 3, \pm 4 \text{ or } \pm 6 \pmod{13}$. This still leave the case of ${\rm PSL}_{2}(\mathbb{F}_{125})$ unresolved. And now the smallest order finite simple group which is not yet known to occur as a Galois group over $\mathbb{Q}$ is $^{2}B_{2}(8)$, the smallest Suzuki group, which has order $29120$. (An approach to solving the inverse Galois problem for this group is discussed on page 32 of William Chen's paper here.)<|endoftext|> -TITLE: Framings for 2-surgeries on 4-manifolds -QUESTION [6 upvotes]: I'm interested in doing $2$-surgeries to $\sharp^k S^1 \times S^3$. That is to the manifold obtained from applying $1$-surgeries to $S^4$. - -Since $\pi_1(O(3)) = \mathbb{Z}_2$, there are two possible framings up to equivalence. Is it possible to distinguish between them in a natural way (similar to how orientability of the resulting manifold works in some other cases)? Maybe something coming from complex geometry? -One more or less natural way I'm thinking about is using Akbulut's convention in the following way. Let $M$ be the manifold obtained by $1$ and $2$ surgeries applied to $S^4$. The handle decomposition of $M$ can be described as follows. There are $k$ $1$-handles. There is a $2$-handle for every $2$-surgery. Then we double this $2$-handlebody to obtain $M$, thus getting twice as many $2$-handles: for every original $2$-handle there will be another handle whose attaching circle (nullhomotopic in $1$-handlebody) is linked with the attaching circle of the original handle. Using Akbulut's dotted circle convention we can assign an integer to every framing of $2$-handles. The handles corresponding to doubling will have framing $0$. If there are no $1$-handles, then even integers will correspond to the same manifold, and odd to the other one (as explained in Gompf and Stipsicz). Does the same hold in the presence of $1$-handles? -If we obtain the answer, can we see which framing corresponds to the manifold obtained by taking the boundary of a neighbourhood of a $2$-complex embedded in $\mathbb{R}^5$? Like in the following construction: -finite generated group realized as fundamental group of manifolds , Constructing 4-manifolds with fundamental group with a given presentation. - -REPLY [5 votes]: You are right that there are two possible ways to perform a 1-surgery, but I don't think that there is a way to choose between them. Think for instance about the simple case where you do a 1-surgery to $S^3 \times S^1$ along the circle $\{p\} \times S^1$. There are two possible choices, but there is no way to choose between them, since there is a self-diffeomorphism of $S^3 \times S^1$ that sends one to the other. They both yield the same manifold $S^4$. -However, you can still resolve this ambiguity in an elegant and simple way by using cohomology and Stiefel - Whitney classes. -You are constructing a 4-manifold $M$ as the boundary of a 5-manifold $W$ obtained with 0-, 1-, and 2-handles, which is in turn obtained by thickening a 2-complex $X$. -The 5-dimensional thickenings $W$ of a 2-complex $X$ are in natural 1-1 correspondence with the elements of $H^2(X, \mathbb Z/_{2\mathbb Z})$ via the second Stiefel - Whitney class (for a proof, see this paper of Hambleton, Kreck, and Teichner). -That is, for every $\alpha \in H^2(X, \mathbb Z/_{2\mathbb Z})$ there is precisely one 5-dimensional thickening $W$ of $X$ such that $w_2(W) = \alpha$. The boundary $M=\partial W$ of course will have $w_2(M) = i^*(w_2(W))$. Note that $i^*\colon H^2(W) \to H^2(M)$ is injective (often not surjective). This is a simple picture to remember. -In particular there is always precisely one thickening $W$ that is spin, that is with $w_2(W)=0$, and also precisely one boundary 4-manifold $M = \partial W$ obtained in this way that is spin. This is the one that you obtain from any embedding of $X$ in $\mathbb R^5$, since in that case $W$ is parallelizable. -When there are no 1-handles, the 2-complex is a bouquet of spheres and hence $H^2(X)$ is a product of one $\mathbb Z/_{2\mathbb Z}$ for each 2-handle. This is the simple case.<|endoftext|> -TITLE: Quadratic Nonresidue -QUESTION [5 upvotes]: Suppose $r$ integers $n_1, \ldots, n_r$ are given such that $0<|n_i| -TITLE: Noetherian local ring and the growth of $\dim_k \operatorname{Ext}^i(k,k)$ -QUESTION [12 upvotes]: Let $A$ be a noetherian local ring with residue field $k$, one can consider $\operatorname{Ext}^i(k,k)$ for every natural number $i$. If it is zero for large $i$, then $A$ is regular and the converse is also true. Then, for which rings the dimension of $\operatorname{Ext}^i(k,k)$ is bounded with respect to $i$ ? -Example: $\mathbb Z /p^n$ -Non-example: $k[x,y]/(x^2,y^2)$ -What about polynomial growth? - -REPLY [11 votes]: To add to Greg's answer and to address zzy's question in the comment: there is an exponential bound. First, the Betti numbers are unchanged by completion, and we can therefore write $R$ as a quotient of a regular local ring $S$ in a minimal way. Then Serre proved long ago that there is a term-wise inequality of power series: -$$\sum \dim Ext^i_R(k,k)t^i \leq \frac{p(t)}{q(t)} $$ -Where $p,q$ are polynomial whose coefficients depends on the minimal resolution of $R$ as a module over $S$. For details, google "Golod rings" (those where the equality is achieved). -To see an easy example of exponential behavior, let $S = k[[x_1,...,x_d]]$ and $R=S/m^2$ where $m$ is the maximal ideal of $S$. Then the first syzygy of $k$ is $m_R= m_R/m_R^2 \cong k^{d}$. So the i-th Betti number is $d^i$. By the way, when $d\leq 2$, and quotient of $S$ is Golod unless if they are complete intersections!<|endoftext|> -TITLE: Understanding the functoriality of group homology -QUESTION [7 upvotes]: EDIT: I've decided to rephrase my question in order for it to be more concise and to the point. -Let $G$ be a group, and let $F_\bullet\rightarrow\mathbb{Z}$ be a free $\mathbb{Z}[G]$-resolution of the trivial $G$-module. In Ken Browns book "Cohomology of groups", given a $G$-module $M$, he defines the homology of $M$ to be the homology of the complex $F_\bullet\otimes_G M$, and defines functoriality as follows: -If $\alpha : G\rightarrow G'$ is a homomorphism of groups, let $F_\bullet'$ be a free $\mathbb{Z}G'$-resolution of $\mathbb{Z}$, and let $\tau : F_\bullet\rightarrow \alpha^\# F_\bullet'$ be a morphism of complexes of $G$-modules, extending the identity on $\mathbb{Z}$. -For $M\in Mod_G, M'\in Mod_{G'}$ and $f : M\rightarrow \alpha^\#M'$ a homomorphism of $G$-modules, then the the map $H_*(G,M)\rightarrow H_*(G',M')$ induced by $(\alpha,f)$ is itself by the map $\tau\otimes f : F_\bullet\otimes_G M\rightarrow \alpha^\#F'_\bullet\otimes_G \alpha^\#M' = F'_\bullet\otimes_{G'}M'$ on chains. -My confusion stems from the following. I am used to the definition of group homology as the derived functors of the $G$-coinvariants functor : $Mod_G\rightarrow Ab$ sending $M\mapsto M_G = M\otimes_{\mathbb{Z}G}\mathbb{Z}$, which identifies $H_*(G,M) = Tor_*^{\mathbb{Z}G}(\mathbb{Z},M)$, which is why the complex $F_\bullet\otimes_G M$ computes the homology of $M$. What's unclear is the functorial dependence of the homology on the complex $F_\bullet\otimes_G M$. -From the derived functor definition of group homology, one has the following notion of functoriality in $G$ and $M$: -Given $\alpha : G\rightarrow G'$ as above, $M\in Mod_G, M'\in Mod_{G'}$, and $f : M\rightarrow\alpha^\# M'$, the usual derived functor functoriality gives us a map -$$f_* : H_*(G,M)\rightarrow H_*(G,\alpha^\#M')$$ -Next, the fact that $H_*(G',-)$ is a universal $\delta$-functor implies that the natural map $(\alpha^\#(-))_G\rightarrow (-)_G$ extends functorially to a morphism of $\delta$-functors: $H_*(G,\alpha^\#(-))\rightarrow H_*(G',-)$, whence a map: -$$cor : H_*(G,\alpha^\#M')\rightarrow H_*(G',M')$$ -My question can be phrased as follows: - -Why is $f_*$ computable using $id\otimes f : F_\bullet\otimes_G M\rightarrow F_\bullet\otimes_G \alpha^\#M'$? (Normally one would take a projective resolution of $M$, then tensor by $\mathbb{Z}$) -Why is the corestriction map $cor$ computable using $\tau\otimes id : F_\bullet\otimes_G \alpha^\# M'\rightarrow \alpha^\#F_\bullet'\otimes_G \alpha^\#M'$? - -I've tried to convince myself of the equivalence, to which I've had a little success, though I'd appreciate it if experts could sketch how they would see that these two definitions are equivalent (different perspectives are always valuable), or perhaps explain why it's obvious from some standard facts which I'm missing. References which explain this would also be appreciated. -BEGIN OLD QUESTION -I'm trying to understand Weibel's description of the action of conjugation on group homology (this is p190 in his "Introduction to Homological Algebra"). I'm having trouble understanding exactly what makes his description valid. -In general, for a morphism of groups $\rho : H\rightarrow G$ and a $G$-module $A$, we may form the $H$-module $\rho^\#A$ with $H$-module structure given via $\rho$, and this functor $\rho^\#$ is exact. -In our case, let $H\le G$ be a subgroup, and let $c_g : H\rightarrow gHg^{-1}$ be the conjugation isomorphism $h\mapsto ghg^{-1}$. -If $A$ is a $G$-module, the abelian group map $\mu_g : A\rightarrow A$ given by $a\mapsto ga$ is actually an $H$-module map from $A$ to $c_g^\#A$. -Thus, $\mu_g$ determines a natural map $(\mu_g)_* : H_*(H,A)\rightarrow H_*(H,c_g^\#A)$. -Moreover, the functors $Mod_H\rightarrow Ab$ given by $M\mapsto H_*(H,c_g^\#A)$ is a homological $\delta$-functor (since $c_g^\#$ is exact), and hence the natural abelian group homomorphism -$$H_0(H,c_g^\#A) = (c_g^\#A)_H\rightarrow A_{gHg^{-1}} = H_0(gHg^{-1},A)$$ -extends to a morphism of $\delta$-functors -$$\zeta_* : H_*(H,c_g^\#A)\rightarrow H_*(gHg^{-1},A)$$ -(This $\zeta_*$ is normally called the corestriction map) -I'd like to understand the composition: -$$\zeta_*\circ(\mu_g)_* : H_*(H,A)\rightarrow H_*(gHg^{-1},A)$$ -Weibel gives the following way of understanding this composition. He picks a projective $\mathbb{Z}G$-resolution $P_\bullet\rightarrow\mathbb{Z}$ (with $\mathbb{Z}$-the trivial $G$-module), and notes that it is simultaneously projective for $\mathbb{Z}G,\mathbb{Z}H$, and $\mathbb{Z}[gHg^{-1}]$. Then he notes that $\mu_g : P_i\rightarrow P_i$ sending $p\mapsto gp$ is an $H$-module chain map $P_i\rightarrow c_g^\# P_i$ lying over the identity map on $\mathbb{Z}$. He deduces that $\zeta_*\circ(\mu_g)_*$ is induced from -$$P\otimes_{\mathbb{Z}H}A\rightarrow P\otimes_{\mathbb{Z}[gHg^{-1}]}A\qquad x\otimes a\mapsto gx\otimes ga$$ -Why does this calculate the morphism $\zeta_*\circ(\mu_g)_*$? -My model of what's going on is the following. Suppose you have a right-exact functor $F : \mathcal{A}\rightarrow\mathcal{B}$ where $\mathcal{A}$ has enough projectives. Then for a morphism $f : X\rightarrow Y$ in $\mathcal{A}$, one can calculate $L_iF(f)$ by picking projective resolutions $P_\bullet\rightarrow X$ and $Q_\bullet\rightarrow Y$, extending $f$ to a morphism of complexes $f_\bullet : P_\bullet\rightarrow Q_\bullet$ (unique up to homotopy), applying $F$ to $f_\bullet$, and taking the induced maps on homologies. (Roughly speaking this works because in the derived category one computes everything by replacing what we care about by projective resolutions.) -However, this doesn't seem to directly apply to Weibel's situation, since we have multiple $\delta$-functors at work. I would very much appreciate it if an expert could describe the situation more clearly for me. - -REPLY [3 votes]: We are basically spelling out that group (co)homology is a functor in both variables. For conjugation by $g\in G$, we have two maps $c_g:H\to gHg^{-1}$ and $\mu_g:A\to A$ which can be composed on the homology level, but it is better to think of the pair $(c_g,\mu_g)$ as a morphism in the category of pairs {(group, coefficient module)}. To compute the induced map on (co)homology, you need a chain map $\tau$ between two projective resolutions (over $H$ and $gHg^{-1}$) which are compatible with $c_g$ (here you take the same resolution $P$ as in your post), and then $\tau\otimes \mu_g$ is a chain map (it's the diagonal one you wrote in your post) that defines the induced map! -By the way, not bashing on Weibel's book at all, but Ken Brown's bible "Cohomology of groups" is the clearest description in my opinion (and this discussion about conjugation is explained in Chapter III.8 in lieu of II.6).<|endoftext|> -TITLE: Number defined by a recursive binary sequence -QUESTION [5 upvotes]: In a math column in Scientific American many years ago, I encountered a peculiar binary sequence I describe below. Unfortunately I can't find a reference on this, so I would be grateful for any pointers or references. -Let $\mathbb{N}$ be the set of positive integers and let $T = \{2^n: n\in \mathbb{N}\cup \{0\}\}$ denote the set of powers of $2$. Let $\text{m}:\mathbb{N}\to T\cup\{0\}$ be defined by $n\mapsto \max\big(\{0\}\cup \{t\in T: t -TITLE: "Exactness" of operadic cohomology -QUESTION [9 upvotes]: There are two somewhat widely known theorems which say - -if $A$ is a nonnegatively graded commutative algebra in char $0$, then forgetful map on operadic cohomology $H^*_{Harr}(A, A) \to H^*_{Hoch}(A, A)$ is injective -if $L$ is a Lie algebra in char $0$, then $H^*(L, L) \to H^*(L, UL)$ is injective - -Are they related at all? Is it a part of more general phenomenon occuring for exact sequences of (probably Koszul) operads? -(Also, it's obvious that latter theorem does not care about grading at all, just because $L \to UL$ is split injection of modules. Is it true for former one? I gess not.) - -REPLY [13 votes]: The second result you mention is significantly easier than the first. Indeed from the PBW theorem we know that $L$ is a direct summand of $UL$ as an $L$-module, and the second result follows. But in fact there is a strong connection between the two results, since the first theorem may also be seen as a consequence of a sufficiently abstract version of the PBW theorem. This is much less well known and I haven't seen it written down anywhere. -(In particular the phenomenon is not something general for any sequence of Koszul operads - you really need a PBW theorem in the background.) -Here is how it goes. Via the map $\mathsf{Lie} \to \mathsf{Ass}$ we may consider $\mathsf{Ass}$ as a bimodule over the Lie operad. Recall that if $P$ is an operad, then a $P$-bimodule is the same thing as a $P$-algebra in the category of right $P$-modules, so we may think of $\mathsf{Lie} \to \mathsf{Ass}$ as a morphism of Lie algebras in a certain tensor category. In fact we may identify $\mathsf{Ass}$ with the universal enveloping algebra of $\mathsf{Lie}$ in this category, considered as a Lie algebra. By a general form of the PBW theorem (for Lie algebras in abstract symmetric tensor categories over a field of characteristic zero) it then follows that there exists a splitting $\phi \colon \mathsf{Ass} \to \mathsf{Lie}$ in the category of modules over the Lie operad, considered as an algebra over itself. Unraveling this, this means that $\phi$ is a map of infinitesimal bimodules. -Explicitly, the fact that $\phi$ is a morphism of infinitesimal bimodules means that we have two commutative pentagons. The first says that the composition -$$ \mathsf{Lie} \circ_{(1)} \mathsf{Ass} \to \mathsf{Ass} \circ_{(1)} \mathsf{Ass} \to \mathsf{Ass} \stackrel\phi\to \mathsf{Lie}$$ -coincides with -$$ \mathsf{Lie} \circ_{(1)} \mathsf{Ass} \stackrel{\mathrm{id}\circ_{(1)} \phi}\to \mathsf{Lie} \circ_{(1)} \mathsf{Lie} \to \mathsf{Lie}.$$ -The other says the same thing except with $\mathsf{Lie}$ acting on $\mathsf{Ass}$ on the right. -Now let $A$ be a $C_\infty$-algebra. The $C_\infty$-structure is given by a Maurer--Cartan element $\mu$ in the pre-Lie algebra $\mathfrak g := \mathrm{Hom}_{\mathbb S}(\mathsf{coLie},\mathsf{End}_A)$, which is essentially the Harrison chain complex. The pre-Lie product $f \star g$ of two Harrison chains is given by the composition -$$ \mathsf{coLie} \to \mathsf{coLie} \circ_{(1)} \mathsf{coLie} \stackrel{f \circ g}\to \mathsf{End}_A \circ_{(1)} \mathsf{End}_A \to \mathsf{End}_A.$$ -We also have the associative version $\mathfrak h := \mathrm{Hom}_{\mathbb S}(\mathsf{coAss},\mathsf{End}_A)$ with analogously defined pre-Lie product. Via the map $\mathsf{coAss} \to \mathsf{coLie}$ we can think of $\mathfrak g$ as a pre-Lie subalgebra of $\mathfrak h$. Now the dual of $\phi$ induces a map $\phi^\ast \colon \mathfrak h \to \mathfrak g$, which is not in general a morphism of pre-Lie algebras. However, the fact that $\phi$ is a map of infinitesimal bimodules is exactly the condition needed for $\phi^\ast$ to satisfy a "projection formula": for $f \in \mathfrak g \subset \mathfrak h$ and $g \in \mathfrak h$ we have $\phi^\ast(f \star g) = f \star \phi^\ast(g)$ and $\phi^\ast(g \star f) = \phi^\ast(g) \star f$. Indeed if we write out the definitions of the pre-Lie product then the conditions we need for the projection formula to hold become exactly that we have a map of infinitesimal bi-comodules $\mathsf{coLie} \to \mathsf{coAss}$. -In particular we have the Harrison differential on $\mathfrak g$ given by $df = f \star \mu - (-1)^{\vert f\vert}\mu \star f$ and the analogous Hochschild differential on $\mathfrak h$. The previous paragraph says in particular that the splitting $\mathfrak h \to \mathfrak g$ is compatible with these differentials, so that Harrison chains are a direct summand of Hochschild chains.<|endoftext|> -TITLE: $f,g \in \mathbb{Z}[x,y]$ satisfying: $\operatorname{Jac}(f,g)=0$ and $f,g \notin \mathbb{Z}[h]$ for every $h \in \mathbb{Z}[x,y]$? -QUESTION [6 upvotes]: Is it possible to find $f,g \in \mathbb{Z}[x,y]$ (with $\deg(f),\deg(g) \geq 1$) such that the following two conditions are satisfied: -(1) $\operatorname{Jac}(f,g)=f_xg_y-f_yg_x = 0$. -(2) There exist no $h \in \mathbb{Z}[x,y]$ such that $f,g \in \mathbb{Z}[h]$. - -Please see the answer to this question, in which it is shown that, -if in the above question we replace $\mathbb{Z}$ by some non-normal integral domain, then the answer is positive. -Any comments are welcome! - -REPLY [7 votes]: $\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}$No. As explained in this question, in $\mathbb{Q}[x,y]$, the condition $\operatorname{Jac}(f,g)=0$ implies that there exists an $h \in \mathbb{Q}[x,y]$ such that $f$ and $g$ are in $\mathbb{Q}[h]$. We now need some lemmas that are basically variants of Gauss's lemma, with multiplication replaced by composition. -Recall that a polynomial with coefficients in $\ZZ$ is called primitive if the set of its coefficients have $GCD=1$. I'll also define a polynomial to be very primitive if the GCD of the coefficients other than the constant term is $1$. -Lemma 1 Let $a \in \ZZ[t]$ be primitive and $b \in \ZZ[x,y]$ be very primitive. Then $a \circ b$ is primitive. -Proof: Suppose to the contrary that $p$ is a prime dividing every coefficient $a \circ b$. Let $\bar{a}$ and $\bar{b}$ denote the reductions modulo $p$, so these are polynomials in $(\ZZ/p)[t]$ and $(\ZZ/p)[x,y]$ respectively. The polynomial $\bar{a}$ is nonzero, $\bar{b}$ is not a constant, and $\ZZ/p$ is a field, so $\bar{a} \circ \bar{b}$ is nonzero. But the hypothesis is that $a \circ b$ is $0$ modulo $p$, and composition commutes with reduction modulo $p$. $\square$ -Lemma 2: Let $b \in \ZZ[x,y]$ be very primitive, let $c \in \QQ[t]$ and suppose that $c \circ b \in \ZZ[x,y]$. Then $c \in \ZZ[t]$. -Proof: Write $c(t) = \tfrac{p}{q} a(t)$ with $a$ primitive and $p$ and $q \in \ZZ$ relatively prime. Then $c(b(x,y,)) = \tfrac{p}{q} a(b(x,y))$ and, by Lemma 1, $a(b(x,y))$ is primitive. So $q$ divides every coefficient of a primitive polynomial, and we deduce that $q=1$. So $c(t) = p a(t) \in \ZZ[t]$. $\square$ -We now prove your result. Let $Jac(f,g)=0$. So there is $h \in \QQ[x,y]$ and $a$ and $b \in \QQ[t]$ such that $f=a \circ h$ and $g = b \circ h$. The case where $h$ is constant is clear, so we assume it is not. -Subtracting a constant from $h$, (and changing $a$ and $b$ appropriately) we may assume that the constant term of $h$ is $0$. Rescaling $h$ by an appropriate element of $\QQ$ (and rescaling the coefficients of $a$ and $b$ correspondingly), we may assume that $h$ is primitive. A primitive polynomial with constant term $0$ is very primitive. So Lemma 2 tells us that $a$ and $b \in \ZZ[t]$, and we are done. $\square$ -This argument generalizes immediately to any UFD, and with a bit more work to any normal ring.<|endoftext|> -TITLE: Is there a name of semidirect product of a group with its automorphism group? -QUESTION [13 upvotes]: Consider the construction $G \rtimes \text{Aut}(G)$. Here $ -G$ is a group, $\text{Aut}(G)$ is the automorphism group and the semidirect product is over the most obvious action. -1) Is there any name for such a general construction? To me, it seems like the most straight-forward example of a semi-direct product. -2) Are there any surveys over such constructions or any big theorems about the structure of such groups? -3) I'm specifically interested in finding torsional elements when $G = F_2$, the free group of two generators. Is there any result about this special scenario? -Edit: More thoughts about question 3 are below. -As pointed in the comments, if you have any torsional automorphism $\phi$ of the free group (there are good classification theorems for such automorphisms), then a torsional element of the semidirect product will be of the form $(g,\phi)$ such that $g\phi(g)\phi^2(g)... \phi^{k-1}(g)=1$, $k$ being the order of $\phi$. -You can check that any element of the form $(\alpha^{-1} \phi(\alpha),\phi)$ works where $\phi$ is a torsional automorphism and $\alpha \in F_2$. Are these all such elements? Is there a general form of such elements? - -REPLY [6 votes]: As a first remark, note that if $\tilde{H}\leq G\rtimes \operatorname{Aut}(G)$ is a finite subgroup and $G$ is torsion-free, then the projection $p: G\rtimes \operatorname{Aut}(G)\to \operatorname{Aut}(G)$ maps $\tilde{H}$ isomorphically to some finite subgroup $H=p(\tilde{H})\leq \operatorname{Aut}(G)$. -Now for each finite subgroup $H\leq \operatorname{Aut}(G)$, you can ask yourself what are the subgroups $\tilde{H}\leq G\rtimes \operatorname{Aut}(G)$ projecting to $H$? These correspond to splittings of the semi-direct product $G\rtimes H$, which correspond to $1$-cocycles (or crossed homomorphisms) $f:H\to G$, that is, functions satisfying -$$ -f(ab) = f(a){}^af(b),\quad a,b\in H. -$$ -Here ${}^ag$ denotes the action of $a$ on $g$. These $1$-cocycles represent elements of the first non-abelian cohomology set $H^1(H;G)$, the trivial element of which is represented by any cocycle of the form $f(a) = (g^{-1}){}^ag$ for some $g\in G$. -So in some sense the answer to your follow-up question lies in the non-abelian cohomolgy of finite subgroups of $\operatorname{Aut}(G)$.<|endoftext|> -TITLE: What is the "analytic" analogue of the valuative criterion of properness -QUESTION [10 upvotes]: Let $X$ be a Hausdorff complex analytic space. Below, let $D$ be the open unit disc in $\mathbb{C}$. Let $D^*$ be the punctured open unit disc. -I am looking for an analogue of the valuative criterion of properness in complex analysis. -Is the following correct? - - -The complex analytic space $X$ is compact if every holomorphic map $D^*\to X$ extends to a holomorphic map $D\to X$. - - -The converse implication is not true, because there are non-extendable maps from $D^*$ to $\mathbb{P}^1$, e.g., $z\mapsto \exp(-1/z^2)$. -I am thinking of $D^*$ as Spec $K$ and $D $ as Spec $R$, where $R$ is a dvr with fraction field $K$. - -REPLY [11 votes]: No, the open disk in $\mathbb C$ is a counterexample (removable singularity theorem plus maximum principle).<|endoftext|> -TITLE: Gluing hexagons to get a locally CAT(0) space -QUESTION [21 upvotes]: I believe that there are four ways to glue (all) the edges of a regular Euclidean hexagon to get a locally CAT(0) space: - -The first two give the torus and the Klein bottle, respectively. What are the last two? In particular, do their fundamental groups have another name? Do they have the same fundamental group? - -EDIT: HJRW points out that I missed the several ways to glue a hexagon to get a non-orientable surface of Euler characteristic -1. I count 8 possibilities up to symmetries. - -REPLY [3 votes]: The fourth example was studied by Brady and Crisp in their CMH paper CAT(0) and CAT(-1) dimensions of torsion-free hyperbolic groups, so it would be reasonable to call its fundamental group the "Brady--Crisp group". (Brady and Crisp also note that it belongs to a family studied by Haglund and Ballmann--Brin.) -They study a one-parameter family of CAT(0) metrics on the complex, and prove the very nice fact that any CAT(-1) model for this group has to have dimension at least 3. (And they exhibit a 3-dimensional CAT(-1) model.)<|endoftext|> -TITLE: Is the complement of an affine open in an abelian variety ample? -QUESTION [14 upvotes]: Let $U$ be an affine open subscheme of an abelian variety $A$ over $\mathbb{C}$. Is $A-U$ an ample divisor? -If $\dim A =1$ this is true. -If $\dim A = 2$, the complement is a divisor $D_1+\ldots + D_n$. If all of these are elliptic curves, then the complement is not affine (as it will contain the translate of one of these elliptic curves). So, wlog $D_1$ is not an elliptic curve. But then $D_1$ is ample (by Nakai-Moishezon). And this implies that $D_1 +\ldots +D_n$ is also ample. -I couldn't figure it out for abelian threefolds. - -REPLY [18 votes]: Welcome new contributor. Yes, that is true. Let $k$ be any field, let $A$ be an Abelian variety over $k$, and let $U\subset A$ be a dense open affine. Denote by $D\subset A$ the complementary divisor with its induced reduced structure. Denote the invertible sheaf of this divisor by $\mathcal{L}:=\mathcal{O}_A(D).$ Denote by $s$ the global section of $\mathcal{L}$ whose zero locus equals $D.$ -Ampleness can be proved after faithfully flat base change over $\text{Spec}\ k$, thus assume that $k$ is algebraically closed. By Lemma II.5.14 of Hartshorne, for every $g\in \mathcal{O}_A(U)$, there exists an integer $n_0$ and a section $\widetilde{g}\in \mathcal{L}^{\otimes n_0}(A)$ such that $\widetilde{g}|_U$ equals $g\cdot s^{n_0}|_U$. Notice that for every integer $r\geq 0$, also $s^r\widetilde{g}|_U$ equals $g\cdot s^{n_0+r}|_U$. Since the $k$-algebra $\mathcal{O}_A(U)$ is finitely generated, there exists an integer $n_0>0$ such that the image of $(s|_U)^{-n}\cdot \mathcal{L}^{\otimes n}(A) \to \mathcal{O}_A(U)$ generates $\mathcal{O}_A(U)$ as a $k$-algebra for every $n\geq n_0$. In particular, the base locus $B_n$ of the complete linear system of $\mathcal{L}^{\otimes n}$ is disjoint from $U$, and the induced morphism to projective space, $$\phi_n:A\setminus B_n \to \mathbb{P}^{d_n}_k,$$ restricts as a locally closed immersion on $U$. -The set $A(k)_{\text{tor}}$ of torsion $k$-points $a$ of $A$ is dense. Thus, denoting by $$\tau_a:A\to A$$ the morphism of translation by $a$, for every $k$-point $p$ of $A$, the set $\{\tau_a(p)| a\in A(k)_{\text{tor}} \}$ is dense in $A$. In particular, not all of these points can lie in the proper, Zariski closed subset $D$. Thus, there exists an integer $m>0$ and an $m$-torsion point $a$ such that $\tau_a^*D$ does not contain $p$. In other words, the point $p$ is contained in the open affine subset $\tau_a^{-1}(U)$. -Since $a$ is $m$-torsion, $\tau_a^*\mathcal{L}^{\otimes n}$ is isomorphic to $\mathcal{L}^{\otimes n}$ for every positive integer $n$ that is divisible by $m$. For such $n$, also the base locus $B_n$ is disjoint from $\tau_a^{-1}(U)$, and $\phi_n$ restricts to a locally closed immersion on $\tau_a^{-1}(U)$. -The set of translates $\tau_a^{-1}(U)$ for $a\in A(k)_{\text{tor}}$ is an open affine covering of $A$. Since $A$ is quasi-compact, there exist finitely many torsion translates of $U$ that cover $A$. For $m>1$ equal to the least common multiple of the orders of those finitely many torsion points, for every positive integer $n$ that is divisible by $m$, the base locus $B_n$ is disjoint from each of these open translates in this open covering. Thus, the base locus $B_n$ is empty. Moreover, the $k$-morphism $\phi_n$ restricts as a locally closed immersion on each of these open translates. -If $\phi_n$ had a positive dimensional fiber, that fiber would intersect one of these open translates in a positive dimensional subvariety, and that would contradict that $\phi_n$ is an immersion on that open translate. Thus, every fiber of $\phi_n$ is finite. Since $\phi_n$ is a morphism between proper $k$-schemes that has finite fibers, the morphism $\phi_n$ is a finite morphism. Since the pullback of an ample invertible sheaf by a finite morphism is ample, the invertible sheaf $\mathcal{L}^{\otimes n} \cong \phi_n^*\mathcal{O}(1)$ is ample on $A$. Finally, since $\mathcal{L}^{\otimes n}$ is ample, also $\mathcal{L}$ is ample.<|endoftext|> -TITLE: Reference for parabolic root systems -QUESTION [5 upvotes]: Let $G$ be a connected reductive group with maximal split torus $A_0$, and $P = MN$ a parabolic subgroup with Levi $M$ containing $A_0$. Let $A_M$ be the split component of $\mathfrak a_M^{\ast} = X(A_M) \otimes \mathbb R$. Then the linear span of $\Phi(A_M,G) \subseteq \mathfrak a_M^{\ast}$ is usually not a genuine root system. But we still have a notion of positive roots, namely $\Phi(A_M,N)$, and a set of simple roots $\Delta_P \subset \Phi(A_M,N)$. -There is some nice affine geometry one can relate to these sorts of systems. For example, one can consider the hyperplanes $H_{\alpha} = \{ h \in \mathfrak a_M : \langle h,\alpha \rangle = 0\}$ for $\alpha \in \Phi(A_M,G)$, and the chambers of $\mathfrak a_M$ cut out by these hyperplanes are in bijection with the parabolic subgroups of $G$ which having $M$ as a Levi subgroup. -Is there any general reference for these "parabolic" root systems? It seems one must deal with these systems when learning about the trace formula or Harish Chandra's results about harmonic analysis on reductive groups. - -REPLY [6 votes]: Try section "Relative roots" here: http://math.stanford.edu/~conrad/249BW16Page/handouts/249B_2016.pdf -They originate from the paper of Borel and Tits, I guess: -http://www.numdam.org/item/PMIHES_1965__27__55_0/<|endoftext|> -TITLE: On $\prod^{(p-1)/2}_{i,j=1\atop p\nmid 2i+j}(2i+j)$ and $\prod^{(p-1)/2}_{i,j=1\atop p\nmid 2i-j}(2i-j)$ modulo a prime $p>3$ -QUESTION [7 upvotes]: QUESTION: Is my following conjecture true? -Conjecture. Let $p>3$ be a prime and let $h(-p)$ be the class number of the imaginary quadratic field $\mathbb Q(\sqrt{-p})$. Then -$$\frac{p-1}2!!\prod^{(p-1)/2}_{i,j=1\atop p\nmid 2i+j}(2i+j)\equiv -\begin{cases}(-1)^{(p-1)/4}\pmod p&\text{if}\ p\equiv 1\pmod4, -\\(-1)^{(h(-p)+1)/2}\pmod p&\text{if}\ p\equiv 3\pmod 4.\end{cases}$$ -Also, -$$\frac{p-3}2!!\prod^{(p-1)/2}_{i,j=1\atop p\nmid 2i-j}(2i-j)\equiv -\begin{cases}1\pmod p&\text{if}\ p\equiv1\pmod4,\\(-1)^{(p-1+2h(-p))/4}\pmod p&\text{if}\ p\equiv3\pmod4.\end{cases}$$ -I have checked the conjecture via a computer. It should be true in my opinion. Your comments are welcome! - -REPLY [6 votes]: Here is a respectively short way to write down what we came up with Dmitry Krachun tonight. -Denote $p=2m+1$. -The idea is very simple: calculate the product $$\prod_{j\in\{s,s+1\}, 1\leqslant i\leqslant m,\atop p\nmid 2i+j} (2i+j).$$ -Note that this is a product of all non-zero residues modulo $p$ except $s+1$, thus it equals $-1/(s+1)$. Now apply this observation for $s=1,3,\dots,m-1$ (if $m$ is even) and $s=2,4,\dots,m-1$ (if $m$ is odd) and multiply, you almost get your double product. -Namely, if $m$ is even (so $p\equiv 1\pmod 4$) you get the whole double product, which appears to be congruent $\pmod p$ to $(-1)^{m/2}/m!!$ as conjectured. -If $m$ is odd, you get that the double product is congruent $\pmod p$ to $\prod_{i=1}^{m-1} (1+2i)\cdot (-1)^{(m-1)/2}/m!!$ (the first product corresponds to the case $j=1$), and since the right hand side of your formula is just $m!$ by Mordell, we need only to prove that $\prod_{i=1}^{m-1} (1+2i)\cdot (-1)^{(m-1)/2}\equiv m!$, that is easy: write each $1+2i$ as $-2(m-i)$, you get $2^{m-1}(m-1)! (-1)^{(m-1)/2}$ in the left hand side and $m!=-(m-1)!/2$ in the right hand side, so we need $2^m (-1)^{(m+1)/2}\equiv 1$ which is clear as $2^m\equiv (-1)^{(p^2-1)/8}=(-1)^{(m+1)/2}$.<|endoftext|> -TITLE: $T_2$-spaces in which no two open sets are homeomorphic -QUESTION [6 upvotes]: This question was about spaces in which all non-empty open sets "look alike". -Now I am interested in the opposite: Is there a $T_2$-space $(X,\tau)$ with $|X|>1$ such that whenever $U\neq V$ are open subsets of $X$, they are not homeomorphic when endowed with the subspace topology? - -REPLY [9 votes]: There is an example in this paper, A method for constructing ordered continua, of an ordered continuum in which no two intervals are homeomorphic. -Because an open set is a union of a disjoint family of intervals a homeomorphism between two open sets it has to map constituents intervals to constituent intervals; as that can only be done by the identity we see that homeomorphic open sets are equal. -And to answer @bof's question: yes there is also a subset of $\mathbb{R}$ with this property. To see that enumerate the set of triples $\langle f,A,B\rangle$, where $A$ and $B$ are disjoint uncountable $G_\delta$-sets and $f:A\to B$ is a homeomorphism as $\bigl<\langle f_\alpha,A_\alpha,B_\alpha\rangle:\alpha<\mathfrak{c}\bigr>$. Recursively, using that the sets $A_\alpha$ and $B_\alpha$ have cardinality $\mathfrak{c}$, choose $x_\alpha\in A_\alpha$ such that $x_\alpha$ and $f_\alpha(x_\alpha)$ are both not in the union $\{x_\beta:\beta<\alpha\}\cup\{f_\beta(x_\beta):\beta<\alpha\}$. -In the end let $X=\{x_\alpha:\alpha<\mathfrak{c}\}$. Claim: $X$ is as required. Assume $f:U\to V$ is a homeomorphism between open subsets and assume $f(x)\neq x$ for some $x$. Pick an open interval $I$ around $x$ such that $f[I\cap X]\cap I=\emptyset$. Apply Lavrentieff's theorem to find $G_\delta$-sets $A$ around $I\cap X$ and $B$ around $f[I\cap X]$ and a homeomorphism $\tilde f:A\to B$ that extends $f$. We can assume $A\subseteq I$, so $A\cap B=\emptyset$. Then $\langle f,A,B\rangle = \langle f_\alpha,A_\alpha,B_\alpha\rangle$ for some $\alpha$ and we find that $x_\alpha\in X$ but $f(x_\alpha)=f_\alpha(x_\alpha)\notin X$. This contradiction shows that $f$ must be the identity and hence that $U=V$.<|endoftext|> -TITLE: Find the tight upper bound of $\sum_{i=1}^n \frac{i}{i+x_i}$, where the $x_i$'s are distinct in $\{1,2,...,n\}$ -QUESTION [10 upvotes]: What is the tight upper bound of $\sum_{i=1}^n \frac{i}{i+x_i}$, where the $x_i$'s are distinct integers in $\{1,2,...,n\}$? - -REPLY [9 votes]: I doubt that there is an exact formula for this maximum, and unfortunately Wolfgang's guess is incorrect. Indeed, let -$$ -a_n = \mathrm{max}_{\sigma \in \mathfrak{S}_n} \sum_{i=1}^n \frac{i}{i + \sigma(i)}. -$$ -Then by considering the standard embedding $\mathfrak{S}_n \times \mathfrak{S}_m \hookrightarrow \mathfrak{S}_{n+m}$, one checks that -$$ -a_{n+m} \geq a_n + a_m, -$$ -so that the sequence $\frac{a_n}{n}$ converges to the number $c = \mathrm{sup}_n \frac{a_n}{n} \in [\frac{1}{2},1]$. Now, I claim that $c > \frac{1}{2}$, so that neither $\mathrm{id}$ nor $(n, 1, 2,...,n-1)$ grant the maximum when $n$ is large. -For $r = r_n = \lfloor \alpha n \rfloor$ with $\alpha \in ]0,1[$, let us consider the cycle $(n-r+1,n-r+2,...,n-1,n,1,2,...,n-r)$. Then we get -\begin{align*} -c &\geq \lim_n \frac{1}{n} \left( \sum_{j=1}^r \frac{j}{n-r+2j} + \sum_{j=r+1}^n \frac{j}{2j-r} \right) \\ -&= \frac{1}{2} + \frac{\alpha}{4} \log \left( \frac{2-\alpha}{\alpha} \right) - \frac{1-\alpha}{4} \log \left( \frac{1+\alpha}{1-\alpha} \right) . -\end{align*} -This is $> \frac{1}{2}$ for $\alpha \in ]0, \frac{1}{2}[$. The maximum is around $\alpha = 0.14868$, where the bound is $c >0.529$.<|endoftext|> -TITLE: Almost graceful tree conjecture -QUESTION [10 upvotes]: The graceful tree conjecture is the following statement: for any tree $T = (V, E)$ with $|V| = n$ there is a bijective map $f: V \to [n]$ such that $D = \{|f(x) - f(y)| \mid xy \in E\} = [n - 1]$. -There are some positive results about narrow classes of trees, as well as computational results for small $n$ (the best I could find is positive for all $n \leq 35$). One could, however, ask for unconditional results about largest $|D|$ achievable for all trees with a given $n$. An easy greedy algorithm yields $|D|$ of size $n / 3$ for any tree of size $n$. Are there better results, for instance with $|D| \geq cn$ for $c > 1/3$, or even $|D| \geq n - o(n)$? - -REPLY [9 votes]: I know no reference, but here is an easy way of achieving $|D|\geq\lceil n/2\rceil$ (for $n\geq 2$, surely). -The vertices of every tree can be decomposed into stars (i.e., graphs of te form $K_{1,d}$ with $d\geq 1$): if a tree is not a star, merely find a non-leaf vertex having just one non-leaf neighbor and single out this vertex with all its leaves; then proceed by induction. -Now, given such star decomposition, we assign to the vertices some $n$ consecutive integers (this suffices due to shift-invariance) in the following way. Take the stars in turn; for every star, assign the largest unassigned negative integer to its center, and the smallest unassigned nonnegative integers to the other vertices. This way, all stars' edges get different labels, and there are at least $\lceil n/2\rceil$ of them. -Remarks. There are some improvements of the algorithm, but they seem to add $o(n)$. E.g.,one may choose a `path of stars' and make the labels of the edges of this path also distinct from each other and from the labels of stars' edges. Alternatively, each star can be equipped with one additional path starting from a non-central vertex: - -REPLY [4 votes]: For the class of trees with maximum degree $o(n/\log n)$, https://arxiv.org/abs/1608.01577 shows that we can achieve a slightly different version of 'almost graceful': if $T$ has not $n$ vertices but only $(1-o(1))n$, we can have an injective $f$ such that your set $D$ has size $|E(T)|$. -It follows that for $n$-vertex trees with maximum degree $o(n/\log n)$ we can have $|D|=(1-o(1))n$. To see this, remove $o(n)$ leaves, use the AAGH result to label the remaining tree, and then replace the leaves, labelling them in any way which completes $f$ to a bijection. This might fail to add any new entries to $D$, but it was already big enough for the claim.<|endoftext|> -TITLE: What is the definition of a $\mathcal{U}$-category? -QUESTION [5 upvotes]: Let $\mathcal{U}$ be a universe. The adaptation of the concept of a locally small category to universes is a $\mathcal{U}$-category. -There are two definitions of $\mathcal{U}$ category I've met. - -$(1)$ A category $\mathsf{C}$ is a $\mathcal{U}$-category if $\forall X,Y \in \mathsf{C}, \mathsf{Hom_C}(X,Y) \in \mathcal{U}$. -$(2)$ A category $\mathsf{C}$ is a $\mathcal{U}$-category if - -$\mathsf{Ob(C)} \subseteq \mathcal{U}$, - -$\forall X,Y \in \mathsf{C}, \mathsf{Hom_C}(X,Y) \in \mathcal{U}$. - - - -(SGA takes a different approach because they define a $\mathcal{U}$-small set not as an element of $\mathcal{U}$, but as a set which is equinumerous to some element of $\mathcal{U}$, but this is not something I would like to discuss here) -As you can see, these two definitions differ in whether we require the set of objects of a $\mathcal{U}$-category to be a subset of $\mathcal{U}$. If I'm not mistaken, the straightforward adaptation of the concept of a locally small category seems to be the second approach: with universes, we treat elements of a universe $\mathcal{U}$ as "sets" and subsets of a universes $\mathcal{U}$ as "classes". -What I want to know is which definition is more useful? In particular, is there any reason to restrict the definition of a $\mathcal{U}$-category by requiring the set of object to be a subset of $\mathcal{U}$? It appears to create some problems with functor categories (for example, given a $\mathcal{U}$-small category $\mathsf{C}$ and a $\mathcal{U}$-category $\mathsf{D}$, $[\mathsf{C,D}]$ is not a $\mathcal{U}$-category). - -REPLY [8 votes]: I asked a similar question (note that there I called "$\mathcal{U}$-locally small categories" what you call "$\mathcal{U}$-categories"). I still don't have any strong opinion about this, but here are a few relevant points. - -One generally wants to put some kind of bound on the objects of a locally small for the same reason one normally uses universes in category theory: in order to make the collection of "all" locally small categories into a legitimate 2-category, for example, so that one can talk about its relationship to other 2-categories. -The question is then: what bound? A natural alternative is to require that the collection of objects belongs to the successor universe of $\mathcal{U}$; or one could parameterize the definition on two universes. - -A minor advantage to requiring the collection of objects of a category $C$ to be a subset of $\mathcal{U}$ is that, if you write "set" for "element of $\mathcal{U}$", then a "set of objects" of $C$ automatically has the correct meaning for settings in which the distinction between sets and classes of objects is important, like the theory of locally presentable categories. If the objects of $C$ don't have to belong to $\mathcal{U}$, then technically $C$ might not have any nonempty "sets" of elements in the above sense; in that case you may find yourself needing a word for "a set which is equinumerous to some element of $\mathcal{U}$"... - -I used to believe something like your final parenthetical, but I think it is actually false. If $x$, $y \in \mathcal{U}$, then $(x, y) \in \mathcal{U}$; and then the key point is that if $A \in \mathcal{U}$ and $B \subset \mathcal{U}$, then each function $f : A \to B$ is actually an element of $\mathcal{U}$ under the standard encoding as a set of ordered pairs, because this set has the same cardinality as $A$ and each member $(a, f(a))$ belongs to $\mathcal{U}$. Then, functors from a $\mathcal{U}$-small category $C$ to a $\mathcal{U}$-category $D$ are also elements of $\mathcal{U}$, and so $[C, D]$ has as set of objects a subset of $\mathcal{U}$. -This argument is somewhat dependent on the encodings of ordered pairs and functions and you may not care for it much. In type-theoretic foundations, an argument like this may not even be possible; that is perhaps one philosophical reason to prefer the requirement that the collection of objects belongs to the successor universe of $\mathcal{U}$.<|endoftext|> -TITLE: Degree of the variety of independent matrices of rank $\leq r$? -QUESTION [9 upvotes]: Consider an $m$-by-$n$ matrix $A$ with entries in a field $k$; we can see $A$ as a point in the affine space $\mathbb{A}^{m n}$. The rank of $A$ will be $\leq r$ (where $r<\min(m,n)$) if and only if every $(r+1)$-by-$(r+1)$ minor of A is $0$. That tells us that the set of all such matrices $A$ forms a variety $V$. Moreover, by higher-dimensional Bézout, we obtain a bound on the sum of the degrees of the components $V_i$ of $V$: it is at most -$$(r+1)^{k_{r,m,n}},$$ -where $k_{r,m,n} = \binom{n}{r+1} \binom{m}{r+1}$ is the number of $(r+1)$-by-$(r+1)$ minors. -Unfortunately, that's quite a large upper bound. Is there another way to characterize $V$, resulting in a better upper bound for $\sum_i \deg(V_i)$? - -REPLY [10 votes]: The affine variety you described are called determinantal rings, and just about everything is known about them: dimension, singular loci, etc. The degree is also called the Hilbert-Samuel multiplicity, and it was computed as determinant of the matrix $a_{ij}= \binom{m+n-i-j}{m-i}$ for $i,j=1,...,r$. See this paper. - -REPLY [7 votes]: One can compute the degree explicitly, by using a natural resolution of singularities of the associated projective variety. Indeed, to give a matrix of a rank $r$ one needs to fix an $r$-dimensional subspace of $k^n$, i.e., a point of $Gr(r,n)$, and a surjective map from $k^m$ to this subspace, i.e., a full-rank vector in the fiber of the vector bundle $k^m \otimes U \cong U^{\oplus m}$, where $U$ is the tautological bundle of the Grassmannian. If we relax the surjectivity (full-rank) condition and projectivize, we obtain a map -$$ -P_{Gr(r,n)}(U^{\oplus m}) \to P^{mn-1}, -$$ -which is birational onto the projectivization of the variety of matrices of rank at most $r$. Now, if $H$ is the relative hyperplane class for the projective bundle, the required degree is equal to $H^N$, where -$$ -N = r(n-r) + mr - 1 = r(n + m - r) - 1 -$$ -is the dimension. The integer $H^N$ can be explicitly computed using the Grothendieck description of the cohomology of a projective bundle and Schubert calculus on $Gr(r,n)$.<|endoftext|> -TITLE: When can a function be made positive by averaging? -QUESTION [37 upvotes]: Let $f: {\bf Z} \to {\bf R}$ be a finitely supported function on the integers ${\bf Z}$. I am interested in knowing when there exists a finitely supported non-negative function $g: {\bf Z} \to [0,+\infty)$ (not identically zero) such that the convolution $f * g$ is also non-negative. In other words, is there a convex combination of translates of $f$ that is non-negative? -From the Laplace transform identity -$$ \sum_n f*g(n) e^{nt} = (\sum_n f(n) e^{nt}) (\sum_n g(n) e^{nt})$$ - there is the obvious necessary condition that the Laplace transform of $f$ must be everywhere positive: -$$ \sum_n f(n) e^{nt} > 0 \hbox{ for all } t \in {\bf R}.$$ -But is this condition also sufficient? That is, if a function $f$ has positive Laplace transform, can it always be averaged to be non-negative? -A possibly more general necessary condition is that for any positive function $h: {\bf Z} \to (0,+\infty)$, one has -$$ \sum_n f(n) h(n-m) > 0$$ -for at least one integer $m$, since if $\sum_n f(n) h(n-m) \leq 0$ for all $m$ then $\sum_n f*g(n) h(n) \leq 0$ for all non-negative $g$, and then one could not make $f*g$ non-negative. Duality suggests that this more general necessary condition is essentially sufficient. The Laplace transform condition corresponds to the special case when $h(n) = e^{nt}$, but I don't know if this case already is strong enough to cover all the others. -EDIT: an equivalent question is to ask when a polynomial of one variable can be written as the quotient of two other polynomials with non-negative coefficients. The obvious necessary condition is that the polynomial has to be positive on $(0,+\infty)$; is this also sufficient? - -REPLY [17 votes]: Here is the divisibility theorem for polynomials (and thus for Laurent polynomials in several variables), chapter 3 in Positive polynomials, convex integral polytopes, and a random walk problem SLN 1282 (1986 or so) [either this, or Positive polynomials and product type actions of compact groups in Memoirs AMS 320; I am out of town, so do not have access to the references]. This can of course be translated back via Fourier transforms, if you want. -Let $Q$ be a polynomial in $d$ variables, let $K$ be its Newton polytope (the convex hull of its exponents with nonzero coefficients). We can reduce to the case that $K$ contains interior in $R^d$ (by changing variables). For each face of $K$, $F$, define $Q_F$ to be the subpolynomial obtained from $Q$ by throwing away all the terms whose exponents don't belong to $F$. Then (assuming $Q$ has at least one positive value on the positive orthant) $Q$ can be multiplied by a polynomial (and additionally we can assume it has no negative coefficients) so that the product has no negative coefficients if and only if (a) $Q$ is strictly positive on the strictly positive orthant and (b) for every face $F$ of dimension one or more, $Q_F$ is strictly positive on the positive orthant. -From this there is an easy corollary, that if merely $Q$ is strictly positive on the positive orthant, then an arbitrarily small perturbation with Newton polytope equalling $K$ and with positive coefficients will render it of this form. -A fine variation is given in Deciding eventual positivity of polynomials, Ergodic theory and dynamical systems -(1986) 6, 57-79, which determines, given $P$ with no negative coefficients, when there exists $n$ such that $P^nQ $ has no negative coefficients. This generalizes the results of Polya (simplex) and Meissner (cube). -A variation occurs in Iterated multiplication of characters of compact connected Lie groups -Journal of Algebra (1995) 173, 67-96, which deals with random walks arising from characters of compact groups, rather than just characters of tori (Laurent polynomials). There are others, including convolution with general measures, etc (write to me, this answer is getting too long). -See also Reference request: one of Poincare's theorems about positive functions<|endoftext|> -TITLE: Function as sum of distances over a connected, compact metric space -QUESTION [6 upvotes]: If $X$ is a connected, compact metric space with distance function $d : X^2 \rightarrow \mathbb{R}^+$, is it true that there exists a positive real number $a$, dependent on $X$ and $d$, such that for any $n$ and for any $x_1, x_2, \cdots, x_n \in X$, there exists $y$ such that $$\frac{1}{n}\left(d(x_1,y) + d(x_2,y) + \cdots + d(x_n,y)\right) = a?$$ -Motivation: trivial when $X$ is the boundary of a circle, a tricky contest problem when $X$ is the boundary of a square (these examples all use Euclidean distance) - -REPLY [5 votes]: It's a classic theorem of O. Gross from 1964. The number $a$ is also unique for a given space $(X,d)$. -There is an exposition at https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/Cleary-Morris-Yost260-275.pdf -The original paper is: -O. Gross, "The rendezvous value of a metric space", Advances in Game Theory, Ann. of Math. Studies no. 52, Princeton, 1964, 49-53<|endoftext|> -TITLE: Does every sheaf embed into a quasicoherent sheaf? -QUESTION [14 upvotes]: Question. Let $X$ be a scheme. Let $\mathcal{E}$ be a sheaf of $\mathcal{O}_X$-modules. Is there always a quasicoherent sheaf $\mathcal{E}'$ together with a monomorphism $\mathcal{E} \to \mathcal{E}'$? -Remark. The coherator yields a way to find a quasicoherent sheaf together with a morphism to $\mathcal{E}$. But I'm interested in finding a quasicoherent sheaf together with a monomorphism from $\mathcal{E}$. -Motivation. There is a way to set up the theory of sheaf cohomology for quasicoherent sheaves without injective or flabby resolutions. If any sheaf of modules would embed into a quasicoherent one, we might be able to extend this development to arbitrary (not necessarily quasicoherent) sheaves of modules. - -REPLY [33 votes]: That already fails for $X$ equal to $\text{Spec}\ R$, where $R$ is a DVR with generic point $\eta = \text{Spec}\ K$. Since there are only two nonempty open subsets of $X$, namely all of $X$ and $\{\eta\}$, there is a straightforward equivalence between the category of $\mathcal{O}_X$-modules and the category of triples $(M,V,\phi)$ of an $R$-module $M$, a $K$-module $V$, and an $R$-module homomorphism $$\phi:M\to V.$$ This is quasi-coherent if and only if $\phi$ induces an isomorphism $$M\otimes_R K \xrightarrow{\cong} V,$$ i.e., the $\mathcal{O}_X$-module is equivalent to $$(M,M\otimes_R K,\iota_M).$$ In particular, consider the $\mathcal{O}_X$-module $$(R,\{0\},0).$$ -For every $\mathcal{O}_X$-module homomorphism of this $\mathcal{O}_X$-module to a quasi-coherent $\mathcal{O}_X$-module, $$(\psi_R,\psi_\eta):(R,\{0\},0) \to (M,M\otimes_R K,\iota_M),$$ the composite $\iota_M\circ \psi_R$ equals $0$. Thus, the image $\psi_R(R)$ is contained in the torsion submodule of $M$. Every torsion quotient of $R$ has nonzero kernel. Thus, $(\psi_R,\psi_\eta)$ is not a monomorphism.<|endoftext|> -TITLE: Minkowski sum of polytopes from their facet normals and volumes -QUESTION [5 upvotes]: By Minkowski's work in the early 1900s, every polytope $P\subset\mathbb R^n$ is determined up to translation by its unit facet normals $u_1,\dots,u_k$ and facet volumes $\alpha_1,\dots,\alpha_k$. Associating to each polytope the set of normal vectors $N(P) = \{\alpha_i u_i\}$ this describes a bijection between polytopes up to translation and balanced vector configurations. (This generalizes to a bijection between convex bodies up to translation and their surface measures.) -When $n=2$ and $P,Q\subset\mathbb R^2$ are polytopes, the set $N(P+Q)$ associated to the Minkowski sum is the union $N(P)\cup N(Q)$ with vectors in the same direction added together. (The sum of the surface measures.) -For $n>2$ these are two different operations: the union of the vector configurations with vectors in the same direction summed up defines the Blaschke sum $P \# Q$. -Is any description of $N(P+Q)$ in terms of $N(P)$ and $N(Q)$ known for $n>2$? - -REPLY [7 votes]: There is no simple description. -A face of the Minkowski sum is the Minkowski sum of faces of the summands. More exactly, if $F_u(P)$ denotes the face of $P$ with outer normal $u$, then -$$ -F_u(P+Q) = F_u(P) + F_u(Q). -$$ -In dimension $2$ this yields the description that you gave. In higher dimensions it can happen that $F_u(P+Q)$ is a facet while $F_u(P)$ and $F_u(Q)$ are faces of smaller dimensions. (Take for example the octahedron and perturb its support numbers in two different ways; the Minkowski sum of these two polytopes might have more than eight faces: new faces will be parallel to the "equators" of the octahedron.) -The normal fan of the Minkowski sum is the coarsest common refinement of the normal fans of the summands. Thus, in order to know the normals to the facets of $P+Q$ you need to know the face structures of $P$ and $Q$, which is not easy to read off the facet volumes $\alpha_i$, $\beta_i$. -Even if you know the normals to the facets, it is not easy to determine the facet volumes (you need to compute the volume of a Minkowski sum). This is not easy already in the combinatorially simplest case, when the face structures (the normal fans) of $P$ and $Q$ are the same.<|endoftext|> -TITLE: Number of permutations that are products of disjoint cycles of distinct length -QUESTION [5 upvotes]: What is the number of permutations $\pi\in S_n$ that are products of disjoint cycles of distinct length? What is the number of permutations that are products of disjoint cycles such that no more than $k$ cycles are of any given length? -I am really interested in the asymptotics of the proportion of permutations in $S_n$ with these properties as $n\to \infty$. What is a good reference for this sort of statistics? - -REPLY [12 votes]: If we denote by $c_i(\sigma)$ the number of cycles of length $i$ in $\sigma$, we can write the exponential generating function of permutations with cycle statistics as -$$\sum_{n\geq 1}\sum_{\sigma\in S_n}\left(\frac{x^n}{n!}\prod_{i\geq 1}t_i^{c_i(\sigma)}\right)=\exp\left(\sum_{i\geq 1} \frac{t_ix^i}{i}\right) =\prod_{i\geq 1} \left(1+\frac{t_ix^i}{i}+\frac{t_i^2x^{2i}}{2i^2}+\cdots\right)$$ -From here we see that the exponential generating function of permutations with distinct cycle sizes can be obtained by removing all terms where any $t_i$ has exponent $\geq 2$, and then setting all $t_i=1$. So we get -$$\prod_{i\geq 1}\left(1+\frac{x^i}{i}\right)$$ -From here the methods of A Hybrid of Darboux's Method and Singularity Analysis in Combinatorial Asymptotics by P. Flajolet, E. Fusy, X. Gourdon, D. Panario, N. Pouyanne show that the coefficient of $x^n$ is asymptotically equal to -$$e^{-\gamma}+\frac{e^{-\gamma}}{n}+O\left(\frac{\log n}{n^2}\right)$$ -which means that our desired number of permutations is asymptotically given by $$n!\left(e^{-\gamma}+\frac{e^{-\gamma}}{n}+O\left(\frac{\log n}{n^2}\right)\right).$$ -Notice that in section 3 they actually provide much more refined asymptotics, in case you wanted more terms. Moreover, I believe their method should let you compute the asymptotics for permutations where no more than $k$ cycles are of any given length. In this case the generating function is given by -$$\prod_{i\geq 1}\left(1+\frac{x^i}{i}+\frac{x^{2i}}{2i^2}+\cdots +\frac{x^{ki}}{k!i^k}\right)$$ -from the same considerations as above.<|endoftext|> -TITLE: Is $j(\tau)^{1/3}$ the hauptmodul for the congruence subgroup generated by $\tau\rightarrow\tau+3, \tau\rightarrow-1/\tau$? -QUESTION [10 upvotes]: The 3rd root of the modular invariant $j$ is -$$ j(\tau)^{1/3}=q^{-1/3}(1+ 248q+ 4124q^2+ 34752q^3+\cdots),$$ -where $q=e^{2\pi i \tau}$. -I was wondering if $j(\tau)^{1/3}$ the hauptmodul for the congruence subgroup generated by $\tau \rightarrow \tau+3, \tau \rightarrow-1/\tau$. -If this true, can we say the following assertion? -If a function $f(\tau)$ that takes the form $f(\tau)=q^{-1/3}(1+\sum_{n=1}^{\infty} a_n q^n)$ with $a_n \geq 0$ and is invariant under $\tau \rightarrow \tau+3, \tau \rightarrow-1/\tau$, then $f(\tau)=j(\tau)^{1/3}$. -Thanks a lot! - -REPLY [11 votes]: The function $x(\tau) = j(\tau)^{1/3}$ is a hauptmodul, just not for the group that you indicate. This function is also invariant under $\tau \mapsto \frac{2 \tau + 1}{\tau + 1}$ and $\tau \mapsto \frac{\tau+1}{\tau + 2}$, and this -means that $x(\tau)$ is a hauptmodul for an index $3$ subgroup of ${\rm SL}_{2}(\mathbb{Z})$, which is the normalizer of a non-split Cartan modulo 3. (This has been known for quite a while. If you want to read a proof of this, see the paper of Imin Chen titled "On Siegel's modular curve of level 5 and the Class Number One Problem" published in the Journal of Number Theory in 1999. Look in Section 4.) The subgroup you specify has index at least $18$ (and it may not even be congruence). -I think that the natural modification of your second question also has a negative answer. (Edited to fix the example.) In particular, -$$ - x(\tau) \frac{j(\tau)}{j(\tau)-1728} = q^{-1/3} (1 + 1976q + 2133020q^{2} + \cdots) -$$ -is a modular function for the same group as $x(\tau)$ and also has non-negative Fourier coefficients. With some thought, one can see that $j(\tau)$ has non-negative Fourier coefficients, and so does $\frac{1}{j(\tau) - 1728} = \frac{E_{4}^{3}/E_{6}^{2} - 1}{1728}$. Here $E_{4} = 1 + 240 \sum_{n=1}^{\infty} \sigma_{3}(n) q^{n}$ and $E_{6} = 1 - 504 \sum_{n=1}^{\infty} \sigma_{5}(n) q^{n}$ are the usual Eisenstein series.<|endoftext|> -TITLE: Making binary matrix positive semidefinite by switching signs -QUESTION [9 upvotes]: Let $A \in \{\pm 1\}^{n \times n}$ be a symmetric matrix whose diagonal entries are $+1$. Let $f(A)$ be the smallest number of signs we need to change in $A$ so that it becomes positive semidefinite (while preserving symmetry). My questions are: - -Let $A_n$ be an $n \times n$ matrix with all off-diagonal entries equal to $-1$. What is the value of $f(A_n)$? -Let $f_{\max}(n)$ be the maximum of $f(A)$ over all suitable $n \times n$ matrices. Is $f_{\max}(n) = f(A_n)$ true? What is the value of $f_{\max}(n)$? - -I am also interested in the same questions for weighted $f(A)$, where changing any element by $x$ costs us $|x|$, and we want to make $A$ positive semidefinite as cheaply as possible. - -REPLY [4 votes]: The answer to part 1 is $\lceil \frac{n^2}{2}\rceil-n$. Any fewer than that and the vector $v$ of all $1$'s will satisfy $v^\top Av< \lfloor\frac{n^2}{2}\rfloor-\lceil \frac{n^2}{2}\rceil\le 0$. Let $w$ be the $n$-by-$1$ vector with $i$ entry $(-1)^i$. Then $w w^\top$ is $\lceil \frac{n^2}{2}\rceil-n$ moves away from $A_n$. It has $n-1$ eigenvalues equal to $0$ and one eigenvalue equal to $n$ corresponding to the eigenvector $w$.<|endoftext|> -TITLE: Need software for subject classification to stick to, for personal library purposes? -QUESTION [6 upvotes]: This question obviously applies to not only mathematics, so maybe I should post it on academia.SE or somewhere else; but then again, mathematical literature has its own specifics related to existing sources of mathematical data on MR, Zentralblatt, arXiv, etc., so I still decided to post it here. -I've got, by now, around 4gb of various mathematical texts in electronic form (pdf, djvu, ps files of papers, books, talk slides, etc.). Clearly I try to keep them under control by sorting them out in subfolders, but sometimes it is hard to figure out where a specific file belongs. So it would be extremely helpful to use some existing classification like the AMS one. Even if none of the existing classifications suits my tastes completely, it is very convenient to use choices made by somebody else already instead of thinking about such choice everytime by myself. -Does anyone know any existing software that would facilitate such a thing? That is, given 4gb of files, look up the titles, tie them to some online classification source, and create a searchable queryable (?) database. - -REPLY [3 votes]: I would try out Papers. You'll need to build your own work flow, it's not like you can throw in the entire library and come back later for a complete classification database, but the software will handle the bookkeeping for you. -It can read both pdf and djvu formats and will automatically -query a variety of repositories (ACM, ADS, Google Books, Google Scholar, JSTOR, MathSciNet, Project Muse, Scopus, Web of Science) for bibliographic information. The database is fully searchable. -A possible work flow might go like this: - -import the entire library into Papers -have Papers search the -MathSciNet repository, and other repositories of choice, to automatically add -metadata to each paper -add key words to each paper by hand -define sets based on the AMS classification scheme -add papers to sets based on key words - -Somewhat time consuming, true, but not too tedious I would think.<|endoftext|> -TITLE: Klein Gordon equation - references -QUESTION [5 upvotes]: The Klein Gordon equation of the form: -$\Delta u+ \lambda u^p=0$ -is been studied for $p = 2$? -(i.e.$\Delta u+ \lambda u^2=0$) -If yes are there references? - -REPLY [5 votes]: Perhaps I should mention what I did in my PhD thesis. I studied the non-homogeneous case, that is $\Delta u=\lambda u^2+f$, where $f=f(x)$ is the data. It is particularly interesting to consider the dependence of the solution in terms of the parameter $\lambda$. I worked in a bounded domain $\Omega$, with Dirichlet boundary condition $u=0$ on $\partial\Omega$. -Let me call a solution stable if the linear operator $-\Delta+2u$ is positive, that is -if $w\mapsto\int_\Omega(|\nabla w|^2+2uw^2)dx$ is equivalently to the (square) norm of $H^1_0(\Omega)$. - -Theorem : There exist two numbers $-\infty\le\lambda_-(f)<\lambda_+(f)\le+\infty$ such that the boundary value problem admits a stable solution if and only if $\lambda\in(\lambda_-(f),\lambda_+(f))$. Moreover, this stable solution is unique (denote it $u_\lambda$). For such parameters, there exists at least one non-stable solution. The map $\lambda\mapsto u_\lambda$ is increasing. - -It happens that $\lambda_+(f)=+\infty$ iff the solution of the linear equation $\Delta U=f$ is non-positive. At a finite extremity of the interval, the solution still exists (it is the limit of $u_\lambda$) but is marginally stable. -Even more interesting is: - -Theorem: The function $\lambda\mapsto u(\lambda)$ is analytic and extends to the upper and lower halves of the complex plane. This extension is continuous up to the real line. When $\lambda\in{\mathbb R}\setminus[\lambda_-(f),\lambda_+(f)]$, this extension provides a (non-real) complex solution which has a stability property, and it is unique in this class, up to complex conjugacy. - -See: Prolongement analytique et nombre de solutions d'une équation aux dérivées partielles elliptique non linéaire paramétrée, C. R. Acad. Sci. Paris Sér. A (1978) 287, pp 1021-1023.<|endoftext|> -TITLE: Can free rational curves lift to ramified covers of Fano varieties? -QUESTION [6 upvotes]: Does there exist $X$ a smooth Fano manifold, $f: Y \to X$ a nontrivial ramified finite cover, $C \subseteq X$ a smooth very free rational curve, such that $f$ is étale over a neighborhood of $C$? -Without the Fano condition, an example is $X = \operatorname{Hilb^2} (\mathbb P^2)$, $Y = \mathbb P^2 \times \mathbb P^2$ blown up at the diagonal, $C$ is the image of a strict transform of a general very free rational curve on $\mathbb P^2 \times \mathbb P^2$, which doesn't intersect the diagonal, so its image doesn't intersect the branch locus and hence the covering is etale over it. -A minimal non-Fano example is the surface $\mathbb P^1 \times \mathbb P^1$ blown up at the four torus fixed points $(0,0),(0,\infty),(\infty,0),(\infty,\infty)$. The ramified cover $\sqrt{xy}$ is ramified at the four divisors at zero and infinity but not at the four exceptional divisors. There is nodal $(2,2)$ curve which intersects each of the exceptional divisors once (and thus the four branched divisors zero times) passing through two general points, so this is an example. -However, there is no Fano surface example. We can fix a point on the branch divisor and a transverse tangent vector. We can degenerate $C$ to a curve going through that point with that tangent vector, which must have an irreducible component positive intersection with the branch divisor. Since $C$ has zero intersection, some other component must have negative intersection with the branch-divisor. Because the branch divisor is a curve, this component must in fact be a component of the branch divisor, with negative self-intersection. So because $X$ is Fano it is a $-1$ curve and we can blow it down, and doing so does not affect the curve class $C$ because $C$ does not intersect the branch divisor. -The motivation is trying to understand Emmanuel Peyre's philosophy in Liberté et accumulation - specifically whether free rational curves, and their associated free rational points, can lie in the exceptional thin subsets in Manin's conjecture. - -REPLY [4 votes]: After some helpful conversations with Johan de Jong, I came up with an example. After finding it I realize I probably should have figured it out earlier. -In fact, $\operatorname{Hilb}^2(\mathbb P^n)$ is an example for any $n>2$. -The blow-up of $\mathbb P^n \times \mathbb P^n$ at the diagonal is a two-to-one cover of $\operatorname{Hilb}^2(\mathbb P^n)$, ramified only at the exceptional divisor. If we take a general rational curve in $\mathbb P^n \times \mathbb P^n$, it will avoid the diagonal, hence lift to the blow-up and descend to $\operatorname{Hilb}^2(\mathbb P^n)$, where it will remain very free, and then lift back to the covering. -It's easier to calculate on this blow-up. Its canonical bundle is $\mathcal O(-n-1,-n-1) \otimes \mathcal O((n-1) E)$. The branch divisor is $E$, so the pullback of the canonical bundle of $\operatorname{Hilb}^2(\mathbb P^n)$ is $\mathcal O(-n-1,-n-1) \otimes \mathcal O((n-2) E)$. It suffices to show that the inverse $\mathcal O(n+1,n+1) \otimes \mathcal O(-(n-2) E)$ is ample. Sections of the line bundle $\mathcal O(1,1) \otimes \mathcal O(- E)$ are given by symplectic bilinear forms. From this, one can see that they define a two-to-one projective embedding and thus are ample (in fact, they are the Plucker coordinates of the Grassmanian of two-dimesnional quotients of $H^0(\mathbb P^n, \mathcal O(1))$ which $\operatorname{Hilb}^2(\mathbb P^n)$ naturally maps to), and then $\mathcal O(3,3)$ is nef, so combining them we get an ample line bundle.<|endoftext|> -TITLE: How to visualize the Microsupport of a Sheaf? -QUESTION [12 upvotes]: I am looking through Persistent homology and microlocal sheaf theory to learn a bit on barcodes. They are require the notion of a microsupport of a sheaf, looks like it could be a rather concrete geometric object. In order to study this particular notion of "barcode" they require microlocal sheaves and derived categories. -The reference book of Kashiwara-Shapira Sheaves on Manifolds has a through general discussion, but now it's hard to extract only the relevant parts to this one application. As for the barcodes paper they say - -(p. 2) The main goal of this paper is to describe constructible $\gamma$-sheaves on $\mathbb{V}$ -(p. 9) The aim of this paper is to describe the category $D^b_{\mathbb{R}c, \gamma^{\circ a}}(\mathbf{k}_{\mathbb{V}})$. - -Before we do any of that, we try to understand the notion of microsupport for a sheaf. - -$F \in D^b(\mathbf{k}_M)$ write as $\mu \text{supp}(F)$ it's microsupport, a closed conic isotropic subset of $T^*M$. -Let $M$ be a real manifold of dimension $\text{dim} M$. Let $\text{Mod}(\mathbf{k}_M)$ be the abelian category of sheaves of $\mathbf{k}$-modules on $M$, and let $\text{D}^b(\mathbf{k}_M)$ denote it's bounded derived category. - -And I think we need slightly more, the notion of a $\gamma$-sheaf: -$$ \text{D}^b_{\gamma^{\circ a}}(\mathbf{k}_{\mathbb{V}}) = \{ F \in \text{D}^b(\mathbf{k}_\mathbb{V}) ; \mu\text{supp} \subset \mathbb{V} \times \gamma^{\circ a} \} $$ -Here $\mathbb{V}$ is a finite dimensional vector space (such as $\mathbb{R}$ or $\mathbb{R}^2$) and $\gamma^\circ$ define a certain cone in a vector bundle $\pi: E \to M$ -So it seems in order to learn this particular theory of barcodes, we need to understand microlocal sheaves a bit better (and I do not). However, there is also other points of view: - -Robert Ghrist Barcodes: The Persistent Topology of Data -Gunnar Carlson Topology and Data - -REPLY [5 votes]: This might be too late, or out-of-date for you, but I've always understood the microsupport best in terms of Morse theory (or microlocal Morse theory, or the stratified Morse theory of Goresky-MacPherson). -In classical Morse theory, if $f : M \to \mathbb{R}$ is a smooth proper function with no critical values in some closed interval $[a,b]$, then the sub-level sets $M_{\leq a} = f^{-1}(-\infty,a]$ and $M_{\leq b} = f^{-1}(-\infty,b]$ are diffeomorphic, with diffeomorphism given by flowing along the one-parameter family of diffeomorphisms induced by (the dual of) $df$. Homologically, this says $H^k(M_{\leq b},M_{\leq a};\mathbb{Z}) = 0$ for all $k$. Sheaf-theoretically, the derived pushforwards $R^k f_*\mathbb{Z}_M^\bullet$ are locally constant sheaves over the interval $[a,b]$. -Microlocally speaking, we'd say (in the first case, with no critical points) that $SS(Rf_*\mathbb{Z}_M^\bullet)$ is just the zero section (away from the endpoints), not saying anything there. -If $f$ has a single (non-degenerate) critical point $p$ of index $\lambda$ over the interval $(a,b)$, then $M_{\leq b}$ is obtained from $M_{\leq a}$ by smoothly attaching a $\lambda$-handle. Homologically, this says $H^{\dim M-k}(M_{\leq b},M_{\leq a};\mathbb{Z}) = 0$ except for $k = \lambda$, where it is a single copy of $\mathbb{Z}$. Sheaf theoretically, changes in the topology of sub-level sets are detected by the failure of (the cohomology sheaves of) $Rf_*\mathbb{Z}_M^\bullet$ to be locally constant over $[a,b]$. -Microlocally, we'd detect $(p,d_pf) \in SS(Rf_*\mathbb{Z}_M^\bullet)$. -It isn't really incorrect to think of the microsupport of a general complex $F^\bullet \in D^b(\mathbb{Z}_M)$ in this way, as a subset of the cotangent bundle that sees "generalized critical points of functions with coefficients in $F^\bullet$". In fact, in the complex-constructible setting, this is an equivalent way of thinking about the microsupport: -Theorem (somewhere near the end of section 8.6 of "Sheaves on Manifolds"): -$(p,d_pf) \in SS(F^\bullet)$ if and only if $\phi_f F^\bullet_p \neq 0$. -where $\phi_f$ is the vanishing cycles functor along $f$. -This is also perhaps the best way to understand Kashiwara-Schapira's "local type" construction for a complex $F^\bullet \in D_{\mathbb{C}-c}^b(\mathbb{Z}_M)$, where purity of local type is equivalent to (middle) perversity.<|endoftext|> -TITLE: Quantum groups and deformations of the monoidal category of $U(\frak{g})$-modules -QUESTION [11 upvotes]: In the first answer for this question is writen, about the braided category of representation of the enveloping algebra $U(\frak{g})$, for $\frak{g}$ a semisimple Lie algebra: - -The space of deformations of the braided tensor structure on these particular categories is one dimensional (I believe this is a result of Drinfeld), so we get a universal family of braided deformations of U(g)-mod by varying q in U_q(g). - -What is the precise reference for this result? I guess here by tensor, the answerer means equivalently monoidal? -What happend if we take out "braided"? Does there exist deformations of the monoidal category of $U(\frak{g})$-modules which are not equivalent to the cateogry of modules of $U_q(\frak{g})$, for some $q$? - -REPLY [5 votes]: The following is not really an answer but a rather too-long comment, with respect to your second question: - -Does there exist deformations of the monoidal category of $U(\frak{g})$-modules which are not equivalent to the cateogry of modules of $U_q(\frak{g})$, for some $q$? - -2-parameter deformations of the universal enveloping algebras $U_{pq}(\frak{g})$ have been studied such as for example: $U_{pq}[sl(2)]$, $U_{pq}[sl(3,C)]$, $U_{pq}[u(2)]$, $U_{pq}[u(1,1)]$, etc. Most of these are based on the two-parameter deformation function -$$ -[x]_{pq}=\frac{q^{x}-p^{-x}}{q-p^{-1}} -$$ -I am not really sure as to whether the representation categories of such deformed algebras are braided or even monoidal or if they may be described as deformations of the symmetric monoidal categories of the representations of $U(\frak g)$. -However, in the first of the above references, the authors claim that: - -the two-parametric quantum group denoted as $U_{pq}[sl(2)]$ admits a class of infinite-dimensional representations which have no classical (non-deformed) and one-parametric deformation analogues, even at generic deformation parameters - -which may be considered -in my understanding- as an indication that the representation categories of $U_q[sl(2)]$ and $U_{pq}[sl(2)]$ cannot be equivalent. -Thus, if i am correct in this and given the excerpt from the OP, the categories of $U_{pq}[sl(2)]$-modules cannot be braided. -Also, given Noah Snyder's answer, such categories of representations of two-parameter deformed algebras cannot even be monoidal. -On the other hand, in the last of the above cited articles, a $pq$-deformed coproduct $\Delta_{pq}$ is given for $U_{pq}[u(2)]$ and an element $R_{pq}$ such that: -$$ -\Delta_{qp}=R_{qp}\Delta_{pq}R_{qp}^{-1} -$$ -For $p=q^{-1}$, $R_{qp}$ reduces for the universal $R$-matrix of Drinfeld. However, i do not generally know if categorical implications (such as an associated braiding or deformed braiding) have been explicitly studied for $R_{qp}$ and more generally for deformations of this kind.<|endoftext|> -TITLE: Is the Euclidean topology on $\mathbb{R}$ contained in a maximal connected topology? -QUESTION [5 upvotes]: If $(X,\tau)$ is a connected space, then $\tau$ need not be contained in a maximal connected topology. -Is the Euclidean topology on $\mathbb{R}$ contained in a maximal connected topology? - -REPLY [8 votes]: Yes, see Maximal connected expansions of the reals by J. A. Guthrie, H. E. Stone and M. L. Wage, Proc. Amer. Math. Soc. 69 (1978), 159-165.<|endoftext|> -TITLE: What is the complexity of counting Hamiltonian cycles of a graph? -QUESTION [6 upvotes]: Since deciding whether a graph contains a Hamiltonian cycle is $NP$-complete, the counting problem which counts the number of such cycles of a graph is $NP$-hard. -Is it also $PP$-hard in the sense that $PP\subseteq P^{\#HAM-CYCLE}$? - -REPLY [3 votes]: In this paper, Liskiewicz et al. state their Lemma 4 as follows: - -Lemma 4: The problem of counting Hamiltonian paths in planar graphs of max-deg $\Delta=3$ is $\#P$-complete under $\leq^p_{r-shift}$-reductions. - -And, the definition of $\leq^p_{r-shift}$-reductions is as follows: - -DEFINITION: Polynomial-time Right-bit-Shift Reduction -Let $f,g:\Sigma^*\rightarrow\mathbb{N}$, $f$ is poly-time right-bit-shift reducible to $g$, denoted $f\leq^p_{r-shift}g$, if there exists a poly-time computable function $R_3:\Sigma^*\rightarrow\mathbb{N}-\{0\}$ and a polynomial-time computable function $R_1:\Sigma^*\rightarrow\Sigma^*$, such that $f(x)=g(R_1(x))\mathrm{div}\ 2^{R_3(x)}$, for all $x$. - -So, yes, $PP\subseteq P^{\#HAM-CYCLE}$.<|endoftext|> -TITLE: Is $\in$-induction provable in first order Zermelo set theory? -QUESTION [6 upvotes]: Are there models of first order Zermelo set theory (axiomatized by: Extensionaity, Foundation, empty set, pairing, set union, power, Separation, infinity) in which $\in$-induction fail? -I asked this question on Mathematics Stack Exchange here 4 days ago, to receive no answer yet, I thought this question already has a well known answer that is somehow evading me. Seeing that nobody answered thus far, makes me wonder if this is really an elementary issue? The whole issue is whether the proof that Foundation implies $\in$-induction necessitate existence of transitive closures for all sets, and since first order Zermelo doesn't prove the existence of transitive closures for all sets, then this raised this issue in my mind. - -REPLY [11 votes]: The answer is no. Take the standard model of Z and add in a $\mathbb{Z}$-sequence of objects, each of whose only element is the previous one. I.e., define -$M=\bigcup_{n<\omega} \bigcup_{m<\omega}\mathcal{P}^n(V_{\omega} \cup (\{\omega + \omega\}\times[-m, \infty))),$ -where we adjust the $\mathcal{P}$ operator to replace each singleton of the form $\{(\omega+\omega,m)\}$ with $(\omega+\omega,m+1).$ -We define the relation $E$ on -$V_{\omega} \cup (\{\omega + \omega\}\times \mathbb{Z})$ by $E \restriction V_{\omega}=\in \restriction V_{\omega}$ and $E \restriction \mathbb{Z}=\{((\omega+\omega,n),(\omega+\omega,n+1)): n \in \mathbb{Z}\},$ with no relations between "integer objects" and "set objects." Extending $E$ to the iterated power sets is done in the natural way. -It's easy to see $(M,E)$ satisfies Z, and furthermore an object has a (set-sized) transitive closure iff its transitive closure class has no integer objects. In particular, every counterexample to "everything has a transitive closure" contains another counterexample, and such counterexamples exist, contradicting $E$-induction.<|endoftext|> -TITLE: How should I think about the Grothendieck-Springer alteration? -QUESTION [10 upvotes]: Given a simple complex Lie algebra $\mathfrak{g}$, recall the Springer resolution of its nilpotent cone $\widetilde{\mathcal{N}}\to \mathcal{N}$. Several times I have seen someone explaining Springer theory in terms of perverse sheaves, and each time the existence of the Grothendieck-Springer alteration $\pi: \widetilde{\mathfrak{g}}\to\mathfrak{g}$ along with the diagram -$$\require{AMScd}\begin{CD} -\widetilde{\mathcal{N}} @>>> \widetilde{\mathfrak{g}} @>>>\mathfrak{t}\\ -@VVV @VVV @VVV\\ -\mathcal{N} @>>> \mathfrak{g}@>>>\mathfrak{t}/W -\end{CD}$$ - is magically pulled out of a hat (eg: There also exist this other thing that...), where $\mathfrak{t}$ is the universal Cartan and $W$ is the Weyl group. Then one uses the fact that $\pi$ is a small map, giving an IC sheaf $\pi_\ast\underline{\mathbb{Q}}_\widetilde{\mathfrak{g}}$ with a natural $W$-action, which in turn induces a $W$-action on the Springer sheaf by some functoriality. I find this unsatisfying because it seems like $\widetilde{\mathfrak{g}}$ is kept mysterious. -My vague question is how should I think about the Grothendieck-Springer resolution and what is its role in modern representation theory? I know this is not a good question, so let me try to refine it by asking the two following questions. - -1) Is there a broader theoretical context to fit the above diagram into where I am given a resolution of singularities $X_0\to Y_0$ (maybe with $X$ symplectic?), and can find a smooth family $X\to T$ and a proper map of smooth varieties $X\to Y$ fitting into the diagram - $$\require{AMScd}\begin{CD} -X_0 @>>> X \\ -@VVV @VVV\\ -Y_0 @>>> Y, -\end{CD}$$ - or is the Springer map special in a sense I don't understand? - -Aside from applications to proving a generalized Springer correspondence, are there other examples where the existence and properties of this remarkable space are used in representation theory? - -2) What are other applications of the Grothendieck-Springer resolution? - -For example, the Springer resolution can be interpreted as a moment map, and David Ben-Zvi's answer to this question shows how this may be interpreted as the semiclassical shadow to Beilinson-Bernstein localization. Is there an analogous quantization of $\widetilde{\mathfrak{g}}\to \mathfrak{g}$? EDIT: I would be particularly interested in applications which are not so closely connected with the Springer resolution. -I'll stop here, since I have probably already asked too many questions. I would greatly appreciate any references to a modern understanding of $\widetilde{\mathfrak{g}}$. - -REPLY [3 votes]: For question 1, the precise statement is due to Namikawa, but perhaps best summarized in Proposition 2.7 of the paper by Braden-Proudfoot-Webster, https://arxiv.org/abs/1208.3863. To fit with the notation of the question, let $ X_0 \rightarrow Y_0 $ be a conical symplectic resolution. (This means that the map is a projective resolution, that $ X $ is symplectic, and we have an action of $ \mathbb C^\times $ on the pair which acts with positive weight on the symplectic form.) -In this case, there exists a universal Poisson deformation $ X \rightarrow H^2(X_0, \mathbb C) $. We can then define $ Y $ to be the affinization of $ X$ and we obtain the diagram -$$\require{AMScd}\begin{CD} -X_0 @>>> X @>>> H^2(X_0, \mathbb C) \\ -@VVV @VVV @VVV\\ -Y_0 @>>> Y @>>> H^2(X_0, \mathbb C)/W -\end{CD}$$ -where $ W $ is Namikawa's Weyl group. -In the case where $ Y_0 = \mathcal N $, this recovers the Grothendieck-Springer diagram above. -There are lots of nice examples of this diagram in general. Of course, the one I like best is when $ Y_0 $ is an affine Grassmannian slice.<|endoftext|> -TITLE: How many maximal length Bruhat paths from $u$ to $w$ can there be? -QUESTION [8 upvotes]: I've been doing some work with saturated Bruhat paths in a Coxeter group between two elements $u\leq w$. It seems to me that if $\ell(u) =0$, then there are at most $\ell(w)! $. I haven't tried to prove this, it's more of an empirical observation. (Now that I think about it, it's pretty easy to prove, because if you fix a reduced word a downward path is uniquely determined by the deletion order of the letters in the word.) -Taking this as a given, if $\ell(u)>1$ there should be fewer than $\ell(w)! $ paths, but I've noticed it's possible that there are more than $(\ell(w) - \ell(u))!$. Is there a known (or easily knowable) bound for the number of paths from $u$ to $w$ that involves $\ell(u) $? -Let's restrict to finite groups, because I'm aware short Bruhat intervals can be large in infinite groups. This doesn't preclude the possibility of the kind of bound I'm asking for, but I'm more interested in finite groups anyway and I would guess we can do better in that case. - -REPLY [3 votes]: If you'll allow me to ignore the restriction to finite groups, there is a conjectured upper bound, phrased in terms of polytopes, depending on both $\ell(u)$ and $\ell(w)$: Conjecture 7.3 of the paper "The cd-index of Bruhat intervals" (Electronic journal of combinatorics 11 (2004), #R74) is that the cd-index is maximized on certain intervals that are isomorphic to "dual stacked polytopes". (If this is the maximum, it is attained: see Proposition 7.2 of the same paper.) -The cd-index is a noncommutative generating function encoding chain enumeration by ranks. The conjecture would in particular imply that the number of saturated chains is maximized on these dual stacked polytopes. -As far as I know, the conjecture is still wide open.<|endoftext|> -TITLE: Which partitions realise group algebras of finite groups? -QUESTION [14 upvotes]: Fix an algebraically closed field $K$ (maybe of characteristic zero first for simplicity, like $\mathbb{C}$). -Given a partition $p=[a_1,...,a_m]$ of an integer $n$. We can identify $p$ with the semisimple algebra $A_p:=M_{a_1}(K) \times \cdots \times M_{a_m}(K)$. -Questions: - -Given a partition $p$, which finite groups $G$ have their group algebra over $K$ being isomorphic to $A_p$? - -Given a natural number $n$, how many partitions p with sum $n$ are there such that $A_p$ is isomorphic to a group algebra? - -How does the sequence of numbers of such partitions begin depending on $K$ (Does it in general depend on the field or perhaps just the characteristic of the field?)? -Probably the answer is very complicated, but can something be said about the rough behavior of the sequence? - - -It should start as follows for $n \geq 1$ and $K=\mathbb{C}$, using GAP: 1,1,1,2,1,4,1,4,2,4,1,9,1,5,4 - -$\textbf{Are those numbers always powers of primes?}$. (The next term takes forever to calculate, but maybe I should wait longer before asking this question....) - -I dont know how complicated those questions are, so partial answers are also welcome. - -REPLY [6 votes]: Working in characteristic zero, let $M(G)$ denote the maximum multiplicity of a character degree $a_i$ appearing in the partition $[a_1,a_2,\ldots, a_m]$. Moretó conjectured that $|G|$ is bounded by a function of $M(G)$; or equivalently, that $M(G)$ tends to infinity with $|G|$. He proved his conjecture for all non-alternating simple groups and reduced it to the case when $G$ is a symmetric group. Craven then proved Moretó's Conjecture for symmetric groups, using a very clever argument with hook lengths.<|endoftext|> -TITLE: What are the 'wonderful consequences' following from the existence of a minimal dense subspace? -QUESTION [15 upvotes]: In Peddechio & Tholens Categorical Foundations they quote PT Johnstone in their chapter on Frames & Locales: - -...the single most important fact which distinguishes locales from spaces: the fact that every locale has a smallest dense sublocale. If you want to 'sell' locale theory to a classical topologist, it's a good idea to begin asking him to imagine a world in which any intersection of dense subspaces would always be dense. Once he has contemplated some of the wonderful consequences that would follow from this result you can tell him this world is exactly the category of locales. - -Q. My classical topology in somewhat rusty and neither Johnstone nor the book in which this quote is embedded in expand upon 'the wonderful consequences'. What might they be? -It seems to me that one obvious result that would be a triviality is Baires Category theorem. Also, given that position and momentum observables are represented by densely defined unbounded operators, is this result useful there, directly or indirectly? - -REPLY [3 votes]: The reason why the minimal dense sublocale of a frame is important is because it is a complete Boolean algebra and complete Boolean algebras are very special kinds of frames. Furthermore, the nucleus corresponding to a minimal dense sublocale gives an example of a frame homomorphism $f:L\rightarrow B$ and we shall see why such homomorphisms are essential to point-free topology. In this post, I will mainly talk about the importance of frame homomorphisms $f:L\rightarrow B$ for the sake of generality. -Complete Boolean algebras make point-free topology much more elegant than it would otherwise be, and the minimal dense sublocale of a frame is an important part of the relation between complete Boolean algebras and point-free topology. -Complete Boolean algebras satisfy some of the highest separation axioms, and they even satisfy some other notable peculiar point-free topological properties. Complete Boolean algebras are always regular, completely regular, normal, zero-dimensional, paracompact, ultraparacompact, extremally disconnected, $P$-frames, etc. Complete Boolean algebras are also notable because all atomless complete Boolean algebras are completely point-free (the points in a complete Boolean algebra are precisely the generic ultrafilters on that complete Boolean algebra). -If $\kappa$ is an uncountable regular cardinal, then we say a regular space $(X,\mathcal{T})$ is a $P_{\kappa}$-space if the intersection of less than $\kappa$ many open sets is still open. We say that a frame $L$ is weakly $\kappa$-distributive if it satisfies the property $x\vee\bigwedge_{i\in I}y_{i}=\bigwedge_{i\in I}(x\vee y_{i})$ whenever $|I|<\kappa$. Therefore, the notion of a weakly $\kappa$-distributive frame is a generalization of the notion of a $P_{\kappa}$-space (Caution: We need to be careful since there are multiple inequivalent but natural ways of generalizing the notion of a $P_{\kappa}$-space to point-free topology). -$\textbf{Theorem:}$ A regular space $(X,\mathcal{T})$ is a $P_{\kappa}$-space if and only if the frame $\mathcal{T}$ is weakly $\kappa$-distributive. -$\textbf{Theorem:}$ A regular frame $L$ is a complete Boolean algebra if and only if it is weakly $\kappa$-distributive for all uncountable regular cardinals $\kappa$. -One should think of a complete Boolean algebra as therefore like a space where the arbitrary intersection of open sets is open, and complete Boolean algebras in a sense behave like discrete spaces. In fact, every sublocale of a complete Boolean algebra is both an open and a closed sublocale. A regular frame is a complete Boolean algebra precisely when it has no proper dense sublocale. -If $B_{L}$ is the minimal dense sublocale of a frame $L$, then there is a surjective frame homomorphism $L\rightarrow B_{L}$ defined by $x\mapsto x^{**}$. -Let $\mathfrak{b}(a)$ denote the smallest dense sublocale of $L$ containing $a$. Then there is a surjective frame homomorphism -$\phi_{a}:L\rightarrow\mathfrak{b}(a)$ defined by $\phi_{a}(x)=(x\rightarrow a)\rightarrow a.$ Then $\mathfrak{b}(a)$ is a complete Boolean algebra and if $S\subseteq L$ is a sublocale which is a complete Boolean algebra, then $S=\mathfrak{b}(a)$ for some $a$. -Fact: Every frame $L$ embeds as a subframe of a product of complete Boolean algebras. In particular, the mapping $\phi:L\rightarrow\prod_{a\in L}\mathfrak{b}(a)$ defined by $\phi(x)=(\phi_{a}(x))_{a\in L}$ is an embedding. -The above fact in a sense of the extension of the fact that every topology $(X,\mathcal{T})$ embeds as a subframe of $\{0,1\}^{X}$. One could also show that every regular frame $L$ is a subframe of a product of complete Boolean algebras using forcing, but such an argument is more difficult and less efficient in terms of the cardinality of the complete Boolean algebras required. -The congruence tower -We shall now describe a construction that allows us to extend a frame $L$ to a much larger frame $M$ but where the frame homomorphisms $f:L\rightarrow B$ are in a one-to-one correspondence with the frame homomorphisms $g:M\rightarrow B$. This construction is only interesting because for each frame $L$ there are many interesting frame homomorphisms $f:L\rightarrow B$ including the frame homomorphism from $L$ onto the minimal dense sublocale of $L.$ -Suppose that $L$ is a frame. Then a congruence on $L$ is an equivalence relation $\simeq$ such that if $v\simeq w,x\simeq y$, then $v\wedge x\simeq w\simeq y$ and if $x_{i}\simeq y_{i}$ for $i\in I$, then $\bigvee_{i\in I}x_{i}\simeq\bigvee_{i\in I}y_{i}$. Let $\mathfrak{C}(L)$ denote the lattice of all congruences on the frame $L$. Then $\mathfrak{C}(L)$ is itself a zero-dimensional frame. Define a mapping $\nabla_{L}:L\rightarrow\mathfrak{C}(L)$ by letting $(x,y)\in\nabla_{L}(a)$ if and only if $x\vee a=y\vee a$. Then $\nabla_{L}$ is an injective frame homomorphism such that each $x\in L$ is complemented in $\mathfrak{C}(L)$. - -$\textbf{Theorem:}$ Suppose that $L$ is a frame, $B$ is a complete - Boolean algebra, and $f:L\rightarrow B$ is a frame homomorphism. Then - there is a unique frame homomorphism $\overline{f}:\mathfrak{C}(L)\rightarrow B$ such that $f=\overline{f}\nabla_{L}$. - -The above theorem would not be as interesting if frame homomorphisms $f:L\rightarrow B$ with $B$ Boolean were rare or difficult to construct. -Suppose that $L$ is a frame. Then define $\mathfrak{C}^{\alpha}(L)$ for each ordinal $\alpha$ by letting $\mathfrak{C}^{0}(L)=L$, $\mathfrak{C}^{\alpha+1}(L)=\mathfrak{C}(\mathfrak{C}^{\alpha}(L))$ and where if $\gamma$ is a limit ordinal, then $\mathfrak{C}^{\gamma}(L)$ is the direct limit of $(\mathfrak{C}^{\alpha}(L))_{\alpha<\gamma}$ where the transitional mappings are produced by the mappings of the form $\nabla_{\mathfrak{C}^{\alpha}(L)}$ and direct limits of such mappings. Let $\nabla_{L}^{\alpha}:L\rightarrow\mathfrak{C}^{\alpha}(L)$ be the canonical embedding. - -$\textbf{Theorem:}$ Suppose that $L$ is a frame, $B$ is a complete - Boolean algebra, $\alpha$ is an ordinal, and $f:L\rightarrow B$ is a - frame homomorphism. Then there is a unique frame homomorphism - $\overline{f}:\mathfrak{C}^{\alpha}(L)\rightarrow B$ such that $f=\overline{f}\nabla_{L}^{\alpha}$. - -We now have examples of towers of structures that, unlike the automorphism group tower, do not stop growing. - -$\textbf{Theorem:}$ There is a frame $L$ such that the congruence - tower $(\mathfrak{C}^{\alpha}(L))_{\alpha}$ never stops growing. - -$\textbf{Forcing and the congruence tower}$ -The existence of many frame homomorphisms $\phi:L\rightarrow B$ play a very important role in how frames behave when you put them into forcing extensions and they are the basis of a field which I have worked on called Boolean-valued point free topology which is essentially about considering frames in forcing extensions (for the set theorists, since frames are complete Boolean algebras, the Boolean-valued model approach to forcing works quite well.). The frame homomorphisms $\phi:L\rightarrow B$ should be thought of as the points that you add to the frame $L$ when you pass $L$ to turn it into a frame a forcing extension using the Boolean-valued model approach to forcing. -Let $\textrm{Sp}_{B}(L)$ be the set of all frame homomorphisms $\phi:L\rightarrow B$. Then $\textrm{Sp}_{B}(L)$ is a $B$-valued structure where we set $\|\phi=\theta\|=b$ when $b$ is the largest element in $B$ with $\phi(x)\wedge b=\theta(x)\wedge b$ for each $b\in B$. Since $\textrm{Sp}_{B}(L)$ is a $B$-valued structure, one should consider $\textrm{Sp}_{B}(L)$ as an object in the Boolean-valued universe $V^{B}$. - -$\textbf{Theorem}:$ Suppose that $L$ is a frame. Then the mapping - $(\nabla_{L}^{\alpha})^{*}:\textrm{Sp}_{B}(\mathfrak{C}^{\alpha}(L))\rightarrow\textrm{Sp}_{B}(L)$ is a bijection. In particular, - $$V^{B}\models\text{There is a bijective continuous function from $\textrm{Sp}_{B}(\mathfrak{C}^{\alpha}(L))$ to $\textrm{Sp}_{B}(L)$.}$$ - -This correspondence is startling since the frame $\mathfrak{C}^{\alpha}(L)$ could have arbitrarily large cardinality.<|endoftext|> -TITLE: Spectra with "finite" homology and homotopy -QUESTION [15 upvotes]: As known, any non-trivial finite spectrum $X$ can not have non-zero homotopy groups $\pi_i(X)$ only for finite number of $i$. As I understand, the same is true for any spectrum $X$ with finitely generated homology groups $H\mathbb{Z}_i(X)$ which are non-zero only for finite number of $i$. -There is a "dual" statement (?) that a spectrum $Y$ with finitely generated homotopy groups $\pi_i(Y)$ which are non-zero only for finite number of $i$ can not has the same property for homology groups $H\mathbb{Z}_i(Y)$. -If all this true, is there a conceptual reason for such a "duality" in complexity? Can these statements being proof uniformly? - -REPLY [16 votes]: Here are two ways of thinking about it. The first comes from the way one proves the final statement you cited: if $X$ has finitely many nonzero homotopy groups which are all finitely generated, then it is an extension of Eilenberg-MacLane spectra, which necessarily have infinitely many cells to kill most of their homotopy groups. This necessitates $X$ having huge (i.e., not finitely generated) homology, etc. Using this observation, one can also prove your first statement that any nonzero finite spectrum must have infinitely many nonzero homotopy groups: every finite spectrum is harmonic (i.e., $\bigvee_{n\geq 0} K(n)$-local, where $K(n)$ is the $n$th Morava $K$-theory), but torsion Eilenberg-MacLane spectra are all dissonant. One now concludes using the observation that $\pi_\ast X\otimes\mathbf{Q} = \mathrm{H}_i(X;\mathbf{Q})$ must be bounded. One can similarly prove the second statement you made. This shows where the duality between Postnikov and cellular decompositions of spectra comes from. -The second way of thinking about this comes from Brown-Comenetz duality. Let us $p$-localize everywhere. Let $I_{\mathbf{Q/Z}}$ denote the Brown-Comenetz dualizing spectrum, defined so that $\pi_\ast F(X, I_{\mathbf{Q/Z}}) = \mathrm{Hom}(\pi_{-\ast} X, \mathbf{Q/Z})$. In his answer to my question at https://mathoverflow.net/a/301928/102390, Strickland argues that if $X$ is a spectrum, then $X$ is $I_{\mathbf{Q/Z}}$-acyclic if and only if $F(X, S^0)$ is contractible. If $X$ is finite, then this happens if and only if $X$ is itself contractible. In other words, nontrivial finite spectra are never $I_{\mathbf{Q/Z}}$-acyclic. However, almost every non-finite spectrum you can concoct (e.g., any spectrum with unbounded cohomology) will be $I_{\mathbf{Q/Z}}$-acyclic; see Strickland's answer for more. (One interesting non-example is the infinite stunted projective space $P^\infty_{-\infty}$, which doesn't look like a finite spectrum, but is in fact equivalent to the $2$-complete sphere by Lin's theorem.) In fact, Drew's answer to my question discusses the dichotomy conjecture, which states that if $E$ is any spectrum which is not $I_{\mathbf{Q/Z}}$-acyclic, then there is a finite spectrum $X$ such that $\langle E \rangle \geq \langle X \rangle$. (One interesting example of a non-finite spectrum $E$ which satisfies this is $\mathbf{C}P^\infty$; an unpublished theorem of Hopkins says that it satisfies $\langle \mathbf{C}P^\infty\rangle = \langle S^0 \rangle$.) -By the way, note that most of your claims are also true unstably. For example, the claim that any nonzero finite complex must have infinitely many nonzero homotopy groups is also true unstably: this is the McGibbon-Neisendorfer theorem. Also note that the distinction you talk about disappears rationally: in the stable world, the rational sphere is just the Eilenberg-MacLane spectrum $\mathbf{Q}$, and unstably, I'm pretty sure that the only spaces whose Hurewicz map is an isomorphism are given by the $K(\mathbf{Q},n) \simeq S^n_\mathbf{Q}$ for odd $n$. Finally, you should also look up Eckmann-Hilton duality (and, in particular, Fuks duality).<|endoftext|> -TITLE: Number of irreducible representations of a finite group over a field of characteristic 0 -QUESTION [28 upvotes]: Let $G$ be a finite group and $K$ a field with $\mathbb{Q} \subseteq K \subseteq \mathbb{C}$. -For $K=\mathbb{C}$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of $G$. -For $K=\mathbb{R}$ the number of irreducible representations of $KG$ is equal to $\frac{r+s}{2}$, where $r$ denotes the number of conjugacy classes of $G$ and $s$ the number of classes stable under inversion. -For $K=\mathbb{Q}$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of cyclic subgroups of $G$. -(You can find quick proofs of those results in the very recent book "A Journey Through Representation Theory: From Finite Groups to Quivers via Algebras" by Caroline Gruson and Vera Serganova, where a nice quick overview of representation theory of finite groups in characteristic 0 is given in chapters 1 and 2.) -Question: - -Are there such nice closed forumlas for other fields $K$? For example quadratic, cubic or cyclotomic field extensions of $\mathbb{Q}$. - -REPLY [30 votes]: There is a characterization due to Berman for any field. In your case let $n$ be the least common multiple of the orders of elements of $G$. Let $\zeta$ be a primitive $n^{th}$ root of unity. Let $H=Gal(K(\zeta)/K)$. We can identify $H$ with a subgroup of $(\mathbb Z/n\mathbb Z)^*$. Call $a,b\in G$ $K$-conjugate if $b^j=gag^{-1}$ for some $g\in G$ and $j\in H$. This is an equivalence relation and the number of irreducible $KG$-modules is the number of $K$-conjugacy classes. -In characteristic $p$ you do the analogous thing but only look at this equivalence relation on $p$-regular elements.<|endoftext|> -TITLE: Physical (GR) Differential Geometry? -QUESTION [6 upvotes]: I am looking for problem lists or books which contain open problems in the area of mathematics motivated by physics. Ideally, I am looking for questions asking about which reduce to some calculation from General Relativity, which might shed light on a purely differential geometric question/conjecture. -If it helps, I have a bachelors degree in mathematics, but am currently doing a masters in Theoretical Physics. My primary motivation is to write my dissertation discussing such a topic. Within my masters studies, I have completed a course on Riemannian Geometry based on do Carmo's book, and a course on GR based on Schutz's book. - -REPLY [4 votes]: See -Alan Coley, "Mathematical General Relativity" (arXiv:1807.08628) - -Abstract. We present a number of open problems within general relativity. After a brief introduction to some technical mathematical issues and the famous singularity theorems, we discuss the cosmic censorship hypothesis and the Penrose inequality, the uniqueness of black hole solutions and the stability of Kerr spacetime and the final state conjecture, critical phenomena and the Einstein-Yang--Mills equations, and a number of other problems in classical general relativity. We then broaden the scope and discuss some mathematical problems motivated by quantum gravity, including AdS/CFT correspondence and problems in higher dimensions and, in particular, the instability of anti-de Sitter spacetime, and in cosmology, including the cosmological constant problem and dark energy, the stability of de Sitter spacetime and cosmological singularities and spikes. Finally, we briefly discuss some problems in numerical relativity and relativistic astrophysics. - -$\,$ -Particularly interesting (and pressing) is the problem of inhomogeneous cosmology that Coley's text considers in section 3.5. On p. 28 he writes: - -An important open question in cosmology is whether averaging of - inhomogeneities can lead to significant backreaction effects on very - large scales. - -This innocent sounding statement acknowledges that the "Backreaction Debate" in mathematical general relativity has remained inconclusive: -The standard model of cosmology assumes that it is accurate to neglect matter inhomogeneity on large cosmic scales. Under this assumption, the model famously finds a positive "cosmological constant". But various informal arguments as well as computer simulations suggests (not fully conclusively so far) that cosmic inhomogeneity may have a non-negligible effect on cosmic evolution which may account for some or even all of the effective cosmological constant. -This shows that there is immense phenomenological impact behind the problem of inhomogeneous cosmology in mathematical general relativity. -Of course there are other open problems, too. See Coley's survey!<|endoftext|> -TITLE: What is known about sufficient conditions for the rigidity of a convex surface? -QUESTION [16 upvotes]: A convex surface is a connected open subset of the boundary of a convex body in $\mathbb{R}^3$. -An "infinitesimal bending" of a convex surface $S$ is a deformation of $S$ given by a velocity field $v:S\rightarrow \mathbb{R}^3$ such that the length of every curve on the surface is preserved. Vector field $v$ is called the "bending field." A bending field is "trivial" if it corresponds to a simple translation or rotation of the surface. A surface is called "rigid" if all of it's bending fields are trivial. -A regular convex surface is one where every point has a neighborhood which can be parametrized by $r(u, v)$ with $r_u \times r_v \neq 0$. -Blaschke proved that a closed regular convex surface is rigid. Cauchy proved that all convex polyhedra (which are irregular) are rigid. -I've begun dabbling in A.V. Pogorelov's "Extrinsic Geometry of Convex Surfaces," where he proves and discusses both of these results in one of the chapters. However, the translation was published in 1973, so I'm curious about more recent research in this area. Namely, what is known about the rigidity of 1) regular convex surfaces which are not closed and 2) irregular convex surfaces which are not polyhedral? - -REPLY [6 votes]: Answer to question 1: If a convex surface is not closed, then generally it is far from rigid as it might admit infinitely many isometric deformations; however, if the surface has $\mathcal{C}^{2,1}$ regularity and positive curvature, then it becomes rigid as soon as one fixes an arbitrarily small curve segment on that surface. This has been proved very recently in a joint work with Joel Spruck, which has been accepted for publication in International Math. Research Notices (IMRN): - -Rigidity of nonnegatively curved surfaces relative to a curve, - arXiv:1805.02580. - -This result may also be extended to nonnegatively curved surfaces under some additional conditions. The proof uses Hormander's unique continuation principle for elliptic PDEs. Our methods also yield a very short proof of Cohn-Vossen's rigidity theorem for smooth closed convex surfaces, via Hopf's maximum principle, which is included in the appendix to the paper.<|endoftext|> -TITLE: Harmonic oscillator in spherical coordinates -QUESTION [9 upvotes]: It is probably the most well-known result in quantum mechanics that the harmonic oscillator can be solved by supersymmetry. -More precisely, the operator -$$-\frac{d^2}{dx^2}+x^2$$ -can be decomposed as -$$-\frac{d^2}{dx^2}+x^2 = \left(-\frac{d}{dx}-x\right)\left(\frac{d}{dx}-x\right)=:a^*a.$$ -When I tried to see whether the same holds true in spherical coordinates I noticed a complete suprise -$$\left(-\frac{d}{dr}-\frac{n-1}{r}-r\right)\left(\frac{d}{dr}-r\right) = -\frac{d^2}{dr^2} -\frac{n-1}{r} \frac{d}{dr} +r^2+(n-1).$$ -The first three terms comprise the harmonic oscillator in spherical coordinates that is -$$-\Delta_r + r^2.$$ -For convenience, I discarded the angular part in this calculation. -But for some reasons I seem to get this superficial term $(n-1)$. Why is that? And can I do anything smarter to avoid this term? - -REPLY [7 votes]: Indeed, the supersymmetric operators do not factorize the Hamiltonian of the three-dimensional harmonic oscillator, there is an additional term. See Creation and annihilation operators, symmetry and supersymmetry of the 3D isotropic harmonic oscillator, equation 16. - -REPLY [4 votes]: Supplementing @CarloBeenakker's answer: a different sort of "factorization" of the Laplacian can be done in terms of the Dirac operator, which in some regards better preserves the nature of the situation, but/and has scalar coefficients in an appropriate Clifford algebra, which in general is not commutative. E.g., on $\mathbb R$, it is $\mathbb C$, which is convenient. But already in two dimensions the relevant Clifford algebra is the Hamiltonian quaternions... -Depending what you want to do, this might be a useful direction to consider, in addition to the "scalarization" of immediate reduction to the spherically symmetric case.<|endoftext|> -TITLE: Inductive folk model structure on strict ω-categories -QUESTION [5 upvotes]: There is a paper of Lafont, Metayer, and Worytkiewicz [1] that constructs a model structure on the category of strict $\omega$-categories that they call the folk model structure. This model structure has many nice properties, but it has a strange class of weak equivalences, which are maps $X\to Y$ that are coinductive equivalences. -The way that these equivalences are defined are as follows: -First, we say that two parallel $n$-cells of a strict $\omega$-category $X$ are equivalent, written $x \sim x^\prime$ if there exists a reversible $n+1$-cell $x\xrightarrow{\sim} x^\prime$. -We say that an $n$-cell $f:a\to b$ is reversible if there is an $n+1$-cell $g:b\to a$ such that $g\circ f \sim \operatorname{id}_a$ and $f\circ g \sim \operatorname{id}_b$. -This gives a coinductive notion of equivalence of $n$-cells, such that specifying an equivalence requires the specification of an infinite tower of reversible arrows witnessing the invertibility. -Then we say that a map $f:X\to Y$ of strict $\omega$-categories is a coinductive weak equivalence if the following two properties hold: - -For every $0$-cell $y \in Y$, there exists a $0$-cell $x$ in $X$ such that $fx \sim y$ -For any parallel pair of $n$-cells $x,x^\prime$ in $X$ and any $n+1$-cell $v:fx\to fx^\prime$ in $Y$, there exists an $n+1$-cell $u:x\to x^\prime$ such that $fu \sim v$. - -Using this definition of weak equivalence, the authors give a set of generating cofibrations: -$$I=\{\partial O^n \hookrightarrow O^n | n\geq 0\}$$ where $O^n$ denotes the globular $n$-disk, and $\partial O^n = O^{n-1} \cup O^{n-1}$ is the union of two parallel $n-1$ disks along their boundaries. -The authors then verify the requirements of Jeff Smith's theorem and give a combinatorial model structure on the category of strict $\omega$-categories where the weak equivalences are as above and the cofibrations are given by $I-\operatorname{Cof}$. -The trouble with this model structure is that it models the projective limit of $n-\operatorname{Cat}$ along the cotruncation functors $n+1-\operatorname{Cat} \to n-\operatorname{Cat}$ given by collapsing all $n+1$-cells to identities. -In the homotopical models for weak $\omega$-categories, we can also exhibit an inductive model structure, where the equivalences are similar to the above ones, except that we also require that all towers of equivalence data terminate after a finite number of steps. -The problem with trying to find such a model on strict $\omega$-categories is that the trivial fibrations with respect to the set of generating cofibrations $I$ as above need not necessarily be inductive equivalences, as can be seen by taking, for example, the cofibrant replacement of the terminal strict $\omega$-category (which has no nontrivial isomorphisms of cells but is coinductively contractible). This means that to find an inductive model structure, we must change the generating cofibrations, and I am not aware of any obvious candidates. -The only possible idea I had was to adjoin to the generating cofibrations the set of maps $$I^\prime = \{\Sigma^n(C(G_2)) \to \Sigma^n(G_2)|n\geq 0\},$$ -where $C(G_2)$ is a cofibrant replacement in the coinductive model structure of the contractible groupoid on two objects $G_2$ and $\Sigma^n$ denotes the $n$th power of the $2$-point suspension functor, where $\Sigma(X)$ is the strict $\omega$-category whose objects are $0,1$ and whose Hom-objects are given by -$$\Sigma(X)(i,j)=\begin{cases}\ast, &\text{if}\qquad i=j\\ X, &\text{if}\qquad ij\end{cases}.$$ -But this seems somehow too strong a requirement, since it means that all cells in trivially fibrant objects with respect to this set of generating cofibrations should have completely strict inverses. -So the question: Is there any known model structure on strict $\omega$-categories where the weak equivalences are the inductive ones? Does the proposed set of generating cofibrations above seem plausible? - -REPLY [3 votes]: Denote by $tr_n$ the "intelligent truncation functor" which quotients an $\omega$-category by its $n$-cells. If I understand you correctly, the model structure you are looking for on strict $\omega$-categories would still have the property: - -the maps $tr_{n+1} C (G_2) \to tr_n C(G_2)$ are trivial fibrations, and so in particular the maps $tr_n C(G_2) \to G_2$ too. - -The issue is that the map $C(G_2) \to G_2$ is a trivial fibration for any model structure satisfying this property, which I believe contradicts what you are looking for (it is not an "inductive" trivial fibration). -This follows form the fact that $C(G_2)$ is the colimit of the sequence of projections $tr_{n+1} C(G_2) \to tr_n C(G_2)$<|endoftext|> -TITLE: Differentiability of Fourier series -QUESTION [11 upvotes]: Consider the function defined by the Fourier series -$$ f(x;\alpha) = \sum_{n=1}^\infty \frac{1}{n^\alpha} \exp(i n^2 x ) , $$ -where $\alpha >1 $. -For what values of $\alpha $ is $f$ differentiable? Based on numerics, it is conjectured that $\alpha = 2$ is a critical value. For $\alpha <2 $, the function is nowhere differentiable; while for $\alpha >2 $, the function is differentiable almost everywhere. - -REPLY [5 votes]: The critical value is $a= 5/2$. -Chamizo and Cordoba showed that the fractal dimension of $f(x,a)$ is $2 + \frac{1}{2}(1/2 - a), \ \ \ 1 \leq a \leq \frac52$.Thus $f(x,a)$ is not differentiable for $a < 5/2$ (differentiable functions have dimension 1). On the other hand, for $\alpha > 5/2$, $f(x,a)$ is differentiable almost everywhere. They also include pictures for the $a = 3/2 case. -The point is that $\sum_{N \leq n \leq 2N} e^{itn^2}$ typically has size around $\sqrt{N}$ (for instance Weyl-differencing) and so at scale $N$ on the Fourier side, we see a decay of about $N^{1/2 - a}$. For differentiability, we want decay of $N^{-2}$ and so the critical value is $5/2$. -See Jaffard's "The spectrum of singularities of Riemann's function" for precise results on the case $\alpha = 2$. He shows the set of points for which $f(x,2)$ is differentiable has Hausdorff dimension zero and gives a very precise formulation of the regularity at each point in terms of the continued fraction expansion of $x$.<|endoftext|> -TITLE: What happened to Suren Arakelov? -QUESTION [113 upvotes]: I heard that Professor Suren Arakelov got mental disorder and ceased research. However, a brief search on the Russian wikipedia page showed he was placed in a psychiatric hospital because of political dissent. -Since in Soviet Union days a healthy person can get into psychiatric hospital because of "sluggish schizophrenia", it is unclear to me whether he was really sick. Perhaps he was tortured and that is why he stopped research? Is he still alive? Does he still give lectures in Moscow? -The whole Arakelov theory obviously owe its foundation to Arakelov's ground-breaking work in 1970s. The invention of Faltings height is stemed from Arakelov theory. I am wondering what is the current situation of Prof. Arakelov (no email, no physical address, only some dubious reports on wikipedia). The whole situation sounds like John Nash, except Arakelov have not recovered. -I ask at here because I honestly do not know who else to ask (I do not know anyone else graduated from Gubkin Russian State University of Oil and Gas in my university). - -Update: -I received an email forwarded from Prof. Beilinson, written by Prof. Bogomolov, which clarified the matter completely. The alleged event did not happen, even though Prof. Arakelov was warned by the government for his actions. Instead Prof. Arakelov was sick due to private personal reasons. As a result I am voting the post to close. -Thanks for everyone's help in this matter. - -REPLY [124 votes]: There are memoirs by Mikhail Zelikin in Russian. He knew Arakelov personally and quite explicitly describes what happened to him. - -Через несколько дней произошло следующее. Был арестован Солженицын. Сурен Аракелов, верный ученик и последователь своего великого учителя, решил вступить в борьбу с режимом. Он изготовил два плаката — на грудь и на спину — с надписью: “Свободу Александру Солженицыну” и отправился на Красную площадь. Там его и арестовали и отправили прямехонько в институт Сербского. Выписали его через пару лет. Игорь Ростиславович приехал его навестить и поразился произошедшей перемене. Внутренний огонь его души был наглухо затоптан и погашен. Его не интересовали ни математика, ни политика, ни даже внимание его некогда обожаемого учителя. Через некоторое время он женился, нашел какую-то рутинную работу и превратился в добросовестного обывателя. Специалисты из института Сербского на этот раз блестяще продемонстрировали свою профессиональную состоятельность. Они превратили гениального мальчишку в “нормальную” посредственность. - -In English: - -A few days later the following happened. Solzhenitsyn was arrested. Suren Arakelov, a loyal disciple and follower of his great teacher [Shafarevich], decided to fight the regime. He made two posters - on his chest and on his back - with the inscription: “Freedom for Alexander Solzhenitsyn” and went to Red Square. There he was arrested and sent straight to the Serbsky Psychiatry Institute. He was discharged a couple of years later. Igor Rostislavovich [Shafarevich] came to visit him and was amazed at the change that had occurred. The inner fire of his soul was trampled and extinguished. He was not interested in mathematics, or politics, or even the attention of his once beloved teacher. After a while he got married, found some routine job and turned into an average man. On this occasion the specialists from the Serbsky Institute have brilliantly demonstrated their professional competence. They turned a genius into a “normal” mediocrity.<|endoftext|> -TITLE: Fundamental representations and weight space dimension -QUESTION [8 upvotes]: For the Lie algebra $\frak{sl}_n$, its fundamental representations can be realised as the exterior powers of the first fundamental representation. From this we can see that their weight spaces are all $1$-dimensional. Is this true for the other series - are the weight spaces of the fundamental representations always $1$-dimensional. If this is true, does it identify the fundamental representations, that is are the fundaental representation precisely those with $1$-dimensional weight spaces? - -REPLY [10 votes]: Let $\mathfrak{g}$ be a simple Lie algebra over an algebraically closed field of characteristic $0$ and denote the fundamental highest weights by $\varpi_1, \ldots, \varpi_l$, where $l$ is the rank of $\mathfrak{g}$ and the ordering of the $\varpi_i$ is the usual one (Bourbaki). -The cases where an irreducible representation of highest weight $\lambda$ has all weight spaces $1$-dimensional are the following: - -Type $A_l$: $\lambda = \varpi_i, c \varpi_1$, or $c\varpi_l$. -Type $B_l$ ($l \geq 2$): $\lambda = \varpi_1$ or $\varpi_l$. -Type $C_l$ ($l \geq 3$): $\lambda = \varpi_1$. For type $C_3$, also $\lambda = \varpi_3$. -Type $D_l$ ($l \geq 4$): $\lambda = \varpi_1$, $\varpi_{l-1}$ or $\varpi_{l}$. -Type $G_2$: $\lambda = \varpi_1$. -Type $E_6$: $\lambda = \varpi_1$ or $\varpi_6$. -Type $E_7$: $\lambda = \varpi_7$. - -So most fundamental representations do not have $1$-dimensional weight spaces. Also, in type $A$ you have non-fundamental representations which have all weight spaces $1$-dimensional. These can be realized as symmetric powers of the natural irreducible representation and its dual. -For a reference, see Chapter 6 in Seitz, The maximal subgroups of classical algebraic groups (Memoirs of the AMS). Seitz works with algebraic groups, but this will of course give you the result for Lie algebras as well. -Seitz actually proves a result over an algebraically closed field of characteristic $p \geq 0$. In characteristic $p > 0$ we get some additional examples, such as $\lambda = d \varpi_i + (p-d-1) \varpi_{i+1}$ in type $A_l$ (for $1 \leq i < l$ and $0 \leq d < p$).<|endoftext|> -TITLE: Bounding the eigenvalues of $B A B^T$ with the eigenvalues of $A$ -QUESTION [6 upvotes]: Given a Hermitian positive semi-definite $n \times n$ matrix $A$ and a rectangular $m \times n$ matrix $B$, is there anything that can be said about the eigenvalues of the matrix $B A B^T$? -It seems to me like one can regroup the product with a test vector $x$ to show that $(x^T B)A(B^T x)$ is at least the smallest eigenvalue of $A$ and at most the largest eigenvalue of $A$. However, this seems like it’s too easy of a solution... - -REPLY [7 votes]: The following paper studies relations between $\lambda(BAB^T)$ and $\lambda(A)$: -Li, Mathias (1999). The Lidskii-Mirsky-Wielandt theorem – additive and multiplicative versions. Numerische Mathematik. January 1999, Volume 81, Issue 3, pp 377–413.<|endoftext|> -TITLE: When can one continuously prescribe a unit vector orthogonal to a given orthonormal system? -QUESTION [58 upvotes]: Let $1 \leq k < n$ be natural numbers. Given orthonormal vectors $u_1,\dots,u_k$ in ${\bf R}^n$, one can always find an additional unit vector $v \in {\bf R}^n$ that is orthogonal to the preceding $k$. My question is: under what conditions on $k,n$ is it possible to make $v$ depend continuously on $u_1,\dots,u_k$, as the tuple $(u_1,\dots,u_k)$ ranges over all possible orthonormal systems? (For my application I actually want smooth dependence, but I think that a continuous map can be averaged out to be smooth without difficulty.) -When $k=n-1$ then one can just pick the unique unit normal to the span of the $u_1,\dots,u_k$ that is consistent with a chosen orientation on ${\bf R}^n$ (i.e., take wedge product and then Hodge dual, or just cross product in the $(k,n)=(2,3)$ case). But I don't know what is going on in lower dimension. Intuitively it seems to me that if $n$ is much larger than $k$ then the problem is so underdetermined that there should be no topological obstructions (such as that provided by the Borsuk-Ulam theorem), but I don't have the experience in algebraic topology to make this intuition precise. -It would suffice to exhibit a global section of the normal bundle of the (oriented) Grassmannian $Gr(k,n)$, though I don't know how to calculate the space of such sections. - -REPLY [67 votes]: $\def\RR{\mathbb{R}}$ This problem was solved by -Whitehead, G. W., Note on cross-sections in Stiefel manifolds, Comment. Math. Helv. 37, 239-240 (1963). ZBL0118.18702. -Such sections exist only in the cases $(k,n) = (1,2m)$, $(n-1, n)$, $(2,7)$ and $(3,8)$. -All sections can be given by antisymmetric multilinear maps (and thus, in particular, can be taken to be smooth). The $(2,7)$ product is the seven dimensional cross product, which is octonion multiplication restricted to the octonions of trace $0$. -The $(3,8)$ product was computed by -Zvengrowski, P., A 3-fold vector product in $R^8$, Comment. Math. Helv. 40, 149-152 (1966). ZBL0134.38401 -to be given by the formula -$$X(a,b,c) = -a (\overline{b} c) + a (b \cdot c) - b (c \cdot a) + c (a \cdot b)$$ -where $\cdot$ is dot product while multiplication with no symbol and $\overline{b }$ have their standard octonion meanings. Note that, if $(a,b,c)$ are orthogonal, the last $3$ terms are all $0$, so the expression simplifies to $- a (\overline{b} c)$; writing in the formula in the given manner has the advantage that $X(a,b,c)$ is antisymmetric in its arguments and perpendicular to the span of $a$, $b$ and $c$ for all $(a,b,c)$.<|endoftext|> -TITLE: Is there a formula for the determinant of a block matrix of this kind? -QUESTION [5 upvotes]: I am looking for an expression that gives the determinant of a matrix of the form -\begin{bmatrix} A & B & 0 & \dots & 0 & C \\ -B & A & B & & 0 & 0 \\ -0 & B & A & \ddots & 0 & \vdots \\ -0 & & & \ddots & A & B \\ -C & 0 & \dots & \dots & B & A \\ -\end{bmatrix} -Just to clarify. The above matrix is a block tridiagonal matrix with "extra" block entries in the "corners" of the matrix. All block entries are of the same size. They are all square matrices of the same size. -My first port of call was to recursively apply the block formula given in the following link under the heading "Block Matrices". -https://en.wikipedia.org/wiki/Determinant -The expression gets quite messy. I was wondering if there was a paper or well known formula among mathematicians which would make easy work of such a task. -Each of the blocks $A,B,C$ are tridiagonal, toeplitz matrices which are invertable. -EDIT: attempting to implement the suggested solution given by Federico -I can write the above matrix as -$$\mathcal{A} = \bar{\mathcal{A}} + UWV^T \in \mathbb{R}^{ml \times ml},$$ where $$U = [e_l \otimes\mathbb{I}_m,e_1 \otimes\mathbb{I}_m], W = diag(C,C), V = [e_1 \otimes\mathbb{I}_m,e_l \otimes\mathbb{I}_m]$$ -and -$$ -\bar{\mathcal{A}} = -\begin{bmatrix} A & B & & & & \\ -B & A & B & & & \\ - & B & A & \ddots & & \\ - & & & \ddots & A & B \\ - & & & & B & A \\ -\end{bmatrix} -$$ -This allows me to say -$$\det\left(\bar{\mathcal{A}} + {UWV}^T\right) = \det (W^{-1} + V^T\bar{\mathcal{A}}^{-1}U)\det{W}\det \bar{\mathcal{A}}$$ -The problem for me now is to somehow find an expression for the determinant of $\bar{\mathcal{A}}$ and the determinant of $W^{-1} + V^T\bar{\mathcal{A}}^{-1}U$. - -REPLY [4 votes]: Just a sketch of an idea that seems to work: - -You can get rid of the corrections $C$ using the matrix determinant lemma (or, better, replace them with $B$, which makes the matrix block circulant). -Once you have made those corrections, you can change basis using the generalized Schur decomposition and reduce to a case in which $A$ and $B$ are upper triangular. -After you have done this reduction, you can reorder the entries to obtain a block triangular matrix, in which every diagonal block is tridiagonal Toeplitz or circulant. Those matrices have explicit formulas for their determinants (and, actually, even for their eigenvalues).<|endoftext|> -TITLE: smooth homotopy 4-balls with sphere boundary in dimension 4 -QUESTION [6 upvotes]: What follows is, as far as I can tell, totally standard folklore. I have one particular point of confusion, other than that, I wanted to confirm that I am uttering the incantations correctly. -The smooth 4-dimensional Poincare conjecture (SPC4) is the statement that any smooth 4-dimensional manifold $\Sigma$ that is homeomorphic to $S^4$ is diffeomorphic to $S^4$. Is SPC4 equivalent to the conjecture that any smooth 4-manifold $B$ with $\partial B = S^3$ that is homotopy equivalent to $B^4$ is diffeomorphic to $B^4$? -Going one direction: Assume SPC4 is true. -If $B$ is a homotopy 4-sphere with $\partial B = S^3$ then we can fill the boundary with $B^4$ and we obtain a simply connected (by van Kampen) homology $S^4$ (by Mayer-Vietoris), which (by Hurewicz and Whitehead) must be a homotopy $S^4$, which (by Freedman) must be homeomorphic to $S^4$, which (assuming SPC4) is then diffeomorphic to $S^4$. -I imagine that we are about ready to conclude that $B$ must be standard - since it is now sitting inside of $S^4$ with its boundary bounding a standard $B^4$ on the other side. How do we finish up? Can we ambiently isotope $S^4$ so as to have the image of the $S^3$ be just the usual $S^3 \subset S^4$? -Going the other direction: Assume that every smooth homotopy $B^4$ with boundary $S^3$ is diffeomorphic to $B^4$. -Suppose that $\Sigma$ is a smooth homotopy $S^4$. Take a small 4-ball $B^4$ in $\Sigma$ and remove it. What is left (by van Kampen) is a smooth homotopy $B^4$ with boundary $S^3$ which (by assumption) is then diffeomorphic to $B^4$. Now (by Cerf) since $\Sigma$ is just the union of two copies of $B^4$, it is in fact diffeomorphic to $S^4$. - -REPLY [6 votes]: Yes, you can perform that ambient isotopy: any oriented embedding $i: B^n \to M^n$ is isotopic to any other. (This is a lemma proven independently by Cerf and Palais1, but the idea is quite clear: shrink the image of $i$ until it's contained in the chart, then take the limit that defines the derivative of a map.) -In particular, if $h$ is your diffeomorphism $B \cup_{\partial B} B^4 \to S^4$, you may isotope the embedding of $h(B^4)$ so that it is the standard inclusion of the north hemisphere. Then $h$ restricts to a diffeomorphism $h: B \to B^4$, where this $B^4$ is the southern hemisphere of $S^4$. -1The references given on this Manifold Atlas page are Palais, Theorem 5.5 (the essential content is Lemma 5.2) and somewhere in Cerf's treatise on embedding spaces.<|endoftext|> -TITLE: Calculating the Ext-algebra with a computer -QUESTION [6 upvotes]: Given a finite dimensional quiver algebra $A$ over an arbitrary field and a module $M$ of finite injective dimension or finite projective dimension. -Let $B$ be the Ext algebra of $M$, that is $B:=\bigoplus_{k=0}^{\infty}{Ext_A^i(M,M)}$. Note that $B$ is finite dimensional because of the assumptions on $M$. -Questions: -1.Is there a computer software that can check whether $B$ is again a quiver algebra and can calculate the quiver and relations of $M$ in case it is a quiver algebra? -I am just aware of the GAP-package QPA, which is able to calculate minimal generators of $B$. - -Is there a precise condition when $B$ is a quiver algebra? - -REPLY [4 votes]: Q2: By Auslander-Reiten-Smalø, Theorem 1.9 page 65, says: - -If $A$ is a finite dimensional elementary algebra over a field $k$ (that is, $A/\operatorname{rad}(A) \simeq k^n$ for some $n$), then $A\simeq kQ/I$ for some finite quiver $Q$ and some admissible ideal $I$ in $kQ$. - -If $A\simeq kQ/I$ for a finite quiver $Q$ and an admissible ideal $I$ in $kQ$, the radical of $A$ is $\langle\textrm{arrows}\rangle/I$ and $A/\operatorname{rad}(A) \simeq kQ/\langle\textrm{arrows}\rangle \simeq k^{|Q_0|}$. Hence $A$ is an elementary algebra, and the above characterizes admissible quotients of path algebras. -This applied to $B=\operatorname{Ext}^*_A(M,M)$ gives that $M$ must be a basic module, that is, $M \simeq \oplus_{i=1}^n M_i$ with $M_i$ indecomposable and $M_i\not\simeq M_j$ for $i\neq j$. Furthermore, for $B$ to elementary $\operatorname{End}_A(M_i)/\operatorname{rad}\operatorname{End}_A(M_i)$ must be isomorphic to a fixed field $K$ for all $i$. This ensures that $B$ is a quiver algebras. -For Q1: Given the above, the software needs to be able compute the radical of $\operatorname{End}_A(M)$ and determine the structure of $\operatorname{End}_A(M)/\operatorname{rad}\operatorname{End}_A(M)$. For QPA this is to my knowledge only possible over finite fields. The command $\textrm{IsElementaryAlgebra}$ checks if a finite dimensional algebra over a finite field is elementary. To understand the structure of $\operatorname{End}_A(M)/\operatorname{rad}\operatorname{End}_A(M)$ is basically to decompose $M$ into indecomposable modules. The command $\textrm{DecomposeModule}(M)$ does this over finite fields. This gives a complete set of primitive idempotents of $\operatorname{End}_A(M)$, which correspond to the vertices in the quiver one wants to construct. -The arrows of degree $0$ is given by basis of $\operatorname{rad}\operatorname{End}_A(M)/\operatorname{rad}^2\operatorname{End}_A(M)$. The command $\textrm{EndOfModuleAsQuiverAlgebra( M )}$ would do this if the field $K$ is the same field $A$ is an algebra over. -Given that $\operatorname{rad}B = \operatorname{rad}(A) \oplus \operatorname{Ext}^{\geq 1}_A(M,M)$, we need to compute - -$\operatorname{Ext}^n_A(M,M)/(\operatorname{rad}\operatorname{End}_A(M)\operatorname{Ext}^n_A(M,M) + \operatorname{Ext}^{n-1}_A(M,M)\operatorname{Ext}^1_A(M,M) + \cdots + \operatorname{Ext}^1_A(M,M)\operatorname{Ext}^{n-1}_A(M,M) + \operatorname{Ext}^n_A(M,M)\operatorname{rad}\operatorname{End}(M))$ - -to find the arrows of degree $n$. The command $\textrm{ExtAlgebraGenerators}(M, n)$ would find $\operatorname{Ext}^n_A(M,M)$ modulo everything (among other things) except for the two outer terms (so the documentation on this command is misleading as it is only correct if the degree zero part is a semisimple ring). However all of this is possible to do in QPA, but there is no readily available commands to preform these tasks. These necessary commands are something we want to implement in QPA version 1 or version 2. -I hope that these are useful comments. -Best regards, The QPA-team.<|endoftext|> -TITLE: Can scalar curvature and diameter control volume? -QUESTION [12 upvotes]: Scalar curvature can control the volume of geodesic ball locally, however, it can not bound the diameter. As far as I know, the example for a manifold with a large scalar curvature and volume has large diameter. -Comparing with the n sphere with the standard metric, if a smooth manifold has scalar curvature larger than this sphere, and the diameter of the manifold is smaller, should the volume of the manifold be smaller than the sphere? - -REPLY [8 votes]: In general, the scalar curvature does not give enough control to get these sorts of volume estimates. However, if you use the Ricci curvature instead, you can get these kinds of estimates using the volume comparison theorem or something similar. -For an example of why scalar curvature bounds and diameter bounds are not enough to compare the volume to a sphere, consider the manifold $M_1$ which is the metric product of a ball of radius r in hyperbolic space $\mathbb{H}^2$ (with sectional curvature 1) and the sphere $\mathbb{S}^2(r)$. For the second manifold, consider the $4$-sphere $\mathbb{S}^4(s)$. -The volume of $M_1$ is $\sinh(t) \times 4 \pi r^2$ while its diameter is $\sqrt{\pi^2 r^2+ 4t^2} < \pi r +2 t$. The scalar curvature is $2(\frac{1}{r^2}-1)$. Whenever $r<1$, the scalar curvature of $M_1$ is positive. Meanwhile, the volume of the $4$-sphere is $\frac{8}{3}\pi^2 s^4$, its diameter is $\pi s$, and its scalar curvature is $\frac{12}{s^2}$. -To force our first manifold to have much larger volume then the second, we set $s$ very large so that the scalar curvature is less than $1$. -We also set $r=1/2$ and $t=s-1/2$. In this case, the scalar curvature of $M_1$ is 6 and the diameter of $M_1$ is strictly smaller than that of $\mathbb{S}^4(s)$. However, since hyperbolic sine grows exponentially in $t$ whereas the volume of the $4$-sphere grows as $s^4$, for sufficiently large $s$ the volume of $M_1$ is larger than that of the sphere. By setting $s$ arbitrarily large, we can force $M_1$ to have much larger volume than $\mathbb{S}^4$.<|endoftext|> -TITLE: Length minimizing graphs between a finite set of points -QUESTION [5 upvotes]: Consider a set of $n$ points in the plane. Among all the connected graphs (trees) $T$ in the plane that have these $n$ points among their vertices, I am looking to find one such that the sum of its edge lengths is minimum. Note that this question is different from Minimum Spanning Tree as we allow $T$ to have vertices other than the given $n$ points. For example, if $n=3$, there are 2 possibilities: Let $A,B,C$ be the three given points. If one of the angles in the triangle $ABC$, say $ABC$ is bigger than $120^\circ$ then AB+BC is the minimizing tree. If all the angles are less than $120^\circ$, let $P$ be the point within the triangle $ABC$ such that the angles $APB$, $BPC$, $CPA$ are all $120^\circ$. Then the tree $T$ with the set of vertices $A,B,C,P$ and edges $AP,BP,CP$ is the length minimizing one. -Is this problem worked out before? - -REPLY [7 votes]: This is the so-called Steiner Tree Problem.<|endoftext|> -TITLE: Hasse principle and its failure for a special class of plane cubics -QUESTION [5 upvotes]: Let $X$ be a variety defined over a number field $K$. The Hasse principle, or the local-to-global principle, asserts that $X(K) \ne \emptyset$ if and only if for each completion $K_v$ of $K$, we have $X(K_v) \ne \emptyset$. This is known to hold for every quadric hypersurface, by the work of Hasse and Minkowski. In general, this does not hold. -For curves, there is a famous example of Selmer: the plane cubic curve $C: 3x^3 + 4y^3 + 5z^3 = 0$ has a point over $\mathbb{F}_p$ for every prime $p$ which can be lifted to a point in $\mathbb{Q}_p$, and as a cubic curve it has a real point. Nevertheless, it has no rational points. -By now it is known, due to the following paper of Bhargava which builds upon the previous work of Bhargava and Shankar on computing the average sizes of $\text{Sel}_2(E), \text{Sel}_3(E), \text{Sel}_5(E)$, that a positive proportion of plane cubics satisfy the Hasse principle (that is, they either fail to have a local point at some completion or they have a rational point), and a positive proportion fails the Hasse principle (that is, they have a local point everywhere but no global point). -However, there is a different question that perhaps should have been inspired by Selmer's original example. The Selmer curve is an example of a generalized Fermat cubic curve $ax^3 + by^3 + cz^3$, and even more generally, an example of a plane cubic where the Arnholdt invariant $S$ vanishes, or in the language of this paper of Bhargava and Shankar, the $I$-invariant. Such ternary cubic forms are also characterized by the property that their Hessian determinant (also a ternary cubic form) splits into three linear factors over $\mathbb{C}$. -My question is: what is the behaviour of the Hasse principle among generalized Fermat cubics $ax^3 + by^3 + cz^3$ with $a,b,c \in \mathbb{Z}$ varying over a box? What about plane cubics with vanishing $I$-invariant? - -REPLY [10 votes]: The case of plane cubics is actually easier than the family $ax^3 + by^3 + cz^3$. The reason being that a positive proportion of plane cubics are everywhere locally soluble, when ordered by the height of their coefficients. -However the problem, which one would naively expect to be simpler, of determining the number of Selmer curves which are everywhere locally soluble, turns out to be incredibly difficult. The number of such curves with coefficients bounded by $T$ was shown to be -$$\ll \frac{T^3}{( \log T)}, \quad T \to \infty,$$ -in T. Browning, R. Dietmann - Solubility of Fermat equations. This upper bound is expected to be sharp, but is far from being proved. -So we don't even understand the distribution of everywhere locally soluble Selmer curves, let alone those which fail/satisfy the Hasse principle.<|endoftext|> -TITLE: Why are finite cell complexes also finite as infinity-categories? -QUESTION [14 upvotes]: A quasicategory ($\infty$-category) $\mathcal{C}$ is finite if there is a finite simplicial set $K$ and a categorical equivalence $K\rightarrow\mathcal{C}$. -On the other hand, a Kan complex (space) $X$ is finite if there is a finite simplicial set $K$ and a weak homotopy equivalence $K\rightarrow X$. Finite Kan complexes are precisely (up to equivalence) finite CW complexes. -Question: Now suppose $X$ is a finite Kan complex. It is also a quasicategory. In Higher Topos Theory (1.2.14.2), Lurie takes for granted that $X$ is also finite as a quasicategory. However, this doesn't seem obvious to me. Is there an easy proof? -To see what I mean, take the standard simplicial structure for a circle, with one 0-simplex and one 1-simplex. This describes a finite simplicial set $K$ and a weak homotopy equivalence $f:K\rightarrow S^1$. But $f$ is not a categorical equivalence. To build a finite model for the $\infty$-category $S^1$, we need a second 1-simplex, along with two 2-simplices which declare that the 1-simplices are inverse to each other. - -REPLY [12 votes]: Start from a finite simplicial set $K$ which is homotopicaly equivalent to a Kan complex $X$. -Then by applying a finite number of pushout of outer horn inclusion to $K$, you can build homotopy equivalences $K \hookrightarrow K' \rightarrow X$ such that all the $1$-cells of $K'$ are "invertible" (in the sense that "for all $1$-cell $f$ there exists $2$-cells attesting homotopies $g \circ f => 1$ and $f \circ h => 1$ " see the "edit" below though ). $K'$ is still a finite simplicial set. -I claim that $K' \rightarrow X$ is now an equivalence in the Joyal model structure, which conclude the proof. -Indeed, as all the $1$-cells of $K'$ have homotopy inverses, the homotopy category of $K'$ (in the sense of the left adjoint to the nerve functor) is a groupoid. -So if I take $K' \hookrightarrow Y \rightarrow X$ a factorization as a Joyal trivial cofibration followed by a Joyal fibration, $Y$ is a quasi-category whose homotopy category is equivalent to the homotopy category of $K'$, hence is a groupoid, hence $Y$ is a Kan complex. -And $Y \rightarrow X$ is a homotopy equivalences between Kan complexes, hence it is a Joyal equivalence. So as announced, $K' \rightarrow X$ is a Joyal equivalence. -Edit: small correction and answering your comment. You are indeed right that it is not exactly possible to get what I said. What we need to do precisely is the following: -For each $1$-cell of $K$ you use a pushout by a $\Lambda^0 [2] \hookrightarrow \Delta[2]$ and one by a $\Lambda^2 [2] \hookrightarrow \Delta[2]$ to add a cells $g$ and $h$ with $2$-cells $f \circ g => 1$ and $h \circ f => 1$. -And you stop there, we don't add any new cells (no right inverse for $g$, or left inverse for $h$) -This is enough to ensure that the homotopy category of $K'$ is a groupoids: the original cells of $K$ , like $f$, will be invertible because they have both a left inverse and a right inverse, and the new cells $g$ and $h$ are invertible because they are either right or left inverse to an invertible cell. -As every arrow the homotopy category of $K'$ is a composite of $1$-cell of $K'$ they will all be invertible.<|endoftext|> -TITLE: Finite groups which have trivial outer automorphism group -QUESTION [7 upvotes]: I was wondering if it is possible to classify the finite groups which have no outer automorphisms? -I am currently only aware of the Symmetric Groups ($n \neq 6$) as an infinite class of examples. If there is no classification I would still appreciate any further example of groups which have no outer automorphisms. -By asking GAP to go through the groups of order (up to 110) I found the following groups with trivial outer automorphism: -1, S2, S3, C5 : C4, S4, C7 : C6, C9 : C6, C11 : C10 -(These are how GAP describes these groups when I used the function "StructureDescription" and I believe : indicates some sort of semidirect product. In these cases I think the smaller cyclic group may be isomorphic to the automorphism group of the larger one and the action is the action of automorphisms.) From this data, it looks as though there could be a class of groups $C_n \rtimes Aut(C_n)$ which have no outer automorphisms but I am unsure for exactly which $n$ this will be the case. -[Edit: ran through groups of order 1 .. 255 on GAP: -1, C2, S3, C5 : C4, S4, C7 : C6, C9 : C6, S5, S3 x (C5 : C4), (C3 x C3) : QD16, S3 x S4, C13 : C12, (C2 x C2 x C2) : (C7 : C3), (C2 x C2) : (C9 : C6), S3 x (C7 : C6)] - -REPLY [7 votes]: Wielandt's automorphism tower theorem states that if $G$ is any finite group with $Z(G) = 1$, then the sequence of groups $G_{n}$ with $G_{0}= G$ and $G_{n+1} = {\rm Aut}(G_{n})$ is eventually stable (and it follows that the sequence stabilizes with a complete group- ie a group without outer automorphisms). -Later edit, rewording after @YCor's comment. -Hence each finite group with trivial center gives rise to a uniquely determined finite group with trivial center AND with trivial outer automorphism group.<|endoftext|> -TITLE: A symmetric bilinear form and a Plücker identity -QUESTION [8 upvotes]: It turns out that a special case of something I'm working on gives, as a corollary, a rather 19th-century-looking elementary statement about the rank of a certain symmetric matrix. I thought I would post it here in the hope that someone might recognize it as something familiar. -Let $A$ be a $2\times n$ matrix over a field $\mathbb k$, where $n\geq 3$. Let $|A_{ij}|$ denote its minors, for $1\leq i,j\leq n$, taking the convention that $|A_{ii}|=0$. Let $B$ be the $n\times n$ symmetric matrix with $B_{ij}=|A_{ij}|^2$, for $1\leq i,j\leq n$. It can be shown that, if at least three columns of $A$ are pairwise linearly independent, then $B$ has rank exactly $3$. -In other words, over the polynomial ring $R:={\mathbb k}[x_{ij}\colon 1\leq i \rank C + \rank D$ equals $0$ (because if we let $A$ be the corresponding minor of $C$ and $B$ be the corresponding minor of $D$, then all the terms in the sum will be $0$). Thus, $\rank\tup{C + D} \leq \rank C + \rank D$. $\blacksquare$ - -Lemma 4. Let $C_1, C_2, \ldots, C_p$ be any $p$ matrices of a given size. Then, $\rank\tup{C_1 + C_2 + \cdots + C_p} \leq \rank\tup{C_1} + \rank\tup{C_2} + \cdots + \rank\tup{C_p}$. - -Proof of Lemma 4. This follows by induction on $p$, using Lemma 3. $\blacksquare$ -Proof of Proposition 1. Let $a_1, a_2, \ldots, a_n$ be the $n$ entries of the first row of $A$. Let $b_1, b_2, \ldots, b_n$ be the $n$ entries of the second row of $A$. Then, the column vector $A_i$ can be written as $A_i = \tup{a_i, b_i}^T$ for each $i \in \ive{n}$. Hence, for any $i, j \in \ive{n}$, we have -$A_{i, j} = \begin{pmatrix} a_i & a_j \\ b_i & b_j \end{pmatrix}$ and thus -\begin{align} - \tup{\det\tup{A_{i,j}}}^2 - &= \tup{\det \begin{pmatrix} a_i & a_j \\ b_i & b_j \end{pmatrix}}^2 - = \tup{a_i b_j - b_i a_j}^2 \\ - &= a_i^2 b_j^2 - 2 a_i b_j b_i a_j + b_i^2 a_j^2 - = a_i^2 b_j^2 - 2 a_i b_i a_j b_j + b_i^2 a_j^2 . -\end{align} -Now, define three $n \times n$-matrices $C$, $D$ and $E$ over $k$ as follows: - -Let $C$ be the $n \times n$-matrix whose $\tup{i, j}$-th entry is $a_i^2 b_j^2$ for all $i, j \in \ive{n}$. -Let $D$ be the $n \times n$-matrix whose $\tup{i, j}$-th entry is $- 2 a_i b_i a_j b_j$ for all $i, j \in \ive{n}$. -Let $E$ be the $n \times n$-matrix whose $\tup{i, j}$-th entry is $b_i^2 a_j^2$ for all $i, j \in \ive{n}$. - -Then, for all $i, j \in \ive{n}$, the $\tup{i, j}$-th entry of the matrix $C + D + E$ is -$a_i^2 b_j^2 - 2 a_i b_i a_j b_j + b_i^2 a_j^2 = \tup{\det\tup{A_{i,j}}}^2$; -but this is the same as the $\tup{i, j}$-th entry of the matrix $B$. -Thus, $C + D + E = B$. -Lemma 2 (applied to $p_i = a_i^2$, $q_j = b_j^2$ and $X = C$) yields $\rank C \leq 1$. -Lemma 2 (applied to $p_i = - 2 a_i b_i$, $q_j = a_j b_j$ and $X = D$) yields $\rank D \leq 1$. -Lemma 2 (applied to $p_i = b_i^2$, $q_j = a_j^2$ and $X = E$) yields $\rank E \leq 1$. -Now, Lemma 4 (applied to $p = 3$, $C_1 = C$, $C_2 = D$ and $C_3 = E$) yields -\begin{equation} - \rank\tup{C + D + E} \leq \underbrace{\rank C}_{\leq 1} + \underbrace{\rank D}_{\leq 1} + \underbrace{\rank E}_{\leq 1} \leq 1 + 1 + 1 = 3. -\end{equation} -In view of $C + D + E = B$, this rewrites as $\rank B \leq 3$. This proves Proposition 1. $\blacksquare$ -Next, we claim: - -Proposition 5. Let $p, q, r$ be three elements of $\ive{n}$ such that $p < q < r$. Let $\overline{B}$ be the submatrix of $B$ formed by removing all rows except for the $p$-th, $q$-th and $r$-th rows and all columns except for the $p$-th, $q$-th and $r$-th columns. (This is a $3 \times 3$-matrix.) Then, - \begin{equation} - \det \overline{B} = 2 \tup{\det\tup{A_{q, r}}}^2 \tup{\det\tup{A_{p, r}}}^2 \tup{\det\tup{A_{p, q}}}^2 . -\end{equation} - -This will follow from the following lemma, which is easily checked by hand: - -Lemma 6. Let $u, v, w \in k$. Then, - \begin{equation} - \det \begin{pmatrix} - 0 & w & v \\ w & 0 & u \\ v & u & 0 - \end{pmatrix} - = 2 u v w . -\end{equation} - -Proof of Proposition 5. The definition of $\overline{B}$ yields -\begin{equation} -\overline{B} = -\begin{pmatrix} - \tup{\det\tup{A_{p,p}}}^2 & \tup{\det\tup{A_{p,q}}}^2 & \tup{\det\tup{A_{p,r}}}^2 \\ - \tup{\det\tup{A_{q,p}}}^2 & \tup{\det\tup{A_{q,q}}}^2 & \tup{\det\tup{A_{q,r}}}^2 \\ - \tup{\det\tup{A_{r,p}}}^2 & \tup{\det\tup{A_{r,q}}}^2 & \tup{\det\tup{A_{r,r}}}^2 -\end{pmatrix} -= -\begin{pmatrix} - 0 & \tup{\det\tup{A_{p,q}}}^2 & \tup{\det\tup{A_{p,r}}}^2 \\ - \tup{\det\tup{A_{p,q}}}^2 & 0 & \tup{\det\tup{A_{q,r}}}^2 \\ - \tup{\det\tup{A_{p,r}}}^2 & \tup{\det\tup{A_{q,r}}}^2 & 0 -\end{pmatrix} -\end{equation} -(since any $i, j \in \ive{n}$ satisfy $\tup{\det\tup{A_{i,j}}}^2 = \tup{\det\tup{A_{j,i}}}^2$, and since each $i \in \ive{n}$ satisfies $\det\tup{A_{i,i}} = 0$). Hence, -\begin{equation} - \det \overline{B} - = \det - \begin{pmatrix} - 0 & \tup{\det\tup{A_{p,q}}}^2 & \tup{\det\tup{A_{p,r}}}^2 \\ - \tup{\det\tup{A_{p,q}}}^2 & 0 & \tup{\det\tup{A_{q,r}}}^2 \\ - \tup{\det\tup{A_{p,r}}}^2 & \tup{\det\tup{A_{q,r}}}^2 & 0 - \end{pmatrix} - = 2 \tup{\det\tup{A_{q, r}}}^2 \tup{\det\tup{A_{p, r}}}^2 \tup{\det\tup{A_{p, q}}}^2 -\end{equation} -(by Lemma 6, applied to $u = \tup{\det\tup{A_{q, r}}}^2$, $v = \tup{\det\tup{A_{p, r}}}^2$ and $w = \tup{\det\tup{A_{p, q}}}^2$). This proves Proposition 5. $\blacksquare$ - -Proposition 7. Assume that there exist $p, q, r \in \ive{n}$ such that $p < q < r$ and $2 \tup{\det\tup{A_{q, r}}}^2 \tup{\det\tup{A_{p, r}}}^2 \tup{\det\tup{A_{p, q}}}^2 \neq 0$. Then, $\rank B = 3$. - -Proof of Proposition 7. Consider these $p, q, r$ whose existence we have assumed. Consider the $3 \times 3$-matrix $\overline{B}$ defined in Proposition 5. Then, Proposition 5 yields -\begin{equation} - \det \overline{B} = 2 \tup{\det\tup{A_{q, r}}}^2 \tup{\det\tup{A_{p, r}}}^2 \tup{\det\tup{A_{p, q}}}^2 \neq 0. -\end{equation} -But $\overline{B}$ is a submatrix of $B$, and thus $\det \overline{B}$ is a minor of $B$ of size $3$. Hence, there exists a nonzero minor of $B$ of size $3$ (since $\det \overline{B} \neq 0$). Thus, $\rank B \geq 3$. Combining this with $\rank B \leq 3$ (which follows from Proposition 1), we obtain $\rank B = 3$. This proves Proposition 7. $\blacksquare$ - -Proposition 8. Assume that $k$ has characteristic $2$. (If $k$ is just a commutative ring, then we should instead assume that $2 = 0$ in $k$.) Then, $\rank B \leq 2$. - -Proof of Proposition 8. This is analogous to the proof of Proposition 1, but we additionally need to observe that $\rank D \leq 0$ (instead of $\rank D \leq 1$), because all entries of $D$ are $0$ (since they contain the factor $2$). $\blacksquare$<|endoftext|> -TITLE: Representability of Hom of two finite flat group schemes -QUESTION [5 upvotes]: I am reading a note at Page 63 -ftp://ftp.math.ethz.ch/users/pink/FGS/CompleteNotes.pdf -It says whenever $G$ is finite and flat over $S$ the functor ${\rm Hom}(G,H)$ is representable. But it does not give the proof or any references. -Can anybody help me with this question? -The original statement - -REPLY [8 votes]: The typical way statements like this are proven is by deducing them from the representability of Hilbert schemes. For example, we can use: - -Theorem (Grothendieck). Let $S$ be a Noetherian scheme, let $X$ and $Y$ be $S$-schemes of finite type, and assume $X$ is flat and projective, and $Y$ is quasi-projective. Then the functor - \begin{align*} -\operatorname{\underline{Sch}}\!/S &\to \operatorname{\underline{Set}}\\ -T &\mapsto \operatorname{Mor}_T(X_T,Y_T) -\end{align*} - is representable by a separated $S$-scheme $\operatorname{\underline{Mor}}(X,Y)$ that is a rising union of quasi-projective $S$-schemes. - -Proof. See for example [FGA TDTE IV, §4c], or [FGAE, Thm. 5.23]. $\square$ -Thus, if $G$ is finite and $H$ is quasiprojective, then this guarantees existence of -$$\operatorname{\underline{Mor}}(G,H).$$ -This is the functor of morphisms as schemes; to get the functor of morphisms of group schemes, just note that a map $f \colon G \to H$ is a morphism of group schemes if and only if the diagram -$$\begin{array}{ccc}G \times G & \stackrel{f \times f}\longrightarrow & H \times H \\ \!\!\!\!{\scriptsize{\mu}}\downarrow & & \downarrow{\scriptsize{\mu}}\!\!\!\! \\ G & \stackrel{f}\longrightarrow & H\end{array}$$ -commutes. Therefore, the homomorphisms of group schemes are given by the fibre product -$$\begin{array}{ccc}\operatorname{\underline{Hom}}(G,H) & \longrightarrow & \operatorname{\underline{Mor}}(G,H) \\ \downarrow & & \downarrow \\ \operatorname{\underline{Mor}}(G \times G, H) & \stackrel{\Delta}\longrightarrow & \operatorname{\underline{Mor}}(G \times G, H) \underset S\times \operatorname{\underline{Mor}}(G \times G, H),\! \end{array}$$ -where the right vertical morphism maps $f$ to $(\mu_H \circ f \times f, f \circ \mu_G)$. Since $\operatorname{\underline{Mor}}(G \times G, H)$ is separable, the diagonal is a closed immersion, hence the same goes for the natural inclusion -$$\operatorname{\underline{Hom}}(G, H) \to \operatorname{\underline{Mor}}(G, H).$$ -Therefore, $\operatorname{\underline{Hom}}(G,H)$ is representable by a separated $S$-scheme that is a rising union of quasi-projective $S$-schemes. $\square$ -Remark. I don't know how strong an assumption the quasi-projectivity of $H$ is, but in most examples one cares about this is true. There are examples of abelian schemes that are not projective, but in that case there are other arguments that still show that $\operatorname{\underline{Hom}}(A, B)$ is representable by a scheme. It always works if you're willing to work with algebraic spaces; see for example [Tag 0D1C]. -Exercise. If $G$ is a finite (constant) group, then show that $\operatorname{\underline{Hom}}(G, H)$ is a closed subset of a suitable power of $H$, hence is of finite type over $S$. (The proof I gave above a priori only gives an $S$-scheme that is locally of finite type.) -Hence, by a descent argument, the same holds when $G$ is étale-locally isomorphic to a finite constant group, or equivalently if $G \to S$ is finite étale. This gives an easier proof of representability in this case, but I don't know how to generalise this method to the general finite flat case. I'm not sure if $\operatorname{\underline{Hom}}(G,H)$ is always of finite type over $S$. -Exercise. Why is $\operatorname{\underline{Hom}}(\mathbb G_a, \mathbb G_m)$ not representable? (Hint: use your intuition over fields to write down what its set of points should be. Use more and more non-reduced schemes to show that there is no appropriate scheme structure.) - -References. -[FGA TDTE IV] A. Grothendieck, Techniques de descente et théoremes d’existence en géométrie algébrique. IV: Les schemas de Hilbert, Sem. Bourbaki 13, No. 221 (1961). ZBL0236.14003. -[FGAE] B. Fantechi, L. Göttsche, L. Illusie, S. L. Kleiman, N. Nitsure, A. Vistoli, Fundamental algebraic geometry: Grothendieck’s FGA explained. Mathematical Surveys and Monographs 123. AMS, Providence, RI (2005). ZBL1085.14001.<|endoftext|> -TITLE: Syzygies in Steinberg module of genus 2 mapping class group $\mathrm{MCG}(\Sigma_2)$ -QUESTION [7 upvotes]: $\DeclareMathOperator\MCG{MCG}$Consider the mapping class group $\MCG(\Sigma_2)$ of the closed genus 2 oriented surface $\Sigma_2$. The algebraic-duality theory of $\MCG_2:=\MCG(\Sigma_2)$ is explicitly described by Nathan Broaddus' very useful paper (arXiv link). -Broaddus constructs a beautiful homologically-nontrivial $2$-sphere $B$ in the curve complex of the genus two closed surface with marked curves α1, α2, α3, α4, α5, α6, β1, β2, β3 as labelled in the image below. ((we are indebted to N.Broaddus for his work and graphics.)) -Question: I seek three distinct nontrivial elements $φ1,φ2,φ3$ of the mapping class group $\MCG(\Sigma_2)$ with the property that the following chain sum vanishes over $\mathbb{Z}/2$-coefficients: -$$α1 + α2 + α3 + α4 + α5 + α6 + β1 + β2 + β3$$ -$$+ α1.φ1 + α2.φ1 + α3.φ1 + α4.φ1 + α5.φ1 + α6.φ1 + β1.φ1 + β2.φ1 + β3.φ1$$ -$$+ α1.φ2 + α2.φ2 + α3.φ2 + α4.φ2 + α5.φ2 + α6.φ2 + β1.φ2 + β2.φ2 + β3.φ2$$ -$$+α1.φ3 + α2.φ3 + α3.φ3 + α4.φ3 + α5.φ3 + α6.φ3 + β1.φ3 + β2.φ3 + β3.φ3$$ - -By "vanishing over $\mathbb{Z}/2$-coefficients" we mean something utterly trivial like $\alpha+\beta + \alpha +\beta = 2 \alpha + 2\beta = 0$ (mod 2). Of course if we took $\phi_1 = \phi_2 =\phi_3=id$ or $\phi_1=\phi_2$, $\phi_3=id$ then the chain sum would vanish (mod 2). -Possibly a suitable finite subgroup $G'$ of $\MCG_2$ will have the desired property. For the once-punctured torus $\Sigma_{1,1}$, indeed the order 3 finite subgroup of $PGL(\mathbb{Z}^2)$ "closes" the necessary (Steinberg) symbol. -Believe it or not, but finding such a triple of elements would yield a $\MCG$-equivariant codimension-two spine $Z \hookrightarrow \mathrm{Teich}(\Sigma_2)$ of the Teichmuller space. -The problem is related to stitching footballs from uniform hexagonal panels, or uniform pentagonal panels, or combinations of both. To stitch a football from panels $\{P_i| i\in I\}$ means finding a finite subset $I' \subset I$ for which the singular chain sum $\sum_{i\in I'} P_i$ has singular chain boundary which vanishes mod $2$, so $$\partial(\sum_{i\in I'} P_i)=\sum_{i\in I'} \partial P_i=0$$ over $\mathbb{Z}/2$-coefficients. When $P$ is two-dimensional hexagon or pentagon, the panels have singular boundary $$\partial P= \sum_{e\text{~edge~of~}P} e.$$ We denote the closed convex hull of the football $F:=conv\{P|\text{~panels}\}$. The panels then become closed subsets of the boundary $\partial F$. For instance since the 1960's, the standard football is stitched after Adidas' ``Telstar" design, having twenty white hexagon panels, and twelve black pentagon panels. But in our applications we assume the patches $\{P_i\}_I$ are pairwise isometric to some regular geodesically-flat polygon $P$. - -REPLY [2 votes]: I recently returned to this question, and have found a formal solution to Closing the Steinberg symbol in genus two using Mark C. Bell's wonderful curver program. -In low genus it was expected that an interesting finite order subgroup could yield formal solutions. This is the case in $PGL(\mathbb{Z}^2)$. For the mapping class group of genus two, there is an interesting order five subgroup $I=\{Id, \mu, \mu^2, \mu^3, \mu^4\}$ which formally closes the Steinberg symbol $\xi:=[\alpha_1]+[\alpha_2]+[\alpha_3]+[\alpha_4]+[\alpha_5]+[\alpha_6]$. (My notation follows the above image of Broaddus' two sphere). To formally close the specific original Steinberg symbol $\xi'$ given by Broaddus' two-sphere would require finding an appropriate conjugate $x\mu x^{-1}$ for $x\in MCG$. For the applications this is not essential. -We used a jupyter notebook to import Bell's curver program and are strictly studying the multiplicative adjoint action in $MCG$. -https://github.com/jhmartel/MCG/blob/master/ClosingSteinbergGenusTwo.ipynb -In python code, the "closing condition $d\xi=0$ mod 2" for the symbol $\xi$ is equivalent to the vanishing of an iterated symmetric difference. -%% Now the next step has always been to use the formal solution to define a convenient chain model $X=\underline{F}=\sum_{\phi \in MCG} \phi.F$, where $F$ is a compact convex "horospherically truncated convex hull" defined over the ``vertices" $I.\xi$. My claim has always been that the singularities constructed from "repulsive" optimal transport programs are excellent candidate equivariant spines of $MCG$.<|endoftext|> -TITLE: Prerequisites for reading papers of arithmetic such as Ribet, Mazur, Faltings, Wiles -QUESTION [6 upvotes]: I've studied some fundamentals of algebraic geometry and number theory, and now I want to read papers which seem to be the "main stream" of frontier research on arithmetic. -I've heard that Mazur's "Modular curves and the Eisenstein ideal" is one of such papers (and I've also heard that it is good for people who have finished reading Hartshorne here), so I'm about to read it. -But glancing through it, I feel that it goes far beyond Hartshorne, and that it needs the modular forms of moduli stack (I don't know what this is at all). -And many papers on arithmetic seem to need this theory. -However these papers refer to Katz's paper and Deligne, Rapoport's paper on the theory of modular forms, I feel these two papers are very difficult (and too long to start reading, not knowing if these are really good). -So my question is: Please suggest me some references on the theory of modular forms (of moduli stack? Sorry, I know nothing, except that this modular forms are not one which I'm studying in Diamond, Shurman.). -If Katz and Deligne, Rapoport are the best, or these are enough to read Mazur, I tried these. -Thank you very much! - -REPLY [11 votes]: I don't know what you mean by Modular forms of moduli stack, I think maybe you mean modular forms on moduli stacks. Either way, you should probably have a look at the book by Katz--Mazur titled "Arithmetic Moduli of Elliptic Curves". It should explain how to think about modular curves in the correct setting you want. In order to understand the modular forms rephrased in this language, the reference to Katz's paper "p-adic properties..." is a really good one, but nowadays there are lots of notes online which explain the "Katz definition of modular forms". I think some really good notes on this are Frank Calegari's AWS notes which you can find here: http://swc.math.arizona.edu/aws/2013/ They should take you from the modular forms you know and love to Katz's version. -Also, if its your first time seeing stacks and are scared of such things (like I am) then you are probably fine just ignoring the word stack and just think about a scheme, since, in most cases you can just rig it so that the moduli problem is representable by some scheme (i.e. by making sure you have a nice enough level (a crucial term being sufficiently small)) -Lastly, I don't know what you kids these days mean by main stream, but looking at books like Cornell--Silverman--Stevens on the proof of Fermat's Last Theorem would give you an idea of the popular tools are used in number theory and arithmetic geometry (although some might even say this is now old stuff and if you want to see the future then you need to look at things people like Scholze are doing..)<|endoftext|> -TITLE: Lifting a diffeomorphism into a spinor bundle automorphism -QUESTION [8 upvotes]: I know several papers that treat this, but it seems that most of these papers do things very differently with quite different conclusions, so I am confused. -Basically, when one tries to do classical field theory (as in, the branch of physics) in a mathematically precise manner, one considers a field $\psi$ to be a section of some fiber bundle $\pi_E:E\rightarrow M$ over the spacetime $M$, and writes up a lagrangian $\hat L\in \Omega^n_H(J^1(E))$, which is a horizontal $n$-form on $J^1(E)$ (sometimes $J^2(E)$) such that the action is $$ S[\psi]=\int_M (j^1\psi)^\ast\hat L. $$ -If one wishes to involve gravity, one seeks to consider field theories that are diffeomorphism invariant. -If $\pi_E:E\rightarrow M$ is a fiber bundle whose sections are some kind of matter fields, then to have a well-defined concept of diffeomorphism-invariance, for any diffeomorphism $\phi:M\rightarrow M$ one must be able to lift this diffeomorphism into a fiber bundle automorphism $\phi_E\in\text{Aut}(E)$ in a consistent manner. -For the tensor bundles (or indeed any natural bundle), there is a functorial lift given by the tangent map (at least in the case of tensor bundles). For example, if $X\in\Gamma(TM)$ is a smooth vector field, and $\phi:M\rightarrow M$ is a diffeo, then $$ \phi_E=T\phi:TM\rightarrow TM $$ is this vector bundle automorphism. -On the other hand, in physics, very important fields are spinor fields, sections of the spinor bundle $\pi_S:S\rightarrow M$. This, however is not a natural bundle (to my knowledge), and I do not know if there is any "canonical" way to lift diffeomorphisms into $\text{Aut}(S)$. -Since general relativity heavily involves diffeomorphism-freedom, it is extremely important to be able to do that. In particular, I have no idea how to define the stress-energy tensor of a spinor field without representing diffeomorphisms on $S$ somehow. -This situation is further confusing me, since so far I have been an ardent defender of the viewpoint that in the usual local tensor calculus-based formalism, there is no essential difference between "active diffeos" (point transformations) and "passive diffeos" (coordinate transformations). -However the behaviour of a "traditional" spinor field under a coordinate transformation is simple and clear, a spinor field transforms as a scalar under coordinate transformations, and "as a spinor" under changes of orthonormal frames. -However in the modern, invariant viewpoint, one cannot afford this approach. For example, if at $x\in M$, one is given a spinor $\psi\in S_x$ and a vector $v\in T_xM$, and one considers the tensor product $\psi\otimes v\in S_x\otimes T_xM$, then a diffeo will move $v$ to $T\phi(v)\in T_{\phi(x)}M$, but it will not move $\psi$ at all, so this tensor product under a diffeo would become $\psi\otimes T\phi(v)\in S_x\otimes T_{\phi(x)}M$, a product of fibers taken over different base points - clearly undesirable. -Is there any agreed-upon method of dealing with the action of spinors under diffeos? If so, is there a simple way of writing it down/stating it? - -REPLY [4 votes]: There is an intrinsic ambiguity in lifting diffeomorphisms to spinors, since as you say spinor bundles are not natural bundles. But the ambiguity is not large and has more to do with the global properties of spin structures. -First, recall that the oriented orthogonal frame bundle $\mathcal{P}_g M$ of a pseudo-Riemannian manifold $(M,g)$ is an $\mathrm{SO}(p,q)$-principal bundle, where $(p,q)$ represents the signature of the metric $g$. Then recall that a spin structure is a $\mathrm{Spin}(p,q)$-principal bundle $\mathcal{S}M$ together with a principal bundle morphism $\mathcal{S}M \to \mathcal{P}_gM$, which fiberwise restricts to the usual projection homomorphism $\mathrm{Spin}(p,q) \to \mathrm{SO}(p,q)$. Now, given a linear representation of of $\mathrm{Spin}(p,q)$ on a vector space $\Sigma$, in particular a spinor representation, the associated vector bundle $\Sigma_{\mathcal{S}} M = (\mathcal{S} M \times \Sigma)/\mathrm{Spin}(p,q)$ is the corresponding spinor bundle over $M$. -Now, a diffeomorphism $\phi\colon M \to M'$ (at least one that preserves metrics and orientations of oriented pseudo-Riemannian manifolds $(M,g)$ and $(M',g')$, which is all that we will consider for now) induces a morphism of oriented frame bundles $\phi^*\colon \mathcal{P}_g M \to \mathcal{P}_{g'}M$. This is a principal bundle morphism, meaning that it is $\mathrm{SO}(p,q)$-equivariant. Suppose that there exists also a $\mathrm{Spin}(p,q)$-principal bundle morphism $\tilde{\phi}^*\colon \mathcal{S}M \to \mathcal{S}'M'$ of spin structures that covers $\phi^*$. The existence of $\tilde{\phi}^*$ is not automatic, possibly it doesn't exist at all. But if it does exist, then it is not unique, since the $\mathbb{Z}_2$ center of $\mathrm{Spin}(p,q)$ acts non-trivially on $\tilde{\phi}^*$. But that's the only freedom you have; if $\tilde{\phi}^*$ exists, then there exist exactly two possibilities. Basically, it is sufficient to fix the lift of $\phi^*$ at a single point $x\in M$ (there are only two possibilities) and the rest is determined by continuity. -Finally, all that remains to observe is that a morphism of principal bundles induces a morphism of associated bundles. Hence, a spin-lift $\tilde{\phi}^*$ of our diffeomorphism $\phi$ induces a vector bundle morphism $\tilde{\phi}^*_\Sigma \colon \Sigma_{\mathcal{S}} M \to \Sigma_{\mathcal{S}'} M'$. This morphism can be constructed as follows. First, consider the map -$$ - (\tilde{\phi}^*, \mathrm{id}) \colon \mathcal{S}M \times \Sigma \to \mathcal{S}'M' \times \Sigma . -$$ Here, the OP's original intuition that spinors should be treated as "scalar fields" whose pointwise values are unchanged by push-forward along diffeomorphisms is implemented by the fact that the above map acts as the identity on the $\Sigma$ factor. Next, note that $(\tilde{\phi}^*, \mathrm{id})$ is $\mathrm{Spin}(p,q)$-equivariant, and hence induces a map of the associated bundles, which is precisely our desired $\tilde{\phi}^*_\Sigma$ push-forward of spinor fields. -It remains only to recall the OP's final piece of intuition, that spinor fields transform as a "spinor" under changes of orthogonal frames and see how it is implemented in the above construction. Locally, a spin structure is often specified by a local choice of a specific frame field. Say that we have chosen specific frame fields $e^i$ on $(M,g)$ and $e'^i$ on $(M',g')$. Obviously, upon pushing forward $\phi^* e^i = \Phi^i_j e'^j$, where $\Phi$ is an $\mathrm{SO}(p,q)$ valued function, which need not be the identity. So, if $\sigma^\alpha(x)$ is a spinor field, that is, a section of $\Sigma_\mathcal{S} M$, then the push-forward $\sigma' = \tilde{\phi}^*_\Sigma \sigma$ spinor field is defined by the formula -$$ - \sigma'^\alpha(\phi(x)) = \Phi^\alpha_\beta \sigma^\beta(x) , -$$ -where $\Phi^\alpha_\beta$ is the matrix representing the pointwise action of $\Phi$ on $\Sigma$. The rigid frame rotation (or "Lorentz transformation") $\Phi$ takes into account the difference between the orthogonal frame $e'^i$ on $(M',g')$ and the push-forward $\phi^* e^i$ from $(M,g)$. -I have one more small remark to make about the meaning of the components $\sigma^\alpha(x)$ on a manifold. On flat space, they would be components with respect to some basis in the vector space $\Sigma$. But, on a manifold, these components obviously need to be interpreted with respect to some (spin-)frame in the spinor bundle $\Sigma_{\mathcal{S}} M$, but a global frame need not always exist, and even locally there is huge freedom in choosing in choosing a spin-frame as a function on $M$. What is important for the above push-forward of spinor fields to make sense, the spin-frames on $\Sigma_{\mathcal{S}} M$ and $\Sigma_{\mathcal{S}'} M'$ need to be chosen such that the $\gamma$-matrices which implement the Clifford multiplication and the action of $TM$ on $\Sigma_{\mathcal{S}} M$ must be fixed. That is, there must be some constant matrix $(\Gamma^i)^\alpha_\beta$ such that $(\gamma^i)^\alpha_\beta(x) = (\Gamma^i)^\alpha_\beta$ with respect to the chosen spin-frame on $M$, while at the same time also $(\gamma'^i)^\alpha_\beta(x') = (\Gamma^i)^\alpha_\beta$ with respect to the chosen spin-frame on $M'$. So the above push-forward rule only works when considering spin-frames of this kind.<|endoftext|> -TITLE: What is the $\mathbb{Z}_2$ cohomology of an oriented grassmannian? -QUESTION [17 upvotes]: Let $\operatorname{Gr}(k, n)$ and $\operatorname{Gr}^+(k, n)$ denote the unoriented and oriented grassmannians respectively. -The $\mathbb{Z}_2$ cohomology of the unoriented grassmannian is -$$H^*(\operatorname{Gr}(k, n); \mathbb{Z}_2) \cong \mathbb{Z}_2[w_1(\gamma), \dots, w_k(\gamma)]/(\overline{w}_{n-k+1}, \dots, \overline{w}_n)$$ -where $\gamma$ is the tautological bundle, $\deg\overline{w}_i = i$ and $\overline{w} = 1 + \overline{w}_1 + \dots + \overline{w}_n$ satisfies $w(\gamma)\overline{w} = 1$. -I was under the impression that for $1 < k < n - 1$, the $\mathbb{Z}_2$ cohomology of the oriented grassmannian is -$$H^*(\operatorname{Gr}^+(k, n); \mathbb{Z}_2) \cong \mathbb{Z}_2[w_2(\gamma), \dots, w_k(\gamma)]/(\overline{w}_{n-k+1}, \dots, \overline{w}_n)$$ -but this is false, as can be seen by considering $\operatorname{Gr}^+(2, 4) = S^2\times S^2$. - - -What is $H^*(\operatorname{Gr}^+(k, n); \mathbb{Z}_2)$? Can it be expressed in terms of $w_i(\gamma)$? - - -What I would really like to know is the answer to the following question (which might be easier to answer than the first question): - - -When is $w_i(\gamma) \in H^*(\operatorname{Gr}^+(k, n); \mathbb{Z}_2)$ non-zero? - -REPLY [3 votes]: This is a partial answer to the second question. -In the question, I intended $\gamma$ to denote the tautological bundle over $\operatorname{Gr}^+(k, n)$, whereas in Mark Grant's answer, he uses $\gamma$ to denote the tautological bundle over $\operatorname{Gr}(k, n)$. To make things absolutely clear, let $\gamma^+$ denote the tautological bundle over $\operatorname{Gr}^+(k, n)$ instead. As $\pi^*\gamma \cong \gamma^+$, where $\pi : \operatorname{Gr}^+(k, n) \to \operatorname{Gr}(k, n)$ is the natural double cover, we see that $w_i(\gamma^+) = w_i(\pi^*\gamma) = \pi^*w_i(\gamma)$ is non-zero if and only if $w_i(\gamma)$ is not in the ideal generated by $w_1(\gamma)$. -As the ideal $(\overline{w}_{n-k+1}, \dots, \overline{w}_n)$ is generated by elements of degree at least $n - k + 1$, the elements $w_{i_1}(\gamma)\dots w_{i_t}(\gamma)$ with $i_1 + \dots + i_t \leq n - k$ are independent. In particular, for $1 < i \leq \min\{k, n-k\}$, $w_i(\gamma)$ does not belong to the ideal generated by $w_1(\gamma)$ and hence $w_i(\gamma^+) \neq 0$. Therefore, if $k \leq n - k$, then $w_2(\gamma^+), \dots, w_k(\gamma^+) \in H^*(\operatorname{Gr}^+(k, n); \mathbb{Z}_2)$ are all non-zero. -This explains the observation at the end of Mark Grant's answer: $w_2(\gamma^+)$ is non-zero in $H^*(\operatorname{Gr}^+(2, n); \mathbb{Z}_2)$ for $n \geq 4$, and $w_2(\gamma^+), w_3(\gamma^+)$ are non-zero in $H^*(\operatorname{Gr}^+(3, n); \mathbb{Z}_2)$ for $n \geq 6$; in addition, $w_2(\gamma^+)$ is non-zero in $H^*(\operatorname{Gr}^+(3, 5); \mathbb{Z}_2)$.<|endoftext|> -TITLE: What is $\mathrm{O}_q/\mathrm{SO}_q$ if $q$ is a quadratic $\mathbb{Z}$-form which is degenerate? -QUESTION [5 upvotes]: Any binary quadratic $\mathbb{Z}$-form $q$ induces a symmetric bilinear form -$$ B_q(u,v) = q(u+v) - q(u) -q(v) \ \ \forall u,v \ \in\mathbb{Z}^2 $$ -and it is considered non-degenerate (over $\mathbb{Z}$) if its discriminant -$$ \text{disc}(q) := \det(B_q(e_i,e_j)_{1 \leq i,j \leq 2}) $$ -where $e_1 = (1,0)$ and $e_2 = (0,1)$ is invertible in $\mathbb{Z}$, i.e., equals $\pm 1$: see (2.1) in: http://math.stanford.edu/~conrad/papers/redgpZsmf.pdf . -Suppose $q$ is degenerate, but still $\text{disc}(q) \neq 0$ (so it is non-degenerate over $\mathbb{Q}$). So its special orthogonal group scheme $SO_q$ defined over $\text{Spec} \ \mathbb{Z}$, does not have to be smooth, but it is flat as $\mathbb{Z}$ is Dedekind (loc. sit. Definition 2.8 and right after), and it is closed in the full orthogonal group $O_q$, whence the quotient $Q:=O_q/SO_q$ is representable. -My question is: Is $Q$ a finite group of order $2$ over $\text{Spec} \ \mathbb{Z}$ ? -Apparently, when applied to any integral domain $R$ which is an extension of $\mathbb{Z}$, the elements of $O_q(R)$, in some matrix realization, must be of $\det \pm 1$, so we could think of $Q$ as a $\mathbb{Z}$-group of order $2$, but as a functor of points, $O_q$ can be applied to any $\mathbb{Z}$-algebra $R$, for which we may find elements of $O_q(R)$ which are not of $\det = \pm 1$. -For example, let $q(x,y)=x^2+y^2$. One can verify it is degenerate. -We get $SO_q = \text{Spec} \ \mathbb{Z}[x,y]/(x^2+y^2-1)$. Consider its matrix realization $$ \left \{ A=\left( \begin{array}{cc} - x & -y \\ - y & x \\ - \end{array}\right): \det(A)=1 \right \}. $$ -Then the component of $\det = -1$ elements in $O_q$ is obtained by $\text{diag}(1,-1)SO_q$. -So apparently, $Q = \mu_2$ (which unlike the other order $2$ group $(\mathbb{Z}/2)_\mathbb{Z}$, it has a double point at the reduction at $(2)$, not two distinct ones). -So far everything is good. But $A=\text{diag}(3,1)$ belongs to $O_q(R)$ where $R=\mathbb{Z}/8$ (as $A^T \cdot A = I_2$ in $R$, where $I_2$ represents $q$), but $\det(A)\neq \pm 1$ in $R$ ! -Does it mean that $O_q$ is more than these two connected components ? -I thought to avoid this problem by considering $O_q$ and $SO_q$ as flat sheaves (in the small site of flat extensions of $\mathbb{Z}$, since what I really care is of $H^1_\text{fppf}(\mathbb{Z},O_q)$), but we may still find extensions such as $\mathbb{Z} \times \mathbb{Z}$ containing a square root of unity other than $-1$ ?! -Could you please help ? -Thank you ! -Rony - -REPLY [3 votes]: We asked: -If indeed $\textbf{O}_q = \textbf{O}_q^+ \cup \textbf{O}_q^-$ defined over $\text{Spec} \, \mathbb{Z}$, then applied as a functor of points to some ring $R$ (being a $\mathbb{Z}$-algebra), how could exist a point in $\textbf{O}_q(R)$ which is neither in $\textbf{O}_q^+(R)$ nor in $\textbf{O}_q^-(R)$ ? -The answer, more generally, is that given a union of schemes $X = X_1 \cup X_2$ -which is not disjoint, and a ring $R$ which is not a domain (having some zero-divisors), then $X(R)$ should not be equal to $X_1(R) \cup X_2(R)$, because a union of schemes corresponds to the multiplication of their underlined varieties. -This equality holds for domains, however. -The simplest example is the group $\mu_2$ defined over $\text{Spec}\,\mathbb{Z}$ -being the union of the two components whose varieties are $x=1$ and $x=-1$. -These coincide at the prime $(2)$. The point $3$ belongs to $\mu_2(R=\mathbb{Z}/8)$ though not to any of the components $R$-points, since $(3-1)(3+1) \equiv 0 (\text{mod}8)$. -This means that not any point of $\mu_2(R)$, considered as a morphism $\text{Spec}\,R \to \mu_2$, can be lifted to a morphism from $\text{Spec} \, R$ to the disjoint union of $\text{Spec} \,\mathbb{Z}[t]/(t-1)$ and $\text{Spec} \,\mathbb{Z}[t]/(t+1)$. -Since $q$ is assumed to be non-degenerate over $\mathbb{Q}$ and invertible matrices over $\mathbb{Z}$ can only have determinant $\pm 1$, the above union is the all orthogonal group. As $\textbf{O}_q^+$ is a closed flat subgroup, $\textbf{O}_q/\textbf{O}_q^+$ is a finite $\mathbb{Z}$-group of order $2$ (can be either $\mu_2$ or $(\mathbb{Z}/2)_\mathbb{Z}$). -Thanks to Prof. Joseph Bernstein.<|endoftext|> -TITLE: Can discriminant polynomials become perfect powers on hyperplanes? -QUESTION [8 upvotes]: Let -$$\displaystyle f(x) = a_d x^d + a_{d-1} x^{d-1} + \cdots + a_0.$$ -Consider the discriminant of $f$, denoted by $\Delta(f)$, defined as -$$\displaystyle \Delta(f) = a_d^{2d-2} \prod_{i < j} (\theta_i - \theta_j)^2,$$ -where $\theta_1, \cdots, \theta_d$ are the roots of $f(x) = 0$ (over some algebraic closure, say). -It is well-known that $\Delta(f)$ is a homogeneous polynomial of degree $2d-2$ in the coefficients $a_d, \cdots, a_0$. -We say that a homogeneous polynomial $F \in \mathbb{C}[x_0, \cdots, x_n]$ of degree $m$ ramifies completely on a hyperplane if there exists a hyperplane $P$ in $\mathbb{P}^n$ such that $F |_P$ is a perfect $k$-th power for some $k > 1$ dividing $m$ (as a polynomial). For example, the cubic polynomial $F(x,y,z) = x^3 + yz^2$ ramifies completely on the lines (hyperplanes in $\mathbb{P}^2$) $y = 0, z = 0$. -For $d = 2$, we have that $\Delta(f) = a_1^2 - 4 a_2 a_0$ ramifies completely on $a_2 = 0, a_0 = 0$. Does this happen for $d > 2$? That is, does there exist $d > 2$ and a hyperplane $P \in \mathbb{P}^d$ such that $\Delta(f) |_P$ is a perfect $k$-th power, for some $k | 2d - 2$? - -REPLY [4 votes]: For any $d \geq 2$, there are hyperplanes on which $\Delta_d$ ramifies, but for $d \geq 3$, it never ramifies completely. I guess there are many proofs of this fact, let me give one based on projective duality. -First note that $\Delta_d$ parametrizes polynomials of degree $d$ having a multiple root. Let me give a geometric interpretation of $\Delta_d$. Let $X = v_d(\mathbb{P}^1) \subset \mathbb{P}(S^d \mathbb{C}^2)$ be the $d$-th Vernoese embedding of $\mathbb{P}^1$. The equation $\Delta_d = 0$ gives in the dual projective space $\mathbb{P}(S^d \mathbb{C}^2)^*$ the variety which parametrizes singular hyperplane sections of $X$. This variety is known as the projective dual of $X$. -For any $Z \subset \mathbb{P}^N$, which is irreducible, the reflexivity Theorem tells you that $(Z^*)^* = Z$ and for any $z \in Z_{smooth}$, the tangency locus of $z^{\perp}$ with $Z^*$ is identified as a scheme to $\mathbb{P}(N_{Z/\mathbb{P}^N,z}^*)$. -Going back to your situation, we have $X = v_d(\mathbb{P}^1) \subset \mathbb{P}(S^d \mathbb{C}^2)$ is smooth, its projective dual $X^* \subset \mathbb{P}(S^d \mathbb{C}^2)^*$ is the hypersurface which equation is $\Delta_d = 0$. You want to know if there exists $x \in \mathbb{P}(S^d \mathbb{C}^2)$ such that $x^{\perp} \cap X^*$ is completely non-reduced. This is equivalent to saying that the reduced space underlying the singular locus of $x^{\perp} \cap X^*$ is equal to the reduced space underlying $(x^{\perp} \cap X^*)$. -Two cases occcur: -1) if $x \notin X$, then $x^{\perp}$ is not tangent to $X^*$ (this is because $(X^*)^*= X$). Hence the singular locus of $x^{\perp} \cap X^*$ has codimension at least one in $x^{\perp} \cap X^*$ and the equation defining $x^{\perp} \cap X^*$ can't be a $k$-perfect power. -2) if $x \in X$, then $x^{\perp}$ is tangent to $X^*$ (this is again because $(X^*)^*= X$). Since $X$ is smooth, the reflexivity Theorem ensures that the tangency locus of $x^{\perp}$ with $X^*$ is scheme-theoretically a hyperplane in $x^{\perp}$. As a consequence, the singular locus of $x^{\perp} \cap X^*$ is generically scheme-theoretically a hyperplane. Now, if $x^{\perp} \cap X^*$ was a $k$-perfect power, then the fact that the singular locus of $x^{\perp} \cap X^*$ is generically scheme-theoretically a hyperplane in $x^{\perp}$ would imply that $k=2$ and $\deg \Delta_d =2$. This is only possible if $d=2$. -As far as non-complete ramification is concerned, the above argument shows that for any $x \in X$, the equation of $x^{\perp} \cap \{\Delta_d = 0\}$ can be written as $L^2Q$ where $L$ is a linear factor and $Q$ a polynomial of degree $2d-4$ which is not a power.<|endoftext|> -TITLE: Motivic cohomology is universal with respect to what (co)homology theories? -QUESTION [9 upvotes]: I have been told several times, at least implicitly, that motivic cohomology should be universal with respect to Bloch-Ogus cohomology theories. Is it proved somewhere or is it just some folk theorem? -More generally, what can be said if one allows general schemes over a sufficiently good base $S$ (probably something like geometrically unibranch?) or change the topology (e.g., étale motivic cohomology)? -Still more generally, what can be said if one uses Fulton-Macpherson's bivariant theories or Mixed Weil cohomology theories? - -REPLY [5 votes]: I think a proof that motivic Borel-Moore homology is a universal Borel-Moore homology theory could be deduced from work of Bloch, Geisser-Levine, and Voevodsky, at least for varieties over fields. I have never seen the details of this written down. -Bloch sketches the fact that higher Chow groups have a cycle class map to certain cohomology theories in "Algebraic cycles and the Beilinson conjectures." A paper of Geisser-Levine elaborate on this, in the case of etale cohomology. I think that these methods can be used to show that higher Chow groups are the universal Borel-Moore homology theory. -Voevodsky proves that motivic Borel-Moore homology for smooth varieties over a field are isomorphic to higher Chow groups. See "Lectures on Motivic Cohomology" by Mazza-Weibel-Voevodsky. It should be possible to verify that this isomorphism is induced by the map from higher Chow groups constructed via the method of Bloch and Geisser-Levine.<|endoftext|> -TITLE: Landau's theorem using nth roots -QUESTION [5 upvotes]: This question was asked earlier at MSE . -Let $\omega$(n) denote the number of distinct primes dividing $n$. The Mobius function is defined as $\mu(n) = (-1)^{\omega(n)}$ if $n$ is squarefree and $\mu(n) = 0$ otherwise . Next let $S(n) = \sum_{k=1}^n \frac{\mu(k)}{k}$. It is known that $S(n)$ approaches zero as $n$ approaches infinity and that this is -equivalent to the prime number theorem (von Mangoldt, Landau). -What happens if we replace powers of $(-1)$ in the Mobius function with other roots of unity? To focus on a specific case, let's use fourth roots and -define $f(n) = i^n$ if $n$ is squarefree and $f(n) = 0$ otherwise. Then $f(n)$ is a -multiplicative function whose initial values are $(1,i,i,0,i,-1,i,0,0,-1,...)$. Finally, let $T(n) = \sum_{k=1}^n \frac{f(k)}{k}$. Does $T(n)$ have properties analogous to those of $S(n)$? -Questions: $(1)$ Does $\sum_{k=1}^{\infty}$ $\frac{f(k)}{k}$ converge? [The corresponding infinite product $\prod_p (1 + i/p)$ does not converge since -$\sum\frac{1}{p^2} < \infty$ while $\sum \frac{1}{p} = \infty$. ] -$(2)$ The partial sums $S(n)$ are known to satisfy $|S(n)| \leq 1$ for all $n$. -Are the partial sums $|T(n)|$ also bounded by some constant independent of $n$? [Over the initial stretch $1 \leq n \leq 20$ , one finds that the max -$T$ value is $|T(19)| = 1.57 ...$ ]. -Thanks - -REPLY [4 votes]: Theorem 1 in Section 6.1 of Tenenbaum's Introduction to Analytic and Probabilistic Number Theory states (after taking its $N=0$ for simplicity) that for any $|z|<2$, -$$ -\sum_{n\le x} z^{\omega(n)} \sim \frac1{\Gamma(z)} \prod_p \bigg( 1 + \frac z{p-1} \bigg) \bigg( 1-\frac1p \bigg)^z \cdot x\, (\log x)^{z-1}. -$$ -The same method would show that -$$ -\sum_{n\le x} \mu^2(n) z^{\omega(n)} \sim \frac1{\Gamma(z)} \prod_p \bigg( 1 + \frac zp \bigg) \bigg( 1-\frac1p \bigg)^z \cdot x\, (\log x)^{z-1}. -$$ -From here, partial summation gives (for $z\ne 0$) -$$ -\sum_{n\le x} \mu^2(n) \frac{z^{\omega(n)}}n = \frac1{\Gamma(z)} \prod_p \bigg( 1 + \frac zp \bigg) \bigg( 1-\frac1p \bigg)^z \cdot \frac{(\log x)^z}z + c(z) + o(1) -$$ -for some constant $c(z)$. So when $|z|<2$ and $\Re z<0$, the sum converges to this $c(z)$; when $|z|<2$ and $\Re z>0$, the sum grows to infinity in modulus. When $|z|<2$ and $\Re z=0$, the sum oscillates asymptotically around a circle in the complex plane. So the interesting case $z=i$, strangely, the answer to your question (1) is no while the answer to your question (2) is yes!<|endoftext|> -TITLE: A conjecture on the coefficient of a special term in the expansion of the graph polynomial? -QUESTION [10 upvotes]: Recently, I am interested in the graph polynomial of the product of cycles. -Let $G = (V , E)$ be an undirected multi-graph with vertex set $\{1,\cdots,n\}$. The graph polynomial of $G$ -is defined by -$$f_G(x_1,x_2,\cdots,x_n)=\prod_{1\leq i 0$. -It remains to prove that $M$ is antihermitian ($M$ is certainly non-zero). That is the point why we have chosen such a set of colors. -Let $u=[u_1,\ldots,u_{2n+1}]\in \mathcal{U}$ be a proper 3-coloring of $C_{2n+1}$ with 3 colors $1,w,w^2$. Denote by $u_i^*$ the unique element of the set $\{1,w,w^2\}\setminus \{u_i,u_{i-1}\}$. -Assuming that the sign of $f_{H}$ is chosen so that $f_H(u)=\prod_i (u_i-u_{i-1})$ (the indices are cyclic modulo $2n+1$), applying the obvious relation -$$\frac1{u_i-u_{i}^*}=\frac{u_i-u_{i-1}} -{\prod_{b\in A\setminus u_i} (u_i-b)}$$ -we get -$$ -M_{u,v}=\prod_{i=1}^{2n+1}\frac{u_i-v_i}{u_i-u_i^*}.\quad \quad (2) -$$ -So the relation to check is -$$ -M_{v,u}=-\overline{M_{u,v}}.\quad \quad \quad \quad (3) -$$ -Applying (2) and substituting $\bar{z}=1/z$ for roots of unity $z=u_i,v_i,u_i^*$ we simplify (3) to the following: -$$ -\prod_i \frac{u_i-u_i^*}{u_i}=-\prod_i \frac{v_i^*-v_i}{v_i^*} \quad \quad (4) -$$ -provided that $u_i\ne v_i$ for all $i$ (if $u_i=v_i$ for some $i$, then $M_{u,v}=M_{v,u}=0$). -Denote $\varepsilon_i=u_i/u_{i-1}$, then $\varepsilon_i\in \{w,w^2\}$ and $u_i^*=u_i \varepsilon_i$. Analogously write $v_i^*=v_i \delta_i$ so we rewrite (4) as -$$ -\prod_i(1-\varepsilon_i)=-\prod_i (1-\overline{\delta_i})=(-1)^{1+(2n+1)}\prod \delta_i^{-1}\prod(1-\delta_i). -$$ -Obviously $\prod \varepsilon_i=\prod \delta_i=1$ so this in turn rewrites as -$$ -\prod_i(1-\varepsilon_i)=\prod_i (1-\delta_i). \quad \quad (5) -$$ -Now we have $1-\varepsilon=\pm i\sqrt{3}\varepsilon^2$ for $\varepsilon\in \{w,w^2\}$ (the signs are distinct for $w$ and $w^2$). Substituting this for $\varepsilon_i$'s and $\delta_i$'s and using $\prod \delta_i=\prod \varepsilon_i=1$, we reduce (5) to the following fact: the total number of $\varepsilon_i$'s and $\delta_i$'s which are equal to $w$ is even. Call the index $i$ white if $u_i=w\cdot v_i$ and black if $u_i=w^2\cdot v_i$. Then $\varepsilon_i=\delta_i$ if $i-1,i$ have the same color and $\varepsilon_i\ne \delta_i$ if $i-1,i$ have different color. Obviously there are even number of indices $i$ of the second type, thus the result.<|endoftext|> -TITLE: Schur's Theorem about immanants -QUESTION [12 upvotes]: $\DeclareMathOperator\Imm{Imm}$I am looking for a proof in English or French of Schur's theorem that, for every $H$ in the space $\mathbb H_n^+$ of positive semi-definite Hermitian matrices, and every irreducible character $\chi$ of $\mathfrak S_n$, $\chi(e)\det H\le\Imm_\chi(H)$, where the immanant $\Imm_\chi$ is defined by -$$\Imm_\chi(H):=\sum_\sigma\chi(\sigma)\prod_{i=1}^nh_{i\sigma(i)}.$$ -Notice that the original paper I. Schur, "Über endlicher Gruppen und Hermiteschen Formen" Math. Z., 1 (1918) pp. 184–207, is in German. -By the way, it seems that many authors relate Schur's theorem to symmetric polynomials. Is there any purely representation-theoretic proof of the inequality above? Let $(\rho,V)$ be a unitary representation whose character is $\chi$. We may associate to $\Imm_\chi(H)$ a Hermitian matrix over $V$ by -$$K_\rho:=\sum_\sigma\left(\prod_{i=1}^nh_{i\sigma(i)}\right)\rho(\sigma).$$ -It would be sufficient to prove that $K\ge(\det H)I_V$, where $I_V$ denotes the matrix of the scalar product. Because of Frobenius's theorem about the orthogonal decomposition of the regular representation, this amounts to proving that the analogous sum, where $\rho$ is replaced by the regular representation, satisfies the same estimate. In other words, Schur's theorem would be implied by the inequality -$$\forall \xi\in{\mathbb C}^{\frak S_n},\,\forall H\in{\mathbb H}_n^+,\qquad |\xi|^2\det H\le\sum_{\sigma,\theta}\bar\xi_\sigma\xi_\theta\prod_ih_{\sigma(i)\theta(i)}.$$ -Is this inequality true? - -REPLY [7 votes]: Many thanks to Denis for pointing out my erroneous initial "proof". This time around the proof is correct, and directly proves the assertion in line 3 of the OP, i.e., $\chi(e)\det(A)\le d_\chi(A)$ (I will write $d_\chi(I)$ instead of $\chi(e)$ for uniformity). -The explicit notation is cumbersome, so I am just writing a proof sketch. - -First, recall that $d_\chi(A)=z^T(\otimes^n A)z$ for a suitable vector $z$ -Next, use Cauchy-Schwarz to obtain $$|z^T(\otimes^n (X^TY))z|^2 = |z^T(\otimes^n X^T)(\otimes^n Y)z|^2\le z^T(\otimes^n X^TX)z \cdot z^T(\otimes^n Y^TY)z$$ -Now write $A=C^TC$ for some upper triangular matrix $C$ (since $A$ is PSD we can do this). Then, put $X=C$ and $Y=I$ above, to obtain -$|z^T(\otimes^n C)z|^2 = |z^T(\otimes^n I)z|^2|\det C|^2 \le |z^T(\otimes^n C^TC)z|\cdot |z^T(\otimes^n I)z|$, where we used the upper triangular nature of $C$ for the first step. In other words, we have shown that -$d_\chi(I)^2 \det(A) \le d_\chi(A)d(I)$, since $|\det C|^2=\det(C^TC)=\det(A)$.<|endoftext|> -TITLE: Find the maximum of the value $c(n)$ (similar to Hardy's inequality) -QUESTION [8 upvotes]: This problem has been posted on Math.SE for seven days, without a solution. - -Let $n\ge 2$ be a given positive integer, and $a_{1},a_{2},\cdots,a_{n}>0$, such that $$a_{1}a_{2}\cdots a_{n}=1$$ - Find the maximum of the value of $C(n)$ satisfying - $$\sum_{k=1}^{n}\left(\dfrac{1}{k}-\dfrac{2}{n(n+1)}\right)a_{k}\ge C(n)\left(\sum_{k=1}^{n}\dfrac{k^2}{a_{k}}\right)^{\frac{1}{n-1}}$$ - -This inequality is similar to Hardy's inequality on Math.SE, Various proofs of Hardy's inequality. -I have tried some methods but not solved it, such as Cauchy-Schwarz inequality or induction. - -REPLY [9 votes]: At first, we denote $c_k=\frac1k-\frac2{n(n+1)}$, take $C(n)=(n-1) \biggl( \frac{2}{n(n+1)!} \biggr)^{1/(n-1)}$ as in Brendan McKay's answer, and make the inequality homogeneous: $$\sum_{k=1}^n c_k a_k\geqslant C(n) \left(\sum_{k=1}^n k^2 \prod_{j\ne k} a_j\right)^{\frac1{n-1}}.$$ -Now we do not need the condition $\prod a_j=1$. Denote $a_k=kx_k$, using Brendan McKay's guess on where is the maximum. The inequality rewrites as -$$\sum_{k=1}^n kc_k x_k\geqslant C(n) (n!)^{\frac1{n-1}} \left(\sum_{k=1}^n k \prod_{j\ne k} x_j\right)^{\frac1{n-1}}.$$ -Now we fix LHS and maximise RHS. If the maximum point is on the boundary, i.e. some $x_i$ equals 0, the inequality reduces to AM-GM for $n-1$ numbers $kc_kx_k$, $k\ne i$ (and comparing the constants - this part is quite boring and I skip it). Otherwise by Lagrange multipliers the partial derivatives of $\sum_{k=1}^n k \prod_{j\ne k} x_j$ in the maximum point should be proportional to the numbers $kc_k$. We have $$\frac{\partial}{\partial x_i} \sum_{k=1}^n k \prod_{j\ne k} x_j=P\sum_{k\ne i} \frac{k}{x_k x_i},$$ -where $P=\prod_k x_k$. So we get, denoting $y_k=\frac1{x_k}$, the following proportionality relations: $\sum_{k\ne i} ky_ky_i=\lambda c_k=\lambda (1-\frac{2i}{n(n+1)})=\frac{\lambda}{n(n+1)/2}\sum_{k\ne i} k$. Now denote $M=\max y_i$, let $M=y_a$, $m=\min y_i$, let $m=y_b$. For $i=a$ we have $$\frac{\lambda}{n(n+1)/2}\sum_{k\ne a} k=\sum_{k\ne a} ky_ky_a\geqslant Mm\sum_{k\ne a} k,$$ for $i=b$ we have $$\frac{\lambda}{n(n+1)/2}\sum_{k\ne b} k=\sum_{k\ne b} ky_ky_b\leqslant Mm\sum_{k\ne b} k.$$ -Therefore $\frac{\lambda}{n(n+1)/2}=Mm$ and we have equalities everywhere. For $n\geqslant 3$ it implies that all $y$'s, thus all $x$'s are equal and the equality occurs. For $n=2$ we always have the equality.<|endoftext|> -TITLE: Stationary Navier-Stokes solutions -QUESTION [5 upvotes]: Are there known nontrivial ($u\neq0$) stationary solutions to Navier-Stokes equations in $\mathbb R^3$ ? Not square integrable of course (that's impossible), but with self-similar amplitudes of Fourier modes such as $|\hat u(c\xi)|=c^{-2}|\hat u(\xi)|$ ? -That would be $\Delta u=B(u,u)$ with $\nabla\cdot u=0$, where $B(u,v):=\sum_{i=1}^3\partial_i(u_iv)+\nabla p$ and $p$ is such that $\nabla\cdot B(u,v)=0$. -Same question for the particular class of cylindrically symmetric solutions. In cylindrical coordinates: $(r,\theta,z)$ and $(u_r,u_\theta,u_z,p)$ not depending on $\theta$, with $\frac1r\frac\partial{\partial r}(ru_r)+\frac{\partial u_z}{\partial z}=0$ (incompressibility). -I wonder if a solution with $u(c\mathbb x)=c^{-1}u(\mathbb x)$ is possible... - -REPLY [2 votes]: Tai-Peng Tsai's book Lectures on Navier-Stokes Equations (2018) cites as Theorem 8.3 (p.149) a theorem of V. Sverak (2011) that excludes the existence of minus one homogeneous solutions on $\mathbb R^3$. Technically, the only such solutions on $\mathbb R^3-\{0\}$ are the so-called Landau or Slezkin-Landau solutions, which in $\mathbb R^3$ satisfy the stationary Navier-Stokes equation with an exterior force $\mathbb b\delta$ (and vanish if $\mathbb b=0$). -What remains from the question is the possible existence of solutions with some other scaling, at infinity only, like $u\propto U(\alpha,\theta)r^{-2/3}$ in spherical coordinates. Such a scaling allows kinetic energy input from infinitely large scales, that would compensate viscous dissipation at small scales.<|endoftext|> -TITLE: A ridiculous combinatorial cardinal characteristic of the continuum? -QUESTION [7 upvotes]: This question assumes familiarity with combinatorial cardinal characteristics of the continuum. It is abstracted out of a question in a joint research with Jialiang He. I hope we've got the abstraction right. -A family of subsets of $\mathbb{N}$ is centered if every finite subfamily -has an infinite intersection. A pseudointersection of a family is an infinite set that is almost contained in every member of the family. -Let $\mathfrak{ridiculous}$ be the the minimal cardinality of a centered family -of subsets of $\mathbb{N}$ with no 2 to 1 image that has a pseudointersection. -By 2 to 1 image of a family $\mathcal{A}$ -we mean the family $\{f[A] : A\in\mathcal{A}\}$, -for some 2 to 1 function $f\colon \mathbb{N}\to \mathbb{N}$. -($f[A]:=\{f(n):n\in A\}$). -We know that $\mathfrak{p}\le\mathfrak{ridiculous}\le \operatorname{add}(\mathcal{M})$. (We see this using selection principles; direct arguments of course must exist, too.) -Question. Is $\mathfrak{ridiculous}=\mathfrak{p}$? -A negative answer (i.e., consistently ``no'') would be ridiculous. - -REPLY [13 votes]: The cardinal $\mathfrak{ridiculous}$ is equal to $\mathfrak p$ (which is equal to the smallest character of a free filter without infinite pseudointesection on $\omega$). It suffices to prove that a free filter $\mathcal F$ on $\omega$ has infinite pseudointersection if $\mathcal F$ has a base $\mathcal B$ of cardinality $|\mathcal B|<\mathfrak{ridiculous}$. -The latter inequality implies that there exists a sequence $(\{x_n,y_n\})_{n\in\omega}$ of pairwise disjoint doubletons such that each - basic set $B\in\mathcal B$ (and consequently, each set in the filter $\mathcal F$) intersects all but finitely many doubletons $\{x_n,y_n\}$. -If $\{x_n\}_{n\in\omega}$ is not a pseudointersection of $\mathcal F$, then there exists a set $E\in\mathcal F$ such that the set $\Omega=\{n\in\omega:x_n\notin E\}$ is infinite. We claim that the set $\{y_n\}_{n\in\Omega}$ is a pseudointersection of $\mathcal F$. Indeed, for any $F\in\mathcal F$ the set $F\cap E$ intersects all but finitely many doubletons $\{x_n,y_n\}$, $n\in\Omega$, and contains no points $x_n\in\Omega$. Then $F\cap E$ contains all but finitely many point $y_n$, $n\in\Omega$, and so does the larger set $E$, which means that $\{y_n\}_{n\in\Omega}$ is an infinite pseudointersection of the filter $\mathcal F$. -So, in both cases, $\mathcal F$ has infinite pseudointersection: $\{x_n\}_{n\in\omega}$ or $\{y_n\}_{n\in\Omega}$ for some infinite set $\Omega\subset\omega$. - -Remark. This method easily generalizes to prove that $\mathfrak p$ is equal to the smallest cardinality $\mathcal F$ of a centered family $\mathcal F$ of subsets of $\omega$ such that for any $n$-to-1 map $\varphi:\omega\to\omega$ the image $\varphi(\mathcal F)$ has no infinite pseudointersection. -On the other hand, the smallest cardinality $\mathcal F$ of a centered family $\mathcal F$ of subsets of $\omega$ such that for any finite-to-one map $\varphi:\omega\to\omega$ the image $\varphi(\mathcal F)$ has no infinite pseudointersection is equal to $\mathfrak b$, see Theorem 9.10 in the survey "Combinatorial Cardinal -Characteristics of the Continuum" by Andreas Blass.<|endoftext|> -TITLE: Why $K(X) \longrightarrow G (X)$ is a Poincaré duality for K-theory? -QUESTION [14 upvotes]: It's well known that for Noetherian separated regular schemes the canonical map $$K(X) \longrightarrow G(X)$$ (Quillen uses $K'$ instead of $G$, though) is a weak equivalence. -This statement is usually called Poincaré duality. -One can also define $K$-theory with compact support (for sufficiently nice schemes $X$) by choosing a compactification $X \hookrightarrow \overline{X}$ and setting $K_c(X)$ as the homotopy kernel of $K (X) \rightarrow K (\overline{X}\setminus X)$. I have no idea on whether such $K_c$ and $K$ enjoy some kind of Poincaré duality. -When I hear something like Poincaré duality I expect some kind of cap product map with some fundamental virtual class $$H^{\bullet} \longrightarrow H_{d -\bullet}^{BM}$$ or, dually, $$H_c^{\bullet} \longrightarrow H_{d - \bullet}$$. Of course, there's a cap product $$K(X) \wedge G(X) \longrightarrow G(X)$$ induced by tensor product which when restricted to tensoring with $\mathscr{O}_X$ gives the Poincaré duality. -However, I'm not satisfied with such analogy. I, hence, ask the following. -1) Is there any sense in which $G$ is a $K$-theory with compact support? Or maybe it's even the opposite: $K$ is a $G$-theory with compact support? -2) If yes, is there any relation between $K_c (X)$ and $G(X)$? -3) If no, is there any kind of duality between $K (X)$ and $K_c (X)$? -4) If I'm actually sounding silly since in ordinary Poincaré duality both sides of the isomorphism are always simultaneously of the same kind (compact or not compact, for instance, $H_{\bullet}^{BM}$ is somehow non compact as $H^{\bullet}$), how can I see the duality as some isomorphism from a cohomology to a homology? In other words, why $K(X)$ should be a cohomology theory and $G(X)$ a homology theory? -5) If one uses some Atiyah-Hirzebruch spectral sequence for $G$-theory, would it be the case that the graded pieces of the $\gamma$ filtration define a motivic cohomology with compact support up to torsion? -6) What about 5 for $K_c$ instead of $G$? What about $G_c$? -7) After applying the Atiyah-Hirzebruch sequence to all the possibilities ($K$, $K_c$, $G$, $G_c$) what sort of Poincaré duality one acquires? -Thanks in advance. -EDIT -I've added new questions in order to correct my lack of attention to concordance of the "kind" (compact or noncompact) of the domain and codomain in the duality. -EDIT2 -Given the comments below by Marc Hoyois and Gasterbiter, $K_c (X)$ should be defined as the homotopy colimit over $r$ of $K (\overline{X}, r (\overline{X}\setminus X))$, where the prefix $r$ denotes the infinitesimal thickening of order $r$ (following the notation of https://arxiv.org/abs/1211.1813). -Also, as noted below, $G$ should behave as a Borel-Moore homology. The analogy, therefore, is that -$$K(X) \longrightarrow G(X)$$ is the analogous of the first duality expressed above (cohomology-BM homology), whereas $$K_c(X) \longrightarrow G_c(X)$$ should correspond to the second duality (the compact version), where $G_c (X) := G (\overline{X}, \overline{X}\setminus X)$ (Btw, how do I state these dualities using the six functor formalism instead of using this "underline $c$"?). -Therefore, only the last questions remain. I will restate them here. -1) If one applies he Atiyah-Hirzebruch spectral sequence to $K$, $K_c$, $G$ and $G_c$, then what will be the graded pieces of the $\gamma$-filtration up to torsion (take $X$ as general as possible)? Or even better, in the level of spectra, what kind of decomposition one acquires? -For instance, in the case of smooth $X$, $K(X) \wedge \mathbb{S}_{\mathbb{Q}} \cong \bigvee_i H \mathbb{Q} \wedge (\mathbb{P}^1)^{\wedge i}$ (I have no idea what happens when $X$ is not smooth, though). -2) Does one recover some kind of Poincaré duality from the graded pieces mentioned in 1? - -REPLY [15 votes]: To my knowledge, one can only make this analogy fully consistent with Weibel's homotopy invariant $K$-theory $KH$ and $G$-theory (although the proofs of what I claim below rely heavily on our understanding of classical algebraic $K$-theory). Then, using the canonical map $K(X)\to KH(X)$, the pairing relating $KH$ and $G$ induce the pairing relating $K$ and $G$ which fits in the folkloric description of Poincaré duality relating $K$ and $G$. -Indeed, classical Poincaré duality is a particular instance of Grothendieck-Verdier duality (i.e. Grothendieck duality for ordinary sheaves of abelian groups). Indeed, if $D(X)$ denotes the derived category of sheaves on a (nice locally compact) space $X$, then we have, for any continuous map $f:X\to Y$ a pullback functor $f^*:D(Y)\to D(X)$ which has a right adjoint $f_*: D(X)\to D(Y)$, and there is push-forward functor $f_!:D(X)\to D(Y)$, the right derived functor of the direct image with compact support functor, which has a right adjoint $f^!: D(Y)\to D(X)$. There is natural map $f_!\to f_*$ which is invertible for $f$ proper, and so on. -Now, if $X$ is a space and $a:X\to \{pt\}$ denotes the canonical maps to the point, - -the cohomology of $X$ with coefficients in $\mathbf Z$ is $a_* a^*(\mathbf Z)$ -the cohomology with compact support is $a_! a^*(\mathbf Z)$ -Borel-Moore homology is $a_* a^!(\mathbf Z)$ -Homology is $a_! a^!(\mathbf Z)$ - -For nice enough spaces (e.g. algebraic varieties), these are perfect complexes of abelian groups (whence dualizable objects in the derived category of abelian groups), and taking the dual in the derived category exchanges $*$ and $!$. In particular, the dual of homology $a_! a^!(\mathbf Z)$ is cohomology $a_* a^*(\mathbf Z)$, while the dual of homology with compact support $a_! a^*(\mathbf Z)$ is Borel-Moore homology $a_* a^!(\mathbf Z)$. Poincaré duality consists in identifying, when $X$ is smooth complex orientable of dimension $d$, $a_! a^* (\mathbf Z)$ and $a_! a^!(\mathbf Z)(-d)[-2d]$ (where $A(-n)=A\otimes H^2(\mathbf P^1(\mathbf C),\mathbf Z)^{\otimes n}$). -Using Morel-Voevodsky's motivic stable homotopy category $SH$, we can extend this to schemes: to simplify, I will restrict to schemes of finite type over a field $k$. Then, given a a commutative motivic ring spectrum $E$ in $SH(k)$, we may define $D(X)$ as the category of $E$-modules in $SH(X)$ (to be precise, of $a^*(E)$-modules in the $(\infty,1)$-category $SH(X)$, where $a:X\to\mathrm{Spec}\, k$ denotes the structural map). And we have most of the features above, replacing $\mathbf Z$ by $E$ (e.g. we have cohomology $a_* a^*(E)$ and so forth). In the case where $E=KGL$ is the object which represents Weibel's $KH$ in $SH$, this gives a context in which Grothendieck's six operations apply. -Here, Poincar�� duality identifies $a_! a^* (KGL)$ and $a_! a^!(KGL)(-d)[-2d]$ (for $X$ smooth of dimension $d$). Dually, it corresponds to -$$a_* a^!(KGL)\simeq a_* a^*(KGL)(d)[2d]$$ -but Bott-periodicity also says that $KGL\simeq KGL(d)[2d]$. -Furthermore, for possibly singular $X$, one can check that the global sections of $a_* a^!(KGL)$ really give back $G$-theory: -$$\Gamma(\mathrm{Spec} \,k,a_* a^!(KGL))=G(X)$$ -(this is essentially a reformulation of $K$-theoretic Poincaré duality as formulated in the question above, of Quillen's localization theorem for $G$-theory and of homotopy invariance for $G$-theory). And we have: -$$\Gamma(\mathrm{Spec} \,k,a_* a^*(KGL))=KH(X)$$ -We could define homotopy invariant $K$-theory with compact support: -$$KH_c(X):=\Gamma(\mathrm{Spec} \,k,a_! a^*(KGL))$$ -and $KH$-homology as $\Gamma(\mathrm{Spec} \,k,a_! a^!(KGL))$. -As for the coniveau spectral sequence (a.k.a the motivic Atiyah-Hirzebruch spectral sequence), applied to $G$-theory, Marc Levine has showed that the $E_2$-term will be motivic cohomology defined through Bloch's cycle complexes (also for $X$ singular): this is why the cohomology defined through Bloch cycle complex should not be called "motivic cohomology" but rather "motivic Borel-Moore homology". -Rationally, $KGL$ is naturally a $H\mathbf Q$-algebra, where $H\mathbf Q$ denotes the $\mathbf Q$-linear motivic Eilenberg-MacLane spectrum. In fact, $KGL$ becomes the free Bott-periodic $H\mathbf Q$-algebra. Furthermore, the category of $H\mathbf Q$-modules in $SH(X)$ is then equivalent to $DM(X,\mathbf Q)$ (the category of motivic sheaves over $X$), and the change of scalars functor from $H\mathbf Q$-modules to $KGL$-modules commutes with the six operations (at least if we restric to compact objects). In particular, one recovers Poincaré duality in $KGL$-modules from the one in motives, but with a "Todd-twist": the classical formulations of Poincaré duality as above involve Thom isomorphisms, which themselves rely on a choice of an orientation. Poincaré duality on homotopy $K$-theory as described above corresponds to the orientation of $KGL$ defined by the multiplicative formal group law, while the one coming from seeing rationalized $KGL$ as a $H\mathbf Q$-algebra corresponds to the additive formal group law. Relating the two through an explicit isomorphism is exactly the purpose of Grothendieck-Riemann-Roch theorems.<|endoftext|> -TITLE: Contravariant internal hom -QUESTION [6 upvotes]: Let $\mathcal{C}$ a symmetric monoidal category. By using symmetry, it is very easy to show that the contravariant internal hom functor $[-,A]\colon\mathcal{C}^{op}\longrightarrow\mathcal{C}$ is adjoint on the right to itself. -My question is: when is this functor additionally $[-,A]$ a left-adjoint? Do such objects $A$ have a name? In the category of $\textbf{Set}$ it is easily seen that $[-,A]$ preserves the terminal object $\textbf{Set}^{op}$ iff $A$ is the empty set. - -REPLY [8 votes]: Suppose that everything is dualizable (as in the category of finite-dimensional vector spaces over a field, for example), and write $TX=[X,A]=X^*\otimes A$. Then -\begin{align*} - \mathcal{C}(TX,Y) &= - \mathcal{C}(1,X\otimes Y\otimes A^*) = - \mathcal{C}(Y^*\otimes A,X) = - \mathcal{C}^{\text{op}}(X,TY) \\ - \mathcal{C}(W,TX) &= - \mathcal{C}(1,W^*\otimes X^*\otimes A) = - \mathcal{C}(X,W^*\otimes A) = - \mathcal{C}^{\text{op}}(TW,X) -\end{align*} -so $T$ is self-adjoint on both sides.<|endoftext|> -TITLE: Is it still an open problem whether $\mathbb{R}^\omega$ is normal in the box topology? -QUESTION [31 upvotes]: On page 205 of his Topology textbook, James Munkres made an interesting remark: - -It is not known whether $\mathbb{R}^\omega$ is normal in the box topology. Mary-Ellen Rudin has shown that the answer is affirmative if one assumes the continuum hypothesis. - -That's a reference to this paper by Mary Ellen Rudin. However, both Munkres and Rudin were writing decades ago. So my question is, what is the state of research on this problem? -Has it been proven in $ZFC$, or has it been proven to be independent of $ZFC$, or is it still an open problem whether it's a theorem of $ZFC$ or not? - -REPLY [9 votes]: I'm transcribing here some things from the comments section, so that this question can be marked as answered: -The problem is still open. -This survey by Roitman and Williams, from 2015, said that it was still open whether $(\omega+1)^\omega$ is normal in the box topology. $\mathbb R^\omega$ has a closed subspace homeomorphic to $(\omega+1)^\omega$, so if $\mathbb R^\omega$ is normal then so is $(\omega+1)^\omega$. -MathSciNet shows that survey as the most recent citation of Rudin's 1972 paper, and presumably anyone who solved the problem would have cited Rudin.<|endoftext|> -TITLE: Lie group actions on $S^n$ with some invariant hypersphere but no totally geodesic ones -QUESTION [8 upvotes]: Does there exist a compact connected Lie group $G$ acting smoothly as isometries on the standard sphere $S^n$ for some $n\ge 3$, so that no totally geodesic hypersphere $S^{n-1}$ is $G$-invariant, but there exists embedded $G$-invariant diffeomorphic $S^{n-1}$? -By $G$-invariant, I mean the submanifold is a union of orbits. I've searched the literature, and a direct search using some combination of the keywords didn't give anything immediately relevant. My guess is that if the principal orbits are of low enough dimension, then we would have enough degree of freedom to produce an invariant diffeomorphic hypersphere. However, I don't know whether this is the case. -I'm aware that Hsiang and Lawson have classified some low cohomogeneity actions back in the 1970s in their paper Minimal Submanifolds of Low Cohomogeneity. However, a quick look at those classifications doesn't seem to produce an example. - -REPLY [3 votes]: The answer below is by no means complete, but at least it treats the case when $G$ is a torus, and proposes an idea of what might be tried in the general case. The following fact is useful: -Remark. $G$ preserves a geodesic sphere on $S^n$ iff it fixes a point on $S^n$. -Statement 1. If the compact Lie group is a torus, then such an action doesn't exist. -Proof. Indeed, suppose that $\mathbb T^k$ preserves a sphere $S^{n-1}$ in $S^n$. We will deduce, that it fixes a point on $S^n$. By Schoenflies theorem $S^{n-1}$ cuts $S^n$ into two manifolds $B_1$ and $B_2$ both homeomorphic to a $n$-ball, in particular they have Euler characteristic $1$. So $\mathbb T^k$ must fix a point $x_1$ and $x_2$ in both. Indeed, any orbit of $\mathbb T^k$ of dimension $\ge 1$ has zero Euler characteristic, so in case there were no fixed points in $B_1$ and $B_2$ they would both have Euler characteristic zero. QED. -Other cases. I think, that this reasoning has a potential to be extended to other cases. Indeed, recall the following classical statement. -Statement. Let $G$ be a compact Lie group and $H$ be a subgroup. Then the homogeneous space $G/H$ has Euler characteristic $0$ if and only if $G$ and $H$ have the same rank. Moreover, in case of the same rank the Euler characteristic can be explicitly calculate. See the beginning of -http://moroianu.perso.math.cnrs.fr/tex/2014crelle.pdf -So I wonder when is the Euler characteristic of the quotient is equal to $1$ exactly. Probably this doesn't happen very often. On the other hand, if we want this strange action, so that $G$ doesn't fix any point on $S^n$ but fixes a hyper-sphere, it looks like one of the orbits should be of Euler characteristic exactly $1$.<|endoftext|> -TITLE: Gauss sums for general number fields -QUESTION [9 upvotes]: There is a wide litterature for the classical Gauss sums. For $\chi$ a primitive Dirichlet character modulo $N$, it is given by -$$\tau(\chi) = \sum_{n \text{ mod } N} \chi(n) \exp(2i\pi n/N).$$ -An interesting fact is that primitive Dirichlet characters are essentially their own Fourier transform, up to this associated Gauss sum: -$$\sum_{n \text{ mod } N} \chi(n) \exp(2i\pi nm/N) = \bar{\chi} (m) \tau(\chi). \qquad (\star)$$ -I am interested in similar properties in general number fields $F$. Let $\omega$ be a nonzero element of $F$ and write $\mathfrak{a}$ for its denominator. Hecke defined some analogous Gauss sums by -$$C(\omega) = \sum_{n \text{ mod } \mathfrak{a}} \exp(\mathrm{tr}(n^2 \omega)).$$ -(here $n \text{ mod } \mathfrak{a}$ means that $n$ is an integer ideal in $\mathfrak{o}_F/\mathfrak{a}\mathfrak{o}_F$, where $\mathfrak{o}_F$ is the ring of integers of $F$). I have not found anything concerning the analogous Gauss sums as above, namely for a finite order character $\chi$ on integer ideals which is a character modulo $\mathfrak{a}$, -$$\tau(\omega, \chi) = \sum_{n \text{ mod } \mathfrak{a}} \chi(n)\exp(\mathrm{tr}(n^2 \omega)).$$ -Is there anything of this kind in the litterature? In particular, do we have an analogous result to $(\star)$? - -REPLY [3 votes]: Hecke Gauss sums are quadratic Gauss sums, and (*) is correct for quadratic characters (Hecke, Satz 155). Hecke Gauss sums were studied later by - -Siegel, Über das quadratische Reziprozitätsgesetz in algebraischen -Zahlkörpern, Nachr. Akad. Wiss. Göttingen, Math. Phys. Kl. II -(1960), 1-16; Ges. Abh. 3 (1966), 334-349 -Shiratani, On the decomposition laws of rational primes in certain -class 2 extensions, Investigations in number theory, -Advances in Pure Math. 13 (1988), 345-411 -Shiratani, On the Gauss-Hecke-sums, J. Math. Soc. Japan 16 (1964), 32-38 -Boylan, Skoruppa, A quick proof of reciprocity for Hecke Gauss sums, -J. Number Theory 133 (2013), 110-114.<|endoftext|> -TITLE: The Hausdorff dimension of the union of singular orbits and exceptional orbits -QUESTION [5 upvotes]: Suppose we have a compact connected Lie group $G$ acting as isometries on a compact manifold $M^n.$ Then is it necessarily true that the Hausdorff dimension of the union of singular and exceptional orbits are no larger than the Hausdorff dimension of individual principal orbits? If not so, what about the union of singular orbits only? -It's known back in the 1950s through the works of Montgomery and Yang that in some cases, $\dim M^n\setminus M_{\text{principal}}\le n-2,$ the dimension of the complement of the union of principal orbits is at most $n-2.$ However, I'm not sure if they refer to Hausdorff dimension here. A search in the literature seems to indiate that there is no direct result on controlling $\dim M^n\setminus M_{\text{principal}}$ by the dimension of principal orbits. - -REPLY [2 votes]: According to the paper "ORBITS OF HIGHEST DIMENSION" by Montgomery and Yang, by "singular orbits" one means orbits that have dimension less than the generic (or principle) ones. -For this reason, the answer to your question is rather negative. Indeed, take the round $S^n$ and consider on it the action of $S^1$ that fixes pointwise a geodesic $S^{n-2}$. Then the dimension of a generic orbit is $1$, but the union of singular orbits is the fixed $S^{n-2}$, i.e., it has dimension $n-2$. -PS. There is also a way to use this example to get a similar cohomogenity $4$ counter-example. Namely consider $S^4\times S^n$ and take the group $S^1\times SO(n+1)$ that acts diagonally. The first $S^1$-factor fixes $S^2\subset S^4$ (as previously) and the second acts transitively on $S^n$. Then a generic orbit has dimension $n+1$, but $S^2\times S^n$ is composed of obits of dimension $n$.<|endoftext|> -TITLE: A character identity -QUESTION [18 upvotes]: This is related to my question, but it concerns a specific point of the proof of Schur's Theorem. -Let $G$ be a finite group and $\chi$ an irreducible character of $G$. Is it true that -$$\forall g\in G,\qquad\sum_{h\in G}\overline{\chi(h)}\chi(gh)=\frac{|G|}{\chi(e)}\,\chi(g)\quad?$$ -This does not seem to be related to the orthogonality properties of the table of characters. This obviously true if $\chi$ is a linear character, or if $g=e$. I checked it for $G=\frak S_3$ and for one nonlinear character of $\frak S_4$. - -REPLY [3 votes]: It is perhaps interesting to note that the Schur relations themselves do follow from the second orthogonality relations alone, in fact from orthogonality with the identity column. Start from that relation in the form $$\sum_{\chi\in\mathrm{Irr}(G)}\chi(1)\chi(gh)=\delta_{gh,1}|G|,$$ where $g,h\in G.$ Fix a matrix representation affording each $\chi\in\mathrm{Irr}(G)$ and label its matrix entries $\chi_{ij}(g)$ for $g\in G$ and $1\le i,j\le \chi(1).$ (The notation is slightly abusive as $\chi_{ij}$ depends on the choice of representation.) Let $I\subseteq\mathrm{Irr}(G)\times\mathbb N^2$ be the set of triples $(\chi,i,j)$ where $\chi\in\mathrm{Irr}(G)$ and $1\le i,j\le\chi(1),$ noting $|I|=|G|$ (from the above orthogonality relation with $gh=1$). Then $$\sum_{(\chi,i,j)\in I,(\psi,k,l)\in I}\frac{\chi(1)}{|G|}\chi_{ij}(g)\delta_{i,l}\delta_{j,k}\delta_{\chi,\psi}\psi_{kl}(h)=\delta_{gh,1}$$ for $g,h\in G,$ where $\delta_{a,b}$ is 1 if $a=b$ and 0 otherwise. This is just the above orthogonality relation again, as can be seen on collapsing the deltas. Define $I\times G,$ $I\times I$ and $G\times G$ matrices $A,B$ and $\Delta$ by $$A_{(\chi,i,j),g}=\chi_{ij}(g), B_{(\chi,i,j),(\psi,k,l)}=\delta_{i,l}\delta_{j,k}\delta_{\chi,\psi}\frac{\chi(1)}{|G|}\text{ and }\Delta_{g,h}=\delta_{gh,1}.$$ Note $A,$ $B$ and $\Delta$ are all $|G|\times|G|$ matrices. The above is equivalent to $$A^TBA=\Delta.$$ Since $\Delta^2=1_{|G|\times |G|},$ all the matrices are invertible. Hence $B^{-1}=A\Delta A^T.$ But -$$B^{-1}_{(\chi,i,j),(\psi,k,l)}=\delta_{i,l}\delta_{j,k}\delta_{\chi,\psi}\frac{|G|}{\chi(1)}\text{ and }A\Delta A^T_{(\chi,i,j),(\psi,k,l)}=\sum_{g\in G}\chi_{ij}(g)\psi_{kl}(g^{-1}).$$ Equating these gives the Schur relations: $$\sum_{g\in G}\chi_{ij}(g)\psi_{kl}(g^{-1})=\delta_{\chi,\psi}\delta_{il}\delta_{jk}\frac{|G|}{\chi(1)}.$$ (If we choose the matrices $\psi_{kl}$ to be unitary, as we always may for a finite group, we can replace $\psi_{kl}(g^{-1})$ with $\bar\psi_{lk}(g).$) -As noted by Geoff Robinson, the idempotent formula follows from the Schur relations by multiplying by $\chi_{ji}(h^{-1}),$ summing over $i$ and $j$: $$\sum_{g\in G}\chi(gh)\psi_{kl}(g^{-1})=\delta_{\chi,\psi}\chi_{kl}(h)\frac{|G|}{\chi(1)}$$ and finally taking the trace on both sides.<|endoftext|> -TITLE: Celestial mechanics and Runge Kutta methods -QUESTION [6 upvotes]: I am working on an example here to simulate the orbit of Earth for one year. -As you can see in the notebook, RK45 doesn't conserve energy, and after one simulated year it has spiraled in substantially. That's as expected. -I also don't expect RK23 to conserve energy; however, I have seen it do better with this kind of system, so I gave it a try. -It turns out to do a lot better: with fewer total function evaluations, it loses much less energy and ends up pretty close to the start. -Does anyone know why RK23 seems to do unreasonably well for this example? -I am using the SciPy function solve_ivp, if you want more detailed info on the implementations: https://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.solve_ivp.html -[And just to be clear, I know that there are other methods that conserve energy; that's not what this question is about.] - -REPLY [5 votes]: According to the documentation for the SciPy function solve_ivp, RK23 is based on the Bogacki-Shampine method, which is implemented in the MATLAB function ode23. Below are numerical results obtained from applying ode23 to a long-time integration of two Hamiltonian systems: a simple double-well example and the OP's earth orbit example. -Double-Well Example -Here I run ode23 on a double-well Hamiltonian system with Hamiltonian function $H(q,p) = (1/2) p^2 + (1/4) (q^2 - 1)^2$ using the following MATLAB code -ff=@(t,y) [y(2); y(1)-y(1).^3]; -opts=odeset('RelTol',tol); -[t,y]=ode23(ff,[0 5000],[0; 2],opts); - -for the tol values indicated in the figure titles below starting with the default relative tolerance of $10^{-3}$. Basically this code numerically integrates $$ -\dot{q} = p \;, \quad \dot{p} = q - q^3 \;, \quad (q(0),p(0)) = (0,2) \;, -$$ for a (long) time span of $5000$. ode23 outputs a vector of times $t$ and a matrix $y$ whose rows are the corresponding numerical approximation. -Below are the outputted discrete trajectories in phase space. The dots are made a bit lighter with time. The solid red curve is the level set of the Hamiltonian corresponding to the initial point. The actual solutions lie on this red curve for all time since they preserve $H$. In contrast, the outputted dots seem to converge to the right well, and the ones with smaller tol seem to take longer to converge. - - - -OP's Earth Orbit Example -Following the OP's description, one can similarly simulate the earth revolving around the sun using -m1=1.989e30; -m2=5.972e24; -G=6.674e-11; -ff=@(t,y) [y(3); y(4); ... - -G*m1*y(1)/(y(1)^2+y(2)^2)^(3/2); ... - -G*m1*y(2)/(y(1)^2+y(2)^2)^(3/2)]; -T=100*31556925.9747; % time-span is 100 years! -opts=odeset('RelTol',tol); -[t,y]=ode23(ff,[0 T],[147*1e9; 0; 0; -30300],opts); - -The following figure shows the relative energy error after 100 years for two different tol values. - -Discussion -Since we see a systematic energy drift in both of these relatively simple Hamiltonian test problems, these numerical counterexamples illustrate that ode23 is probably not a geometric integrator. For example, a geometric integrator that preserves the symplectic form of the Hamiltonian system (called symplectic integrators), typically do not preserve energy, but they do have bounded energy errors over long-time simulations. One can construct adaptive geometric integrators, but this is a bit tricky. See, e.g., -Calvo, M. P.; López-Marcos, M. A.; Sanz-Serna, J. M., Variable step implementation of geometric integrators, Appl. Numer. Math. 28, No. 1, 1-16 (1998). ZBL0930.65136..<|endoftext|> -TITLE: Splitting the injection that you get from the Poincaré-Birkhoff-Witt theorem -QUESTION [17 upvotes]: Let $\mathfrak g$ be a Lie algebra over a field of characteristic zero, with universal enveloping algebra $U\mathfrak g$. By the Poincaré-Birkhoff-Witt theorem one knows that $i:\mathfrak g \to U\mathfrak g$ is injective. In fact there is $s :U\mathfrak g \to \mathfrak g$ such that $s \circ i = \mathrm{id}$ and $s$ is a morphism of $\mathfrak g$-modules. -Can one choose such a splitting $s$ which is a morphism of Lie algebras? -My guess would be "no" (since if this were possible I should find something about it by googling). But I would appreciate a counterexample. -A remark is that there is a retraction of Lie algebras $\operatorname{Gr} U\mathfrak g \to \mathfrak g$. Indeed, the PBW isomorphism $\operatorname{Gr} U\mathfrak g \cong S\mathfrak g$ of algebras is compatible with the Poisson structure on both sides, where the Poisson bracket on $S\mathfrak g$ is $\{x_1\ldots x_n,y_1\ldots y_m\} = \sum_{i,j} [x_i,y_j] x_1 \ldots \widehat x_i \ldots x_n y_1 \ldots \widehat y_j \ldots y_m$ and the bracket $\operatorname{Gr}^n U\mathfrak g \otimes \operatorname{Gr}^m U\mathfrak g \to \operatorname{Gr}^{n+m-1} U\mathfrak g$ is given by $x \otimes y \mapsto [x,y]$, and $S\mathfrak g$ retracts onto $\mathfrak g$. - -REPLY [11 votes]: Edit: here's now a computation-free way to prove the existence for $\mathfrak{sl}_n$ and a few more, and discussion. -First, let me start with a simple observation, for an arbitrary Lie algebra $\mathfrak{g}$ over an arbitrary (associative unital) commutative ring $R$, and $\mathfrak{g}\to U(\mathfrak{g})$ the universal enveloping algebra: we have the equivalence of -(i) There is a Lie $R$-algebra retraction $U(\mathfrak{g})\to\mathfrak{g}$; -(ii) there is a unital associative $R$-algebra $A$ and Lie $R$-algebra homomorphisms $\mathfrak{g}\to (A,[\cdot,\cdot])\to\mathfrak{g}$ composing to $\mathrm{Id}_{\mathfrak{g}}$. -Indeed, (i)$\Rightarrow$(ii): just take $A=U(\mathfrak{g})$. For the converse, use the universal property to get a $R$-algebra homomorphism $U(\mathfrak{g})\to A$, and composing with $A\to\mathfrak{g}$ yields the desired retraction. -So all we need is to find $A$. Fix a field $K$ and denote by $p\ge 0$ its characteristic. Namely for $\mathfrak{g}=\mathfrak{sl}_n(K)$ when $p$ does not divide $n$, we take $A=M_n(K)$ and the retraction there is given by $\nu(v)=v-\frac1{n}\mathrm{Tr}(v)$. (The lengthy computation of my initial post is actually what this retraction $U(\mathfrak{sl}_2)\to M_2\to \mathfrak{sl}_2$ looks like!) -To obtain more examples, one observes that if $\mathfrak{g}$ is endowed with a Cartan grading then any graded Lie subalgebra also inherits a retraction. Let me explain in the case of a diagonalizable Cartan grading. The assumption is that we have a certain subspace $\mathfrak{g}_0$ (with linear dual denoted $\mathfrak{g}_0^*$) Lie algebra grading $\mathfrak{g}=\bigoplus_{\alpha\in\mathfrak{g}_0^*}\mathfrak{g}_\alpha$, such that $\mathfrak{g}_\alpha=\{x\in\mathfrak{g}:[h,x]=\alpha(h)x,\forall h\in\mathfrak{g}_0\}$ for each $\alpha\in\mathfrak{g}_0^*$. If so, for any (unital associative) algebra $A$ endowed with a Lie algebra homomorphism $i:\mathfrak{g}\to (A,[\cdot,\cdot])$ one can define: $A_\alpha=\{x\in A:i(h)x-xi(h)=\alpha(h)x,\forall h\in\mathfrak{g}_0\}$ for each $\alpha\in\mathfrak{g}_0^*$; this is multiplicative in the sense that $A_{\alpha}A_\beta\subset A_{\alpha+\beta}$ for all $\alpha,\beta$, and hence $\bigoplus_{\alpha\in\mathfrak{g}_0^*}A_\alpha$ is a unital subalgebra: in particular, if $i(\mathfrak{g})$ generates $A$ as a subalgebra (which is the only case of interest here since we consider quotients of the universal enveloping algebra), this is an algebra grading of $A$. The retraction has to be grading-preserving. Hence it maps every graded subalgebra to itself. -This applies to graded subalgebras of $\mathfrak{sl}_n$ (with $n$ not divisible by the characteristic $p$): for instance, the 2-dimensional non-abelian Lie algebra for $p\neq 2$, the 3-dimensional Heisenberg Lie algebra (viewed inside $\mathfrak{sl}_3$, or inside $\mathfrak{sl}_4$ to remove the restriction $p\neq 3$), etc, and all products $\prod_i\mathfrak{sl}_{n_i}(K)$. -This also applies under disguised occurrences of $\mathfrak{sl}_n$: for instance, over $\mathbf{R}$, $\mathfrak{so}_3\simeq \mathfrak{sl}_1(\mathbf{H})$ and we have a similar retraction. -This does not apply to other semisimple Lie algebras (i.e., not of type $A_n$ or products thereof). Indeed, assume for simplicity that $K$ is algebraically closed of characteristic zero and that $\mathfrak{g}$ is simple. Then if $\mathfrak{g}$ has the property that some $A$ as in (ii) exists with $A$ finite-dimensional, then $\mathfrak{g}$ is isomorphic to $\prod_i\mathfrak{sl}_{n_i}(K)$ for some family $(n_i)$. Indeed, since $\mathfrak{g}$ is semisimple, one easily deduces that $A$ is semisimple, so the underlying Lie algebra is isomorphic to a direct product $K^\ell\times\prod\mathfrak{sl}_{m_j}(K)$, and all its semisimple quotients have the required form. -So a test-case would be the 10-dimensional Lie algebra $\mathfrak{sp}_4\simeq\mathfrak{so}_5$. As I just said, $A$ in (ii) should be infinite-dimensional. But possibly just a few computations are enough to show that there is no retraction at all. -The question is also reasonable for $\mathfrak{g}$ nilpotent (say, over an algebraically closed field of characteristic zero. Here (ii'), defined as (ii) but without "unital" is a convenient criterion. -((ii) trivially implies (ii') and (ii') implies (ii) by adding a unit: $A'=A\oplus R$, observing that the projection onto $A$ is a Lie algebra homomorphism.) The question is then rephrased as: when is a Lie algebra retract of a Lie algebra whose law is the commutator bracket of some associative law? - -Initial post: - -Yes, there's such a retraction when $\mathfrak{g}=\mathfrak{sl}_2$. - -Write the basis $(h,x,y)$, $[h,x]=2x$, $[h,y]=-2y$, $[x,y]=h$. Denote the Casimir element $c=(h+1)^2+4yx$. I only assume that the ground field $K$ has characteristic $\neq 2$. -The enveloping algebra $U$ has its usual grading $U=\bigoplus_{n\in 2\mathbf{Z}}$. Here $U_0$ is the unital subalgebra generated by $h$ and $c$: it is commutative, and actually a polynomial algebra $K[h,c]$, freely generated by $h$ and $c$, $U_{2n}=x^nU_0$ and $U_{-2n}=y^nU_0$ for $n\ge 0$. In characteristic zero, one can characterize $U_{2n}$ as the $2n$-eigenspace for the derivation $v\mapsto hv-vh$ of $U$. (In arbitrary characteristic, one has a similar description using a 1-dimensional torus of automorphisms instead, but this does not matter.) It is known that there is no zero divisor in $U$, and in particular the multiplication by $x^n$ or $y^n$ from $U_0$ to $U_{\pm 2n}$ is a linear isomorphism. -Define a linear map $r:U\to\mathfrak{sl}_2(K)$ by - -on $U_0$ by $r(P(c,h))=\frac{P(4,1)-P(4,-1)}2h$; -on $U_2$ by $r(xP(c,h))=P(4,-1)x$; -on $U_{-2}$ by $r(yP(c,h))=P(4,1)y$; -on $U_{2n}$, $|2n|\ge 4$, as zero. - -It maps each of $x,y,h$ to itself, so it is a linear retraction. -Theorem. This is a Lie algebra homomorphism. -Before checking it, let me insist that the value 4 in the evaluation at the $c$ variable is absolutely not random. The above retraction actually factors through the quotient $U/(c-4)U$, which therefore retracts onto $\mathfrak{sl}_2$, but this is not the case of other quotients $U/(c-t)U$ for $t\neq 4$. Also notice that $c=4$ corresponds to the 2-dimensional representation... -Proof of the theorem. I'll use the formula $[A,BC]=[A,B]C+B[A,C]$, and its consequence $$=[A,C]BD+A[B,C]D-C[D,A]B+CA[B,D].$$ -Observe that $r$ vanishes on the 2-sided ideal $(c-4)U$, so that $r$ factors through $V=U/(c-4)U$. -Since $r$ preserves the grading, by linearity, we have to show that it preserves the bracket at every given degree. This is trivial in degree $\notin\{-2,0,2\}$. -In degree zero, by linearity (and using the vanishing on $(c-4)U$) we have to check that $$[r(x^\ell h^n),r(y^\ell h^m)]=r([x^\ell h^n,y^\ell h^m])$$ -for all $\ell,n,m\ge 0$. -This is clear if $\ell=0$ since both brackets lie in degree 0 where everything commutes. If $\ell\ge 2$, the left-hand term is clearly $0$. If $\ell=1$, the left-hand term is -$$[r(xh^n),r(yh^m)]=[(-1)^nx,y]=(-1)^nh.$$ -The right-hand term is the evaluation of $r$ at -$$[x^\ell h^n,y^\ell h^m]=[x^\ell,y^\ell]h^{n+m}+x^\ell[h^n,y^\ell]h^m-y^\ell[h^m,x^\ell]h^n+0.$$ -We have $hx^\ell=x^\ell(h+2\ell)$, and hence $h^mx^\ell=x^\ell(h+2\ell)^m$, and thus $[h^m,x^\ell]=x^\ell((h+2\ell)^m-h^m)$. Similarly, $[h^n,y^\ell]=y^\ell((h-2\ell)^n-h^n)$. Hence -$$[x^\ell h^n,y^\ell h^m]=[x^\ell,y^\ell]h^{n+m}+x^\ell y^\ell((h-2\ell)^n-h^n)h^m-y^\ell x^\ell((h+2\ell)^m-h^m)h^n$$ -$$=x^\ell y^\ell (h-2\ell)^nh^m-y^\ell x^\ell (h+2\ell)^mh^n.$$ -A computation yields -$$4^\ell x^\ell y^\ell=(c-(h-1)^2)(c-(h-3)^2)\dots (c-(h-2\ell+1)^2),$$ -and similarly -$$4^\ell y^\ell x^\ell=(c-(h+1)^2)(c-(h+3)^2)\dots (c-(h+2\ell-1)^2).$$ -At $(c,h)=(4,1)$, evaluation of the polynomial $4^\ell y^\ell x^\ell$ vanishes (because of the term $(c-(h+1)^2)$, because $\ell\ge 1$, while evaluation of the polynomial $4^\ell x^\ell y^\ell$ vanishes, because of the term $(c-(h-3)^2)$... as soon as $\ell\ge 2$. If $\ell\ge 2$, similarly both vanish at $(4,-1)$, and we have $r([x^\ell h^n,y^\ell h^m])=0$ as required. -Now concentrate on $\ell=1$, and write $4xy=c-(h-1)^2$, $4yx=c-(h+1)^2$. We have -$$4[x h^n,y h^m]=(c-(h-1)^2)(h-2)^nh^m-(c-(h+1)^2) (h+2)^mh^n.$$ -Evaluation at both $(c,h)=(4,1)$ yields $4(-1)^n$, and at $(4,-1)$ yields $-4(-1)^n$. Thus $r([xh^n,y h^m])=(-1)^nh=[r(xh^n),r(y h^m)]$. -Now let us turn to degree $-2$ (degree $2$ is similar). We have to show that -$$[r(x^{\ell} h^n),r(y^{\ell+1} h^m)]=r([x^{\ell} h^n,y^{\ell+1} h^m])$$ -for all $\ell,n,m\ge 0$. The left-hand term is $0$ for $\ell\ge 1$, and for $\ell=0$ equals $-(1-(-1)^n)y$. -Let us pass to the right-hand term. We have -$$[x^\ell h^n,y^{\ell+1}h^m]=[x^\ell,y^{\ell+1}]h^{n+m}+x^\ell[h^n,y^{\ell+1}]h^m-y^{\ell+1}[h^m,x^\ell]h^n+0$$ -$$=[x^\ell,y^{\ell+1}]h^{n+m}+x^\ell y^{\ell+1}((h-2\ell-2)^n-h^n)h^m-y^{\ell+1}x^\ell((h+2\ell)^m-h^m)h^n$$ -$$=x^\ell y^{\ell+1}(h-2\ell-2)^nh^m-y^{\ell+1}x^\ell(h+2\ell)^mh^n.$$ -We need to write everything as $yP(c,h)$. Anticipating on the result, we define $w_\ell=(c-(h+3)^2)(c-(h+5)^2)\dots (c-(h-2\ell+1)^2)$ and $w'_\ell=(c-(h-5)^2)(c-(h-7)^2)\dots (c-(h-2\ell-1)^2)$, when $\ell\ge 1$. Both are products of $\ell-1$ terms. -Using that $P(c,h)y=yP(c,h-2)$ for every polynomial $P$, one has -$$(x^\ell y^\ell)y=(c-(h-1)^2)(c-(h-3)^2)\dots (c-(h-2\ell+1)^2)y$$ -$$=y(c-(h-3)^2)(c-(h-5)^2)\dots (c-(h-2\ell-1)^2),$$ -which for $\ell\ge 1$ equals $$yw'_\ell (c-(h-3)^2);$$ -for $\ell=0$ this is $y$. Also, one writes $$y^{\ell+1}x^\ell=yw_\ell(c-(h+1)^2).$$ -Then, for $\ell\ge 1$, one gets -$$[x^\ell h^n,y^{\ell+1}h^m]=y\Big(w'_\ell (c-(h-3)^2)(h-2\ell-2)^nh^m+w_\ell(c-(h+1)^2)(h+2\ell)^mh^n\Big).$$ -Then, because of the factors $(c-(h-3)^2)$ and $(c-(h+1)^2)$, evaluation at $(c,h)=(4,1)$ yields zero. Hence $r([x^\ell h^n,y^{\ell+1}h^m])=0$ for all $\ell\ge 1$. It remains $\ell=0$. -$$[h^n,yh^m] -=y\big((h-2)^n-h^n\big)h^m.$$ -Then $r([h^n,yh^m])=((-1)^n-1)y$. This is the desired value. -(Note that by restriction, we also obtain a retraction for the 2-dimensional Lie algebra.)<|endoftext|> -TITLE: Tensor product of a DGA and an $A_\infty$ algebra -QUESTION [7 upvotes]: In general there seems no way to naturally define the tensor product of two $A_\infty$ algebras $A$ and $B$. But, if $(A, m^A_1,m^A_2)$ is only a DGA(differential graded algebra) and $(B, m^B_k, k\ge 1) $ is an $A_\infty$ algebra, then is there a natural way to get an $A_\infty$ algebra structure on the tensor product $A\otimes B$? -I guess this should be correct. But for the safety, I was wondering if there is a standard reference for this fact. -Moreover, if this is right, what I really want is an explicit formula of the $A_\infty$ algebra structure on $A\times B$ in terms of $m^A$ and $m^B$. Thank you! - -REPLY [11 votes]: In fact the tensor product of two $A_\infty$ algebras can be made into an $A_\infty$ algebra in an explicit way: there are two constructions, one by Saneblidze-Umble and one by Loday. See the paper https://arxiv.org/abs/0710.0572 -(For cofibrancy reasons one also knows abstractly that there is such a tensor product, but this of course doesn't give a formula.)<|endoftext|> -TITLE: $GL_n(\Bbb Z_p)$ conjugacy classes in a $GL_n(\Bbb Q_p)$ conjugacy class -QUESTION [20 upvotes]: It is easy to classify conjugacy classes in $GL_n(\mathbb Q_p)$ by linear algebra. How to classify $GL_n(\Bbb Z_p)$ conjugacy classes in a $GL_n(\Bbb Q_p)$ conjugacy class? For example, for general matrix $A \in GL_n(\mathbb Z_p)$, how many $X \in GL_n(\mathbb Z_p)$ are $GL_n(\mathbb Q_p)$ conjugated to $A$ up to $GL_n(\mathbb Z_p)$ conjugaction? In some cases it is finite, is it always finite? -Motivation: when counting invariant lattices, one comes across such problems. -Edit: could we find some numerical invariants that distinguish different $\Bbb Z_p-$ conjugacy classes? - -REPLY [7 votes]: The answer is yes if $A \in \mathrm{GL}_n(\mathbb{Z}_p)$ is semisimple. -We may think of a matrix $A \in M_n(\mathbb{Z}_p)$ as a $\mathbb{Z}_p$-lattice of rank $n$ endowed with a $\mathbb{Z}_p$-linear endomorphism: take the standard lattice $L=\mathbb{Z}_p^n$ endowed with the endomorphism $\varphi$ associated to $A$. Clearly, two matrices $A,A' \in M_n(\mathbb{Z}_p)$ are $\mathrm{GL}_n(\mathbb{Z}_p)$-conjugated if and only if the associated pairs $(L,\varphi)$ and $(L',\varphi')$ are isomorphic. -More formally, consider the algebra $R = \mathbb{Z}_p[A]$ inside $M_n(\mathbb{Z}_p)$. Then $A$ gives rise to an $R$-module $M$ which is $\mathbb{Z}_p$-free of rank $n$. Note that if $A' \in M_n(\mathbb{Z}_p)$ is another matrix which is $\mathbb{Q}_p$-conjugated to $A$ then the algebras $\mathbb{Z}_p[A]$ and $\mathbb{Z}_p[A']$ are isomorphic, as they are both isomorphic to $R = \mathbb{Z}_p[X]/(\mu)$ where $\mu$ denotes the minimal polynomial. However, the associated $R$-modules $M$ and $M'$ need not be isomorphic. In fact, they are if and only if $A$ and $A'$ are $\mathbb{Z}_p$-conjugated. -So the problem reduces to classifying the $R$-modules which are $\mathbb{Z}_p$-free of rank $n$ up to isomorphism. In the semisimple case, the following theorem of Jordan-Zassenhaus gives a positive answer to your third question. - -Theorem (Jordan-Zassenhaus). Let $K$ be a local or global field with ring of integers $\mathcal{O}_K$. Let $L$ be a (commutative) semisimple $K$-algebra, and let $R$ be an $\mathcal{O}_K$-order in $L$. Then for any integer $n \geq 1$, the set of isomorphism classes of $R$-modules which are $\mathcal{O}_K$-lattices of rank $\leq n$ is finite. - -The case at hand follows by taking $K=\mathbb{Q}_p$, $L=\mathbb{Q}_p[X]/(\mu)$ and $R=\mathbb{Z}_p[X]/(\mu)$ where $\mu$ is a squarefree polynomial. A proof of the Jordan-Zassenhaus theorem is given in Reiner, Maximal orders, (26.4), p. 228. It is stated only for global fields, but you can check it also holds for local fields. -EDIT. The proof is not difficult: first one treats the case where $R$ is a maximal order, which is clear since it is a product of discrete valuation rings (hence PIDs). If $R' \subset R$ is an arbitrary order and $M'$ is an $R'$-lattice then there are only finitely many possibilities for $M=M' \otimes_{R'} R$, and since $p^N M \subset M' \subset M$ there are only finitely many possibilities for $M'$. So we get the finiteness. -Regarding the question of an explicit classification, you may be interested by Marseglia's article Computing the ideal class monoid of an order, where he gives algorithms to compute explicit representatives (in the global case). -EDIT 2. Another article giving an algorithm in the global case appeared on the arXiv today: https://arxiv.org/abs/1811.06190<|endoftext|> -TITLE: Frobenius eigenvalues algebraic numbers -QUESTION [5 upvotes]: Let $X$ be a smooth projective variety over $\mathbf{F}_q$ and $\overline{X}$ its base change to $\overline{\mathbf{F}_q}$. -By Deligne’s Weil I, the eigenvalues of the geometric Frobenius acting on $H^{2j}(\overline{X},{\mathbf{Q}}_{\ell})$ are all algebraic numbers. -Is it true that the eigenvalues of the geometric Frobenius acting on $H^{2j}(\overline{X},{\mathbf{Q}}_{\ell}(j))$ are also algebraic numbers? Are they not just the eigenvalues of geometric Frobenius acting on $H^{2j}(\overline{X},{\mathbf{Q}}_{\ell})$, renormalized by $q^{-j}$? -This feels wrong because otherwise geometric Frobenius would act on $H^{2j}(\overline{X},\mathbf{Q}_{\ell}(j))$ in a unipotent way, since its eigenvalues would all be algebraic and of complex absolute value $1$, then roots of unity. This cannot be the case, and my question is “why?”: - -why do Tate twists mess up algebraicity of geometric Frobenius eigenvalues? - -Edit: it’s possible that my confusion is about “all absolute value one algebraic numbers are roots of unity”, and Tate twists are not guilty. -The sentence in quotation marks is false: only the absolute value one roots of a monic polynomial with integer coefficients are roots of unity, and the characteristic polynomial of geometric Frobenius on $H^{2j}(\overline{X},\mathbf{Q}_{\ell}(j))$ may well not be with integer coefficients but only rational coefficients (or in $\mathbf{Z}[q^{-j}]$). Is this the problem? - -REPLY [5 votes]: It seems you have mostly figured this out by yourself. I can happily confirm you are on the right track. - -Are they not just the eigenvalues of geometric Frobenius acting on $H^{2j}(\overline{X},{\mathbf{Q}}_{\ell})$, renormalized by $q^{-j}$? - -They are indeed. So they are in fact algebraic numbers. - -it’s possible that my confusion is about “all absolute value one algebraic numbers are roots of unity”, and Tate twists are not guilty. - The sentence in quotation marks is false: only the absolute value one roots of a monic polynomial with integer coefficients are roots of unity, and the characteristic polynomial of geometric Frobenius on $H^{2j}(\overline{X},\mathbf{Q}_{\ell}(j))$ may well not be with integer coefficients but only rational coefficients (or in $\mathbf{Z}[q^{-j}]$). Is this the problem? - -Yes, this is the problem. -For a simple example you could work with a surface that is the product of two elliptic curves. $H^2( \overline{E_1 \times E_2},\mathbb Q_\ell (-1))$ has dimension $6$ and it is possible for $2$, $4$, or $6$ of the eigenvalues to be algebraic integers (and therefore roots of unity ), with the remainder algebraic numbers, but not algebraic integers.<|endoftext|> -TITLE: I conjecture inequalities $\sum_{k=1}^{n}\{kx\}\le\frac{n}{2}x$ -QUESTION [13 upvotes]: I conjecture the following inequality: - -For $x > 1$, and $n$ a positive integer, - $$\sum_{k=1}^{n}\{kx\}\le\dfrac{n}{2}x.$$ - -For $n=1$, the inequality becomes -$$\{x\}\le\dfrac{x}{2}\Longleftrightarrow \{x\}\le [x];$$ -and, for $n = 2$, it becomes -$$\{x\}+\{2x\}\le x\Longleftrightarrow \{2x\}\le [x],$$ -which is obvious since $[x]\ge 1>\{2x\}$. -If $x\ge 2$, the inequality is obvious, since -$$\dfrac{n}{2}x\ge n\ge\sum_{k=1}^{n}\{kx\}.$$ -However, I cannot prove it for $1 < x < 2$. - -REPLY [7 votes]: This is a proof smilar to Will Sawin's but with no induction. -Set $y=x-1$. We need to prove that the average of $\{y\},\dots,\{ny\}$ is at most $x/2$. The numbers $y,2y,\dots,ny$ split into contiguous groups, each group with the same integer part. It suffices to show that the average of the fractional parts for each group is at most $x/2$. Those fractional parts form an arithmetical sequence, so their average is half the sum of the first and the last term. As the fractional part of the first term in the group is at most $y$, this average is bounded by $\frac{y+1}2=x/2$, as desired.<|endoftext|> -TITLE: Deligne's theorem on finite flat group schemes and generalizations -QUESTION [14 upvotes]: Recall Deligne's theorem that for a finite flat commutative group scheme $G$ of order $n$, the multiplication by $n$ map $[n]: G \to G$ is the zero map. -I have seen the proof a few times but I can't really say I understand the idea behind it. In particular: -1) What is the general idea/strategy behind the proof? -2) Where does the proof break down in the non commutative case? -3) How much do we know about generalizations? Are there conditions weaker than non commutativity under which we can prove it? -4) Related to 2 - what is the conceptual reason for why the non commutative case is so hard? It is after all a very innocuous sounding statement but to be unproven after so many years, I guess there must be something interesting going on. - -REPLY [4 votes]: This is just expanding on user19475's comment, and is taken entirely from Schoof's notes, but as this question is the first thing to come up when googling, I thought it might be helpful to give my novice's interpretation of the key points of the proof. -First, by changing the base if needed, we are reduced to showing that the $m$th power map is trivial on the $R$ valued points of $G$, where $G=Spec(A)$ is a finite, free, affine group scheme of rank $m$ over $Spec(R)$. -Then, using Cartier duality, we may identify $G(R)\subset A^{\vee}$, since:$$G(R)=\text{Hom}_{R-Alg}(A,R)\subset Hom_{R-Mod}(A,R)=A^{\vee}$$ -Further unravelling definitions, the points of $A^{\vee}$ that belong to $G(R)$ are exactly those $x$ which satisfy $$\Delta(x)=x\otimes x$$ -For $\Delta$ the comultiplication on $A^{\vee}$. -The key tool now is the norm map, for any finite, free $R$ algebra $S$, we have the norm map $N:S^*\rightarrow R^*$ on units, given by $x\mapsto \det(\mu_x)$ where $\mu_x$ is the $R$ linear map of multiplication by $x$ on $S$ as an free $R$ module. -The core technical part of the proof is that for any $R\rightarrow S$, the norm map from $(A^{\vee}\otimes S)^*\rightarrow (A^{\vee})^*$ takes $G(S)$ into $G(R)$, and when we set $S=A$, that this norm map is invariant under the action of the translation automorphism $\tau(\alpha)$ on $G(A)$ for any $\alpha$ in $G(R)$. -Once we have these two compatibilities, we simply verify that for the element $\text{Id}_A$ in $G(A)$, we have $$N(\text{Id}_A)=N(\tau(\alpha)\text{Id}_A)=N(\alpha)N(\text{Id}_A)=\alpha^m N(\text{Id}_A)$$ -Which gives the result. -It appears to me that in the proof, there doesn't seem to be any specific place where commutativity is required, as we are really only looking to prove the assertion at a single element of $G(R)$. The commutativity is important insofar as it provides the core tool used for the proof, the norm map, via the interpretion of points of $G(R)$ as elements of the Cartier dual.<|endoftext|> -TITLE: A counter-example for the reversed direction of Casson-Gordon's theorem -QUESTION [8 upvotes]: For a knot $K$, let $\Sigma_K$ be the double cyclic branched cover of a knot. -By the classical work of Casson and Gordon, we know that if $K$ is smoothly slice, then $\Sigma_K$ bounds a rational homology ball. -Is there any well-known counter-example for the reversed direction? -EDIT More general statement is true due to the work of Casson & Gordon, just seen in ACP18. -For any prime $p$ and positive integer $r$, the $p^r$-fold cyclic branched cover of a knot $ K$, denoted by $\Sigma_{p^r}(K)$, is a $\mathbb Z_p$-homology sphere. -Theorem: If $K$ is smoothly slice, then $\Sigma_{p^r}(K)$ bounds $\mathbb Z_p$-homology ball. - -REPLY [10 votes]: Here's a particularly subtle counterexample, from the work of Kirk and Livingston (Topology Vol. 38, No. 3, pp. 663--671, 1999). They show that the pretzel knots $J = P(-3,5,7,2)$ and $K = P(5,-3,7,2)$ are not concordant (even locally flat). These two knots are related by mutation (switch the first two pairs of twists in the standard picture of a pretzel knot). On the other hand, they have the same double branched cover, say $\Sigma$, a certain Brieskorn sphere. (This is a general fact about mutations, but is clear in this setting.) -Then the knot $J \# -K$ is not slice, but its double branched cover $\Sigma \# -\Sigma$ bounds an integer homology ball (in fact $(\Sigma - \text{int}\ B^3) \times I$). -There are many earlier examples in the literature for pairs of knots with the same double branched covers. Presumably it's not hard to check that for some of these, the knots are not concordant, using eg knot signatures. So one could construct many more examples. Likewise there are pairs of knots with the same n-fold branched covers (for a fixed $n$) and I'm sure you could construct counterexamples for those as well. - -REPLY [3 votes]: Probably, I found some more explicit examples: -Akbulut and Larson recently showed in AL18 that the family of Brieskorn spheres $\Sigma(2,4n+1,12n+5)$ and $\Sigma(3,3n+1,12n+5)$ bound rational homology balls. -These Brieskorn spheres are respectively $2$- and $3$-fold cyclic branched cover of torus knots $T_{4n+1,12n+5}$ and $T_{3n+1,12n+5}$ in $S^3$. But these knots are not smoothly slice due to for example Milnor slice genus proven by Kronheimer and Mrowka in KM93.<|endoftext|> -TITLE: The Parovichenko cardinal, is it equal to $\max\{\aleph_2,\mathfrak p\}$? -QUESTION [11 upvotes]: Let us define the Parovichenko cardinal $\mathfrak{P}$ as the largest cardinal $\kappa$ such that each compact Hausdorff space $K$ of weight $w(K)<\kappa$ is the continuous image of the remainder $\beta\mathbb N\setminus\mathbb N$ of the Stone-Cech compactification of the discrete space of positive integers $\mathbb N$. - -By a classical theorem of Parovichenko, $\mathfrak P\ge\aleph_2$. -On the other hand, Theorem 2.7 in this paper of van Douwen and Przymusinski implies that $\mathfrak P\ge\mathfrak p$ where $\mathfrak p$ is the well-known pseudointersection number. - -These two results yield the inequality $\mathfrak P\ge\max\{\aleph_2,\mathfrak p\}$. - -So, under CH we have $\mathfrak P=\aleph_2>\mathfrak c=\mathfrak p=\aleph_1$. -By a result of Kunen of 1968 in the Cohen model $\mathfrak P=\aleph_2=\mathfrak c>\mathfrak p=\aleph_1$. -Finally, PFA implies $\mathfrak P=\aleph_2=\mathfrak c=\mathfrak p>\aleph_1$, see Corollary 4.6 in Baumgartner's survey "Applications of the Proper Forcing Axiom" in the "Handbook of Set-Theoretic Topology". -Let us also mention a result that follows from Theorems 2.1 and 2.2 of van Douwen and Przymisinski: -$\mathfrak P\le\mathfrak c$ if one of the following conditions holds: -$\bullet$ $\aleph_2\le\mathfrak c<2^{\aleph_1}=\aleph_{\omega_2}$ or -$\bullet$ $\mathfrak c=2^{\aleph_1}$ and $\mathfrak q_0=\aleph_1$. -Here $\mathfrak q_0$ is the largest cardinal $\kappa$ such that each subset $X\subset \mathbb R$ of cardinality $|X|<\kappa$ is a $Q$-set (which means that each subset of $X$ is an $F_\sigma$-set in $X$). - -Problem. Is it consistent that $\mathfrak P>\max\{\aleph_2,\mathfrak p\}$? - -REPLY [9 votes]: No -- it is consistent that $\mathsf{CH}$ fails, and that every compact Hausdorff space of weight $\leq\!\mathfrak{c}$ is a continuous image of $\beta \mathbb N \setminus \mathbb N$. (This is due to Baumgartner, who mentions it off-hand in his article in the Handbook of Set Theoretic Topology; the mutual consistency with $\mathsf{MA}_{\sigma\text{-linked}}$ was established later by Baumgartner, Frankiewicz, and Zbierski in this paper.) In such a model, $\mathfrak{P} = \mathfrak{c}^+ > \max\{\aleph_2,\mathfrak{p}\}$.<|endoftext|> -TITLE: Schoof's Algorithm for Hyperelliptic curves over $\mathbb{F}_q$ : Question regarding computation of resultant: Gaudry -QUESTION [7 upvotes]: I am new to StackExchange and I am currently going through Gaudry's paper on counting points on hyperelliptic curves (see https://hal.inria.fr/inria-00512403/document). As a part of the generalization of Schoof s algorithm for genus 2, we have to compute l-torsion points. For this we use Cantor s division polynomials, and the problem reduces to solving the following two equations: -$$ \begin{align} -E_1(x_1,x_2) &= (d_1(x_1)d_2(x_2)-d_1(x_2)d_2(x_1))/ \langle x_1-x_2 \rangle = 0 \\ -E_2(x_1,x_2) &= (d_0(x_1)d_2(x_2)-d_0(x_2)d_2(x_1))/\langle x_1-x_2 \rangle =0 -\end{align} $$ -where deg($d_0$)=$2l^2-1$, deg($d_1$)=$2l^2-2$, deg($d_2$)=$2l^2-3$. We eliminate $x_2$ by computing the resultant wrt to $x_2$ denoted as $R(x_1)$. -I understood that the maximum degree of $R(x_1)$ is $ 2(2l^2-3)(2l^2-2)$. Then the author claims that a high power of $d_2(x_1)$ denoted as $\delta$ divides $R(x_1)$ and $\delta = 2l^2 - 2$ and in some current papers $\delta = 2l^2 -3$. -I fail to understand how they could compute $\delta$ and is the degree of $R(x_1)$ its maximum degree? - -REPLY [4 votes]: Answer: -The resultant $R(x_1)$ vanishes at $x_1=X_1$ iff $E_1(X_1,x_2)=0=E_2(X_1,x_2)$ for some $x_2=X_2$. Take $X_2$ to be one of the $\delta={\rm deg}\,(d_2)=2l^2-3$ roots of the polynomial $d_2$. Then $E_1(x_1,X_2)$ and $E_2(x_1,X_2)$ vanish if $d_2(x_1)=0$, so $d_2(x_1)$ is a factor of $R(x_1)$ with multiplicity $\delta$. - -Discussion: -This answer $\delta=2l^2-3$ differs from the value $2l^2-2$ given in the cited paper by Gaudry and Harley, so as a test I used Mathematica to calculate the resultant for a simple case of $l=2$: -$$d_0(x)=x^7+x^6+x^5+x^4+x^3+x^2+2 x+1$$ -$$d_1(x)=x^6+x^5+x^4+x^3+x^2+x+1$$ -$$d_2(x)=x^5+x^4+x^3+x^2+2 x+1$$ -These three polynomials are irreducible, of degrees ${\rm deg}(d_0)=2l^2-1=7$, ${\rm deg}(d_1)=2l^2-2=6$, and ${\rm deg}(d_2)=2l^2-3=5$. The resultant over $x_2$ of -\begin{align} -E_1(x_1,x_2) &= [d_1(x_1)d_2(x_2)-d_1(x_2)d_2(x_1)] (x_1-x_2 )^{-1} \\ -E_2(x_1,x_2) &= [d_0(x_1)d_2(x_2)-d_0(x_2)d_2(x_1)](x_1-x_2 )^{-1} -\end{align} -equals - -$$R(x_1)=\left(x_1^5+x_1^4+x_1^3+x_1^2+2 x_1+1\right)^5 \left(7 x_1^{30}+19 x_1^{29}+43 x_1^{28}+79 x_1^{27}+127 x_1^{26}+188 x_1^{25}+286 x_1^{24}+384 x_1^{23}+491 x_1^{22}+591 x_1^{21}+675 x_1^{20}+736 x_1^{19}+805 x_1^{18}+804 x_1^{17}+774 x_1^{16}+713 x_1^{15}+630 x_1^{14}+531 x_1^{13}+455 x_1^{12}+342 x_1^{11}+250 x_1^{10}+175 x_1^9+117 x_1^8+72 x_1^7+52 x_1^6+24 x_1^5+12 x_1^4+6 x_1^3+3 x_1^2+x_1+2\right)$$ - -So indeed, $R(x_1)$ contains a factor $d_2(x_1)$ to the power $\delta=2l^2-3$. -The quotient -$$\tilde{R}(x_1)=\frac{R(x_1)}{[d_2(x_1)]^\delta}$$ -is an irreducible polynomial of degree $30$, in accord with the value $4l^4-10l^2+6$ in Gaudry & Harley. I would conclude that their value of $\delta$ is simply a misprint. Note that the degree of $\tilde{R}$ is one half the maximum degree of $R$ mentioned in the OP.<|endoftext|> -TITLE: Does the notion of a compactly generated space (or $k$-space) depend on the choice of universe? -QUESTION [5 upvotes]: We recall the notion of a $k$-space (or compactly generated space) to fix our notations. For every topological space $X$, we can define a category $\mathfrak{M}_X$. The class of objects of $\mathfrak{M}_X$ is the class of continuous mappings $u: K \to X$ from a compact Hausdorff space $K$ to $X$. Let $v : C \to X$ be another object in $\mathfrak{M}_X$, the set of morphisms from $u$ to $v$ is the continuous mappings $f: K \to C$ such that $u = v\circ f$. Let $\tau$ be the collection of open sets in $X$. Define $k(\tau)$ to be the final topology on $X$ with respect to the class $\mathrm{Obj}(\mathfrak{M}_X)$. In other words, for any subset $U$ of $X$, $U \in k(\tau)$ if and only if $u^{-1}(U)$ is open in $K$ for any object $u : K \to X$ in $\mathfrak{M}_X$. Let $k(X)$ be the space $X$ equipped with the topology $k(\tau)$ which is finer than $\tau$ by definition. Then we call $X$ a $k$-space if $k(X) = X$, or more precisely, $\tau = k(\tau)$. -If we adopt the approach of the Grothendieck's axiom on universe to approach the above categorical setting. We can fix a universe $\mathcal{U}$ with $X \in \mathcal{U}$. The category $\mathfrak{M}_X$ is redefined as the category of $\mathcal{U}$-small continuous mappings $u : K \to X$ from a compact Hausdorff $K$ which is also $\mathcal{U}$-small. To emphasize this dependence on our universe $\mathcal{U}$, we denote this category by $\mathfrak{M}_{X,\mathcal{U}}$. Similarly, the final topology on $X$ with respect to $\mathrm{Obj}(\mathfrak{M}_{X, \mathcal{U}})$ is denote by $k_{\mathcal{U}}(\tau)$. -My question is, if we have another universe $\mathcal{V}$ with $X \in \mathcal{V}$, does $k_{\mathcal{U}}(\tau) = k_{\mathcal{V}}(\tau)$? -A sufficient condition would be there is a small (small is for any universe containing $X$) final subcategory $\mathfrak{S}_X$ of $\mathfrak{M}_X$, as is the case when $X$ is assumed to be weakly Hausdorff (we can take $\mathfrak{S}_X$ to be the full subcategory of $\mathfrak{M}_X$ generated by all embeddings $K \hookrightarrow X$ where $K$ is a closed compact Hausdorff subspace of $X$). But for general topological space $X$, can we find a small final subcategory $\mathfrak{S}_X$ of $\mathfrak{M}_X$? - -REPLY [4 votes]: The following part of 5.9.1 of Topology and Groupoids shows for a particular space $X$ how to reduce the role of a universe. -If $X$ is a k-space, there is a set $\mathcal C_{X}$ of maps $t : C_{t} \to X$ for compact Hausdorff -spaces $C_t$ such that a set $A$ is closed in $X$ if and only if $t^{-1}(A)$ is -closed in $C_{t}$ for all $t \in \mathcal C_{X}$. -The proof goes as follows: for each -non-closed subset $B$ of $X$ there is a compact Hausdorff space $C_{B}$ and map $t: C_{B} \to X$ such that $t^{-1}[B]$ is not closed in $C_B$. Choose one such -$C_{B}$ and one such $t$ for each non-closed $B$, and let $\mathcal C_{X}$ be -the set of all these $t$. That this set has the required property is clear. -This leads to standard properties of such k-spaces. -The following paper uses various classes of compact Hausdorff spaces to construct and study convenient categories: -"Monoidal closed, Cartesian closed and convenient categories of topological spaces" -Peter I. Booth and J. Tillotson, 88 (1980), No. 1, 35–53 -DOI: 10.2140/pjm.1980.88.35<|endoftext|> -TITLE: Noetherian spectral space comes from noetherian ring? -QUESTION [10 upvotes]: Let $X$ be a spectral space (en.wikipedia.org/wiki/Spectral_space), i.e. a space of the form $\textrm{Spec}(A)$ for some commutative ring $A$. If $X$ is noetherian, does there also exist a noetherian ring $B$ such that $X=\textrm{Spec}(B)$? - -REPLY [9 votes]: Graph $N_5$ with poset order topology (i.e. poset $M=\{p,q,r\}, P_2=\{p,q\}, P_1=\{p\}, Q=\{r\}, N=\phi$) is not Spec($A$) for Noetherian $A$ because if $a \in Q-P_2$ then 1 = dim$(A/a)$ = dim$(A)-1$ = 2 by the principal ideal theorem.<|endoftext|> -TITLE: Where to seek translations of research articles -QUESTION [14 upvotes]: I am collecting a large number of research articles from the historical record on a particular topic (comparative prime number theory). A good handful of them—about 10-15 or so spanning the 20th century—are written in German. -How does one go about (seeking or) soliciting translations into English of full papers, on this scale? -The translations don't have to be utterly perfect, but they should be professional-level translations—meaning we should be confident that mathematicians who read the English translations should get out of it the same mathematical content that someone reading the German original would get. - -REPLY [7 votes]: Like you, I struggle to decipher articles in German. The older influential articles (before 1980) were all translated in Russian by top Russian mathematicians . If you can read Russian I highly recommend these translations. Often they are better than the original since the translations occurred several years after the original publication and they often include as appendices surveys of what happened since the publication. Many typos and mathematical errors in the original were corrected, and sometime in the footnotes you can find sketches of different arguments. -Another approach I am using relies on Google Translate. It has improved considerably and I have used it successfully to read German articles, one paragraph at a time. The translation is not perfect but close enough so you can figure out yourself the mathematical arguments.<|endoftext|> -TITLE: $G(k)/H(k)$ as a submanifold of $G/H(k)$ -QUESTION [8 upvotes]: Let $k$ be a local field (if necessary, assume characteristic zero). In general, if $X$ is a smooth variety of finite type over $k$ of dimension $n$, then the set of $k$-rational points $X(k)$ is an analytic manifold over $k$ of dimension $n$. I was thinking about the passage $X \mapsto X(k)$ from smooth varieties to manifolds in the context of coset spaces, and had a couple of questions. -Let $H$ be a closed subgroup of a linear algebraic group $G$ over $k$. Assume $H$ is defined over $k$. Then the coset space $G/H$ has the structure of a smooth quasi-projective variety over $k$, which tells us that $G/H(k)$ is an analytic manifold over $k$ of dimension $n = \operatorname{Dim}G - \operatorname{Dim}H$. -On the other hand, $H(k)$ is a closed subgroup of the analytic Lie group $G(k)$, and the quotient $G(k)/H(k)$ has the structure of an $n$ dimensional analytic manifold (the quotient map $G(k) \rightarrow G(k)/H(k)$ is a principal fiber bundle with $H(k)$ as a fiber). This just follows from the general theory of Lie groups. -Of course, $G(k)/H(k)$ injects into $G/H(k)$, but they don't have to be equal. They are the same if $H^1(\operatorname{Gal}(k_s/k),H) $ is trivial. -What can we say about the inclusion $G(k)/H(k) \rightarrow G/H(k)$ from the perspective of manifolds? Is the inclusion an analytic map? Does it make $G(k)/H(k)$ into an open submanifold? How different can these two manifolds be? - -REPLY [4 votes]: From the number of votes, it seems that the useful comment of Laurent Moret-Bailly has been under appreciated, so I thought it would be useful to explicitly record what the main theorem in the linked paper says (in community wiki mode, since this is really his answer). -Theorem 1.2 of GGMB14 (in the special case of $k$ a non-archimedean local field) The map $G(k)/H(k)\to (G/H)(k)$ -$\bullet$ is always a homeomorphism onto its image, -$\bullet$ has closed image if $H$ satisfies the condition ($\ast$), -$\bullet$ has an open image if $H$ is smooth. -The condition ($\ast$) appearing in this theorem is technical (see Definition 2.4.3 of the paper), but $H$ satisfies the condition ($\ast$) if it has either one of the following properties: smooth, unipotent, commutative, or being a normal subgroup of a smooth group. Interestingly, there are examples of $H$ not satisfying $(\ast)$ for which the map $G(k)/H(k)\to (G/H)(k)$ has non-closed image (see example 7.1 of the paper, taking for example $k=\mathbf{F}_p(\!(T)\!)$). -Note that in characteristic $0$, (affine) group schemes (of finite type) are always smooth by a result of Cartier. Also, this puts in perspective the comment of YCor on the non-oppeness of $\text{SL}_p(K)\to \text{PGL}_p(K)$ for $K = \mathbf{F}_p(\!(T)\!)$. Finally, let me remark that the implicit function theorem should prove in all characteristic the openness of $G(k)/H(k)\to (G/H)(k)$ when $H$ is smooth (as suggested in Venkataramana's answer in characteristic $0$).<|endoftext|> -TITLE: Almost-prime values attained by a product of quadratic polynomials -QUESTION [5 upvotes]: Let $F(x) = \prod_{i=1}^{k} (a_i x +b_i)$ be a product of $k$ linear polynomials, where $a_i,b_i$ are integers. Under very reasonable conditions, it is known that a constant $C_k$ exists with the following property: $F(n)$ is divisible by at most $C_k$ primes, for infinitely many $n$-s. (This is proven in Chapter 10.3 of `Sieve Methods' by Halberstam and Richert). In fact, $C_k$ is a very explicit constant, and one can require in addition that $F(n)$ will be squarefree and coprime to a given $M$. - -Question: Consider a polynomial $F(x) = \prod_{i=1}^{k} (a_i x^2 + b_i x +c_i)$, a product of $k$ quadratic polynomials. Suppose that the $k$ polynomials are pairwise coprime, and that $\gcd(F(1),F(2),F(3),\ldots)=1$. Is there a constant $D_k$ such that $F(n)$ is divisible by at most $D_k$ primes, for infinitely many $n$-s? - -I am only aware about results on the case $k=1$, due to Iwaniec and Lemke-Oliver. I am interested in larger $k$, though. Even results for special $F$-s interest me, e.g. $F(x) = \prod_{i=1}^{k} ((a_ix)^2+1)$. - -REPLY [4 votes]: A statement of this type follows from Selberg's sieve, -details are in Halberstam and Richert, Sieve methods, section 10.3 and 10.5. -(In the meantime there may be numerically somehwat stronger estimates, -but the flavour might still be the same.) -Let me quote Theorem 10.11. (hence $r$ is your $C_k$ and $g$ is your $k$). -Let $F_1(n), \ldots, F_g(n)$ be distinct irreducible polynomials with integral coefficients, -and let $F(n)$ denote their product. Also let $h_i$ denote the degree of $F_i$ and $G$ the degree of $F$. -Let $\rho(p)$ denote the number of solutions of the congruence $F(n)\equiv 0 \bmod p$, -and suppose that $\rho(p)G-1+ g \sum_{j=1}^g {\frac{1}{j}} + g \log \left(\frac{2G}{g}+ \frac{1}{g+1}\right)$, or ( a more complicated one). -Corollary 10.11.1 says that if the degree of all $F_i=h$ is constant -the condition on $r$ is (in a simplified form) -$r>g \log g +O_h(g)$.<|endoftext|> -TITLE: Definitions of enriched monoidal category -QUESTION [13 upvotes]: This question is about two definitions of enriched monoidal categories I have: -Let $\mathcal{V}$ be a symmetric monoidal closed category. -The first definition: a $\mathcal{V}$-enriched category $\mathcal{C}$ is a pseudomonoid object in the Day-convolution monoidal category $(\mathcal{V}\text{-}\mathbf{Cat}, \otimes_{\mathrm{Day}}, y(1))$. This is a straightforward generalization of the definition of monoidal categories but everything has to be worked externally and I don't really feel like to work in this definition (for example, I am pretty sure that a kind of coherence theorem holds but I don't know if I will be able to write down the proof). -The second definition can be found here (Bar constructions for topological operads and the Goodwillie derivatives of the identity, Definition 1.10): https://projecteuclid.org/download/pdf_1/euclid.gt/1513799607 -In this paper, an enriched monoidal category $(\mathcal{C}, \overline{\wedge}, S)$ over $(\mathcal{V}, \wedge, I)$ is a $\mathcal{V}$-enriched category $\mathcal{C}$ which is tensored (denoted by $\otimes$) and cotensored, together with a monoidal structure on the underlying category $\mathcal{C}_0$ and the distribution natural transformation $d: (X\wedge Y)\otimes (C \overline{\wedge} D)\to (X\otimes C)\overline{\wedge}(Y\otimes D)$ satisfying certain associativity and unitality condition, which is a natural requirement for bar constructions. -So my question is: What is the relation between these two? It will be pleasing to be able to replace the first definition by the second one in other situations but I could not see any immediate implication from one to another or confluence under some conditions. - -REPLY [8 votes]: The two definitions are equivalent, for monoidal structures on $\mathcal{V}$-categories that are tensored over $\mathcal{V}$. I'll describe how the tensor product corresponds to the distributivity map: -The tensor product for a monoidal $\mathcal{V}$-category $\mathcal{C}$ in the first sense is given by specifying a tensor product on objects, and compatible maps $\mathcal{C}(C,D) \wedge \mathcal{C}(C',D') \to \mathcal{C}(C\overline{\wedge}C', D\overline{\wedge} D')$ in $\mathcal{V}$. So to show the second definition implies the first, we need to construct these maps. If $\mathcal{C}$ is tensored over $\mathcal{V}$, we can use the tensor product in $\mathcal{C}_0$ to get a natural map of sets -$$ \mathcal{V}(X, \mathcal{C}(C,D)) \times \mathcal{V}(Y, \mathcal{C}(C',D')) \cong \mathcal{C}_0(X \otimes C, D) \times \mathcal{C}_0(Y \otimes C', D') \to \mathcal{C}_0((X \otimes C) \overline{\wedge} (Y \otimes C'), D \overline{\wedge} D').$$ -Using the distributivity transformation we get a map -$$ \mathcal{C}_0((X \otimes C) \overline{\wedge} (Y \otimes C'), D \overline{\wedge} D') \to \mathcal{C}_0((X \wedge Y) \otimes (C \overline{\wedge} C'), D \overline{\wedge} D') \cong \mathcal{V}(X \wedge Y, \mathcal{C}(C \overline{\wedge} C', D \overline{\wedge} D').$$ Now set $X = \mathcal{C}(C,D)$ and $Y = \mathcal{C}(C',D')$, then the pair of identity maps gets sent to the required morphism $\mathcal{C}(C,D) \wedge \mathcal{C}(C',D') \to \mathcal{C}(C\overline{\wedge}C', D\overline{\wedge} D')$. -On the other hand, if we have these maps then for $X,Y \in \mathcal{V}$, $C,D \in \mathcal{C}$, we can form the composite -$$ X \wedge Y \to \mathcal{C}(C, X \otimes C) \wedge \mathcal{C}(D, Y \otimes D) \to \mathcal{C}(C \overline{\wedge} D, (X \otimes C) \overline{\wedge} (Y \otimes D)),$$ -where the first map is the tensor product of unit maps for the cotensoring adjunction. This is then adjoint to the distributivity morphism $(X \wedge Y) \otimes (C \overline{\wedge} D) \to (X \otimes C) \overline{\wedge} (Y \otimes D))$.<|endoftext|> -TITLE: Is it known how the Sigma Algebra generated by Jordan measurable sets compares to universally measurable sets and analytic sets? -QUESTION [15 upvotes]: Unlike the collection $L$ of Lebesgue measurable sets, the collection $J$ of Jordan measurable sets do not form a Sigma algebra. (A set is Jordan measurable if and only if its characteristic function is Riemann integrable, and the characteristic function of a singleton is Riemann integrable, but the characteristic function of the rational numbers is not.) But we can still talk about $\sigma(J)$, the Sigma algebra generated by $J$. -Now in this 1993 journal paper, K.G. Johnson shows that the Borel Sigma algebra is a proper subset of $\sigma(J)$ which is a proper subset of the Lebesgue Sigma algebra, and that a set is in $\sigma(J)$ if and only if it can be written as a union of a Borel set and a subset of a measure $0$ $F_\sigma$ set. But Johnsons says there are two questions he doesn’t know the answer to: - -Where does $\sigma(J)$ stand relative to the analytic sets ... and relative to the universally measurable sets[?] - -My question is, have either of these questions been answered in the two decades since this paper was published? Is it known whether there are analytic sets which are not in $\sigma(J)$ or vice versa? Is it known whether there are universally measurable sets which are not in $\sigma(J)$ or vice versa? Or are these still open problems? - -REPLY [7 votes]: Regarding the collections of analytic sets and Jordan measurable sets, a simple cardinality argument shows there exist Jordan measurable sets that are not analytic. Any subset of the Cantor middle thirds set is Jordan measurable (indeed, any such subset has Jordan measure zero), and thus there are $2^c$ many Jordan measurable sets. This result was essentially proved by Philip E. B. Jourdain around 1903. However, there are only $c$ many analytic sets, a fact that follows easily from each of several characterizations of analytic sets. -喻 良 has mentioned a paper by Mauldin in which results stronger in certain ways are obtained. Slightly more information about this paper and a related paper by Mauldin can be found by searching for the phrase "In [28] Mauldin proves" in my 30 April 2000 sci.math essay (see [1]) on $F_{\sigma}$ Lebesgue measure zero sets (this may also be of interest). -[1] https://groups.google.com/forum/#!searchin/sci.math/HISTORICAL%2420ESSAY%2420ON%2420F_SIGMA%7Csort:date/sci.math/ZBCCYjtCSMA/19OUtU88RwAJ<|endoftext|> -TITLE: Interesting behaviour of binomial coefficients -QUESTION [6 upvotes]: Let $\binom{n}{k}:=\frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(1-k+n)}$ be the generalized binomial coefficient then I noticed by playing around with Mathematica that the function $f:[0,n/2] \rightarrow \mathbb R$ -$$f(x) = \log\left(\binom{n}{n/2+x} \right)-n \alpha(1-(2x/n)^2)$$ -has very interesting properties. -For $\alpha\le \tfrac{1}{2}$ the global maximum of this function is attained at $x_n=0$ for sufficiently large $n.$ -Yet, for $\alpha>\tfrac{1}{2}$ the global maximum seems to be attained at some $x_n>0.$ -My question is: Is it possible to analytically verify this property $(x_n>0)$ and in particular can anybody shed some light on what is so special about $\alpha =1/2$? (The first comment below this thread seems to indicate already that $\alpha=1/2$ is indeed the right threshold) -In particular, Mathematica gave me the following equation for the derivative of $f$ -$$f'(x)= \frac{8 \alpha x}{n} + \operatorname{HarmonicNumber}(n/2 - x) - \operatorname{HarmonicNumber}(n/2 + x).$$ - -REPLY [6 votes]: We have -\begin{equation} - f'(x)=\frac{8 \alpha x}{n}+\psi\left(\frac{n}{2}-x+1\right)-\psi\left(\frac{n}{2}+x+1\right), -\end{equation} -where $\psi:=(\ln\Gamma)'$. -By the Gauss formula, Theorem 1.6.1, page 26 -\begin{equation} - \psi(x)=\int_0^\infty\Big(\frac{e^{-z}}{z}-\frac{e^{-xz}}{1-e^{-z}}\Big)dz, -\end{equation} -we have -\begin{equation} - f'(x)=\frac{8 \alpha x}{n}-2\int_0^\infty\frac{dz}{1-e^{-z}}\,e^{-(1+n/2)z}\sinh xz. -\end{equation} -Since $\sinh$ is strictly convex on $(0,\infty)$, $f'$ is strictly concave on $(0,n/2)$. Also, $f'(0)=0$ and $f'(\frac n2)\to-\infty$ as $n\to\infty$. -So, eventually (i.e., for all large enough $n$), either $f'<0$ on $(0,n/2)$ (which will be the case if $f''(0)\le0$) or $f'$ changes sign only once on $(0,n/2)$, from $+$ to $-$, at some $x_n\in(0,n/2)$. -So, either $f$ is decreasing on $(0,n/2)$ (which will be the case if $f''(0)\le0$) or increasing-decreasing, attaining its only maximum at some $x_n\in(0,n/2)$. -As $x\downarrow0$, -\begin{equation} - f'(x)=\Big(\frac{8\alpha}{n}-2\psi'(n/2+1)\Big)x+O(x^3), -\end{equation} -and -\begin{equation} - \frac{8\alpha}{n}-2\psi'(n/2+1)=\frac{4\alpha-2}{n}+\frac2{n^2}+O(n^{-3}) -\end{equation} -as $n\to\infty$. -So, if $\alpha<1/2$, then $f''(0)<0$ eventually, and so, $f$ is decreasing on $(0,n/2)$, with the maximum at $0$. -If $\alpha=1/2$, then $f'(c\sqrt n)=\left(4 c-\frac{16 c^3}{3}\right) - \left(\frac{1}{n}\right)^{3/2}+O\left(\left(\frac{1}{n}\right)^{5/2}\right)$ for each real $c>0$ as $n\to\infty$, and so, eventually $f$ attains its only maximum at some $x_n\sim c\sqrt n$, where $c=\sqrt3/2$. -Finally, if $\alpha>1/2$ and $c\in(0,1/2)$, then $f'(cn)=g(c)+O(1/n)$ as $n\to\infty$, where $g(c):=8 \alpha c+\ln \frac{1-2 c}{2 c+1}$ for each real $c>0$. We have $g(c)=0$ iff $\alpha=a(c):=-\frac1{8c}\,\ln\frac{1-2 c}{2 c+1}$. Note that $a(c)$ increases from $1/2$ to $\infty$ as $c$ increases from $0$ to $1/2$. So, eventually $f$ attains its only maximum at some $x_n\sim cn$, where $c=a^{-1}(\alpha)$.<|endoftext|> -TITLE: On proof-verification using Coq -QUESTION [45 upvotes]: So i recently learnt that there is now a certain software called ''Coq'' by which one can check the validity of mathematical proofs. My questions are: - -Are there limitations on the kinds of proofs that Coq can verify? -How long on average does Coq take to verify a proof? -Do math journals use Coq? -How do I go about it if I want to verify a proof using Coq? - -REPLY [2 votes]: In my answer I am referring to other systems like HOL Light, but if the formal verification mentioned below, would be implemented in Coq the situation would not be much different. -Do math journals use Coq? -No, they do not, but there is one exceptional example that needs to be remembered and I will mention it below. Usually, formal verification is applied to results that have previously been checked by humans. However, there is one amazing result that the only way we can be sure it is true is because it was formally verified. I am quoting after Wikipedia: -In 1998 Thomas Hales, following an approach suggested by Fejes Tóth (1953), announced that he had a proof of the Kepler conjecture. Hales' proof is a proof by exhaustion involving the checking of many individual cases using complex computer calculations. Referees said that they were "99% certain" of the correctness of Hales' proof, and the Kepler conjecture was accepted as a theorem. In 2014, the Flyspeck project team, headed by Hales, announced the completion of a formal proof of the Kepler conjecture using a combination of the Isabelle and HOL Light proof assistants. In 2017, the formal proof was accepted by the journal Forum of Mathematics, Pi. -The original proof of the Kepler conjecture was submitted to Annals of Mathematics and the panel consisted of 12 referees! It was published after a very long referee process. The reason why the referees could not check correctness of the proof was because it involved computer code for the verification of thousands of cases. This was the reason why Hales decided to write a formal proof that was verified by a computer. Note that all numerical computations have also been verified formally. This was possible because he was using the interval arithmetic that allowed for a rigorous estimates of the approximation. -How long on average does Coq take to verify a proof? -One may expect that while a proof was written by humans who are rather slow, a computer should be able to check it quickly. Not necessarily. Regarding the formal proof of the Kepler conjecture the time needed for a formal verification was astonishing. Here is a quote from https://code.google.com/archive/p/flyspeck/wikis/AnnouncingCompletion.wiki: -The term the_nonlinear_inequalities is defined as a conjunction of several hundred nonlinear inequalities. The domains of these inequalities have been partitioned to create more than 23,000 inequalities. The verification of all nonlinear inequalities in HOL Light on the Microsoft Azure cloud took approximately 5000 processor-hours. -The verification was in HOP Light, but I would not expect that in Coq verification would be much faster.<|endoftext|> -TITLE: Functoriality of the Hopf dual -QUESTION [10 upvotes]: Given Hopf $\mathbb{C}$-algebra $H$, it's Hopf dual $H^o$ is the largest Hopf algebra contained in $H^*$, the $\mathbb{C}$-linear dual of $H$. (This is well known to be well-defined, see for example Sweedler book.) -If $j:G \to H$ is a linear map, then we have dual linear map -$$ -j^*:H^* \to G^*, ~~~~~~~~~ f \mapsto f \circ j. -$$ -If we restrict $j^*$ to $H^o$, then does $j^*(H^o)$ be contained in $G^o$? -If $j$ is algebra map, then it is easy to show that -$$ -\Delta(j^*(f)) = j^*(f_{(1)}) \otimes j^*(f_{(2)}) \in G^o \otimes G^o, -$$ -where we use Sweedler notation. However, it $j$ is not algebra map, then it is not clear what happens. My guess is that the dual is only "functorial" for algebra maps, but it is not clear for me. - -REPLY [3 votes]: -as suggested after the discussion in the comments- -i am understanding that the OP is asking whether a linear map $j:G \to H$ is functorial, in the sense that the image of the dual map $j^*:H^* \to G^*$ preserves the finite dual hopf algebra $H^\circ$, i.e. -$ -j^*(H^\circ)\subseteq G^\circ -$ -The answer is generally no -at least for an arbitrary linear map $j$- and in my understanding this has already been shown in the counterexample proposed by user66288 in the comments to the OP. -However, it will be the case, i.e. $j$ will be functorial if it vanishes on an ideal of finite codimension in $G$, since then the image of the finite dual $H^\circ$ will be inside the finite dual $G^\circ$, i.e. the image $j^*(f)$ of an arbitrary element $f\in H^\circ$ will be vanishing on an ideal of finite codimension in $G$: -$ -j^*(f)=f\circ j\in G^\circ -$ -for all $f\in H^\circ$. -(Notice that in this situation we actually have that $j^*(f)\in G^\circ$ -for all $f\in H^*$, and this is something which makes me wonder whether it is too restrictive, as i have already mentioned in the comments to the OP). -On the other hand, this is not an IFF statement, in the sense of the first of my comments above: -Consider an ideal $R$ of $G$ of finite codimension, not necessarily contained in the kernel of $j$. Then, those $f\in H^\circ$ for which $j(R)\subseteq ker f$ will be mapped in $G^\circ$ under $j^*$. If this happens for all $f\in H^\circ$ then $j^*(H^\circ)\subseteq G^\circ$, so we cannot necessarily conclude the converse statement: -"if $j$ is functorial then it vanishes on an ideal of finite codimension in $G$" -Edit (Nov 17): i was thinking that an iff characterization of the functoriality of $j$, is still possible, in the sense of the last paragraph above: - -a linear map $j:G\to H$, between two hopf algebras $H$, $G$ is functorial if and only if for any $f\in H^\circ$, there is an ideal $R_f$ of finite codimension in $G$, which is mapped inside the kernel of $f$ under $j$, i.e. $j(R_f)\in ker f$ - -(I am not sure if this implies that $j$ should be an algebra map then).<|endoftext|> -TITLE: Determinant of a block matrix with many $-1$'s -QUESTION [9 upvotes]: For an array $(n_1,...,n_k)$ of non-negative integers and non-zero reals $a_1,...,a_k$, define a block matrix $M$ of size $n=n_1+\cdots+n_k$ as follows: -The main diagonal has blocks of sizes $n_i$ and shapes $$M_i=J_{n_i}+a_i I_{n_i}=\begin{pmatrix} -a_i+1&1&\cdots&1\\ -1&a_i+1&\ddots&\vdots\\ -\vdots&\ddots&\ddots&1\\ -1&1&1&a_i+1\\ -\end {pmatrix}$$ and all the other entries are $-1$. -Experimentally I have found $$\det(M)= \prod_{i=1}^k a_i^{n_i}\sum_{j=0}^k(2-j)2^{j-1}e_j =\prod a_i^{n_i}(1+e_1-4e_3-16e_4-48e_5-\cdots),$$ -where $e_j$ is the $j^{th}$ elementary symmetric function of $\dfrac{n_1}{a_1},\dots,\dfrac{n_k}{a_k}$. -It should be a bit technical but not too hard to prove that by induction, but is there a more elegant way? - -REPLY [2 votes]: Alternatively, one may just inspect the eigenvalues (whose product is the determinant). -Clearly, all the vectors whose support lies in the $i$th block, and whose coordinates sum up to $0$, are eigenvectors with eigenvalue $a_i$; hence $a_i$ is the eigenvalue with multiplicity (at least) $n_i-1$. -An invariant complement to the sum of already found subspaces is the set $V$ of all block-constant vectors; so we need to check the determinant of the restriction onto $V$. A natural base in $V$ consists of the vectors havig ones in the $i$th block, and zeroes elsewhere. The matrix in this base is -$$ - A=\left( - \begin{array}{ccccccccc} - n_1+a_1& -n_2& -n_2& \cdots& -n_k\\ - -n_1& n_2+a_2& -n_3& \cdots& -n_k\\ - -n_1& -n_2& n_3+a_3& \cdots& -n_k\\ - \vdots& \vdots& \vdots& \ddots& \vdots\\ - -n_1& -n_2& -n_3& \cdots& n_k+a_k - \end{array} - \right), -$$ -whose determinant is $n_1n_2\dots n_k$ times the determinant of -$$ - B=\left( - \begin{array}{ccccccccc} - 1+a_1/n_1& -1& -1& \cdots& -1\\ - -1& 1+a_2/n_2& -1& \cdots& -1\\ - -1& -1& 1+a_3/n_3& \cdots& -1\\ - \vdots& \vdots& \vdots& \ddots& \vdots\\ - -1& -1& -1& \cdots& 1+a_k/n_k - \end{array} - \right). -$$ -Its determinant may be computed either directly, or via the same Matrix determinant lemma, yielding -$$ - \det B=\prod_{i=1}^k\left(2+\frac{a_i}{n_i}\right) - -\sum_{i=1}^k - \prod_{\textstyle{1\leq j\leq k\atop j\neq i}} - \left(2+\frac{a_j}{n_j}\right). -$$ -After substituting into -$$ - \det M=\det B\cdot \prod_{i=1}^k n_ia_i^{n_i-1} -$$ -and expanding the brackets, we get the desired result. (Although, I would admit, the form it has been obtained in does not look much worse for me.)<|endoftext|> -TITLE: Given any finite relation $R$ what is the cardinality of $\langle R\rangle=\{\underbrace{R\circ R\cdots \circ R}_{n\text{ times}}:n\in\mathbb{N}\}$? -QUESTION [8 upvotes]: Given any finite relation $R$ if we let $\circ$ denote relation composition and define $R^n=\underbrace{R\circ R\cdots \circ R}_{n\text{ times}}$ then does there exist an explicit formula for the cardinality of the set $\langle R\rangle=\{R^n:n\in\mathbb{N}\}$? If not, then are there any decent bounds for $|\langle R\rangle|$? I mean clearly we have that: $$\{R^n:n\in\mathbb{N}\}\subseteq\wp(\text{dom}(R)\times\text{rng}(R))\implies |\{R^n:n\in\mathbb{N}\}|\leq 2^{|\text{dom}(R)||\text{rng}(R)|}\leq 2^{|R|^2}$$ But this is a terrible upper bound. How can it be improved? -Also I can prove if $R$ is functional and we define the digraph $D=(\text{dom}(R)\cup\text{rng}(R),R)$ then if we let $m$ be the length of any longest directed path in the condensation of $D$ and we let $n$ be the least common multiple of the lengths of every directed cycle in $D$, then we have: -$$|\langle R\rangle|=\begin{cases}n&\text{ if }D\text{ is the graph of a permutation}\\m+1&\text{ if }D\text{ is a directed acyclic graph}\\m+n-1&\text{ otherwise}\end{cases}$$ -However again this is just a special case, so to reiterate given any finite relation $R$ does there exist a general expression or formula for the cardinality of $\langle R\rangle$? - -REPLY [5 votes]: $\newcommand{\N}{\mathbb{N}}$ -Recall that Landau function $g(n)$ is the biggest possible $\mbox{lcm}$ of numbers wich sum up to $n$. It's asymptotic is well-studied. -I'll prove the following -$\textbf{Theorem 1.}$ There is some $C > 0$ such that we have $|\{ R^n : n\in \N\}| \le g(|R|) + C|R|^2$, moreover we can find $0 < k \le g(|R|)$ such that $R^{n + k} = R^n$ for $n = C|R|^2$. -Instead of talking about relations I prefer to talk about finite oriented graphs and nondeterministic finite automata over unary language. Let us prove the following -$\textbf{Theorem 2.}$ Let $G$ be a directed graph and let $C_1, \ldots , C_m$ be its strongly connected components. Then for every vertex $v\in G$ we have that set of vertices that one can reach from $v$ in exactly $M$ steps is up to some pre-period of length $O(|G|^2)$ is something wich is periodic with period $\mbox{lcm}(c_1, \ldots , c_m)$, where $c_i$ is $\mbox{gcd}$ of length of all cycles in $C_i$. -Let us prove Theorem 1 from Theorem 2: -Using cycles of different length one can easily construct example where period is at least $g(|R|)$. -On the other hand we have that up to very small pre-period we have that everything is periodic with period $\mbox{lcm} (c_1, \ldots , c_m)$. Since $c_i \le |C_i|$ we have that $c_1 + \ldots + c_m \le |G|$. Now it is obvious that $|G| \le 2|R|$ (we do not consider vertices of degree $0$). But we can say a bit more: note that any vertex of total degree $1$ can not be a part of any strongly connected component thus we may not consider these vertices as well. And number of other vertices is at most $|R|$ so we got that length of the period is at most $g(|R|)$. -It remains to prove Theorem 2. -To do so we need the (proof of) Lemma 4.3 from Chrobak's paper [1]. -Consider vertices $v, u\in G$. Let us build NFA: it's graph will be just $G$, every edge corresponds to the same letter, we begin at $v$ and the only accepting state is $u$. It is enough to prove that set of all $m$ such that we can reach $u$ from $v$ in exactly $m$ steps is up to pre-period of size $O(|G|^2)$ is something with period $\mbox{lcm}(c_1, \ldots , c_m)$. But that is exactly what Chrobak did! So the Theorem 2 is also proved. -Now it remains to use asymptotic of Landau function to get that maximum possible $k = \exp( (1 + o(1))\sqrt{|R|\log |R|})$. -As a final remark note that everyting here depends mostly on the number of elements we have our relation on (that is $|G|$) rather than $|R|$ (that is number of edges).<|endoftext|> -TITLE: Does every index $p$ subgroup of $SL(2,\mathbb{Z}_p)$ contain $\Gamma(p)$? -QUESTION [5 upvotes]: Does every index $p$ subgroup of $SL(2,\mathbb{Z}_p)$ contain the principal congruence subgroup $\Gamma(p)$? -Equivalently, must it be the preimage of an index $p$ subgroup of $SL(2,\mathbb{Z}/p\mathbb{Z})$? - -REPLY [9 votes]: Yes. -Let $H$ be the subgroup, and let $N$ be the normal closure. The index of $N$ in $\Gamma = \mathrm{SL}(2,\mathbb{Z}_p)$ has index dividing $p!$ which is not divisible by $p^2$. Hence either: - -$N$ contains the principal congruence subgroup $\Gamma(p)$ and you win, -$N \cap \Gamma(p)$ has index $p$ inside $\Gamma(p)$. - -If $p > 2$, one can now apply the following observations: - -The abelianization of $\Gamma(p)$ is $V = \Gamma(p)/\Gamma(p^2)$. (It's easy to write down topological generators of $\Gamma(p^2)$ using commutators when $p > 2$.) -As a module for $\Gamma/\Gamma(p) = \mathrm{SL}(2,\mathbb{F}_p)$ under conjugation, $V$ is the adjoint representation and is irreducible. - -This gives a contradiction, since, if $N$ is normal, then $\Gamma(p)/(N \cap \Gamma(p))$ will be a proper quotient of $V$ under the action of $\mathrm{SL}(2,\mathbb{F}_p)$. -If $p = 2$, then you can work a little harder with explicit computations and use a similar argument (now the abelianization of $\Gamma(2)$ is something close to $\Gamma(2)/\Gamma(8)$) or instead simply make the stupid observation that $\mathrm{SL}(2,\mathbf{Z})$ is dense in $\Gamma$ and thus there is an injection from subgroups of $\Gamma$ (of any index) to subgroups of $\mathrm{SL}(2,\mathbf{Z})$ of the same index (not a bijection because of the failure of the congruence subgroup property). However, the latter is well known to have abelianization $\mathbf{Z}/12 \mathbf{Z}$ and so has a unique index $2$ subgroup.<|endoftext|> -TITLE: Does the Banach space $( \ell ^2 \oplus \ell ^2 )$ have F.P.P? -QUESTION [9 upvotes]: The space $( \ell^2 ,\lVert \cdot \rVert _2 )$ is a Hilbert space. The space - $X=(\ell^2 \oplus \ell^2 , \lVert \cdot \rVert_\infty )$ is a Banach space. Does X have fixed point property? (For any closed convex bounded subset $C\subseteq X $ and any nonexpansive map $T:C\to C $ there is a $x\in C$ such that $T(x)=x$) -The space $X$ isn't uniformly convex so I can't use theorems about uniformly convex. There is no theorems about F.P.P in products of spaces and I have no another idea. Does someone have any idea? - -REPLY [8 votes]: I think that the answer is yes, and that it should follow from the following facts: - -every Hilbert space is uniformly convex, hence it has normal structure; -the direct sum of two Banach spaces with normal structure, endowed with the infinity norm, has again normal structure (Belluce-Kirk-Steiner, Pacific Journal Math. 1968); -the finite direct sum of separable Banach spaces (with any of the possible equivalent norm on it) is itself reflexive, in particular $X = \ell^2 \oplus \ell^2$ is reflexive; -normal structure on a reflexive Banach space implies FPP (Kirk, Amer. Math. Monthly 1965).<|endoftext|> -TITLE: Bounds for the size of arrays with distinct subarray sums -QUESTION [5 upvotes]: Consider an array $A$ of length $n$ with $A_i \in \{1,\dots,s\}$ for some $s\geq 1$. For example take $s = 6$, $n = 5$ and $A = (2, 5, 6, 3, 1)$. Let us define $g(A)$ as the collection of sums of all the non-empty contiguously indexed subarrays of A. In this case -$$g(A) = [2,5,6,3,1,7,11,9,4,13,14,10,16,15,17]$$ -In this case all the sums are distinct. However, if we looked at $g((1,2,3,4))$ then the value $3$ occurs twice as a sum and so the sums are not all distinct. -For $s \geq 1$, I would like to understand what the largest $n$ is such that that there exists an array $A$ of length $n$ with all distinct $g(A)$. This question arose orginally as a coding competition and the answers for $s = 1,\dots, 21$ are $n = 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 8, 9, 10, 11, 11, 11, 12, 13, 13, 14, 14$. - -On the assumption that an exact formula is hard to come by, is it - possible to show its asymptotics? For example, is it true that - $n$ is $\Theta(s)$? - -REPLY [8 votes]: Let $0=:x_02$ be a prime number and consider the set of numbers $x_k=2pk+\{k^2\}_p$, where $\{x\}_p$ denotes a remainder of $x$ modulo $p$ and $k$ varies from 0 to $p-1$. Obviously $x_{k+1}-x_k<3p$ for all $k=0,1,\dots,p-1$ and the equation $x_i+x_j=x_u+x_v$ implies $i+j=u+v$, $\{i^2\}_p+\{j^2\}_p=\{u^2\}_p+\{v^2\}_p$ (considering remainders and quotients of both parts modulo $2p$), thus modulo $p$ we have $i+j=u+v$, $i^2+j^2=u^2+v^2$, $(i-u)(i+u)=(v-j)(v+j)$, $(i-u)(i+u-v-j)=0$, either $i=u,v=j$ or $i+u=v+j=v+u+v-i$, $i=v,u=j$. Therefore this is indeed a Sidon sequence and for any $s\geqslant 3p-1$ we get an array of length $p-1$. Since the primes number are frequent enough, this yields that $n=(1/3+o(1))s$ is possible. -The upper bound $n\leqslant s$ also may be improved using the idea of Erdos and Turan on Sidon sets. Namely, we may consider the differences $x_j-x_i$ for $0 -TITLE: Does a complex Fano manifold have simplicial Mori cone when all extremal contractions are fiber type? -QUESTION [15 upvotes]: Let $X$ be a complex Fano manifold such that each extremal ray of $\overline{\text{NE}(X)}_{\mathbb{R}}$ is generated by a primitive class in $H_2(X;\mathbb{Z})$ of a free rational curve. Thus, the extremal rays are all fiber type. (The "primitive" hypothesis rules out, e.g., conic bundles where "half" of the fiber class is integral.) -Question. Is the cone of effective curves in $H_2(X;\mathbb{Z})$ a free $\mathbb{Z}_{\geq 0}$-semigroup, $\mathbb{Z}_{\geq 0}^r$? -There are many positive examples, e.g., Fano manifolds with a transitive (algebraic) action of a complex Lie group and complete intersections in these of low degree. However, for the general question, the best results that I have found are in the following article of Jaroslaw Wisniewski. -MR1639552 (2000e:14018) -Wiśniewski, Jarosław A. -Cohomological invariants of complex manifolds coming from extremal rays. -Asian J. Math. 2 (1998), no. 2, 289–301. -https://arxiv.org/abs/math/9803010 - -REPLY [9 votes]: I think that this is open in general, and that one would expect the answer to be positive. Related references are: - -MR1103910 Wiśniewski, Jarosław A., -On contractions of extremal rays of Fano manifolds. -J. Reine Angew. Math. 417 (1991), 141–157 - -He shows in Theorem 2.2 that if X is a smooth Fano where every extremal ray is of fiber type, then the number of extremal rays is at most dim(X), hence also rho(X) is at most dim(X), where rho(X) is the Picard number - -MR3533196 -Druel, Stéphane, -On Fano varieties whose effective divisors are numerically eventually free. -Math. Res. Lett. 23 (2016), no. 3, 771–804 - -He classifies the cases where rho(X)=dim(X), and NE(X) turns out to be simplicial. - -Ou, Wenhao, Fano varieties with Nef(X)=Psef(X) and rho(X)=dim(X)-1, Manuscripta Math. 2018 - -He studies the next case rho(X)=dim(X)-1, I think that probably NE(X) turns out to be simplicial too, but I did not check - -MR2474316 by myself, Quasi-elementary contractions of Fano manifolds. Compos. Math. 144 (2008), no. 6, 1429–1460 - -It follows from Theorem 4.1 that NE(X) is always simplicial if dim(X) is at most 5. -I hope this helps. Please let me know if you find more general results, I am interested in this question!<|endoftext|> -TITLE: Maps which are both completely positive and positive -QUESTION [8 upvotes]: Definition:A linear map $f:\mathbb C^n\to \mathbb C^n$ is called positive if $\langle fa,a\rangle\ge0$ for all $a\in \mathbb C^n$. Equivalently, $f\in M_{n}(\mathbb C)$ is positive if it can be written in the form $g^*g$ for some $g\in M_{n}(\mathbb C)$. The unique positive element $g$ satisfying $g^2=f$ is denoted $\sqrt f$, and is called the square root of $f$. -Definition:A linear map $\phi:M_{n}(\mathbb C)\to M_{m}(\mathbb C)$ is called completely positive if it sends positive elements to positive elements, and the same holds for $\phi\otimes id_{M_{k}(\mathbb C)}:M_{n\cdot k}(\mathbb C)\to M_{m\cdot k}(\mathbb C)$ for every $k\in\mathbb N$. - -Let $e_{ij}\in M_{n}(\mathbb C)$ be the elementary matrix with a $1$ at $(i,j)$ and all other entries zero. Using the basis elementary matrices $\{e_{ij}\}$ to identify $M_{n}(\mathbb C)$ with $\mathbb C^{n^2}$, we can talk about a linear map $f:M_{n}(\mathbb C)\to M_{n}(\mathbb C)$ being positive (this has nothing to do with sending positive elements to positive elements). - -Question:Let $\phi:M_{n}(\mathbb C)\to M_{n}(\mathbb C)$ be a map which is both positive and completely positive. Is its square root $\sqrt\phi:M_{n}(\mathbb C)\to M_{n}(\mathbb C)$ completely positive? - -Remark: -Maps which are both positive and completely positive are frequent. Indeed, for every completely positive map $\phi:M_{n}(\mathbb C)\to M_{m}(\mathbb C)$, its adjoint $\phi^{*}:M_{m}(\mathbb C)\to M_{n}(\mathbb C)$ is also completely positive. So $\phi^{*}\phi:M_{n}(\mathbb C)\to M_{n}(\mathbb C)$ is both positive and completely positive. - -REPLY [6 votes]: No. I use facts about Schur multipliers which can be found, for example in Paulsen's monograph on completely bounded maps. Basically you are searching for a positive semidefinite matrix with positive entries such that the pointwise square root of the matrix is not positive semidefinite. Specifically, let $A=\left[ \begin{array}{ccc} 1 & 2^{-1/2} & 0\\ - 2^{-1/2} & 1 & 2^{-1/2}\\ - 0 & 2^{-1/2} & 1 \end{array} \right].$ -Let $S_A:M_3(\mathbb{C})\rightarrow M_3(\mathbb{C})$ be the Schur multiplier associated with $A$ (i.e. $S_A$ takes a matrix to its Schur product with $A$). Since $A$ is positive semidefinite, the Schur multiplier is completely positive. Since $S_A(e_{i,j})=r_{i,j}e_{i,j}$ for some $r_{i,j}\geq 0$ and the set $(e_{i,j})$ forms an orthonormal basis, this is a positive map. The positive square root is clearly $S_B$ where $B=\left[ \begin{array}{ccc} 1 & 2^{-1/4} & 0\\ - 2^{-1/4} & 1 & 2^{-1/4}\\ - 0 & 2^{-1/4} & 1 \end{array} \right].$ - But $B$ is not a positive semidefinite matrix and therefore $S_B$ is not completely positive.<|endoftext|> -TITLE: Maximal inequality for the average of i.i.d. random variables -QUESTION [11 upvotes]: Let $Z_i$ be i.i.d. random variables with $\mathbb{E}[Z_i] = 0$ and $\mathbb{E}|Z_i|^p< \infty$ for $p=1,2,3,\cdots$. I am looking for the following type of estimate if possible, and it is not like the concentration inequalities that one normally sees. - -There exists $N_0$ sufficiently large and $t_0$ sufficiently small - such that for all $N\geq N_0$ and $1/N0$ be any number such that the probability that the standard Gaussian random variable is less than $\delta$ is strictly less than $q=\frac 23$, say. Fix $b\in(0,\frac 12)$ and an integer $A\ge 2$ to be chosen later. Split the natural numbers $1,2,3,\dots$ into intervals $J_k$ of length $|J_k|=A^k$ from left to right (so $J_1=\{1,\dots,A\}, J_2=\{1+A,\dots, A+A^2\}, J_3=\{1+A+A^2,\dots,A+A^2+A^3\}$ and so on). Consider -$$ -W_k=\sum_{j\in J_k}Z_j -$$ -where $k$ runs from $1$ to $K$ with $K$ being the largest number satisfying -$$ -2 A^K t\le b A^{K/2} -$$ -Notice that then $A^K\approx t^{-2}$ so we have all the corresponding intervals $J_k$ in the range $[1,N]$ if $t\ge N^{-1/2}$. However, if we cover this range, we can then easily extend it to the required range $t>\frac 1N$ by just decreasing the value of $\alpha$ twice. Also $K\approx \frac{\log\frac 1t}{\log A}$ for small $t$. -Now, the event we are interested in is a subset of the event -$$ -W_1\le bA^{1/2}, W_1+W_2\le b A^{2/2},\dots, W_1+\dots+W_K\le bA^{K/2}\,. -$$ -We shall prove by induction that the event -$$ -E_k=\{W_1\le bA^{1/2}, W_1+W_2\le b A^{2/2},\dots, W_1+\dots+W_k\le bA^{k/2}\} -$$ -has probability at most $CQ^k$ with $Q=\frac 34$, say, provided that $b$ and $A$ are chosen appropriately. Since $K$ is of order $\log\frac 1t$, this will yield the desired power bound immediately. -By the CLT, $W_k$ get close to normals with mean $0$ and variance $A^k$ for large $k$. All we need is that -$$ -P\{W_k<\delta A^{k/2}\}\le q\text{ for }k>k_0\,. -$$ -Note that $k_0$ depends only on the distribution of $Z$ here (which, of course, should play some role). -Now choose $C>0$ so that $CQ^{k_0}\ge 1$, so we have nothing to prove for $k=1,\dots,k_0$. -Assume now that $k>k_0$ and the desired estimate holds for $1,\dots,k-1$. The event $E_k$ is covered by the events -$$ -A_m=E_{m-1}\cap\{W_m\le -b^{k-m} A^{k/2}\},\quad m=1,2,\dots,k-1 -$$ -and -$$ -A=E_{k-1}\cap \{W_1+\dots+W_{k-1}\ge -\tfrac b{1-b}A^{k/2}\}\cap \{W_k\le (b+\tfrac{b}{1-b})A^{k/2}\}\,. -$$ -Here $E_0$ is the entire probability space. -By the independence of $W_m$, the induction assumption, and Chebyshev's inequality, we have -$$ -P(A_m)\le CQ^{m-1}\frac 1{(b^2A)^{k-m}} -$$ -while -$$ -P(A)\le CQ^{k-1}q -$$ -provided that $b+\frac b{1-b}\le\delta$. -Adding all these bounds, we get -$$ -CQ^{k-1}\left[q+\sum_{\ell=1}^{k-1}(Qb^2A)^{-\ell}\right]\le CQ^{k-1}\left(q+\frac 1{Qb^2A-1}\right)\le CQ^k\,, -$$ -provided that $A$ was chosen large enough.<|endoftext|> -TITLE: Expressions for the inverse function of f(x) = ln(x)e^x -QUESTION [9 upvotes]: Can the inverse of $ ln(x)e^x $ be finitely expressed in terms of the Lambert-W function or any other well-known transcendental functions? It is clear that a closed-form elementary function expression is unreachable. -The reason I ask is in pondering on the links between the inverse Lambert-W and some naturally arising functions of similar forms. Recall that the Lambert-W, a transcendental function, is defined as $ W(xe^x) = x. $ -It is natural then to consider the inverse of functions such as $ g(x) = xe^{e^x} $ and those with further exponentiation. With a simple transformation $ z= e^x $ we can reduce $ g(x) $ to the form $ ln(z)e^z $ as originally posed. So the broader question arises: are there tangible algebraic links between the inverses of the set -$$ {xe^x, xe^{e^x},xe^{e^{e^x}}}... $$ - -REPLY [8 votes]: These are so-called hyper-Lambert functions, see -On some applications of the generalized hyper-Lambert functions.<|endoftext|> -TITLE: Classification of compact connected abelian groups -QUESTION [5 upvotes]: It is known that torsion-free compact abelian groups are exactly the product of the maximal solenoid group $\Sigma_{(2,3,\cdots)}$ (which is the Pontryagin dual of the additive group $\mathbb{Q}$ of rational numbers equipped with the discrete topology) and the additive $p$-adic integers $\Delta_p$ (cf. 25.4 and 25.8 of Abstract Harmonic Analysis by Hewitt & Ross). This makes me wonder if a similar classification result exists for connected (instead of torsion free) compact abelian groups. -As in the torsion-free case, this reduces to the problem of classifying all discrete abelian groups for which the Pontryagin dual is connected, which seems a little intractable for me. Perhaps a simpler, more tractable problem is, does there exist any compact connected abelian group which is not a product of $\mathbf{a}$-adic solenoids $\Sigma_\mathbf{a}$ for various sequence $\mathbf{a} = (a_1, a_2, \cdots)$ of integers greater than $1$ (cf. 10.12 and 10.13 ibid) and the circle group $\mathbb{T}$ (with products of $\mathbb{T}$ accounts for the torsion part and solenoids for the rest, if such an idea can be made precise)? In particular, is it true that any compact connected abelian group with a dense torsion subgroup a product of $\mathbb{T}$? More particularly, is it true that any compact connected abelian Lie group a finite product of $\mathbb{T}$? I think the last question has an affirmative answer (it is related to this question), but I don't know a rigorous proof. - -REPLY [7 votes]: Perhaps a simpler, more tractable problem is, does there exist any compact connected abelian group which is not a product of $\mathbf{a}$-adic solenoids $\Sigma_\mathbf{a}$ for various sequence $\mathbf{a} = (a_1, a_2, \cdots)$ of integers greater than $1$ (cf. 10.12 and 10.13 ibid) and the circle group $\mathbb{T}$ - -Following YCor's comments, the Pontryagin dual question is whether there exists a torsion-free abelian group which is not a direct sum of localizations of $\mathbb{Z}$. And the answer is yes: for example, you can consider the $p$-adic integers $\mathbb{Z}_p$ as a discrete group. Cf. the accepted answer at this MO question about simply presented abelian groups; YCor's answer is also educational and suggests simpler examples such as suitable subgroups of $\mathbb{Z}[1/p]^2$. -Edit: Let me see if I can flesh out YCor's claim about subgroups of $\mathbb{Z}[1/p]^2$. The interesting ones are the ones of rank $2$, and for simplicity I'll restrict my attention to subgroups containing $\mathbb{Z}^2$ (I thought I had an argument that every subgroup of rank $2$ is isomorphic to such a subgroup but now I'm not so sure). These correspond to subgroups of the quotient $\mathbb{Z}[1/p]^2/\mathbb{Z}^2 \cong \mu_{p^{\infty}}^2$, the product of two copies of the Prüfer $p$-group. -We can describe such subgroups using Goursat's lemma, which says in this case that if $H \subseteq \mu_{p^{\infty}}^2$ is a subgroup such that the two projections $\pi_1, \pi_2 : H \to \mu_{p^{\infty}}$ are surjective (this is the interesting case), then there exist subgroups $N_1, N_2 \subseteq \mu_{p^{\infty}}$ and an isomorphism $\varphi : \mu_{p^{\infty}}/N_1 \cong \mu_{p^{\infty}}/N_2$ such that -$$H = \{ (q_1, q_2) \in \mu_{p^{\infty}}^2 : \varphi(q_1) \equiv q_2 \bmod N_2 \}.$$ -The proper subgroups of $\mu_{p^{\infty}}$ take the form $\mu_{p^n}$ for $n \in \mathbb{Z}_{\ge 0}$ (if we consider all of $\mu_{p^{\infty}}$ then $H$ is all of $\mu_{p^{\infty}}^2$). The quotient $\mu_{p^{\infty}}/\mu_{p^n}$ is isomorphic to $\mu_{p^{\infty}}$ again, and its automorphism group is the group of $p$-adic units $\mathbb{Z}_p^{\times}$, of which there are uncountably many. -So there are uncountably many choices for $H$, and hence uncountably many subgroups of $\mathbb{Z}[1/p]^2$ containing $\mathbb{Z}^2$. Of these, the subgroups isomorphic to $\mathbb{Z}^2, \mathbb{Z} \times \mathbb{Z}[1/p]$, or $\mathbb{Z}[1/p]^2$ are determined by the image of the copy of $\mathbb{Z}^2$ in each of them, and there are countably many choices for this image. So as promised, at most countably many subgroups can be isomorphic to a direct sum of localizations of $\mathbb{Z}$. Unfortunately I don't quite see how to explicitly exhibit a particular choice of $H$ which is not such a direct sum. -Edit #2: Let's take $N_1 = N_2 = 0$ to be trivial above, and $\varphi : \mu_{p^{\infty}} \cong \mu_{p^{\infty}}$ to be multiplication by a $p$-adic unit $u \in \mathbb{Z}_p^{\times}$. Then -$$H = \{ (q_1, q_2) \in \mu_{p^{\infty}}^2 : u q_1 = q_2 \}$$ -is a subgroup of $\mu_{p^{\infty}}^2$ isomorphic to $\mu_{p^{\infty}}$, and it lifts to a subgroup -$$\widetilde{H} = \{ (q_1, q_2) \in \mathbb{Z}[1/p]^2 : u (q_1 \bmod 1) \equiv q_2 \bmod 1 \}.$$ -Thinking of $H$ as a "line with slope $u$" suggests that $\widetilde{H}$ is isomorphic to a direct sum of localizations of $\mathbb{Z}$ (in fact to $\mathbb{Z} \times \mathbb{Z}[1/p]$) iff $u$ is rational. Indeed, if $u = \frac{a}{b}$ is rational, so that $a, b$ are integers relatively prime to $p$ and each other, then the condition that $u (q_1 \bmod 1) \equiv q_2 \bmod 1$ is equivalent to the condition that $aq_1 - bq_2 \in \mathbb{Z}$, and consequently the map -$$\widetilde{H} \ni (q_1, q_2) \mapsto (q_1, aq_1 - bq_2) \in \mathbb{Z}[1/p] \times \mathbb{Z}$$ -is an isomorphism. -Now suppose that $u$ is irrational. First let's show that $\widetilde{H}$ has no subgroup isomorphic to $\mathbb{Z}[1/p]$: equivalently, no nonzero element is $p$-divisible. If $(q_1, q_2) \in \widetilde{H}$ is a nonzero element, for it to be $p$-divisible would require that -$$u \left( \frac{q_1}{p^k} \bmod 1 \right) \equiv \frac{q_2}{p^k} \bmod 1$$ -for all $k$, or equivalently that -$$u(q_1) - q_2 \in p^k \mathbb{Z}$$ -for all $k$. Taking $k \to \infty$ gives $u = \frac{q_2}{q_1}$, but this contradicts $u$ irrational. -Hence in this case, if $\widetilde{H}$ is isomorphic to a direct sum of localizations of $\mathbb{Z}$ then it can only be isomorphic to $\mathbb{Z}^2$. But the projection to either coordinate gives a surjection onto $\mathbb{Z}[1/p]$, which $\mathbb{Z}^2$ does not possess. -Some additional comments. Projection onto the first coordinate shows that $\widetilde{H}$ is an extension -$$0 \to \mathbb{Z} \to \widetilde{H} \to \mathbb{Z}[1/p] \to 0.$$ -The argument above shows that this sequence splits iff $u$ is rational. In general, this extension is classified by a class in $\text{Ext}^1(\mathbb{Z}[1/p], \mathbb{Z})$, which can be computed from $\text{Ext}^1(\mu_{p^{\infty}}, \mathbb{Z}) \cong \mathbb{Z}_p$ and the short exact sequence $0 \to \mathbb{Z} \to \mathbb{Z}[1/p] \to \mu_{p^{\infty}} \to 0$ to give -$$\text{Ext}^1(\mathbb{Z}[1/p], \mathbb{Z}) \cong \mathbb{Z}_p/\mathbb{Z}.$$ -The point of this computation is to show that for general homological reasons there are interesting nontrivial extensions of torsion-free abelian groups by torsion-free abelian groups; taking Pontryagin duals, there are interesting nontrivial extensions of compact connected abelian groups by compact connected abelian groups. -It should be possible to match up this $\mathbb{Z}_p$ with the $\mathbb{Z}_p$ that $u$ lives in above but I'm not entirely sure how.<|endoftext|> -TITLE: Reference request for K-Theory linearization -QUESTION [7 upvotes]: I posted this question on math.se here:https://math.stackexchange.com/q/2996787/482732, but I think it may be more appropriate here, sorry if I am wrong about that. -In Waldhausen's paper Algebraic K theory of Spaces(the long one) he proves the following: -$$A(X)\simeq \mathbb{Z}\times B\widehat{Gl}(\Omega^{\infty}\Sigma^\infty |G|)$$ -Where $|G|$ is a loop group of $X$. My problem is that to apply $B$ we need $\widehat{Gl}(\Omega^{\infty}\Sigma^\infty |G|)$ to be $A^\infty$, but since $\Omega^{\infty}\Sigma^\infty |G|$ is only a ring up to homotopy this isn't obvious to me. Waldhausen just brushes past this point, so I was hoping to get a reference to somewhere that shows this in detail. Thanks. - -REPLY [9 votes]: I claim that for every $A_\infty$-space $A$, there is a canonical $A_\infty$-ring structure on $\Omega^\infty\Sigma^\infty_+A$. -First, $\Sigma^\infty_+$ from spaces to spectra is symmetric monoidal. So it sends an $A_\infty$-space $A$ to an $A_\infty$-algebra in spectra $\Sigma^\infty_+A$, that is an $A_\infty$-ring spectrum. The fact that $\Sigma^\infty_+$ is symmetric monoidal (at the model category/∞-category level) can be found in any modern book treating the smash product of spectra (e.g. it is proposition 4.7 in Elmendorf-May-Kriz-Mandell and corollary 4.8.2.19 in Lurie's Higher Algebra). -Secondly I claim that if $E$ is an $A_\infty$-ring spectrum, then $\Omega^\infty E$ has a canonical $A_\infty$-ring space structure. Exactly how this works will depend on your preferred definition of $A_\infty$-ring space, but it can be proven, e.g., with the same technique that May uses to prove the analogous statement for $E_\infty$-ring spaces (Corollary 7.5 in May's What precisely are $E_\infty$-ring spaces and $E_\infty$-ring spectra). - -If all you care for is a construction of the $A_\infty$-structure on $GL_1(\Sigma^\infty_+A)$, I particularly like the approach in - -Matthew Ando, Andrew J. Blumberg, David Gepner, Michael J. Hopkins, Charles Rezk An ∞-categorical approach to R-line bundles, R-module Thom spectra, and twisted R-homology - -where they identify $GL_1(R)$ with the automorphism group of $R$ as an $R$-module (and so it has an $A_\infty$-structure, since all automorphism groups in an ∞-category "trivially" do).<|endoftext|> -TITLE: "Scott completion" of dcpo -QUESTION [12 upvotes]: If $A$ is poset with all directed suprema, it is common to consider the Scott topology on $A$, whose open subsets are the $U \subset A$ such that $U$ is upward closed and if $\bigcup_I a_i \in U $ for some directed supremum then $\exists i, a_i \in U$. -It is a classical fact that the specialization order induced by this topology on $A$ is exactly the order relation on $A$. -There are some examples (See P.Johnstone's 'Scott is not always sober') where the Scott topology is not sober. But one can always consider $\mathcal{O}(A)$ the frame of open subsets for the Scott topology, and look at: -$$ \overline{A} = pt(\mathcal{O}(A)) $$ -the poset of 'points' (i.e. frame homomorphisms $\mathcal{O}(A) \rightarrow \{0,1\}$) of this frame. The fact that the Scott topology is a topology on $A$ immediately implies that there is a map $A \rightarrow \overline{A}$, and the observation above implies that this is an order embeding, moreover $\overline{A}$ is again directed complete and this inclusion preserves directed suprema. The question of soberness of the Scott topology boils down to whether this map is an isomorphism. -I want to know if this construction $A \rightarrow \overline{A}$ can be seen as a "completion", i.e. whether it is idempotent. The following are (I think) equivalent ways of formulating this question: - -is the map $\overline{A} \rightarrow \overline{\overline{A}}$ a bijection. -When $B$ is the poset of points of a frame, do we always have that $B \rightarrow \overline{B}$ is an equivalence. -Does the restriction to $A$ induce a bijection between the Scott topology of $\overline{A}$ and the Scott topology of $A$. - -I found some papers that seem to suggest that the question has been studied, but I couldn't find any precise claim anywhere. - -REPLY [10 votes]: I believe the paper by Johnstone linked to in the question contains the answer, and it is negative. -In that paper, Johnstone constructs a Scott topology that is not sober as a byproduct of answering in the negative the following question (marked by (a) in the paper): - -if $(X,\leqslant)$ has directed joins, is there a sober topology on $X$ inducing $\leqslant$? - -(A counterexample $(X,\leqslant)$ to (a) evidently has non-sober Scott topology since the latter induces $\leqslant$.) -But further, he answers in the negative also the following question (marked (b)): - -if $(X,\leqslant)$ is induced by some sober topology, is the Scott topology on $X$ sober? - -And, what is crucial for the present question, his counterexample to (b) is obtained as the sobering of a non-sober Scott topology. -Thus, he has got $(X,\leqslant)$ with directed joins such that taking in succession Scott, then sobering, then specialization gives another poset with directed joins, and for this another poset, again taking Scott, sobering, specialization gives still something different. -In more detail - -Notation, terminology: I will call a poset with all directed joins/suprema a dcpo. For a dcpo $P$ I will denote by $S(P)$ the topological space with the underlying set $P$ and with the Scott topology of $P$, i. e. topology with closed sets = those downsets of $P$ which are sub-dcpos. For a topological space $X$ I will denote by $P(X)$ the specialization preorder of $X$, and by $s(X)$ the sobering of $X$, i. e. the space of irreducible closed subsets of $X$ with its sober topology. Note that $P(s(X))$ is given by the inclusion order of irreducible closed subsets, and $X$ maps to $s(X)$ by sending a point to its closure. -Thus, in the notation of the question, $\overline A$ is $P(s(S(A)))$, and the question asks whether $P(s(S(A)))\to P(s(S(P(s(S(A))))))$ is a bijection for every dcpo $A$. -Johnstone constructs a dcpo $X$ such that $X$ is the only irreducible closed subset of $S(X)$ that is not closure of any point. Hence the specialization order of its sobering, $X^+=P(s(S(X)))$, is obtained by adding a top element $\infty$ to $X$, with $X^+$ the closure of $\{\infty\}$ in $s(S(X))$. At the same time, $S(X^+)$ is not sober since in the Scott topology of $X^+$, the subset $X$ is closed (being a sub-dcpo downset of $X^+$) and irreducible (since a topology inducing a dcpo order is contained in the Scott topology of that order). -Look now at the sobering $s(S(X^+))$ of $S(X^+)$. Because of the above, it contains one extra point $\infty'$, such that $X\cup\{\infty'\}$ is the closure of $\infty'$ in $s(S(X^+))$. Then in $P(s(S(X^+))$, the point $\infty'$ is an upper bound of $X$ and $\infty'<\infty$ (since as irreducible closed subsets of $S(X^+)$, $\infty'$ is $X$ and $\infty$ is $X^+$ ($=$ closure of $\{\infty\}$)). -So the map $P(s(S(X)))\to P(s(S(P(s(S(X))))))$, i. e. $X^+\to P(s(S(X^+)))$, is not a bijection: $\infty'$ is not in its image.<|endoftext|> -TITLE: continuity of certain map which is defined on a Stonean space -QUESTION [5 upvotes]: Let $G$ be a discrete group which acts continuously on a Stonean space $\Omega$. Consider the map $f\colon \Omega\to \{0,1\}^G$ sending $x\in \Omega$ to $\chi_{G_x}$, where $\chi_{G_x}$ denotes the characteristic function of the stabilizer $G_x$. -Why is $f$ continuous? -We can restrict to the subbasis-sets of the product topology and for $g\in G$ fixed, it is enough to consider the preimages of $\{0\}\times \{0,1\}^{G\setminus \{g\}}$ and $\{1\}\times \{0,1\}^{G\setminus \{g\}}$ under $f$, which must be open in $\Omega$. That $f^{-1}(\{0\}\times \{0,1\}^{G\setminus \{g\}}) $ is open in $\Omega$ can be seen using that in the Hausdorff space $\Omega$, the diagonal $\{(x,x)\in \Omega\times \Omega\mid x\in \Omega\}$ is closed in $\Omega\times \Omega$, but without using that $\Omega$ is Stonean. -My question therefore reduces to: Why is $ f^{-1}(\{1\}\times \{0,1\}^{G\setminus \{g\}}) $ open in $\Omega$ ? I think to prove this, it must be used that $\Omega$ is Stonean and probably it helps to know that in a Stonean space, the closure of disjoint open sets is again disjoint. -The background of my question is the following: I am currently reading the paper https://arxiv.org/pdf/1509.01870.pdf and that $f$ is continuous is used in the proof of Theorem 4.1, direction $<=$. - -REPLY [5 votes]: It's indeed true: whenever a discrete group $G$ acts continuously on a Hausdorff, extremally disconnected space $X$, then the map $x\mapsto G_x$ is continuous, where the set of subgroups of $G$ is endowed with its compact topology given by inclusion in $2^G$. -One has to show that for any $g\in G$, the map $u_g:x\mapsto \chi_{G_x}(g)$, $X\to\{0,1\}$ is continuous. That is, that $u_g^{-1}(\{0\})$ and $u_g^{-1}(\{1\})$ are both closed. -That $u_g^{-1}(\{0\})=\{x:gx=x\}$ is closed holds for whenever $G$ acts continuously on a Hausdorff topological space. (It's false when $X$ is not assumed Hausdorff.) -The claim is that $u_g^{-1}(\{1\})=\{x:gx\neq x\}$ is closed for every $g$. -Assume otherwise: this means that there exists $x_0\in X$ such that $gx_0=x_0$ and $x_0$ is in the closure of $\{x:gx\neq x\}$. Let $V\subset X$ an open subset, maximal for the property that $V\cap gV=\emptyset$. -Claim: if $X$ is Hausdorff, then $x_0$ belongs to the closure of $V$. Indeed, let $U$ be the complement of this closure. If by contradiction $x_0\in U$, define $U'=U\cap g^{-1}U$: this is an open neighborhood of $x_0$. By assumption, there exists $x\in U'$ with $gx\neq x$. Using that $X$ is Hausdorff, one can find an open neighborhood $V'$ of $x$, with $V'\subset U'$ and $V'\cap gV'=\emptyset$. Then taking $V\cup V'$ contradicts the maximality of $V$. -But then $x_0$ also belongs to the closure of $gV$. Assuming that $X$ is extremally disconnected, this contradicts that $V$ and $gV$ are disjoint. -The latter argument shows more generally that for any continuous self-map $g$ of a Hausdorff extremally disconnected space $X$, the closed subset $\{x:g(x)=x\}$ is also open.<|endoftext|> -TITLE: Whitehead products in homotopy groups of spheres -QUESTION [12 upvotes]: Here is what I know about Whitehead products in homotopy groups of spheres: - -$[\mathrm{id}_{S^{2n}},\mathrm{id}_{S^{2n}}]$ has Hopf invariant (EDIT: $\pm$) two. -No element that survives into the stable range can be a Whitehead product, since the suspension of a Whitehead product is trivial. -If $\alpha \in \pi_m(S^n)$ with $m$ odd, then $[\alpha,\alpha]$ has order at most 2. This is because the Whitehead product is a graded Lie bracket. - -I have two questions, one open-ended and one more specific. -The open-ended question: What are some known examples of nontrivial Whitehead products in finite homotopy groups of spheres? Are there other situations, besides the ones mentioned, where Whitehead products must be trivial or have order at most 2? -The specific question: Is there an example of $\alpha \in \pi_m(S^n)$, for some $m>n$, such that $2\cdot[\Sigma\alpha,\Sigma\alpha] \neq 0$? - -REPLY [7 votes]: James proves a great number of things about the Whitehead product in his paper On the suspension sequence (though a number of results in that paper are stated in terms of cases rather than the stronger results that hold 2-locally). For example, he shows that there is a 2-local pairing -$$ -\{-,-\}: \pi_p(S^k) \times \pi_q(S^k) \to \pi_{p+q+1}(S^{2k+1}) -$$ -such that $\{\alpha, \beta\} = (-1)^{pq + k} \{\beta, \alpha\}$ and such that the composite with the "Whitehead product" map -$$ -P: \pi_{p+q+1}(S^{2k+1}) \to \pi_{p+q-1}(S^k) -$$ -that appears in the EHP sequence is the ordinary Whitehead product. In particular, since the Whitehead product is graded-commutative we get an identity $2 [\alpha, \alpha] = 0$ whenever $k$ is odd, complementing the fact that $2[\alpha, \alpha] = 0$ by graded-commutativity whenever the source degree $p$ of $\alpha$ is odd. (In particular, this shows that when $n$ is even, $2 [\Sigma \alpha, \Sigma \alpha] = 0$ for any element $\alpha$ in the homotopy groups of $S^n$.) -James also shows naturality for the EHP sequence in a sense, and this has the following consequence. Writing the element $[\Sigma \alpha, \Sigma \alpha]$ as a composite of $[id_{m+1}, id_{m+1}]: S^{2m+1} \to S^{m+1}$ with the map $\Sigma \alpha: S^{m+1} \to S^{n+1}$, James' naturality shows that -$$ -\{\Sigma \alpha, \Sigma \alpha\} = \Sigma^{m+1} \alpha \circ \Sigma^{n+1} \alpha \circ \{id_{m+1}, id_{m+1}\} = \pm 2 (\Sigma^{m+1} \alpha \circ \Sigma^{n+1} \alpha) = \pm 2 \Sigma^{n+1}(\Sigma^{m-n} \alpha \circ \alpha). -$$ -and thus that -$$ -[\Sigma \alpha, \Sigma \alpha] = \pm 2 P(\Sigma^{n+1} (\Sigma^{m-n} \alpha \circ \alpha)). -$$ -In particular, for this element to not be 2-torsion, the element $4\Sigma^{n+1} (\Sigma^{m-n} \alpha \circ \alpha)$ must not be in the kernel of $P$. However, the EHP spectral sequence tells you that this is the same as asking that the element $4\Sigma^{n+1} (\Sigma^{m-n} \alpha \circ \alpha)$ must not be in the image of the Hopf invariant map $H$. (Stably, this element is $4 \alpha^2 = (2 \alpha)^2$.) -(This is the point where I say that I've exhausted most of my knowledge of unstable theory. I don't know if there are any examples of non-2-torsion self-Whitehead squares.)<|endoftext|> -TITLE: Primes arising from permutations -QUESTION [7 upvotes]: Recently, Paul Bradley proved in arXiv:1809.01012 that for any positive integer $n$ there is a permutation $\pi_n$ of $\{1,\ldots,n\}$ such that $k+\pi_n(k)$ is prime for every $k=1,\ldots,n$ (cf. http://arxiv.org/abs/1809.01012). Motivated by this, here I pose the following question. -QUESTION: Is my following conjecture true? -Conjecture. (i) For any positive integer $n$, there is a permutation $\sigma_n$ of $\{1,\ldots,n\}$ such that $k\sigma_n(k)+1$ is prime for every $k=1,\ldots,n$. -(ii) For any integer $n>2$, there is a permutation $\tau_n$ of $\{1,\ldots,n\}$ such that $k\tau_n(k)-1$ is prime for every $k=1,\ldots,n$. -I have checked the conjecture for $n$ up to $11$. For example, -$(1, 3, 2, 9, 6, 5, 10, 11, 4, 7, 8)$ is a permutation of $\{1,\ldots,11\}$ -with -\begin{gather}1\times1+1,\ 3\times2+1,\ 2\times3+1,\ 9\times4+1,\ 6\times5+1, \ 5\times 6+1, -\\10\times 7+1,\ 11\times8+1,\ 4\times9+1,\ 7\times10+1,\ 8\times11+1 -\end{gather} -all prime, and $(3, 2, 1, 5, 4, 7, 6, 9, 8, 11, 10)$ is a permutation of $\{1,\ldots,11\}$ -with -\begin{gather}3\times1-1,\ 2\times2-1,\ 1\times3-1,\ 5\times4-1,\ 4\times5-1, \ 7\times 6-1, -\\6\times 7-1,\ 9\times8-1,\ 8\times9-1,\ 11\times10-1,\ 10\times11-1 -\end{gather} -all prime. -Remark. I also conjecture that for any integer $n>2$ there is a permutation $\pi_n$ of $\{1,\ldots,n\}$ such that the $2n$ numbers $k+\pi_n(k)\pm1\ (k=1,\ldots,n)$ are all prime. This is stronger than the twin prime conjecture. - -REPLY [11 votes]: Let us look at one "difficult" sub-case, which completely avoids the permutation issue. -With $k=n$ it is required that there exists some small $x \leq n$, such that -$nx+1$ is prime. Therefore this necessary condition is related to Linnik's theorem. -https://en.wikipedia.org/wiki/Linnik%27s_theorem -By Linnik's theorem it is known that $nx+1 \ll n^5$ exists. -(The constant 5 was proved by Xylouris, (PhD Thesis, Bonn, 2011) -\"{U}ber die Nullstellen der Dirichletschen L-Funktionen und die kleinste Primzahl in einer arithmetischen Progression. -https://bib.math.uni-bonn.de/downloads/bms/BMS-404.pdf ) -Let $p(a,n)$ denote the least prime in the arithmetic progression $a+nx$. -It is known that on GRH one has $p(a,n)\leq \varphi(n)^2 (\log n)^2$, which is just a bit larger than $n^2$. ($\varphi$ denotes Euler's totient function.) -In view of this, it seems, one would need to make considerable progress on the difficult topic -"least prime in arithmetic progression", at least in the special case of the residue class $1 \bmod n$. -As I am not aware that for $1\bmod n$ significantly better results are known than in the general case, I would assume -that even assuming GRH we do not quite get the necessary condition $x \leq n$, for this one prime! -(Even with a very strong result on least primes in progressions the above comment says nothing towards the existence of the -permutation).<|endoftext|> -TITLE: Asymptotic Expansion of Bessel Function Integral -QUESTION [6 upvotes]: I have an integral: -$$I(y) = \int_0^\infty \frac{xJ_1(yx)^2}{\sinh(x)^2}\ dx $$ -and would like to asymptotically expand it as a series in $1/y$. Does anyone know how to do this? By numerically computing the integral it appears that -$$I(y) = \frac 12 - \frac 1 {\pi y}+ \frac {3\zeta(3)}{4y^3\pi^3} + O(y^{-5}) $$ -but this is just (high precision) guesswork and I would like to understand the series analytically. - -REPLY [7 votes]: Inserting the Mellin-Barnes representation for the square of the Bessel function (DLMF), -\begin{equation} - J_{1}^2\left(xy\right)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i -\infty}\frac{\Gamma\left(-t\right)\Gamma\left(2t+3\right)}{\Gamma^2\left(t+2\right)\Gamma% -\left(t+3\right)}\left(\frac{xy}{2}\right)^{2t+2}\,dt -\end{equation} -where $-3/2<\Re (c)<0$, and changing the order of integration, one obtains -\begin{equation} - I(y)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i -\infty}\frac{\Gamma\left(-t\right)\Gamma\left(2t+3\right)}{\Gamma^2\left(t+2\right)\Gamma% -\left(t+3\right)}\left(\frac{y}{2}% -\right)^{2t+2}\,dt\int_0^\infty \frac{x^{2t+3}}{\sinh^2x}\,dx -\end{equation} -From G. & R. (3.527.1) -\begin{equation} -\int_0^\infty \frac{x^{2t+3}}{\sinh^2x}\,dx=\frac{1}{2^{2t+2}}\Gamma\left( 2t+4 \right)\zeta\left( 2t+3 \right) -\end{equation} -valid for $t>-1$. Thus we choose $-1<\Re(c)<0$ -and thus -\begin{equation} - I(y)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i -\infty}\frac{\Gamma\left(-t\right)\Gamma\left(2t+3\right)\Gamma\left( 2t+4 \right)\zeta\left( 2t+3 \right)}{\Gamma^2\left(t+2\right)\Gamma\left(t+3\right)}\left(\frac{y}{4}\right)^{2t+2}\,dt -\end{equation} -To evaluate asymptotically this integral, one can close the contour by the large left half-circle. Poles are situated at $t=-1$ and $t=-\frac{2n+1}{2}$, with $n=1,2,3\ldots$. With the help of a CAS, the first corresponding residues are: -\begin{equation} - R_{-1}=\frac{1}{2}\quad ;\quad R_{-3/2}=-\frac{1}{4\pi}\quad ;\quad R_{-5/2}=-\frac{3}{64\pi}\zeta'(-2)\quad ;\quad R_{-7/2}=\frac{15}{8192\pi}\zeta'(-4) -\end{equation} -(General expressions can probably be found, if necessary). The derivative of the Riemann Zeta function at even integer values are involved and can be simply expressed. We obtain finally -\begin{equation} - I(y)=\frac{1}{2}-\frac{1}{\pi}y^{-1}+\frac{3\zeta(3)}{4\pi^3}y^{-3}+\frac{45\zeta(5)}{32\pi^5}y^{-5}+O\left( y^{-7} \right) -\end{equation}<|endoftext|> -TITLE: The $\mathbb{Q}$-rational cuspidal group of $J_0(N)$ -QUESTION [5 upvotes]: Let $N$ be a positive integer and consider the modular curve $X_0(N)$ over $\mathbb{Q}$. Also, consider the Jacobian variety $J_0(N)$ of $X_0(N)$, which is an abelian variety defined over $\mathbb{Q}$. -Let $\mathsf{Cusp}$ denote the group of cuspidal divisors, namely, the group of divisors supported only on cusps and -let $\mathsf{Cusp}^0$ denote the group of degree-0 cuspidal divisors. Let $\mathcal{C}(N)$ denote the image of $\mathsf{Cusp}^0$ in $J_0(N)$, which is called the cuspidal group of $J_0(N)$. -By Manin and Drinfeld, the group $\mathcal{C}(N)$ is finite. Let $\mathcal{C}(N)_\mathbb{Q}$ be the $\mathbb{Q}$-rational cuspidal group of $J_0(N)$, which is defined by the subgroup of $\mathcal{C}(N)$ consisting of the elements fixed by the action of the absolute Galois group $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ of $\mathbb{Q}$. -Here is my question: -Is the group $\mathcal{C}(N)_\mathbb{Q}$ generated by the images of the degree-0 $\mathbb{Q}$-rational cuspidal divisors? -(Here, by the degree-0 $\mathbb{Q}$-rational cuspidal divisors we mean the degree-0 cuspidal divisors which are fixed by the action of $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$.) A priori, the group generated by the images of the degree-0 $\mathbb{Q}$-rational cuspidal divisors is only a subgroup of $\mathcal{C}(N)_\mathbb{Q}$. -In the paper by Ling, "On the $\mathbb{Q}$_rational cuspidal subgroup and the component group of $J_0(p^r)$" published in Israel Journal of Mathematics 99 (1997), 29--54, he says that -it is easy to see that the $\mathbb{Q}$-rational cuspidal subgroup $\mathcal{C}(N)_\mathbb{Q}$ of $J_0(N)$ is generated by divisors coming from divisors of the kind $\phi((d, N/d))P_1-(P_d)$ as $d$ runs through the positive divisors of $N$. -(This is on page 34.) Can anyone prove this statement? -(This is equivalent to my question.) - -REPLY [3 votes]: I more or less make the same claim as Ling in a paper that I wrote. I think the key thing that one needs to know is how the Galois group acts on the cusps of $X_{0}(N)$, and for this I found the book "Arithmetic on modular curves" by Glenn Stevens to be helpful. In particular, Theorem 1.3.1 from that book says that if $d | N$ and $\begin{bmatrix} c \\ d \end{bmatrix}$ represents a cusps of $X_{0}(N)$ and $\tau_{s} \in Gal(\mathbb{Q}(\zeta_{N})/\mathbb{Q})$ sends $\zeta_{N} = e^{2 \pi i / N}$ to $\zeta_{N}^{s}$, then -$$ -\tau_{s}\left(\begin{bmatrix} c \\ d \end{bmatrix}\right) = \begin{bmatrix} c \\ s' d \end{bmatrix} -$$ -where $s s' \equiv 1 \pmod{N}$. From this, it takes a little bit of thought to see that $Gal(\mathbb{Q}(\zeta_{N})/\mathbb{Q})$ acts transitively on all $\phi((d,N/d))$ cusps with "denominator" $d$. Therefore the divisor $P_{d}$ (from Ling's paper) is a single Galois orbit, and from this it follows that any element of $\mathcal{C}(N)_{\mathbb{Q}}$ is an integer linear combination of the divisors $\phi((d,N/d)) P_{1} - P_{d}$. -I feel like there ought to be a simpler pure thought reason why the answer to your first question is yes, but I can't quite see it right now.<|endoftext|> -TITLE: Is there an accessible exposition of Gelfand-Tsetlin theory? -QUESTION [24 upvotes]: I'm hoping to start an undergraduate on a project that involves understanding a bit of Gelfand-Tsetlin theory, and have been tearing my hair out looking for a good reference for them to look at. Basically what I would want is something at the level of Vershik-Okounkov - A new approach to the representation theory of symmetric groups. 2 (or the book Ceccherini-Silberstein, Scarabotti, and Tolli - Representation theory of the symmetric groups based on it) but for finite dimensional representations of $\mathrm{GL}_n$. I feel like such a book or at least some expository notes should exist, but I have had zero luck finding any. - -REPLY [15 votes]: Good question. I've found this to be a difficult subject to get into myself. An abstract approach can seem arcane, but concrete constructions can be complicated and messy. You might try the paper Hersh and Lenart - Combinatorial construction of weight bases: The Gelfand–Tsetlin basis as a starting point. They take a concrete approach, which has the advantage that you can start computing with small examples relatively quickly. A disadvantage is that your student might miss the big picture of how all this fits into the general representation theory of classical Lie algebras. For that, perhaps the work of Molev, such as Molev - Gelfand–Tsetlin bases for classical Lie algebras, might be helpful.<|endoftext|> -TITLE: Is a measurable solution continuous? -QUESTION [8 upvotes]: Let $f: \mathbb R\to \mathbb R$ be a Borel measurable function. Suppose that for each $q\in \mathbb Q$, the function $f(q+x)-f(x)$ is continuous on $\mathbb R$. Is it true that there is a continuous function $g: \mathbb R\to \mathbb R$ such that $f(x)=g(x)$ for Lebesgue almost every $x\in \mathbb R$? -If the answer to the above question is negative, how about assuming in addition that $f(x+m)=f(x)$ for all $m\in -\mathbb Z$. Namely, we ask the same question for a function defined on $\mathbb R/\mathbb Z$. - -REPLY [7 votes]: Consider the 1-periodic function $f$ with Fourier series $$\sum n^{-1}\cos (2\pi n! x).$$ -Note that it satisfies your property, since all but finitely many summands are $h$-periodic for any rational $h$. On the other hand, if it were a Fourier series of a continuous function $F$, its partial sums would be Cesàro convergent to the values of $F$, but for $x=0$ the sum of series is infinite (both Cesàro or usual). -Well, strictly speaking the above function $f$ is defined only on a set of full measure (not pointwise) and $f(x+h)-f(x)$ is equivalent to a continuous function, but not genuine pointwise continuous. Is it ok for you? If not, you may carefully correct it on a set of measure 0. For example, define $f(x)$ as a sum of above series when it converges (by Carleson's theorem the series converges almost everywhere to the initial function). After that it remains to define the values of $f$ on the exceptional set. This exceptional set has measure 0 and is invariant under shifts by rational numbers, we must force the difference $f(x+h)-f(x)$ take the necessary values $f_N(x+h)-f_N(x)$, where $N$ is chosen so that $N!h$ is integer, and $f_N$ denotes the corresponding finite sum. This is possible since these values agree in a natural sense.<|endoftext|> -TITLE: reverse mathematics of the Lebesgue measurability of analytic sets -QUESTION [13 upvotes]: Can the fact that all analytic sets are Lebesgue measurable be proven in $Z_2$, or in some weak subsystem such as $\Pi^1_1\text{-CA}_0$? Conversely, can certain set existence axioms be derived from the assumption that all analytic sets are Lebesgue measurable? - -REPLY [6 votes]: I believe $\Delta^1_2$-CA$_0$ does indeed suffice; however, I don't see a way to pull this down to $\Pi^1_1$-CA$_0$. -This is contrary to my previous claim; my error was with respect to the strength of the relevant choice principle. - -Let me briefly outline the classical proof: - -First, we show a version of Lebesgue regularity: that for any analytic set $A$, the outer measure of $A$ exists and that there is a $G_\delta$ (and hence measurable) set $B$ with $A\subseteq B\subseteq cl(A)$ and $m(X)=0$ for all measurable $X\subseteq B\setminus A$. -Next, let $A$ be an analytic set. Fixing a continuous function $f$ whose image is $A$, we let $A_\sigma$ be the image of $f$ restricted to the set of reals beginning with $\sigma$ (for $\sigma\in\omega^{<\omega}$) and we let $B_\sigma$ be the set corresponding to $A_\sigma$ per the bulletpoint above. -Looking to the $B_\sigma$s for a moment, it's not hard to show that for all $\sigma\in\omega^{<\omega}$, the set $$Z_\sigma:=B_\sigma\setminus\bigcup_{n\in\omega}B_{\sigma n}$$ is measurable and is contained in $B_\sigma\setminus A_\sigma$. But from this it follows that $Z_\sigma$ is null. -Elementary set-juggling shows that $$B\setminus A\subseteq\bigcup_{\sigma\in\omega^{<\omega}}Z_\sigma.$$ Since the union of countably many null sets is null, we have that $B\setminus A$ is measurable; since $B$ is measurable, this implies that $A$ is also measurable, and so we're done. - - -Now, translating this into the context of reverse mathematics, we run into three important points: - -We need to talk about codes for sets of reals, rather than sets of reals themselves. Luckily, we're basically looking here at the first couple levels of the Borel hierarchy and at analytic sets, so we have reasonable coding notions. -We need to be able to do basic "set-juggling" with these codes. But this winds up being straightforward. -Finally, we need to be able to prove basic facts about Lebesgue measure: in particular, that the measurable (coded) sets are closed under countable Boolean combinations and that the union of countably many null sets is null. This is the difficult bit: it involves choosing "nearly optimal" covers in a concrete way: e.g. to show that $\bigcup_{n\in\omega} X_n$ is null if each $X_n$ is null, we really want a sequence $(\mathcal{O}_{n,k})_{n,k\in\omega}$ such that $\mathcal{O}_{n,k}$ is an open cover of $X_n$ with total measure ${1\over k+1}$. When the $X_n$s are given uniformly by analytic codes - as in our situation here - the set of legal choices for $\mathcal{O}_{n,k}$ is a nonempty $\Pi^1_1$ set uniformly in $n,k$. - -So the question now is: how strong is $\Pi^1_1$ choice? Unfortunately (and contra my original claim) it turns out that it is rather strong indeed. In fact, it is equivalent to $\Delta^1_2$-CA$_0$; this follows from Theorem VII.$6.9(1)$, page $298$, in Simpson's book together with Carl's observation that $\Pi^1_1$-choice is equivalent to $\Sigma^1_2$-choice. -So the above argument gives a proof of the result in a weak-but-not-too-weak subsystem of Z$_2$.<|endoftext|> -TITLE: Decide if a matrix is transposable -QUESTION [11 upvotes]: A matrix $M$ is called transposable if it can be transformed into its transpose $M^t$ via row and column permutations. - -Is there an efficient a way/algorithm to decide if a given matrix is - transposable and gives us a certificate? - -REPLY [10 votes]: There are polynomial-time reductions from your problem to Graph Isomorphism and vice-versa. -As a quick definition, when I speak of 'subdividing' an edge, I mean to replace each edge $u, v$ with a path $u, w, v$ where $w$ is a new 'midpoint' vertex. -Transposable $\rightarrow$ GI: -Convert your $n \times n$ matrix into a bipartite graph $H$ with coloured edges (using up to $n^2$ colours). Then let $G_1$ be the graph obtained from $H$ by adjoining an extra vertex connected to each element of the vertex-class $V_1$, and analogously for $G_2$. Then the transposability of the original matrix is equivalent to finding an isomorphism between $G_1$ and $G_2$. -If you don't like colouring edges, you can subdivide each edge and hang a motif (let's say, a large complete graph whose number of vertices encodes the colour of the edge) from the newly-created midpoint. -GI $\rightarrow$ Transposable: -Suppose $G_1$ and $G_2$ are two graphs, such that we wish to determine whether they're isomorphic. This question is trivial if $G_1$ and $G_2$ have different number of vertices/edges, and also trivial if one of them is a complete graph, so assume they're incomplete graphs with equal numbers of vertices ($n$) and equal numbers of edges ($m$). -We now subdivide each edge in each of $G_1$ and $G_2$, obtaining two bipartite graphs $B_1, B_2$ each with $m + n$ vertices. We adjoin another vertex in the 'edge' class of each bipartite graph, connected to every vertex in the 'vertex' class, so that each $B_i$ now has vertex-classes of sizes $m + 1$ and $n$. Observe that each $B_i$ is connected, and has a distinguished vertex $v_i$. -Then we take the disjoint union of $B_1$ and $B_2$, and connect -$v_1$ and $v_2$ by an edge. This graph is still bipartite, and has $m + n + 1$ vertices in each class. Its biadjacency matrix $M$ is a $0-1$ matrix which I claim is transposable if and only if $G_1$ and $G_2$ were isomorphic. Clearly the 'if' direction is true, so let's tackle the 'only if'. -Suppose $M$ is transposable. This induces an automorphism of the bipartite graph which exchanges the two parts. The distinguished vertices $v_1$ and $v_2$ must be exchanged (because they're the endpoints of the only bridge which, when cut, divides the graph into two equally-sized connected components). But this means that the automorphism gives us an isomorphism between the bipartite graphs $B_1$ and $B_2$, which are isomorphic only if the original graphs $G_1$ and $G_2$ were. -The result follows.<|endoftext|> -TITLE: The structure of complex cobordism cohomology of the Eilenberg-Maclane spectrum -QUESTION [14 upvotes]: Let $MU$ be the complex bordism spectrum and let $H\mathbb{Z}$ be the Eilenberg-Maclane spectrum. - -Is it know what the structure of the complex cobordism cohomology $MU^{*}(H\mathbb{Z})$ is? - -EDIT: What if instead $H\mathbb{Z}$, one consider $H\mathbb{Z}/(p)$ for a prime $p$? - -REPLY [16 votes]: One can prove that $\mathrm{Map}(H\mathbf{F}_p,MU)$ is contractible. We know that $H\mathbf{F}_p$ is dissonant (Theorem 4.7 of Ravenel's "Localization with Respect to Certain Periodic Homology Theories"), but $MU$ is harmonic (Theorem 4.2 of that paper). Since dissonant spectra (resp. harmonic spectra) are by definition $E$-acyclic (resp. local) for the spectrum $E = \bigvee_p E_p$, where $E_p = \bigvee_{0\leq n<\infty} K(n)$, the claim follows. (I just realized that this is Corollary 4.10 of Ravenel's paper.) -It is, however, not the case that $\mathrm{Map}(H\mathbf{Z},MU)$ is contractible. (What I wrote previously was incorrect.) The spectrum $H\mathbf{Z}$ is not dissonant, so we cannot immediately apply the above argument. Since $MU$ is harmonic, there is, however, an equivalence between $\mathrm{Map}(H\mathbf{Z},MU)$ and $\mathrm{Map}(L_E H\mathbf{Z},MU)$. We therefore need to understand $L_E H\mathbf{Z}$. By the discussion at this question, we can conclude that $L_{E_p} H\mathbf{Z} \simeq H\mathbf{Q}_p$. It therefore suffices to understand $MU^\ast(H\mathbf{Q})$. But $H\mathbf{Q}$ is the colimit of multiplication by $2,3,5,7,\cdots$ on the sphere, so $MU^\ast(H\mathbf{Q})$ admits a description in terms of $\lim^0$ and $\lim^1$ of multiplication by $2,3,5,7,\cdots$ on $\pi_\ast MU$. In particular, the $\lim^1$ term is $\mathrm{Ext}^1_\mathbf{Z}(\mathbf{Q,Z}) \cong \widehat{\mathbf{Z}}/\mathbf{Z}$.<|endoftext|> -TITLE: Looking for “Set theory for a small universe” by Ketonen -QUESTION [10 upvotes]: In the paper Partition theorems for systems of finite subsets of integers, Pudlák and Rödl show a Ramsey-type result. The main feature of this result is that the sizes of sets in such systems are not fixed in advance (as in Ramsey's original theorem and Erdős and Rado's generalization to arbitrary partitions of $[\mathbb{N}]^k$). -The proof of the main theorem is by induction on $\omega_1$, and the authors say that the idea of doing that came from an unpublished manuscript by Ketonen, Set theory for a small universe, I. The Paris-Harrington Axiom. -Does anyone here has access to this manuscript? -Thanks in advance. - -REPLY [6 votes]: A pdf version of the 27-page manuscript by Jussi Ketonen, "Set Theory for a Small Universe, I. The Paris-Harrington Axiom", is here on Google Drive. The date of manuscript is perhaps 1979, or 1978. -The left margin in my hard-copy is not the best: on many lines, the first letter is truncated. But it is still readable. The scanned version captures everything in the hard-copy. -From Ketonen's introduction: - -"We will give a purely combinatorial framework for dealing with set-theoretic relations of the Harrington-Paris type; the situation will turn out to be highly analogous to the theory of large cardinals. For example the notion of 'largeness' corresponds to 'Mahloness' and the Harrington-Paris axiom transforms into a statement concerning the n-subtle cardinals of Baumgartner. ... [And conversely,] the various known large cardinal axioms naturally suggest new number-theoretical statements."<|endoftext|> -TITLE: A mysterious connection between primes and squares -QUESTION [19 upvotes]: Motivated by two previous questions of mine (cf. Primes arising from permutations and Primes arising from permutations (II)), here I ask a curious question which connects primes with squares. -QUESTION: Is my following conjecture true? -Conjecture. Let $n$ be any positive integer, and let $S(n)$ denote the number of permutations $\tau$ of $\{1,\ldots,n\}$ with $k^4+\tau(k)^4$ prime for all $k=1,\ldots,n$. Then $S(n)$ is always a positive square. -Via a computer, I find that -$$(S(1),\ldots,S(11))=(1,1,1,4,4,4,4,64,16,144,144).$$ -For example, $(1,3,2)$ is a permutation of $\{1,2,3\}$ with -$$1^4+1^4=2,\ \ 2^4+3^4=97,\ \ \text{and}\ 3^4+2^4=97$$ -all prime. - -REPLY [25 votes]: The fact on squareness is easy. -If $n$ is odd, among the sums of the form $i^4+\tau_i^4$ there is an even one; it should equal to 2, hence $\tau_1=1$. All other odd numbers should map to even ones, and vice versa. Hence the number of permutations under the question is the square of the number of bijections from odds (from 3 to $n$) to evens satisfying the same constraint. -If $n$ is even, we cannot have $\tau_1=1$, otherwise there would be another even sum. Hence a similar reasoning applies. -It remains to show that there exists at least one such permutation.. (Notice that for the constraint on $i+\tau_i$ this would follow easily from the Bertrand postulate...)<|endoftext|> -TITLE: Is there a version of the "infinitary" disjunctive normal form theorem for topoi and slice categories? -QUESTION [8 upvotes]: According to nLab - -Proposition 2.3. A complete Boolean algebra is completely distributive iff it is atomic (a CABA), i.e., is a power set as a Boolean algebra. - -This is basically an infinitary version of the statement that full disjunctive normal forms exist and are unique. -For example, let $B$ denote the boolean algebra freely generated by $X = \{x,y\}$. Since $X$ is finite, hence $B$ is a completely distributive boolean algebra. Hence by (a slight refinement of) the above theorem, $B$ is isomorphic to the powerset of its atoms. -The atoms of $B$ are $$\{x \wedge y, \neg x \wedge y, x \wedge \neg y, \neg x \wedge \neg y\}.$$ By the above theorem, every element of $B$ can be expressed as a join of these four elements in a unique way. In other words, we've obtained the existence and uniqueness of full disjunctive normal form as a special case of Proposition 2.3 above. -I'd like to know if there's anything like this in which power sets are replaced by slice categories and, and complete Boolean algebras are replaced by topoi (or some variant thereof). The theorem should read something like: - -A bicomplete locally cartesian closed category satisfying $P$ satisfies $Q$ iff it is a slice category. - -where $P$ is a technical condition and $Q$ is the condition "completely distributive" or some variant on that, like the statement that every epimorphism splits (which is a version of the axiom of choice). - -Question. Do there exist conditions $P$ and $Q$ that make this true, and if so, what are they? - -REPLY [3 votes]: Edit: I just saw that you were willing to take "every epi split" as an assumption. In this case there is a considerably more classical approach: -Lemma A topos in which every epimorphism split is equivalent to the topos of sheaves on a complete boolean algebra. -This a relatively classical result. If I remember well it can be found in MacLane and Moerdijk "Sheaves in geometry and logic". SO if you simply add to this condition the usual "complete distributiviy" assumption on the lattice of sub-object of the terminal you can apply the result you quoted and you get a topos of sheaves over a complete atmoic boolean algebra, i.e. a slice of the category of sets. -I'm leaving the original answer below, which I think is also interesintg. - -I'm proposing the following statement. I'm relatively sure it is true. But I have to admit that a full proof might require a bit of work to fill all the gaps. What I mean is that if you really need to use it somewhere, I wouldn't quote this post as a proof ! -Claim: A completely distributive boolean topos is the topos of $G$-set for $G$ a groupoid. -So $G$-set for $G$ a groupoid is the next best thing to slice of the category of sets: it is a category that is locally a slice of the category of sets. Also the result mentioned in the question can essentially be recovered (without too much work) as a special case of this one. -I' don't think I need to explain what is a Boolean Topos, nor why this is a reasonable candidate to repalce boolean algebras. The tricky notion is "completely distributive". -For a regular cardinal $\kappa$, $\kappa$-distributive Grothendieck topos or $\kappa$-topos, is a $\kappa$-exact localization of a presheaf category. Usual formal arguement should implies that this is equivalent to the assumption that the "colim" functor from the category of small presheaves of the topos $Psh(\mathcal{T}) \rightarrow \mathcal{T}$ commutes to $\kappa$-small limits. -Note that commutation of this functor to finite limits is a known way to encode all of the usual Giraud's axioms of a topos except accessibility, see for example Lack-Garner -By totally distributive I mean $\kappa$-distributive for all $\kappa$, which seems to also be the definition of the papers mentioned by Mike Shulman in the comments. -$\kappa$-distributivity can also be rephrased in a way that is a little closer to distributivity of lattices: C.Espindola has some papers on infinitary categorical logic where he studied a bit these $\kappa$-toposes. He has observed that this condition of $\kappa$-distributivity on a topos can be rephrased as, informally: -"a $\kappa$-small transfinite composition of covering sieve is a covering sieve". -this has to be formalized using trees. i.e. given a $S$ a tree where each branch has height less than $\kappa$, and $D:S^{op} \rightarrow \mathcal{T}$ a functor such that for each node of the tree $s$, the childrens of $s$ form a covering of D(s) (a jointly epimorphic familly). -For each branch $b$ of the tree, you can form its limit $S_b$, then his condition is that for each such diagram, the $S_b \rightarrow S_0$ (with $S_0$ the root) form a covering of $S_0$. -For a topos this is equivalent to $\kappa$-distributivity. -As we are assuming this for all $\kappa$ there might be further simplification on this condition (and one can maybe get rid of these trees in this case) But I'm not sure about it at this point. -Now: -Claim 2 A totally distributive Grothendieck topos is a presheaf topos. -This seems to follow from lemma 1 in the second paper linked by Mike, though it is late and I have some trouble following this paper that I discovered a few minutes ago only. The real reason I believe this claim is true is because of an unpublished result of C.Espindola in which I'm relatively confident and which implies this. -It is then a classical result (exercise ? ) that a presheaf topos $Prsh(I)$ is boolean if and only if $I$ is a groupoid. Hence the first claim follows from this second one.<|endoftext|> -TITLE: Sum of the reciprocals of radicals -QUESTION [16 upvotes]: Recall that the radical of an integer $n$ is defined to be $\operatorname{rad}(n) = \prod_{p \mid n } p$. -For a paper, I need the result that -$$\sum_{n \leq x} \frac{1}{\operatorname{rad}(n)} \ll_\varepsilon x^{\varepsilon} \tag{$*$},$$ -for all $\varepsilon > 0$. I have a proof of this using complex analysis and Perron's formula, but this seems a bit overkill given that I'm looking for a weak upper bound for a problem in elementary number theory. - -Does anyone know of a short elementary proof of the bound $(*)$? Or better yet, a reference? - -REPLY [9 votes]: de Bruijn studies this sum in "On the number of integers $\le x$ whose prime factors divide $n$", which was published in a 1962 volume of the Illinois J. Math; see -https://projecteuclid-org.proxy-remote.galib.uga.edu/euclid.ijm/1255631814 -He proves there (see Theorem 1) that $$\sum_{n \le x} \frac{1}{\mathrm{rad}(n)} = \exp((1+o(1)) \sqrt{8\log{x}/\log\log{x}}),$$ as $x\to\infty$. Of course, this implies the $O(x^{\epsilon})$ bound you were after. However, his proof (which uses a Tauberian theorem of Hardy and Ramanujan) is not as elementary as some others that have been suggested here (but gives a more precise result). - -REPLY [2 votes]: sIt seems that this argument hasn't been presented yet, so I might as well include it. -We can sort the integers $n \in [1, X]$ by their radicals, which is necessarily a square-free integer $m$. Thus we have -$$\displaystyle \sum_{n \leq X} \frac{1}{\text{rad}(n)} = \sum_{\substack{m \leq X \\ m \text{ square-free}}} \frac{1}{m} \sum_{\substack{n \leq X \\ \text{rad}(n) = m}} 1.$$ -Now, $\text{rad}(n) = m$ if and only if $p | n \Rightarrow p | m$. If we write $m = p_1 \cdots p_k$, then -$$\displaystyle \sum_{\substack{n \leq X \\ \text{rad}(n) = m}} 1 = \#\{(x_1, \cdots, x_k) : x_i \in \mathbb{Z} \cap [0,\infty), p_1^{x_1} \cdots p_k^{x_k} \leq X/m\}.$$ -The inequality defining the right hand side is equivalent to -$$\displaystyle x_1 \log p_1 + \cdots + x_k \log p_k \leq \log(X/m),$$ -and this is just counting integer points with non-negative entries bounded by a simplex, and it is easy to see that -$$\displaystyle \# \{(x_1, \cdots, x_k) : x_1 \log p_1 + \cdots + x_k \log p_k \leq \log(X/m)\} \ll \frac{\log(X/m)}{\prod_{1 \leq i \leq k} \log(p_i)} \ll \log X.$$ -EDIT: This last step is wrong, but it can be fixed. Indeed, we can arrange the $p_i$'s so that $p_1 < p_2 < \cdots < p_k$. It then follows from Davenport's lemma that -$$\displaystyle \# \{(x_1, \cdots, x_k) : x_1 \log p_1 + \cdots + x_k \log p_k \leq \log(X/m)\} = O \left(\sum_{i=0}^k \frac{(\log X/m)^{k-i}}{\prod_{1 \leq j \leq k-i} \log p_i} \right).$$ -It then follows that -$$\displaystyle \sum_{n \leq X} \frac{1}{\text{rad}(n)} \ll \sum_{\substack{p_1 < \cdots < p_k \\ p_1 \cdots p_k \leq X}} \sum_{i=0}^k \frac{(\log X)^{k-i}}{\prod_{1 \leq j \leq k -i} p_i \log p_i}.$$ -From here I think it is possible to get the bound $O_\epsilon(X^\epsilon)$, but it requires a somewhat more refined analysis on the interaction between the number of primes and the size of the primes.<|endoftext|> -TITLE: Hyperbolic $3$-manifold groups that embed in compact Lie groups -QUESTION [9 upvotes]: Is there a closed hyperbolic $3$-manifold whose fundamental group is isomorphic to a subgroup of some compact Lie group? -It is known that every surface group can be embedded into any semisimple connected Lie group. -I would be interested in similar results for fundamental groups of closed nonpositively curved manifolds of any dimension, but this is probably much harder. All I know is how to embed the fundamental groups of tori, closed surfaces, and their products. - -REPLY [19 votes]: All closed hyperbolic 3-manifold groups embed into a compact Lie group. -To prove this, note first of all that given a hyperbolic 3-manifold $M$, it suffices to show that a finite-index subgroup $G\leq \pi_1(M)$ of index $m$ embeds into a compact Lie group. Then the representation $\rho: G\hookrightarrow O(n)$ will induce a represenation $Ind_G^{\pi_1(M)} \rho : \pi_1(M) \hookrightarrow O(nm)$. -Now, by the proof of the virtual Haken conjecture, $\pi_1(M)$ has a finite-index subgroup $G$ which is the fundamental group of a special cube complex, which implies that $G$ embeds into a right-angled artin group $A$, and hence into a right-angled Coxeter group $\Gamma$. -Finally, right-angled Coxeter groups have faithful embeddings into $O(n)$. This follows from a result of Vinberg, which gives a faithful linear action by reflections on $\mathbb{R}^n$. We'll review this construction following section 7 of this paper. -Fix a right-angled Coxeter group -$$\Gamma = \langle \gamma_1,\dots,\gamma_k ~|~ (\gamma_i \gamma_j)^{m_{i,j}} = 1\quad \forall i,j\rangle,$$ -where $m_{i,i}=1$ and $m_{i,j}\in\{ 2,\infty\}$ for all $i\neq j$. -For $t \in \mathbb{R}$, the matrix $M_t=(M_t(i,j))_{1\leq i,j\leq k}$ where -$$M_t(i,j) = \left\{ \begin{array}{cl} -1 & \text{if }m_{i,j}=1, \text{ i.e. $i=j$},\\ -0 & \text{if }m_{i,j}=2,\\ --t & \text{if }m_{i,j}=\infty -\end{array}\right.$$ - defines a symmetric bilinear form $\langle\cdot,\cdot\rangle_t$ on $\mathbb{R}^k$. -Note that $\mathrm{det}(M_t)$ is a nonzero polynomial in $t$ (take $t=0$), hence it is nonzero outside of some finite set $F$ of exceptional values of $t$. -For any $t\in \mathbb{R}-F$, the form $\langle\cdot,\cdot\rangle_t$ is nondegenerate. -Define the representation $\rho_t : \Gamma\to\mathrm{Aut}(\langle\cdot,\cdot\rangle_t)\leq \rm{GL}(k,\mathbb{R})$ by -$$\rho_t(\gamma_i) : v \mapsto v - 2\langle v, e_i \rangle_t \, e_i $$ -for all $i$. Each generator is a reflection in a hyperplane perpendicular to $e_i$ with respect to the metric $\langle\cdot,\cdot\rangle_t$. -For $t>1$, the convex cone -$$\widetilde{\Delta}_t = \{ v\in\mathbb{R}^k ~|~ \langle v, e_i\rangle_t \leq 0 \ \,\forall i\}$$ -descends to a convex polytope $\Delta_t$ in an affine chart of $\mathbb{P}(\mathbb{R}^{k})$. -By Theorem 2 of Vinberg, the representation $\rho_t$ is discrete, faithful, the set $\rho_t(\Gamma)\cdot\Delta_t$ is convex in $\mathbb{P}(\mathbb{R}^k)$, and the action of $\Gamma$ on the open set -$$\mathcal{U}_t:=\mathrm{Int} \left ( \rho_t(\Gamma)\cdot\Delta_t \right )$$ is properly discontinuous. -Now, let $t$ be close to $0$ and transcendental, so that $M_t$ is positive definite. Then $\rho_t : \Gamma\to \mathrm{Aut}(\langle\cdot,\cdot\rangle_t) \cong O(k)$ is faithful (it is Galois conjugate to a representation for some transcendental $t>1$). -Remark: I think the first step of taking an induced representation can be eliminated. If we assume $G\lhd \pi_1(M)$, and $K=\pi_1(M)/G$, then one may embed $G$ into a right-angled Coxeter group which admits an action of $K$ by permuting its generators, and so that the embedding of $G$ is $K$-equivariant. Then $\pi_1(M)$ should embed in this Coxeter group extended by $K$, which clearly still also has a faithful representation. -Remark 2: Regarding your second question, as should be clear from the discussion, this holds for any cubulated hyperbolic group. This includes uniform arithmetic hyperbolic lattices of simple type, and the examples of Gromov-Piatetskii-Shapiro.<|endoftext|> -TITLE: Positive Ricci curvature on fiber bundles -QUESTION [10 upvotes]: My advisor and I are working on Ricci curvature and an anonymous referee pointed out the following conjecture: - -Let $F\hookrightarrow M\stackrel{\pi}{\to}B$ be a fiber bundle from a compact manifold $M$ with fiber $F$, compact structure group $G$ and base $B$. Suppose that: - i) $B$ has a metric of positive Ricci curvature; - ii) $F$ has a $G$-invariant metric of positive Ricci curvature. - Then $M$ carries a metric of positive Ricci curvature. - -Both the referee and us are not sure if it is in literature or not. It happens that the conjecture is easily proved using classical arguments. -Therefore, we would like to ask if someone knows a reference, or if the conjecture is well known to be true among specialists. -Partial results we could find are: -1) (Nash) https://projecteuclid.org/euclid.jdg/1214434973 -2) W. A. Poor, Some exotic spheres with positive Ricci curvature, Math. Ann. 216 (1975) -245-252. - -REPLY [4 votes]: This conjecture is already proved in -Gromoll, Detlef; Walschap, Gerard, Metric foliations and curvature, Progress in Mathematics 268. Basel: Birkhäuser (ISBN 978-3-7643-8714-3/hbk). viii, 174 p. (2009). ZBL1163.53001. -(page 100, Theorem 2.7.3).<|endoftext|> -TITLE: Serre spectral sequence degeneration in homology vs cohomology -QUESTION [6 upvotes]: Let $\pi\colon E \rightarrow B$ be a fiber bundle with fiber $F$. I am not assuming that $B$ is simply-connected. We then have Serre spectral sequences in both rational homology and rational cohomology: -$$E^2_{pq} \cong H_p(B;H_q(F;\mathbb{Q})) \Longrightarrow H_{p+q}(E;\mathbb{Q})$$ -$$E_2^{pq} \cong H^p(B;H^q(F;\mathbb{Q})) \Longrightarrow H^{p+q}(E;\mathbb{Q})$$ -Since $B$ is not simply connected, the coefficient systems have to be regarded as local (twisted) coefficient systems on $B$. Assume that I know that one of these spectral sequences degenerates at the second page. Does it follow that the other one does as well? - -REPLY [3 votes]: The answer is yes, and this follows from the comments, but it's worth spelling out the (non-obvious) details. -By Will Sawin's comment, the claim will follow if we can show that -$$ -\dim_\mathbb{Q} H^p(B;H^q(F;\mathbb{Q})) = \dim_\mathbb{Q} H_p(B;H_q(F;\mathbb{Q})) -$$ -for all $p,q\ge0$. The Universal Coefficient Theorem does not hold with twisted coefficients. However, note that the vector spaces on the right are by definition the homology groups of the chain complex -$$ -C_*(\widetilde{B})\otimes_\pi H_q(F;\mathbb{Q}), -$$ -with differential induced by that of the chain complex $C_*(\widetilde{B})$ of the universal cover. Here $\pi=\pi_1(B)$ and we are taking tensor product of $\mathbb{Z}\pi$-modules. -Following archipelago's comment, let's see what happens if we dualize this chain complex by taking Hom into the rationals. By the tensor-hom adjunction there are natural isomorphisms -$$ -{\rm Hom}_\mathbb{Z}\big(C_*(\widetilde{B})\otimes_\pi H_q(F;\mathbb{Q}),\mathbb{Q}\big)\cong {\rm Hom}_\pi\big(C_*(\widetilde{B}),{\rm Hom}_\mathbb{Z}(H_q(F;\mathbb{Q}),\mathbb{Q})\big) \cong {\rm Hom}_\pi\big(C_*(\widetilde{B}), H^q(F;\mathbb{Q})\big),$$ -and so we get precisely the cochain complex whose cohomology gives the vector spaces on the left. Since everything in sight is a rational vector space, the usual algebraic Universal Coefficient Theorem implies the claim.<|endoftext|> -TITLE: The Locus of Complete Intersection Points -QUESTION [10 upvotes]: Let $X$ be an algebraic variety over an algebraically closed field. Consider the two subsets $X_0\subseteq X_1 \subseteq X$: -$$X_0 = \{a\in X| a \mbox{ is a scheme-theoretic complete intersection in }X\},$$ -$$X_1 = \{a\in X| a \mbox{ is a set-theoretic complete intersection in }X\}.$$ -Question: what do we know about these sets? -I am looking for any positive information, possibly for some restricted classes of varieties. Here are some precise questions: -(1) Is $X_k$ open, closed, locally closed, constructible, etc.? -(2) What are the varieties with $X_k=\emptyset$? -(3) What are the varieties with $X_k=X$? - -REPLY [5 votes]: This should be difficult in general. However there are some easy remarks to get going: -First, if $p$ is a CI point, then $X_p$ is regular. That is because when you localize, the number of generators can only drops, and it is still have to be at least $n=\dim X$. So they are equal. -Now let's try to answer number 3), when $X_0=X$? The above remark says that $X$ is non-singular. But it is more, it says that the Chow group of points is trivial. -Consider $X$ projective. Clearly the property $X_0=X$ depends on the embedding. For example, with $P^1= \text{Proj} \ k[x,y,z]/(xy-z^2)$, the point $(x,z)$ is not CI. So we just consider $X= \text{Proj}\ S/I$ with $S=k[x_0,...,x_d]$. Then a point $p$ in $\text{Proj} S$ is defined by $d$ linear forms. Modulo $I$, the number of generators drops to $n$, so $I$ must contains $d-n$ forms, and by dimension reasons, $I$ is generated by those. So $X$ must be $P^n$. (indeed, here we only needs to assume that $X_0$ is non-empty, so this also answers Question (2)). -It is harder when $X$ is affine. Except in dimension one, then we are looking for a smooth affine curve with trivial Picard group, so it must be rational curve. -I don't know much about the sCI case, or when $X_1=X$. If $X$ is smooth, we are forcing the Chow group of points to have rank one, and this should be restrictive. -As for number (1), there is a paper by Weibel, where he conjectured that for affine $X$, the set $X_0$ is a countable union of closed subsets, and solved it for dimension at most $3$.<|endoftext|> -TITLE: Show that if $p\neq 2$, then $\mathbb{Z}_p$ cannot act freely on $\mathbb{C}P^n$ -QUESTION [5 upvotes]: If $p\neq 2$, then the cyclic group $\mathbb{Z}_p$ has no free continuous action on $\mathbb{C}P^n$. My question is how to prove the above fact using Leray-Serre spectral sequence associated to the Borel fibration $ \mathbb{C}P^n\hookrightarrow X_{\mathbb{Z}_p}\rightarrow B_{\mathbb{Z}_p}$. -From the Euler Characteristic argument, $p$ divides $n+1$. Also, $\pi_1(B_{\mathbb{Z}_p})={\mathbb{Z}_p}$ acts trivially on $H^*(\mathbb{C}P^n;\mathbb{Z}_p)$ by using Lefschetz fixed point theorem. $H^*(\mathbb{C}P^n;\mathbb{Z}_p)=\mathbb{Z}_p[b]/\langle b^{n+1}\rangle$ and $H^*(\mathbb{Z}_p;\mathbb{Z}_p)=\bigwedge(s)\otimes\mathbb{Z}_p[t]$. So the only possibility is $d_3(b)=st$. It follows $d_3(sb)=0$ and $d_3{(tb)}=st^2$. After that, I am unable to deduce any contradiction. -Thank you so much in advance. - -REPLY [9 votes]: Consider the cohomology with $\mathbb{Z}$ coefficients (and reduce the 0-th term modulo $p$ to get uniform description of it). Then we have a spectral sequence starting from $\mathbb{F}_p[x,y]/x^{n+1}$ with $deg(x)=deg(y)=2$ and converging to $H^*(\mathbb{C}P^n/\mathbb{Z}_p,\mathbb{Z})$. Since the $E^2$ term is concentrated in even degrees, the spectral sequence degenerates at the $E^2$-term, and so $H^*(\mathbb{C}P^n/\mathbb{Z}_p)$ has arbitrarily high non-zero cohomologies. But is is a finite dimensional manifold (being the quotient of a manifold by a free action), and this is a contradiction.<|endoftext|> -TITLE: Tensor product of bimodules -QUESTION [6 upvotes]: Im not sure whether this question is appropriate for MO, but I do not have much experience with bimodules (or I forgot many things). -Let $A$ be a finite dimensional (connected) algebra over a field $k$ and $M$ an indecomposable $A$-bimodule (finite dimensional prefered). For simplicity we can assume that $A$ is even a quiver algebra and maybe that the field is algebraically closed in case this is needed. - -Is $M^{\otimes n}$ also indecomposable (or zero) for any $n \geq 1$ as an $A$-bimodule? If not, is it true under some extra conditions? - What about when $M=D(A)$ ? (it seems to be this should be true at least for acyclic quiver algebras) -What can be said about the endomorphism ring of - $M^{\otimes n}$ as an $A$-bimodule when the endomorphism ring of $M$ is known? - -Here the tensor product is always over the algebra $A$. You can also give an answer for general rings $A$ and modules $M$ but I have no feeling whether this question is interesting or trivial in such generality. - -REPLY [3 votes]: Let $Q=( 2_{\circlearrowleft c}\xleftarrow a 1 \xrightarrow b 3_{\circlearrowleft d})$, that is, 3 vertices with 4 arrows, one loop at vertex 2 and 3 and two arrows from vertex 1 to 2 and 3. Consider the relations $\{ac, c^2, d^2, bd\}$ on $Q$. Define $A = \mathbb{Z}_7Q/\langle ac, c^2, d^2, bd\rangle$. -Let $D(A)$ be the dual of $A$ as a bimodule over $A$. Then given that QPA2 doesn't do any mistakes, $DM^{\otimes 2}$ is decomposable. Here are the computations done in QPA2: -gap> Q := RightQuiver( "Q(3)[a:1->2,b:1->3,c:2->2,d:3->3]" ); -Q(3)[a:1->2,b:1->3,c:2->2,d:3->3] -gap> KQ := PathAlgebra(GF(7), Q); -GF(7) * Q -gap> rels := [ KQ.ac,KQ.cc,KQ.dd,KQ.bd ]; -[ Z(7)^0*(a*c), Z(7)^0*(c*c), Z(7)^0*(d*d), Z(7)^0*(b*d) ] -gap> A := KQ/rels; -(GF(7) * Q) / [ Z(7)^0*(a*c), Z(7)^0*(c*c), Z(7)^0*(d*d), Z(7)^0*(b*d) ] -gap> M := AlgebraAsBimodule(A); -<1,1,1,0,2,0,0,0,2> -gap> DM := DualOfModule(M); -<1,1,1,0,2,0,0,0,2> -gap> DM2 := TensorProductOfModules(DM,DM); -<0,1,1,0,1,0,0,0,1> -gap> DM3 := TensorProductOfModules(DM,DM2); -<0,1,1,0,1,0,0,0,1> -gap> IsIndecomposableModule(DM2); -false -gap> DecomposeModule(DM2); -[ <0,0,0,0,0,0,0,0,1>, <0,0,0,0,1,0,0,0,0>, <0,0,1,0,0,0,0,0,0>, <0,1,0,0,0,0,0,0,0> ] -gap> IsomorphicModules(DM2,DM3); -true - -Here QPA2 claims that $D(A)^{\otimes 2}$ decomposes in 4 simple bimodules. Furthermore, the bimodules $D(A)^{\otimes n}$ are all isomorphic for $n\geq 2$, and $\operatorname{End}(D(A)^{\otimes 2}) \simeq \mathbb{Z}_7^4$. -I hope that these comments are helpful. -The QPA-team.<|endoftext|> -TITLE: Explaining patterns in modular multiplication graphs -QUESTION [6 upvotes]: Let the multiplication graph $n/m$ be the graph with $m$ points distributed evenly on a circle and a line between two points $a$, $b$ when $an \equiv b\operatorname{mod} m$. -These graphs often look somehow random but by carefully choosing $n$ and $m$ one finds intricate patterns. Let $n = 20, 40, 60, 80, 100, 120$ and -$m = n + 19$ - -$m = 2n + 19$ - -$m = 3n + 19$ - -Note that it seems essential that $19 = 20 -1$ which guarantees that $m,n$ are coprime. -Let $A, B, C$ be arbitrary positive numbers. - -Why does the graph $n/m$ have $(C-1)A+1$ petals when $n = AB$ and $m = Cn + B - 1 = B(AC +1) - 1$? - -(Note that the number of petals doesn't depend on $B$!) -For example with $A=2$, $B=19$, $C=6$, the graph $38/246$ with $38 = 2\cdot 19$ and $246 = 6\cdot 38 + 18 = 2\cdot 3\cdot 41$ has $11 = 5\cdot 2 + 1$ petals: - -REPLY [3 votes]: The following does not answer the question about the number of cusps, but might be useful. -Consider a continuous version of your construction: -draw a line through the points $(\cos t, \sin t)$ and $(\cos nt, \sin nt)$ for every $t$ and look at its envelope. If $F(x,y,t)=0$ is an equation of this line, then the envelope is found from resolving the system of equations -$$ -F(x,y,t)=0, \quad \frac{\partial F}{\partial t}(x,y,t)=0 -$$ -with respect to $t$. The result is -$$ -\gamma(t) = \frac{1}{n+1} -\begin{pmatrix} -\cos nt + n\cos t\\ \sin nt + n\sin t -\end{pmatrix} -$$ -which parametrizes the trajectory of a point on a circle of radius $\frac{1}{n+1}$ rolling on a circle of radius $\frac{n-1}{n+1}$. -(That this trajectory has the $t$-$nt$ lines as tangents can also be proved geometrically by looking at the instantaneous motion of the point: it rotates about the point of contact between the two circles.) -The curve is called epicycloid, and the construction is attributed to Cremona, see for example this webpage. -It has $n-1$ cusp, which is different from your result. Probably the reason is that you consider a discrete set of lines. By the way, your pictures look similar to those for epicycloids for rational non-integer ratio of the circles radii, see the Wikipedia page. -Also I found a page discussing Mathematica codes for Cremona construction.<|endoftext|> -TITLE: Probably true, but provably unprovable -QUESTION [35 upvotes]: I'm wondering if anyone has found, or can find, a sequence of statements $P(n)$ ($n \in \mathbb{N}$) such that: - -Heuristic arguments using probability theory suggest that all the statements $P(n)$ are true. -One can prove in some widely accepted axiom system $X$, preferably by making the heuristic arguments rigorous, that "$P(n)$ holds for infinitely $n$". -One can prove in some widely accepted axiom system $Y$ that "$X$ cannot prove $\forall n P(n)$". - -The goal here would be to find statements that are true 'just because they are probably true', not because we can put our finger on a reason why any individual one must be true. -An example of such a sequence of statements might be 'if $p_n$ is the $n$th prime, there are infinitely many repunit primes in base $p_n$'. However while this example meets condition 1 we seem far from having enough understanding of mathematics to prove 2, and 3 seems hopeless. I think we need a less charismatic sequence of statements that are more carefully crafted to the task at hand. - -REPLY [4 votes]: I will take the question as reworded in the comments: 1) asks for a plausible probabilistic argument for $\forall nP(n)$ using any framework you like, 2) asks for a proof in some widely accepted axiom system $X$ of "$P(n)$ for infinitely many $n$", and 3) asks for a proof in some widely accepted axiom system that $X$ cannot prove $\forall nP(n)$. -Let us call a function $f$ from the positive integers to the positive integers Collatz-like if it is piecewise linear, the pieces being finite in number and depending only on the congruence class of the argument modulo some $m$. The ur-Collatz function is given by $C(n)=n/2$ if $n\equiv0\bmod2$, $C(n)=(3n+1)/2$ if $n\equiv1\bmod2$. Another example is given by $D(n)=n/2$ if $n\equiv0\bmod2$, $D(n)=(5n+1)/2$ if $n\equiv1\bmod2$. -We make ourselves interested in the question, given a Collatz-like function $f$, is there a positive integer $s$ such that the sequence $s,f(s),f(f(s)),f(f(f(s))),\dots$ goes to infinity. -Heuristically, $C(n)$ is, on average, much like $3n/4$, hence, decreasing, while $D(n)$ is much like $5n/2$, hence, increasing. Henceforward, we restrict our attention to those functions that are (heuristically, on average) decreasing. We enumerate them, and let $P(n)$ be the statement that there is no input to the $n$th such function such that the iterates of the function go to infinity. I find $\forall nP(n)$ plausible, on heuristic grounds. -Now, infinitely many of our functions $f$ satisfy the condition $\forall mf(m)\le m$. The heuristic argument becomes a trivial proof that for all these infinitely many functions, $P(n)$. -Finally, John Conway, Unpredictable Iterations, in Proc. 1972 Number Theory Conference, University of Colorado, Boulder, CO. 1972 , pp 49-52 (MR 52 #13717), proved that in the set of Collatz-like functions there are those for which $P(n)$ is algorithmically undecidable. So, we can't prove $\forall nP(n)$.<|endoftext|> -TITLE: Integral Tate-Sen theory -QUESTION [6 upvotes]: Let $K$ be a finite extension of $\mathbb{Q}_p$, and let $C=\widehat{\overline{K}}$ be the completion of the algebraic closure of $K$. Let $\mathscr{O}_C$ be the ring of integers in $C$, and let $G_K$ be the absolute Galois group of $K$. I would like to know the following: - -What is the annihilator of $H^1_{\mathscr{O}_{C}-mod}(G_K, \mathscr{O}_C(i))$, for $i\in \mathbb{Z}, i\not=0$? - -Here $\mathscr{O}_C$-mod is the category of continuous $\mathscr{O}_C$-semilinear representations of $G_K$. -The (proofs of) Tate-Sen theory tell us that this is a torsion module annihilated by some element $\mathscr{O}_K$. I've convinced myself that in principle one could extract such an element from the proofs, though not that the existing proofs would give the best possible result. That said, I'm hoping that someone has already worked this out. In particular, I would like to know: - -Is the answer (for fixed $i$) independent of $K?$ - -Remarks: -I think it's not too hard to reduce this to computing $$H^1_{\mathscr{O}_{K_\infty}-mod}(\Gamma, \mathscr{O}_{\widehat{K_\infty}}(i))$$ where $K_\infty$ is a cyclotomic $\mathbb{Z}_p$-extension of $K$ with $\mathbb{Z}_p\simeq \Gamma:=\text{Gal}(K_\infty/K)$. In other words, if $\sigma$ is a topological generator of $\Gamma$, this boils down to computing the annihilator of the cokernel of $$\sigma-\text{Id}: \mathscr{O}_{\widehat{K_\infty}}(i)\to \mathscr{O}_{\widehat{K_\infty}}(i),$$ if I'm not mistaken. -The annihilator of this cokernel is not quite the same as the annihilator of $H^1_{\mathscr{O}_{C}-mod}(G_K, \mathscr{O}_C(i))$, but if I've worked things out correctly, they differ by an amount which is independent of $K$. - -REPLY [5 votes]: I think the answer is "no", and that the minimal power of $p$ annihilating this module can be unbounded for fixed $i$ and varying $K$. -More precisely, fix a choice of a Galois-equivariant continuous $\mathcal{O}_K$-linear surjection $\widehat{\mathcal{O}_{K_\infty}}\to \mathcal{O}_K$ -(e.g. by playing with Tate's normalized traces). This induces a surjection of $\mathcal{O}_K$-modules $H^1(\Gamma,\widehat{\mathcal{O}_{K_\infty}}(i)) \to H^1(\Gamma,\mathcal{O}_K(i))$. Now the point is that $H^1(\Gamma,\mathcal{O}_K(i))$ is easy to compute in most cases (probably all cases but I am too lazy right now). In particular, let $n=n_K$ be the largest integer such that $K$ contains a primitive $p^n$th root of unity, and assume that $n>0$. Then $H^1(\Gamma,\mathcal{O}_K(i))$ is isomorphic to $\mathcal{O}_K/((1+p^n)^i-1)$; this is an easy computation, using the fact that the cyclotomic character sends any generator of $\Gamma$ to something of the form $1+p^nu$ for some $p$-adic unit $u$. Since the $p$-adic valuation of $(1+p^n)^i-1$ is clearly unbounded for any fixed $i$ and varying $n$, this proves my claim.<|endoftext|> -TITLE: $p_n(x,y)=\sum_{i=0}^{n-1}x^{n-1-i}y^{i}$ is always an integer -QUESTION [6 upvotes]: Does anyone know if the following problem has ever been studied? - -Let $a$ and $b$ be two real numbers and consider the polynomial: $$p_n(x,y)=\sum_{i=0}^{n-1}x^{n-1-i}y^{i}$$ - where $n$ is a positive integer. -Does there exist a value of $k$ such that if $p_n(a,b)$ is an integer for $k$ consecutive values of $n$ then $p_n(a,b)$ is an integer for every $n$? If so what is the minimum value of $k$? - What happens if we change the 'integer' condition to 'rational' values at the previous question? - -It's not difficult to establish some recurrence relation among the values of $p_n$ but none of them seem to be promising. -I would like to know any reference for this problem or how it could be solved. -Any help would be appreciated. - -REPLY [11 votes]: I will do the rational case and assume $a,b\neq 0$ otherwise the problem is trivial. You just need four consecutive values. Note that $p_n(a,b)=\cfrac{a^n-b^n}{a-b}$. -Say you have $p_k$, $p_{k+1}$, $p_{k+2}$, $p_{k+3}$ are all rational. -Note that $p_{k+1}^2-p_kp_{k+2}=(ab)^k$ and $p_{k+2}^2-p_{k+1}p_{k+3}=(ab)^{k+1}$. -Thus $ab$ is rational.Now $p_{k+1}(a+b)=p_{k+2}+abp_{k}$ so it follows that $a+b$ is also rational or $p_{k+1}=0$. But similarly $p_{k+2}(a+b)=p_{k+3}+abp_{k+1}$ so if $p_{k+2}=0$ then $a=b$ and again the problem is trivial. -Thus we have $a+b,ab \in \mathbb{Q}$ and now note that $p_n(a,b)$ is a symmetric polynomial so it can be expressed as a polynomial with rational coefficients in $a+b,ab$ so it is always rational.<|endoftext|> -TITLE: Positive-definiteness of radial sinc function in three dimensions -QUESTION [6 upvotes]: In dimension one, it is well known that $\mathcal{F}\chi_{(-1,1)}=\frac{\sin{x}}{x}$. This implies, in particular, that $\frac{\sin{x}}{x}$ is a definite positive function. I wonder if a similar result holds in dimension three. Moreover, it would be nice to actually get strictly positive definiteness. So my question is: -Is it true that the function $f:\mathbb{R}^3\to\mathbb{R}$ given by $f(x)=\frac{\sin{|x|}}{|x|}$ is the Fourier transform of a positive (or at least non-negative) function? - -REPLY [5 votes]: It’s the Fourier transform of the rotation-invariant probability measure on the unit sphere, and as such is positive definite. -Strict positive definiteness also holds, by the theorem in zu Castell, Filbir and Szwarc (2005).<|endoftext|> -TITLE: What is the current status on the corank conjecture for Selmer groups? -QUESTION [5 upvotes]: Let $E$ be an elliptic curve over $\mathbb{Q}$ and $p$ a prime. It is conjectured in the book of Coates and Sujatha "Galois Cohomology of Elliptic Curves" (Conjecture 2.5) that the corank of the Selmer group of $E$ over the cyclotomic $\mathbb{Z}_p$ extension is $1$ when $E$ has potentially supersingular reduction at $p$ and $0$ otherwise. There is a really nice proof of this following Theorem 2.14 which is subsequently given in the case in which the Selmer group of $E$ over $\mathbb{Q}$ is finite (so in particular only when $E$ has rank zero and the $p$ part of the Tate Shafarevich group is assumed to be finite). -What is the current state of affairs on this conjecture? - -REPLY [5 votes]: Yes, the corank conjecture is a theorem for elliptic curves over $\mathbb{Q}$. The key to the proof is the following: - -Theorem (Kato, 2004): For any $E$ and any $p$, the "fine Selmer group" $Sel_p^0(E / \mathbb{Q}_\infty) = \operatorname{ker}\Big(Sel_p(E / \mathbb{Q}_\infty) \to H^1(\mathbb{Q}_{p, \infty}, E[p^\infty])\Big)$ is cotorsion. - -One of the great things about this theorem is that both its statement and its proof are completely independent of the local behaviour of $E$. The dependence on local behaviour comes when you try to use this to deduce things about the classical Selmer group $Sel_p(E / \mathbb{Q}_\infty)$ from Kato's theorem. -Combining Kato's theorem, Poitou--Tate duality, and a theorem in local Iwasawa theory due to Berger, one gets the following consequence: - -Corollary: The corank of $Sel_p(E / \mathbb{Q}_\infty)$ is 1 if $T_p(E) |_{G_{\mathbb{Q}_p}}$ is irreducible, and 0 otherwise. - -So one needs to check that $T_p(E) |_{G_{\mathbb{Q}_p}}$ is irreducible if and only if $E$ has potentially supersingular reduction, which is a fun exercise.<|endoftext|> -TITLE: Translating Grothendieck axiom UB into ZFC -QUESTION [12 upvotes]: In SGA 4, Grothendieck introduced a set-theoretic device called a Grothendieck universe. He and his collaborators worked in Bourbaki set theory, which is practically similar to $\mathsf{ZFC}$ but fundamentally different. To ease the work with universes, he introduced two addition axioms to the theory: they are mostly referred to as UA and UB. -While UA is widely known and doesn't specifically depend on Bourbaki set theory (it basically states that for any set $X$ there is a universe $\mathscr{U}$ containing it), UB apparently heavily relies on that version of set theory. In particular, it uses such concepts as relations and and the operator $\tau_x$, which $\mathsf{ZFC}$ lacks. -My question is the following one: is it possible to get a version of UB with respect to $\mathsf{ZFC}$ which would serve the same purposes for universes? -Here's the axiom UB in the language of Bourbaki set theory: - -Let $R\{x\}$ be a relation and $\mathscr{U}$ a universe. If there is $y \in \mathscr{U}$ so that we have $R\{y\}$, then $\tau_xR\{x\} \in \mathscr{U}$. - -P.S. I apologize if the question is not the best quality, I suspect it could even be a trivial one, but I personally don't understand Bourbaki set theory very well. Still, MO favors questions which a mathematical researcher potentially can ask and I can imagine a research who doesn't understand Bourbaki set theory but is interested in using universes à la Grothendieck in SGA. - -REPLY [8 votes]: If we work with some given universe $U$, we have to make sure that we do not leave it accidentally. The definition of a universe does this for most operations. But there is still a way to leave the universe, namely by the (global) axiom of choice, i.e., Hilbert's symbol $\tau$. -Consider a relation $R$ with a variable $x$. If there does not exist an object fulfilling this relation, then $\tau_x(R)$ is an arbitrary set of which we may say nothing (in particular not whether or not it is contained in $U$). If there exists an object fulfilling this relation, then $\tau_x(R)$ denotes such an object. Without any further axiom we cannot say whether or not $\tau_x(R)$ is contained in $U$. This is precisely the role played by the axiom scheme UB: It makes sure that if there exists an object fulfilling $R$ in $U$, then $\tau_x(R)$ lies in $U$. Using the fact that intersections of universes are again universes we can formulate UB as follows: - -$\tau$ always chooses in the smallest possible universe. - -Now, we do not have a global choice operator in ZFC, and as far as I understand it is not possible to leave a given universe with the usual axiom of choice in ZFC. Therefore, there is no need for an axiom similar to UB in ZFC.<|endoftext|> -TITLE: Do solenoids embed into Möbius strips? -QUESTION [11 upvotes]: I found a strange attractor which looks a lot like a solenoid. The attractor continuum is the closure of a continuous line which limits onto itself, and it is locally homeomorphic to Cantor set times Reals. It sits in the Möbius strip. - -Does the Dyadic solenoid embed into the Möbius strip? What about solenoids in general? Could the strange attractor above actually be a solenoid? - -REPLY [8 votes]: No solenoid can be embedded into the Mobius strip. To derive a contradiction, assume that some solenoid $S$ embeds into the Mobius strip $M$. Let $\pi:C\to M$ be a 2-fold covering map of the cylinder $C$ onto the Mobius strip. It is well-known that the solenoid $S$ contains a dense subset $D$ which is the image of the real line under a continuous map $\phi:\mathbb R\to D$ (and this image of $\mathbb R$ is called a composant of the solenoid). By the lifting property of the covering map $\pi$, there exists a continuous map $\varphi:\mathbb R\to C$ such that $\pi\circ\varphi=\phi$. Then the closure $K$ of the connected set $\phi(\mathbb R)$ in $C$ is a continuum such that $\pi(K)=\bar D=S$. Taking into account that the cylinder $C$ embeds into the plane, we conclude that $K$ is a planar continuum and hence the solenoid $S$ is a continuous image of a planar continuum. -On the other hand, by a result of Krasinkiewicz in his paper Mappings onto circle-like continua, no solenoid is a continuous image of a planar continuum (as solenoids have infinitely divisible first cohomology group whereas the first cohomology group of any planar continuum is finitely divisible). This is a desired contradiction.<|endoftext|> -TITLE: Is $\frac{\sin |\xi|}{|\xi|}$ in range of Fourier Transform for $n \ge 3$? -QUESTION [7 upvotes]: Does there exist $f \in L^1(\mathbb{R}^n)$ s.t., $\displaystyle \widehat{f}(\xi) = \frac{\sin |\xi|}{|\xi|}$ in case of dimension $n \ge 3$? -It is known that for $n = 2$, the function $\displaystyle f(x) = \frac{\chi_{\{|x| < 1\}}}{\sqrt{1-|x|^2}}$ curiously has some constant multiple of $\dfrac{\sin |\xi|}{|\xi|}$ as Fourier transform. Or at least is there a way of finding the Fourier transform $\frac{\sin |\xi|}{|\xi|}$ in sense of tempered distributions for $n \ge 3$? - -REPLY [10 votes]: Here's a formula from Duistermaat and Kolk book Distributions: Theory and Applications, Chapter 17, Eq, (17.13). We denote by $\newcommand{\eF}{\mathscr{F}}$ $\eF$ the Fourier transform. Then $\newcommand{\ii}{\boldsymbol{i}}$ $\newcommand{\ve}{\varepsilon}$ -$$ -\eF^{-1}\left(\frac{e^{\ii t\Vert \xi\Vert}}{\Vert\xi\Vert}\right)=p_n\lim_{\ve\searrow 0} \Big(\; \Vert x\Vert^2-(t+\ve\ii)^2\;\Big)^{-\frac{n-1}{2}}, -$$ -where $p_n$ is a certain universal constant, -$$p_n=\frac{\Gamma\big(\frac{n+1}{2}\big)}{(n-1)\pi^{(n+1)/2}}. $$<|endoftext|> -TITLE: How do we formally construct the successor universe $\mathscr{U}^+$ of a universe $\mathscr{U}$ in $\mathsf{ZFC}$? -QUESTION [6 upvotes]: A set $\mathscr{U}$ is a universe if the following conditions are met: - - -For any $x \in \mathscr{U}$ we have $x \subseteq \mathscr{U}$ - -For any $x,y \in \mathscr{U}$ we have $\{x,y\} \in \mathscr{U}$, - -For any $x \in \mathscr{U}$ we have $\mathcal{P}(x) \in \mathscr{U}$, - -For any family $(x_i)_{i \in I}$ of elements $x_i \in \mathscr{U}$ indexed by an element $I \in \mathscr{U}$ we have $\bigcup_{i \in I} x_i \in \mathscr{U}$. - - - -Grothendieck introduced an addition axiom $\mathscr{U}$A which says that every set $x$ is contained in some universe $\mathscr{U}$. -I've seen some authors use the concept of the successor universe $\mathscr{U}^+$ of a given universe $\mathscr{U}$. It is the smallest universe which contains $\mathscr{U}$. However, I'm not sure how to prove that such a thing exists in $\mathsf{ZFC}$ (provided that $\mathscr{U}$ exists in the first place). If we knew that for any two universes $\mathscr{U}$ and $\mathscr{V}$ we have either $\mathscr{U} \in \mathscr{V}$ or $\mathscr{V} \in \mathscr{U}$, it would be easy. But I'm not sure if we can prove that latter without showing first that universes are equivalent to $V_\kappa$ for inaccessible cardinals $\kappa$. -Edit. The question, as evident from the accepted answer is turned out to be quite trivial. I apologize for that. - -REPLY [14 votes]: ZFC doesn't even prove that universes exist in the first place. However: - -ZFC proves that universes are exactly sets of the form $V_\kappa$ for $\kappa$ inaccessible. In particular, "any two universes are $\in$-comparable" is provable in ZFC alone; what's not provable is that there are many universes, or indeed any. -The successor universe of $\mathcal{U}$ is therefore a well-defined concept, and every universe has a successor universe assuming enough universes exist in the first place: ZFC proves that there is exactly one universe of height $\kappa$ for each inaccessible cardinal $\kappa$, so to get the successor universe we just "go up to the next inaccessible." ZFC can't prove that there always is a next inaccessible, but it does prove that "every set is contained in a universe" is equivalent to "there is a proper class of inaccessibles," and that each of these implies "every universe has a successor universe." - -REPLY [11 votes]: The intersection of a nonempty set of universes is a universe. Now, let $U$ be a universe, and suppose that there exists a universe $V$ with $U\in V$. Then, the set of all universes $W$ with $U\in W\subseteq V$ exists and is nonempty. Its intersection is a universe, and this is readily checked to be the smallest universe containing $U$. - -REPLY [10 votes]: The question confounds definition of an object and its existence. These are two different notions. - -Definition: The universe $\mathcal{V}$ is the successor of the universe $\mathcal{U}$ iff $\mathcal{U} \in \mathcal{V}$, and for all universes $\mathcal{W}$, if $\mathcal{U} \in \mathcal{W}$ then $\mathcal{V} \subseteq \mathcal{W}$. - -There is no mystery in the definition. I take it that the question is really asking about existence of successor universes, in which case the title of the question should be modified. -As is well-known, in ZFC a set $\mathcal{U}$ is a universe if, and only if, $\mathcal{U} = V_{\kappa}$ for some (unique) inaccessible cardinal $\kappa$. Here $V_\kappa$ is the $\kappa$-th level of the cummulative hierachy. -Because in ZFC the the cardinals are well-ordered, it follows that a universe $V_\kappa$ has a successor if, and only if, there exists an inaccessible larger than $\kappa$, in which case the successor universe is $V_\lambda$, for $\lambda$ the least inaccessible above $\kappa$. -Nowhere in the answer did we have to assume that inaccessibles or universes exist. We observed that existence of universes is equivalent to existence of inaccessible cardinals. ZFC does not prove that inacessible cardinals exist. -The void is psychologically scary but is mathematically quite tame.<|endoftext|> -TITLE: Index of the endomorphism ring of an abelian surface -QUESTION [8 upvotes]: For an abelian surface $A/\mathbb{Q}$ such that $R:=\mathrm{End}_{\mathbb{Q}}(A)$ is an order in a real quadratic field $K$ (so a $\mathrm{GL}_2$-type surface), is there a bound on the index $[O_K : R]$ of the ring inside the maximal order $O_K$? - -REPLY [2 votes]: There is a conjecture of Coleman asserting that up to isomorphism, there are only finitely many possible rings $\operatorname{End}_{\mathbf{Q}}(A)$ where $A$ varies among the abelian surfaces defined over $\mathbf{Q}$ (see Bruin, Victor Flynn, Gonzalez, Rotger, On finiteness conjectures for endomorphism algebras of abelian surfaces). This would imply that the answer to your question is yes. On the other hand, this is a very difficult conjecture, so there might be an easier way to tackle your question.<|endoftext|> -TITLE: Discrete Pin structures -QUESTION [10 upvotes]: It is clear that an oriented manifold $M^n$ (with dimension $n$) admits spin structures if and only if its second Stiefel-Whitney class $[w^2]\in H^2(M,\mathbb Z_2)$ vanishes. In the construction of the simplicial complex upon triangulation of $M$, one can also use the $(n-2)$-th Stiefel-Whitney homology class $[w_{n-2}]$, which is the Poincare dual of $[w^2]$. -There is a relation between the discrete spin structure and Kasteleyn orientations, given in Kasteleyn (Ref 1). -For a spatial 2-manifold $M^2$ ($n=2$) with triangulation $T$, the Stiefel-Whitney homology class $[w_0]$ has a representative that is the summation of all vertices $v$ with some (mod 2) coefficients as follows Goldstein- Turner (Ref 2): -$$ -w_0 = \sum_{v\in T} \# \{\sigma | v \subseteq \sigma \text{ is regular} \} \cdot v. -$$ -Here, $v \subseteq \sigma$ means that $v$ is a sub-simplex of simplex $\sigma$. The subsimplex $v \subseteq \sigma$ is called regular if $v$ and $\sigma$ satisfy the certain relative positions. The $\# \{\sigma | v \subseteq \sigma \text{ is regular} \} \cdot v$ denotes the formal product of the (mod 2) number of regular pairs $v \subseteq \sigma$ and the vertex $v$. -One can call vertex $v$ singular if $\# \{\sigma | v \subseteq \sigma \text{ is regular} \}$ is odd. Then $w_0$ is the formal summation of all singular vertices. $w_0$ is a vector (0-th singular chain) in the vector space (of 0-th singular chains) spanned by the formal bases of all vertices with $\mathbb Z_2$ coefficients. -Note any 2D oriented manifold allow spin structures. - -Question: -(1) Are there analogous discrete Pin structures (say Pin$^+$ or Pin$^-$) that we can define as "Stiefel-Whitney homology classes" for non-orientable $n$-manifolds $M$. Say $n=3, 4, 5$? -where Pin structure is given by: -$$ -1\to \mathbb{Z}_2 \to \text{Pin}^{\pm}(n) \to \text{Spin}(n) \to 1 -$$ - -In terms of cohomology class, for Pin$^+$, $w^2(M)=0$; and for Pin$^-$, $w^2(M)+(w^1)^2(M)=0$. (Here the usual notation shall be, for Pin$^+$, $w_2(M)=0$; and for Pin$^-$, $w_2(M)+w_1^2(M)=0$. -) - -(2) What is the counterpart of Kasteleyn orientations, in -$$\text{Kasteleyn orientations v.s. discrete Spin structures}$$ -$$\simeq \text{??? v.s. discrete Pin structures?}$$ - -References are welcome. - -P. W. Kasteleyn, Dimer Statistics and Phase Transitions Journal of Mathematical Physics 4, 287 (1963); https://doi.org/10.1063/1.1703953 - -Richard Z Goldstein and Edward C Turner, “A formula for Stiefel-Whitney homology classes,” Proceedings of the American -Mathematical Society 58, 339–339 (1976). - - -see also: - -David Cimasoni and Nicolai Reshetikhin, “Dimers on Surface Graphs and Spin Structures. i,” Communications in Mathematical -Physics 275, 187–208 (2007). - -David Cimasoni and Nicolai Reshetikhin, “Dimers on Surface Graphs and Spin Structures. II,” Communications in Mathematical -Physics 281, 445–468 (2008). - -REPLY [5 votes]: Here are some partial answers. -For your first question: there are combinatorial formulas for all Stiefel-Whitney homology classes $w_k$, due to -Whitney and rediscovered by Cheeger. -Specifically, on a manifold with a triangulation $\Pi$, $w_k$ is represented by the sum of all $k$-simplices in -the barycentric subdivision of $\Pi$. -So the story you have for spin 2-manifolds and $w_0$ should generalize for a pin$+$ $n$-manifold, -where you want a trivialization of $[w_{n-2}]$, and can represent it using Whitney's combinatorial formula. For a -pin$-$ $n$-manifold, though, you'd need a trivialization of the Poincaré dual of $(w^1)^2$, which is not -calculated by Whitney's formula. It may be possible to imitate Whitney's argument for $(w^1)^2$, but I don't think it's -trivial to do so. -For your second question: in “Dimers on graphs in non-orientable -surfaces”, §§4–5, David Cimasoni describes a generalization of -Kastelyn orientations to pin$-$ surfaces, and proves in Theorem 5.3 that given a cell decomposition -and a dimer configuration on a closed surface $\Sigma$, generalized Kastelyn orientations are equivalent to -pin$-$-structures on $\Sigma$. -As far as I know, however, nothing has been written generalizing this past dimension 2, nor about combinatorial -pin$+$-structures in any dimension. -This question is related to another MathOverflow question from a few -years ago.<|endoftext|> -TITLE: Criterion for acyclicity of flag complexes -QUESTION [5 upvotes]: Let $\Delta$ be a flag complex on $n$ vertices. Let $r$ be the smallest size of the facets of $\Delta$. Suppose that $2r>n$. Must $\Delta$ be acyclic? - -REPLY [3 votes]: It looks to me like you can prove the stronger property of contractibility by induction, as follows. Let $\Delta$ be the independence complex of graph $G$, as guaranteed by the flag property. -If $\Delta$ is a cone, then $\Delta$ is contractible. This will be the base case, along with dimension 0 (where $\Delta$ has a single point) and dimension 1 (where $\Delta$ is a 1-simplex). -Otherwise, every vertex $v$ has degree at least 1 in $G$. Consider a fixed pair of vertices $v,w$ that are adjacent in $G$. We'll use that the link of $v$ in $\Delta$ is the independence complex of $G \setminus N[v]$ (and similarly for $w$). Now $\operatorname{link}_\Delta v$, $\operatorname{link}_\Delta w$ are contained in the independence complex of $G \setminus \{v,w\}$. -Since we remove at most one vertex from each maximal face in each case, $r$ goes down by at most one in each considered subcomplex, while the number of vertices goes down by at least 2. So by induction, $\operatorname{link}_\Delta v$, $\operatorname{link}_\Delta w$, and the independence complex of $G \setminus {v,w}$ are all contractible. -Now $\Delta$ is the union of the faces that contain $v$, those that contain $w$, and those that contain neither. So $\Delta$ is the union of the subcomplexes $\Delta_1 = v*\operatorname{link} v$, $\Delta_2 = w*\operatorname{link} w$, and $\Delta_0$ the independence complex of $G \setminus \{v,w\}$. It now follows by e.g. Lemma 10.4(ii) of Björner's Topological methods that $\Delta$ is contractible. -I didn't immediately see the answer to the question of whether there's a sensible generalization of some sort to non-flag complexes. The flag property is used above in where $\operatorname{link}_\Delta v$ is the independence complex of $G \setminus N[v]$; also in finding a pair of vertices that are in no common face. -(Updated over initial version to fix a problem with the facet sizes in the induction, in response to a comment of and off-MO discussion with @Hailong Dao.)<|endoftext|> -TITLE: Permutations $\pi\in S_n$ with $\sum_{k=1}^n\frac1{k+\pi(k)}=1$ -QUESTION [21 upvotes]: Let $S_n$ be the symmetric group of all the permutations of $\{1,\ldots,n\}$. -Motivated by Question 315568 (http://mathoverflow.net/questions/315568), here I pose the following question. - -QUESTION: Is it true that for each integer $n>5$ we have $$\sum_{k=1}^n\frac1{k+\pi(k)}=1$$ for some odd (or even) permutation $\pi\in S_n$? - -Let $a_n$ be the number of all permutations $\pi\in S_n$ with $\sum_{k=1}^n(k+\pi(k))^{-1}=1$. Via Mathematica, I find that -\begin{gather}a_1=a_2=a_3=a_5=0,\ a_4=1,\ a_6=7, -\\ a_7=6,\ a_8=30,\ a_9=110, \ a_{10}=278,\ a_{11}=1332.\end{gather} -For example, $(1,4,3,2)$ is the unique (odd) permutation in $S_4$ meeting our requirement for $n=4$; in fact, -$$\frac1{1+1}+\frac1{2+4}+\frac1{3+3}+\frac1{4+2}=1.$$ -For $n=11$, we may take the odd permutation $(4,8,9,11,10,6,5,7,3,2,1)$ since -\begin{align}&\frac1{1+4}+\frac1{2+8}+\frac1{3+9}+\frac1{4+11}+\frac1{5+10}\\&+\frac1{6+6}+\frac1{7+5}+\frac1{8+7}+\frac1{9+3}+\frac1{10+2}+\frac1{11+1}\end{align} has the value $1$, we may also take the even permutation $(5, 6, 7, 11, 10, 4, 9, 8, 3, 2, 1)$ to meet the requirement. -I conjecture that the question has a positive answer. Your comments are welcome！ -PS: After my initial posting of this question, Brian Hopkins pointed out that A073112($n$) on OEIS gives the number of permutations $p\in S_n$ with $\sum_{k=1}^n\frac1{k+p(k)}\in\mathbb Z$, but A073112 contains no comment or conjecture. - -REPLY [33 votes]: Claim: $a_n>0$ for all $n\geq 6\quad (*)$. -Proof: We use induction to prove $(*)$. -We have $a_6,a_7,a_8,a_9,a_{10},a_{11}>0$. Assume $(*)$ holds for all the integers $\in [6,n-1]$. We want to show that $a_n>0$ for all $n\geq12$. -If $n$ is an odd, let $n=2m+1$, we have $m\geq6$, so by induction hypothesis, there exists $\pi\in S_m$ such that $\sum\limits_{k=1}^{m}\frac{1}{k+\pi(k)}=1$. -Let -\begin{align*} -\sigma(2k+1)&=2m+1-2k\quad\text{for}\quad k=0,1,2\ldots, m,\\ -\sigma(2k)&=2\pi(k)\quad\text{for}\quad k=1,2,\ldots, m. -\end{align*} -Then $\sigma\in S_{n}$, and -$$\sum\limits_{k=1}^{n}\frac{1}{k+\sigma(k)}=\sum\limits_{k=1}^{m}\frac{1}{2k+\sigma(2k)}+\sum\limits_{k=0}^{m}\frac{1}{2k+1+\sigma(2k+1)}\\ -=\frac{1}{2}\sum\limits_{k=1}^{m}\frac{1}{k+\pi(k)}+\sum\limits_{k=0}^{m}\frac{1}{2k+1+(2m-2k+1)}=\frac{1}{2}+(m+1)\frac{1}{2m+2}=1.$$ -If $n$ is an even, let $n=2m$, also we have $m\geq6$ and there exists $\pi\in S_m$, such that $\sum\limits_{k=1}^{m}\frac{1}{k+\pi(k)}=1$. -Let -\begin{align*} -\sigma(2k-1)&=2m+1-2k\quad\text{for}\quad k=1,2,\ldots,m, \\ -\sigma(2k)&=2\pi(k)\quad\text{for}\quad k=1,2,\ldots,m. -\end{align*} -Then $\sigma\in S_{n}$ and -$$\sum\limits_{k=1}^{n}\frac{1}{k+\sigma(k)}=\sum\limits_{k=1}^{m}\frac{1}{2k+\sigma(2k)}+\sum\limits_{k=1}^{m}\frac{1}{2k-1+\sigma(2k-1)}=\frac{1}{2}\sum\limits_{k=1}^{m}\frac{1}{k+\pi(k)}+ -\sum\limits_{k=1}^{m}\frac{1}{2k-1+(2m-2k+1)}=1.$$ -Hence $a_{n}>0$. -By induction $(*)$ holds for all the $n\geq 6$.<|endoftext|> -TITLE: If $\text{dim}(X \times X) = 2\text{dim}(X)$, does $\text{dim}(X^n) = n\text{dim}(X)$? -QUESTION [32 upvotes]: I have been learning some (topological) dimension theory and have gotten through most of the basic material, at this point, and am about to start looking at papers. In particular, I want to get familiar with the standard counterexamples regarding dimension of products, but I haven't noticed the question in the title addressed. -So my question would be what conditions on $X$ (e.g. separable metric, compact metric, continuum, homology manifold) are sufficient to ensure that if $\text{dim}(X \times X) = 2\text{dim}(X)$, then $\text{dim}(X^n) = n\text{dim}(X)$? The initial assumption could also be weakened from $\text{dim}(X \times X) = 2\text{dim}(X)$ to $\text{dim}(X^m) = m\text{dim}(X)$ for some $1 < m < n$, or for various subsets of $\lbrace 2, 3, \dots, n - 1 \rbrace$. The case $m = n - 1$ seems especially interesting. -In particular, have the dimensions of all the powers of the Pontryagin Surface been computed? -Edit after a couple of days: Dranishnikov (arguably the top living expert) says it is true for compact metric spaces. He didn't mention any counterexample for more general spaces. It was already known to Hurewicz in the 1930's to be true for compact metric spaces of dimension one. -As noted in the comments, it is known that for any compact Hausdorff or any metrizable space, the limit (called the stable dimension of $X$) $\lim \frac{\text{dim}(X^n)}{n}$ exists. The quantity is positive if $X$ is compact metric and $\text{dim}(X) \geq 1$. Thus to prove the titular proposition for compact metric spaces it's sufficient to prove the case $n = 4$, because we can then iterate and apply the uniqueness of the limit, using the fact that for reasonable spaces it is true that $\text{dim}(X \times Y) \leq \text{dim}(X) + \text{dim}(Y)$. So if there was some pathology it would carry all the way to the limit. -This method assumes that if $\text{dim}(X) > 0$ then the stable dimension is positive; this is true for compact metric spaces, but is not true in general for separable metric spaces. For example, let $E$ be Erdos Space. It is known that $E = E \times E$ and $\text{dim}(E) = \text{dim}(E \times E) = 1$, so $\text{stabdim}(E) = 0$. To see that it is true for (locally) compact metric spaces in dimension larger than $1$, use the small inductive dimension to obtain closed subspaces of dimension at least $1$ as boundaries of neighborhoods of a point, then apply the classical Hurewicz result mentioned above to see that in fact $\text{stabdim}(X) \geq 1$, since the product of these boundaries is closed and dimension is monotone wrt closed subsets. -This would extend the proof method to general compact Hausdorff spaces if we knew that the conjecture in the title were true for one-dimensional compact Hausdorff spaces (that it follows is due to the fact that for compact Hausdorff spaces $X$ we have $\text{dim}(X) \leq \text{ind}(X)$). I am not sure if that is true or false, but I assume it's known. Anybody have a reference? Extending to LCH spaces is tricky in dimensions $0$ and $1$ so I have no idea about that, and I don't know if the stable dimension of LCH spaces is even well-defined. That would also be good to know. - -REPLY [12 votes]: As John Samples noted in his comment, Dranishnikov's Theory of cohomological dimension implies the positive answer to this problem for compact (even $\sigma$-compact) metrizable spaces. Namely, according to a Definition on page 15 of the paper "Cohomological dimension theory of compact metric spaces" a compact metrizable space $X$ is called of basic type if $\dim X^2=2\dim X$ and of exceptional type if $\dim X^2=2\dim X-1$ (which is equivalent to $\dim X^2<2\dim X$ by Theorem 4.16). After this definition Dranishnikov writes that each compactum $X$ of basic type has $\dim X^n=n\dim X$ and each compactum $X$ of exceptional type has $\dim X^n=n\dim X-n+1$, for every $n\in\mathbb N$. -This implies that if some compactum $X$ has $\dim X^n=n\dim X$ for some $n\ge 2$, then it is of basic type and hence $\dim X^k=k\dim X$ for all $k\in\mathbb N$. Also the Dranishnikov's result implies that the stable dimension $\lim_{n\to\infty}\frac{\dim X^n}n$ of a compactum $X$ equals $\dim X$ if $X$ is of basic type and equals to $\dim X-1$ if it is of exceptional type. -In the first paragraph of Section 1 Dranishnikov writes: "Actually everywhere -in this paper one can replace compact spaces by σ-compact".<|endoftext|> -TITLE: Bernstein type theorems for CMC hypersurfaces in $\mathbb{R}^{n+1}$ -QUESTION [5 upvotes]: Is there any Bernstein type theorems for CMC hypersurfaces in $\mathbb{R}^{n+1}$ in the literature? -More precisely I would like to know if there is an answer to the following -QUESTION: Let $f : \mathbb{R}^n \to \mathbb{R}$ be a smooth function such that $\mathrm{graph}(f)$ is a constant mean curvature hypersurface of $\mathbb{R}^{n+1}$. Is it true that $\mathrm{graph}(f)$ must be an affine hyperplane? -I don't know much about CMC hypersurfaces and I don't know where to look for an answer. Even if the question has a negative answer, I would like to know if there are counterexamples or if one can get an affirmative answer under some volume growth condition. -Any help will be very much appreciated! -Thanks! - -REPLY [7 votes]: This was solved in a series of articles in the 1960s. -De Giorgi, Almgren, and Simons have shown that in $\mathbb{R}^{\le 8}$ every CMC graph is a hyperplane. Then Bombieri - De Giorgi - Giusti have shown that in $\mathbb{R}^{\ge 9}$ there are minimal graphs which are not hyperplanes. -Here is a link to the latter article: -Bombieri, E.; De Giorgi, E.; Giusti, E., Minimal cones and the Bernstein problem, Invent. Math. 7, 243-268 (1969). ZBL0183.25901.<|endoftext|> -TITLE: Relationship between Hilbert-Samuel multiplicity and polar multiplicity -QUESTION [7 upvotes]: Let $f \in \mathbb{C}[[x,y]]$ be the germ of an isolated plane curve singularity. Then the Hilbert-Samuel multiplicity $e_f$ of $f$ is given as follows: -$$e_f = \lim_{s \to \infty}\frac{1}{s} \cdot \dim_{\mathbb{C}} \mathbb{C}[[x,y]]\bigg/\left(f, \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)^s\right)$$ -I've noticed that in various examples (like the $A_{n-1}$-singularities $f = y^2 - x^n$, or more generally, the hypercuspidal singularities $f = y^a - x^b$) that $e_f$ is equal to the intersection multiplicity of $f$ and a ``generic polar'' of $f$. More precisely, I've observed that the following equality holds in many examples: -$$e_f = \dim_{\mathbb{C}} \mathbb{C}[[x,y]]\bigg/\left(f, \alpha \cdot \frac{ \partial f}{\partial x} - \beta \cdot \frac{\partial f}{ \partial y}\right) \qquad (*)$$ -where $[\alpha : \beta] \in \mathbb{P}_{\mathbb{C}}^1$ is generic. My question: is it well-known (or obvious) whether the equality $(*)$ holds for an arbitrary isolated plane curve singularity germ $f$? -(Note: the colength of the polar (the quantity on the right of $(*)$) is of interest because it is related to Teissier's notion of ``polar invariant.'') -What I have so far: Note that the polar is itself a complete intersection, so the colength of the polar in the ring $\mathbb{C}[[x,y]]/(f)$ is equal to the Hilbert-Samuel multiplicity of the polar. Since the Hilbert-Samuel multiplicity of an ideal is equal to that of any reduction, it suffices to show that the ideal $I = \left(\alpha \cdot \frac{\partial f}{\partial x} - \beta \cdot \frac{\partial f}{\partial y}\right)$ generated by the polar is a reduction of the ideal $J = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$ (here, we are regarding $I,J$ as ideals of the ring $\mathbb{C}[[x,y]]/(f)$). I.e., it suffices to show that for some $s \geq 0$, the obvious inclusion -$$I(J^s) \subset J^{s+1}$$ -of ideals of $\mathbb{C}[[x,y]]/(f)$ is in fact an equality. Now restrict to the case where $f \in \mathbb{C}[x,y]$ is homogeneous of degree $d$. Then it is not hard to check by counting generators that $IJ^{d-1} = (x,y)^{d(d-1)} \supset J^d$ (as ideals of $\mathbb{C}[[x,y]]/(f)$). I do not currently know how to generalize this argument to handle germs $f$ that are not homogeneous. - -REPLY [4 votes]: First, the quantity you defined is the Hilbert-Samuel multiplicity of the ideal $J= (f_x,f_y)$ in $R=\mathbb C[[x,y]]/(f)$. The multiplicity of $f$ usually refers to the multiplicity of the maximal ideal $m$ of $R$. -As you noted, it is enough to show that a generic combination of the generators of $J$ is a reduction. This is true much more generally, and basically the point is to consider the fiber cone $S= F(I) = k\oplus \frac {J}{mJ}\oplus \frac {J^2}{mJ^2}\oplus... $. This algebra has dimension $1$, and is generated in degree one. Then, for a generic linear form $l \in S_1$, $S/lS$ is $0$-dimensional, so $lS_n=S_{n+1}$ for $n\gg 0$. But since $l \in S_1= J/mJ$, this says exactly that $J^nc = J^{n+1}$ where $c$ is a lift of $l$.<|endoftext|> -TITLE: Integral of product of Schur functions -QUESTION [10 upvotes]: Schur functions are irreducible characters of the unitary group $\mathcal{U}(N)$. This implies the integration formulae -$$ \int_{\mathcal{U}(N)}s_\lambda(AUA^\dagger U^\dagger)dU=\frac{|s_\lambda(A)|^2}{s_\lambda(1)}$$ -and -$$ \int_{\mathcal{U}(N)}s_\lambda(AU)\overline{s_\mu(A U)}dU=\frac{\delta_{\lambda\mu}s_\lambda(AA^{\dagger})}{s_\lambda(1)},$$ -where the overline means complex conjugation. -My question is whether we can compute the integral -$$ \int_{\mathcal{U}(N)}s_\lambda(AUA^\dagger U^\dagger)\overline{s_\mu(AUA^\dagger U^\dagger)}dU$$ - -REPLY [3 votes]: As a first step towards a solution, using the identity -$$\int_{\mathcal{U}(N)} U_{\alpha a}U_{\alpha' a'}\bar{U}_{\beta b}\bar{U}_{\beta' b'}\,dU=\frac{1}{N^{2}-1}\bigl( \delta_{\alpha\beta}\delta_{ab}\delta_{\alpha'\beta'}\delta_{a'b'}+ \delta_{\alpha\beta'}\delta_{ab'}\delta_{\alpha'\beta}\delta_{a'b}\bigr)$$ -$$\qquad\qquad\mbox{}-\frac{1}{N(N^{2}-1)}\bigl( \delta_{\alpha\beta}\delta_{ab'}\delta_{\alpha'\beta'}\delta_{a'b}+ \delta_{\alpha\beta'}\delta_{ab}\delta_{\alpha'\beta}\delta_{a'b'}\bigr),$$ -I computed this integral for the trace, -$$\int_{\mathcal{U}(N)}{\rm tr}\, (AUA^\dagger U^\dagger)\,\overline{{\rm tr}\,(AUA^\dagger U^\dagger)}\,dU=\frac{1}{N^2-1}\left(|{\rm tr}\, A|^4+|{\rm tr}\,AA^\dagger|^2\right)-\frac{2}{N(N^2-1)}|{\rm tr}\,A|^2({\rm tr}\,AA^\dagger).$$ -As a check, take $A=1$ and see that the right-hand-side reduces to $(N^4+N^2)/(N^2-1)-2N^3/(N^3-N)=N^2$.<|endoftext|> -TITLE: Does there exist a free action of $\mathbb Z_p$ on this space? -QUESTION [9 upvotes]: Is it true that for prime $p\neq 2 $, $k > 1$ and $n_1,n_2,\dots,n_k\geq 1$, the cyclic group $\mathbb{Z}_p$ has no continuous free action on $ \mathbb{C}P^{n_1} \times \mathbb{C}P^{n_2} \times \cdots \times \mathbb{C}P^{n_k}$? -How to prove it? -Thank you so much in advance. - -REPLY [8 votes]: Consider the action of the automorphism on $H^2(\prod_i \mathbb C P^{n_i} , \mathbb Z)$ by linear automorphisms and $H^*(\prod_i \mathbb C P^{n_i} , \mathbb Z)$. -Inside $H^2(\prod_i \mathbb C P^{n_i} , \mathbb Z)$ consider the set of nonzero integral multiples of hyperplane classes of the different factors. This set is consists exactly of the elements of $H^2$ that cannot be written as a sum of two elements of smaller nilpotence order in the ring $H^*$. (This follows from the fact that the nilpotence index of an element $x$ is the sum of $n_i$ over all $i$ such that $x$ contains a nonzero multiple of the hyperplane class of $\mathbb CP^{n_i}$, which can be checked by binomial coefficients). -Because this set has a characterization, it is preserved by any automorphism. Hence any automorphism of the product of projective spaces acts by permutation of the hyperplane classes and scalar multiplication. Thus the action on $H^2$ is by the group of signed permutation matrices. Conjugacy classes in their group are characterized by their cycle type, and, for each cycle, the product of the nonzero entries, which is $\pm 1$. -Hence elements of order $p$ consist of a union of $p$-cycles and fixed points. The products of the nonzero entries should have order p and thus should be $+1$. Moreover, the $p$-cycles consist must permute tuples of $i$ where $n_i$ is constant. -The Lefschetz number of such an automorphisms (i.e. its trace on cohomology) is the product over the cycles and fixed points of the trace on the corresponding tensor factor of the cohomology. For a fixed $\mathbb CP^{n_i}$, this is simply $n_i+1$. For a $p$-cycle acting on $(\mathbb CP^{n_i})^p$, we can form a basis of the cohomology consisting of products of powers of the hyperplane classes of the individual $\mathbb CP^{n_i}$s, and the only monomials fixed by this action are those with equal powers of each class, of which there are $n_i+1$. -Hence the product is positive, so the Lefschetz number is nonzero, so the number of fixed points is nonzero.<|endoftext|> -TITLE: Fundamental group of an open subscheme of a normal scheme -QUESTION [6 upvotes]: Let $X$ be an irreducible normal projective scheme over $\mathbb{C}$. Let $U$ be the open subscheme of smooth points of $X$. Consider the closed subscheme $Z = X \setminus U$. Suppose that the codimension of $Z$ in $X$ is at least $2$. Is it true that the fundamental group of $U$ and $X$ are isomorphic? -Edit: Is it true for $X$ an integral normal projective scheme over $\mathbb{C}$? - -REPLY [3 votes]: Although $X$ and $U$ need not have isomorphic fundamental group (as the accepted answer shows), the induced map from the inclusion $U\hookrightarrow X$ is generally $\pi_1$-surjective. -Precisely, if $X$ is a normal projective variety, and $A\subset X$ is a proper closed subvariety, then the natural homomorphism $\pi_1(X-A) \to \pi_1(X)$ is surjective. -You can find this result in On the fundamental groups of normal varieties by Donu Arapura, Alexandru Dimca, Richard Hain. -Interestingly, although natural inclusions only give $\pi_1$-surjections, it turns out that natural projections do give $\pi_1$-isomorphisms. -Precisely, let $G$ be a connected reductive algebraic affine group over an algebraically closed field $k$ (arbitrary characteristic). Assume $G$ is acting on a smooth connected projective variety $M$ (and there is an appropriate ample line bundle $\mathcal{L}$). Then the homomorphism (induced by the GIT projection) of algebraic fundamental groups $\pi_1(M)\to \pi_1(M/\! /_{\mathcal{L}}G)$ is an isomorphism. If $k = \mathbb{C}$, then there is also an isomorphism between the topological fundamental groups. -You can find this result in Fundamental group of a geometric invariant theoretic quotient by Indranil Biswas, Amit Hogadi, A. J. Parameswaran.<|endoftext|> -TITLE: Can one recover an algebraically closed field $k$ from the dots and arrows of its category of finitely generated $k$-algebras? -QUESTION [12 upvotes]: You are gracious enough to host me for a few days while I attend a conference. After I leave, you're surprised to see a gift on the kitchen table. It's a box with a category inside! The objects aren't labeled so it's a little hard to tell what's going on with it, but you can see, for example, that there's a terminal object. There's also a note: -Dear X, -I've always been very fond of this category, and I thought you might like it too. It's a category of finitely generated algebras over some algebraically closed field - unfortunately I've forgotten which one! But I'm sure you'll figure it out. -Regards, -Sarah -Can you determine what the field in question is? -More generally, given just the dots and arrows of a category (no taking sections!) and the information that for some unknown ring $R$ the category is the category of - -Classical varieties over $R$ -Affine schemes over $R$ -Schemes over $R$ -etc. - -in what cases is there a unique ring $R$ (up to isomorphism) which produces my category (up to equivalence)? -(I posted this on stack exchange, but it seems like a reasonable overflow question so I am crossposting it: https://math.stackexchange.com/questions/3002833/can-one-recover-an-algebraically-closed-field-k-from-its-category-of-finitely) - -REPLY [16 votes]: I think the following works for any commutative ring: Consider the category of abelian group objects in the overcategory above the initial element. This is equivalent to the category of finitely generated $R$-Modules (see for example https://ncatlab.org/nlab/show/module). By the general theory of Morita equivalence, this determines $R$. - -REPLY [6 votes]: Here is a crude attempt. I am working with the opposite category: affine schemes of finite type over $k$. We make the following definitions. -Let $T$ be the terminal object. Let us call a map $X\to Y$ injective/surjective/bijective/constant if it becomes so after taking ${\rm Hom}(T, -)$. Note that bijective does not imply isomorphism. -Say that $X$ is irreducible if whenever we have a map $Y\sqcup Z\to X$ which is surjective then one of $Y\to X$, $Z\to X$ is surjective. -Say $X$ is a fat point if $X\to T$ is bijective. -Say $X$ is an irreducible curve if whenever we have an injective $Y\to X$ with $Y$ irreducible, then either $Y$ is a fat point or there exist finitely many maps $T_i\to X$ ($i=1, \ldots, r$) with $T_i\simeq T$ such that -$$ Y\sqcup T_1 \sqcup \ldots \sqcup T_r \to X$$ -is bijective. -Finally, say $A$ is an affine line if it is an irreducible curve and has the property that every irreducible curve $Y$ admits a map $Y\to A$ which is not constant. -Now, let $A$ be an affine line. Fix two different points $0\colon T\to A$ and $1\colon T\to A$. There exists a unique structure of a ring object on $A$ with $0$ as zero and $1$ as one. -Then $k = {\rm Hom}(T, A)$ as rings. -Added later. One can extend this to perfect fields: -Say that $S$ is a separable field extension if it is not the disjoint union of two non-initial objects, and if $S\times S$ is isomorphic to a finite disjoint union of copies of $S$. -We change the first definition for $k$ algebraically above to read: -Let us call a map $X\to Y$ injective/constant if it becomes so after taking ${\rm Hom}(S, -)$ for every separable field extension. Call it surjective if for every $S\to Y$ from a separable field extension there exists a separable field extension $S'$ and a commutative square -$$ S'\to S, \quad S'\to X \quad X\to Y, \quad S\to Y $$ -(forgot how to draw diagrams). Call it bijective if it is both injective and surjective. -Now the rest of the definitions is unchanged, except that in the definition of an irreducible curve we allow the $T_i$ to be separable field extensions.<|endoftext|> -TITLE: Grassmannians of planes isotropic with respect to general tensors -QUESTION [7 upvotes]: In symplectic geometry, the Grassmannian of isotropic planes for a symplectic vector space is a well known and well studied object; for example, one can realize it as a homogeneous space with a known stabilizer subgroup of the symplectic group, and one can also realize a Schubert cell decomposition. -However, one can relax the symplectic condition in two ways: by relaxing its degeneracy condition, and by allowing it to take vector values (i.e., allowing for general tensors). A good example generalizing both of these situations can be found from the study of distributions: for any given distribution $D$ on a tangent bundle $TM$ of some smooth manifold $M$, there is a natural (O'Neill?) torsion tensor $T \in \Gamma(\wedge^2D^* \otimes TM/D)%$ defined by: -\begin{equation} -T(X,Y) = [X,Y] \text{ mod }D -\end{equation} -Clearly, $T$ is a skew symmetric bilinear map. However, its degeneracy could vary depending on the distribution. On one hand, $T = 0$ identically gives that $D$ is integrable. On the other, it being nondegenerate gives that $D$ is maximally non-integrable and is a contact structure. One can still make sense of planes isotropic with respect to $T$, as the planes on which $T$ vanish identically. - -Is there anything known about the isotropic Grassmannian of planes - with respect to the above tensor (e.g., is it homogeneous with respect - to a nice known Lie group? Can one describe its Schubert cell - decomposition?)? With respect to more general settings/tensors as above? - -I imagine in the case of the above particular tensor, one treats each component of the vector as a (possibly degenerate) symplectic form but I have not seen anything about it in the literature, so I'm hesitant to declare everything should be exactly the same as for the standard Grassmannian of isotropic planes with respect to an ordinary symplectic form. - -REPLY [11 votes]: What you are asking about is very classical in the theory of exterior differential systems. The subspaces of $D$ that you are calling `isotropic' are what Élie Cartan called the integral elements of the differential ideal $\mathcal{I}$ generated by the sections of $D^\perp\subset T^*M$ (the annihilator subbundle of $D$). The geometry and topology of these spaces of integral elements can be quite subtle, and a significant part of the basic theory of exterior differential systems is devoted to developing ways to describe these subspaces of the Grassmann bundles, particularly deriving necessary and sufficient conditions that a given integral element $E\subset D_x\subset T_xM$ should be a tangent space of an integral manifold, i.e., a submanifold $N\subset M$ all of whose tangent spaces are sub-planes of the plane field $D$. This leads to the Cartan-Kähler theory and beyond. -For references, you can look in our book Exterior Differential Systems (Bryant, et. al) or in Cartan for Beginners (Ivey and Landsberg).<|endoftext|> -TITLE: Divisors of the regular character of a finite group -QUESTION [8 upvotes]: Recall that the regular character $\rho=\hspace{-.2cm}\sum\limits_{\chi\in\operatorname{Irr}(G)}\hspace{-.2cm}\chi(1)\chi$ of a finite group $G$ takes values -$$ -\rho(g)= -\left\{\begin{array}{cl} - |G|,&\quad\mbox{if $g=1$.}\\ - 0,&\quad\mbox{if $g\in G\backslash\{1\}$.} - \end{array} -\right. -$$ -Given $\alpha\in\operatorname{Irr}(G)$, we shall say that $\alpha$ divides $\rho$ if $\alpha\beta=\rho$, for some generalized character $\beta$. -Now $\alpha\rho=\alpha(1)\rho$. So linear characters of $G$ divide $\rho$. Notice that if $\alpha$ is faithful, then there is an integer polynomial $p$ such that $\alpha p(\alpha)=n_\alpha\rho$, for some positive integer $n_\alpha$. But I don't know anything about $n_\alpha$. I have found only a small number of cases of $G$ and $\alpha$ where $\alpha$ does not divide $\rho$ (see below). My question is a small variant on https://math.stackexchange.com/questions/2933550/tensor-complement-of-representations-of-finite-groups: -Question: Does there exist a $p$-group $G$ and $\alpha\in\operatorname{Irr}(G)$, such that $\alpha$ does not divide $\rho$? -Note that in my original question solvable group was in place of $p$-group. As Jeremy Rickard has pointed out, there are 3 groups of order $72$ with irreducible characters which do not divide $\rho$ (SmallGroups(72,n), for $n=22,23,24$, in GAP notation). -Non-solvable example: Let $\alpha$ be one of the two degree 4 irreducible characters of $\operatorname{SL}(2,5)$. Then $\alpha$ has a single conjugacy class of zeros $4a$. Let $\beta$ be a class function such that $\alpha\beta=\rho$. Then $\beta(1a)=30$, $\beta(4a)$ is odd and $\beta$ vanishes on all other classes. Let $\psi\in\operatorname{Irr}(\operatorname{SL}(2,5))$, with $\psi(1)=2$. Then $\psi(4a)=0$. So $\langle\beta,\psi\rangle=\frac{1}{2}$, showing that $\beta$ is not a generalized character. -A similar example is provided by a degree $9$ irreducible character of the group $3.A_6.2_2$ (the second degree 2 extension of the triple cover of the alternating group $A_6$, in Atlas notation). The groups $\operatorname{SL}(2,5)$ and $3.A_6.2_2$ are in the small list of non-solvable groups with an irreducible character which vanishes on only one conjugacy class. See S.~Madanha, On a question of Dixon and Rahnamai Barghi, arXiv:1811.03972 [math.GR]. -On a positive note, $\alpha$ divides $\rho$ in the following cases: - -$\alpha(1)=|G|_p$, for some prime $p$ (take $\beta=1_S^G$, for $S\in\operatorname{Syl}_p(G)$). -$\alpha$ is totally ramified with respect to an irreducible character of $Z(G)$. -$\alpha$ is induced from a linear character of a normal subgroup of $G$ with cyclic quotient (R. Gow). -$G$ is a $2$-group with $|G|\leq256$ or a $3$-group with $|G|\leq729$ (GAP). In fact for all such $G$ and $\alpha$, it seems that there is a character $\beta$ with $\alpha\beta=\rho$. -$G=A_n$ or $S_n$, for $n\leq15$ (GAP). -$G=\operatorname{SL}(2,3),\operatorname{GL}(2,3)$ or the binary octahedral group (`fake $\operatorname{GL}(2,3)$'). -$G$ is a small finite simple group (GAP). - -REPLY [2 votes]: I am not sure if it helps, but I think in a minimal $p$-group $P$ which has an irreducible character $\alpha$ not being a divisor of $\rho,$ we may suppose that $\alpha$ vanishes identically outside $\Phi(P).$ -For suppose there is $x \in P \backslash \Phi(P)$ with $\alpha(x) \neq 0.$ We may choose a maximal subgroup $M$ of $P$ with $x \not \in M.$ Then ${\rm Res}^{P}_{M}(\alpha)$ is irreducible ( for otherwise $\alpha$ is induced from an irreducible character of $M$ and vanishes identically outside $M$ (so, in particular, $\alpha(x) = 0,$ contrary to assumption)). -By the minimal choice of $P,$ there is a generalized character $\gamma$ of $M$ with ${\rm Res}^{P}_{M}(\alpha) \gamma = \rho_{M},$ the regular character of $M$. -Now for each $y \in P \backslash M,$ both ${\rm Res}^{P}_{M}(\alpha)$ and $\rho_{M}$ are $y$-stable, so we also have ${\rm Res}^{P}_{M}(\alpha) \gamma^{y} = \rho_{M}.$ -Now it follows that $\alpha {\rm Ind}_{M}^{P}(\gamma) = \rho,$ since this generalized character certainly vanishes identically outside $M,$ and agrees with $p\rho_{M}$ on $M.$ Hence $\alpha$ is a divisor of $\rho,$ contrary to assumption.<|endoftext|> -TITLE: Effective Bertini -QUESTION [7 upvotes]: Let $X$ be a smooth complex projective manifold, and $L$ an ample line bundle. By Bertini's Theorem, for every integer $q$ big enough there exists an open dense subset $U_q\subset |qL|$ such that every divisor $D$ in $U_q$ is smooth. -Warm up question: is the complement of $U_q$ always a divisor? -We can define a bigger open subset $V_q\subset |qL|$ as -$$ -V_q:=\{D\in |qL| \; \textrm{s. t.} \; (X,\frac{1}{q}D) \; \textrm{ is klt} \} -$$ -My question is: what is the dimension of the complement of $V_q$ ? (or at least can we bound its asymptotic in $q$ ? e.g. is it upper-bounded by $aq^{\dim X}$ with $a$ a constant which is strictly smaller than the volume of $L$? ) - -REPLY [6 votes]: Regarding the warm up question: No (although for $q$ sufficiently large the answer is yes, as Jason Starr comments.) Let $X = \mathbb{P}^1 \times \mathbb{P}^2$ and let $L= \mathcal{O}(1,1)$. Write homogeneous coordinates on the first factor as $(u:v)$ and on the second factor as $(x:y:z)$. A divisor $D$ in $H^0(L)$ is of the form -$$u (ax+by+cz) + v (dx+ey+fz)=0.$$ -$D$ is singular if and only if the matrix -$$\begin{bmatrix} -a & b& c \\ d & e & f \\ \end{bmatrix}$$ -has rank $1$. (If this matrix has rank $2$, $D$ is isomorphic to $\mathbb{P}^2$ blown up at a point, when the matrix drops rank, $D$ turns into the union of a $\mathbb{P}^2$ and a $\mathbb{P}^1 \times \mathbb{P}^1$.) The condition that this matrix drops rank is codimension $2$. -The complement of $U_q$ is called the dual variety to $X$. It is usually (no precise meaning attached) true that the dual of a smooth variety is a divisor. For a while, it was an open problem to characterize smooth projective toric varieties whose duals were not divisors; this was solved by Dickenstein, Feichtner and Sturmfels.<|endoftext|> -TITLE: Distribution of $\{cn^a\}$ -QUESTION [8 upvotes]: Assume that $10$ or $f''(x)\le-\rho<0$ for $x\in[a,b]$. Then $$\left|\sum_{n=a}^be^{2\pi if(n)}\right|\le(|f'(b)-f'(a)|+2)\left({4\over\sqrt{\rho}}+3\right)$$<|endoftext|> -TITLE: Why is Voevodsky's motivic homotopy theory 'the right' approach? -QUESTION [42 upvotes]: Morel and Voevoedsky developed what is now called motivic homotopy theory, which aims to apply techniques of algebraic topology to algebraic varieties and, more generally, to schemes. A simple way of stating the idea is that we wish to find a model structure on some algebraic category related to that of varieties or that of schemes, so as to apply homotopy theory in an abstract sense. -The uninitiated will find the name of the theory intriguing, and will perceive the simple idea presented above as a very fair approach, but upon reading the details of the theory, he might get somewhat puzzled, if not worried, about some of its aspects. We begin with the following aspects, which are added for completeness. - -The relevant category we start out with is not a category of schemes, but a much larger category of simplicial sheaves over the category of smooth schemes endowed with a suitable Grothendieck topology. A relevant MathOverflow topic is here. -The choice of Grothendieck topology is the Nisnevich topology, the reasons being discussed right over here. -The choice of smooth schemes rather than arbitrary schemes has been discussed here. - -Let us now suppose the uninitiated has accepted the technical reasons that are mentioned inside the linked pages, and that he will henceforth ignore whatever aesthetic shortcomings he may still perceive. He continues reading through the introduction, but soon finds himself facing two more aspects which worries him even more. - -The resulting homotopy theory does not behave as our rough intuition would like. In particular, what we might reasonable want to hold, such as the fact that a space ought to be homotopy equivalent to the product of itself with the affine line, is false. We solve the issue by simply forcing them to be homotopy equivalent, and hope that whatever theory rolls out is more satisfactory. -There are two intuitive analogues of spheres in algebraic geometry, and the theory does not manage to identify them. We solve the issue by just leaving both of them into the game, accepting that all homology and cohomology theories will be bigraded, and we hope that this doesn't cause issues. - -The fact that the resulting theory is satisfactory, has proved itself over time. But hopefully the reader will not find it unreasonable that the uninitiated perceives the two issues mentioned above as a warning sign that the approach is on the wrong track, if not the 'wrong' one altogether; moreover, he will perceive the solutions presented as 'naive', as though it were but a symptomatic treatment of aforementioned warning signs. - -Question. How would you convince the uninitiated that Voevodsky's approach of motivic homotopy theory is 'the right one'? - -REPLY [20 votes]: I just want to point out, with regards to your second question, that the fact that the $\mathbb P^1$ is not equivalent to a simplicial complex homotopic to the sphere built out of affine spaces (or whichever model for an un-Tate-twisted sphere) is exactly what you would expect from Grothendieck et. al.'s theory of motives. -In fact having a homology or cohomology theory without a notion of Tate twist would be very strong evidence that we are on the wrong track! -More carefully one might point out that we want a map to etale cohomology and a notion of Chern class, which requires the Tate twisting because Tate twists show up in the Chern class map in etale cohomology.<|endoftext|> -TITLE: What is the fairest order for stage-striking (and is it the Thue-Morse sequence)? -QUESTION [19 upvotes]: Here's a fair-sequencing problem that doesn't quite match the usual fair-division problems. I think that, like those, the answer should also be the Thue-Morse sequence ("balanced alternation"), because the same heuristic reasoning that suggests it's the fairest way there works here as well, but the problem doesn't seem to reduce to them, so it's not obvious. (See here for more on using Thue-Morse for fair division, or this earlier MathOverflow question.) -Anyway, the problem is stage-striking (as is used in certain competitive video games for stage selection :) ). There are two players and $n+1$ objects ("stages"). The two players have different preferences regarding the stages (ideally, opposite preferences, but you'll see below why we don't assume that). The two players will take turns (in some order -- thus the question) removing ("striking") stages that have not already been struck; once only a single stage is left, that stage is selected (both players "get" it). The question, then, is what is the fairest order for stage striking; as mentioned above, I suspect it should be Thue-Morse (one player strikes on 0, the other player strikes on 1), for similar reasons that this is the answer to the old problem of what order to take turns in for fair division. -Of course this raises the question of how we're formalizing this and what we mean by "fair". I'll present here the formalization of the problem that (after discussing this with some other people) I think is best, but answers to other ways of formalizing it would also be OK so long as they don't trivialize the problem. -So -- note that if the players assign the stages opposite values (i.e. they agree about which stages give how much of an advantage to who), as you would expect, then the striking order becomes irrelevant, so long as both players get the same number of strikes; regardless of order, the median stage will be selected. So instead we have to assume the players may disagree about which stages advantage who. Also, since we can only really deal with the order of the stages here, we won't allow them to have arbitrary numeric values as in the fair-division problem; rather we'll assume each player assigns the $n+1$ stages the values $0, 1, \ldots, n$, so that the value of a stage to a player depends only on where it falls in their preference ordering. -Now, since perfect information makes the problem trivial, we'll go all the way in the opposite direction -- each player's preferences are uniformly random; or rather, each player sees the other's preferences as uniformly random. What we want to compare, then, and to make as equal as possible, is the expected value that player 1 gets (when player 2 strikes randomly), vs the expected value that player 2 gets (when player 1 strikes randomly). -(I'm pretty sure that, in this formulation, we can assume that each player always strikes their least-preferred stage at each step, and that there is no advantage from deviating from this. But obviously correct me if I'm wrong there...) -So, for instance, in this model, if $n=2$, then the first player to strike gets an expected value of $3/2$ (they eliminate their least preferred stage and get one of the remaining two at random), while the second player to strike gets an expected value of $5/3$ (they have a $2/3$ chance their most-preferred stage is not eliminated, and a $1/3$ chance they have to settle for the median). So we get a difference of $1/6$. You see? -So the question then is, is the Thue-Morse striking order the fairest? Or is it something else? Is it at least the fairest when $n$ is a power of $2$, even if it might not be otherwise? -EDIT: Actually, a thought -- maybe it should be reverse Thue-Morse? (As in, if $n=12$, you would go $011001101001$ rather than $011010011001$; you just reverse the sequence, and then, if necessary, swap the roles of the players so as to start with a $0$.) This seems possible because here it's going later, rather than going earlier, that seems to confer an advantage. Of course, if $n$ is a power of two, this distinction is irrelevant, as reversing the sequence would merely swap the roles of the players. - -REPLY [8 votes]: Here is a counter-intuitive result that gets us as far as possible from the Thue-Morse sequence. Infinitely many sequences which are alternating only once are among the fairest sequences of all. Their score differences are 0. -Let's use free-monoid notations and write $1^40^2$ for $111100$. Then -$v_0(1^40^2) = v_1(1^40^2) = 5$ -$v_0(1^{12}0^4) = v_1(1^{12}0^4) = 14$ -$v_0(1^{24}0^6) = v_1(1^{24}0^6) = 27$ -$v_0(1^{40}0^8) = v_1(1^{40}0^8) = 44$ -$v_0(1^{60}0^{10}) = v_1(1^{60}0^{10}) = 65$ -$v_0(1^{84}0^{12}) = v_1(1^{84}0^{12}) = 90$ -More generally, -$$v_0(1^{2k(k+1)}0^{2k}) = v_1(1^{2k(k+1)}0^{2k}) = k(2k+3)$$ -for any integer $k\ge 0$. -EDIT: for exactly the same lengths $n = 2k(k+2)$ as above, there are perfectly fair sequences with as many $0$'s as $1$'s, provided you allow two alternations instead of only 1. -$v_0(0^2 1^3 0) = v_1(0^2 1^3 0) = 5$ -$v_0(0^6 1^8 0^2) = v_1(0^6 1^8 0^2) = 14$ -$v_0(0^{12} 1^{15} 0^3) = v_1(0^{12} 1^{15} 0^3) = 27$ -$v_0(0^{20} 1^{24} 0^4) = v_1(0^{20} 1^{24} 0^4) = 44$ -$v_0(0^{30} 1^{35} 0^5) = v_1(0^{30} 1^{35} 0^5) = 65$ -$v_0(0^{42} 1^{48} 0^6) = v_1(0^{42} 1^{48} 0^6) = 90$ -and more generally -$$v_0(0^{k(k+1)}1^{k(k+2)}0^k) = v_1(0^{k(k+1)}1^{k(k+2)}0^k) = k(2k+3)$$ -for any integer $k\ge 0$. -$Proof$: As RaphaelB4 commented, his insight about the simple multiplicative form of his recursion, -$$v_1(0b_1\dots b_n)+1 = \frac{n+3}{n+2}\left(v_1(b_1\dots b_n)+1\right)$$ -iteratively yields -$$v_1(0^{i}b_1\dots b_n)+1 = \frac{n+2+i}{n+2}(v_1(b_1\dots b_n)+1)$$ -so that -$$v_0(1^{2k(k+1)}0^{2k}) + 1 = v_1(0^{2k(k+1)}1^{2k}) +1$$ -$$ = \frac{2k+2+2k(k+1)}{2k+2} (v_1(1^{2k})+1)$$ -$$ = (k+1) (2k+1) = k(2k+3) + 1$$ -and -$$v_1(1^{2k(k+1)}0^{2k}) + 1 = 2k(k+1) + v_1(0^{2k})+1$$ -$$ = 2k(k+1) + \frac{0+2+2k}{0+2} = k(2k+3) + 1$$ -which proves the first equality. The second equality with two alternations is similar. It is also easy to prove there isn't any other solutions with one or two alternations.<|endoftext|> -TITLE: Linear equations in primes -QUESTION [8 upvotes]: Quoting from Green-Tao, "Linear equations in primes" (especially Cor. 1.9 in https://arxiv.org/pdf/math/0606088.pdf), any system of linear forms of finite complexity and without any local obstructions will assume simultaneous prime values infinitely often. -(E.g., $(X,Y,X+Y)$ is of finite complexity, but with a local obstruction at 2, since one of these is always even. On the other hand $(X,Y,X+2Y)$ has no local obstruction, so there are infinitely many primes $x,y$ such that $x+2y$ is also prime.) -My question is: Does this result remain true when, instead of asking for prime values, one asks for values in some positive density subset of the primes? This kind of strengthening is true e.g. for the famous special case $X, X+Y, X+2Y,...$ of arithmetic progressions in primes, but I couldn't find it in the general case. -Again, to illustrate what I mean: Let's take the set $S$ of all primes $\equiv 1$ mod $4$. Then the system $(X,Y,3X+4Y)$ will be obstructed, since for $x,y\equiv 1$ mod $4$, one has $3x+4y\equiv 3$ mod $4$. On the other hand, $(X,Y,5X+4Y)$ would not be obstructed, so should be expected to have infinitely many values in $S^3$. Is this known in generality? -[I should add that for the last example, one may of course just replace X and Y by 4X+1 and 4Y+1; but of course there are positive density subsets of primes which are not just residue classes. I'm interested in those as well.] -[EDIT: To be more explicit: Let S be a Chebotarev set of primes, i.e. the set of all primes with some given Frobenius in some given Galois extension $K/\mathbb{Q}$. Then there will usually be a congruence condition (but not only!) on $S$ coming from the largest cyclotomic subfield of $K$. So there is some $N$ and some smallest possible finite union $U$ of mod $N$ residue classes containing all of $S$. A local ostruction for a system of linear forms now definitely occurs if there are no integer values simultaneously lying in $U$. I'm asking if that is the only obstruction for values to simultaneously lie in $S$. -Example: if S is the set of primes which decompose completely in the splitting field of $x^3-x^2+1$, then the largest abelian subfield is $\mathbb{Q}(\sqrt{-23})$. To be decomposed in this field means to be a square mod 23 (leaving 11 of 23 possible residue classes). I'm tempted to conjecture that a system of linear forms of finite complexity which has integer values simultaneously lying in one of these 11 classes (i.e. not obstructed at 23), and which for any other prime $p$ has integer solutions simultaneously coprime to $p$ (i.e. not obstructed at $p$) will have prime values simultaneously lying in $S$.] - -REPLY [10 votes]: Roughly speaking, the transference principle used in my work with Ben shows that the obstructions to solving (finite complexity) linear equations in dense sets of primes are the same as the obstructions to solving linear equations in dense sets of integers (modulo a technical issue involving dilating the equations by a medium-size integer $W$, coming from the "$W$-trick"). Thanks to the inverse conjecture for the Gowers norms, these obstructions are classified: local obstructions such as congruence classes are certainly obstructions, but in fact every nilsystem could potentially produce an obstruction. -For instance, the circle shift $x \mapsto x + \sqrt{2} \hbox{ mod } 1$ on ${\bf R}/{\bf Z}$ provides a family of obstructions based on the distribution of values of $\sqrt{2} n \hbox{ mod } 1$ in the set $S$ of interest. If for example $S$ happens to be contained in the Bohr set $\{ n: \{ \sqrt{2} n \} \in [0, 0.01] \}$ (where $\{x\} := x - \lfloor x \rfloor$ is the fractional part function), then there will be no patterns of the form $x, x+y, x+2y+2$ in the set $S$, despite the lack of congruence obstructions, simply because $\sqrt{2} x - 2 (\sqrt{2} (x+y)) + \sqrt{2}(x+2y+2) = 2\sqrt{2} \hbox{ mod } 1$ is too far away from zero. Intersecting this Bohr set with the primes would then give a positive density set of primes that had no patterns of the form $x,x+y,x+2y+2$. -For more complicated patterns one can cook up similar examples involving for instance the quadratic Bohr set $\{ n: \{ \sqrt{2} n^2 \}\in [0,0.01]\}$ or the more exotic ``bracket quadratic'' Bohr set $\{ n: \{ \lfloor \sqrt{2} n \rfloor \sqrt{3} n \} \in [0,0.01] \}$ that is associated to a Heisenberg nilflow. For a pattern of complexity $s$ (as defined in our paper, or by using the somewhat smaller notion of "true complexity" studied in later papers by Gowers-Wolf and others), every nilsystem of nilpotency class $s$ or less is a potential obstruction. (Congruence class obstructions can be thought of as corresponding to finite systems, which by convention can be viewed as "complexity 0" obstructions.) -On the other hand, if your set $S$ is equidistributed with respect to all such Bohr sets (after taking into account the local irregularities of distributions of the primes, for instance by using the "W-trick" of our paper), then the theory in my papers with Ben (together with one paper as well with Tamar Ziegler) will allow one to compute asymptotics for these patterns.<|endoftext|> -TITLE: Nontrivial p-divisible groups over $\mathbb Z$ for general prime $p$ -QUESTION [9 upvotes]: In Tate's famous paper about $p$-divisible groups, for a prime number $p$ he asks whether there exists a $p$-divisible group $G$ over $\mathbb Z$ such that $G$ is not a direct sum of $\mu_{p^\infty}$ and $\mathbb Q_p/ \mathbb Z_p $, i.e a direct sum of a constant group and a diagonalizable group. He calls such $p$-divisible groups nontrival. -This question has some good applications. For instance we know there is no abelian scheme over $\mathbb Z$, whose key idea involves analyzing $p$-divisible groups for small $p$ using discriminant bound. In particular Tate's question is true for $p=3,5,7,11,13,17$ (see theorem $4$ in Fontaine's paper Il n'y a pas de variété abélienne sur $\mathbb Z$ for a proof). -For the negative side, the case $p=2$ is bad but also good up to isogeny. In an old paper $p$-divisible group over $\mathbb Z$ by V. A. Abrashkin, he claims that there also exist nontrivial $p$-divisible groups for some small irregular primes, but this seems to be wrong (see one answer below). -Odlyzko's discriminant bound is not good for large $p$. But time has passed for more than $30$ years, what about the general case now? Is there any progress towards Tate's question? With the technique of mordern $p$-adic Hodge theory, maybe such question could be solved. Indeed, some generalizations of Fontaine's result for abelian schemes rely on the Fontaine-Laffaille theory. - -REPLY [5 votes]: From works of V. A. Abrashkin, we know there exist nontrivial $p$-divisible groups for $p=2$ and some irregular primes. - -This is not true, and Tate's conjecture is very much still expected to hold. That said, the only known general cases are those which proceed via discriminant bounds, which might just include $p = 2$ and $p = 3$ (and maybe $p = 5$ or $7$ i'm not sure). - -Suppose that $V$ is a $p$-divisible group over $\mathbf{Z}$, and assume that the corresponding Galois representation is absolutely irreducible. If you could prove that $V = \mathbf{Q}_p/\mathbf{Z}_p$ or $\mu_{p^{\infty}}$ then you would know all $p$-divisible groups over $\mathbf{Z}$, because you know how to compute extensions between such objects (there are none if $p > 2$ and easy to understand when $p = 2$ (there is an extension killed by $2$ and split over $\mathbf{Q}(\sqrt{-1})$.) -The main thing that has changed since Tate's original conjecture is the Langlands program. In this context, $V$ conjecturally gives rise to a cuspidal automorphic form $\pi$ for $\mathrm{GL}(n)/\mathbf{Q}$ of level one. By the Fontaine-Mazur conjecture and the standard conjectures, it should come from a pure motive $M$ with Hodge Tate weights $0$ and $1$. Possibly $M$ might have coefficients, but by taking the direct sum of the conjugates of this compatible system we may assume that $M$ is a pure motive over $\mathbf{Z}$ (though maybe now no longer absolutely irreducible). If $M$ is of weight $0$, then $M$ is etale and then it is easy to see that $M$ comes from the trivial motive $\mathbf{Z}$, and $V = \mathbf{Q}_p/\mathbf{Z}_p$. Similarly, if $M$ has weight $2$, then it is Cartier dual to something etale, and then $M$ has to be $\mathbf{Z}(1)$ and $V = \mu_{p^{\infty}}$. So it remains to consider the case when $M$ has weight one. The infinity type of $\pi$ is then determined by the Hodge--Tate weights, since complex conjugation on the Hodge structure sends $H^{1,0}$ to $H^{0,1}$ and is thus completely determined. In particular, the $L$-function $L(M,s) = L(\pi,s)$ can then be shown not to -exist precisely in the same way that Mestre proves that the $L$-function of an abelian variety $A$ over $\mathbf{Z}$ does not exist -(the same argument works). -So the real "modern work" on this problem is the work of modularity in the Langlands program. For example, if $V$ is irreducible and is assumed to come from a $2$-dimensional representation (possibly with coefficients), then the work of Chandrashekar Khare (Serre's modularity conjecture: the level one case, DUKE) proves that $V$ does not exist. -I tried to look at Abrashkin's paper (in cyrilic) to see what was going on. This is just a guess, because I don't speak Russian. -On page 1006 or so, there seems to be exact sequences of finite flat group schemes of the form: -$$0 \rightarrow (\mathbf{Z}/p^n \mathbf{Z}) \rightarrow \Gamma \rightarrow \mu_{p^n} \rightarrow 0$$ -(or rather the generic fibre of such finite flat group schemes). For this to actually come from a non-split extension of finite flat group schemes, there would have to be a non-trivial extension of finite flat group schemes of $\mu_p$ by $\mathbf{Z}/p \mathbf{Z}$. Since, (by the connected-etale sequence) such sequence would locally split, this would give rise to an unramified extension of $\mathbf{Q}(\zeta_p)$ and imply that $p$ is irregular. However, the converse is not true. For the converse to apply, there would have to be an extension of a very specific form, namely a $p$-quotient of the class group with a precise action of the Galois group of $\mathbf{Q}(\zeta_p)$. Such things do not exist, by Herbrand's theorem. Already the lack of such extensions was proved by Mazur in his Eisenstein ideal paper (See proposition 2.1, page 49). But again I can't be sure if this is exactly what is going on in the cited paper. -tl;dr: This problem "reduces" to the Langlands conjecture. Our best people are currently working on it. If you want to do something useful, read Mazur's Eisenstein Ideal paper and then everything on the Langlands program in the last 50 years.<|endoftext|> -TITLE: Is there a pair of non-isomorphic structures each of which is isomorphic to an ultrapower of the other? -QUESTION [16 upvotes]: Does there exist a pair of non-isomorphic structures $\mathfrak{A}$ and $\mathfrak{B}$ as well as sets $I$ and $J$ and ultrafilters $\mathcal{U}$ on $I$ and $\mathcal{F}$ on $J$ such that $\mathfrak{A}^I/\mathcal{U}\cong\mathfrak{B}$ and $\mathfrak{B}^J/\mathcal{F}\cong\mathfrak{A}$? - -This question is inspired by the Keisler-Shelah theorem, which says that you can arrange $\mathfrak{A}^I/\mathcal{U} \cong \mathfrak{B}^J/\mathcal{F}$ (moreover you can arrange that $\mathfrak{A}^I/\mathcal{U}\cong \mathfrak{B}^I/\mathcal{U}$). The answer to this question very likely depends on set theoretic assumptions, so a reasonable weakening would be mere consistency relative to some strong set theoretic assumptions or in some forcing extension. Another reasonable weakening would be to ask about iterated ultrapowers rather than just ultrapowers. -On the other hand assuming a positive answer an obvious follow-up would be the question of the existence of a triple of pairwise non-isomorphic structures, each of which is isomorphic to ultrapowers of the other two. Another obvious follow-up is whether or not it can be arranged that $\mathfrak{A}^I/\mathcal{U}\cong\mathfrak{B}$ and $\mathfrak{B}^I/\mathcal{U\cong{\mathfrak{A}}}$, i.e. whether or not there exists a structure and an ultrafilter such that iterating ultrapowers by that ultrafilter alternates between two isomorphism classes. - -REPLY [8 votes]: An example assuming large cardinals: suppose $U$ and $W$ are normal ultrafilters on a measurable cardinal $\kappa$ such that $(V_{\kappa+2})^{M_U}\neq (V_{\kappa+2})^{M_W}$, where $M_U$ denotes the transitive collapse of the ultrapower of $V$. Let $Z = U\times W$. Let $M$ be the iterated ultrapower of length $\omega$ hitting $Z$ and its images. Then $M = j_Z(M) = j_U(j_W(M))$. But $j_W(M)$ is not isomorphic to $M$ since $(V_{\kappa+2})^M = (V_{\kappa+2})^{M_U}\neq (V_{\kappa+2})^{M_W} = (V_{\kappa+2})^{j_W(M)}$. For any structure $N$ in the language of set theory and any ultrafilter $D$, $j_D(N)$ is isomorphic to the (external) ultrapower of $N$ by $D$. So $M^\kappa / W$ is not isomorphic to $M$ but $(M^{\kappa}/W)^\kappa/U$ is isomorphic to $M$, as desired. If you want to replace $M$ with a set sized structure, consider instead $(V_\lambda)^M$ where $\lambda$ is a fixed point of $j_Z$.<|endoftext|> -TITLE: Examples of Mathematical Papers that Contain a Kind of Research Report -QUESTION [48 upvotes]: What are examples of well received mathematical papers in which the author provides detail on how a surprising solution to a problem has been found. -I am especially looking for papers that also document the dead ends of investigation, i.e. ideas that seemed promising but lead nowhere, and where the motivation and inspiration that lead to the right ideas came from. -By "surprising solution" I mean solutions that feel right at first reading and it isn't clear why they haven't been found earlier. - -REPLY [3 votes]: I enjoy The genesis of the Macdonald polynomial statistics, complete with journal entries, and detailed descriptions of the experimental method. -This paper describes how the researchers came up with a nice formula for the combinatorial (aka modified) Macdonald polynomials.<|endoftext|> -TITLE: Killing vector fields of a conformally flat Riemannian metric -QUESTION [5 upvotes]: Let $f: \mathbb{R}^n \to \mathbb{R}$ be a smooth function and let's consider the conformally flat Riemannian metric $g = e^f \delta_{ij} dx^idx^j$ on $\mathbb{R}^n$. -Is it true that the Killing vector fields of the manifold $(\mathbb{R}^n, g)$ are exactly the Killing fields of the Euclidean space such that $f$ is constant along their flow? -For sure if $K$ is an Euclidean Killing vector field with such property, then it must be a Killing field also for the manifold $(\mathbb{R}^n, g)$, but I'm not completely sure that those are exactly all the possible Killing fields of $(\mathbb{R}^n, g)$. I checked some simple cases and it seems to be the case. -Is it true in general? Or is it possible to find a counterexample where $(\mathbb{R}^n, g)$ has some symmetries that do not arise from symmetries of $f$? -Any help will be very appreciated! Thanks a lot! - -REPLY [4 votes]: Hint -Assume that $h=e^fg$ is a conformal change of $g$ on a manifold $M$. Let $X$ be a vector field on $M$. Then -$${\cal{L}}_{X}h={\cal{L}}_{X}(e^fg)\\ -={\cal{L}}_{X}(e^f)g+e^f{\cal{L}}_{X}g\\ -=e^fX(f)g+e^f{\cal{L}}_{X}g$$ -where ${\cal{L}}_{X}$ denotes the Lie derivative in direction of $X$ on $M$. This leads to say "Killing fields of both metrics are the same if and only if $f$ is constant along the flow lines of these vector fields"<|endoftext|> -TITLE: "Closed bicategories" -QUESTION [8 upvotes]: I am interested in the following property that a bicategory may or may not have. -Let $\mathbf{B}$ be a bicategory. Every one-morphism $f\colon x\rightarrow y$ defines a functor $\mathbf{B}(y,z)\rightarrow \mathbf{B}(x,z)$ for an arbitrary object $z$ via precomposition $g\mapsto g\ast f$. Similarly, there is a postcomposition functor. -We claim that -($\ast$) The pre- and postcompositions with any $1$-morphism $f$ (with respect to an arbitrary object $z$) both have right adjoints $f_{!}$ and $f^{!}$. -Examples: 1. A bicategory with one object aka monoidal category has property ($\ast$) if and only if it is closed (sometimes called biclosed). Thus the title of the question. - -The bicategory with rings as objects, $(R,S)$-bimodules as $1$-morphisms (composition: tensor product) and $(R,S)$-linear maps as $2$-morphisms also has this property. If $M$ is an $(R,S)$-bimodule and $N$ is an $(R,T)$-bimodule, then $M_{!}(N)$ is given by $\mathrm{Hom}_R(M,N)$ which is an $(S,T)$-bimodule. -The bicategory of profunctors $\mathrm{Prof}(\mathcal{V})$ where $\mathcal{V}$ is a cosmos (similar construction to 2). - -Question: Has this notion been studied? - -REPLY [6 votes]: Section 4 of Street's 1974 paper Elementary cosmoi is about "extension systems", which are like bicategories but have only one of the "adjoints to composition" (not composition itself), just like a closed category has "only" the internal-hom of a closed monoidal category and not the monoidal structure itself. He remarks as a special case that a bicategory in which all right extensions exist (a "right (Kan) extension" is another name for one of the adjoints to composition, the other one being a "right (Kan) lifting") yields an extension system, and that a bicategory of this sort is called a "closed bicategory" by "some authors". So the notion is at least that old (Wood's paper mentioned in Fosco's answer is from 1982), although unfortunately this doesn't help find the oldest usage. -Such a use of "closed bicategory" for a bicategory with only one sided adjoints to composition is in line with the use of "biclosed bicategory" when both adjoints exist, although I personally don't really like using the prefix "bi-" with two different meanings in the same phrase, and the plain "closed bicategory" is ambiguous as to which adjoint is meant. I prefer the terminology used e.g. in May-Sigurdsson, section 16.3 where a "closed bicategory" has both adjoints, and when only one is present we speak of "left closed" or "right closed" bicategories (although there's probably no canonical way to decide which is "left" and which is "right").<|endoftext|> -TITLE: An upper bound for the largest Laplacian eigenvalue of a graph in terms of its diameter -QUESTION [10 upvotes]: Let $G$ be a simple graph with $n$ vertices and $\lambda$ be the largest eigenvalue of its Laplacian operator $L=D-A$. I have some evidence for the following conjecture: - -Conjecture: If G has diameter $\delta>3$ then $\lambda\leq n-1$. - -I need a proof or a counterexample for this conjecture. Does there exist a good general upper-bound for $\lambda$ in terms of $\delta$ which includes the above conjecture (if it is true)? - -REPLY [4 votes]: I have not been able to conclude, but I think the following is a good start. (EDIT: the proof should be complete now) -Because the diameter is at least $4$, there exist $x,y$ with $d(x,y)\geq4$. -In particular for all $z$ , $d(x,z)+d(y,z)\geq4$ We recall that -$$ -\lambda=\frac{1}{2}\max_{\sum|u(i)|^{2}=1}\sum_{i\sim j}(u(i)-u(j))^{2} -$$ -and therefore is monotone increasing in the edges. To consider the -maximum of $\lambda$, it is enough to study the maximal graphs $G=(V,E)$ -with $d(x,y)=4$. We can then suppose that for all $z$: $d(x,z)+d(y,z)=4$. -(Indeed if for example $d(x,z)\geq3$ (resp. $d(y,z)$) let $z'$ -such that $d(x,z')=2$, we can add the edges $(z,y)$ and $(z,z')$ -to our graph). -Let us call -$$ -A=\text{\{}z:d(x,z)=1\text{\}} -$$ -$$ -B=\text{\{}z:d(x,z)=2\text{\}} -$$ -$$ -C=\text{\{}z:d(x,z)=3\text{\}} -$$ -Because of the maximality of $G,$ we have -$$ -V=A\cup B\cup C\cup\text{\{}x,y\text{\}} -$$ -$$ -E=\text{\{}(x,z):z\in A\text{\}}\cup\text{\{}(z_{1},z_{2}):z_{1},z_{2}\in A\cup B\text{\}}\cup\text{\{}(z_{1},z_{2}):z_{1},z_{2}\in B\cup C\text{\}}\cup\text{\{}(y,z):z\in C\text{\}}. -$$ -Let us now consider the following orthonormal family : $1_{\text{\{}x\text{\}}},1_{\text{\{}y\text{\}}},u_{A}=\frac{1_{A}}{\sqrt{|A|}},u_{B}=\frac{1_{B}}{\sqrt{|B|}},u_{C}=\frac{1_{C}}{\sqrt{|C|}}$. -We have -$$ -\begin{cases} -L(1_{\text{\{}x\text{\}}})=|A|1_{\text{\{}x\text{\}}}-1_{A}\\ -L(1_{A})=-|A|1_{\text{\{}x\text{\}}}-|A|1_{B}+(1+|B|)1_{A}\\ -L(1_{B})=-|B|1_{A}-|B|1_{C}+(|A|+|C|)1_{B}\\ -L(1_{C})=-|C|1_{\text{\{}y\text{\}}}-|C|1_{B}+(1+|B|)1_{C}\\ -L(1_{\text{\{}y\text{\}}})=|C|1_{\text{\{}y\text{\}}}-1_{C}. -\end{cases} -$$ -This gives in the family $(1_{\text{\{}x\text{\}}},u_{A},u_{B},u_{C},1_{\text{\{}y\text{\}}})$ -the following matrix -$$ -M=\begin{pmatrix}|A| & -\sqrt{|A|} & & 0 & 0\\ --\sqrt{|A|} & |B|+1 & -\sqrt{|A||B|} & & 0\\ - & -\sqrt{|A||B|} & |A|+|C| & -\sqrt{|B||C|}\\ -0 & & -\sqrt{|B||C|} & |B|+1 & -\sqrt{|C|}\\ -0 & 0 & & -\sqrt{|C|} & |C| -\end{pmatrix} -$$ -Actually we can write $L$ as the block matrix -$$ -L=\begin{pmatrix}M & 0\\ -0 & L' -\end{pmatrix} -$$ -EDIT (2) :We note $p$ the orthonormal projector on $(1_{\text{\{}x\text{\}}},u_{A},u_{B},u_{C},1_{\text{\{}y\text{\}}}) )^\perp$ and with a small abuse on notation we denote -$L'=pLp$. Let $a\in A$ then $L(1_{a})=-1_{x}-1_{B}-1_{A}+(|A|+|B|)1_{a}$. We have $p(1_{a})=1_{a}-\frac{1}{\sqrt{A}}u_{A}$ -and therefore $pLp(1_{a})=(|A|+|B|)(1_{a}-\frac{1}{\sqrt{A}}u_{A})$. Similar for $b\in B$ and $c\in C$. Finally we have for any $v$ -$$ \langle v, pLp v\rangle=(|A|+|B|)\big(\sum_{a\in A} |v(a)|^2-\frac{1}{|A|} (\sum_{a\in A} v(a) )^2\big)+(|A|+|B|+|C|-1)\big(\sum_{b\in B} |v(b)|^2-\frac{1}{|B|} (\sum_{b\in B} v(b) )^2\big)+(|B|+|C|)\big(\sum_{c\in C} |v(c)|^2-\frac{1}{|C|} (\sum_{c\in C} v(c) )^2\big) $$ -which is smaller than $(|A|+|B|+|C|-1)\|v\|^2$ -So we only have to deal with $M$. The -problem can be state as follow :Do we have $\|M\|\leq|A|+|B|+|C|+1$? -Here is where I get stuck. It should be possible to do a complete analysis of $M$ and calculate explicitely the largest eigenvalue but it seems a bit tedious. We can also make some numerical simulations : no counter example appear for $|A|+|B|+|C|\leq 100$ -EDIT (end of the proof) -To finish the proof, we show that -$$ -|A|+|B|+|C|+1-M=\begin{pmatrix}|B|+|C|+1 & \sqrt{|A|} & 0 & 0 & 0\\ -\sqrt{|A|} & |A|+|C| & \sqrt{|A||B|} & 0 & 0\\ -0 & \sqrt{|A||B|} & |B|+1 & \sqrt{|B||C|} & 0\\ -0 & 0 & \sqrt{|B||C|} & |A|+|C| & \sqrt{|C|}\\ -0 & 0 & 0 & \sqrt{|C|} & |A|+|B|+1 -\end{pmatrix} -$$ -is a positive matrix, computing all its minors determinants. Computations -gives (notation :$|A|=a,|B|=b,|C|=c$): -$$ -m_{1}=b+c+1 -$$ -$$ -m_{2}=ab+ac+bc+c^{2}+c -$$ -$$ -m_{3}=ac+b^{2}c+bc^{2}+2bc+c^{2}+c -$$ -$$ -m_{4}=a^{2}c+2abc+2ac^{2}+ac+bc^{2}+c^{3}+c^{2} -$$ -$$ -m_{5}=a^{3}c+3a^{2}bc+2a^{2}c^{2}+2a^{2}c+2ab^{2}c+3abc^{2}+3abc+ac^{3}+2ac^{2}+ac -$$ -and they are all positive.<|endoftext|> -TITLE: Surjectivity of differential operators with constant coefficients -QUESTION [8 upvotes]: I would like a proof or a reference (or a counter-example...) for the following fact. Let $P\in \mathbb{C}[x_1,\ldots ,x_n]$ and $D\in \mathbb{C}[\frac{\partial }{\partial x_1} ,\ldots ,\frac{\partial }{\partial x_n}]$ be nonzero homogeneous polynomials. Then there exists a homogeneous polynomial $Q\in \mathbb{C}[x_1,\ldots ,x_n]$ such that $D(Q)=P$. - -REPLY [2 votes]: Too long for a comment. I want to use a version of the Lojaciewicz theorem of division of distributions by an analytic function (in fact Hörmander's result of division by a polynomial). We may assume that $P(x) =x^\alpha=x_1^{\alpha_1}\dots x_n^{\alpha_n}$ a monomial homogeneous with degree $\vert \alpha\vert=\alpha_1+\dots +\alpha_n$. Let us replace your notation $D$ by an operator $$A(D)=\sum_{\vert \beta\vert =m}a_\beta D^\beta,\quad D=-i\nabla.$$ -The question at hand is to find an homogeneous polynomial $Q$ such that -$ -A(D) Q= x^\alpha. -$ -By Fourier transformation, it is equivalent to solve -$$ -A(\xi)\widehat Q(\xi)=i^{\vert \alpha\vert}\delta^{(\alpha)}, -$$ -which means divide a derivative of the Dirac mass by an homogeneous ($A$ is assumed to be non-zero) polynomial. It is indeed possible by the aforementioned results of division and we find that $\widehat Q$ is homogeneous with degree $-n-\vert \alpha\vert-m$, so that $Q$ is homogeneous with degree $\vert \alpha\vert+m$. Moreover $\widehat Q$ is supported at the origin, proving that $Q$ is a polynomial.<|endoftext|> -TITLE: example of a non-amenable l.c. group such that $C_r^*(G)$ satisfies the UCT -QUESTION [5 upvotes]: Are there known any examples of non-amenable locally compact (or more restrictive, non-amenable discrete) groups $G$ for which the reduced group $C^*$-algebra $C_r^*(G)$ satisfies the universal coefficient theorem (UCT)? In this case, $C_r^*(G)$ is non-nuclear. For example,$G=\mathbb{F}_2$ the free non-abelian group in 2 generators is known to be non-amenable. However, $C^*(\mathbb{F}_2)$ is exact. However, I don't know whenever or not this $C^*$-algebra satisfies the UCT. -I am happy about any references, comments etc... Thanks - -REPLY [6 votes]: Both $C^\ast(\mathbb F_2)$ and $C^\ast_r(\mathbb F_2)$ satisfy the UCT. This is the special case of the following: -$\mathbf{Theorem}$. If $G$ and $H$ are countable, discrete, amenable groups, then $C^\ast(G\ast H)$ and $C^\ast_r(G \ast H)$ are $KK$-equivalent and satisfy the UCT. -$\mathbf{Proof}$. By Theorem 2.4 (c) in Cuntz' paper "$K$-theoretic amenability for discrete groups" it follows that $G\ast H$ is $K$-amenable and thus $C^\ast(G\ast H)$ and $C^\ast_r(G\ast H)$ are $KK$-equivalent. In the beginning of Section 3 of the same paper, Cuntz shows/remarks that $C^\ast(G\ast H)$ is $KK$-equivalent to the pull-back -\begin{equation} -C^\ast(G) \oplus_{\mathbb C} C^\ast(H) = \{ (x,y) \in C^\ast(G) \oplus C^\ast(H) : t_G(x) = t_H(y)\} -\end{equation} -via the trivial representations $t_G$ and $t_H$, so it suffices to show that this $C^\ast$-algebra satisfies the UCT. It fits into a short exact sequence -\begin{equation} -0 \to I(G) \to C^\ast(G) \oplus_{\mathbb C} C^\ast(H) \to C^\ast(H) \to 0 -\end{equation} -where $I(G) = \mathrm{ker} \, t_G$ is the augmentation ideal. As $C^\ast(H)$ is nuclear the sequence is semi-split, so $C^\ast(G) \oplus_{\mathbb C} C^\ast(H)$ satisfies the UCT provided that $C^\ast(H)$ and $I(G)$ satisfy the UCT, by the 2-out-of-3-property for satisfying the UCT. $C^\ast(H)$ and $C^\ast(G)$ satisfy the UCT by Tu's theorem. As $I(G)$ fits into the split short exact sequence -\begin{equation} -0 \to I(G) \to C^\ast(G) \to \mathbb C \to 0, -\end{equation} -and as $C^\ast(G)$ and $\mathbb C$ satisfy the UCT, so does $I(G)$ which completes the proof. QED -Note that for the free group $\mathbb F_2 = \mathbb Z \ast \mathbb Z$ one does not need to use Tu's deep theorem to obtain UCT. -To my knowledge, the only known group $C^\ast$-algebras which do not satisfy the UCT, are $C^\ast_r(G)$ when $G$ is a countable, discrete group with property (T) and the Akemann-Ostrand property, see Skandalis' "Une Notion de Nuclearite en K-Theorie". By also assuming that the group is residually finite, one can show that $C^\ast(G)$ does not satisfy UCT.<|endoftext|> -TITLE: $*$-algebras, completions, and $K$-theory -QUESTION [7 upvotes]: What is an example of a $*$-algebra $\cal{A}$, which admits two non-equivalent norms $\| \cdot \|_1$ and $\| \cdot \|_2$, with respect to which we can complete $\cal{A}$ to give two $C^*$-algebras $A_1$ and $A_2$, such that the two associated $K$-theory groups are non-isomorphic, that is: -$$ -K(A_1) \not\simeq K(A_2). -$$ - -REPLY [11 votes]: Any infinite discrete group $\Gamma$ with Kazhdan's property (T) gives an example. Since it is not amenable, the full and reduced C*-algebras (which are both completions of the group algebra) do not coincide. Moreover, the full C*-algebra contains a projection with non-trivial K-theory class (so-called Kazhdan projection) which is mapped to $0$ in the reduced C*-algebra. -As for free groups, they are $K$-amenable, meaning that the canonical surjection between the full and reduced C*-algebras induces an isomorphism in $K$-theory. -The basic reference for this is J. Cuntz's article K-theoretic amenability for discrete groups, J. reine angew. Math. 1983 (where he also acknowledges overlapping work of E.C. Lance).<|endoftext|> -TITLE: Koszul-Tate Resolution for Subvarieties of $\mathbb P^n$ -QUESTION [8 upvotes]: All varieties appearing below are assumed smooth projective over $\mathbb C$ and all vector bundles, sections etc are assumed to be algebraic/holomorphic. We use the word resolution to mean quasi-isomorphism. -Suppose $X\subset\mathbb P^n$ is a closed subvariety of codimension $r$. Suppose we can a find a locally free sheaf $E$ on $\mathbb P^n$ of rank $r$ and a section $s$ such that $s^{-1}(0)$ is cut out transversely and equals $X$ (for example, if $X$ is a complete intersection). Then, we get the Koszul resolution -$0\to\wedge^rE^\vee\to\cdots\to\wedge^2E^\vee\to E^\vee\to\mathcal O_{\mathbb P^n}\to\mathcal O_X\to 0$ -I would like to view this as a resolution of $\mathcal O_X$ by a differential graded commutative $\mathcal O_{\mathbb P^n}$-algebra $K(s)$ with $E^\vee$ placed in degree $-1$ (call $K(s)$ the Koszul algebra of $s$). Notice that as a graded commutative algebra $K(s)$, is simply the (graded commutative) symmetric algebra $\text{Sym}^\bullet_{\mathcal O_{\mathbb P^n}}(E^\vee[1])$. Now, suppose we had an extension $0\to E\to E'\to F\to 0$ of vector bundles, then we could consider the section $s'$ of $E'$ which is the image of $s$. Now, the Koszul algebra $K(s')$ is no more a resolution of $\mathcal O_X$, in fact its cohomology algebra is $\mathcal O_X\otimes_{\mathcal O_{\mathbb P^n}}\wedge^\bullet F^\vee$. Thus, by adding the free generator $F^\vee$ in degree $-2$ to $K(s')$, we get a new commutative differential graded algebra with the differential mapping $F^\vee$ to $E'^\vee$ by the dual of the map $E'\to F$. The evident map from this new algebra to $K(s)$ is a quasi-isomorphism and thus, we again have a resolution of $\mathcal O_X$. -More generally, there is the notion of Koszul-Tate resolution of $\mathcal O_X$ which consists of a commutative differential graded algebra $A$ over $\mathcal O_{\mathbb P^n}$ resolving $\mathcal O_X$ and having the following additional property. Locally on $\mathbb P^n$, $A$ is given (as a graded algebra) by a free (graded commutative) polynomial ring (in possibly infinitely many variables) over $\mathcal O_{\mathbb P^n}$ with finitely many generators in each negative degree. We'll say that $A$ is locally finitely generated if, locally on $\mathbb P^n$, it is a polynomial algebra over $\mathcal O_{\mathbb P^n}$ in finitely many variables (with the variables placed in negative degrees). -We have seen above that $K(s)\to\mathcal O_X$ is a resolution with $K(s)$ finitely generated. Are there any other examples of locally finitely generated resolutions besides the Koszul algebras $K(s)$ (and quasi-isomorphic modifications coming from exact sequences of the form $0\to E\to E'\to \cdots\to F\to 0$)? -In particular, are there examples of smooth $X\subset\mathbb P^n$ which are not complete intersections, for which $\mathcal O_X$ admits a Koszul-Tate resolution by a locally finitely generated differential graded commutative $\mathcal O_{\mathbb P^n}$-algebra? It seems pretty easy to show that Koszul-Tate resolutions exist if we do not require them to be locally finitely generated. -Note that not every $X\subset\mathbb P^n$ admits such a resolution, since a Koszul-Tate resolution of $\mathcal O_X$ would give a resolution (using the theory of the cotangent complex) of $\Omega^1_X$ by vector bundles on $X$ pulled back from $\mathbb P^n$, and therefore, if the algebra resolution is finitely generated, then so is the resolution of the cotangent bundle. Thus, taking determinants, we see that the canonical bundle $K_X$ lies in the image of the restriction map $\text{Pic}(\mathbb P^n)\to\text{Pic}(X)$. For instance, this shows that the twisted cubic in $\mathbb P^3$ doesn't admit a locally finitely generated Koszul-Tate resolution. There are also intrinsic conditions on $X$, for instance: any $X$ admitting a finite resolution must necessarily have $K_X$ to be either trivial, very ample or negative of very ample (since every line bundle on $\mathbb P^n$ is such). This, for example, rules out any $X$ which is a product of a Fano variety with a Calabi-Yau variety. - -REPLY [2 votes]: Horrocks-Mumford bundle on $\mathbb{P}^4$ gives an abelian surface as the zero locus of its general section, which is not a complete intersection.<|endoftext|> -TITLE: Why did Voevodsky abandon his work on "singletons"? -QUESTION [9 upvotes]: In an interview (I link the Google translation), Voevodsky talks about how, in the late 2000s, he worked on the problem of "restoring the history of populations according to their modern genetic composition". Some of his unpublished papers on this topic are now available online. For example, a paper titled "Singletons" is available on the IAS website. Why did Voevodsky abandon the subject of this rather fleshed-out paper so suddenly? - -REPLY [9 votes]: Dan Grayson features this paper in his talk at the Voevodsky memorial conference. He points to an interview by Mikhailov, where VV says (Google translate): - -As a result, I chose, as I understand correctly now, the problem of restoring the history of populations according to their modern genetic composition. I was busy with this task for a total of about two years and in the end, already in 2009, I realized that what I had invented was useless. In my life, for now, this was probably the biggest scientific failure. A lot of work was invested in the project, which completely failed. Some benefit, of course, nevertheless, was - I learned a lot from probability theory, which I did not know well, and I also learned a lot about demography and demographic history.<|endoftext|> -TITLE: Congruence subgroups in arithmetic lattices of $\mathrm{SO}(n,1)$ -QUESTION [5 upvotes]: I am currently reading the paper Deformation Spaces Associated to Hyperbolic Manifolds by Johnson and Millson, Section 7, and the highlighted bit below has been giving me difficulty: - -Specifically, how do we define this ideal $\frak{a}$ that allows us to get this group $\Gamma=\Gamma(\frak{a})$? I have tried following up the reference to Geometric construction of cohomology for arithmetic groups by Millson and Raghunathan, but unfortunately despite my best efforts I have really struggled to follow the latter paper and extract anything useful, since I am not only new to hyperbolic geometry, but also completely unfamiliar with number theory. -If someone could concretely describe how to define such an ideal and the group $\Gamma$, I would really appreciate it! - -REPLY [6 votes]: I think the reference to Millson-Raghunathan may be a little misleading. All that is needed is that if $\Gamma \subset G(\mathbb Z)$ is an arithmetic group in a linear algebraic group $G$ defined over $\mathbb Q$, then there exists an ideal $m\mathbb Z$ of $\mathbb Z$ such that the group $\Gamma (m)$ of matrices in $\Gamma$ congruent to identity modulo $m$, is ${\bf torsion-free}$. -In your notation, the group $\Phi$ is an arithmetic subgroup of $SO(Q)$ but defined over a number field $K$. By a standard restriction of scalars argument, you may think of $\Phi=SO(Q)(O_K)$ as commensurable to $G(\mathbb Z)$, where $G=R_{K/\mathbb Q}(SO(Q))$ and where $R$ denotes the restriction of scalars. You may choose the ideal $\mathfrak a$ of $O_K$ to be generated by a rational integer $m$ as in the preceding paragraph. - -REPLY [4 votes]: I want to extend the answer of Venkataramana a bit in order clarify how -you can choose a suitable ideal $\mathfrak{a} \subseteq \mathcal{O}$. -Here is an abstract version of Minkowski's classical theorem. -Let $G$ be an affine group scheme of finite type over the ring of integers $\mathcal{O}$ of some algebraic number field. -If $\mathfrak{a} \subseteq \mathcal{O}$ is an ideal such that for every -prime number $p$ we have $ p\mathcal{O} \not\subset\mathfrak{a}^{(p-1)}$, -then the principal congruence subgroup -$$ \Gamma(\mathfrak{a}) = \ker\bigl( G(\mathcal{O}) \to G(\mathcal{O}/\mathfrak{a}) \bigr)$$ is torsion-free (cf. III.2.3 in this thesis). -In particular, here is a concrete example: The ideal -$\mathfrak{a} = m \mathcal{O}$ for any integer $m \geq 3$ always yields a torsion-free principal congruence subgroup.<|endoftext|> -TITLE: "Almost-ideals" in the (simple) Lie algebra of an algebraic group? -QUESTION [6 upvotes]: Let $G<\mathrm{GL_n}$ be a simple linear algebraic group defined over a finite field $K$. Let $\mathfrak{g}$ be its Lie algebra. Assume $\mathfrak{g}$ is simple. -Is it necessarily the case that there is no subspace $\mathfrak{v}\subset \mathfrak{g}$ with $0<\dim(\mathfrak{v})<\dim(\mathfrak{g})$ such that $\mathfrak{v}$ is invariant under $\mathrm{Ad}_g$ for every $g\in G(K)$? -Note: it is clear that there is no $\mathfrak{v}\subset \mathfrak{g}$ with $0<\dim(\mathfrak{v})<\dim(\mathfrak{g})$ such that $\mathfrak{v}$ is invariant under $\mathrm{Ad}$ for every $g\in G(\overline{K})$. It is also clear that the answer to the question above is "yes" when the number of elements of $K$ is larger than a constant depending only on $n$: since $\mathfrak{v}$ is not an ideal, there is a $v\in\mathfrak{v}$ such that all $g\in G(\overline{K})$ such that $\mathrm{Ad}_g(v)\in \mathfrak{v}$ lie in a proper subvariety of $G$. -Note 2: A friend has just proposed over the breakfast table that there are linear algebraic groups with no non-trivial rational points over $K$. That would obviously imply an answer of "no" to my question. I am not convinced that such a thing is really possible, at least not when we are talking about the group $G(K)$, $G$ simple (as opposed to more exotic groups of Lie type). -EDIT: as per Venkataramana's comment below, this situation cannot, in fact, occur for $G$ simple. - -REPLY [2 votes]: You are looking for the following theorem of Steinberg: -Let q=|K|. An irreducible algebraic representation of $G(\overline{K})$ of highest weight λ remains irreducible when restricted to G(K) if $\langle \lambda,a^\vee\rangle -TITLE: Can natural section/retraction be checked pointwise? -QUESTION [6 upvotes]: Analogously to -this old question, -I was asking myself if it is possible to describe left/right invertible natural transformations by their components. Obviously this property is inherited by the components of a transformation. -For the converse, I don't think that every choice of left inverses of the components yields a natural transformation, or even that such a choice always has to exist. My thoughts until now have been: -Given a natural transformation $\varepsilon: F \Rightarrow G$ between functors $F, G: C \to D$, if its components are right invertible with sections $\eta_C: GC \to FC$, naturality of $\eta$ comes down to $Gc \circ \eta_C = p_{C'} \circ Gc \circ \eta_C$ for any morphism $c: C \to C'$, where I defined the idempotents $p_C := \eta_C \circ \varepsilon_C$. At this point I don't see under which conditions there is a choice of $\eta_C$ and $\eta_{C'}$ such that the above equation is fulfilled (apart from $p_C = \text{id}_{FC}$, i.e. natural isomorphisms, of course). -I also don't see a way to apply the solution of the question mentioned above, since being a section/retraction is not a property that can be detected using (co)limits as far as I know (apart from "boring" categories, where the sections are exactly the effective monomorphisms, or even all monomorphisms). -Another approach might be that the split monics/epics are exactly the absolute monics/epics, but I didn't get very far this way... -If it turns out that not every natural transformation with left/right invertible components is left/right invertible, I'd be interested if there is an additional criterion on the components that guarantees the existence of a left/right inverse of the transformation (and is ideally equivalent to it). - -REPLY [5 votes]: Here's an example I find easier to think about than the examples given so far. Let $G$ be a group and $k$ a field, let $C = BG$ be the category with one object with automorphisms $G$, and let $D = \text{Vect}(k)$ be the category of $k$-vector spaces. Then the functor category $[C, D]$ is the category of linear representations of $G$ over $k$. -Your question in this special case, for retracts, is equivalent to asking whether every subrepresentation $V \subseteq W$ is a direct summand as a representation (it is always a direct summand as a vector space). This is true iff the group algebra $k[G]$ is semisimple, which is true iff $G$ is finite and the characteristic of $k$ does not divide $|G|$. So for an explicit counterexample we can either take $G = \mathbb{Z}$ and consider a nontrivial Jordan block, or take $G = C_p, k = \mathbb{F}_p$ and, well, again consider a nontrivial Jordan block.<|endoftext|> -TITLE: Smooth functions on sphere -QUESTION [21 upvotes]: Let $u$ be a smooth function defined on the unit sphere $S^2$. Assume $u$ has two local maxima, two local minima, and two saddle points (a total of 6 critical points). Does there exist a plane $P$ passing through the origin such that $P\cap S^2$ contains at least three points $x_1,x_2,x_3$ with $\nabla u(x_i) \cdot n =0$, $i=1,2,3$, where $n$ is a vector normal to the plane $P$? - -REPLY [3 votes]: EDIT As DimitriPanov points out Claim 1' is false and this leads to an unfixable gap. I'll leave the answer in case someone finds it useful. -Ironically, it looks like I can salvage the proof only for any Morse functions $u$ with at most 4 critical points. -First some notation. -For any $\mathbf{v}\in \mathbb{S}^2$ let $D(\mathbf{v})=\{\mathbf{y}\in \mathbb{S}^2: \mathbf{y}\cdot \mathbf{v}\leq 0\}$ be the hemisphere on the side of the plane normal to $\mathbf{v}$ that does not contain $\mathbf{v}$. Let $C(\mathbf{v})=\partial D(\mathbf{v})$ be the great circle at the boundary of $D(\mathbf{v})$ whose (outward) normal is $\mathbf{v}$. Let $P=\{p: \nabla u (p)=0\}$ be the set of critical points of $u$. As $u$ is Morse, this is a finite set with an even number of points. -Assumption: Let us now assume that $\nabla u \cdot \mathbf{v}$ has at most two zeros on $C(\mathbf{v})$ for all $\mathbf{v}$. -I claim this assumption will lead to a contradiction and hence prove what you wanted, but for strictly fewer critical points than you requested. -Let $\Omega\subset \mathbb{S}^2$ be the set of $\mathbf{v}$ so that $\nabla u \cdot \mathbf{v}$ is never zero on $C(\mathbf{v})$ and let $\Omega'$ be the set of $\mathbf{v}$ so that $\nabla u \cdot \mathbf{v}$ is both positive and negative on $C(\mathbf{v})$. Notice both $\Omega$ and $\Omega'$ are open sets. -The assumption implies that, for $\mathbf{v}\in \Omega'$, $\nabla u\cdot \mathbf{v}$ vanishes at exactly two points on $C(\mathbf{v})$. It also ensures that if $\mathbf{v}\not\in \Omega\cup \Omega'$ then $\nabla u\cdot \mathbf{v}$ has a sign away from either one or two points. -Claim 1: (This has been corrected thanks to DmitriPanov's observation). Suppose $\mathbf{v}$ is such that $C(\mathbf{v})\cap P=\emptyset$. If $D(\mathbf{v})\cap P$ contains an even number of points, then $\mathbf{v}\in \Omega'$. -Proof: This follows from the Poincare-Hopf index theorem applied to the vector field $\nabla u$. Basically, the standing assumption ensures the boundary contributes $1$ (when $\mathbf{v}\not\in\Omega'$) and critical points contribute either $1$ (for maximum or minima) or $-1$ (for saddle points). The claim follows by (for instance) working mod 2. -Claim 1': Suppose $\mathbf{v}$ is such that $C(\mathbf{v})\cap P=\emptyset$. If $\mathbf{v}\in \Omega'$, then either $D(\mathbf{v})\cap P$ contains an even number of points or both $D(\mathbf{v})\cap P$ and $D(-\mathbf{v})\cap P$ contains at least three points of $P$. -Proof: Poincare-Hopf tells us that if $\nabla u$ contributes $0$ mod 2 along $C(\mathbf{v})$, then $D(\mathbf{v})\cap P$ contains an even number of points. Now suppose, $\nabla u$ contributes $1$ mod $2$ along the boundary. Let $\mathbf{T}:C(\mathbf{v})\to \mathbb{R}^3$ be the unit tangent to $C(\mathbf{v})$ compatible with the orientation. Write along $C(\mathbf{v})$ -$$ -\nabla u(p)=r(p) \cos \theta(p) \mathbf{T}(p)+r(p)\sin \theta(p) \mathbf{v} -$$ -where here $\theta:C(\mathbf{v})\to \mathbb{S}^1$ and $r>0$. -The standing assumption ensures that $\sin \theta(p)=0$ only at two points $p_-$ and $p_+$ and $\mathbf{v}\in \Omega'$ means that, up to relabelling $\sin(\theta(p))<0$ on $C_-(\mathbf{v})$, the region between $p_-$ to $p_+$, (relative to the orientation) and $\sin(\theta(p))>0$ on $C_+(\mathbf{v})$, the region between $p_+$ and $p_-$. One verifies (e.g., by drawing a graph) that $\nabla u$ contributes $0$ to Poincare-Hopf if and only if $\cos(\theta(p_+))=-\cos(\theta(p_-))$ (e.g. $\theta(p_+)=0, \theta(p_-)=\pi$ and) while $\nabla u$ contributes $1$ when $\cos(\theta(p_+))=\cos(\theta(p_-))$ (e.g. $\theta(p_+)=\theta(p_-)=0$). -In this latter case (which is the one we are interested in), the fact that $\nabla u$ is a gradient vector field, implies there are at least two points in both $C_-(\mathbf{v})$ and $C_+(\mathbf{v})$ where $\sin \theta$ vanishes. In fact, one the two points will be a local maximum of $u|_{C_\pm (\mathbf{v})}$ and the other will be a local minimum of $u|_{C_\pm (\mathbf{v})}$. Let $q_+$ be the local maximum in $C_-(\mathbf{v})$ and $q_-$ be the local minima in $C_+(\mathbf{v})$. By analayzing level sets one obtains corresponding local maxima and minima of $u$, $q_-'$ and $q_+'$ in $D(\mathbf{v})$ ** Edit: This reasoning only works if $q_-$ and $q_+$ are actually absolute maximam on $C(\mathbf{v})$ -- i.e. if $q_-$ only a local minima can't guarantee existence of $q_-'$. **. That is, $D(\mathbf{v})$ contains at least two critical points of $u$ and so, by Poincare-Hopf, at least three points. This argument is symmetric in $\mathbf{v}$ and $-\mathbf{v}$ so $D(-\mathbf{v})$ also contains at least three critical points of $u$. -Returning to the main argument: -Let $\Gamma=\bigcup_{p\in P} C(\mathbf{v}(p))$ where $\mathbf{v}(p)$ is chosen so $\mathbf{w}\in C(\mathbf{v}(p))$ if and only if $p\in C(\mathbf{w})$. Our standing assumption and the lower bound on the number of cirtical points ensures $C(\mathbf{v}(p))\neq C(\mathbf{v}(p'))$ for $p \neq p'$ as if $C(\mathbf{v}(p))=C(\mathbf{v}(p'))$ then $p$ and $p'$ are equal or antipodal. In the later case, one can find a $C(\mathbf{w})$ containing $p,p'$ and a third element $p''$ so $\nabla u(p'')\cdot \mathbf{w}=0$ -- contradicting the standing assumption (see User4966's argument below). Moreover, the standing assumption implies that $C(\mathbf{v}(p_1))\cap C(\mathbf{v}(p_2))\cap C(\mathbf{v}(p_3))=\emptyset$ for $p_1, p_2, p_3$ distinct. Indeed, if $\mathbf{w}\in C(\mathbf{v}(p_1))\cap C(\mathbf{v}(p_2))\cap C(\mathbf{v}(p_3))$ then $p_1,p_2, p_3\in C(\mathbf{w})$ so $\mathbf{w}$ violates the standing assumption. -Let $\Omega''=\mathbb{S}^2 \backslash \overline{\Omega'}$ so $\Omega''$ is open. Clearly, $\Omega\subset \Omega''$ (though they might not be equal). -Claim 2: If $P$ has at most $4$ points then, both $\Omega'$ and $\Omega''$ are non-empty and $\mathbb{S}^2\backslash \Gamma=\Omega'\cup \Omega''$ and each component of $\Omega'$ is contained in an open hemi-sphere. -Proof: -Pick a $C(\mathbf{w})$ so $C(\mathbf{w})\cap P=\{p\}$. That is, so $\mathbf{w}\in C(\mathbf{v}(p))\subset \Gamma$ but $\mathbf{w}\not\in C(\mathbf{v}(p'))$ for any other $p'\in P$. By perturbing $\mathbf{w}$ slightly to $\mathbf{w}'$ one can ensure $C(\mathbf{w}')\cap P=\emptyset$ and $D(\mathbf{w}')\cap P=D(\mathbf{w})\cap P$ or $D(\mathbf{w}')\cap P=(D(\mathbf{w})\cap P)\backslash \{p\}$. Depending on which of the two cases occur $D(\mathbf{w}')\cap P$ contians either an even or odd number of points. Furthermore, as $P$ has at most $4$ elements $D( \mathbf{w}')$ and $D(-\mathbf{w}')$ can't both contain at least three points of $P$. Hence, by Claim 1 and Claim 1', we can make $\mathbf{w}'$ lie in either $\Omega'$ or in $\Omega''$. That is $\mathbf{w}\in \partial \Omega'$ and also $\mathbf{w}\in \partial \Omega''$. It follows that $\Omega'$ and $\Omega''$ are non-empty and $\mathbf{w}\not\in \Omega'\cup \Omega''$. The second part of the claim follows as each component of $\Omega'$ lies in one of the components of $\mathbb{S}^2 \backslash C(\mathbf{v}(p))$ for some $p\in P$. -However we also have -Claim 3: There exists a connected component of $\Omega'$ that contains two antipodal points. -Proof: -Let $F:\mathbb{S}^2 \to \mathbb{R}$ be defined by -$$ -F(\mathbf{v})=\int_{C(\mathbf{v})} \nabla u \cdot \mathbf{v}. -$$ -Clearly, $F$ is a continuous odd function. Our standing assumption ensures that $F^{-1}(0)\subset \Omega'$. I claim that there is a connected component $Z$ of $F^{-1}(0)$ that contains two antipodal points -- see the second answer to this question for a proof (I didn't check this in detail but it seems intuitively clear). This implies that there is such a component for $\Omega'$ and proves the claim. -Since Claim 2 and Claim 3 are in contradiction the standing assumption is false, which gives what you want when $u$ has at most 4 critical points.<|endoftext|> -TITLE: A question on the fundamental group of a compact orientable surface of genus >1 -QUESTION [9 upvotes]: Let $G=\pi(X,x)$ be the fundamental group of a compact orientable -surface of genus $g\ge 2$. It is well known that a presentation of -$G$ is -$$G=\langle x_1,y_1,\dots,x_g,y_g \ | \ [x_1,y_1]\cdots -[x_g,y_g]=1\rangle$$ (where $[x,y]=xyx^{-1}y^{-1}$ is the -commutator). -Denote by $F$ be the free group with $2g$ generators -$x_1,y_1,\dots,x_g,y_g$ and by $R$ be the normal closure of the -relation $r=[x_1,y_1]\cdots [x_g,y_g]$, so $G=F/R$. -It is clear that $r\in [F,F]$. - -Question: Is there an elementary proof that $r=[x_1,y_1]\cdots [x_g,y_g]\not\in [F,[F,F]]$? - -This result appears when one considers the Stallings exact -sequence associated to -$$1\to [G,G]\to G\to G^{ab}\to 1$$ -to get -$$ H_2(G,\mathbb{Z})\to H_2(G^{ab},\mathbb{Z})\to [G,G]/[G,[G,G]]\to -H_1(G,\mathbb{Z})\to H_1(G^{ab},\mathbb{Z}) \to 0$$ -Since $H_1(G,\mathbb{Z})\cong H_1(G^{ab},\mathbb{Z})\cong G^{ab}$ we -obtain a short exact sequence -$$ H_2(G,\mathbb{Z})\to H_2(G^{ab},\mathbb{Z})\to [G,G]/[G,[G,G]]\to -0$$ which should be injective at the left (see at the end some argument why). -Now, Hopf's formula gives $$H_2(F/R,\mathbb{Z})=(R\cap -[F,F])/[R,F]=R/[R,F]$$ since $R\subset [F,F]$, hence -$H_2(F/R,\mathbb{Z})$ is cyclic and the generator is given by (the -class of) $r$. So the map $\psi:H_2(G,\mathbb{Z})\to -H_2(G^{ab},\mathbb{Z})$ is either injective or zero. But -$$H_2(G^{ab},\mathbb{Z})\cong [F,F]/[F,[F,F]]$$ -since $G^{ab}\cong F/[F,F]$, and so the map $\psi$ is given by the -natural map -$$\psi:R/[R,F]\to [F,F]/[F,[F,F]]$$ -coming from the inclusion $R\hookrightarrow [F,F]$, hence $\psi$ is -injective if and only if $r\not\in [F,[F,F]]$. -One possibility is to use another description of the map $\psi$ as -$$H_2(G,\mathbb{Z})\to -\bigwedge H_1(G^{ab},\mathbb{Z})$$ that should correspond -to the dual of the cup product in cohomology via Poincaré duality -(i.e. dual universal coefficient theorem) (but I am not sure if to consider this -approach as really elementary). - -REPLY [3 votes]: The dual to the map -$\psi\colon H_2(G,\mathbb{Z}) \to H_2(G^{\operatorname{ab}},\mathbb{Z})$ is the cup-product map $\cup\colon H^1(G,\mathbb{Z})\wedge H^1(G,\mathbb{Z}) \to H^2(G,\mathbb{Z})$; see e.g. Lemma 1.10 in arXiv:math/9812087. Clearly, the latter map is surjective; hence, the former map must be injective.<|endoftext|> -TITLE: Polymath type websites for specialized areas -QUESTION [7 upvotes]: This question is inspired by the success and more importantly, the democratizing affect of the polymath projects and Mathoverflow. I put the idea in my NSF proposals a few times, but the panels don't seem to like it these days. So this year, instead of asking the government for help, I will appeal to my fellow MO readers! (-: -So, my dream is to build up a website where people who like to think about, say, commutative algebra and related questions, can gather, propose and work on different research questions, from undergraduate level all the way up. So, anyone can start a thread about some question/project they like to do, and people can discuss, develop and hopefully produce research-level results. Or, if you want to set up a MRC type summer school but don't want to bother applying for it. Or, if you want to study a paper but don't want to do it alone. -Other features may include: - -Links to good survey articles/freely available text/insightful MO anwers/blog posts on various topics. - -Articles on the history of the area. - -News on interesting connections to related areas. - - -I think such a resource would be quite useful and as mentioned above, make math more democratic. For instance, it is important for young people to take part in research as soon as they can, but if you come from a small school or a poor country, opportunity for such experience might be non-existent. Even for more senior researchers, funding/invitation to conferences are increasingly difficult to obtain and depend sometimes on who you are friendly with. -So my questions are: - -Question 1: Do you know of any such attempt in any area (mathematics or other)? -Question 2: Do you have any technical advice on where to host the website, what new (hopefully cheaply available) software to help making it, potential funding partners? Any experience to share, or potential problems you can foresee? - -PS: I was away from MO for a while, where is the community wiki button now? -PPS: if you are interested in such a project for commutative algebra, let me know in the comments or email. - -REPLY [4 votes]: There's a dispersive PDE wiki: -https://dispersivewiki.org -Also Tricki: http://www.tricki.org/<|endoftext|> -TITLE: Regular sequence from prime ideal -QUESTION [5 upvotes]: Let $I$ be a prime ideal in $\mathbb{C}\{x_1, \ldots, x_n\}_0$ (the localization at the maximal ideal that defines $0$) and suppose that the height of $I$ is $h$. Then, there is a standard trick to extract a regular sequence of length $h$ from $I$. And so one can always see $V:=V(I)$ (which has codimension $h$) as an irreducible component of a complete intersection of codimension $h$. - -Can you choose the regular sequence $g_1, \ldots, g_h$ so that the intersection of $V$ with each of the other irreducible components of $V(g_1,\ldots,g_h)$ is just $\{0\}$? What if we assume that the local ring of $(V(I),0)$ is Cohen-Macaulay? - -It seems to me that this is a strong condition on the ideal $g:=(g_1, \ldots, g_h)$, however, you have some freedom to choose the elements within $I$. Observe, that this would imply that there exists a primary decomposition of $g$ such that the sum of the primary ideal corresponding to $V$ and any other primary ideal contains (a power of) the maximal ideal of $\mathbb{C}\{x_1, \ldots, x_n\}_0$. - -REPLY [5 votes]: It is impossible to do this if $\dim V(I)\geq 2$. Because then $g$ defines a complete intersection of dimension at least $2$. But for any Cohen-Macaulay local ring of dimension at least $2$, the punctured spectrum is connected. (Unless of course if $V(g)$ has only one component, whence $I$ is a set-theoretic complete intersection).<|endoftext|> -TITLE: Mode of a sum of Bernoulli random variables -QUESTION [12 upvotes]: Let $S_n=\tau_1+\cdots+\tau_n$ be a sum of independent Bernoulli random variables such that $\mathbb{P}(\tau_i=1)=p_i$. Is it true that the mode of $S_n$ is either its mean rounded up or rounded down? - -REPLY [5 votes]: Darroch's theorem is the following. Let $p = \sum a_i x^i$ be a polynomial with positive coefficients, and suppose that all the roots of $p$ are real (hence negative or zero) [the corresponding distribution of coefficients is called PF, for Polyà frequency]. Then the mean of the distribution $(a_i)$ differs from the mode by less than $1$. [The theorem also gives a partial result on which of the two integers nearest the mean is the mode when the mean is not an integer; but in cases where I wanted to use it, it was just too crude.] -Here the polynomial whose distribution we are interested in is $\prod (1-p_i + p_i x)$, which obviously has only negative real roots. Hence the answer to the question is yes. -I became aware of Darroch's theorem from a preprint of J Pitman, eventually published, Probabilistic bounds on the coefficients of polynomials with only real zeros, J. Combin. Theory Ser. A, 77(2):279–303, 1997 (unfortunately behind an Elsivier paywall), although the preprint might be available somewhere. It contains lots of examples using Darroch's theorem. Fedja found Darroch's paper at projecteuclid.org/euclid.aoms/1177703287.<|endoftext|> -TITLE: Hadamard theorem about embedding -QUESTION [14 upvotes]: The following theorem is commonly attributed to Jacques Hadamard. - -Assume $\Sigma$ is a smooth locally convex immersed surface in the Euclidean space. Then $\Sigma$ is embedded and bounds a convex set. - -Many authors refer to Hadamard's Sur certaines propriétés des trajectoires en Dynamique (1897) -(for example, James Stoker in his Über die Gestalt der positiv... (1936)). -Likely the statement is there, but the paper is long, it is in French and often the statements are not clearly marked; I was searching for it for several days. I asked a friend and she said that it was there 20 years ago, but she could not find it; she also said that it was not easy to extract it from what is written ( = one has to think). [For sure the word immersion is not there.] -I hope someone here knows this paper and can help me. -P.S. Now I see it this way: Stoker was the first who had formulated and proved the theorem; at the beginning of his paper he attributed the theorem to Hadamard because it almost follow from item 23 in his paper. After Stoker everyone did the same. - -REPLY [11 votes]: I think the relevant location is item 23, page 352, but what Hadamard aims to is stated as follows: - -A smooth, co-orientable surface of $\mathbb{R}^3$ with Gauss curvature bounded below by some $\kappa >0$ is simply connected. (implicitly, the surface is compact without boundary) - -("Or une surface à deux côtés et sans points singuliers, à courbure partout positive (la valeur zéro et les valeurs infiniment petites étant exclues) est toujours simplement connexe.") -The goal is to use the Gauss-Bonnet Formula to deduce that when curvature is positive, any two closed geodesics must meet (otherwise they would together bound a total curvature 0 region of the surface). -What is not clear from the text of item 23 is whether the surface assumed to be immersed or embedded. He basically says that the normal map is a global diffeomorphism, because positive curvature makes it a covering of the sphere. -It seems the argument does provide the statement attributed to this paper, although it seems not explicitly stated. Second edit: Mohammad Ghomi gives an argument to that effect in comment.<|endoftext|> -TITLE: A peculiarity of Henkin's 1950 proof of completeness for higher order logic -QUESTION [14 upvotes]: My question concerns Henkin's original (1950) completeness proof in Completeness in the theory of types for classical higher order logic and type theory relative to so-called general models. -His 1950 proof seems quite different to his (1949) completeness proof of first order logic in The completeness of the first-order functional calculus. -In particular, the 1950 completeness proof for classical type theory does not, at first sight, seem to employ the famous construction whereby the language is expanded by adjoining constants to it thereby providing witness to existential formulae, whereas his 1949 completeness proof does. (The main body of his 1950 proof is on p.85-88 of Completeness in the theory of types, which constructs a maximally consistent set of formulae, but surprisingly does not mention constants and their role as witnesses for existential formulae) -I mentioned this to a colleague who expressed surprise/incredulity that the completeness proof for classical type theory does not employ this construction. He suggested that it might be hidden somewhere in the proof, and that perhaps the $\iota$ operator employed by Henkin plays the role of supplying fresh constants with which to witness existential formulae. -Indeed, it is clear from Henkin's comments in that he views the $\iota$ as playing an important role in the completeness proof, and this passage suggests he views it as a way of providing witnesses for existential formulae: - -"The very first question I asked myself was whether I could use the method -that gave me completeness for type theory, to get a new proof of Godel's -completeness for first-order logic. It seemed, at first, that there was no -possibility to do so. The reason is that the axiom of choice, and in particular Church's neat formulation of it via the constants $i_{a(Oa)}$, played a crucial role in the proof for type theory, while in first order logic there is no axiom of choice, and no way to formulate one. -I decided to analyze carefully the role of the axiom of choice in the completeness proof, to see whether there was some other way of accomplishing it in first-order logic. The role that I saw first was performed in the proof by induction, sketched above, that domains $D_{\alpha}^{'}$ satisfying conditions (i)-(iii), could be constructed. But when I wrote down details of the proof that the resulting hierarchy of domains $D_{\alpha}^{'}$ satisfy the maximal consistent set of (added) axioms, I saw that the axiom of choice is needed there in a more general way, of which the earlier use is just a special case. The more general need is to show that whenever we have a wff $\textbf{M_0}$ such that $\vdash (\exists x_b)M_0$, then we also have $\vdash (\lambda x_b. \textbf{M_0 }) (\iota_{b,(0b)}(\lambda x_{\beta}.\textbf{M_{0}}))$. The fact that this condition holds is a direct consequence of having Axiom Schema $11^b$ in the deductive system that Church had set up for the theory $\mathscr{T}$, as that schema is trivially equivalent to $(\exists x_b f_{0b} x_b) \supset f_{0b}(\iota_{b(0b)} f_{0b})$...I did not altogether hide the symbols $i_{b(0b)}$ from the reader of my dissertation, -for in passing from Theorem VI to Theorem VII, in which the -generalized completeness theorem is extended to languages that can be nondenumerable or have additional constants, I cited Church's system in which the axiom schemas of description and choice are formulated with the symbols $\iota{b(0b)}$, as an example. With that example is a brief note to the effect -that when this formulation of the axiom of choice is made, the adjunction of special constants $u_b, u_{b}^{'}, u_{b}^{''},...$ for each type symbol $b$ becomes unnecessary in the completeness proof, as their role can be taken over by the symbols $\iota_{b(0b)}$. Still, it would be a very sharp-eyed historian who could detect in that brief note appended to Theorem VII, the origin of the proof of Theorem I!" (from http://www2.mat.ulaval.ca/fileadmin/Cours/MAT-10391/Henkin_BSL_1996.pdf) - -So is my colleague right about this? -Does the 1950 proof implicitly contain the relevant construction? - -Edit: -The mystery is made worse by the fact that Henkin in his 1949 paper writes that his proof "suggests a new approach to the problem of completeness for functional calculi of higher order" to be "taken up" in a further paper (presumably his 1950 paper). In the 1950 paper he writes (ft6, p.82) that a completeness proof of second order logic relative to general models "can be carried out along the lines of the author's recent paper" (i.e his 1949 paper). So what precisely is the link between the earlier paper and the later (presuming it is not the famous construction involving adjoining constants to the language)? - -REPLY [10 votes]: Henkin's completness proof for first order logic (using his method of constants) and Henkin's work on the completeness of type theory were both carried out in his doctoral 1947 dissertation written under the direction of Alonzo Church. The dissertation was never published, but Henkin wrote a fascinating and detailed article about the genesis of the intertwining ideas in his thesis in 1996 (The Discovery of My Completeness Proofs, The Bulletin of Symbolic Logic, Vol. 2, No. 2 (Jun., 1996), pp. 127-158). -The following quote from the above paper (starting in the bottom of page 154 and continued to the next page) seems to answer your question (and vindicate your colleague). Henkin uses "cwff" for closed well-formed formulae; i.e., those with no free variables. - -As indicated, in addition to putting the first-discovered proof into Part - III of the dissertation, I hid the discovery process in another way. Namely, - in setting forth the formal language for type theory, in Part III, I deleted - from Church's system the symbols $\iota_b(0b)$ and the axiom of choice for which they were used! Those symbols, and that axiom, which quintessential role in the discovery process, were omitted in Theorem VI, and the role of the symbols $\iota_b(0b)$ in forming "witnesses" cwffs of the form $\exists x_bM_0$ was taken over by the adjunction of constants $u_b$, $u'_{b}$, $u''_{b}$ for each type symbol $b$.<|endoftext|> -TITLE: Representations of degenerate Clifford algebras -QUESTION [11 upvotes]: Given a real finite-dimensional vector space $V$ with a symmetric bilinear form $b$, we define the Clifford algebra $Cl(V,b)$ as the quotient of the tensor algebra $\bigotimes V$ by the two sided ideal generated by $v\otimes v + b(v,v) \mathbb{1}$ for all $v \in V$. $Cl(V,b)$ is a $\mathbb{Z}_2$-graded unital real associative algebra. -Every such $(V,b)$ is isomorphic to $\mathbb{R}^{r+s+t}$ with bilinear form defined by polarisation from the quadratic form -$$ q(x) = \sum_{i=1}^{r + s + t} \varepsilon_i x_i^2, $$ -where -$$ \varepsilon_i = \begin{cases} 0 & i = 1,\dots,r\\ -1 & i = r+1,\dots,r+s\\ -1 & i = r+s+1, \dots, r+s+t .\end{cases} $$ -Let $Cl(r,s,t)$ denote the Clifford algebra of $\mathbb{R}^{r+s+t}$ and the above bilinear form. -As $\mathbb{Z}_2$-graded unital real associative algebras, -$$ Cl(r,s,t) \cong \Lambda \mathbb{R}^r \hat\otimes Cl(s,t), $$ -where $\hat\otimes$ is the $\mathbb{Z}_2$-graded tensor product and where $Cl(s,t):= Cl(0,s,t)$ are the standard Clifford algebras associated to non-degenerate bilinear forms. -The representations of $Cl(s,t)$ are well-known: there are either one or two simple modules (up to isomorphism) depending on $s,t$ and every finite-dimensional module is a direct sum of simples. -I am interested in the representations of $Cl(r,s,t)$ for $r=1$, but more generally for $r>0$. -For $(s,t) = (1,0)$ and $(0,1)$, it is easy to work this out "by hand". The resulting category of representations is no longer semisimple, but it is not hard to show that any finite-dimensional module is a direct sum of indecomposable (but not simple) modules. -But before attempting to study the case of general $(s,t)$, I wonder whether there is some technology out there which can be brought to bear on this problem. -More concretely, I have a couple of -Questions - -Would a knowledge of the indecomposable modules of $\Lambda \mathbb{R}$ and $Cl(s,t)$ be sufficient to determine the indecomposable modules of their $\mathbb{Z}_2$-graded tensor product? If so, how? -Is there a classification of indecomposable finite-dimensional modules of the exterior algebra $\Lambda \mathbb{R}^r$ for $r>1$? If so, where? - -REPLY [3 votes]: To question 2: -The exterior algebra is a finite dimensional local quiver algebra with loops $x_1,...,x_n$ and relations $x_i^2,x_i x_j + x_j x_i (i -TITLE: Stably equivalent but not homotopy equivalent -QUESTION [16 upvotes]: What are some examples of (compact, say) manifolds $X$ and $Y$ that are stably equivalent, i.e. $\Sigma^{\infty}X_+\simeq\Sigma^{\infty}Y_+$, but are not homotopy equivalent? - -REPLY [13 votes]: Maybe it is worth adding some simply-connected examples. -Every simply connected closed 4-manifold may be described as $X = D^4 \cup_f (S^2 \vee \cdots \vee S^2)$, where $f$ is a map $S^3 \to S^2 \vee \cdots \vee S^2$; $\pi_3$ of this wedge is known to be generated by Hopf maps and Whitehead products of two factors, so we may represent such a map by a symmetric integer matrix $A$; this integer matrix may just as well be interpreted as the cup product pairing $H^2(X;\Bbb Z) \otimes H^2(X;\Bbb Z) \to H^4(X;\Bbb Z) = \Bbb Z$. The resulting homotopy type is determined up to isomorphism by $A$ up to isomorphism of symmetric bilinear forms over $\Bbb Z$. See, for instance, Example 4.52 in Hatcher's algebraic topology book. -On the other hand, if $n > 2$, then $$\pi_{n+1}(\vee^k S^n) = (\pi_{n+1} S^n)^k = (\Bbb Z/2)^k.$$ The Whitehead product factors vanish. This follows by an inductive argument using the Hilton-Milnor theorem, as stated in this answer. -Furthermore, any map in $GL_k(\Bbb Z)$ may be realized as an automorphism of $H_n(\vee^k S^n)$ by some autoequivalence of $\vee^k S^n$, and the map $GL_k(\Bbb Z) \to GL_k(\Bbb Z/2)$ is surjective (check at the level of the generating set of elementary matrices). Because every two nonzero vectors in $(\Bbb Z/2)^k$ are related by some matrix in $GL_k(\Bbb Z/2)$, the homotopy type of $D^{n+2} \cup_f (\vee^k S^n)$ is determined entirely by whether or not $f$ is nontrivial. -Suspending the presentation given for $X$ takes the diagonals of the matrix $A$ mod 2, and so the homotopy type of the suspension is dictated by whether or not $X$ was spin (that is, whether or not its intersection form was even). If $X$ was spin, then $$\Sigma X \simeq S^5 \vee^{b_2} S^3;$$ if $X$ was not spin, then $$\Sigma X \simeq \Sigma \Bbb{CP}^2 \vee^{b_2-1} S^3.$$ In particular, the stable homotopy type is determined entirely by "even-ness" and rank of the intersection form.<|endoftext|> -TITLE: Going up of an amalgamated decomposition of a subgroup of finite index -QUESTION [7 upvotes]: Let $G$ be a finitely presented group and H a subgroup of index $n$ in $G$. Suppose that H has a non-trivial decomposition as amalgamated product, say $H = A \ast_U B$. I am wondering about the following two questions: -(i) If $n = 2$, does then $G$ also have a non-trivial amalgamated decomposition? -(ii) More generally, does there exists sufficient conditions (for example on $A,B,U$) such that $G$ also have non-trivial amalgamated decomposition? - -REPLY [7 votes]: Yes, there are many examples: start from any group $A$, and consider $A\wr C_2=A^2\rtimes C_2$, $C_2$ permuting both copies. Consider the inclusion of index 2 $$A^2\subset A\wr C_2.$$ -a) If $A$ has a nontrivial amalgam decomposition, so does the group $A^2$ (since it has $A$ as quotient group). -b) It remains the question when $A\wr C_2$ has Property FA. This is answered in Theorem 1.1 of my paper with Aditi Kar (arxiv link, J. Group Theory publisher link). Assuming that $A$ is countable, and $F$ a nontrivial finite group, $A\wr F=A^F\rtimes F$ has Serre's Property FA if and only if $A$ is finitely generated and has finite abelianization. ($\star$) -So, for (a) let's assume that $A$ has a nontrivial amalgam decomposition, and for (b) let's assume that $A$ is finitely generated with finite abelianization. In this case, $A^2$ admits a nontrivial amalgam decomposition, but not its overgroup of index two $A\wr C_2$. -To satisfy both fulfillments, one can take $A$ infinite dihedral, or any Coxeter group with an amalgam decomposition (such as $\langle a,b,c:a^2=b^2=c^2=(ab)^k=(bc)^m=1\rangle$), or many other examples. -The same holds with the cyclic group $C_2$ replaced with $C_n$ for arbitrary $n\ge 2$. - -To be self-contained here's a proof of ($\star$). Let $A$ be a finitely generated group with finite abelianization, $F$ a nontrivial finite group, and let $G=A\wr F$ act on a tree $T$ without edge inversion. Write $A^F=\prod_{i\in F}A_i$. -Suppose that the action is unbounded. Then it is unbounded on $A_i$ for some $i$, and hence on each $A_i$ since they are all conjugate. -Since $A$ is finitely generated, choose some element $f_i$ in $A_i$ acting as a loxodromic isometry; its axis $D_i$ is then invariant by $A_j$ for all $j\neq i$. For $j\neq i$, $D_j$ is $f_j$-invariant and hence $D_j=D_i$. This proves (using that $|F|\ge 2$ to ensure the existence of $j$) that $D_i$ is $A_i$-invariant. Also it does not depend on $i$; call it $D$: $D$ is the unique minimal $A_i$-invariant subtree $T_i$ of $T$. Since $F$ permutes the $T_i$, it thus preserves the axis $D$. -Since the actions of each $A_i$ on the axis $D$ commute with each other and are all unbounded and $|F|\ge 2$, they cannot be orientation reversing (since the centralizer of an orientation-reserving unbounded action on the line is trivial); so the action of $A_i$ on $D$ is given by a nontrivial homomorphism $A_i\to\mathrm{Aut}^+(D)\simeq\mathbf{Z}$. But since $A$ has a finite abelianization, we deduce a contradiction.<|endoftext|> -TITLE: $\mathscr{U}$-categories and $\mathsf{Hom}$-functors -QUESTION [8 upvotes]: Let $\mathscr{U}$ be a universe. Call a set $X$ $\mathscr{U}$-small if there is a set $Y \in \mathscr{U}$ so that $X \cong Y$. Call a category $\mathsf{C}$ a $\mathscr{U}$-category if for any $X,Y \in \mathsf{C}$, $\mathsf{Hom_C}(X,Y)$ is $\mathscr{U}$-small. -Assume $\mathsf{ZFC}$ as our foundational system (not Bourbaki set theory). -Let $\mathsf{C}$ be a $\mathscr{U}$-category and let $\mathscr{U}\text{-}\mathsf{Set}$ be the category of all sets which belong to $\mathscr{U}$. -How do we construct a $\mathsf{Hom}$-functor $\mathsf{Hom_C}(X,-)\colon\mathsf{C}\to\mathscr{U}\text{-}\mathsf{Set}$? Note for every $Y \in \mathsf{C}$, $\mathsf{Hom_C}(X,Y)$ doesn't belong to $\mathscr{U}\text{-}\mathsf{Set}$, but rather is isomorphic to a set in there. Grothendieck in SGA uses Bourbaki set theory and $\tau$ choice operator (also axiom $\mathscr{U}$B), while in $\mathsf{ZFC}$ we don't have that. -Is it even possible to work with these definition in $\mathsf{ZFC}$? - -REPLY [7 votes]: This depends on what definition of category you have taken: do you assume the objects form a set? -(I’ll avoid the terminology “small category”, since while this is sometimes used to mean “objects form a set”, it’s also used to mean “objects form a $U$-set”, for some universe $U$ whose sets have been earlier designated as “small”.) -If you’ve taken the definition where objects of a category are always assumed to form a set, then there’s no problem: just use the axiom of choice. For each $X, Y$ in $\operatorname{ob} C$, you know there exist some set in $U$ and isomorphism $C(X,Y) \cong U$; so by the axiom of choice, there’s a function which chooses, for each $X$, $Y$, some such set and isomorphism. Given this, the construction of the hom-functor is straightforward. -If you’re not assuming that the objects of a category form a set — i.e. you define a category to have a class of objects — then life gets rather subtle, because categories in this sense aren’t something you can really talk about inside ZFC. You can represent classes by predicates, and so talk about individual categories (or parametrised families of categories) schematically — so any theorem about arbitrary categories is really a theorem schema. Or you can extend your language to something like NBG set theory, where you really can talk about classes (and hence categories) directly. -In either of those setups of “large categories”, I don’t see how you can define the hom-functor for an arbitrary $U$-category without invoking something approaching the axiom of global choice, which essentially gives you choice functions on classes instead of sets. Given global choice, we can construct hom-functors essentially by re-running the argument used for the small case above. -Using the language of NBG, one can ask whether the existence of hom-functors for all such categories is (a) independent of NBG (which is conservative over ZFC), or (b) equivalent over NBG to the axiom of global choice. It’s equivalent to the statement “there is a global choice function for non-empty $U$-small sets”, if I’m not mistaken. My guess would be that this statement is independent of NBG, but strictly weaker than full global choice. I can’t substantiate either part of this, off the top of my head, but I think for someone less rusty than me on the relevant techniques, it should be fairly standard.<|endoftext|> -TITLE: Finite groups containing no subgroups of a given order or index -QUESTION [8 upvotes]: The classical Lagrange's Theorem says that the order of any subgroup of a finite group divides the order of the group. For abelian groups this theorem can be completed by the following simple fact: Abelian groups contain subgroups of any order that divides the order of the group. For non-abelian groups this is not true: the alternating group $A_4$ has cardinality 12 but contains no subgroup of cardinality 6; the alternating group $A_5$ has order 60 but contains no subgroup of order 30. On the other hand, these alternating groups contain subgroups of index 6 and 30, respectively. - -Problem. Is there a finite group $G$ and a divisor $d$ of $|G|$ such that $G$ contains no subgroup of order or index, equal to $d$? Is there such a group $G$ among finite simple groups? - -REPLY [10 votes]: The alternating group $A_{9}$ has no subgroup of order $35$ and no subgroup of index $35$. This can be checked from the Atlas ( or probably with GAP, etc). Note that if there were a subgroup of index $35,$ it would have to be maximal and its order would be $2^{6}.3^{4}$, so it is only necessary to check maximal subgroups, and there are none of that order.<|endoftext|> -TITLE: How many sigma algebras exist on $\mathbb{R}$? -QUESTION [22 upvotes]: On the one hand, there are at least $2^\mathfrak{c}$ sigma algebras on $\mathbb{R}$: one can take any subset $A$ of $\mathbb{R}$ and consider a sigma algebra $\{\emptyset, A, \bar A, \mathbb{R}\}$ -On the other hand, number of sigma algebras can be bounded as $2^{2^\mathfrak{c}}$ since a sigma algebra is a family of subsets of $\mathbb{R}$. -Let's assume generalized continuum hypothesis to simplify. In this case there are only two possible answers. Unfortunately I can't either find $2^{2^\mathfrak{c}}$ different sigma algebras, or prove their number is exactly $2^\mathfrak{c}$. - -REPLY [5 votes]: The other answer gives many $\sigma$-ideals, but I don't think it settles the question about $\sigma$-algebras. -Assume GCH. Find a family $\mathcal{F}$ of $\aleph_2$ many subsets of $\mathbb{R}$, any two of which have countable intersection. There are $\aleph_3 = 2^{2^c}$ distinct subsets of $\mathcal{F}$ which contain more than one element. -Say that $A \subseteq \mathbb{R}$ essentially contains $B$ if $B \setminus A$ is countable. Then for each $\mathcal{F}_0 \subseteq \mathcal{F}$ with more than one element the family of subsets of $\mathbb{R}$ which either essentially contain every set in $\mathcal{F}_0$ or no set in $\mathcal{F}_0$, is a $\sigma$-algebra which contains no set in $\mathcal{F}_0$ and every set in $\mathcal{F}\setminus\mathcal{F}_0$. So there are as many distinct $\sigma$-algebras as there are subsets of $\mathcal{F}$ with more than one element. -(To construct $\mathcal{F}$, identify $\mathbb{R}$ with $\{f: \alpha \to \{0,1\}\, |\, \alpha$ is a countable ordinal $\}$. For each function from $\aleph_1$ to $\{0,1\}$, its restrictions to all countable ordinals is a subset of this set, and any two of these have countable intersection.) - -REPLY [4 votes]: This is a continuation of my previous answer. -Suppose $F$ is a family of $2^{2^{|\mathbb{R}|}}$ sigma-ideals on $\mathbb{R}$. By throwing away the principal ideals, we can assume that they are all non-principal. Let $I, J \in F$ be distinct and towards a contradiction suppose they generate the same sigma-algebra. Note that the sigma-algebra generated by a sigma-ideal is just the ideal together with the dual filter. Say $X \in I \setminus J$. Then $ \mathbb{R} \setminus X \in J$ and $J$ restricted to $X$ is a non-principal prime sigma-ideal on $X$. But this is impossible as there is no measurable cardinal below $|Y|$. Am I missing something obvious?<|endoftext|> -TITLE: Is there a Giambelli identity with dual representations? -QUESTION [13 upvotes]: For natural numbers $a,b$ with $b\leq n-1$, let $V_{ (a|b)}$ be the irreducible representation of $GL_n$ with highest weight vector $(a+1, 1^b, 0^{n-b-1})$ where the exponentiation denotes repetition. -The Giambelli identity states that if $a_1 > \dots > a_r$ and $b_1 > \dots > b_r$ are natural numbers with $b_1 \leq n-1$ then the determinant in the representation ring of the matrix $M$ with entries $M_{ij} = V_{ (a_i |b_j)} $ is an irreducible representation of $GL_n$ with highest weight vector $$( a_1 + 1,\dots, a_r +r, r^{b_r}, (r-1)^{b_{r-1} -b_r-1}, \dots, 1^{ b_1 - b_{2}-1} , 0 ^{n-1-b_1}).$$ -Technically, the Giambelli identity is an identity in the ring of symmetric functions in infinitely many variables. We deduce this using the fact that the representation ring of $GL_n$ is the quotient ring of symmetric functions in $n$ variables, where each Schur function is sent to an irreducible representation or zero depending on the number of parts. - -Let $a_1 > \dots a_r, b_1> \dots > b_{2r}, c_1> \dots c_r$ be natural numbers with $b_1 \leq n-1$. -Let $M$ be the matrix whose entries are $M_{ij} = V_{(a_i|b_j)}$ if $1 \leq i \leq r$ and $M_{ij} = V_{( c_{2r-i} | n-1-b_j )}^\vee$ if $ r+1 \leq i \leq 2r$. -Is $\det M$ an irreducible representation? Is it the one with highest weight vector $$(a_1+1,\dots, a+r+r, r^{b_{2r} }, (r-1)^{b_{2r-1}-b_{2r} -1},\dots, (-r+1)^{ b_1-b_2 -1}, (-r)^{n-1-b_{1}}, -c_r-r, \dots, -c_1-1)?$$ -If not, does some similar formula hold for the class of this irreducible representation in the representation ring? - -The motivation is that this would help generalize my work in arXiv:1810.01303 on the CFKRS conjecture for integral moments, from the moments conjectures to the ratios conjecture. -It doesn't seem possible to deduce this from any nice identity purely in the ring of symmetric functions, so the same approach might not work here. - -REPLY [8 votes]: Yes, your prediction is correct. The determinant identity in this case is theorem 3.5 in Division and the Giambelli Identity, by Wu and Yang (also published at Linear Algebra Appl. 406 (2005), 301-309).<|endoftext|> -TITLE: Anderson localization for fractional Laplacians -QUESTION [5 upvotes]: There is a vast literature on Anderson localization, namely, the study of decay of eigenfunctions of operators on $l^2(\mathbb{Z}^d)$ such as -$$ --\Delta+\lambda V -$$ -where $\Delta$ is the discrete lattice Laplacian and the potential $V$ is random say given by a vector $(V_{\mathbf{x}})_{\mathbf{x}\in\mathbb{Z}^d}$ of iid standard normal variables. The constant $\lambda$ is the strength of disorder. -Did anyone study similar random operators -$$ -(-\Delta)^{\alpha}+\lambda V -$$ -with a fractional Laplacian? -I am in particular interested in references from the physics literature which provide some heuristics, e.g., a scaling theory à la Abrahams et al. - -REPLY [2 votes]: Since fractional Laplacians describe diffusion on a fractal, for a physics application I would focus on Anderson localization on a fractal. The scaling theory mentioned in the OP has been applied to a fractal in An attractive critical point from weak antilocalization on fractals.<|endoftext|> -TITLE: A finite group that has no decomposition of given cardinality -QUESTION [14 upvotes]: Let $a,b$ be two positive integer numbers. A group $G$ is called $a{\times}b$-decomposable if there are subsets $A,B\subset G$ of cardinality $|A|=a$ and $|B|=b$ such that $AB=G$ where $AB=\{xy:x\in A,\;y\in B\}$. - -I am looking for an example of a finite group $G$ which is not $a{\times}b$-decomposable for some numbers $a,b$ with $a\cdot b=|G|$. - -Remark. It is easy to see that a group $G$ is $a{\times}b$-decomposable if $a\cdot b=|G|$ and $G$ contains a subgroup of order or index equal to $a$ or $b$. Consequently, any Abelian group $G$ is $a{\times}b$-decomposable for any numbers $a,b$ with $a\cdot b=|G|$. -According to the answer of Geoff Robinson to this MO-problem the alternating group $A_9$ contains no subgroups of order or index equal to 35. - -Question 1. Is the group $A_9$ $35{\times}5184$-decomposable? - -By the comments of @YCor to the same MO-problem, -$\bullet$ the group $PSL_2(11)$ has cardinality $|PSL_2(11)|=15\times 44$ but contains no subgroups of order or index 15; -$\bullet$ the group $PSL_2(13)$ has cardinality $|PSL_2(13)|=21\times 52$ but contains no subgroups of order or index 21. - -Question 2. Is the group $PSL_2(11)$ $15{\times}44$-decomposable? -Question 3. Is the group $PSL_2(13)$ $21{\times}52$-decomposable? - -REPLY [5 votes]: Question 1 can be solved in a similar manner to @FrançoisBrunault's answer. -Let $A_{i+1}=L_{i+1}A_i=A_iR_{i+1}$ with $|L_{i+1}|=|R_{i+1}|=i+1$. Then -$A_9=A_7R_8R_9=L_7A_6R_8R_9=(L_7L_5)(A_4R_6R_8R_9)$.<|endoftext|> -TITLE: Base zero-dimensional spaces -QUESTION [5 upvotes]: Definition. A zero-dimensional topological space $X$ is called base zero-dimensional if for any base $\mathcal B$ of the topology that consists of closed-and-open sets in $X$, any open cover $\mathcal U$ of $X$ has a disjoint refinement $\mathcal V\subset\mathcal B$. - -It can be shown that -(1) each countable regular space is base zero-dimensional; -(2) the Cantor set is not base zero-dimensional. -Let $\mathfrak z$ be the smallest cardinality $|Z|$ of a subset $Z\subset\mathbb R$, which is not base zero-dimensional. It follows that $\aleph_1\le\mathfrak z\le\mathfrak c$. So, $\mathfrak z$ is a typical small uncountable cardinal. - -Problem 1. Is $\mathfrak z$ equal to some known small uncountable cardinal? Is $\mathfrak z=\mathfrak c$ under MA or PFA? - - -Edit 1 (written following a suggestion of @user64494): I found a (relatively) simple solution to my original question (about the base zero-dimensionality of the Cantor set) and then edited my post asking the next natural question in this context (about the base zero-dimensionality of uncountable sets of the real line). -By the way, a base $\mathcal B$ witnessing that the Cantor cube $2^\omega$ is not base zero-dimensional consists of the sets -$$B_s:=\{x\in 2^\omega:x{\restriction}n=s\mbox{ and }(x(n)\ne s(n)\Rightarrow x(n{+}1)=0)\}$$where $n:=\{0,\dots,n-1\}\in\omega$ and $s\in 2^{n+1}=\{0,1\}^{n+1}$. - -REPLY [3 votes]: After an e-mail communication with Lubomyr Zdomskyy, we came to the conclusion that the base zero-dimensionality is equivalent to the Rothberger property. -So, $\mathfrak z=\mathrm{cov}(\mathcal M)$ by a result of Fremlin and Miller.<|endoftext|> -TITLE: Is there an adjoint to the inclusion of I-adically complete modules to all modules? -QUESTION [7 upvotes]: A module $M$ over a ring $R$ is $I$-adically complete with respect to the ideal $I$, if the canonical map $M \to \lim M/I^nM$ is an isomorphism. There exists a completion functor: $M \mapsto \lim M/I^n M$. However, one can see that for modules that are not finitely generated, this functor is not exact even in the middle (see https://stacks.math.columbia.edu/tag/05JF). Thus completion cannot be left adjoint to the inclusion of complete modules into all modules, as then it would have been right exact. -Does such an adjoint exist? - -REPLY [11 votes]: Contrary to the skepticism expressed in the question, for a finitely generated ideal $I$ in a commutative ring $R$, the completion functor $\Lambda_I\colon M\longmapsto \varprojlim_n M/I^nM$ is, in fact, left adjoint to the embedding of the full subcategory of $I$-adically complete $R$-modules into the category of all $R$-modules. -The key issue here is indeed whether the completion is a complete module. This is not true for infinitely generated ideals $I$, generally speaking. However, for a finitely generated ideal $I$ in a commutative ring $R$, the $I$-adic completion of any $R$-module is $I$-adically complete. This observation goes back (at least) to A. Yekutieli's paper "On flatness and completion for infinitely generated modules over Noetherian rings", Commun. in Algebra 39 #11 (2011); see Section 1 in https://arxiv.org/abs/0902.4378 . -Essentially, the same argument also proves that the completion functor is left adjoint to the inclusion of the full subcategory of complete modules. For an explicit reference, I can suggest my paper L. Positselski, "Contraadjusted modules, contramodules, and reduced cotorsion modules", Moscow Math. Journ. 17 #3 (2017); see Theorem 5.8 in https://arxiv.org/abs/1605.03934 . -The nonexactness issue pointed out in the question is resolved as follows. The counterexample mentioned in the question shows that the $I$-adic completion is not exact as a functor $R{-}mod\longrightarrow R{-}mod$, i.e., from the category of $R$-modules to itself. -Generally speaking, given two categories $\mathcal A$ and $\mathcal B$ and a pair of adjoint functors $F\colon\mathcal A\longrightarrow\mathcal B$ and $G\colon\mathcal B\longrightarrow\mathcal A$, where $F$ is left adjoint to $G$, one can claim that $F$ preserves all colimits that exist in $\mathcal A$ and $G$ preserves all limits that exist in $\mathcal B$. It does not follow that the composition $GF\colon\mathcal A\longrightarrow\mathcal A$ preserves any limits or colimits. -When both the categories $\mathcal B$ and $\mathcal A$ are abelian, this means, in particular, that $F$ preserves the cokernels and $G$ preserves the kernels; in other words, $F$ is right exact and $G$ is left exact. Assuming additionally that the functor $G$ is exact, one can indeed conclude that the composition $GF$ is right exact. -This is not the case in the situation at hand, however. Denoting by $\mathcal A=R{-}mod$ the category of all $R$-modules and by $\mathcal B\subset\mathcal A$ the full subcategory of $I$-adically complete $R$-modules, one observes, first of all, that the category $\mathcal B$ is not abelian (see Example 2.7(1) in my paper in Moscow Math. Journ. cited above). Moreover, the inclusion functor $G\colon\mathcal B\longrightarrow\mathcal A$ does not preserve cokernels. The completion functor $F\colon\mathcal A\longrightarrow\mathcal B$ preserves cokernels, as it has to, being a left adjoint; but the composition $GF\colon R{-}mod\longrightarrow R{-}mod$ doesn't. - -To conclude, let me say that there are, in fact, (somewhat confusingly) two natural abelian full subcategories in $R{-}mod$ for a finitely generated ideal $I$ in a commutative ring $R$, both very similar to the full subcategory of $I$-adically complete $R$-modules, but slightly bigger and much better behaved. One of these two subcategories is contained in the other one, and they very often coincide (e.g., they always coincide for a Noetherian commutative ring $R$). The functors of inclusion of these two subcategories into $R{-}mod$ are exact, and both of them have left adjoints. -Objects of these two full subcategories in $R{-}mod$ (or more precisely, maybe, of the bigger one of the two) are variously called "Ext-complete", "weakly complete", "cohomologically complete", or "derived complete" in the literature. I call objects of the bigger abelian category "$I$-contramodule $R$-modules". The left adjoint functor to the inclusion of the full subcategory of $I$-contramodule $R$-modules into $R{-}mod$ is denoted by $\Delta_I$ in my paper cited above (where it is discussed at length). It is indeed right exact as a functor $R{-}mod\longrightarrow R{-}mod$, as it should be. -The objects of the smaller abelian full subcategory in $R{-}mod$ (which is therefore a closer abelian approximation of the full subcategory of $I$-adically complete $R$-modules) are called "quotseparated $I$-contramodule $R$-modules" in our preprint L. Positselski, A. Slávik "Flat morphisms of finite presentation are very flat", https://arxiv.org/abs/1708.00846 , where we discuss them in Section 5.5. -(Here the terminology is that, in both these papers of mine, what are above called "complete modules" are instead called "separated and complete modules", while it is shown that all contramodules are "complete (but not necessarily separated)"; so there is a nonabelian category of separated complete modules = separated contramodules, a slightly bigger abelian category of quotseparated contramodules, and a yet slightly bigger abelian category of all contramodules; all the three of them full subcategories in the category of modules.) -Finally, the left adjoint functor to the functor of inclusion of the full subcategory of quotseparated $I$-contramodule $R$-modules into $R{-}mod$ is simply the $0$-th left derived functor $\mathbb L_0 \Lambda_I$ of your non-right-exact $I$-adic completion functor $\Lambda_I\colon M\longmapsto \varprojlim_n M/I^nM$. This means that, given an $R$-module $M$, in order to compute the derived functors $\mathbb L_i\Lambda_I(M)$, one needs to choose a projective (or flat) resolution $P_\bullet\longrightarrow M$ of the $R$-module $M$ and put $(\mathbb L_i\Lambda_I)(M)=H_i(\Lambda_I(P_\bullet))$. The functor $\mathbb L_0\Lambda_I$ is right exact as a functor $R{-}mod\longrightarrow R{-}mod$.<|endoftext|> -TITLE: Should computer code be included within publications that present numerical results? -QUESTION [34 upvotes]: Many research papers include numerical results obtained through computation. Most of the time such computations are performed using software that is used by many mathematicians, i.e., Maple, Mathematica, or even C/C++ code. Should such code be included in the body of the published paper? -I've heard arguments from both sides: - -Including such code can greatly decrease the time taken by a referee to replicate the results, -The code can be easily modified by further authors who wish to extend the result, -The reader does not need to spend time searching the journal website or the Internet for any "auxilliary files" containing the code. - -On the other hand, - -Pages of code degrade the aesthetic nature of the publication, -The author might need to spend additional space explaining the coding decisions that were made in the algorithms, -It is likely that there exist (much) better ways to write the same algorithms in the given, or any other, language. - -So what is the standard in mathematical research papers that present numerical results, either as a main or as a side result? Should code be included within the body of the publication, as an auxilliary file, or not at all? - -REPLY [5 votes]: There are some issues that are not emphasised enough in the previous comments and answers. Having the source code used by an author does not let you check that the author's theorems are correct. It only lets you check that the program does what the author claims. Transcribing the program output to the published paper is the step where an error is least likely to have occurred. Much more likely is an error in the program. -So, can you eyeball the program to check if it is correct? Not unless it is a very short simple program. I publish articles that rely on tens of thousands of lines of code that took me and others months of hard work to write and debug. Your chances of looking at it and checking its correctness in a reasonable amount of time are next to zero. One day there will be programs that can check correctness for you; the beginnings exist today but generally useful checkers are still a long way off. -So what to do? If you are an author, get a coauthor and aim for separately implemented programs that get the same result, hopefully using different methods. (An axiom of software engineering is that programmers solving the same problem using the same method tend to make the same mistakes.) Intermediate results are very useful for checking, especially when the final answer has low entropy (like "yes" or "empty set"). -Another fact is that problems which needed very tricky programming and bulk computer time 20 years ago can now be solved in a reasonable time using simpler programs. Presumably that trend will continue. Any computational result that is important enough will eventually be replicated independently without so much effort.<|endoftext|> -TITLE: Basic theorem on induction for representations of $p$-adic groups -QUESTION [5 upvotes]: I know a lot of places where the following is sparsely proved, but I remember there was some paper where I read it in basically the same form I write it, but unfortunately I can't remember where it was. -Let $\mathcal{H}$ be the Hecke algebra of a reductive $p$-adic group, and $K$ be a compact open subgroup of $G$. There is a restriction functor $r : \mathcal{M}(\mathcal{H})\rightarrow \mathcal{M}(\mathcal{H}_K)$ defined by $r(V)=e_K V=V^K$ having the following properties. - -$r$ has a left adjoint, the induction functor $i: \mathcal{M}(\mathcal{H}_K)\rightarrow \mathcal{M}(\mathcal{H})$ (Frobenius reciprocity). -$i$ carries admissible representations to admissible representations. -The pair $(r,i)$ gives a bijection between $\mathcal{M}(\mathcal{H}_K)$ and the subcategory of $\mathcal{M}(\mathcal{H})$ of representations that are generated by $K$-fixed vectors. - -REPLY [4 votes]: The general setting for your question is the theory of types as developped by Bushnell and Kutzko: -Smooth representations of reductive p-adic groups: structure theory via types. Proc. London Math. Soc. (3) 77 (1998), no. 3, 582–634. -First in order that your question make sense, let us clarify a few things. -$\bullet$ The good "induction functor" here is -$$ -i(M) = {\mathcal H}(G)\otimes_{{\mathcal H}(K)} M -$$ -where $M$ is a left ${\mathcal H}(K)$-module. Here you consider ${\mathcal H}(G)$ as a right ${\mathcal H}(K)$-module and left $G$-module. -$\bullet$ It is false that $i$ takes "admissible modules" to "admissible modules". For instance if $M={\mathbb C}$ is the "trivial" character of ${\mathcal H}(K)$, then $i(M)$ is not admissible! Indeed if $L$ is a compact open subgroup of $G$, $i(M)^L$ is the space of $(L,K)$⁻bi-invariant functions on $G$ with finite support modulo $(L,K)$, an infinite dimensional space. -$\bullet$ You refer the pair $(r,i)$ as a "bijection". Of course you mean an equivalence of categories. This is false in general! -In order to get an equivalence of categories you need strong assumptions on $K$. As Peter McNamara writes, you need that $K$ have some nice "Iwahori decompositions". Typically a maximal compact subgroup of $G$ does not fulfill this condition. However this is true for a general reductive group $G$ and: -$\bullet$ $K=I$ an Iwahori subgroup of $G$ (Borel, Admissible representations of a semi-simple group over a local field with vectors fixed under an Iwahori subgroup. Invent. Math. 35 (1976), 233–259). -$\bullet$ $K$ is a certain "congruence subgroup" ("le centre de Bernstein" in Deligne, P.; Kazhdan, D.; Vignéras, M.-F. Représentations des algèbres centrales simples p-adiques. (French) [Representations of central simple p-adic algebras] Representations of reductive groups over a local field, 33–117, Travaux en Cours, Hermann, Paris, 1984. -The aim of Bushnell and Kutzko's theory is to generalize this as follows. Instead of taking the trivial character of $K$, you fix an irreducible smooth representation $\rho$. You define an idempotent $e_\rho$ in ${\mathcal H}(G)$ such that $e_\rho V$ is the $\rho$-isotypic component $V^\rho$ of $V$, for all smooth $G$-modules $V$. You define the Hecke algebra of the pair $(K,\rho )$ by -${\mathcal H}_\rho = e_\rho \star {\mathcal H}(G)\star e_\rho$. You have two functors between the category of $G$-modules $V$ generated by $V^\rho$ and the category of left ${\mathcal H}_\rho$-modules: $r(V)=V^\rho$, and $i(M) = {\mathcal H}(G)\otimes_{{\mathcal H}_\rho} M$. Then by definition $(K,\rho )$ is a type in $G$ if these functors define an equivalence of categories. -Given an irreducible representation $\pi$ of $G$, exhibiting a nice type $(K,\rho )$ such that $\pi_{\mid K}$ contains $\rho$ gives a lot of informations on $\pi$.<|endoftext|> -TITLE: Why only finite morphisms in etale fundamental group? -QUESTION [8 upvotes]: Can one define a version of etale fundamental group which takes into account infinite etale covers? What properties of the usual etale fundamental group would fail for it? -P.S.: here one can find illuminating discussion: - -As finite étale maps of - complex varieties are equivalent to finite topological covering spaces, this definition begs - the question: why have we restricted to finite covering spaces? There are at least two - answers to this question, neither of which is new: the first is that the covering spaces of - infinite degree may not be algebraic; it is the finite topological covering spaces of a complex - analytic space corresponding to a variety that themselves correspond to varieties. - The second is that Grothendieck’s étale $π_1$ classifies more than finite covers. It classifies - inverse limits of finite étale covering spaces [SGA1, Exp. V.5, e.g., Prop. 5.2]. These - inverse limits are the profinite-étale covering spaces we discuss in this paper (see Definition - 2.3). Grothendieck’s enlarged fundamental group [SGA 3, Exp. X.6] even classifies - some infinite covering spaces that are not profinite-étale. - -REPLY [8 votes]: As mentioned in other comments, there is a "pro-étale fundamental group" considered by Bhatt and Scholze. It is introduced in Chapter 7. of their article "The pro-étale topology for schemes". It is a topological group that is a so-called "Noohi group". For a connected (locally topologically noetherian) scheme $X$, it parameterizes the schemes $Y \rightarrow X$ that are étale and satisfy the valuative criterion of properness - the authors call such $Y$ "geometric covers" of $X$. Because we do not assume $Y$ to be of finite type over $X$, the map is not necessarily proper and we get more than just finite covers. -Another words, there is an equivalence of categories between the category of (possibly infinite) discrete sets with a continuous action of $\pi_1^{\mathrm{pro\acute{e}t}}(X)$ and the category of geometric covers of $X$. -The pro-\'etale fundamental group generalizes both the usual étale fundamental group and the "SGA3 fundamental group". The group $\pi_1^{\mathrm{\acute{e}t}}(X)$ is the profinite completion of $\pi_1^{\mathrm{pro\acute{e}t}}(X)$ and $\pi_1^{\mathrm{SGA3}}(X)$ is the pro-discrete completion of $\pi_1^{\mathrm{pro\acute{e}t}}(X)$. The situation is as follows: -If the scheme is normal, all three groups match, i.e. every geometric cover is finite. In the case of the nodal curve there exists an infinite geometric cover. In that case one can show $\pi_1^{\mathrm{pro\acute{e}t}}(X)=\pi_1^{\mathrm{SGA3}}(X)=\mathbb{Z}$ (assuming the base field was algebraically closed). However, for more complicated non-normal schemes $\pi_1^{\mathrm{pro\acute{e}t}}(X)$ gives more than $\pi_1^{\mathrm{SGA3}}(X)$: see Example 7.4.9. of "The pro-étale topology for schemes".<|endoftext|> -TITLE: On the coalgebraic homotopy transfer theorem -QUESTION [11 upvotes]: Let $A$ be a dg algebra, say over a field. The Homotopy Transfer Theorem says that $H(A)$ can noncanonically be given the structure of $A_\infty$-algebra, extending the induced multiplication on $H(A)$, in such a way that $A$ and $H(A)$ are quasi-isomorphic as $A_\infty$-algebras. -If $C$ is instead a dg coalgebra then we can also transfer the coalgebra structure to a quasi-isomorphic $A_\infty$-coalgebra structure on $H(A)$. Unfortunately the relation of quasi-isomorphism is less well behaved for coalgebras than for algebras and often one wants to consider instead the notion of weak equivalence: a coalgebra morphism $C \to C'$ is called a weak equivalence if the induced map on cobar constructions $\Omega C\to\Omega C'$ is a quasi-isomorphism. Is any dg coalgebra weakly equivalent to its cohomology as an $A_\infty$ coalgebra? - -REPLY [7 votes]: We worked out some answers to this question in our paper: arXiv:1904.03585 (Edit: the following answer refers to v1 of the paper on the arXiv!) -Here's the short version. There are two possible natural definitions of an $A_\infty$-coalgebra: -(A) It is a graded vector space $V$ with maps $V \to V^{\otimes n}$, $n \geq 1$, satisfying identities exactly dual to those defining an $A_\infty$-algebra. -(B) It is a graded vector space $V$ and a square zero derivation of the tensor algebra on $V$. -They are not equivalent; (B) is stronger than (A). In the paper we call them naive and genuine $A_\infty$-coalgebras. There is a homotopy transfer theorem for naive $A_\infty$-coalgebras: for any naive $A_\infty$-coalgebra $C$ there is a transfered naive $A_\infty$-structure on $H(C)$ which is quasi-isomorphic to the one on $C$. This is false for genuine $A_\infty$-algebra structures in general. -We say that an $A_\infty$-coalgebra $C$ is conilpotent if it admits an exhaustive filtration of the form -$$ 0 = F_0 C \subseteq F_1C \subseteq F_2C \subseteq \ldots $$ -which is compatible with the coalgebra structure. We call such a filtration a positive filtration. In the conilpotent case, the notions of genuine and naive $A_\infty$-coalgebra coincide. Any filtered quasi-isomorphism of positively filtered $A_\infty$-coalgebras is a weak equivalence. If $C$ is a conilpotent $A_\infty$-coalgebra equipped with a positive filtration, and $C \to V$ is a filtered quasi-isomorphism of chain complexes, then it's possible to transfer the $A_\infty$-coalgebra structure on $C$ to an $A_\infty$-structure on $V$, so that the $A_\infty$-morphism $C \rightsquigarrow V$ is a filtered quasi-isomorphism and hence a weak equivalence. It follows that although $C$ is in general not weakly equivalent to $H(C)$ for any transferred structure, there is always a weak equivalence from $C$ to $H(\operatorname{Gr} C)$ with a transferred $A_\infty$-coalgebra structure on $H(\operatorname{Gr} C)$. If we fix a positive filtration on $C$ then the $A_\infty$-coalgebra $H(\operatorname{Gr} C)$ with its transferred structure is uniquely determined up to a noncanonical filtered $A_\infty$-isomorphism, and it deserves to be called the filtered minimal model of $C$.<|endoftext|> -TITLE: A question on UCS p-groups -QUESTION [5 upvotes]: A $p$-group $G$ is called a ${\it UCS}$ $p$-group if $G$ has precisely three characteristic subgroups, namely $1$, $\Phi(G)$ and $G$. -Let $G$ be a finite UCS $p$-group of order $p^{2n}$ such that $\Phi(G)$ is elementary abelian $p$-group of order $p^n$. As an example of such a group we can give $G=\underbrace{\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p^{2}}\times\dots\times\mathbb{Z}_{p^{2}}}_{n\,\,times}$. Does there exist any nonabelian $p$-group with mentioned properties? - -REPLY [4 votes]: I think there are such examples for all odd primes $p$ and all $n \ge 3$. -There is a $p$-group $P$ of exponent $p$ of class $2$, with $\Phi(P)=Z(P)$ and $P/\Phi(P)$ and $\Phi(P)$ elementary abelian, with $|\phi(P)| = p^{n(n-1)/2}$, $|P/\Phi(P)|=p^n$, such that ${\rm Aut}(P)$ acts on $P/\Phi(P)$ as ${\rm GL}(n,p)$, where the induced action on $\Phi(P)$ is as the exterior square of the natural module for ${\rm GL}(n,p)$. -The group $P$ is defined by the presentation $\langle X \mid R \rangle$, where $$X=\{x_i:1 \le i \le n\} \cup \{y_{ij}^p: 1\le i -TITLE: Deforming metrics from non-negative to positive Ricci curvature -QUESTION [10 upvotes]: Given a closed Riemannian manifold $(M,g)$ with non-negative Ricci curvature and $dim\geq 3$, when can we deform the metric to a positive Ricci curved one? -I know it's impossible in general due to the flat factor in the universal covering. But what about we add some topological restrictions on $M$ like simply connectedness? Are there any positive or negative results on this problem? -( Besides, are there now any examples of simply connected closed manifold with positive scalar curved metric by do not admit a positive Ricci curved metric? ) -------------------------------------------------------------update 1------------------------------------------------------------ -Thanks to the answer by Robert, I may simplify the problem in the following sense, -Given a simply connected flat Einstein manifold $(M,g)$ with $\hat{A}$-genus non-vanishing, and set $(\mathbb{S}^2,h)$ be the standard unit sphere, can $(M\times \mathbb{S}^2, g+h)$ be perturbed to a Ricci-positive manifold? $\ \ $In general, what about changing $(\mathbb{S}^2,h)$ to an arbitrary closed Ricci-positive manifold? - -REPLY [11 votes]: There are obstructions. Perhaps the most famous comes from the theorem that, if a compact spin manifold has a metric of positive scalar curvature, then its $\hat A$-genus must vanish. -If you take a compact Riemannian spin manifold $(M,g)$ with special holonomy $\mathrm{G}_2$ (in dimension $7$), $\mathrm{Spin}(7)$ (in dimension $8$), or holonomy in $\mathrm{SU}(n)$ (in dimension $2n$) whose $\hat A$-genus is nonzero (and there are lots of these, even simply-connected ones), then $g$ will be Ricci-flat and hence will have non-negative Ricci curvature. However, by the above theorem, it cannot carry any metric with positive scalar curvature, let alone a metric with positive Ricci curvature.<|endoftext|> -TITLE: Sectional curvature of leaves of foliation -QUESTION [7 upvotes]: Given a $k$- dimensional foliation $F$ of a riemannian $n$-manifold $M$, with the property that the leaves of the foliation have constant sectional curvature $s$, for some $s$, is it true that $M$ will also have the same constant sectional curvature? -Is the same true if sectional curvature is replaced by the Gaussian curvature? -If it's a well known result, any hint at proving this or a possible reference or a counter example otherwise, will be most welcome. -Thanks. - -REPLY [3 votes]: One more counterexample: the hyperbolic $n$-space is foliated by horospheres with common center. Horospheres have zero curvature, the hyperbolic space curvature $-1$.<|endoftext|> -TITLE: Obstructions to realisation of dual finite spectra as suspension spectra -QUESTION [5 upvotes]: Suppose $X$ is a finite dimensional CW-complex with top cell at dimmension $n$ and consider its S-dual denoted by $DX$. I wonder if there are any obstructions to find a space $Y$ and an interger $k\geqslant n$ so that -$$\Sigma^kD(X)\simeq\Sigma^\infty Y_+ ?$$ -For example, in the case of $X=S^m$ the answer is positive. -EDIT In the case of finite dimensional projective spaces, one may start from spaces $\mathbb{R}P_m^n$ and choosing $m$ and $n$ in accordance with James periodicity, we can actually compute such a $k$. -The question is that is there any such $k$ for a given CW-complex of finite type (which of course after Neil's answer below I realise the answer is positive if one is to choose $k$ enough large, and obstructions are for specific $k$'s) -I would be very grateful for any references. - -REPLY [8 votes]: Firstly, you say that the answer is negative for finite-dimensional projective spaces. However, in this case, and for any finite complex $X$, the answer will be positive if we take $k$ sufficiently large. Perhaps you are just thinking of the case $k=n$? Anyway, I will assume that we have fixed some particular $k\geq n$ and we want to know whether $\Sigma^kDX$ is a suspension spectrum. -The most obvious point is that $H^*(\Sigma^kD(X);\mathbb{F}_p)$ needs to be an unstable module over the Steenrod algebra. This is easy to check if you have a good understanding of $H^*(X;\mathbb{F}_p)$ as a Steenrod module. Similarly, the module $M=K^0(\Sigma^kDX)$ has naturally defined Adams operations $\psi^q\colon M\to M[q^{-1}]$, and if $\Sigma^kDX$ is a suspension spectrum then these will lift in a compatible way to give operations $\psi^q\colon M \to M$. If this does not settle the question then one can consider additive unstable operations in $BP$ theory or $MU$ theory. These are harder to handle explicitly but the basic slogan is as follows: $MU^*(\Sigma^kDX)$ is functorial for isomorphisms of formal groups, and if $\Sigma^kDX$ is a suspension spectrum then this extends to give functoriality for all homomorphisms of formal groups.<|endoftext|> -TITLE: Factorizable groups -QUESTION [12 upvotes]: Definition. A finite group $G$ is factorizable if for any positive integer numbers $a,b$ with $ab=|G|$ there are subsets $A,B\subset G$ of cardinality $|A|=a$ and $|B|=b$ such that $AB=G$. - -Problem 1. Is each finite group factorizable? - -As I understood from these MO-posts (1, 2, 3), this problem is wide open and there is no intuition if it is true or not. So, we can ask a related - -Problem 2. Which finite groups are factorizable? - -The class of factorizable groups has a nice 3-space property that can be formulated in terms of bifactorizable subgroups. -A subgroup $H$ of a group $G$ is called bifactorizable if $H$ is factorizable and for any positive integer numbers $a,b$ with $ab=|G|$ there are sets $A,B\subset G$ such that $AHB=G$, $|A|\cdot |H|\cdot |B|=|G|$, $|A|$ divides $a$ and $|B|$ divides $b$. - -Theorem. A finite group $G$ is factorizable if $G$ contains a bifactorizable subgroup $H$. - -Proof. Given positive integers $a,b$ with $ab=|G|$ use the bifactorizability of $H$ to find subsets $A_1,B_1\subset G$ such that $A_1HB_1=G$, $|A_1|\cdot |H|\cdot |B_1|=|G|$, $|A_1|$ divides $a$, and $B_1$ divides $b$. The factorizability of $H$ yields two sets $A_2,B_2\subset H$ of cardinality $|A_2|=a/|A_1|$ and $|B_2|=b/|B_1|$ such that $A_2B_2=H$. Then the sets $A=A_1A_2$ and $B=B_2B_1$ have cardinality $|A|\le a$, $|B|\le b$ and -$AB=A_1A_2B_2B_1=A_1HB_1=G$. It follows from $ab=|G|=|A|\cdot|B|\le ab$ that $|A|=a$ and $|B|=b$. $\square$ -It is easy to see that a subgroup $H$ of a group $G$ is bifactorizable if it is factorizable and has prime index in $G$. Moreover, as was observed by M.Farrokhi D.G. in his answer to this post, a subgroup $H$ of a group $G$ is bifactorizable if $H$ is factorizable and the index of $H$ in $G$ is a prime power $p^k$ such that $p^{2k-1}$ divides $|G|$. -A normal subgroup $H$ of a group $G$ is factorizable if both groups $H$ and $G/H$ are factorizable. This implies - -Corollary. A finite group $G$ is factorizable if $G$ contains a bifactorizable subgroup $H$ with factorizable quotient $G/H$. - -This corollary reduces Problems 1,2 to studying the factorizability of finite simple groups. According to the classification of finite simple groups, each finite simple group is either cyclic of prime order, or alternating, or belongs to 16 families of groups of Lie type or is one of 27 sporadic groups. -Among these families only the factorizability of finite cyclic groups is trivially true. - -Problem 3. Is each alternating group $A_n$ factorizable? - -It may happen that the argument of Ilya Bogdanov from his answer to this MO-problem can be helpful here. On the other hand, it can be shown that the subgroup $A_n$ is bifactorizable in $A_{n+1}$ if and only if $A_n$ is factorizable and $n\ne 3$ is a power of a prime. It follows from the answer to this MO-question that the subgroup $A_3$ is not bifactorizable in $A_4$. - -Problem 4. Is any hope to prove that some infinite family of simple groups of Lie type consists of factorizable groups? - -REPLY [7 votes]: Definition. We say that a group $G$ of order $n$ is good if it satisfies any of the following equivalent conditions: - -For every divisor $d$ of $n$, there exists a nontrivial proper subgroup $H$ of $G$ such that either $d$ or $n/d$ divides $|H|$. -For every divisor $d$ of $n$, there exists a nontrivial proper subgroup $H$ of $G$ such that $[G:H]$ divides either $d$ or $n/d$. - -In the above definition, we can replace the subgroups with maximal subgroups. A set of (maximal) subgroups appearing in the definition of a good group is called a good set of (maximal) subgroups. -Theorem. Every good finite group with a good set of factorizable subgroups is factorizable. -Proof. Let $X$ be a good set of factorizable subgroups of $G$. If $|G|=ab$, then there exists a subgroup $H\in X$ such that $|H|$ is divisible by $a$ or $b$, say $a$. Let $|H|=aa'$ and $H=AA'$ be such that $|A|=a$ and $|A'|=a'$. Then $G=H(H\backslash G)=A(A'(H\backslash G))$ is a factorization of $G$ with $|A|=a$ and $|A'(H\backslash G)|=b$. Therefore, $G$ is factorizable. -Corollary. Let $q\neq2,4$ be a prime power. If $A_{q-1}$ is factorizable, then so is $A_q$. -Proof. The group $A_q$ has a maximal subgroup $A_{q-1}$ of index $q$, and that $q$ divides either $a$ or $b$ whenever $|A_q|=ab$. Hence, $A_q$ is good with the good set $\{A_{q-1}\}$ of factorizable subgroups so that $A_q$ is factorizable. -As a result, we obtain a new proof that $A_9$ is factorizable. -I think a large class of finite simple groups can be studied using the above theorem and the following criterion of François Brunault: - -If $|G|=ab$ and $G$ has a section of size $a$ or $b$, then $G=AB$ with $|A|=a$ and $|B|=b$.<|endoftext|> -TITLE: Does $\mathsf{MA}^+(\sigma-{\rm closed})$ imply there are no Kurepa Trees? -QUESTION [7 upvotes]: The question in the title is somewhat self contained but let me make some definitions and remarks to clarify. -Recall that $\mathsf{MA}^+(\sigma-{\rm closed})$ is the statement that if $\mathbb P$ is a $\sigma$-closed poset, $\langle D_\alpha \; |\; \alpha < \omega_1\rangle$ is a sequence of dense subsets of $\mathbb P$ and $\dot{S}$ is a name for a stationary set then there is a filter $G \subseteq \mathbb P$ so that $G \cap D_\alpha \neq \emptyset$ for all $\alpha < \omega_1$ and $\{\alpha \; | \; \exists p \in G \; p \Vdash \check{\alpha} \in \dot{S}\}$ is stationary. -A Kurepa tree is a tree of height $\omega_1$, countable levels and $\geq \omega_2$ branches. -My question is simply if $\mathsf{MA}^+(\sigma - {\rm closed})$ implies there are no Kurepa trees. -I have no intuition about this, as on the one hand I don't see a way that $\sigma$-closed forcing suffices to show there are no Kurepa trees and on the other hand I don't see a way to force $\mathsf{MA}^+(\sigma-{\rm closed})$ without killing all Kurepa trees. Specifically: -It's well known that Silver collapsing an inaccessible to $\omega_2$ implies that there are no Kurepa trees so in natural models of $MA^+(\sigma-{\rm closed})$ obtained by iterating all $\sigma$-closed forcing notions below some sufficiently large inaccessible (supercompact?) $\kappa$ there are no Kurepa Trees. -Meanwhile, the proof that $\mathsf{PFA}$ implies there are no Kurepa trees involves specializing an Aronszajn tree and such specialization forcings are not in general $\sigma$-closed. -Thanks! - -REPLY [9 votes]: Yes, it implies no Kurepa trees. First, note that the forcing axiom you consider implies the Weak Reflection Principle, which in turn implies (a strong form of ) Chang's Conjecture. Both of those facts are covered in the Foreman-Magidor-Shelah paper on Martin's Maximum. And Chang's Conjecture implies there are no Kurepa trees. I believe the proof of the latter appears in Foreman's chapter in the Handbook of Set Theory, but here is a sketch: suppose $T$ was a Kurepa tree on $\omega_1$; let $\langle b_i \ : \ i < \omega_2 \rangle$ be a 1-1 list of cofinal branches of $T$. By Chang's Conjecture, there is an $X \prec (H_{\omega_3},\in, \vec{b})$ such that $X \cap \omega_2$ has ordertype $\omega_1$, but $X \cap \omega_1 \in \omega_1$. Consider the collection $\{ b_i \ : \ i \in X \cap \omega_2 \}$. For any distinct $b_i$, $b_j$ in that collection, since $i \ne j$ and both indices are in $X$, it follows that $b_i$ and $b_j$ diverge at some level below $X \cap \omega_1$. This implies that level $X \cap \omega_1$ of $T$ has size $\omega_1$, contradicting that it is a thin tree.<|endoftext|> -TITLE: Poincaré metric on the Riemann sphere minus more than two points -QUESTION [18 upvotes]: If we omit more than two points from the Riemann sphere, we will obtain a hyperbolic Riemann surface endowed with a canonical metric descending from its universal cover which is the Poincaré disk. Let us denote the hyperbolic metric on this surface by $d_h$, and the usual spherical metric on the Riemann sphere by $d$. Here is my question: -Can we find a constant $C>0$ such that for any two points $x$ and $y$ in this punctured sphere we have: -$$d(x,y) -TITLE: "Insanely increasing" $C^\infty$ function with upper bound -QUESTION [16 upvotes]: Let $C^\infty$ denote the collection of functions $f:\mathbb{R}\to\mathbb{R}$ such that for every positive integer $n$, the $n$-th derivative of $f$ exists. For $f\in C^\infty$ we set - -$f^{(0)} = f$, and -$f^{(n+1)} = \big(f^{(n)}\big)'$ for all non-negative integers $n$. - -Is there $f\in C^\infty$ with the following properties? - -for all $x\in (-\infty, 0]$ we have $f(x)=0$. -for all non-negative integers $n$ and $x\in (0,\infty)$ we have $f^{(n+1)}(x) > f^{(n)}(x)$ -there is a function $h:\mathbb{R}\to\mathbb{R}$ such that for all non-negative integers $n$ and all $x\in\mathbb{R}$ we have $f^{(n)}(x)\leq h(x)$. - -(Additional question for curiosity, answering it is not needed for acceptance of answer: can $h$ be chosen to be continous? Or even $h\in C^\infty$?) - -REPLY [22 votes]: By Willie Wong's comment and answer, $f^{(n)}>0$ on $(0,\infty)$ for all $n=0,1,\dots$. Hence, $f^{(n)}\ge0$ on $\mathbb R$. So, by Bernstein's theorem on completely monotone functions (used "right-to-left"; see the "footnote" for details), $f(x)=\int_0^\infty e^{tx}\mu(dt)$ for some nonzero nonnegative measure $\mu$ on $[0,\infty)$ and all real $x\le1$. So, $f>0$ on $(-\infty,1]$, which contradicts condition 1 in the OP. Thus, there exists no $f$ satisfying conditions 1 and 2. (Condition 3 is not needed for the non-existence.) -"Footnote": The condition $f^{(n)}\ge0$ on $\mathbb R$ for all $n=0,1,\dots$ implies $(-1)^n g^{(n)}\ge0$ on $[0,\infty)$ for all $n=0,1,\dots$, where the function $g\colon[0,\infty)\to\mathbb R$ is defined by reading $f$ "right-to-left": -$g(y):=f(1-y)$ for $y\in[0,\infty)$. So, $g$ is completely monotone and hence, by Bernstein's theorem, $g(y)=\int_0^\infty e^{-ty}\nu(dt)$ for some nonnegative measure $\nu$ and all real $y\ge0$. Moreover, the measure $\nu$ is nonzero, because otherwise we would have $g=0$ (on $[0,\infty)$), that is, $f=0$ on $(-\infty,1]$, which would contradict the condition that $f^{(n)}>0$ on $(0,\infty)$ for all $n=0,1,\dots$. -So, $f(x)=g(1-x)=\int_0^\infty e^{-t+tx}\nu(dt)=\int_0^\infty e^{tx}\mu(dt)$ for all real $x\le1$, where $\mu(dt):=e^{-t}\nu(dt)$, and the measure $\mu$ is nonnegative and nonzero. - -REPLY [21 votes]: Combining my comments with that of Terry Tao's: - -First we show that $f^{(n)} > 0$ on $(0,\infty)$ for all $n \geq 1$. The argument is given for $f'$, but, extends easily to all $n \geq 1$. -Start by noticing that $f'(a) = 0 \implies f''(a) > 0$ so that $f'$ changes sign at most once, and that if it is not everywhere positive on $(0,\infty)$ there must be an initial interval $(0,\epsilon)$ on which $f' < 0$. Assume WLOG that $\epsilon < \frac12$. On $(0,\epsilon)$, we have that $f(x) \geq x \inf_{y\in (0,x)} f'(y)$ by the mean value theorem. This is incompatible with $f'(y) > f(y)$. Hence we conclude that $f'$ is always positive on $(0,\infty)$. -Step 1 implies that $f^{(n)}$ is increasing for all $n \geq 1$. Therefore if $h$ is a function as in condition (3) of the question, so is the increasing function -$$ \tilde{h}(y) = \inf_{[y,\infty)} h $$ -This function is locally bounded, and implies (by Taylor's inequality) that $f$ is real analytic. -Real analytic functions can't be vanishing on $(-\infty,0)$ and be non-trivial.<|endoftext|> -TITLE: Is each finite group multifactorizable? -QUESTION [10 upvotes]: Definition. A finite group $G$ is called multifactorizable if for any positive integer numbers $a_1,\dots,a_n$ with $a_1\cdots a_n=|G|$ there are subsets $A_1,\dots,A_n\subset G$ such that $A_1\cdots A_n=G$ and $|A_i|=a_i$ for all $i\le n$. -In this case we shall write that the group $G$ is $a_1{\times}\cdots{\times}a_n$-factorizable. - -It can be shown that each finite Abelian group is multifactorizable. - -Problem 1. Is each finite (simple) group multifactorizable? - -As was observed by Geoff Robinson in his answer to this question, each finite nilpotent group is multifactorizable. - -Problem 2. Is each finite solvable group multifactorizable? - -Added in Edit. It turns out that the alternating (solvable) group $A_4$ is not multifactorizable, more precisely, $A_4$ is not $2{\times}3{\times}2$-factorizable. -Now it remains to find an example of a finite simple group which is not multifactorizable. - -Problem 3. Is the alternating group $A_5$ multifactorizable? In particular, is $A_5$ $2{\times}15{\times}2$-factorizable? - - -Added in Edit 2. By computer calculations, @Fracois Brunault proved that the alternating group $A_5$ is not multifactorizable. More precisely, the group $A_5$ is not $2{\times}3{\times}5{\times}2$-factorizable. -Added in Edit 3. -On the other hand, as was remarked by @Gro-Tsen, it is an open problem (of minimal logarithmic signature) if any finite (simple) group $G$ can be written as the product $G=A_1\cdots A_n$ of subsets $A_i\subset G$ whose cardinality $|A_i|$ is a prime number or 4 such that $|G|=|A_1|\cdots|A_n|$. This problem is resolved for some classes of finite simple groups, see this paper for more information. - -REPLY [6 votes]: I wrote a Magma procedure to test whether $A_5$ is multifactorizable. A brute force search does not seem feasible, so I used the ideas in Taras Banakh's answer and in the comments. -Let $A_5 = ABCD$ with $|A|=2$, $|B|=3$, $|C|=5$, $|D|=2$. We may assume $A,B,C,D$ all contain the identity element. Then $A=\{e,a\}$ and $D=\{e,d\}$ where $a$ and $d$ distinct elements of order 2. Applying an automorphism of $A_5$ and using Taras's argument, we may reduce to the case $a=(1 2)(3 4)$ and $d=(2 3)(4 5)$. -I loop over all possible subsets $B$ of $A_5$ of size 3 containing $e$ such that $\# ABD = 12$ (there are 1248 such subsets). For each such $B$, I compute the list of all subsets of the form $ABcD$ with $c \in A_5$ which are disjoint from $ABD$ and are of size 12 (typically, there are about ten such subsets) and I test whether any of them add up to $A_5$. It turns out that there is no solution, so the group $A_5$ is not multifactorizable.<|endoftext|> -TITLE: Compact simply-connected homogeneous symplectic manifold -QUESTION [8 upvotes]: I was reading a paper in which the authors use the fact that any compact simply-connected homogeneous symplectic manifold has non-zero Euler characteristic. They prove it by quoting a theorem by Kostant which implies that the manifold is symplectomorphic to a coadjoint orbit of a semisimple group, then state that compact coadjoint orbits of semisimple groups have non-zero Euler characteristic. -I am looking for a more direct proof of that fact. Do you know some? - -REPLY [6 votes]: Let your manifold be $X=G/H$. First of all, since it is simply connected, we can write it as $K/U$ where $K$ and $U=K\cap H$ are compact in $G$ (Montgomery’s theorem, 1950). Next, since $K/U$ is homogeneous symplectic, one knows that $U$ is the centralizer of a torus $S\subset K$.1) In particular $U$ contains any maximal torus containing $S$, i.e. $U$ is an equal rank subgroup of $K$. And finally, one knows that equal rank subgroups satisfy $χ(K/U)\ne0$: e.g. Samelson (1958), or Mostow (2005). - -1) That is clear, with $S$ the closure of $\exp(\mathbf Rx)$, if we already know that $X\simeq$ the (co)adjoint orbit of some $x\in\mathfrak k^*\simeq\mathfrak k$. But it can also be proved a priori : Borel–Weil (1954, Thm 1), or in more detail Matsushima (1957, Thm 1).<|endoftext|> -TITLE: Symplectic connections are (locally) Levi-Civita connections -QUESTION [7 upvotes]: I was wondering... Is every symplectic connection $\nabla$ -on some symplectic manifold $(M,ω)$ the Levi-Civita connection of some metric $g$ on $M$? What about the local statement? - -REPLY [2 votes]: I think I've got the answer, and it is "no", at least for the global question. -When I started to try to understand the problem, I realized that to get some structure that is invariant by the connection we can (have to) fix the structure at some point and use parallel transport through picewise smooth paths to extend it. The resulting structure will be well-defined iff the initially fixed structure is invariant by the holonomy group at the point. The idea is then to construct some connection whose holonomy group at some point preserves some symplectic structure on the tangent space of the point but does not preserve any metric on such tanget space. To get the second feature, we may try to construct the connection in such a way that its holonomy group is unbounded. After some tries, I got the following: -On $\mathbb{R}^2$, we consider the connection $\nabla$ given by \begin{align*} \nabla_{\frac{\partial}{\partial x}}\frac{\partial}{\partial x} & = y\frac{\partial}{\partial x} & \nabla_{\frac{\partial}{\partial x}}\frac{\partial}{\partial y} & = 0 \\ -\nabla_{\frac{\partial}{\partial y}}\frac{\partial}{\partial y} & = x\frac{\partial}{\partial y} & \nabla_{\frac{\partial}{\partial y}}\frac{\partial}{\partial x} & = 0. -\end{align*} -The curvature $R$ of $\nabla$ at $(0,0)$ is given by $R(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}) = \begin{bmatrix} --1 & 0 \\ -0 & 1 -\end{bmatrix} \in GL(\mathbb{R}^2) = GL(T_{(0,0)}\mathbb{R}^2)$, and all its covariant derivatives are of the form $\begin{bmatrix} --c & 0 \\ -0 & c -\end{bmatrix}$ at such point. Because of that and of the connectedness of $Hol(\nabla, (0,0))$ (simply-connectedness of $\mathbb{R}^2$), Ambrose-Singer's theorem implies that the canonical symplectic form of $T_{(0,0)}\mathbb{R}^2 = \mathbb{R}^2$ is invariant by $Hol(\nabla, (0,0))$. On the other hand, $Hol(\nabla, (0,0))$ contains the exponential of $tR(\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$ for every $t \in \mathbb{R}$, and is therefore an unbounded subset of $GL(T_{(0,0)}\mathbb{R}^2)$. As such, it doesn't fix any metric on $T_{(0,0)}\mathbb{R}^2$. - -$ \textit{edit:}$ using Levi-Civita's formula, it is easy to conclude that the connection $\nabla$ defined above is not the Levi-Civita connection of any metric, even locally. -With that in mind, Darboux's theorem shows us that for every symplectic manifold there exist some open subset $U$ of the manifold and some symplectic connection $\nabla$ defined on $U$ such that $\nabla$ is not the Levi-Civita connection of any metric on $U$. -By using partitions of unity, we conclude, then, that $\textbf{every symplectic manifold admits}$ $\textbf{a symplectic connection that is not a}$ $\textbf{Levi-Civita connection.}$<|endoftext|> -TITLE: global sections of locally free sheaf on projective space -QUESTION [5 upvotes]: Let $\mathcal{E}$ be a locally free sheaf on $\mathbb{P}^n_A=\mathbb{P}^n\times_{Spec k} Spec A$, where $A$ is a finitely generated algebra over a field $k$. By a well known theorem (see e.g. Hartshorne's Algebraic Geometry, Thm 5.19) $H^0(\mathbb{P}^n_A, \mathcal{E})$ is a finitely generated $A$- module. - -Is it true that $H^0(\mathbb{P}^n_A, \mathcal{E})$ is a projective module? - -Let $B$ be a finitely generated $k$-algebra, $f: Spec B \to Spec A$ a morphism and $f^*\mathcal{E}$ the pullback of $\mathcal{E}$ to $Spec B$. - -Is it true that $H^0(\mathbb{P}^n_A, \mathcal{E})\otimes _A B= H^0(\mathbb{P}^n_B, f^*\mathcal{E})$ ? - -1 and 2 above are true when $\mathcal {E}$ is a direct sum of line bundles of the form $\mathcal O(n)$. I was wondering if they are true for general $\mathcal{E}$. - -REPLY [8 votes]: The answer to both questions is negative, see counterexamples below. -1) Let $A = k[x,y,z]$, $n = 1$. Note that -$$ -H^1(\mathbb{P}^n_A,O(-2)) \cong A. -$$ -Consider the extension -$$ -0 \to O(-2) \to E \to O \oplus O \oplus O \to 0 -$$ -whose extension class is $(x,y,z)$. Then the cohomolopgy exact sequence -$$ -0 \to H^0(\mathbb{P}^n_A,E) \to A \oplus A \oplus A \stackrel{(x,y,z)}\to A -$$ -shows that $H^0(\mathbb{P}^n_A,E)$ is reflexive but not locally free (hence not projective). In fact, this is the simplest example of a reflexive non-locally free sheaf. -2) Take $A = k[x,y,z,w]$ and define $E$ as the extension -$$ -0 \to O(-2) \to E \to O \oplus O \oplus O \oplus O \to 0 -$$ -whose extension class is $(x,y,z,w)$. Let $B = k$ with the morphism $A \to B$ defined by $x,y,z,w \mapsto 0$. Then $f^*E \cong O(-2) \oplus O \oplus O \oplus O \oplus O$, hence -$$ -H^0(\mathbb{P}^n_B,f^*E) = B \oplus B \oplus B \oplus B. -$$ -On the other hand, tensoring -$$ -0 \to H^0(\mathbb{P}^n_A,E) \to A \oplus A \oplus A \oplus A \stackrel{(x,y,z,w)}\to A \to B \to 0 -$$ -by $B$ (over $A$), we deduce -$$ -H^0(\mathbb{P}^n_A,E) \otimes_A B \cong Tor_2^A(B,B) \cong B^{\oplus 6}. -$$<|endoftext|> -TITLE: Approximate homology of a large simplicial complex -QUESTION [6 upvotes]: I can use software to calculate the Betti numbers $\beta_0,\beta_1,\beta_2,\dots$ of a finite simplicial complex. -This is prohibitive for large complexes, built on say > 100,000 nodes. -Is there some way to computationally approximate the ranks of the first $n$ homology groups? Results e.g. Carlson here seem to work only for data points in Euclidean space, where the relations on which the complex is built are interpretations of the data (i.e. persistent homology). I have a set of fixed, deterministic relations on a set of vertices i.e. a graph, and the corresponding clique complex. - -REPLY [3 votes]: You have to find a way to reduce the size of your simplicial complex. Some algorithms based e.g. on discrete Morse theory can do that fairly rapidly, but they don't have guarantees on the amount of size reduction. I don't think there exists faster algorithms for approximate Betti numbers in general, but I believe it can be done if your simplicial complex is "low-dimensional", meaning that for any point, the size of the sub complex formed by vertices at distance less than $r$ from that point grows slowly with $r$.<|endoftext|> -TITLE: Relation between the Casson-Gordon invariants $\sigma(M, \chi)$ and $\sigma_r(M, \chi)$ -QUESTION [10 upvotes]: Setting: There are two objects in knot theory that are commonly referred to as the Casson-Gordon invariants: the invariant $\sigma$, and the invariant $\tau$ (see for example A. Conway’s notes Algebraic Concordance and Casson-Gordon Invariants [3] for an introduction to these invariants). When it comes to the $\sigma$-invariant, usual references in the literature include Casson and Gordon’s original papers Cobordism of Classical Knots [1], where the invariant appears as $\sigma(M, \chi)$, and On Slice Knots in Dimension Three [2], where the invariant appears as $\sigma_r(M, \chi)$. I am currently trying to understand the relation between these two different formulations of the $\sigma$-invariant, but I am having some difficulties with it. -To be more precise, let me quickly outline the construction of the invariant $\sigma(M, \chi)$, as found in [3] on p.14, (resp. [1] on p. 183), and of $\sigma_r(M, \chi)$, as found in [2] on p. 41/42 (readers that are already familiar with the definitions and/or cited papers might want to skip the next two paragraphs). -Definition of $\sigma(K, \chi)$: Given a compact $4$-manifold $W$ and a morphism $\psi: \pi_1(W) \rightarrow \mathbb{Z}_m$, let $\widetilde W \rightarrow W$ be the associated $\mathbb{Z}_m$-covering. Then $H_2(\widetilde W; \mathbb{Z})$ is a $\mathbb{Z}[\mathbb{Z}_m]$-module. By mapping the generator of $\mathbb{Z}_m$ to $\omega := e^{\frac{2\pi i}{m}}$, we get a map $\mathbb{Z}_m \rightarrow \mathbb{Q}(\omega)$, which endows $\mathbb{Q}(\omega)$ with a $(\mathbb{Q}(\omega), \mathbb{Z}[\mathbb{Z}_m])$-bimodule structure. Set -\begin{equation} -H_*(W; \mathbb{Q}(\omega)) = \mathbb{Q}(\omega) \otimes_{\mathbb{Z}[\mathbb{z}_m]} H_*(\widetilde W; \mathbb{Z}). -\end{equation} -Then the $\mathbb{Q}(\omega)$-vector space $H_2(W; \mathbb{Q}(\omega))$ is endowed with a $\mathbb{Q}(\omega)$-valued Hermitian form $\lambda_{\mathbb{Q}(\omega)}$ (whose definition is analogous to the ordinary intersection form on $H_2(W; \mathbb{Z})$). Define the signature of $W$ twisted by $\psi$ as -\begin{equation} -\textrm{sign}^\psi(W) := \textrm{sign}(\lambda_{\mathbb{Q}(\omega)}). -\end{equation} -Now, given a closed $3$-manifold $M$ and a homomorphism $\chi: \pi_1(M) \rightarrow \mathbb{Z}_m$, bordism theory implies that there exists a non-negative integer $k$, a $4$-maifold $W$ and a homomorphism $\psi: \pi_1(W) \rightarrow \mathbb{Z}_m$ such that $\partial(W, \psi) = k(M, \chi)$. Define the invariant $\sigma(M, \chi)$ as -\begin{equation} -\sigma(M, \chi) := \frac{1}{k}(\textrm{sign}^\psi(W) - \textrm{sign}(W)) \in \mathbb{Q}. -\end{equation} -Definition of $\sigma_r(M, \chi)$: Let $M$ be a closed $3$-manifold and $\chi: H_1(M) \rightarrow \mathbb{Z}_m$ an epimorphism. Again, $\chi$ induces an $m$-fold cyclic covering $\widetilde M \rightarrow M$ with a canonical generator of the group of covering translations, corresponding to $1 \in \mathbb{Z}_m$. Suppose that for some positive integer $k$ there exists an $mk$-fold cyclic branched covering of $4$-manifolds $\widetilde W \rightarrow W$, branched over a surface $F \subset \textrm{int }W$, such that $\partial(\widetilde W \rightarrow W) = k(\widetilde M \rightarrow M)$, and such that the covering translation of $\widetilde W$ that induces rotation through $2\pi/m$ on the fibers of the normal bundle of $\widetilde F$ restricts on each component of $\partial\widetilde W$ to the canonical covering translation of $\widetilde M$ determined by $\chi$. Let this covering translation induce $\tau$ on $H = H_2(\widetilde W) \otimes \mathbb{C}$. Further, let $\cdot_H$ denote the intersection form of $H_2(\widetilde W)$ extended to $H$ (which is Hermitian, but in general not non-singular). Then $(H, \cdot)$ decomposes as an orthogonal direct sum of eigenspaces of $\tau$, corresponding to the eigenvalues $\omega^r$, $0 \leq r < m$, where $\omega := e^{\frac{2\pi i}{m}}$. Let $\varepsilon_r(\widetilde W)$ denote the signature of $\cdot_H$ restricted to the eigenspace corresponding to the eigenvalue $\omega^r$. Define, for $0 < r < m$, the invariant $\sigma_r(M, \chi)$ as -\begin{equation} -\sigma_r(M, \chi) := \frac{1}{k}\left(\textrm{sign}(W) - \varepsilon_r(\widetilde W) - \frac{2[F]^2r(m-r)}{m^2}\right) \in \mathbb{Q}, -\end{equation} -where $[F]$ denotes the self-intersection number of the branching surface $F$. -Problem: On pages 185-186 in [1], Casson and Gordon explain the relation of $\sigma(M, \chi)$ to the Atiyah-Singer G-signature. In particular, they conclude on page 186 that (with the notation from above), -\begin{equation} -k\sigma(M, \chi^r) + \textrm{sign}(W) = \varepsilon_r(\widetilde W), -\end{equation} -which reads for $r = 1$ in particular as -\begin{equation} -k\sigma(M, \chi) + \textrm{sign}(W) = \varepsilon_1(\widetilde W). -\end{equation} -Further, Proposition 2.19 on p. 18 in [3] states that the formula for computing $\sigma_r(M, \chi)$ in terms of a surgery description of $M$ (Lemma 3.1 on p. 42 in [2]) is also valid for $\sigma(K, \chi)$. From this, I deduced that we (should) have a relation like $\sigma(M, \chi^r) = -\sigma_r(M, \chi)$ for $0 < r < m$. However, there is the summand $\frac{2[F]^2r(m-r)}{m^2}$ appearing in the definition of $\sigma_r(M, \chi)$, and I don't see where or in what form this summand appears in the definition of $\sigma(M, \chi)$ (I do see however why it appears in the definition of $\sigma_r(M, \chi)$, namely because of the fact that for a branched covering $\widetilde W \rightarrow W$ of closed 4-manifolds, there is the formula $\varepsilon_r(\widetilde W) = \textrm{sign}(W) - \frac{2[F]^2r(m-r)}{m^2}$ (Lemma 2.1 on p. 40 in [2])). So my question is the following: -Question: Is the relation $\sigma(M, \chi^r) = -\sigma_r(M, \chi)$ correct? If so, where does the summand $\frac{2[F]^2r(m-r)}{m^2}$ appear in $\sigma(M, \chi)$? If not, is there another relation that holds for $\sigma(M, \chi)$ and $\sigma_r(M, \chi)$? -I assume that my question has something to do with the branching set $F$ as I don't see it mentioned in [1] or [3]. However, I'm too unexperienced in the subject to come to a conclusion on my own. Thus, any elaboration would be greatly appreciated. Thanks in advance! -References: -[1] A. J. Casson; C. Mc A. Gordon, Cobordism of Classical Knots, À la Recherche de la Topologie Perdue, Progress in Mathematics, Volume 62, pp. 181 - 199, Birkhäuser Boston, 1986. With an appendix by P. M. Gilmer (available here) -[2] A. J. Casson; C. Mc A. Gordon, On Slice Knots in Dimenstion Three, Proceedings of Symposia in Pure Mathematics, Volume 32, pp. 39 - 53, American Mathematical Society, 1978 (available here). -[3] A. Conway, Algebraic Concordance and Casson-Gordon Invariants, notes of a reading group held in Geneva, Spring 2017 (available here). - -REPLY [7 votes]: I will expand on my comment. Since $\sigma_r(M,\chi)$ is independent of $F$, you can take $F=\emptyset$ and therefore $\sigma_r(M,\chi)=\frac{1}{k}(\operatorname{sign}(W)-\epsilon_r(\widehat{W}))$, where I write $\widehat{W} \to W$ for the $m$-fold cover induced by $\psi$ (I will use $\widetilde{W}$ for the universal cover). It remains to show that $\epsilon_r(\widehat{W})=\operatorname{sign}^\psi(W)$, where $\psi$ extends $\chi^r$. -Let $H_{\omega^r}$ be the $\omega^r$-eigenspace of the $\mathbb{C}$-vector space $H_2(\widehat{W};\mathbb{C})$ (untwisted coefficients). I will use $\mathbb{C}$ instead of $\mathbb{Q}(\omega)$, but everything basically works the same. It is not too difficult to show that $H_{\omega^r} \cong \mathbb{C}_{\omega} \otimes_{\mathbb{Z}[\mathbb{C}_m]} H_2(\widehat{W};\mathbb{C})$. Since $\mathbb{Z}_m$ is finite, Maschke’s theorem implies that $\mathbb{C}[\mathbb{Z}_m]$ is a semi-simple ring. Consequently all its (left)-modules are projective and in particular flat. It follows that -$$H_{\omega^r} \cong \mathbb{C}_{\omega} \otimes_{\mathbb{C}[\mathbb{Z}_m]} H_2(\widehat{W};\mathbb{C})=H_2(\mathbb{C}_{\omega} \otimes_{\mathbb{Z}[\mathbb{C}_m]} C_*(\widehat{W}) ).$$ -Writing $C_*(\widehat{W})=\mathbb{Z}[\mathbb{Z}_m] \otimes_{\mathbb{Z}[\pi_1(W)]} C_*(\widetilde{W})$, you can then conclude that -$$H_{\omega^r} \cong H_2(W;\mathbb{C}_{\omega}).$$ -I slightly abused notations: I wrote $\mathbb{C}_{\omega}$ both for the $\mathbb{C}[\mathbb{Z}_m]$-module structure and for the $\mathbb{Z}[\pi_1(W)]$-module structure induced by $\chi^r$. -You can check that this isomorphism preserves the intersection forms and the equality of signatures follows. Finally, note that on page 42 of the paper you cite, Casson and Gordon write "We shall, however, always be in a situation where either $n=1$ or $F=\emptyset$".<|endoftext|> -TITLE: Is Post's tag system solved? -QUESTION [18 upvotes]: Has the 3-tag system investigated by Emil Post $(0\to00, 1\to1101)$ been solved? Is there a decision algorithm to determine which starting strings terminate, which end up in a cycle, and which (if any) grow without bound? -Also, what cycle structures are there? Setting $a=$ '00' and $b=$ '1101', the only cycles I know of begin with $ab, b^2 a^2$, combinations of these, and $a^2 b^3 (a^3 b^3)^n$. Are there any more? - -REPLY [7 votes]: Here are the two irreducible repeating patterns that Liesbeth de Mol discovered, together with a third high-period irreducible repeating pattern discovered by Rich Schroeppel: - -$b^3 a^5 b^5$ (period 40); -$a b^2 a b^3 a^3 b^3 a^2 b^2 a^4 b^2$ (period 66); -$a b^3 a b a b a^2 b^2 a b^9 a^2$ (period 282); - -This was discovered by 2004, and as far as I know no other irreducible repeating patterns are known yet.<|endoftext|> -TITLE: linear recurrence relation for square of sequence given recursively -QUESTION [8 upvotes]: If $a_n$ satisfies the linear recurrence relation $a_n = \sum_{i=1}^k c_i a_{n-i}$ for some constants $c_i$, then is there an easy way to find a linear recurrence relation for $b_n = a_n^2$ ? -For example, if $a_n = a_{n-1} + a_{n-3}$, then $b_n=a_n^2$ seems to satisfy $b_n=b_{n-1}+b_{n-2}+3b_{n-3}+b_{n-4}-b_{n-5}-b_{n-6}$. - -REPLY [9 votes]: Yes. Take the companion matrix $M$ of the characteristic polynomial of your original recurrence. Then the squared recurrence satisfies a recurrence with the characteristic polynomial of the symmetric square $S^2(M)$ of $M$. If the original recurrence has order $n$ this new recurrence has order ${n+1 \choose 2}$, and its coefficients are polynomials (depending only on $n$) in the coefficients of the old recurrence. -To prove this it suffices to consider the case where the characteristic polynomial has distinct roots, by a density argument. Then $a_n$ is a linear combination of the sequences $\lambda_i^n$ where $\lambda_i$ are the roots of the characteristic polynomial. So $a_n^2$ is a linear combination of the sequences $(\lambda_i \lambda_j)^n$ (and we may have $i = j$), and $S^2(M)$ has the $\lambda_i \lambda_j$ as its eigenvalues. In general these are all also distinct, so the characteristic polynomial of $S^2(M)$ is minimal with this property and it's not possible to reduce the order further than this in general. -The same argument shows that for $k^{th}$ powers we can use the symmetric powers $S^k(M)$, and that for general products $a_n b_n$ of sequences satisfying linear recurrences we can use tensor / Kronecker products of the companion matrices of their characteristic polynomials. - -REPLY [4 votes]: T. Brown and P.J. Shiue's paper here might be of interest as a first reference. In the introduction they mention that if $a_n$ is a second-order sequence then the sequence -of squares $a_{n}^2$ is a third-order sequence. They go on to show necessary conditions for the squares sequence to be a second-order sequence when $a_n$ is a homogeneous sequence. -In the paper of Cooper and Kennedy here, (section 5) they give an order six linear recurrence relation for the square of a third order linear recurrence relation (as appears in your example): -$$x^2_n = (a^2 + b)x^2_{n−1} + (a^2b + b^2 + ac)x^2_{n−2} + (a^3c + 4abc − b^3 +2c^2)x^2_{n−3}+(−ab^2c + a^2c^2 − bc^2)x^2_{n−4} + (b^2c^2 − ac^3)x^2_{n−5} − c^4x^2_{n−6}$$ -where $x_n = ax_{n−1} + bx_{n−2} + cx_{n−3}$. - For similar questions where squares have been replaced with higher powers, the paper here by Stinchcombe might be interesting. -Edit: Qiaochu Yuan has provided the correct order for squares. Higher powers are addressed using the same argument in Theorem 3 of Stinchcombe's paper above. For convenience it is stated here: -Question: For what order does $y_{n} =x^l_{n}$ satisfy a linear recurrence relation, for $x_{n}$ a recurrence relation of order $k$? -A recurrence equation exists and the degree of the corresponding characteristic polynomial for the recurrence is counted by the -number of elements in $B_{l}$, where -$$B_l=\{(i_1,...,i_k)\ |\ \text{ each } i_j\in\mathbb{N}\ \text{and}\ i_1 + ... + i_k = l \}$$ -Given a value of $k$, define $S(k, l) =|B_{l}|$, then -Theorem 3: $S(k, l)$ obeys the relations: $S(k, l) = k$ for all $k$, $S(1, l) = l$ for all $l$, and $S(k,l) =S(k-1,l) + S(k, l-1)$ for every $k$ and $l$. Equivalently, $S(k, l)=\binom{k+l-1}{l}$.<|endoftext|> -TITLE: The developing map of conformally flat manifold -QUESTION [7 upvotes]: There is one sentence I don't understand in some paper. -"A simply connected and conformally flat three mainifold can be conformally immersed into $S^3$" by the means of a developing map. -Is any reference about this short argument? Maybe it is a direct consequence from definition. Could anyone explain a little bit to me? - -REPLY [10 votes]: Since $\mathbb{S}^3$ is conformally flat, we can think that any point of our manifold $M$ admits a neighborhood that is conformally equivalent to an open set in $\mathbb{S}^3$. -If two such neighbohoods overlap then the corresponding gluing map between corresponding open sets in $\mathbb{S}^3$ is a composition of inversions (Liouville's theorem). -So after applying a composition of inversions to one of the open sets you can assume that the gluing map is identity. -This way you can extend the parametrization of a neighborhood to an immersed neighborhood of any path. -The composition of inversions at a neighborhhod of the end point of path depends continuously on the path and therefore has to be the same for homotopic paths. -Since $M$ is simply connected it gives a well defined conformal immersion $M\hookrightarrow \mathbb{S}^3$.<|endoftext|> -TITLE: Extension of a von Neumann algebra by a von Neumann algebra -QUESTION [6 upvotes]: I asked this question at MSE now I repeat it at MO: -Let $A,B,C$ be $3$ unital $C^*$ algebras. Assume that we have the following short exact sequence of $C^*$-algebras: -$$0\to A\to C\to B\to 0$$ -Assume that $A,B$ are generated by their projections. Is $C$ necessarily generated by its projections, too? - -Assume that $A,B$ are von Neumann algebras, is $C$ necessarily a von Neumann algebra, too? - -Does the last question has an obvious answer when $A,B$ (hence $C$) are commutative algebras? - -REPLY [6 votes]: This is an extended comment on Nik Weaver's nice answer. (Unless I've made a mistake, this argument shows that $A$ being unital does all the work). -What exactly do we mean by -$$ 0 \rightarrow A \rightarrow C \rightarrow B \rightarrow 0 $$ -is an exact sequence of $C^*$-algebras? I think what is meant is that we have $*$-homomorphisms -$$ \phi : A\rightarrow C, \qquad \psi :C\rightarrow B, $$ -with $\phi$ injective, $\psi$ surjective, and $\ker\psi = \operatorname{im}\phi$. We do not assume that $\phi$ or $\psi$ is unital. -However, in the original question, we do assume that $A,B,C$ are unital. Let $p=\phi(1)\in C$ so that $p=p^2=p^*$ as $\phi$ is a $*$-homomorphism. So $p$ is a projection, so also $1-p$ is a projection. Also $p\phi(a) = \phi(a)p$ for all $a\in A$. -As $\phi(A) = \ker\psi$ is an ideal in $C$, if $c\in C$ with $pc=c$ (or $cp=c$) then $c\in \phi(A)$. So $\phi(A) = \{ c\in C: cp=c \}$. Further, for any $c\in C$, we see that $pc=c$ if and only if $c=cp$. So, for any $c\in C$, we have that $pc=pcp$ and $cp=pcp$. For $c\in C$ let $d=(1-p)c$ so $0 = pd = dp$ so $d(1-p)=d$ so $(1-p)c=(1-p)c(1-p)$. -Define -$$ B' = \{ c\in C : cp=pc=0 \} = \{c\in C: c(1-p)=(1-p)c=c\}. $$ -Then $B'$ is a $C^*$-subalgebra of $C$. If $c\in B'$ with $\psi(c)=0$ then $c=\phi(a)$ for some $a\in A$ so $cp=c=0$. So $\psi$ is injective on $B'$. For any $c\in C$, -$$ c = pc + (1-p)c = pcp + (1-p)c(1-p), $$ -from the discussion above. Then $pcp\in \phi(A)$ and $c'=(1-p)c(1-p)\in B'$ so $\psi(c) = \psi(c')$. So $\psi$ restricts to a $*$-isomorphism between $B'$ and $B$. -We have hence carefully shown that $C$ is isomorphic to $A\oplus B$. -It now immediately follows that if $A,B$ are generated by their projections, then so is $C$; if $A,B$ are von Neumann algebras, then so is $C$.<|endoftext|> -TITLE: Theorem 2.1.2.2 Higher Topos Theory -QUESTION [6 upvotes]: At the page 74 of HTT, there is the following theorem - -Let $S$ be a simplicial set, $\mathcal{C}$ a simplicial category, and $\phi: \mathfrak{C}[S] \rightarrow \mathcal{C}^{op}$ a simplicial functor. The straightening and unstraigntening functors determine a Quillen adjunction - $$ St_{\phi} : (Set_{\Delta})_{/S} \leftrightarrows Set_{\Delta}^{\mathcal{C}} :Un_{\phi}$$ - where $(Set_{\Delta})_{/S}$ is endowed with the contravariant model structure and $Set_{\Delta}^{\mathcal{C}}$ with the projective model structure. [...] - -In then says that the proof is easy, but I can't manage to show that $St_{\phi}$ sends cofibrations to projective cofibrations. I thought that since the the class of morphisms which are sent to projective cofibrations is weakly saturated it is enough to show the result for all inclusions $\partial \Delta^n \subseteq \Delta^n$. -I did not have much success for the simplicial category $\mathcal{C}$ and the map $\phi$ could be anything and I have a hard time dealing with it. -Furthermore there is something else which troubles me: the model structure on the $Set^{\mathcal{C}}_{\Delta}$ makes no use of the simplicial enrichement on both $\mathcal{C}$ and $sSet$ so I was wondering if I was not missing something by believing that the that model structure on $Set^{\mathcal{C}}_{\Delta}$ is really the projective model structure coming from the Kan model structure on $sSet$ and not the one coming somehow from an other model stucture using the simplicial enrichement. - -REPLY [5 votes]: I think what Lurie might have meant when he wrote "It is easy to see that $St_{\phi}$ preserves cofibrations" in the proof of Theorem 2.2.1.2, is that it is easy to see it if you take into account the compatibility of straightening with left Kan extensions as described just above in Proposition 2.2.1.1. Indeed, combining (1) and (2) of the latter proposition we see that given any generating cofibration $\partial \Delta^n \to \Delta^n \to S$ in $({\rm Set}_{\Delta})_{/S}$, in order to show that $St_{\phi}(\partial \Delta^n \to \Delta^n)$ is a projective cofibration in ${\rm Set}_{\Delta}^{{\cal C}}$ it is enough to show that $St_{Id}(\partial \Delta^n \to \Delta^n)$ is a projective cofibration in ${\rm Set}_{\Delta}^{\mathfrak{C}[\Delta^n]^{op}}$. The map $St_{Id}(\partial \Delta^n \to \Delta^n)$ can then be described very explicitly, and one can check that it is a pushout of a map of the form $(i_0)_!\partial ((\Delta^1)^n) \to (i_0)_!(\Delta^1)^n$, where $(i_0)_!$ denotes the left Kan extension functor along the terminal object inclusion $\{0\} \subseteq \mathfrak{C}[\Delta^n]^{op}$ and $\partial ((\Delta^1)^n) \to (\Delta^1)^n$ is the inclusion of the boundary of the $n$-cube inside the full $n$-cube. -Edit: -In response to the comment, here are some more details on the computation. First note that by definition the map $St_{Id}(\partial \Delta^n)(0) \to St_{Id}(\Delta^n)(0)$ can be identified with the inclusion -$$ (*)\quad {\rm Map}_{\mathfrak{C}[\partial \Delta^{n+1}]}(0,n+1) \subseteq {\rm Map}_{\mathfrak{C}[\Delta^{n+1}]}(0,n+1) = {\rm N}({\rm P}(\{1,...,n\})) \cong (\Delta^1)^n ,$$ -where ${\rm P}(\{1,...,n\})$ is the poset of subsets of $\{1,...,n\}$ and ${\rm N}(-)$ is the nerve. Now for $i \in \{1,...,n\}$, image of $\mathfrak{C}[\Delta^{\{0,...,\hat{i},...,n+1\}}] \to \mathfrak{C}[\Delta^{n+1}]$ on the mapping space from $0$ to $n+1$ is exactly the face of ${\rm N}({\rm P}(\{1,...,n\}))$ corresponding to the subposet spanned by those subsets which do not contain $i$. On the other hand, the face of ${\rm N}({\rm P}(\{1,...,n\}))$ corresponding to the subposet spanned by those subsets which do contain $i$ is exactly the image of -$${\rm Map}_{\mathfrak{C}[\Delta^{\{0,...,i\}}]}(0,i) \times {\rm Map}_{\mathfrak{C}[\Delta^{\{i,...,n+1\}}]}(i,n+1) \subseteq {\rm Map}_{\mathfrak{C}[\partial \Delta^{n+1}]}(0,n+1)$$ -in ${\rm Map}_{\mathfrak{C}[\Delta^{n+1}]}(0,n+1)$. This shows that the image of (*) contains all the boundary of the cube. It is also not difficult to check that this image is contained in the boundary of the cube, and hence coincides with it.<|endoftext|> -TITLE: Formality over $\mathbb{R}$ vs formality over $\mathbb{Q}$ -QUESTION [9 upvotes]: On ncatlab page on formality, it is stated that Deligne--Griffiths--Morgan--Sullivan proved that the real homotopy type of a closed Kaehler manifold is formal. Later, Sullivan "improved" this to $\mathbb{Q}$-formality. -My question is: are there some easy examples of closed topological manifolds whose $\mathbb{R}$-homotopy type is formal, but $\mathbb{Q}$-homotopy type isn't? - -REPLY [23 votes]: What Sullivan proved is not just that the $\mathbb R$-formality from Deligne-Griffiths-Morgan-Sullivan can be improved to $\mathbb Q$-formality, but rather that formality over any field of characteristic zero for any space always implies formality over $\mathbb Q$. See Sullivan's paper.<|endoftext|> -TITLE: How is topological André-Quillen homology (TAQ) a "stabilization", exactly? -QUESTION [8 upvotes]: Let $S\to A\to B$ be cofibrations of commutative $S$-algebras. Then the topological André-Quillen $B$-module $TAQ(B|A)$ can be computed as a stabilization. Precisely, I think it means the following: let $I$ be the augmentation ideal functor from augmented commutative $B$-algebras to $B$-modules; it is right Quillen. These categories are enriched and tensored over based spaces: let $\wedge$ denote the tensor of $B$-modules and $\odot$ denote the tensor of augmented commutative $B$-algebras. -There are canonical maps $$X\wedge IC\to I(X\odot C)$$ for any augmented commutative $B$-algebra $C$ and based space $X$. In particular, there are maps $S^1\wedge I(S^n\odot C) \to I(S^{n+1}\odot C)$. Taking $n$-th loops (in $B$-modules) of the adjoint gives maps -$$\Omega^nI(S^n\odot C)\to \Omega^{n+1} I(S^{n+1} \odot C).$$ -The claim is that if we take $C=B\wedge_A B$ (a $B$-algebra on the first variable and the augmentation is the multiplication map), then the homotopy colimit in $B$-modules of this tower is weakly equivalent to $TAQ(B|A)$. Am I saying it right? In symbols, - -Is $TAQ(B|A)\simeq \operatorname{hocolim} (\Omega^nI(S^n \odot (B\wedge_AB)))$? - -It seems the answer can be deduced from the results in Basterra-Mandell, Homology and cohomology of $E_\infty$-ring spectra, but it's not obvious to me. If it's the case, then how? -Another presentation of the "$TAQ$ as stabilization" statement I've seen (up to my own misinterpretations) is the following: the category of commutative $A$-algebras (no augmentation) is tensored over unbased spaces: call this tensor $\otimes_A$. One can take the tensors $S^n\otimes_A B$, and consider the inclusion of the point $*\to S^n$ which gives maps $B\to S^n\otimes_A B$ of which we can take the cofiber in $B$-modules. Denote this cofiber by $S^n\overline{\otimes}_AB$. These also form a direct system by passing to cofibers some maps gotten similarly as above. - -Is $TAQ(B|A)\simeq \operatorname{hocolim}(\Omega^n(S^n\overline\otimes_AB))$? - -Note: There is yet another formulation which might be useful and which is equivalent to the first one. The functor $I$ factors via the category of non-unital commutative $B$-algebras (nucas), as a functor $I_0$ followed by the forgetful functor $U$ (both right Quillen). The category of nucas is also tensored over pointed spaces; call the tensor $\tilde\otimes$. Instead of considering $I(S^n\odot C)$ in the directed system above, we could consider $U(S^n\tilde\otimes I_0C)$. We get a directed system similarly as above, and - -$\operatorname{hocolim} (\Omega^nI(S^n \odot C )) \simeq \operatorname{hocolim}(\Omega^nU(S^n\tilde\otimes I_0 C))$. - -This follows from the fact that $I_0$ is a right Quillen equivalence, so it preserves tensors (properly interpreted; see Does the right adjoint of a Quillen equivalence preserve homotopy colimits?). - -REPLY [6 votes]: These stabilization formulas do indeed follow from the paper of Basterra-Mandell. Fix a commutative $S$-algebra $A$. Then Basterra and Mandell prove the following: -1) [Theorem 3] Given a commutative $A$-algebra $B$, the $(\infty-)$category of $\Omega$-spectrum objects in augmented $B$-algebras is equivalent to the $(\infty-)$category of $B$-modules via the functor that takes an $\Omega$-spectrum object $\{X_n\}$ in augmented $B$-algebras to the augmentation ideal of $X_0$. -2) [Theorem 4] Under the identification of (1), the $B$-module $TAQ(B|A)$ corresponds to (the $\Omega$-spectrum replacement of) the suspension spectrum of the augmented $B$-algebra $B \wedge_A B$. -Combining (1) and (2) we get your first stabilization formula. Indeed, the suspension spectrum of $B \wedge_A B$ is given in degree $n$ by what you denote by $S^n \odot (B \wedge_A B)$. The $\Omega$-spectrum replacement of this suspension spectrum then has in degree $0$ the homotopy colimit ${\rm hocolim}_n \Omega^nS^n \odot (B \wedge_A B)$, where the loop operation $\Omega$ is performed in augmented $B$-algebras. Applying the equivalence of (1) and using the fact that the augmentation ideal functor $I$ commutes with loops we get that the $B$-module corresponding to the suspension spectrum of $B \wedge_A B$ is given by the formula ${\rm hocolim}_n \Omega^n I(S^n \odot (B \wedge_A B))$. By (2) this is also $TAQ(B|A)$. -To get the second stabilization formula from the first note that $B \wedge_A (-)$ is left Quillen from $A$-algebras to $B$-algberas and that homotopy colimits of augmented $B$-algebras are computed in $B$-algebras. This means that $S^n \otimes_A B \simeq S^n \odot (B \wedge_A B)$. Then use the fact that for an augmented $B$-algebra $C$ there is a natural equivalence between $I(C)$ and the cofiber of $B \to C$. The third stabilization formula is also equivalent to the first, as you mention.<|endoftext|> -TITLE: "Overdetermined" Poisson equation -QUESTION [7 upvotes]: Consider the PDE $-\Delta u = f$ on a bounded domain $\Omega \subset \mathbb{R}^n$, where $f \in C^\infty(\bar{\Omega})$. I wish to consider both the boundary conditions $u = 0$ and $\frac{\partial u}{\partial n} = 0$. My question is, are there reasonably well-known "compatibility" conditions under which such equations admit a solution? - -REPLY [2 votes]: We claim the following: a solution $u$ exists if and only if $f$ is orthogonal to the Poisson kernel with pole at every $x \in \partial \Omega$. - -Suppose that $u$ is a solution. Since $u = 0$ on the boundary, we have $$u(x) = \int_\Omega G_\Omega(x, y) f(y) dy,$$ where $G_\Omega(x, y)$ is the Green function. Assuming $\Omega$ is sufficiently regular, one can differentiate under the integral sign and write $$0 = \partial_n u(x) = \int_\Omega \partial_n G_\Omega(x, y) f(y) dy,$$ where $\partial_n$ denotes the derivative in $x$ in the direction normal to the boundary; here of course $x \in \partial \Omega$. The normal derivative of the Green function defines a Poisson kernel. Thus, $f$ is orthogonal to the Poisson kernel with pole at every $x \in \partial \Omega$. -Conversely, if $f$ is orthogonal to the Poisson kernel with pole at every $x \in \partial \Omega$, then $u(x) = \int_\Omega G_\Omega(x, y) f(y) dy$ has all desired properties. Therefore, these two conditions are equivalent, as desired. - -Alternatlively, one could argue as follows: if $u$ is a solution and $h$ is a harmonic function in $\Omega$ which is $C^1$ in $\overline{\Omega}$, then $$\int_\Omega h(x) f(x) dx = \int_\Omega h(x) \Delta u(x) = \int_\Omega \Delta h(x) u(x) dx = 0$$ by Green's first identity and the boundary conditions imposed on $u$. Therefore, $f$ is necessarily orthogonal to all harmonic functions.<|endoftext|> -TITLE: Are the ideles literally a Picard group? -QUESTION [10 upvotes]: I understand that in the number field / function field analogy, the ideles $\mathbb I_K$ of a number field $K$ are supposed to be analogous to the Picard group of a function field. -Question: Is this more than an analogy? Is there an actual geometric setting in which the ideles parameterize "line bundles" over a geometric object attached to $K$? -My first inclination was to see if this was the case in Berkovich geometry, in the case $K = \mathbb Q$. It's encouraging to note that ideles correspond to Cech 1-cocycles for reasonable coverings of the Berkovich space $\mathcal M(\mathbb Z)$ with values in $\mathcal O_\mathbb Z^\times$. But nevertheless, every such cocycle is a coboundary, so it seems to not literally be the case that the ideles parameterize line bundles on $\mathcal M(\mathbb Z)$. - -REPLY [3 votes]: It seems it would be helpful to discuss idealic/adelic view on line/vector bundles, which is quite simple and intuitive. At the end we might arrive at a kind of asnwer to the question. -Step 1: - Bundles are described by transition functions from one chart of the covering to another - in case of line bundles we have invertible elements living on intersection , in vector bundles case - $GL(O_{intersection})$. (Looking forward idels are exactly these transition functions for specific covering). -Do not forget about factorization $GL(O_{U1})$ \ $GL(O_{intersection})$ / $GL(O_{U2})$. -Step 2: Consider 1d-scheme - i.e. curve (over ANY(!) field ) or Spec(Z). -Consider quite specific covering - we take all geometric points $x$ of the curve and "general point" (i.e. Spec(FractionField)). -And consider "infinitesemal neighbourhood" of every point $U_x$ - it is $Spec( K_x [[t]] )$ . Fact 1. Infinitesemal neighbourhoods of different points do not intersect. Fact 2. Intersection of $U_x$ and "generic point" is -$Spec( K_x ((t)) )$. -Comibining step 1 and 2 we arrive to idelic/adelic picture of bundles: -Line bundles = K(x) \ $ \prod_x K_x ((t))^*$ / $\prod_x K_x [[t]]^*$ -You can recognize idels at the middle: $ \prod_x K_x ((t))^*$ -Vector bundles = GL( FractionField) \ $ \prod_x GL( K_x ((t)) ) $ / $\prod_x GL( K_x [[t]] )$ -======================== -I am not sure other steps are necessary, but let me discuss. -If I undestand you question correctly you want to exclude factorization over -/ $\prod_x K_x [[t]]^*$ from the factor above - just to have idels. -That can be done. -Claim: K(x) \ $ \prod_x K_x ((t))^*$ describes line bundles over geometiric object - each point $x$ is glued to all points in its infinitesemal neighbourhood. -Step 3: Gluing points A,B means we consider functions: f(A)=f(B). -Now take $A = B+ \epsilon$ for $\epsilon^n=0$. If we consider f(A)=f(B) we glue A to its n-neighbourhood. The function field will be $f^{(k)}(A)=0$ for all k <= n, -that means series of the form $C+ c_{n+1}t^{n+1}+c_{n+2}t^{n+2}$. -Taking limit n to infinity we kill all the functions. -However the fraction field does not change - i.e. intersection with generic point is still the same. -In that way idels are preserved but the factor is $\prod_x K_x [[t]]^*$ is killed. -PS -Curves with $f^{(k)}(A)=0$ considered in Serre book "Algebraic Groups and Class Fields", despite being singular, one can consider their Jacobians -and it is related to ramified covering in the same way as usual Jacobion related to unramified coverings.<|endoftext|> -TITLE: Covering the finite plane with lines -QUESTION [6 upvotes]: This is, essentially, a geometrically rendered version of the question I asked a week ago, with the emphases slightly shifted; it seems more natural and appealing (to me, at least) in this form. -Let $p\ge 3$ be a prime number. Suppose we are given $N$ lines $l_1,\dotsc,l_N\subset\mathbb F_p^2$, and we want to translate them to get new lines $l_1',\dotsc,l_N'$ (which are either parallel, or identical to the original lines) so as to have the whole vector space $\mathbb F_p^2$ covered by these new lines: $l_1'\cup\dotsb\cup l_N'=\mathbb F_p^2$. This can be impossible if $N\le2(p-1)$, as it follows by considering the system of $p-1$ "vertical" and $p-1$ "horizontal" lines. Is this always possible if $N\ge 2p-1$? -More generally, given $(p-1)n+1$ affine hyperplanes in $\mathbb F_p^n$, can one always translate them so that the resulting translates cover the whole space $\mathbb F_p^n$? - -REPLY [4 votes]: Let $C>0$ be any fixed number. Take $p-3$ horizontal lines and $p-b$ vertical lines where $p\gg b\gg C$. If we want to stay within $2p+C-3$ lines, we should be able to cover some $3\times b$ rectangular configuration by at most $b+C$ lines of any prescribed slopes $a_1,\dots, a_{b+C}\ne 0$. Notice that we have $3b$ points to cover and each line can cover at most $3$ points, so we can afford only $3C$ lines that cover $2$ points or fewer. Thus some $b-2C$ lines should pass through $3$ points. -Let $x_1,x_2,\dots,x_b$ be the base of our $3\times b$ configuration and $u,v,w$ be its "vertical side". Then for each slanted line coming through $3$ points, we have some triple $i,j,k$ such that -$$ -(w-u)(x_j-x_i)=(v-u)(x_k-x_i) \tag {$*$} -$$ -and the slope of the corresponding line is determined by that triple. Notice also that those triples cover at least $b-6C$ indices $1,\dots,b$ (the indices not taken by the exceptional $3C$ lines). Thus we can choose $\frac b3-2C$ linearly independent equations of the type ($*$) (just take an equation including some index not used yet every time until you run out of them). There are some $K(b)$ possible arrangements of those equations and $p^3$ choices of $u,v,w$, so we see that we can have at most $K(b)p^{3+\frac 23b+2C}$ arrangements that are coverable by $b+C$ slanted lines in principle and all slopes (and even lines) except $3C$ are determined by $x_j$ and the ($*$)-equations up to the order. That results in the bound $K(b)(b+C)!p^{3+\frac 23b+5C}$ for the number of choices of $b+C$ slopes that can be used to cover any $3\times b$ configuration, which falls short of $(p-1)^{b+C}$ available choices if $b\gg C$ and $p>p(b)$. -If you do it carefully, you get the lower bound of the type $N(p)\ge 2p+p^\alpha$ for large $p$ with some $\alpha\in(0,1)$, but it is still only a rather pitiful improvement of the trivial lower bound $2p-1$, so I didn't try to be very precise in the estimates.<|endoftext|> -TITLE: Lifting of families of curves to characteristic 0 -QUESTION [9 upvotes]: Let $k$ be a finite field, $X_0$ be a smooth affine variety over $k$ and $C\rightarrow X$ a smooth projective family of curves of genus $\geq 2$. -By a result of Elkik we can always lift $X_0$ to a smooth affine scheme over $W(k)$. -It seems that, up to replace $X_0$ with an open subset, one can always lift $C_0\rightarrow X_0$ to a smooth projective family of curves $C\rightarrow X$. -Does anyone have a reference for this fact? - -REPLY [4 votes]: I am just posting my comment as an answer. This type of lifting is "classical". In the article linked above with Chenyang Xu, we needed bounds on the lift, to deduce height bounds for some rational points. So we worked out in detail the classical result. -The result that I called "Chow's Lemma for Stacks" is not actually Chow's Lemma for Stacks (sorry). Rather, it is Théorème 6.3, p. 49 of the following. -MR1771927 -Laumon, Gérard; Moret-Bailly, Laurent -Champs algébriques. -Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge, Volume 39. -Springer-Verlag, Berlin, 2000. xii+208 pp. -ISBN: 3-540-65761-4 -For each field $k$ of finite characteristic $p$, denote by $W(k)$ a complete, local, Noetherian DVR with uniformizing element $p$ such that $W(k)/pW(k)$ equals $k$. Let $\mathcal{M}$ be a smooth stack over $\text{Spec}\ W(k)$. Let $X_0$ be a connected, smooth, quasi-projective $k$-scheme. By hypothesis, there exists a locally closed immersion of $k$-schemes, $$i:X_0 \to \mathbb{P}^n_k.$$ Let $\zeta:X_0\to \mathcal{M}$ be a $1$-morphism over $\text{Spec}\ W(k)$. By Théorème 6.3, up to replacing $X_0$ by a dense open subscheme, there exists a factorization of $\zeta$, $$X_0\xrightarrow{z} M \xrightarrow{\phi} \mathcal{M},$$ where $M$ is a smooth, quasi-projective $W(k)$-scheme, where $\phi$ is a smooth $1$-morphism over $W(k)$, and where $z$ is a $W(k)$-morphism. -So now we are reduced to lifting over $W(k)$ the locally closed subscheme $X_0$ in the closed fiber of the smooth, quasi-projective $W(k)$-scheme $$\mathbb{P}^n_{W(k)}\times_{\text{Spec}\ W(k)} M.$$ This is "classical". Since $X_0$ is smooth, the locally closed immersion is a regular embedding. Thus, up to shrinking $X_0$ further, this closed subscheme is an open subscheme of a complete intersection of degree $d$ ample hypersurfaces in $\mathbb{P}^n_{W(k)}\times_{\text{Spec}\ W(k)} M$ for all $d\geq d_0$. (You can spend a lot of time trying to optimize degrees in this argument, but the OP did not ask for an "optimal lifting", just some lifting.) The defining equations of those hypersurfaces have coefficients that are elements of $k$. By lifting those coefficients to $W(k)$, we can lift the hypersurfaces over $W(k)$. This lifts the complete intersection over $W(k)$. -As the commenters note, this clashes with intuition that lifts should preserve intersection numbers and other notions that definitely are preserved for lifts of projective schemes. However, this is a well-known "pathology" of lifts of affine schemes. We can find projective models of these lifts, but then the closed fiber will be reducible. Those properties suggested by our intuition typically do occur for the closed fiber of a projective model. However, the property may hold for some irreducible component of the closed fiber different from the component containing $X_0$.<|endoftext|> -TITLE: Is there a minimal extension of $L$ that is not a forcing extension? -QUESTION [9 upvotes]: It's well known that Sacks forcing constructs a real of minimal constructability degree, i.e. a real $x$ such that for any $y\in L(x) \setminus L$, $L(y) = L(x)$. It's also well known that certain objects, such as $0^\sharp$, can never be created by a forcing extension. -Given these two facts there is a natural question: - -Can there exist a real $x\notin L$ such that any inner model $W$ with $L(x) \supseteq W \supseteq L$ either $W=L$ or $W= L(x)$ but such that $L(x)$ is not a (set) forcing extension of $L$? - -Obviously we could also ask the question regarding larger sets than reals or class forcing extensions. - -REPLY [10 votes]: Yes, this is possible and follows from Sy Friedman's paper Minimal coding (you may better look at Fine structure and class forcing). -It follows from the results of the above - paper that there is an $L$-definable class forcing for producing a real $R$ which is minimal over -$L$ but is not set-generic over $L$.<|endoftext|> -TITLE: van der Waerden's theorem in Reverse Mathematics -QUESTION [9 upvotes]: What is known about weak systems of axiomata that allow one to prove van der Waerden's theorem? -van der Waerden's theorem can be used to show that there are infinitely many primes (see below). Is this proof ever not pointless — i.e. is there an axiom system where such a technique would shorten a proof or proof schema? -If there were finitely many primes, then we could associate to each natural number $n$ the tuple of $v_p(n) {\rm mod} 2$ for each prime p, where $v_p(n)$ -is the largest $i$ such that $p^i$ is a factor of $n$. -By van der Waerden's theorem, there arbitrarily long arithmetic progression on which the sequence is constant. -If $a, a+d, a+2d, a+3d$ is an arithmetic progression of length $4$, and $R = \prod_p {p_i}^{e_i}$ where $e_i$ is $v_p(a) {\rm mod} 2$, then we can divide by $R$ to get a sequence of squares. But there is no four term arithmetical progression of perfect squares. QEA -If we also colour each natural number by the list of primes dividing it, we could get a contradiction more easily by looking at an arithmetic progression of length larger than the square of the largest prime; if the progression is $a, a+d, a+2d, \dots$ then: -If for some prime $p$ $v_p(a) < v_p(d)$ then $v_p(a+pd) = v_p(a + d) + 1$ QEA -If for some prime $p$ $v_p(a) = v_p(d)$ then $v(a+kd) = v_p(a) + 1$ if $k$ is chosen so that $k \leq p^2$ and $A + kD \equiv p \pmod {p^2}$, where $a = p^i A$, $d = p^j D$ with $A$ and $D$ prime to $p$. QEA -But if $v_p(a) < v_p(d)$ for every prime $p$ then $v_p(a) = v_p(a+d)$ for all $p$ QEA. -To see that there is no arithmetic progression of four squares: -suppose without loss of generality that all four squares are odd and pairwise relatively prime, and call the elements: $x - 6n,$ $x - 2n,$ $ x + 2n,$ $ x + 6n$. Then: $y^2 = (x^2 - 4n^2)(x^2 - 36n^2) = (x^2 - 20n^2)^2 - 256 n^4$ -Then we have a Pythagorean triple : $(16n^2, y, x^2 - 20n^2)$. So there exist $u$, $v$ such that $2u$ and $v$ are relatively prime and $4uv = 16n^2$, $4u^2 + v^2 = x^2 - 20n^2$. So $u$ is even, and we can write $u = 4A^2$, $v = 4D^2$ with $D$ odd. -So $4A^2 + D^2$ and $16 A^2 + D^2$ are squares. -So we can write $2A = 2UV$, $U^2 - V^2 = \pm D$ and $4U'V' = 4A$ $4U'^2 V'^2 = \pm D$. -This implies $UV = U'V' = A$, so $\pm D$ is a difference of two squares in two different ways, so we can write -$\pm D = 4a^2b^2 - c^2d^2 = 16a^2 c^2 -b^2 d^2$ with $2a,b,c,d$ pairwise relatively prime. -enabling us to create an arithmetic progression of relatively prime squares with smaller difference $4ad$. - -REPLY [15 votes]: There is a powerful combinatorial theorem, known as the Hales–Jewett theorem, which readily implies van der Waerden's theorem. On the other hand, the paper below by Matet exhibits primitive recursive upper bounds for the Hales–Jewett function $HJ(m,n)$, and therefore for the van der Waerden function $W(m,n)$. This was originally shown in a paper of Shelah: Primitive recursive bounds for van der Waerden numbers, J. Amer. Math. Soc. 1 no. 3 (1988), 683–697, doi:10.1090/S0894-0347-1988-0929498-X. -The last paragraph of Matet's paper makes it clear that the Hales–Jewett theorem (in the form $\forall m \forall n\, HJ(m,n) \text{ exists}$), and (therefore also van der Waerden's theorem in the form $\forall m \forall n\, W(m,n) \text{ exists}$) is provable in the fragment of Peano arithmetic known as superexponential function arithmetic (often denoted $\textrm{I}\Delta_0 + \text{Supexp}$). -For $HJ(m,n)$ to exist for concrete $m$ and $n$, one only needs the weaker system known as exponential (or elementary) function arithmetic (often denoted $\textrm{I}\Delta_0 + \text{Exp}$). - -Pierre Matet, Shelah's proof of the Hales–Jewett theorem revisited, European J. Combin. 28 (2007), no. 6, 1742–1745, doi:10.1016/j.ejc.2006.06.021.<|endoftext|> -TITLE: The de Rham complex of the octonionic projective spaces -QUESTION [25 upvotes]: The complex projective space $\mathbb{CP}^n$ is a complex manifold, and hence its de Rham complex carries a representation of the complex numbers in the form of its complex structure. The quaternionic projective space $\mathbb{HP}^n$ is a quaternionic Kähler manifold, and so its de Rham complex carries a local representation of the quaternions. Continuing the analogy, do the octonionic line $\mathbb{OP}^1$ and the octonionic projective plane $\mathbb{OP}^2$ carry some representation of the octonions, or at least have some extra structure reflecting their octonionic construction? - -REPLY [4 votes]: A description of the octonionic projective plane in terms of the octonionic algebra is described by John Baez in his notes on Octonionic projective geometry: -The Jordan algebra of 3×3 Hermitian octonionic matrices, with multiplication rule $x\circ y=(xy+yx)/2$, generates the octonionic projective plane $\mathbb{OP}^2$ if we restrict the matrices $p$ to unit trace and idempotent ($p^2=p$). Lines through the origin containing $(x,y)$ with $x\neq 0$ equal $\{(\alpha(y^{-1}x),\alpha):\alpha\in\mathbb{O}\}$. -The differential forms of the de Rham complex are constructed in terms of "octonionic Pauli matrices" by Piccinni in On the cohomology of some exceptional symmetric spaces (section 4).<|endoftext|> -TITLE: The Lefschetz operator -QUESTION [15 upvotes]: Let $\omega=\sum_{i=1}^n dx_i\wedge dy_i\in\bigwedge^2(\mathbb{R}^{2n})^*$ be a standard symplectic form. The following result is due to Lefschetz: - -For $k\leq n$, the Lefschetz operator -$L^{n-k}:\bigwedge^k(\mathbb{R}^{2n})^*\to - \bigwedge^{2n-k}(\mathbb{R}^{2n})^*$ defined by $$ - L^{n-k}\alpha=\alpha\wedge\underbrace{\omega\wedge\ldots\wedge\omega}_{n-k}=\alpha\wedge\omega^{n-k} -$$ is an isomorphism. - -This follows from Proposition 1.2.30 [1]. The proof is slick and it uses representation theory. - -Question: What are other standard references for this result? I want to quote it properly and I would like to see a reference to a -proof by brute force. A proof by brute force is not difficult but ugly -and perhaps in some reference I can find an elegant presentation of -such a proof. - -This result seems so fundamental that it should be available in many textbooks, but I am not aware of any except [1]. -[1] D. Huybrechts, Complex geometry. An introduction.. Universitext. Springer-Verlag, Berlin, 2005. - -(MathsciNet review). - -REPLY [16 votes]: I find the proof mentioned by Robert Bryant so beautiful that I cannot resist and I have to include it here. The following proof is from [1] Proposition 1.1a. - -Definition. Given $w\in\mathbb{R}^n$ and $1\leq k\leq n$, we define the interior product $$ - \iota_w:\bigwedge\nolimits^k(\mathbb{R}^n)^*\to - \bigwedge\nolimits^{k-1}(\mathbb{R}^n)^* \quad \text{by} \quad - \iota_w(\xi)(v_1,\ldots,v_{k-1})=\xi(w,v_1,\ldots,v_{k-1}). $$ - -For the proof of the next lemma see Proposition 2.12 in [2]. - -Lemma. If $\xi\in \bigwedge\nolimits^k(\mathbb{R}^n)^*$, $\eta\in \bigwedge\nolimits^\ell(\mathbb{R}^n)^*$ and $v\in\mathbb{R}^n$, then - $$ - \iota_v(\xi\wedge\eta)=\iota_v(\xi)\wedge\eta+(-1)^k\xi\wedge\iota_v(\eta). $$ - -This lemma and a simple induction imply that -$$ -\iota_v(\omega^k)=k\omega^{k-1}\wedge\iota_v(\omega), -\quad -\text{for $v\in\mathbb{R}^{2n}$.}\ \ \ \ \ \ \ (*) -$$ -Now we can prove the main result: - -Theorem. For $1\leq k\leq n$, the Lefschetz operator $$ L^k:\bigwedge\nolimits^{n-k}(\mathbb{R}^{2n})^*\to - \bigwedge\nolimits^{n+k}(\mathbb{R}^{2n})^*, \quad L^k(\xi)=\xi\wedge\omega^k - $$ is an isomorphism. - -Proof. -Since the spaces $\bigwedge\nolimits^{n-k}(\mathbb{R}^{2n})^*$ and $\bigwedge\nolimits^{n+k}(\mathbb{R}^{2n})^*$ have equal dimensions, it suffices to show that $L^k$ is injective. We will use a backwards induction starting with $k=n$. If $k=n$, then $\omega^n=n!dVol$ and it is obvious that -$$ -L^n:\bigwedge\nolimits^0(\mathbb{R}^{2n})^*=\mathbb{R}\to \bigwedge\nolimits^{2n}(\mathbb{R}^{2n})^* -$$ -is an isomorphism and hence injective. -Suppose that the result is true for $k$. We need to prove it is true for $k-1$. Thus -we need to prove that if -$L^{k-1}(\xi)=\xi\wedge\omega^{k-1}=0$ for some -$\xi\in\bigwedge\nolimits^{n-(k-1)}(\mathbb{R}^{2n})^*$, then $\xi=0$. -Clearly, $\xi\wedge\omega^k=0$. If $v\in\mathbb{R}^{2n}$, then the Lemma and (*) yield -$$ -0=\iota_v(\xi\wedge\omega^k)= -\iota_v(\xi)\wedge\omega^k\pm \xi\wedge\iota_v(\omega^k)= -\iota_v(\xi)\wedge\omega^k\pm k\underbrace{\xi\wedge\omega^{k-1}}_{0}\wedge\iota_v(\omega) -$$ -so $L^k(\iota_v(\xi) )=\iota_v(\xi)\wedge\omega^k=0$ and hence $\iota_v(\xi)=0$ by the injectivity of $L^k$ (induction hypothesis). -Since $\iota_v(\xi)=0$ for all $v\in\mathbb{R}^{2n}$, it follows that $\xi=0$. The proof is complete. -$\Box$ -[1] R. Bryant, P. Griffiths, D. Grossman, Exterior differential systems and Euler-Lagrange partial differential equations. Chicago Lectures in Mathematics. University of Chicago Press, Chicago, IL, 2003. -[2] F. W. Warner, Foundations of differentiable manifolds and Lie groups. Corrected reprint of the 1971 edition. Graduate Texts in Mathematics, 94. Springer-Verlag, New York-Berlin, 1983.<|endoftext|> -TITLE: A question about HTT Lemma 5.5.2.1 -QUESTION [5 upvotes]: I have a question about the statement of Lemma 5.5.2.1 in Lurie's `Higher Topos Theory'. -``Let $S$ be a small simplicial set, let $f: S\rightarrow \mathcal{S}$ be an object of $\mathcal{P}(S^{op})$, and let $F: \mathcal{P}(S^{op})\rightarrow \widehat{\mathcal{S}}$ be the functor corepresented by $f$. Then the composition -$$S\overset{j}{\rightarrow}\mathcal{P}(S^{op})\overset{F}{\rightarrow}\widehat{\mathcal{S}}$$ -is equivalent to $f$. -" -The Yoneda embedding $j$ should be from $S$ to $\mathcal{P}(S)$, but the last line doesn't seem to be just a typo. Do I miss something? Thanks! - -REPLY [3 votes]: The middle term should be $\mathcal{P}(S^{op})^{op}$. The Yoneda embedding gives a functor $S^{op} \to \mathcal{P}(S^{op})$. The functor corepresented by $f \in \mathcal{P}(S^{op})$ is given by $Hom_{\mathcal{P}(S^{op})}(-,f)$, which is a functor $\mathcal{P}(S^{op})^{op} \to \widehat{\mathcal{S}}$. -As a reality check, the analogous theorem in ordinary category is that the composite -$$ -x \mapsto Hom_S(x,-) \mapsto Nat(Hom_S(x,-),f) -$$ -is naturally isomorphic via the Yoneda lemma to the functor $x \mapsto f(x)$.<|endoftext|> -TITLE: What are the definable sets in Skolem arithmetic? -QUESTION [5 upvotes]: Definable subsets of $\mathbb N$ in the language of Presburger arithmetic are exactly the eventually periodic sets and quantifier free part corresponds to Integer Programming with linear inequalities and variations lead to mixed integer linear programming, convex integer programming with convex constraints. What about - -definable subsets of $\mathbb N$ in the language of Skolem arithmetic and -would it be sensible to seek programming constructs that with 'decidable portions of Skolem' leads to (if I am not wrong then atomic formulae here might be of form $a\prod_{i=1}^nx_i^{b_i}\leq b$ or $a\prod_{i=1}^nx_i^{b_i}=b$)? - -My background is not logic and not sure if I make sense however if there is reasonable way to salvage the post it will be nice. I am trying to see if fixed dimension linear integer programming that runs in polynomial time has an analogy in Skolem arithmetic where variable addition is disallowed? - -REPLY [5 votes]: $\def\mr{\mathrm}$As it happens, quantifier elimination for Skolem arithmetic came up recently in my research. The concise description is that every formula $\phi(x_1,\dots,x_k)$ is in $(\mathbb N^{>0},{\cdot})$ equivalent to a Boolean combination of formulas expressing -$$\tag1\bigl|\{p\in\mathbb P:\psi(v_p(x_1),\dots,v_p(x_k))\}\bigr|\ge n,$$ -where $\psi(y_1,\dots,y_k)$ is a formula of Presburger arithmetic, and $n\in\mathbb N$. -In the special case of formulas in one variable with parameters that you are interested in, this boils down to the following: definable subsets are Boolean combinations of sets defined by - -$v_q(x)=n$, -$v_q(x)\equiv a\pmod m$, -$\bigl|\{p\in\mathbb P:v_p(x)=n\}\bigr|\ge b$, -$\bigl|\{p\in\mathbb P:v_p(x)\ge n,v_p(x)\equiv a\pmod m\}\bigr|\ge b$, - -for some $q\in\mathbb P$, $n,b\in\omega$, $0\le a0},{\cdot})$ are equivalent to Boolean combinations of (1) follows from the results of Mostowski [1]. I will sketch how to prove the other direction, that all sets of the form (1) are first-order definable. -Using $\cdot$, we can define the divisibility, coprimeness, and primality predicates as -$$\begin{align*} -x\mid y&\iff\exists z\,(y=x\cdot z),\\ -x\perp y&\iff\forall z\,(z\mid x\land z\mid y\to z=1),\\ -\mr{Prime}(x)&\iff x\ne1\land\forall z\,(z\mid x\to z=1\lor z=x). -\end{align*}$$ -Then, we can define the set of powers of a prime by -$$\mr{Power}(p,x)\iff\mr{Prime}(p)\land\forall z\,(z\perp p\to z\perp x).$$ -Finally, we can define for a given $x$ and a prime $p$ the power of $p$ that appears in the factorization of $x$ by -$$\mr{Val}(p,x,y)\iff\mr{Power}(p,y)\land\exists z\,(x=y\cdot z\land z\perp p).$$ -Now, for each prime $p$, $(\{x:\mr{Power}(p,x)\},{\cdot})$ is a model of Presburger arithmetic (which I assume to be formulated in a language with just a single binary function symbol $+$). Thus, if $\psi(y_1,\dots,y_k)$ is a formula of Presburger arithmetic, let $\psi^p(y_1,\dots,y_k)$ (with an extra free variable $p$) denote the formula of Skolem arithmetic obtained by replacing all occurrences of $+$ with $\cdot$, and relativizing all quantifiers to $\{x:\mr{Power}(p,x)\}$. Then (1) is defined by the formula -$$\exists^{\ge n}p\,(\mr{Prime}(p)\land\exists y_1,\dots,y_k\,(\mr{Val}(p,x_1,y_1)\land\dots\land\mr{Val}(p,x_k,y_k)\land\psi^p(y_1,\dots,y_k))).$$ -EDIT: I defined $\psi^p$ for formulas written in the language with $+$ only to keep the definition succinct, but in practice, it is more convenient to define it directly for a richer language: specifically, we may translate the constants $0$ and $1$ to $1$ and $p$, respectively, and $x\le y$ to $x\mid y$. -To put it differently, any Presburger formula $\psi(\vec y)$ is equivalent to a Boolean combination of integer inequalities $n+\sum_{i -TITLE: Projective dimension of graded modules -QUESTION [5 upvotes]: Short version: - -Why is the projective dimension of a graded module the same as the projective dimension of its underlying ungraded module? - -Longer version: -Let $G$ be a commutative group, let $R$ be a $G$-graded commutative ring, and let $M$ be a $G$-graded $R$-module. The category of $G$-graded $R$-modules has enough projectives, and so we can define the projective dimension ${\rm pd}(M)$ of $M$ as the infimum of the lengths of all projective resolutions of $M$ in the category of $G$-graded $R$-modules. -Let $U(M)$ denote the ungraded $R$-module underlying $M$. Then, $U(M)$ also has a projective dimension ${\rm pd}(U(M))$, defined in the category of (ungraded) $R$-modules. -At several places in the literature one finds the statement that these two projective dimensions coincide, i.e., that $${\rm pd}(M)={\rm pd}(U(M)).$$ The reason for this is always given by the fact that a $G$-graded $R$-module is projective if and only if its underlying $R$-module is so. But this seems to yield only the inequality $${\rm pd}(M)\geq{\rm pd}(U(M)).$$ -If we wish to show the converse, then we may consider a projective resolution $$0\rightarrow P_n\rightarrow P_{n-1}\rightarrow\cdots\rightarrow P_1\rightarrow P_0\rightarrow U(M)\rightarrow 0$$ in the category of $R$-modules and try to get from this a projective resolution of the same length in the category of $G$-graded $R$-modules. But as the $P_i$ need not be obtained from $G$-graded $R$-modules it is not immediately clear how to proceed. So: - -Is the above equality true? And if so, how do we prove it? - -Note 1: Of course this question (and hopefully also its answer) can be generalised to arbitrary coarsenings, but for the moment the forgetful functor $U$ will suffice. -Note 2: There seems to be something more general going on, for the same claim is found in the literature about the weak dimension; here, one should note that a $G$-graded $R$-module is flat if and only if its underlying $R$-module is so. - -REPLY [3 votes]: The other inequality follows from Schanuel's lemma. -For $M$ as is in your question, consider a truncated resolution of projective $G$-graded modules -$$Q_{n-1}\xrightarrow{f}Q_{n-2}\to\dots\to Q_1\to Q_0\to M.$$ -Then -$$U(Q_{n-1})\xrightarrow{f}U(Q_{n-2})\to\dots\to U(Q_1)\to U(Q_0)\to U(M)$$ -is a truncated projective resolution of $U(M)$. -If $\mathrm{pd}\, U(M)\leq n$, then Schanuel's lemma implies that $U(\ker f)$ is projective. Thus, $\ker f$ is projective as a $G$-graded module and $\mathrm{pd}\, M\leq n$. -This argument works for any exact functor $U$ such that $X$ is projective if and only if $U(X)$ is projective (in the appropriate categories).<|endoftext|> -TITLE: Distribution of Elliptic Curve Generators over Q -QUESTION [5 upvotes]: Considering elliptic curves over Q with positive discriminant and rank>0, are there any results or proposed heuristics regarding the fraction of generators that are located on the identity vs non-identity component? How does this vary with rank? -The family of curves I've been examining (associated with $a/(b+c) + b/(a+c) + c/(a+b) = N$) have the following distributions for N even: -Rank 1: 313 have generator on identity, 601 on non-identity -Rank 2: 107 have 2 on identity, 153 have 1, 16 have 0 -Rank 3: 3 have 3 on identity, 16 have 2, 3 have 1, 0 have 0 - -For N odd, all of the generators appear on the identity component. - -REPLY [6 votes]: I gathered some statistics using the Cremona tables. There are 565803 elliptic curves with rank 1, trivial torsion group, positive discriminant (which means that the group of real points has two connected components) and conductor $\leq 4 \times 10^5$ (the current limit of the tables). Among them 239106 curves have the generator of $E(\mathbf{Q})$ located on the identity component. -Here is the histogram representing the location of the generator $P$ on the identity component for these curves. More precisely, this is the distribution of the (unique) real number $z_P \in (0,1/2)$ such that the Mordell-Weil group $E(\mathbf{Q})$ is generated by the class of $z_P \Omega_E^+$, where $\Omega_E^+$ is the real period of $E$. We can note the behaviour near $z=0,1/4,1/3,1/2$ which might be explained by the fact that the torsion points somehow repel the point of infinite order.<|endoftext|> -TITLE: Destroying Suslin, nothing special -QUESTION [8 upvotes]: Recall that a tree on $\omega_1$ is called Suslin if every chain and antichain are countable. If every level is countable and there are no cofinal branches, then it is called Aronszajn (in particular Suslin is Aronszajn). Moreover, an Aronszajn tree is called special if it is the countable union of antichains. -So special Aronszajn trees and Suslin trees represent some sort of endpoints on the spectrum of Aronszajn trees: either every antichain is countable, or the whole tree is a countable union of antichains. -Given an Aronszajn tree, we can always specialize it using a ccc forcing; and given a Suslin tree it is a ccc forcing by itself, and forcing with it will add a cofinal branch (which is an uncountable chain). -Therefore, given a Suslin tree, we have a choice of how we want to violate its Suslinity: either violate its Aronszajn-ness, or specialize it. - -Can we make a Suslin tree into a non-special Aronszajn tree, while destroying its Suslinity? - -If the answer is positive, what is the "best" way of doing so? (Either consistently or provably, from the existence of a Suslin tree.) -Can we do it with a ccc forcing, or with a $\sigma$-closed forcing, or with just a proper forcing? Do the properties of this forcing somehow depend on the tree (e.g. if the tree is rigid, or homogeneous, etc.)? -(As a minor bonus question, is it consistent that all Aronszajn trees are Suslin or special, but Suslin trees exist?) - -Just a remark on triviality, when I say "tree" I always mean that any point has at least two successors, and that it has successors on every level of the tree. I am also going to assume the tree is Hausdorff, in the sense that at limit levels the points are always determined by their predecessors. - -REPLY [10 votes]: Chapter IX of Proper and Improper Forcing addresses this issue. -Shelah proves that Souslin's Hypothesis does not imply every Aronszajn tree is special, and he does this by investigating weak notions of specialness that are still incompatible with Souslinity. He shows that there are forcings that "specialize" Aronszajn trees in the weak sense, and that they can be iterated while still preserving at least one non-special Aronszajn tree. -Corollary 4.8 spells out more details: -He starts with a Souslin tree $T^*$ and preserves the fact that it is a non-special Aronszajn tree while also specializing all A-trees in a weak sense. He uses an $\aleph_1$-free iteration of proper forcing instead of countable support, but it is not clear this is necessary. -The concept of $(T^*, S)$-preserving (Definition 4.5) seems to be the property of forcing notions that ensures the tree $T^*$ never gets fully specialized. -The entire chapter is pretty technical, but it contains many interesting ideas that could and should be developed further.<|endoftext|> -TITLE: Limit of quotients of elements of special Fibonacci matrices -QUESTION [8 upvotes]: Let $F_n$ be the $n$-th Fibonacci number, started with $F_0=0,F_1=1$, and consider the matrices -$$M_n=\pmatrix{F_{n+3} & F_{n+1} \\ F_{n+2} & F_{n}}.$$ -Let -$$\pmatrix{\alpha_n & \beta_n \\ \gamma_n & \delta_n}=M_1\cdot M_2\cdot \ldots \cdot M_n .$$ -It is easy to see by computer, that the quotients $\frac{\alpha_n}{\gamma_n},\frac{\beta_n}{\gamma_n},\frac{\delta_n}{\gamma_n}$ are converging. I found that -$$ \lim_{n\to\infty} \frac{\delta_n}{\gamma_n}={\phi^2}$$ -where $\phi=\frac{\sqrt{5}-1}{2}$. -Unfortunately I can't find the other two limit, but numerically it seems to be that -$$ \lim_{n\to\infty} \frac{\alpha_n}{\gamma_n}\approx 1.3876267558043602953$$ -$$ \lim_{n\to\infty} \frac{\beta_n}{\gamma_n}\approx 0.53002625701851519880$$ -Can anyone give me a "nice" description of these numbers? Alternatively, it would be enough, if someone can decide whether these numbers are algebraic or not. -(I remark that the limit of $\alpha_n / \gamma_n$ is the most interesting for me, since it has a continued fraction expansion: $[1;2,1,1,2,1,1,1,2,...]$ and so on, the number of ones between twos increasing by one. It is a non-periodic, badly approximable number.) - -REPLY [7 votes]: One can prove by induction that some of given ratios have nice continued fraction expansions: -\begin{gather*} -\frac{\alpha_n}{\beta_n }=[2;1^n,2,1^{n-1},\ldots,2,1,1,2,1]\to 1+\varphi=\varphi^2;\\ -\frac{\gamma_n}{\delta_n }=[2;1^n,2,1^{n-1},\ldots,2,1,1,2]\to 1+\varphi=\varphi^2;\\ -\frac{\alpha_n}{\gamma_n}=[1;2,1,1,2,1,1,1,2,\ldots,1^{n-1},2,1^n,2]\to \psi;\\ -\frac{\beta_n}{\delta_n}=[1;2,1,1,2,1,1,1,2,\ldots,1^{n-1},2,1^n]\to \psi, -\end{gather*} -where $\psi=[1;2,1,1,2,1,1,1,2,\ldots]$ is defined by its continued fraction expansion.<|endoftext|> -TITLE: Stabilizer of Sp(n) and U(n) in GL(n) -QUESTION [7 upvotes]: I would be very grateful for a reference -to the following results (which are, I think, true, -though I never saw it in the literature). -Let $G\subset GL(n,{\Bbb C})$ be $U(n)$, -abd $A\in GL(2n,{\Bbb R})$ an endomorphism which satisfies -$AGA^{-1}=G$. Then $A\in {\Bbb R}^* \times U(n)$. -Let $G\subset GL(n,{\Bbb C})$ be the group -$Sp(n)$ of quaternionic Hermitian matrices, -and $A\in GL(2n,{\Bbb R})$ an endomorphism which satisfies -$AGA^{-1}=G$. Then $A\in {\Bbb R}^* \times Sp(n)\times Sp(1)$. -Many thanks in advance. - -REPLY [13 votes]: First, let me fix a misunderstanding: $\mathrm{Sp}(n)$ does not sit in $\mathrm{GL}(n,\mathbb{C})$, but in $\mathrm{GL}(2n,\mathbb{C})$, so I'll assume that you mean, for the second part that $A$ lies in $\mathrm{GL}(2n,\mathbb{C})$. -These both follow immediately from the facts that all the automorphisms of -$\mathrm{SU}(n)$ are either inner or conjugate-inner while the automorphisms of $\mathrm{Sp}(n)$ are are all inner. -It's a bit easier to deal with the $\mathrm{Sp}(n)$ case first since it has no outer automorphisms: If $A\in \mathrm{GL}(2n,\mathbb{C})$ satisfies $A\mathrm{Sp}(n)A^{-1}\subset \mathrm{Sp}(n)$, then consider the homomorphism $\phi:\mathrm{Sp}(n)\to \mathrm{Sp}(n)$ defined by $\phi(g) = AgA^{-1}$. This is a smooth, injective homomorphism, so it must be a smooth automorphism. Since every automorphism of $\mathrm{Sp}(n)$ is of the form $\phi(g) = hgh^{-1}$ for some $h\in\mathrm{Sp}(n)$, it follows that $AgA^{-1} = hgh^{-1}$, so $h^{-1}A$ lies in the commuting ring of $\mathrm{Sp}(n)$ (intersected with the $\mathbb{C}$-linear isomorphisms of $\mathbb{C}^{2n}$), and this is simply the nonzero complex multiples of the identity (since $\mathrm{Sp}(n)$ acts irreducibly on $\mathbb{C}^{2n}$). Thus, $A = \lambda h$ for some $h\in \mathrm{Sp}(n)$ and some nonzero complex scalar $\lambda$. -For the $\mathrm{U}(n)$ case, notice that the problem is equivalent to finding the conjugations that preserve the subgroup $\mathrm{SU}(n)$, so we might as well ask for the $A\in\mathrm{GL}(n,\mathbb{C})$ such that $A\mathrm{SU}(n)A^{-1}= \mathrm{SU}(n)$. Now, there is a slight complication, because for $n>2$, not all of the automorphisms of $\mathrm{SU}(n)$ are inner. For example, the automorphism $\psi(g) = \bar g$ is not inner. Instead, every automorphism is either of the form $\phi(g) = hgh^{-1}$ or $\phi(g) = h\bar gh^{-1}$ for some $h\in\mathrm{SU}(n)$. However, it's easy to see that, for $n>2$, there is no pair $(A,h)$ such that $AgA^{-1} = h\bar gh^{-1}$ for all $g\in\mathrm{SU}(n)$, so we only need to deal with the case $AgA^{-1} = h gh^{-1}$ with $h\in\mathrm{SU}(n)$ and $A\in \mathrm{GL}(n,\mathbb{C})$. Again, we find that $h^{-1}A\in\mathrm{GL}(n,\mathbb{C})$ must commute with all of the elements of $\mathrm{SU}(n)$, and this can happen only if $h^{-1}A$ is a multiple of the identity (again because of the irreducibility of the action of $\mathrm{SU}(n)$ on $\mathbb{C}^n$). -I think that the reason you haven't seen it in the literature is that it is such a direct consequence of well-known facts about Lie groups and representations.<|endoftext|> -TITLE: Degree of the variety of singular points -QUESTION [9 upvotes]: Let $V\subset \mathbb{A}^n$ be an irreducible affine variety. The set of singular points of $V$ is a subvariety $W$ of $V$; denote its components by $W_i$. How may we bound $\sum_i \deg(W_i)$ in terms of $\deg(V)$ and $n$? -I am satisfied if we can find a proper subvariety $Z$ of $V$ containing $W$ -and a bound for $\sum_i \deg(Z_i)$, where the $Z_i$'s are the components of $Z$. -The one way I can see is what I take to be the obvious one. We can define $V$ by $n+1$ polynomials $F_1,\dotsc,F_{n+1}$ of degree $\leq \deg V$ (by J. Heintz, "Definability and fast quantifier elimination...", MR0716823; thanks to Simon Rydin Myerson for the reference). Then the singular points of $V$ are those in which the Jacobian of $(F_1,\dotsc,F_{n+1})$ has rank less than $n-\dim(V)$. -Since $W$ is a proper subvariety of $V$, there must be some $(n-\dim(V))$-by-$(n-\dim(V))$ minor of the Jacobian that does not vanish on all of $V$. Its intersection with $V$ defines a proper subvariety $Z$ of $V$. By B\'ezout, its degree is at most $\deg(V)^{n+1-\dim(V)}$. -That may be far from tight, however. Can one do better? - -REPLY [5 votes]: Edit. As the OP points out, for his purpose it suffices to take the zero locus of a single (nonzero) partial derivative. So the OP produces a proper closed subset of $V$ containing the singular locus and having degree bounded by $(\text{deg}(V)-1)\text{deg}(V)$. Although this is not what the OP asks, there are cases where we need an upper bound on the degree of the singular locus (or at least the union of all components of the singular locus that have the maximum dimension). This often occurs for bounding the set of "bad characteristic" for some property of schemes over a field of (possibly) finite characteristic. The answer below gives an upper bound on the degree of the singular locus. -I am just rewriting the proof of Lemma 4.2.5 of the following as one answer. I learned of this from Fedor Bogomolov. -Jan Gutt -Hwang–Mok rigidity of cominuscule homogeneous varieties in positive characteristic -PhD. thesis, 2013 -https://arxiv.org/pdf/1305.5296.pdf -The original statement is for projective varieties, but the result for affine varieties follows by intersecting with affine space (a Zariski open subset of projective space). -Lemma [Jan Gutt, 2013 thesis, Lemma 4.2.5] For a purely $r$-dimensional closed subscheme $V$ of projective space $\mathbb{P}^n_k$ with degree $D>1$, if the zero scheme $S$ of the $r^\text{th}$ Fitting ideal of $\Omega_{V/k}$ has dimension $m$, then the corresponding $m$-cycle of $S$ has degree no greater than $D(D-1)^{r-m}$. -Proof. When $m$ equals $r$, then this just says that the $m$-dimensional cycle of $S$ has degree no greater than the degree of the $m$-dimensional cycle of $V$. Thus, without loss of generality, assume that $r>m$. Also, it suffices to prove the result when $k$ is algebraically closed. The proof uses Theorem 1.1 of the following. -MR0282975 (44 #209) -Mumford, David -Varieties defined by quadratic equations. 1970 Questions on Algebraic Varieties -(C.I.M.E., III Ciclo, Varenna, 1969) pp. 29–100 Edizioni Cremonese, Rome -http://www.dam.brown.edu/people/mumford/alg_geom/papers/1970a--CIME-QuadEqns-DAM.pdf -Mumford proves that the ideal sheaf $I$ of $V$ is generated in degree $d$. More precisely, the linear system $H^0(\mathbb{P}^n_k,I(d))$ of sections $g$ of $\mathcal{O}(d)$ on $\mathbb{P}^n_k$ that vanish on $V$ has base locus that equals $V$ set-theoretically and that equals $V$ scheme-theoretically, at least on the dense open subset $V\setminus S$ of $V$. -Thus, the common zero scheme in $V$ of the set of partial derivatives, $\partial g/\partial t$ (for varying homogeneous coordinates $t$) is contained in $S$ set-theoretically, and contains $S$ scheme-theoretically (since the Fitting ideal contains these partial derivatives, locally). -By Bertini’s theorem, for $r-m$ general polynomials $g = (g_1 , \dots , g_{r-m})$ in this linear system, for a general choice of homogeneous coordinates on $P^n_k$ and for a choice $t = (t_1 , \dots , t_{r-m} )$ of $r-m$ of these coordinates, the common zero scheme in $V$ of the $r-m$ partial derivative polynomials $\partial g_i /\partial t_i$ is $m$-dimensional and contains $S$. Since these partial derivatives are global sections of $\mathcal{O}(D − 1)$, the degree bound follows. -QED. -Will Sawin's Examples. Let $V$ be a subvariety that spans a linear subspace $\mathbb{P}^{r+1}_k \subset \mathbb{P}^n_k$ and that equals a degree-$D$ hypersurface in this linear space with defining polynomial $g=t_{m+1}^D + \dots + t_{r+1}^D$. Assume that the integer $D$ is nonzero in $k$. The Fitting ideal is precisely defined by $t_{m+1}^{D-1},\dots,t_{r+1}^{D-1}$ and the linear polynomials $t_{r+2},\dots,t_n$. Thus, the degree equals $(D-1)^{r+1-m}$, which is close to $(D-1)^{r-m}D$.<|endoftext|> -TITLE: Flat limit (of twisted cubic) contained in surfaces -QUESTION [5 upvotes]: Let $H$ denote the irreducible component of $\text{Hilb}^{3t+1}\mathbb{P}^3$ whose general member corresponds to a non-singular twisted cubic. Let $C$ be a subscheme lying in the boundary of $H$ and assume it lies in a surface $S \subseteq \mathbb{P}^3$. -Then why is it possible that we can find families $C_R, S_R \subseteq \mathbb{P}^3_R$ over a DVR $R$ with fraction field $K$ such that -1) $C_R \subseteq S_R$ -2) The generic fiber $C_K$ is a non-singular twisted cubic -3) $C \subseteq S$ are the closed fibers of the family. -The authors in "Hilbert Scheme Compactification of the Space of Twisted Cubics": https://www.uio.no/studier/emner/matnat/math/MAT4230/h10/undervisningsmateriale/Hilbertscheme.pdf make the claim on page 4 (pg 763), line 7 of the proof. Although they are studying embedded points, this claim seems to be something more general about flat limits? -More generally, is it true that if something on the boundary of my component in a Hilbert scheme lied in a hypersurface, then I could find a family over a DVR like above? - -REPLY [4 votes]: As @abx and @ulrich explain in the comments, the original question is equivalent to a question about constancy of Hilbert functions for the universal family restricted over the irreducible component $H$. The Hilbert function is constant on $H$, as I explain below. I believe the OP is confused because, at this point in the article of Piene and Schlessinger, they have not proved constancy of the Hilbert function. Nor do they need this. For instance, Piene and Schlessinger point out that for the purposes of their proof, it is fine to replace a hyperplane that contains the curve by a quadric hypersurface that contains the curve (in fact, as follows from constancy of the Hilbert function, there is no hyperplane containing a curve parameterized by $H$). So my advice to the OP for reading the article is: just read the remainder of the proof and then come back to this issue after a complete read-through. -Anyway, the Hilbert function is constant. -Denote by $p(t)$ the Hilbert polynomial $p(t)=3t+1$. On the Hilbert scheme $\text{Hilb}^{p(t)}_{\mathbb{P}^3_k/k}$, the natural action of $\textbf{PGL}_{4,k}$ on $\mathbb{P}^3_k$ induces an action on -$\text{Hilb}^{p(t)}_{\mathbb{P}^3_k/k}$. Denote by $H_0$ the unique open orbit. Denote by $H$ the closure of $H_0$ in $\text{Hilb}^{p(t)}_{\mathbb{P}^3_k/k}$. Denote the restriction of the universal closed subscheme over $H$ by $$Z_H\subset H\times_{\text{Spec}\ k}\mathbb{P}^3_k.$$ -Claim. Every geometric fiber $Z_t$ of the projection $Z_H\to H$ has Hilbert function, -$$ -h_{Z_t}:\mathbb{Z}_{\geq 0} \to \mathbb{Z}_{\geq 0}, \ \ h_{Z_t}(d) := h^0(\mathbb{P}^3_k,\mathcal{O}(d)) - h^0(\mathbb{P}^3_k,\mathcal{I}_{Z_t}(d)), -$$ -equal to $h_{Z_t}(d) = 3d+1$. -Proof. Probably the fastest way to prove this is to use the stratification of $H$ according to the "type" of $Z_t$. Some version of this is contained in Joe Harris's monograph. -MR0685427 (84g:14024) -Harris, Joe -Curves in projective space. -With the collaboration of David Eisenbud. -Séminaire de Mathématiques Supérieures, 85. -Presses de l'Université de Montréal, Montreal, Que., 1982. -138 pp. ISBN: 2-7606-0603-1 -I have a vague recollection that there is a small mistake in Harris's description of the orbit decomposition. I am more familiar with the honors thesis of Yoon-Ho Alex Lee. The most relevant result is Figure 4.4, p. 47. -Yoon-Ho Alex Lee -The Hilbert scheme of curves in $\mathbb{P}^3$ -https://www.uio.no/studier/emner/matnat/math/MAT4230/h10/undervisningsmateriale/ALee_Hilbertschemes.pdf -Since dimensions of cohomology groups are upper semicontinuous, it suffices to prove that $h_{Z_t}(d)$ equals $3d+1$ for $Z_t$ in the two "deepest" strata, XVI and XVII in Lee's notation. With respect to homogeneous coordinates $[ s,t,u,v ]$, the ideal for XVI is $\langle u^2,ut,uv,v^3\rangle$, and the ideal for XVII is $\langle u^2,uv,v^2 \rangle$. It is straightforward in each of these cases to compute that $h_{Z_t}(d)$ equals $3d+1$. QED<|endoftext|> -TITLE: What is the lower bound for the number of facets that a general convex $d$-polytope with $n$ vertices can have? -QUESTION [5 upvotes]: I am familiar with Barnette's Lower Bound Theorem on the number of facets a $d$-dimensional simplicial convex polytope with $n$ vertices can have. Is there a similar result for a general (i.e. not necessarily simplicial) $d$-polytope with $n$ vertices? - -REPLY [3 votes]: A lower bound on the number of facets can be obtained from McMullen's upper bound theorem. It says that, for a given number of vertices, neighborly polytopes maximize the number of faces in all dimensions (among all polytopes, simplicial or not). In particular, a $d$-polytope $P$ with $m$ vertices has at most $F(d,m)$ facets (an exact formula for $F(d,m)$ can be found in Ziegler's "Lectures on polytopes"). By going to the dual polytope $P^*$ one sees that -if the number $n$ of vertices of a $d$-polytope satisfies $F(d,m) \ge n > F(d,m-1)$, then it has at least $m$ facets. -For $n = F(d,m)$ this bound is exact (attained by the duals of neighborly polytopes). I do not know if for any $n$ inbetween there is a polytope with $m$ facets.<|endoftext|> -TITLE: How to formally connect the log-Euclidean and Riemannian metrics for Symmetric Positive Definite matrices? -QUESTION [7 upvotes]: I'm a statistician working on a research project dealing with metrics on SPD matrices, specifically the log-Euclidean $d_{LE}(X, Y) = \|\log(X) - \log(Y)\|$ and the Riemannian metric $d_{R}(X, Y) = \|\log(Y^{-1/2}XY^{-1/2})\|$, where the norm is the Frobenius norm. -I understand that the two metrics are closely related, and that the LE metric is something like a linearization of the Riemannian metric, but I haven't been able to find a reference that makes this precise. Any comments or suggestions are appreciated. -For some references, [1] is a good intro to different metrics for SPD matrices, and [2] gives some more details on the context I'm working in. - -REPLY [2 votes]: Here is a graphical description of the relation between the Riemannian and log-Euclidean metrics [ source, page 40] - -The "linearization" referred to in the OP is a linearization around the identity matrix:<|endoftext|> -TITLE: Equivalences of categories of sheaves vs categories of $\infty$-Stack -QUESTION [18 upvotes]: Let say I have two different sites $(\mathcal{C},I)$ and $(\mathcal{D},J)$ for an ordinary topos $\mathcal{T}$. I.e. -$$Sh(\mathcal{C},I) \simeq \mathcal{T} \simeq Sh(\mathcal{D},J)$$ -And we want to know if one also have an equivalence of the $\infty$-topos of sheaves of spaces on these sites: -$$ Sh_{\infty}(\mathcal{C},I) \overset{?}{\simeq} Sh_{\infty}(\mathcal{D},J)$$ -This question is about finding an explicit counter example where this is not the case. But I'll give a bit more context. -If my understanding is correct, there are two cases where one can say something: -1) If $ Sh_{\infty}(\mathcal{C},I)$ and $Sh_{\infty}(\mathcal{D},J)$ are hypercomplete, or more generally if one only care about their hypercompletion. Indeed, one always has: -$$ Sh_{\infty}(\mathcal{C},I)^{\wedge} \simeq \ Sh_{\infty}(\mathcal{D},J)^{\wedge}$$ -(the $^{\wedge}$ denotes hypercompletion) this is because one can show that they are both equivalent to the $\infty$-category attached to category of simplicial objects of $\mathcal{T}$ with the Jardine-Joyal model structure on simplicial sheaves. And the equivalences of this model structure only depends on the underlying ordinary topos. -2) If $\mathcal{C}$ and $\mathcal{D}$ have finite limits, then one gets the equivalence: -$$ Sh_{\infty}(\mathcal{C},I) \simeq Sh_{\infty}(\mathcal{D},J)$$ -This follows from J.Lurie Lemma 6.4.5.6 in Higher topos theory. This lemma assert that under these assumptions, there is a natural equivalence between -$$ Geom( \mathcal{Y}, Sh_{\infty}(\mathcal{C},I)) \simeq Geom( \tau_0 \mathcal{Y} , Sh(\mathcal{C},I) ) $$ -Where $\mathcal{Y}$ is any $\infty$-topos, $Geom$ denotes the spaces of geometric morphsisms, either of $\infty$-topos or ordinary toposes, and $\tau_0 \mathcal{Y}$ is the ordinary topos of homotopy sets of $\mathcal{Y}$. -THis in particular gives a universal property to $Sh_{\infty}(\mathcal{C},I)$ which only depends on the $Sh(\mathcal{C},I)$ and so this implies the isomorphisms above. - -So what about the general case ? I have quite often heard that this was not true in general, and I'm willing to believe it. But I would really like to see a counter-example ! - -In some of the places where I have seen asserted that this is not true in general, people were pointing out to the examples where $Sh_{\infty}(C,J)$ is not hypercomplete, and often these examples fall under the assumption of the second case. -In a comment on this old closely related question of mine, David Carchedi mention a counter-example due to Jacob Lurie which indeed avoids both situation... But I havn't been able to understand how it works, and it does not seem to appears in print. If someone can figure it out I'll be very interested to see the details. -Also a counter example to this lemma 6.4.5.6 mentioned above (without the assumption that the site has finite limit) would probably produce an answer immediately. Moreover (assuming such a counterexample exists) there must exists one where the topology is trivial, so I guess there has to be some relatively simple counter examples. - -REPLY [11 votes]: I just found an example, so I thought it would be good to post it here, but if anyone knows other examples, or a more general way to construct some I would be interested to see them as well. -This example comes relatively immediately from the example of $1$-site producing a non Hypercomplete $\infty$-topos in the paper of Dugger, Hollander and Isaksen Hypercovers and simplicial presheaves. -They are looking at the site whose underlying category is the posets $I$: -$$ \require{AMScd} -\begin{CD} -V_0 @<<< U^l_0 \\ - @AAA @AAA \\ -U^r_0 @<<< V_1 @<<< U_1^l \\ -@. @AAA @AAA \\ -@. U^r_1 @<<< V_2 @<<< U^l_2 \\ -@. @. @AAA @AAA \\ -@. @. \vdots @<<< \vdots @<<< \dots -\end{CD} -$$ -And where for each $i$, $U_i^r, U_i^l$ forms a cover of $V_i$. It is not too hard to check that this defines a topology. They show that the $\infty$-topos of sheaves of spaces on this site is not Hypercomplete. -But if one now look at the category of sheaves of sets. Then One can apply the comparison lemma. Let $J \subset I$ be the full subcategory on the $U^{l/r}_i$, then as each $V_i$ is covered by the $U^{l/r}_i$ so by the comparison lemma, the category $Sh(I)$ and $Sh(J)$ (for the induced topology) are equivalent. But the topology induced on $J$ is trivial so one has: -$$Prsh(J) \simeq Sh(I) $$ -One the other hand $Prsh_{\infty}(J)$ is obviously hyperconnected while $Sh_{\infty} (I)$ has been proved to not be hyperconnected in the paper mentioned above, so: -$$ Prsh_{\infty}(J) \not\simeq Sh_{\infty} (I) $$ - -Also note that this example of Dugger-Hollander-Isaksen is closely related to J.Lurie's example involving the Hilbert cube. Indeed if $Q= [0,1]^{\mathbb{N}}$ is the Hilbert Cube, then defining: -$$ U_i^l = ]0,1[^i \times [0,1[ \times Q $$ -$$ U_i^r = ]0,1[^i \times ]0,1] \times Q $$ -$$ V^i = U_{i-1}^r \wedge U_{i-1}^l = ]0,1[^{i} \times Q $$ -Gives a full subcategory of $\mathcal{O}(Q)$ (the category of opens of $Q$) stable under intersection and isomorphic to the $I$ above, with the induced topology on $I$ being the topology described by Dugger-Hollander-Isaksen. So one has a geometric morphisms from $Sh(Q)$ to $Sh(I)$. This somehow show how something along the line of the comment of David Carchedi linked in the question might produces an examples (though as I understand it one might needs to consider more open subset than what the comment said, but I might still be missing something)<|endoftext|> -TITLE: On convergent series - in the spirit of Abel and Dini -QUESTION [6 upvotes]: Nonexistence of boundary between convergent and divergent series? -I'm hoping the following is true: - -Suppose $a_i $ is a positive sequence and $\sum_i a_i < \infty.$ Then there exists a positive sequence $b_i$ s.t $\sum_i b_i < \infty$ and $\sum_i \frac{a_i}{b_i} < \infty$. - -REPLY [15 votes]: This is not correct. Take $a_n=1/n^2$. -Let us show that $b_n$ with -required properties does not exist. Consider the set -$$E=\{ n: b_n\geq 1/n\}.$$ -As $\sum b_n<\infty$, we have $$\sum_E1/n<\infty.$$ -Now on $N\backslash E$ we have $b_n<1/n,$ so $1/b_n>n$ and as $\sum a_n/b_n<\infty$, -we conclude that $$\sum_{N\backslash E}1/n<\infty.$$ -adding the last two inequalities we obtain a contradiction.<|endoftext|> -TITLE: Differences between logic with and without equality -QUESTION [10 upvotes]: By logic without equality I mean those kind of logics where equality is treated as a binary relation satisfying some axioms, as opposed to a logics where equality is a logical symbol satisfying some inference rules. -In my understanding there should be no difference at the syntactic level between logics with or without equality since the syntax should not be able to distinguish a logic-primitive predicate or any other predicate (but in case I am wrong please correct me). -So I am wondering what are the differences at a semantic level between these kind of logics. -In particular I am wondering whether all the definitions and theorems of classical model theory can be transported verbatim to the without-equality logics. -If it is possible I would also appreciate references that treat the subject. -Addendum: After Andrej Bauer's answer I realize it would be better to add some specification. -I am interested basically in logical theories where every theory comes equipped with an axiom-schema for substitution, that is a family of axioms of the form -$$\forall x,y. (x=y) \land \varphi(x) \rightarrow \varphi(y)$$ -where $\varphi$ is a formula of the language. -In short I am curious to know what changes we get if we change the semantics of the symbol of equality. -Addendum 2: also it seems from the answer provided so far that the model theory should remain unchanged but it does not look like it to me. -First of all while quotienting for the equivalence relation should provide a (elementary?) equivalent model it does not (necessarily) preserves the homomorphisms: consider a set with an equivalence relation that identifies all elements, the corresponding quotient structure should be the terminal model (in the sense of category theory) but the first model clearly does not have to be a terminal object in the category. -Also it seems to me that one could lose the characterisation of homomorphisms as mapping that preserve atomic formulas, since, if we drop the requirement of interpreting the equality symbol as the identity relation, then we could have mappings that preserve atomic formulas but for some operations symbols do not respect the external equality: i.e. we could have a mapping $f \colon A \to B$ and an operation symbol $o$ such that $$B \models f(o^A(a)) = o^B(f(a))$$ -but $f(o^A(a))$ and $o^B(f(a))$ are not identical. -It seems rather difficult to find references that deal with this kind of phenomena, that is why I would really appreciate if someone could provide some pointers. -Thanks in advance. - -REPLY [4 votes]: When dealing with model theory without equality, I think it is better not to deal with isomorphisms (functions) but with isomorphism relations. For example, in order to recover Fraïssé's theorem, one may consider partial isomorphism relations because one cannot infer the existence of a partial isomorphism from the satisfaction of the same atomic formulas when equality is absent. -What is gained in doing this? I think that something is gained, and it is not extra generality. I will try to make a point: -1) A natural question in foundational investigations is: what are the minimal resources that a classical logical system must have in order to be a possible vehicle for classical mathematical reasoning? One can, with first-order logic without equality, axiomatize set theory with only one primitive binary relation (membership). So, equality is certainly not part of the minimum. Is this (first-order logic with membership but without equality) the minimal vehicle? For one interpretation of what is meant by classical mathematical reasoning, one can prove, using some model theory without equality, that it is not. -2) For pedagogical purposes, results in logic without equality are usually simpler to prove, one can get more focused on the essential ideas, not being distracted by some technical details. For example, the construction of the canonical structure is simpler, either from Hintikka sets or from maximal Henkin sets. Also, proof theory without equality is simpler. For example, one need not talk about quasi-tautologies in Herbrand's theorem without equality. The proof-theoretic characterizations of logical theorems are clearer in this setting. It is more general and simpler. -The following paper of mine deals with some model-theoretic stuff: -https://link.springer.com/article/10.1007/s11787-015-0126-8<|endoftext|> -TITLE: Upgrade adjunction to equivalence -QUESTION [10 upvotes]: I'm studying category theory by myself and I just came across this sentence from Wikipedia: -An adjunction between categories C and D is somewhat akin to a "weak form" of an equivalence between C and D, and indeed every equivalence is an adjunction. In many situations, an adjunction can be "upgraded" to an equivalence, by a suitable natural modification of the involved categories and functors. -Can someone provide an example of an "upgrade" of an adjunction to an equivalence? I'm interested in understanding why I could intuitively think of an adjunction as a weak form of an equivalence. - -REPLY [9 votes]: I think the author of the wikipedia article probably had in mind Leonid Positselski's first answer, where one restricts to the full subcategory of fixed points of the adjunction. Beware there is no guarantee that the fixed points are nonempty! For example, if $F: Set^\to_\leftarrow Ab: U$ is the free/forgetful adjunction bewteen sets and abelian groups, the fixed points are empty. -Here's an illustrative example to have in mind which is not so degenerate. Let $K/k$ be a Galois extension. Then there is an adjuntion between the poset of intermediate subfields $k \subseteq L \subseteq K$ and the opposite of the poset of subgroups of of $Gal(K/k)$; in one direction we send a group to its field of fixed points and in the other direction we send a field to the group of automorphisms that fix it. -This adjunction is typically not an equivalence, but we can pass to the fixed points of the adjunction to get an equivalence between the poset of normal subgroups of $Gal(K/k)$ and the opposite of the poset of intermediate Galois extensions $k \subseteq L \subseteq K$. -Thus the fundamental theorem of Galois theory may be viewed as calculating the fixed point set of an adjunction, and thus as identifying where an adjunction restricts to an equivalence. - -REPLY [7 votes]: I agree that Leonid Positselski’s first answer seems probably what the writer had in mind: given an adjunction, restricting to the categories of “fixed points” on each side yields an equivalence. Here are two important examples in nature, both involving the category of topological spaces: - -There’s an adjunction between the categories of preordered sets and topological spaces, sending a preordered set $(X,\leq)$ to $X$ with the topology of down-closed sets, and sending a topological space $Y$ to $Y$ with its specialisation order. All preorders are fixpoints; on the other side, the fixpoints are exactly the Alexandrov spaces, i.e. spaces where arbitrary intersections of opens are open. Restricting to this subcategory shows that the category of preorders is equivalent (in fact, isomorphic!) to the category of Alexandrov spaces. -The adjunction between the categories of topological spaces and locales, sending a topological space to its frame/locale of opens and sending a locale to its space of points, restricts to the equivalence between spatial locales and sober spaces.<|endoftext|> -TITLE: Tilting Objects in BGG Categories $\mathcal{O}$ -QUESTION [13 upvotes]: Let $\mathcal{O}$ be the BGG category $\mathcal{O}$ with respect to a finite dimensional semisimple Lie algebra $\mathfrak{g}$ and its Borel subalgebra $\mathfrak{b}$ (as define in this book by Humphreys). If $\mathfrak{h}$ is the Cartan subalgebra of $\mathfrak{g}$ contained in $\mathfrak{b}$, then there exists a unique indecomposable module $D(\lambda)$, where $\lambda\in\mathfrak{h}^*$, such that the weight space of $D(\lambda)$ with weight $\lambda$ has dimension $1$, and $D(\lambda)$ has a finite filtration with standard subquotients as well as a finite filtration with costandard subquotients. The construction of $D(\lambda)$ is discussed in the same book. -However, when I read Ringel's definition of tilting modules, I noticed some discrepancies. For example, a tilting module $D(\lambda)$ does not need to have projective dimension $1$ in $\mathcal{O}$ (this is the first condition in the link). Plus, the enveloping algebra $U(\mathfrak{g})$ is not in $\mathcal{O}$, so it is not the kernel of any epimorphism from a direct sum of finitely many copies of $D(\lambda)$ to another direct sum of finitely many copies of $D(\lambda)$, violating the third condition in the link. (I am not sure anyhow if the third condition makes any sense. After all, $\mathcal{O}$ is not the whole category of $U(\mathfrak{g})$-modules.) The only condition $D(\lambda)$ satisfies is $\operatorname{Ext}_{\mathcal{O}}^1\big(D(\lambda),D(\lambda)\big)=0$ (i.e., the second condition of the link). -The first condition can be fixed by using the definition of generalized tilting modules, but this definition may be not-so-helpful for other abelian categories. However, there doesn't seem to be a fix for the third condition. There is still no exact sequence $0\to U(\mathfrak{g})\to T_0\to T_1\to T_2 \to \ldots \to T_n \to 0$, where each $T_i$ is a finite direct sum of $D(\lambda)$. - -Hence, I would like to ask why are $D(\lambda)$ called tilting modules anyway? Is there a version of tilting theory that works with $D(\lambda)$? Do we have something like Ringel duality here? Is there a tilting theory for abelian categories in general? - -For tilting objects in a general abelian category, I am aware of the work by Colpi and Fuller in 2007, but it requires that the category contains the direct sums of arbitrary, possibly infinite, numbers of copies of some objects. I think this requirement is too restrictive. - -REPLY [17 votes]: Words change their meanings. -The original meaning of “tilting module” is that of Happel and Ringel in the representation theory of finite dimensional algebras, which requires the projective dimension to be one, and a “generating” condition (the third condition in your link). -Fairly soon, it was recognized that generalizing from projective dimension one to finite projective dimension was worthwhile. Originally, this generalization was called a “generalized tilting module”, but nowadays it would be confusing not to specify projective dimension one if that’s what you mean. -Then people from the algebraic groups/Lie algebras community got interested in axiomatizing highest weight categories, and Ringel proved the existence of the objects $D(\lambda)$ that you describe in the first paragraph of your question, and that, in the case where the highest weight category is the module category of a finite dimensional algebra, that the direct sum of the modules $D(\lambda)$ is a tilting module (in the generalized sense). -The algebraic groups/Lie algebras people were apparently not paying enough attention, and started referring to the $D(\lambda)$ as “the tilting modules”, which is wrong on possibly as many as three counts. -But the two communities, finite dimensional algebras on the one hand and algebraic groups/Lie algebras on the other, are still on cordial terms, and gently mock each other about the use of the term “tilting module”, in the same way that Britons and Americans gently mock each other about the use of the word “pavement”. -(Of course, it’s the finite dimensional algebra people and the Brits who are correct, but live and let live, eh?)<|endoftext|> -TITLE: Beilinson-Drinfeld quantization and stable bundles -QUESTION [7 upvotes]: To motivate this question, I'm going to try and explain some background notions. This won't be absolutely necessary for experts, but I want to be vaguely honest about where this question comes from. I also want to say that this is the type of question that I would ask at a coffee break during a conference about this subject. If my question is too disorganized for mathoverflow, I apologize, and there will be no hard feelings if this question is closed because it is too speculative. -Given a manifold $M$, one has a canonical symplectic structure on the cotangent bundle $T^{*}M$. Given a Poisson sub-algebra of functions $A$ on $T^{*}M$, one aim of quantization asks if there exists an algebra of differential operators $A^{'}$ on $M,$ (generally they may be differential operators on sections of line bundles over $M$), for which the principal symbol map $A^{'}\rightarrow A$ is a morphism, where commutator of differential operators corresponds to Poisson bracket of functions on $T^{*}M.$ -This question, which has its roots in the passage from classical to quantum mechanics, was employed by Beilinson-Bernstein in their localization procedure for producing modules over the universal enveloping algebra of a semisimple complex Lie algebra $\mathfrak{g}$ from $D$-modules on the flag variety $G/B$ where $G$ integrates $\mathfrak{g}$ and $B