Recent MathOverflow Questions

Can we realize a graph as the skeleton of a polytope that has the same symmetries?

Math Overflow Recent Questions - Thu, 08/16/2018 - 05:18

Given a graph $G$, a realization of $G$ as a polytope is a convex polytope $P\subseteq \Bbb R^n$ with $G$ as its 1-skeleton.

A realization $P\subseteq \Bbb R^n$ is said to realize the symmetries of $G$, if for each graph-automorphism $\phi\in\mathrm{Aut}(G)$ there is an isometry of $\Bbb R^n$ inducing the same automorphism on the edges and vertices of $P$.

Question: Let $G$ be the 1-skeleton of a polytope. Is there a realization of $G$ that realizes all its symmetries?

Note that if we know $G$ to be the skeleton of an $n$-polytope, then the symmetric realization does not have to be of the same dimension. In fact, this is not always possible. The complete graph $K_n,n\ge 6$ is realized as a neighborly 4-polytope, but its only symmetric realization is the simplex in $n-1\ge5$ dimensions. However, this is the only example I know of.

One way to construct a counter-example would be to find an $n$-polytope with a skeleton $G$ that is not more than $n$-connected (e.g. $n$-regular), but where $\mathrm{Aut}(G)$ is no subgroup of the point group $O(n)$. Its symmetric realization must be in dimension $\ge n+1$, but it cannot be because of Balinski's theorem.

Hopf dual of the Hopf dual

Math Overflow Recent Questions - Wed, 08/15/2018 - 09:12

Given any Hopf algebra $A$ over a field $k$, one can also define the Hopf dual $A^*$ of as follows: Let $A^∗$ be the subspace of the full linear dual of $A$ consisting of elements that vanish on some two-sided ideal of $A$ of finite codimension. Then $A^∗$ has a natural Hopf algebra structure.

Question: Is the Hopf dual of the Hopf dual of $A$ isomorphic to $A$. It is not obvious for me that it is. If not, then do we know in which cases it is true?

Maximal surfaces in pseudo-Riemannian manifolds

Math Overflow Recent Questions - Tue, 08/14/2018 - 10:40

Do you know a reference (with proof) for the second variation of the area of a space-like surface with vanishing mean curvature embedded in a pseudo-Riemannian manifold?

Is it known that if the ambient manifold is negatively curved (for instance $\mathbb{H}^{p,q}$), then these surfaces always maximize the area, among all surfaces with the same boundary? I know a similar result holds in codimension 1.

Relation between left projections

Math Overflow Recent Questions - Tue, 08/14/2018 - 09:21

Let $A$ be a Baer *-ring. Let $x$ be in $A$, $L(x)$ is the left projection of $x$ that is the smallest projection with $L(x)x=x$.

Q. Let $p,q$ are projections in $A$ with $p\leq q$. I feel both of the following points are true but cannot prove them.

1- $L(pe)\leq L(qe)$.

2- $L(ap)\leq L(aq)$.

Can sufficiently symmetric polytopes be uniquely reconstructed from their 1-skeleton?

Math Overflow Recent Questions - Tue, 08/14/2018 - 09:15

General convex polytopes can not be uniquely reconstructed from their 1-skeleton, as explained here. Not even the dimension is known from the skeleton, as e.g. the complete graph $K_n,n\ge 5$ is the 1-skeleton of neighborly polytopes that exist in dimensions $\ge 4$.

But how well works the reconstruction from the 1-skeleton if we restrict to sufficiently symmetric polytopes. Obviously, vertex-transitivity is not enough as seen from the existence of vertex-transitive neighborly polytopes. But what if we add edge-transitivity, uniformity, arc-transitivity or some requirements on the edge-lenghts? E.g. only simplices are vertex- and edge-transitive polytopes with $K_n$ as their 1-skeleton.

Question: Given a "symmetric" polytope $P$ (replace "symmetric" with the sufficiently strong symmetry requirements of your choice). Can there be a different "symmetric" polytope with the same 1-skeleton as $P$?

I asked a similar question on Math.SE, without much success.

Proof of an inequality

Math Overflow Recent Questions - Tue, 08/14/2018 - 09:01

I can't prove one of the inequality in a paper:

$\sum_{w=1}^v w^{\lambda}=O(v^{\lambda+1})$ for $\lambda \neq -1$.

When $\lambda>0$ I can use the AM-QM inequality to prove it, but when it comes to $\lambda<0$, I have no idea.

Automorphy of families of motives

Math Overflow Recent Questions - Tue, 08/14/2018 - 08:55

I have a couple of elementary questions regarding automorphy of Galois representations arising from geometric families.

Suppose we have an algebraic family of varieties over a number field, and Galois representations arising from $l$-adic cohomology of one of the varieties in the family are known to be automorphic. What can one say about automorphy of (Galois representations attached to $l$-adic cohomology of) other varieties in the family?

Alternatively, and more specifically, on the moduli space of algebraic curves (of a fixed genus) over $\mathbb{Q}$, take a small/formal neighborhood of a curve which is automorphic in the above sense. What can be said about automorphy of curves in the neighborhood?

One can also ask a related global question: suppose one somehow knows that the universal curve over the moduli space of algebraic curves of genus $g$ is automorphic. Would it follow that all algebraic curves of genus $g$ are automorphic?

Partition theorems for located words

Math Overflow Recent Questions - Tue, 08/14/2018 - 08:51

In this paper Bergelson, Blass, and Hindman prove the following

Theorem 1.2 Let $W(\Sigma; v)$ be colored with finitely may colors and let $\bar s$ be an infinite sequence from $W(\Sigma; v)$. Then $\bar s$ has a variable extraction $\bar t$ such that the set of extracted words of $\bar t$ is monochromatic.

I would like to know if there is an easy counter example to the claim that the components of $\bar t$ are obtained concatenating those of $\bar s$ in increasing order. (As stated, the theorem also allows substitution of the variable $v$ with elements of $\Sigma$ as long as all components of $\bar s$ contain at least a variable.)

Notation $W(\Sigma; v)$ is the set of words in the alphabeth $\Sigma\cup\{v\}$ that contain at least one occurrence of the variable $v$.

Noether-Lefschetz for Kummer-quartic surfaces with curves of bounded degree

Math Overflow Recent Questions - Tue, 08/14/2018 - 08:16

The Noether-Lefschetz theorem states, that a very general quartic surface $S$ in $\mathbb{P}^3$ has $Pic(S)\cong\mathbb{Z}$ generated by $\mathcal{O}_{\mathbb{P}^3}(1)_{|S}$, so all curves on $S$ are complete intersections. (We work over $\mathbb{C}$).

Do we have similar results for non-very general quartic surfaces, if we restrict to curves with bounded degree?

$\textbf{Question:}$ Given natural numbers $k,d\in \mathbb{N}$ with $2\leq k\leq 4$. Is it possible to find a Kummer-quartic surface $S$ in $\mathbb{P}^3$ with $\rho(S)=k$ such that all curves $C$ in $S$ with $deg(C)<d$ are complete intersections? Here Kummer means $S$ has exactly the maximal number of 16 nodes.

If yes: can we somehow describe all of theses surfaces in the moduli space of Kummer-quartic surfaces in $\mathbb{P}^3$?

For example: Can we find a Kummer-quartic surface $S$ in $\mathbb{P}^3$ with $\rho(S)=2$ such that every curve $C$ in $S$ with $deg(C)<16$ is a complete intersection?

About a pattern on prime numbers

Math Overflow Recent Questions - Tue, 08/14/2018 - 08:15

I have read in that says: "But this doesn't explain the magnitude of the bias the team found, or why primes ending in 3 seemed to like being followed by primes ending in 9 more than 1 or 7. Even when they expanded their sample and examined the first few trillion prime numbers, the mathematicians found that - even though the bias gradually falls more in line with randomness - it still persists."

I did a numerical analysis for the pairs (n3, n7), (n3, n9) where the numbers n3, n7, n9 which are integers ending at 3, 7, 9 respectively. These numbers (n3, n7, n9) may be (A) composites, multiples of 3, (B) composites, but not multiple of 3 or (C) prime numbers p3, p7, p9. We are looking for pairs of prime numbers (1) p3 with next p7 or (2) p7 with previous p3, (3) p3 with next p9 or (4) p9 with previous p3.

We take into account, that the prime numbers (random or not) are ranked in ascending order. ie after p1 series has p3 after p7 and then p9. Αfter 1 cycle again 1.

We will now see how a pair fails to be (p3, p7) or (p3, p9). This occurs in 4 cases.

(1) ((A) -> n3, p7) (#), ((A) -> n3, p9) (##)
(2) ((B) -> n3, p7), ((B) -> n3, p9)
(3) (p3, (A) -> n7), (p3, (A) -> n9)
(4) (p3, (B) -> n7), (p3, (B) -> n9)

(5) (p3, p7), (p3, p9)

We knows for big N have about the same amount of p3's, p7's, p9's in primePi(N).

We see (#) that the primes p7 can have losses in all four categories (1)(#), (2), (3), (4). But the p9's (##) only in 3 categories -> (2), (3), (4) Because the (1)(##) never happens, the reason is that p9 = ((A) -> n3) + 6 is multiple of 3, ie never is prime number. So p9's not consume their prime numbers in this category. For this p9's have more opportunities (since they are limited to 3 categories) than p7's (which spread into four) to reach and pair with p3's, see (5). That's why we find more pairs (p3, p9) than (p3, p7). See example in comments. But this phenomenon also seems to be very intense in a small sample of prime numbers, because they are more dense. In huge samples all categories normalized and approach at 25%. So, I will make a conjecture, that when N -> infinity, then (p3, p1), (p3, p3next), (p3, p7), (p3, p9) have a probability of 25%. As we would expect from random numbers.
Question: Am I right or not?

Functoriality in the group $G$ of the domain of the Baum-Connes map

Math Overflow Recent Questions - Tue, 08/14/2018 - 08:08

Lück claims in his preliminary book, that the left hand side of the Baum-Connes map is functorial in the group $G$. For the right hand side $K(A \rtimes G)$ this is clear for the full crossed product, as he himself points out.

How is this justified?

Are $\varepsilon$-connected components dense?

Math Overflow Recent Questions - Tue, 08/14/2018 - 08:01

Let $X$ be a connected compact metric space. Given a positive $\varepsilon$ and two points $x,y\in X$ we write $x\sim_\varepsilon y$ if there exists a sequence $C_1,\dots,C_n$ of connected subsets of diameter $<\varepsilon$ in $X$ such that $x\in C_1$, $y\in C_n$ and $C_i\cap C_{i+1}\ne\emptyset$ for all $i<n$. It is clear that $\sim_\varepsilon$ is an equivalence relation on $X$. The equivalence class $[x]_\varepsilon:=\{y\in X:y\sim_\varepsilon x\}$ will be called the $\varepsilon$-connected component of $x$.

Problem. What can be said about the $\varepsilon$-connected components of $X$? In particular, is each $\varepsilon$-connected component $[x]_\varepsilon$ dense in $X$? Is each $\varepsilon$-connected component $\sigma$-compact?

Added in Edit. It can be shown that each $\varepsilon$-connected component $[x]_\varepsilon$ is $\sigma$-compact. To prove this fact, fix a countable base $\mathcal B$ of the topology of $X$, which is closed under finite unions and let $\mathcal B_\varepsilon$ be the subfamily of $\mathcal B$ consisting of basic sets of diameger $<\varepsilon$. It can be shown that each connected subset of diameter $<\varepsilon$ is contained in the connected component of some set $\bar B$ with $B\in\mathcal B_\varepsilon$. Now fix an enumeration $\{B_n\}_{n\in\omega}$ if the family $\mathcal B_\varepsilon$. Given any point $x\in X$ let $C_0=\{x\}$ and for every $n\in\mathbb N$ let $C_n$ be the union of $C_{n-1}$ with all connected components of $\bar B_n$ that intersect $C_{n-1}$. It can be shown that the sets $C_n$ are compact and $[x]_\varepsilon=\bigcup_{n\in\omega}C_n$.

this problem is related to algebra [on hold]

Math Overflow Recent Questions - Tue, 08/14/2018 - 07:46

sui mother buy some share of A on day 0. On day 7 share price of A is $\$44.6$.If she sell all share of A and buy 2000 shares of B on day 7 she would receive $\$7400$. On day 12 Share price of A is $\$4.8$ and B is $\$0.5$ less than on day 7. If she sell her all shares of A and buy 5000 shares of B on day 12 she would have to pay $\$5800$. Find share of A and share price of B on day 12

Union of random intervals with total length equal to infinity

Math Overflow Recent Questions - Tue, 08/14/2018 - 07:18

Let $a_1,a_2,\dots$ be a sequence of positive numbers less than $1$, such that $$\sum_{n=1}^\infty a_i= \infty,$$ and $S^1 = \mathbb{R}/\mathbb{Z}$.

Suppose $I_1,I_2,\dots$ be random intervals with respective lengths $a_1,a_2, \dots$in $S^1$ such that the distribution of the centers of $I_n$ (for every $n$) are uniform and independent.

It can be shown that with probability $1$, $I = \cup_{n=1}^\infty I_n$ is a full measure subset of $S^1$. Is it true that "With probability $1$, $I = S^1$"? If this is not always true, does there exist a good characterization of the sequences $\{ a_n\}_{n=1}^{\infty}$ with this property?

Edit. A more precise question: "What happens in the special case $a_n = \frac1n$?"

Closed Semi-Riemannian manifolds with non-compact isometry group

Math Overflow Recent Questions - Tue, 08/14/2018 - 07:10

Are there general results about closed Semi-Riemannian manifolds which have a non-compact isometry group?

Background: By a theorem of Myers and Steenrod the isometry group of a Riemannian manifold is a lie group. The same is true for Semi-Riemannian manifolds where the idea of a possible proof (which also works for Riemannian manifolds) is to embed the isometry group into the bundle of orthonormal frames as a closed submanifold. In the case of Riemannian manifolds this immediately yields that the isometry group of a compact Riemannian manifold has to be compact. But this is in general false for Semi-Riemannian manifolds since the bundle of orthonormal frames is not necessarily compact in this case even if the manifold is. So I ask myself if there are general statements about the phenomenon of a closed Semi-Riemannian manifold with non-compact isometry group.

Edit: My goal is to understand a bit better what the intuition behind compactness / non-compactness of the isometry of a closed Semi-Riemannian manifold is. For example topological restrictions on the underlying space. An example of this kind would be the following theorem for Lorentzian manifolds:

Theorem: Let $(M,g)$ be a closed, simply connected Lorentz manifold and assume $M$ and $g$ are analytic. Then the isometry group of $M$ is compact.

Minimal covers in hypergraphs with finite edges

Math Overflow Recent Questions - Tue, 08/14/2018 - 07:05

Let $H=(V,E)$ be a hypergraph. We say that $C\subseteq E$ is a cover if $\bigcup C = E$. Let $H$ be a hypergraph with the following properties:

  1. $\bigcup E = V$,
  2. all members of $E$ are finite, and
  3. $d,e\in E$ with $d\subseteq e$ implies $d=e$.

Question. Does this imply that there is a minimal cover $C_0$ (that is, $C_0$ has the property that for all $c\in C_0$ the set $C_0\setminus \{c\}$ is no longer a cover)?

Classification of line bundles by second cohomology of a manifold

Math Overflow Recent Questions - Tue, 08/14/2018 - 05:27

In the book Loop spaces, Characteristic classes and geometric quantization by Brylinski I see following result when trying to motivate geometric description of $H^3(M,\mathbb{Z})$.

$H^2(M,\mathbb{Z})$ is the group of isomorphism classes of line bundles over $M$.

I guess they mean there is a natural isomorphism.

Can some one give a rough idea of what obvious second cohomology class we can think of given a line bundle over $M$ and what line bundle can we think of given an arbitrary second cohomology class.

Intuitive comments are also welcome.

I am familiar (not the proof details) with following result:

If $G$ is a group and $M$ is a $G$-module, then the $H^2(G, A)$ is in one-one correspondence with the set of equivalence classes of extensions $E$ of $M$ by $G$, in which the action of $G$ on $M$ induced by conjugation in $E$ is the same as the action defined by the $G$- module $M$.

I am expecting some intuitive explanation that looks similar to this.

Is $K[[x_1,x_2,\dots]]$ an $\mathfrak m$-adically complete ring?

Math Overflow Recent Questions - Tue, 08/14/2018 - 04:51

I asked this question on Mathematics Stackexchange (link), but got no answer.

Let $K$ be a field, let $x_1,x_2,\dots$ be indeterminates, and form the $K$-algebra $A:=K[[x_1,x_2,\dots]]$.

Recall that $A$ can be defined as the set of expressions of the form $\sum_ua_uu$, where $u$ runs over the set monomials in $x_1,x_2,\dots$, and each $a_u$ is in $K$, the addition and multiplication being the obvious ones.

Then $A$ is a local domain, its maximal ideal $\mathfrak m$ is defined by the condition $a_1=0$, and it seems natural to ask

Is $K[[x_1,x_2,\dots]]$ an $\mathfrak m$-adically complete ring?

I suspect that the answer is No, and that the series $\sum_{n\ge1}x_n^n$, which is clearly Cauchy, does not converge $\mathfrak m$-adically.

Proof of A Positive Definite Covariance Matrix

Math Overflow Recent Questions - Tue, 08/14/2018 - 01:34

I would like to prove such a matrix as a positive definite one,

$$ (\omega^T\Sigma\omega) \Sigma - \Sigma\omega \omega^T\Sigma $$ where $\Sigma$ is a positive definite symetric covariance matrix while $\omega$ is weight column vector (without constraints of positive elements)

I would apply an arbitrary $x$ belonging to $R^n$ to the following formula, $$ x^T((\omega^T\Sigma\omega) \Sigma - \Sigma\omega \omega^T\Sigma)x > 0 $$

But how could I go further to prove such a inequality above?


Can we specify the value of harmonic forms at a point?

Math Overflow Recent Questions - Tue, 08/14/2018 - 00:56

Let $M$ be a smooth $d$-dimensional oriented Riemannian manifold, and let $1 < k < d$ be fixed.

Let $p \in M$, and let $\alpha_p \in \bigwedge^k(T_pM)^*$.

Does there exist an open neighbourhood $U$ of $p$, and a closed and co-closed $k$-form $\omega \in \Omega^k(U)$ satisfying $\omega_p=\alpha_p$?

This question is equivalent to the following question:

Do closed and co-closed frames for $\bigwedge^k(T^*M)$ always exist locally?

Indeed, if we can specify the value of a form in a point, we can take a basis for $\bigwedge^k(T_pM)^*$, and so obtain forms which form a frame at $p$. Since "being a frame" is an open condition, we have a local frame. On the other hand, suppose that local closed and co-closed frames exist. Then, by choosing a linear combination with constant coefficients of that frame, we can realize any given value at $p$.

Comment: In general we cannot expect such a frame to be induced from coordinates. Indeed, when we specialize to even dimension $d$, and $k=\frac{d}{2}$, then, for a generic metric $g$, there are no coordinate systems where even one wedge $\mathrm{d}x^{i_1}\wedge\cdots\wedge\mathrm{d}x^{i_n} $ is co-closed.


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