Recent MathOverflow Questions

Embedding into $C\times [0,1]$

Math Overflow Recent Questions - Sun, 04/15/2018 - 11:54

Every totally disconnected separable metric space of dimension $n$ homeomorphically embeds into $C\times \mathbb R ^n$.

Is something like this known? $X$ is totally disconnected means that every point in $X$ is equal to the intersection of all clopen sets containing the point. $C$ is the Cantor set.

It is known that there are totally disconnected spaces of arbitrary dimension. But what about just $n=1$? How might we prove $X$ embeds into $C\times [0,1]$?

We know there is a one-to-one continuous mapping $f:X\to C$ such that the quasi-components of $X$ are the point inverses of $f$. It seems like the $C$-coordinates of the embedding should be determined by $f$, but we need another mapping $g:X\to [0,1]$ determine the second coordinates. At this moment it is unclear how to define $g$. The Erdos space has a norm $\|\;\|$, and the mapping $g$ does something like $$\frac{1}{1+\|x\|}.$$ So maybe just fix any $x'\in X$ and let $$g(x)=\frac{1}{1+d(x,x')}\;\;?$$

But I think this would require that all metric balls around $x'$ have zero-dimensional boundary. With Erdos space this occurs, but it is not clear it always happens with totally disconnected spaces.

A specific weak analog of the Baire category theorem for a 'continuously indexed' family of sets

Math Overflow Recent Questions - Sun, 04/15/2018 - 11:48

(This is a refinement/repost of a question I asked on Stack Exchange.)

Suppose that $C\subseteq [0,1]$ is a Cantor set (i.e. a totally disconnected closed perfect set) and $F\subseteq C\times [0,1]$ is a closed set. Let $F(A)=\{y|(\exists x\in A)((x,y)\in F)\}$. Suppose that the following hold:

  • For every clopen-in-$C$ set $Q$, $Q\subseteq F(Q)^\circ$ (where $A^\circ$ is the interior of $A$).

  • For every cover $\mathcal{Q}$ of $C$ by clopen-in-$C$ sets, the set $\{ F(Q)^\circ|Q\in\mathcal{Q} \}$ covers $[0,1]$.

Does it follow that there is some $x\in C$ such that $F(\{ x\})^\circ$ is non-empty? If it helps at all, in the specific case that I care about $F$ comes from some closed $G\subseteq [0,1]^2$ which is symmetric around the diagonal and contains it.

I'm not really satisfied with the title of the question, but I wasn't sure what to call it. The idea is that the Baire category theorem is equivalent to the statement that any countable closed cover of a complete metric space contains a set with non-empty interior.

The strongest analog (dropping the two bulleted assumptions and just requiring that $F(C)=[0,1]$) is false because there is a continuous surjection of Cantor space onto $[0,1]$ where each point has at most two preimages. In the other direction, if $F$ is the graph of a continuous function then it can't even satisfy the first requirement, since $F$ would need to map some point off of $C$, but then the image of some clopen neighborhood $Q$ of that point needs to be disjoint from $Q$.

Reduction of directed weighted graph by collapsing sub-graphs

Math Overflow Recent Questions - Sun, 04/15/2018 - 11:26

I'm trying to determine if there is some known problem definition or additional / better terminology that can describe this scenario.

I have a graph with these properties:

  • Directed weighted graph (theoretically may have cycles, but we can ignore that for simplicity)
  • The graph vertices can be split into two disjoint subsets: let's call them V+ and V-
  • V+ vertices have starting values, V- vertices do not
  • V+ vertices are really the only vertices whose values we ultimately care for, but V- vertices may retain value in intermediate iterations (not final though)
  • All V- vertices belong to some subgraph where:
    • The subgraph has a single source vertex in V+
    • The subgraph has multiples sink vertices in V+
    • (Another way to see it is the subgraph looks like a tree)

Goal: I want to determine an 'equivalent' graph containing only the V+ vertices, where edges previously connected to members of V- are replaced with equivalent edges and weights connected to members of V+ (resolvable by reducing all subgraphs, which are tree-like, into a set of simple weighted edges). I say 'equivalent' because I don't want to mislead people into thinking I want an equivalent graph. The vertices will get reduced to just V+ and new edges will be introduced.

Does this sound like some known / named problems, or are there better ways I can describe this situation (use of terminology, etc.)? I'm not proficient in graph theory, so I'm mostly trying to find the easiest way to describe my problem with known graph theory problems and terms.

Equivalent strictly convex norms in spaces of small density

Math Overflow Recent Questions - Sun, 04/15/2018 - 11:23

The question I am going to ask is attributed to Antonio Avilés, however I learnt that only having already asked it myself.

Can one construct in ZFC a Banach space of density character $\omega_1$ that does not have an equivalent strictly convex norm?

Maybe one may apply some kind of a Löwenheim–Skolem-type argument to a space that does not have a strictly convex norm?

Notes:

  • every separable space has an equivalent strictly convex norm;
  • the classical examples of spaces without a strictly convex norm include $\ell_\infty / c_0$ (see also this post) and $\ell_\infty(\Gamma)$ for any uncountable set $\Gamma$.

Does every non-locally compact metric space admit a violation of Lebesgue's theorem?

Math Overflow Recent Questions - Sun, 04/15/2018 - 09:22

From the results of Preiss and Tišer, it is known that many natural families of measures on Hilbert spaces violate the Lebesgue Density Theorem. Question: Does every non-locally compact metric space admit a measure (on the Borel $\sigma$-algebra) for which the Lebesgue-Besicovitch Density Theorem fails?

Number theory question from Homotopy groups of spheres

Math Overflow Recent Questions - Sun, 04/15/2018 - 08:55

Let $n$ be some integer.

Is it true that there exists odd prime $p$ such that $4n = (p-1) \cdot k$,

where $k$ is an integer coprime with $p$?

This question asked Roman Mikhailov. This is corresponds with Homotopy groups of spheres. Unfortunately I do not know the details.

Longest simple path through hypercube corners

Math Overflow Recent Questions - Sun, 04/15/2018 - 07:21

This is a variation on a previously answered question, Longest path through hypercube corners. Here I am seeking the longest simple (non-self-intersecting) path through the unit hypercube's vertices, composed of segments between vertices. (The segments need not follow the hypercube edges.) Without the simplicity requirement, MTyson showed that the longest path for the hypercube in $\mathbb{R}^d$ has length $$(2^{d-1}-1)\sqrt{d-1} + 2^{d-1}\sqrt{d} \;.$$ For $d=3$, this length is $3\sqrt{2}+4\sqrt{3}$. In contrast, the longest simple path has length $6\sqrt{2}+\sqrt{3}$, as it can only use the $\sqrt{3}$ long diagonal once, and uses all six face $\sqrt{2}$ diagonals:          
          Path: $(1,3,6,8,2,5,4,7)$. Length: $6\sqrt{2}+\sqrt{3} \approx 10.2$. Motivation: This is in some (very loose) sense a Euclidean version of the Snake-in-a-box problem.

$\mathbb{Z}_p[\zeta]$ is Local Ring

Math Overflow Recent Questions - Sun, 04/15/2018 - 06:47

Let consider the ring $\mathbb{Z}_p$ and $\zeta$ be a $p$-th root of unity. Especially $\zeta \not \in \mathbb{Z}_p$. Denote with $\Phi _p(x)$ the cyclotomical polynomial in $p$. Since $p$ is a prime we know that it has the shape $\Phi _p(x)= 1 + x +x^2 +... +x^{p-1}$. This gives rise for the quotient ring

$$ \mathbb{Z}_p[X]/\langle\Phi_p(x)\rangle \cong \mathbb{Z}_p[\zeta] = \mathbb{Z}_p \oplus \zeta \mathbb{Z}_p \oplus \dots \oplus \zeta^{p-2} \mathbb{Z}_p $$

which is obviously a free $\mathbb{Z}_p$-module of rank $p-1$. Denote $g=\zeta -1$.

My question is how to see that $\mathbb{Z}_p[\zeta]$ isa local ring with maximal ideal $g \mathbb{Z}_p[\zeta]$?

I tried to argue in following way: Obviously observation provides $\Phi _p(g +1) =0$ and the formula above gives $$ \Phi_p(x + 1) = p + \binom{p}{2}x + \binom{p}{3}x^2 + \dots + \binom{p}{p - 1} x^{p - 2} + x^{p - 1} $$.

In light of this I can conclude following inclusions:

$p \mathbb{Z}_p[\zeta] \subset g \mathbb{Z}_p[\zeta]$ and $\pi^{p-1} \mathbb{Z}_p[\zeta] \subset p \mathbb{Z}_p[\zeta]$ which imply $(g\mathbb{Z}_p[\zeta]) \cap\mathbb{Z}_p = p\mathbb{Z}_p$.

From here I'm stuck.

Historically, which came first: the Lie algebras or their classification?

Math Overflow Recent Questions - Sun, 04/15/2018 - 05:52

The classification of the complex simple Lie algebras by their Dynkin diagrams gives rise to five exceptional complex simple Lie algebras: $F_4, G_2, E_6, E_7$ and $E_8$.

I am trying to find out whether the classification was discovered first (attributed to Wilhelm Killing [1888-1890]), or whether some/all of the exceptional complex simple Lie algebras were discovered before the classification.

It's said here that Killing discovered $G_2$ in 1887, which would mean that $G_2$ at least came before the classification. I suspect that, since this discovery came only very shortly before his discovery of the classification, this means that the others were discovered as a consequence of the classification, but I'm hoping for clarification.

Coefficients in the sum $\sum_{k=0}^{n-1}\sum_{j=0}^{m}A_{j,m}(n-k)^jk^j=n^{2m+1}, \ m=1,2,....$

Math Overflow Recent Questions - Sun, 04/15/2018 - 04:35

Consider a sum $$\sum_{k=0}^{n-1}\sum_{j=0}^{m}A_{j,m}(n-k)^jk^j$$ which returns an odd power $n^{2m+1}$ of $n$, for $\ m=0,1,2,...$ given fixed $A_{0,m}, \ A_{1,m}, \ ..., \ A_{m,m}$. The coefficients $A_{0,m}, \ A_{1,m},....$ are solutions of system of equations (refer to .txt-file with mathematica codes for $m=1,2,...12$). Coefficients $A_{0,m}, \ A_{1,m},....$ are arranged to the PDF-table. For example, $$\sum_{k=0}^{n-1}\sum_{j=0}^{2}A_{j,2}(n-k)^jk^j=\sum_{k=0}^{n-1}30k^2(n-k)^2+1=n^5$$ Our coefficients are closely related to to coefficients $\beta_{mv}$ from C. Jordan Calculus of Finite Differences, pp. 448 - 450.

Question: Could the coefficients $A_{0,m}, \ A_{1,m}, \ ..., \ A_{m,m}$ be reached in a recurrent way using $\beta_{mv}$? This question also can be stated as: does there exist such a function $f(\beta_{mv})=A_{v,m}$?

Additional Question How exactly are our coefficients $A_{0,m}, \ A_{1,m}, \ ..., \ A_{m,m}$ connected with $\beta_{mv}$?

Distance from nonnegativity of some orthonormal vectors

Math Overflow Recent Questions - Sun, 04/15/2018 - 02:28

Suppose that $1 < k < n$. Does there exist a constant $\beta > 0$, such that for every $k$ orthonormal vectors $f_1,\ldots,f_k \in \mathbb R^n$, there exist $k$ orthonormal vectors with nonnegative elements, $x_1,\ldots,x_k\in \mathbb R_+^n$, such that

$$\sum_{i=1}^k \|x_i - f_i\|^2_2 \leq \beta \sum_{i=1}^k\|f_i^-\|_2^2$$

where $f_i^- := \max\{-f_i,0\}$ is the negative part of the vectors $f_i$?

In another way, I am interested in the estimation of the distance of a $n\times k$ dimensional matrix $F$ whose columns are orthonormal from the set of $n\times k$ matrices whose columns are nonnegative and orthonormal, i.e.,

$$ \mathrm{dist}(F;St_+(n,k)) $$

Where $St_+(n,k)$ is the set of $n\times k$ matrices whose columns are nonnegative and orthonormal. If we drop orthonormal condition and compute $\mathrm{dist}(F;\mathbb{R}^{n\times k }_+)$, we obtain $\|F^-\|$ as a lower bound for the above distance. In this term, my question is as follows: Is a multiple of $\|F^-\|$ an upper bound for the above distance, i.e.,

Is there a constant $c>0$, such that $$ \|F^-\| \leq \mathrm{dist}(F;St_+(n,k)) \leq c \|F^-\| $$ for every $n\times k$ dimentional matrix $F$ whose columns are orthonormal.

In the special case $k=1$, the above statement is true, with $c = 2$. I'm interested in the special case of small values of $k$, such as $k=2$. Experimentally, for $k>1$ and random matrices $F$ and by using Frobenius norm, I get an upper bound for $\mathrm{dist}(F; St_+(n,k))$ by alternating projection to nonnegative matrices and orthonormal matrices. I guess that the above statement is true for $c \approx 2$, $(\beta \approx 4)$.

The geometric meaning of the sign in the functional equation

Math Overflow Recent Questions - Sun, 04/15/2018 - 01:10

Let $X$ be a smooth projective variety of dimension $n=\dim X$ over a finite field $\mathbb{F}_q$. As is well known, its zeta function satisfies a functional equation of the form $$Z(X,q^{-n}T^{-1})=\pm q^{n E/2}t^{E}\zeta(X,T),$$ where $E$ is the Euler characteristic of $X$. The question is, what is the sign? The obvious guess is that it only depends on $E$ so the equation may be written in a slightly more precise form as $$Z(X,q^{-n}T^{-1})=(-q^{n/2}t)^{E}\zeta(X,T).$$ This is true for curves, projective spaces $\mathbb{P}^n$ and some other simple examples. But I do not know if it is true in general (and if not, what is really the meaning of the sign).

EDIT. As was pointed out in the comments, the sign was calculated by Deligne in "La Conjecture de Weil, I". (I should have read it before asking, but my French is awful and I never got beyond the first page.) The sign is positive for odd $n$ and is equal to $(-1)^N$ for even $n$, where $N$ is the multiplicity of the eigenvalue $q^{n/2}$ of the Frobenius action in the middle cohomology $H^n(X,\mathbb{Q}_l)$.

Now, we have $(-1)^E=(-1)^{N+N'}$ where $N'$ is the multiplicity of $-q^{n/2}$. (It is due to Poincare duality and the fact that all other eigenvalues come in conjugate pairs.) For the above "guess" to be wrong for some $X$ we need odd (in particular, nonzero) $N'$. I would appreciate an explicit example of this.

Anyway, the formula is true for a quadratic extension of the base field because of a simple identity $$Z(X/\mathbb{F}_q, T)\cdot Z(X/\mathbb{F}_q, -T)=Z(X/\mathbb{F}_{q^2}, T^2).$$

EDIT: The surface (suggested by David E Speyer) $x^2-y^2=z^2+w^2$ over $\mathbb{F}_3$ gives $N=N'=1$, so my guess was off. Also, it illustrates that the sign (unfortunately) has no "geometric meaning", it is an arithmetic invariant. Thanks to all for help.

A particular functor on the category of abelian groups?

Math Overflow Recent Questions - Sat, 04/14/2018 - 20:25

Is there a functor $F$ from the category of abelian groups to itself such that for every non trivial group $G$, $F(G)$ can not be embedded in $G$?

Edit: According to the comment by Prof. Goodwillie I change the question as follows:

Is there a functor $F$ on the category of infinite abelian groups which does not increase the cardinality of groups but for every infinite group $G$, the group $F(G)$ can not be embedded in $G$? By "Does not increase the cardinality" we mean $\text {Card}(F(G)) \leq \text{Card}( G)$

English translation of paper: Sur les variétés riemanniennes à groupe d'holonomie G2 ou Spin(7)

Math Overflow Recent Questions - Sat, 04/14/2018 - 19:10

I am interested in the history of $G_2$ manifolds and want to read this paper in english:

Sur les variétés riemanniennes à groupe d'holonomie G2 ou Spin(7)

Does anyone know where I can find a translation?

Who wins two player sudoku?

Math Overflow Recent Questions - Sat, 04/14/2018 - 16:18

Let's say players take turns placing numbers 1-9 on a sudoku board. They must not create an invalid position (meaning that you can not have the same number in within a row, column, or box region). The first player who can't move loses, and the other player wins.

Given a partially filled sudoku board, what is a way to evaluate the winner (or even better, the nimber) of the position (besides brute force)?

Additionally, has this perhaps been analyzed before?

(A natural generalization is to allow certain players to only play certain digits, in which case each position can be assigned a CGT game value.)

Question on Power Spectral Density and Wiener-Khinchin theorem

Math Overflow Recent Questions - Sat, 04/14/2018 - 13:29

I tried asking this question on stackexchange and I also extensively researched it online without results so I will ask here.

In my textbook the Wiener-Khinchin theorem is used to connect the auto-correlation definition of PSD with an "intuitive interpretation" of power spectral density for deterministic signals. It says:

$S_{xx}(\omega) = \lim_{T\rightarrow\infty} \frac{1}{2T}\mathbb E\left[|FT\{X(t)*I_{[-T,T]}\}|^2\right]$

But such a theorem assumes you can take the fourier transform of a(truncated) realization of the random process, which unless I am missing something - may not be the case. Such a realization may not be integrable.

So what is going on? Is there a different definition of integral which allows you to do this?

Longest path through hypercube corners

Math Overflow Recent Questions - Sat, 04/14/2018 - 12:03

Is the longest Hamiltonian path through the $2^d$ unit hypercube vertices known, where path length is measured by Euclidean distance in $\mathbb{R}^d$? The unit hypercube spans from $(0,0,\ldots,0)$ to $(1,1,\ldots,1)$.

For example, for the cube in $\mathbb{R}^3$, I believe the longest path has length $3\sqrt{2}+4\sqrt{3} \approx 11.17$, avoiding all edges of length $1$, and using all $4$ of the long diagonals and $3$ short diagonals:          
          Path: $(1,7,2,8,3,5,4,6)$. This likely has been studied, in which case pointers would be appreciated. If exact values are not known, bounds would be useful.

A new characterisation of hereditary algebras?

Math Overflow Recent Questions - Sat, 04/14/2018 - 09:34

Let $A$ be a quiver algebra with global dimension at most two (or more generally finite global dimension) and $A^e=A^{op} \otimes_K A$ be its enveloping algebra.

Guess:Is $A$ hereditary if and only if the socle of the regular module as a bimodule $soc_{A^e}(A)$ is projective?

I think I can prove one direction in case $A=kQ$ is a path algebra, which would prove this direction for example for general finite dimensional algebra over an algebraically closed field.

Namely because there are no relations, we should have that $soc_{A^e}(A)$ is the direct sum of all simple projective modules of the form $p A^e$ for $p$ a maximal path in $A$ (meaning $p \neq 0$ and $xp=0=px$ for all arrows x).

At least for quiver algebras having the property that between any two points there is at most one path the bimodule $soc_{A^e}(A)$ can be described as the module of maximal paths in that algebra. Im not sure if this is a good general description for algebras with more complicated relations...

I tested the guess mostly for Nakayama algebras and there the guess has a positive answer for all algebras with at most 8 simples modules. But I have no good idea for proving the other direction.

Question :The projective dimension of $soc_{A^e}(A)$ is bounded by one iff $A$ is a tilted (or quasi-tilted?) algebra? (I think it is very surprising that this holds even for Nakayama algebras with at most 8 simples)

edit: Jan Geuenich found an example of an algebra with infinite global dimension that gives negative answers to the posed question/guess. I restrict therefore to algebras of finite global dimension since I think there might still be some positive results with possibly stronger assumptions.

Minimum number of operations necessary to arrive at any configuration

Math Overflow Recent Questions - Sat, 04/14/2018 - 04:05

Let $k \geq 2$ and $N_1, N_2, ..., N_k$ be positive integers.
Let $S=\{(a_1,a_2,...,a_k) \in \mathbb{Z}^k:1 \leq a_i \leq N_i\}$ and $A=\{1,2,...,\prod_{i=1}^{k} N_{i}\}$.

Given a bijective map $f:S \to A$ we define a change as the operation of choosing any two $s_1,s_2 \in S$, such that $s_1$ and $s_2$ differ in only one coordinate, and then we interchange the values of $f(s_1)$ and $f(s_2)$.

Consider the set $F$ whose elements are all bijective maps $f:S \to A$.
Is it possible to find the exact value of $M$ (as a function of $k$ and $N_1,N_2,...,N_k$) such that for any $f,g \in F$ after at most $M$ changes beginning from $f$ the result is $g$?

Does anyone know if this result holds for specific values of $k$, such $k=2$ or $k=3$?

Any help or reference would be appreciated.

matrix-valued differential forms on complex manifolds

Math Overflow Recent Questions - Fri, 04/13/2018 - 15:57

I'm pretty clear in my understanding of scalar-valued differential $(p, q)$-forms (resp. holomorphic $(p,0)$-forms) on a complex manifold $M$ and the related Hodge theory. What I'm not sure about is whether there is a matrix-valued analogue of such differential forms on a complex manifold in the math literature. If there is such a thing, what are some good references that I may find useful?

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