(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 $Q$.

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.

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 $\Gamma$.

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?

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.

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.

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.

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.

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 $\beta_{mv}$?

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 $c \approx 2$, $(\beta \approx 4)$.

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.

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 $\text {Card}(F(G)) \leq \text{Card}( G)$

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.)

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?

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.

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?

I am looking for a general forumla to count prime numbers on which the Meissel and Lehmer formula are based:

$$\pi(x)=\phi(x,a)+a-1-\sum\limits_{k=2}^{\lfloor log_2(x) \rfloor}{P_k(x,a)}$$

Wiki - prime counting - Meissel Lehmer

More precisely, I am looking for the detailed description of the $P_k$ for $k>3$.

$P_k(x,a)$ counts the numbers$\leqslant x$ with exactly $k$ prime factors all greater than $p_a$ ($a^{th}$ prime), but in the full general formula, this last condition is not necessary.

The Meissel formula stops at $P_2$ (and still uses some $\phi$/Legendre parts)

Wolfram - Meissel

The Lehmer formula stops at $P_3$ (and still uses some $\phi$/Legendre parts)

Wolfram - Lehmer

I don't find anything about the general formula (using all the $P_k$ terms). Is there any paper on it? Why stop at $P_3$? is it a performance issue?

Lehmer vaguely talk about it in his 1959 paper

On the exact number of primes less than a given limit

Deleglise talks about performances here

Prime counting Meissel, Lehmer, ...

Thanks

Edit: by "a general formula on which the Meissel and Lehmer formula are based", I meant the one they tend to (with all $P_k$), not the one they started from (Legendre, with no $P_k$). Sorry if it was not clear.

**The setting:**

Let $B_t$ be standard scalar Brownian motion.

I am considering the follow SDE

$$dY(t) = TY(t) \ dt + S Y(t) \ dB_t, \text{ and }Y(0)=Y_0 \in \mathbb R^n.$$

for square (time-independent) matrices $T \in \mathbb R^{n \times n}$ and $S\in \mathbb R^{n \times n}.$

The solution is given in terms of a stochastic flow $\Phi_1: \mathbb R^n \rightarrow \mathbb R^n$ (I completely subpress the randomness in the notation, here) as

$$Y(t):=\Phi_1(t,0)Y_0.$$

Now, consider the following equation where we conjugate $T$ and $S$ and study

$$dZ(t) = T^*Z(t) \ dt + S^* Z(t) \ dB_t, \text{ and }Y(0)=Y_0 \in \mathbb R^n.$$

This one has also a solution $$Z(t):=\Phi_2(t,0)Y_0.$$

**My question:** Is it true that $Z(t)$ has the same law as $\Phi_1(t,0)^* Y_0$?
It almost looks like nothing, but I fail to see why this holds.

Any comment/remark/idea is highly appreciated.

Let $K$ be a field of characteristic not equal to $2$. Let $\text{ad} : \text{GL}_2(K) \to \text{GL}_3(K)$ be the adjoint representation, obtained by $\text{GL}_2(K)$ acting on $2 \times 2$ matrices with trace $0$ by conjugation. Suppose $\rho_1, \rho_2 : G \to \text{GL}_2(K)$ are representations of a group $G$ such that $\text{ad}\rho_1 \cong \text{ad}\rho_2$ over $K$. Is there a character $\eta : G \to K^\times$ such that $\rho_1 \cong \rho_2 \otimes \eta$ over $K$? That is, if $x \in \text{GL}_3(K)$ such that $\text{ad}\rho_1 = x(\text{ad}\rho_2) x^{-1}$, can one produce $y \in \text{GL}_2(K)$ and $\eta$ as above such that $\rho_1 = y \eta (\otimes \rho_2) y^{-1}$?

I am most interested in the case when $K$ is a finite field. In that case, I believe one can treat the case when the projective image of $\rho_i$ is isomorphic to $\text{PSL}_2(K)$ or $\text{PGL}_2(K)$ fairly easily by using that those groups have automorphism group $\text{P}\Gamma\text{L}_2(K)$. But I'd like a proof that does not break into cases based on the subgroups of $\text{PGL}_2(K)$.

Let $L$ be a Lévy process and define $M_t:=L_t-t\mathbb E(L(1)),$ then $M$ is a centred martingale.

Now consider the stochastic integral for $f$ a continuous process

$$\int_0^t f(t-s) \ dM_s,$$

is it then true that this is equal $\int_0^t f(s)\ dM_s$ almost surely?

I believe the answer should be yes, because of a linear change of variables is compatible with the iid increments.