Recent MathOverflow Questions

Shattering/covering finite trees, and a simple looking inequality

Math Overflow Recent Questions - Tue, 06/26/2018 - 16:52

Consider the tree $T$ with the set of its maximal elements (denoted $[T]$) equal to $\prod_{m\leq n} X_m$ for some finite sets $X_0,..., X_n$. Let $p(T)=\{b: b\in \prod_{i\leq m \leq j } X_{m}\text{ for some }i<j\leq n \}$.

$\underline{Definition:}$ We say that $S\subseteq p(T)$ shatters $T$ if $$[T]=\{x\in[T]:(\exists b\in S)\ x\restriction\text{dom}(b)=b\}.$$

Is the following combinatorial inequality true:

$\underline{Conjecture}:$ If $S$ shatters $T$ then $$1\leq\sum_{b\in S}\prod_{m\in\text{dom}(b)}\frac{1}{|X_m|}?$$

Solving an equation with a definite integral

Math Overflow Recent Questions - Tue, 06/26/2018 - 16:46

Are there any techniques, either analytical or numerical, to solve an equation of the following form: $$f(x)=\int_{-\infty}^{\infty}g(x,y)h(y)dy$$ where $f(x)$ and $g(x,y)$ are known functions, and we are trying to solve for $h(y)$?

Is Videla's solution of Hilbert's tenth problem for rational functions over field of characteristic 2 wrong?

Math Overflow Recent Questions - Tue, 06/26/2018 - 16:25

The paper in question.

Quick introduction to the problem: suppose that $F$ is a finite field of characteristic 2 (for purposes of this post $F = \mathbb{F}_2$ will suffice) and let $F[t]$ and $F(t)$ be its ring of polynomials (polynomials over a finite field are clearly finite enumerable objects) and field of rational functions respectively. Let $P_1, P_2, \ldots P_m$ be polynomials of variables $x_1, \ldots, x_n$ with coefficients in $F[t]$.

Hilbert's tenth problem for $F(t)$ asks whether existence of $x_1, \ldots, x_n \in F(t)$ such that $P_1 (x_1, x_2, \ldots, x_n) = P_2 (x_1, x_2, \ldots, x_n) = \ldots = P_m (x_1, x_2, \ldots, x_n) = 0$ is decidable (clearly, $n, m$ and $P_i$ are finite enumerable objects, so the problem is well-posed). Understandably, this problem is often called the diophantine problem for $F(t)$ and equations $P_i (x_1, x_2, \ldots, x_n) = 0$ are called diophantine.

Paper in question supposedly proves that answer is indeed "undecidable" by reducing a undecidable problem in some first-order logic (that is not relevant to the question, I guess) to the diophantine problem over $F(t)$.

Well, preliminaries end here. I believe that lemma 2.1 (and its proof, of course) is incorrect. I will quote its statement here:

Let $x \in F(t)$. Then $x \in \{ t^{2^s} : s \geqslant 1 \} \Leftrightarrow \exists u, v, w, s \in F(t)$ such that: \begin{equation} x + t = u^2 + u \end{equation} \begin{equation} u = w^2 + t \end{equation} \begin{equation} x^{-1} + t^{-1} = v^2 + v \end{equation} \begin{equation} v = s^2 + t^{-1} \end{equation}

Well, following values of $u, v, x, s, w$ seem to be a counterexample to this lemma (clearly, $x$ is not a polynomial, let alone $t^{2^k}$ for some $k \geqslant 0$):

\begin{equation} u = \frac{t^1+t^5+t^6+t^7}{1+t^4+t^6} \end{equation} \begin{equation} v = \frac{1+t^1+t^2+t^6}{t^1+t^3+t^7} \end{equation} \begin{equation} x = \frac{t^2+t^6+t^{14}}{1+t^8+t^{12}} \end{equation} \begin{equation} s = \frac{1}{1+t^1+t^3} \end{equation} \begin{equation} w = \frac{t^3}{1+t^2+t^3} \end{equation}

Moreover, conditions $2$ and $4$ from the lemma look a bit fishy: $x + t = u^2 + u, u = w^2 + t \Leftrightarrow x + t = (w^2 + t)^2 + (w^2 + t) = (w^4 + t^2) + (w^2 + t) = w^4 + w^2 + t^2 + t \Leftrightarrow x = (w^2 + w + t)^2$ (here I use that $(a + b)^2 = a^2 + b^2$ in characteristic 2). In the same way $x^{-1} = (s^2 + s + t^{-1})^2$. Basically, they are rewritten in the form $x = y^2, y = (w^2 + w + t), y^{-1} = (s^2 + s + t^{-1})$ or $x = y^2, y + t = w^2 + w, y^{-1} + t^{-1} = s^2 + s$ (here I use that $a^2 = b^2 \Leftrightarrow a = b$ in $F(t)$). For $y$ we get just conditions $1$ and $3$ from the lemma. As it is said in paper itself (an explicit counterexample is given, on which the counterexample above is based) it is possible to choose $y \notin \{ t^{2^k} | k \geqslant 0 \}$ that it satisfies conditions $1$ and $3$. Then $x = y^2$ also does not lie in $\{t^{2^k} | k \geqslant 0 \}$.

And finally, an error in the proof seems to be in the following line (this requires actually reading the proof though): "if $q \not | ~n$ then there exists nonconstant polynomial $p$ such that $p | q, p | b, p \not | ~n$". This does not sound correct, because if, for example $b | n$ (I can't see why this can't happen), there is no such $p$. Moreover $q \not | ~n$ does not imply that a situation like $b = n, q | n^3$ can't happen ($q$ itself is not necessarily a prime).

You may wonder why I am asking this here. The reason is that I already talked about this issue with my scientific adviser and have written a letter to the author ($1,5$ months ago), who have not responded yet and I am not expecting response at all. So I am asking: are my doubts true? And what should I do about this situation?

Monads inside the Kleisli 2-category of another monad

Math Overflow Recent Questions - Tue, 06/26/2018 - 16:09

$\require{AMScd}$Let $\cal C$ be a 2-category. Every 2-monad $T$ on $\cal C$ induces a Kleisli 2-category where

  • 0-cells are the objects of $\cal C$
  • a 1-cell $X\looparrowright Y$ consists of a 1-cell $X\to TY$ in $\cal C$
  • a 2-cell $\alpha : f\Rightarrow g$ is a 2-cell between the 1-cells $f,g : X\to TY$.

Composition of 1-cells is defined as in 1-dimensional Kleisli categories: $$ g\bullet f := \mu_Z\circ Tg\circ f $$ for $f :X\to TY, g : Y\to TZ$ (and identity of $X$ is the unit $\eta_X$ of the monad).

Vertical composition of 2-cells is obvious to define; horizontal composition takes into account how 1-cells compose, but it's equally rather easy and employs horizontal composition $\boxminus$ of 2-cells in $\cal C$: $$ \beta \boxminus_\text{Kleisli}\alpha := \mu_Z * (T\beta \boxminus \alpha) $$ for every pair of 2-cells $\alpha : u \Rightarrow v : A\looparrowright B$ and $ \beta : s\Rightarrow t : B\looparrowright C$.


I would like to know what is a monad in this 2-category, and if it is something already known under another name.

Such a monad amounts to

  • a 0-cell $A$ in $Kl(T)$;
  • an endo-1-cell $s : A\looparrowright A$, i.e. a 1-cell $s : A\to TA$ in $\cal C$
  • a pair of 2-cells $\sigma : s\bullet s \Rightarrow s$ and $\nu : \eta_A\Rightarrow s$ such that associativity[1] and unit[2.1, 2.2] axioms hold: $$ \begin{CD} \mu_A\cdot Ts\cdot \mu_A\cdot Ts\cdot s @>\mu_A * T\sigma * s>> \mu_A \cdot Ts \cdot s\\ @V\mu_ATs * \sigma VV [1] @VV\sigma V \\ \mu_A \cdot Ts\cdot s @>>\sigma > s \end{CD} \qquad \begin{CD} \mu_A\cdot T\eta_A\cdot s @>\mu_A * T\nu * s >> \mu_A \cdot Ts\cdot s @<\mu_A Ts * \nu<< \mu_A \cdot Ts\cdot \eta_A\\ @| [2.1] @V\sigma VV [2.2] @|\\ s @= s @= s \end{CD} $$ I have no idea what such a thing could be.

Local Langlands Correspondence for unramified principal series representations

Math Overflow Recent Questions - Tue, 06/26/2018 - 15:51

Let $G$ be a connected, reductive group over a $p$-adic field $k$. Assume $G$ is quasi-split with Borel subgroup $B = TU$. Consider those irreducible admissible representations $\pi$ of $G(k)$ which are isomorphic to a subrepresentation of the unramified principal series representation $I(\chi)$ for some unramified character $\chi$ of $T(k)$. Equivalently, $\pi$ has an Iwahori fixed vector.

The version of local Langlands correspondence conjectured by A. Borel in Automorphic L-Functions in the Corvallis proceedings would attach to $\pi$ a homomorphism $\varphi': W_k' \rightarrow \space ^LM$, where $W_k'$ is the Weil-Deligne group of $k$. Is this correspondence known for irreducible subrepresentations of an unramified principal series?

What if we just consider the unramified character $\chi$ itself? Is it known how to attach a homomorphism $\varphi: W_k \rightarrow \space ^LT$ to $\chi$? When $T$ is split, this is just local class field theory.

Is the unit ball in $L_p$, $1<p<2$, contained in a "compact perturbation" of the unit ball in $L_2$?

Math Overflow Recent Questions - Tue, 06/26/2018 - 15:02

For each pair of functions $x,y\in L_2[0,1]$ let us denote by $x\cdot y$ their pointwise product $$ (x\cdot y)(t)=x(t)\cdot y(t),\quad t\in [0,1]. $$ It belongs to $L_1[0,1]$ due to the Cauchy-Bunyakovsky inequality: $$ x,y\in L_2[0,1]\quad\Longrightarrow\quad x\cdot y\in L_1[0,1]. $$

And for each pair of sets $A,B\subseteq L_2[0,1]$ by $A\cdot B$ we denote the corresponding "element-wise product", $$ A\cdot B=\{x\cdot y; \ x\in A, \ y\in B\}, $$ which is contained in $L_1[0,1]$: $$ A,B\subseteq L_2[0,1]\quad\Longrightarrow\quad A\cdot B\subseteq L_1[0,1]. $$ Let us denote by $\overline{\operatorname{absconv}}(A\cdot B)$ the closed absolutely convex hull of $A\cdot B$ in $L_1[0,1]$.

And for each $p>1$ let $B_p$ be the unit ball in $L_p[0,1]$.

I wonder in which case $B_p$ is contained in a set of the form $\overline{\operatorname{absconv}}(K\cdot B_2)$ where $K$ is a compact set in $L_2[0,1]$: $$ B_p\subseteq \overline{\operatorname{absconv}}(K\cdot B_2). $$ For $p\ge 2$ this is trivially true, since in this case we can take $K=\{1\}$, the set consisting of just one function, the constant identity ($1(t)=1$, $t\in[0,1]$): $$ B_p\subseteq B_2\subseteq \overline{\operatorname{absconv}}B_2= \overline{\operatorname{absconv}}(\{1\}\cdot B_2) $$

But for $1<p<2$ this seems to be not true:

If $1<p<2$ then there is no a compact set $K\subseteq L_2[0,1]$ such that $$ B_p\subseteq \overline{\operatorname{absconv}}(K\cdot B_2). $$

Am I right?

Quasi-concavity of $f(x)=\frac{1}{x+1} \int_0^x \log \left(1+\frac{1}{x+1+t} \right)~dt$

Math Overflow Recent Questions - Tue, 06/26/2018 - 12:55

I want to prove that function \begin{equation} f(x)=\frac{1}{x+1} \int\limits_0^x \log \left(1+\frac{1}{x+1+t} \right)~dt \end{equation}

is quasi-concave. One approach is to obtain the closed form of the integral (provided below) and then prove that the result is quasi-concave. I tried this but it seems to be difficult. Do you have any idea how to prove the quasi-concavity?

\begin{align} f(x)&=\frac{1}{x+1} \int\limits_0^x \log \left(1+\frac{1}{x+1+t} \right)~dt\\ &= \frac{1}{x+1} \left[ \log(2x + 2) + (2x+1)\log \left(1+\frac{1}{2x+1} \right) -\log(x+2) - (x+1)\log \left( 1+\frac{1}{x+1} \right) \right] \end{align}

On the Diophantine equation $x^{4}+y^{4}=z^p$

Math Overflow Recent Questions - Tue, 06/26/2018 - 11:56

Do there exist integers $x,y,z$ with $xyz\neq 0$, such that

$$x^4 + y^4 = z^p$$

where $p\geq 5$ is some prime ?

If yes, are there infinitely many of them ? And if there exists infinitely many of them, what are their parametrizations ?

Gaussian free field limiting distribution of additive Stochastic heat eqn bounded domain

Math Overflow Recent Questions - Tue, 06/26/2018 - 10:09

Hairer in his spdes notes on pg.6, says that GFF is the stationary solution of $u_{t}(z)=\Delta u(z)+\xi(z,t)$, where $\xi$ is the space-time white noise $$\xi(x,t)=\sum \sqrt{\lambda_{k}} B_{k}(t)e_{k}(x)$$ for iid Brownian motions $B_{k}$ and L2 basis $e_{k}$. So that means we take $$u(x,t)=\Delta^{-1}\xi(x,t). $$ and that at each fixed time it is a GFF $$h(x)=\sum \frac{1}{\sqrt{\lambda_{k}}} B_{k}(t)e_{k}(x).$$

Q: Does SHE in a bounded domain with zero boundary converge to a steady state? So if $u(x,0)=GFF$ or $=0$, do we have some asymptotic results? Any rates? What is the largest function space over which it makes sense to take limits?

For bounded domains D the formula is $$u(x,t)=e^{t\Delta }u(x,0)+\int_{[0,t]}\int_{D}H(t-s,x-y)dW(s,y),$$

where the second term is a Wiener integral with Heat kernel H for domain D. We will compute the covariance for the bounded domain for zero initial data and $\xi(x,t):=\sum B_{k}(t) e_{k}(x)$. We have $$u(x,t)=\int_{[0,t]}\int_{D}H(t-s,x-y)dW(s,y)$$

$$=\sum_{k\geq 1}\int_{0}^{t}e^{-\lambda_{k}(t-s)}dB_{k}(s) e_{k}(x).$$

Therefore, by Ito isometry we indeed obtain the Green function:

$$E[u(t,x)u(t,y)]=\frac{1}{2}\sum_{k\geq 1}\frac{e_{k}(x)e_{k}(y)}{\lambda_{k}}(1-e^{-\lambda_{k}t})\to G(x,y).$$

Function space weak limit We have that the SHE $u\in C^{-\varepsilon}(\mathbb{R}_{+},D)$ and the GFF $h\in H^{-\varepsilon}(D)$ for all $\varepsilon>0$. By Morrey's inequality $$ H^{2}(D)\subset C^{0,\varepsilon}(D)=C^{\varepsilon}(D), $$ where $H^{2}(D)$ is the Sobolev space where second weak derivatives are also square integrable and so we also have $H^{2}(D)\subset H^{\varepsilon}(D)$ for $\varepsilon\leq 2$. So we will work with functions $f\in H^{2}(D)$.

Now I am trying to see if we can modify theorem 2.3 to obtain $$\left \langle u(\cdot,t),f \right\rangle_{H^{2}} \to \left \langle h(\cdot),f \right\rangle_{H^{2}}.$$ That theorem was proved for tempered distributions over $\mathbb{R}^{d}$ (not bounded domains with good enough boundaries).

Q: Any clean results for checking $u\stackrel{weakly}{\to} h$ i.e. $$\left \langle u(\cdot,t),f \right\rangle_{H} \to \left \langle h(\cdot),f \right\rangle_{H}?$$

For the infinite domain, from the same notes, if we compute and expand the covariance for SHE we have


and so even though for each fixed we have a GFF like object, for $t\to \infty$ the covariance becomes infinite. This is fine because for the whole plane we don't even have the usual GFF (meaning the one whose covariance is the Green function but what is called the "Whole-plane GFF").

The closest thing I found in the literature is about stochastic quantization and a special case from there gives that $$u_{t}=\Delta u-u+\xi(x,t)$$ has the GFF as the limiting distribution.

norm inequalities

Math Overflow Recent Questions - Tue, 06/26/2018 - 09:32

Let $p>2$. I'd like to know the best possible lower and upper bound for $\|x\|_p$ given that $x\in R^n$ and $\|x\|_1$, $\|x\|_2$, and $\|x\|_\infty$ have fixed values.

It is well-known that $$\|x\|_\infty\le \|x\|_p\le \|x\|_2^{2/p}\|x\|_\infty^{1-2/p}~~~ \Big[\le \|x\|_2\le \|x\|_1\Big].$$ But these bounds cannot be sharp when an arbitrary value of $\|x\|_1$ is specified.

In case no closed form solution is known, I'd also appreciate pointers to explicit suboptimal bounds that are stronger than the stated ones. For example, I conjecture that a bound of the form $$ \|x\|_p\le C_p\|x\|_2\Big(\frac{\|x\|_\infty}{\|x\|_1}\Big)^r$$ should hold with $r=(p-2)/(2p-2)$ and a constant $C_p$ depending on $p$ only, but I don't have a proof of such an inequality.

$\pi(x.y) \ge \pi(x).\pi(y)$ holds for $x, y \ge 8$

Math Overflow Recent Questions - Tue, 06/26/2018 - 07:33

I am looking for a proof, or a reference, or a remark or a comment with the conjecture as follows:

Let $\pi(x)$ is Prime-counting function of $x$. For $x$, $y$ be two positive integer numbers for $x, y \ge 8$ then

$$\pi(x.y) \ge \pi(x).\pi(y)$$

Noted: I compute the conjecture is true with $8 \le x, y \le 10^9$

PS: This conjecture inspired from Second Hardy–Littlewood conjecture, for $x, y \ge 2$ then

$$\pi(x+y) \le \pi(x)+\pi(y)$$

How to define transfinite derivatives of a function?

Math Overflow Recent Questions - Tue, 06/26/2018 - 05:38

There are all manners of theories generalizing the notion of derivative. Amongst them is the fractional calculus, a rich theory which gives a sense to the derivation and integration of non-integer (i.e. rational, real, complex) order, that are the real/complex number powers of the differentiation ($D$) and integration ($J$) operators:



Along these lines, one may think about transfinite iteration of $D$ and $J$ as well. While $D^{(n)}$ and $J^{(n)}$ are quite well-defined and well-behaved operators for the natural number $n$, I haven't found any convenient definition of $D^{(\alpha)}$ or $J^{(\alpha)}$ for the ordinal $\alpha\geq \omega$ in the literature if there is any.

Due to the similarity between natural and ordinal numbers, it seems the only difficulty is to formulate a definition of the differentiation operator in the limit steps like $D^{\omega}$ or $D^{\omega+\omega}$. One straightforward (but not necessarily well-defined, natural or fruitful) way to do so is to think about $D^{\omega}$ as a functional limit of $D^{n}$s in a certain function space. Though, I am not sure if it is the most clever approach. Anyway, it is somehow "natural" to expect that any $D^{\omega}$ operator demonstrates certain properties like: $D^{\omega}x^{n}=0$ for every $n\in \omega$.

Also, the spaces of smooth and analytic functions, $C^{\infty}$ and $C^{\omega}$, don't seem to capture the essence of the very notion of the $\omega$-the derivative of a function, particularly because they don't suggest a clear way of calculating transfinite successor differentiation operators, $D^{\omega+1}$, $D^{\omega+2}$, $D^{\omega+3}$, etc.

Question. Is there any paper in which transfinite derivatives of (real/complex) functions are defined/used? If so, what sort of applications do they have?

Update. Due to the answer that Andrés mentioned in his comment, it turned out that defining $D^{\omega}$ operator as the limit of $D^{n}$s gives rise to a trivial notion. So maybe a more direct approach is needed here.

Primes approximated by eigenvalues?

Math Overflow Recent Questions - Tue, 06/26/2018 - 04:25

Let the matrix $T$ be defined by:

$$\displaystyle T(n,k) = -\varphi^{-1}(\operatorname{GCD}(n,k))$$

where $\varphi^{-1}$ is the Dirichlet inverse of the Euler totient function.

$$\varphi^{-1}(n) = \sum\limits_{d|n}\mu(d)d$$

$$\varphi^{-1}(n) = +1, -1, -2, -1, -4, +2, -6,...$$

The matrix $T$ starts:

$$\displaystyle T = -\left( \begin{array}{ccccccc} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array} \right)$$

Does the largest eigenvalue of the matrix $T$ approximate the previous prime number sequence?

The largest eigenvalues rounded start:

{-1, 1, 3, 3, 5, 5, 7, 7, 7, 8, 11, 11, 13, 13, 13, 13, 17, 17, 19, 19, 19, 19, 23, 23, 23, 23, 23, 23, 29, 29, 31, 31, 31, 31, 31, 31, 37, 37, 37, 37, 41, 41}

compared to the previous prime number sequence:

{-2, 2, 3, 3, 5, 5, 7, 7, 7, 7, 11, 11, 13, 13, 13, 13, 17, 17, 19, 19, 19, 19, 23, 23, 23, 23, 23, 23, 29, 29, 31, 31, 31, 31, 31, 31, 37, 37, 37, 37, 41, 41}

The agreement appears to hold for $n>10$.

Link to original question at Mathematics Stack Exchange asked 6 years ago.

Associated Mathematica program:

(*start*) nn = 42; A = Table[Table[If[Mod[n, k] == 0, 1, 0], {k, 1, nn}], {n, 1, nn}]; B = Table[ Table[If[Mod[k, n] == 0, MoebiusMu[n]*n, 0], {k, 1, nn}], {n, 1, nn}]; T = A.B; Round[Table[ Max[Eigenvalues[ Table[Table[N[-T[[n, k]]], {k, 1, nnn}], {n, 1, nnn}]]], {nnn, 1, nn}]] Table[NextPrime[i, -1], {i, 2, nn+1}] (*end*)

$\pi((n+1)^2)-\pi(n^2) \le \pi(n)$ for all $n \ge 370$?

Math Overflow Recent Questions - Mon, 06/25/2018 - 20:11

There are some conjectures of the form: There always exist at least $X$ prime numbers between $A$ and $B$. Examples:

If we denote by $\pi(x)$ the prime-counting function we can rewrite the above conjectures in the following form:

  • Bertrand's postulate: $\pi(2n)-\pi(n) \ge 1$ for $n>1$
  • Legendre's conjecture: $\pi(n+1)^2)-\pi(n^2) \ge 1$
  • Brocard's conjecture: $\pi(p_{n+1}^2)-\pi(p_{n}^2) \ge 4$
  • Oppermann's conjecture: $\pi(n^2)-\pi(n(n-1)) \ge 1$

    • I computed and saw that $f(n) = \pi(n^2)+\pi(n)+2-\pi((n+1)^2)$ is increasing when $n$ increasing and $f(n)\ge 0$ for all $n=1, 2, \dots, 18700$ (equivalent to $n^2=1, 4, 25 \cdots , 3.5\times 10^8)$.

    • Graph of $(n,f(n))$ where $f(n) = \pi(n^2)+\pi(n)-\pi((n+1)^2); \; 370 \le n \le 1.1\times10^4$

  • So I proposed two conjecture as follows:

Conjecture 1: For every positive integer $n$, the number of primes between $n^2$ and $(n + 1)^2$ is less than the number of primes between $1$ and $n$ add $2$:

$$\pi((n+1)^2)-\pi(n^2) \le \pi(n)+2.$$

Conjecture 2: For every positive integer $n$ greater than $369$, the number of primes between $n^2$ and $(n + 1)^2$ is less than the number of primes between $1$ and $n$:

$$\pi((n+1)^2)-\pi(n^2) \le \pi(n).$$

Could you give a remark, comment, reference, or proof?

Noting that if the conjecture is true, it is stronger than a special case of the Second Hardy–Littlewood conjecture but this conjecture is not contradictory with the K-Tuple conjecture.

Every closed and convex subset of a uniformly convex metric space is Chebyshev?

Math Overflow Recent Questions - Mon, 06/25/2018 - 11:57

I came across the statement ``Every closed and convex subset of a uniformly convex b-metric space is Chebyshev'' in [1]. Here, the term `convex' is in the sense of Takahashi. I tried looking up for the proof of the same without success. I would be grateful if someone can direct me to the proof. Thanks.

[1] H. Fukhar-ud-din, One step iterative scheme for a pair of nonexpansive mappings in a convex metric space, Hacet. J. Math. Stat., Vol. 44 (2015), 1023 – 1031.

Instanton configurations of self-dual and anti-self-dual instantons interplay

Math Overflow Recent Questions - Sat, 06/23/2018 - 21:41

Yang-Mills gauge theory is given by the action $$ S_\text{YM}[A] = \int_M\mathrm{Tr}_\mathfrak{g}(F\wedge \star F)$$ whose Euler-Lagrange equations are the classical equations of motion. The classical solutions are stationary points of this functional under variation. The $M$ is the spacetime base manifold. The $\mathrm{Tr}_\mathfrak{g}$ is the trace over the Lie algebra $\mathfrak{g}$ of the principle compact Lie group $G$-bundle. The $F$ is the field strength of the connection $A$.

Equations of motion are: $$ dF=0, \quad d\star F=0 $$

  1. It is known that one way to satisfy the classical equations of motion are equivalent to $$\star F = \pm F,$$ i.e. the classical solutions of instanton equation to the equation of motion.

  2. It is said that the self-dual $F^+$ and an anti-self-dual $F^-$ are orthogonal to each other, say $$\star F_{\pm} = \pm F_{\pm},$$ w.r.t. the inner product defined by $ (f_1,f_2) = \int f_1\wedge\star f_2$.

Altogether this gives $S_\text{YM}[A] = \int \mathrm{Tr}_\mathfrak{g}(F^+\wedge \star F^+) + \int \mathrm{Tr}_\mathfrak{g}(F^-\wedge \star F^-).$ The second Chern class $$C_2(A) :=\frac{1}{8 \pi^2} \int\mathrm{Tr}_\mathfrak{g}(F\wedge F),$$ (I hope I get the normalization correct.) Notice that $$S_\text{YM}[A] \geq 8 \pi^2 \lvert C_2(A)\rvert ,$$ is locally minimized when the equality holds.

The equality holds exactly when either $F^+ = 0$ or $F^- = 0$, i.e. when the full field strength $F$ is itself either self-dual $F^+$ or anti-self-dual $F^-$, or zero field strength.

Question: At the classical level, it looks that the self-dual instanton configuration ($\star F_{+} = + F_+$) and anti-self-dual instanton configuration ($\star F_{-} = - F_-$) are totally decoupled. However, at the quantum level, in terms of the Yang-Mills functional $Z= \int [DA]\exp[- S_\text{YM}[A]] = \int [DA]\exp[- \int_M\mathrm{Tr}_\mathfrak{g}(F\wedge \star F)]$, do the self-dual instanton and anti-self-dual instanton configurations interact with each other in some nontrivial way? Namely, are there some configurations of $F_{+}$ and $F_{-}$, that they interplay with each other, at the even classical or only at quantum level?

If so, what are the examples? How do they alter the classical theory or quantum theory of path integral?

An epllipse through 12 points related to Golden ratio

Math Overflow Recent Questions - Sat, 06/23/2018 - 18:34

I am looking for a proof of the problem as follows:

Let $ABC$ be a triangle, let points $D$, $E$ be chosen on $BC$, points $F$, $G$ be chosen on $CA$, points $H$, $I$ be chosen on $AB$, such that $IF$, $GD$, $EH$ parallel to $BC$, $CA$, $AB$ respectively. Denote $A'=DG \cap EH$, $B'=FI \cap GD$, $C'=HE \cap IF$, then 12 points:

$D$, $E$, $F$, $G$, $H$, $I$ and midpoints of $AB'$, $AC'$, $BC'$, $BA'$, $CA'$, $CB'$ lie on an ellipse if only if


Note: This problem don't appear in AMM, and I don't have a solution for this problem, but there is the same configuration appear in:

Does every $n\times n\times n$ Latin cube contain a Latin transversal?

Math Overflow Recent Questions - Sat, 06/23/2018 - 18:22

In 1967 H. J. Ryser conjectured that every Latin square of odd order has a Latin transversal. Similar to Latin squares, we may consider Latin cubes.

QUESTION: Let $n$ be any positive integer. Does every $n\times n\times n$ Latin cube contain a Latin transversal?

Let $N$ be any positive integer. In 2008, I proved that for the $N\times N\times N$ Latin cube over $\mathbb Z/N\mathbb Z$ formed by the Cayley addition table, each $n\times n\times n$ subcube with $n\le N$ contains a Latin transversal (cf. my paper An additive theorem and restricted sumsets). Motivated by this, in the 2008 paper I conjectured that my above question has a positive answer.

Any comments are welcome!

Compactness of operators and norming sets

Math Overflow Recent Questions - Sat, 06/23/2018 - 16:12

Originally asked on MSE.

Let $T$ be a linear map from a normed space $E$ into a Banach space $F$.

Let $D\subset \overline{B}_{F^{*}}$ be norming, i.e. there is $r>0$ such that $\sup\limits_{v\in D}|\left<f,v\right>|\ge r\|f\|$, for every $f\in F$.

It is easy to see that a linear map $T:E\to F$ is bounded iff $T^{*}D$ is bounded.

If $T$ is (weakly) compact, then $T^{*}$ is a (weakly) compact, and then $T^{*}D$ is relatively (weakly) compact.

I am wondering about the converse of the last statement.

I can show that compactness of $T$ follows from relative compactness of $T^{*}D$ under the assumption that $E^{*}$ is separable, but no progress whatsoever about the weak compactness.

Dirichlet series as rational zeta expressions

Math Overflow Recent Questions - Sat, 06/23/2018 - 15:45

Let $D(f,s):=\sum_{n=1}^\infty \frac{f(n)}{n^s}$, otherwise known as a Dirichlet series. When $f$ is a multiplicative, number theoretic function, $D(f,s)$ tends to be expressed as a rational product of zeta functions. For example, $D(\mu,n) = \zeta^{-1}(s)$, $D(\sigma,n)=\zeta(s)\zeta(s-1)$, $D(\phi,n) = \zeta(s-1)/\zeta(s)$, and many, many more.

Question 1: Is it possible to deduce for which family of $f$, $D(f,s) = \frac{\prod_{a_i}\zeta(s-a_i)}{\prod_{b_i}\zeta(s-b_i)}$?

Question 2: Conversely, given an arbitrary such rational product of zeta functions, does it always correspond to some multiplicitive number-theroteic function? If yes, have any new "interesting" such functions been discovered by such methods?

As I understand, it the answer these questions should likely follow from Perron's formula, but I'm not seeing an obvious way of proceeding. In fact, the questions seem to reduce to basic generating function theory, along with the inclusion-exclusion principle, as many of the above known number-theoretic identities can be derived from it as such.


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