Recent MathOverflow Questions

How do mathematicians and physicists think of SL(2,R) acting on Gaussian functions?

Math Overflow Recent Questions - Tue, 07/24/2018 - 05:25

Let $\mathcal{N}(\mu,\sigma^2)$ denote the Gaussian distribution on $\mathbb{R}$:

$$ \mathcal{N}(\mu,\sigma^2)(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}.$$

A Gaussian distribution is defined by its mean $\mu\in\mathbb{R}$ and its standard deviation $\sigma>0$. Thus the upper-half plane is the parameter space for the normal family of distributions. Let $z = \mu+i\sigma$ denote an element of the upper-half plane, to be interpreted as the parameters for the Gaussian $\mathcal{N}(\mu,\sigma^2)$.

Let $G\in SL(2,\mathbb{R})$, the group of two by two real matrices with unit determinant. Let $g = \begin{pmatrix}a & b\\ c&d\end{pmatrix}$ be an element of $G$. This group acts on the upper-half plane by fractional linear transformations:

$$g\cdot z = \frac{az+b}{cz+d}.$$

Thus $G$ acts on Gaussians via $$ g\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}(Re(g\cdot z),\, (Im(g\cdot z))^2).$$

We now describe this action in more detail. If $g = \begin{pmatrix}a & b\\ 0&d\end{pmatrix}$, then $g$ acts on a Gaussian as follows: $$ g\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left(\frac a d \mu + \frac b d,\, \frac {a^2} {d^2} \sigma^2\right).$$

For an arbitrary $g$, it follows that $$\begin{align*} Re(g\cdot z) &= \frac{ac(\mu^2+\sigma^2) + bd + (ad+bc)\mu}{(c\mu+d)^2+c^2\sigma^2}\\ Im(g\cdot z) &= \frac{\sigma}{(c\mu+d)^2+c^2\sigma^2}, \end{align*} $$ and the action is $$ g\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left(\frac{ac(\mu^2+\sigma^2) + bd + (ad+bc)\mu}{(c\mu+d)^2+c^2\sigma^2},\left(\frac{\sigma}{(c\mu+d)^2+c^2\sigma^2}\right)^2\right).$$

$G$ has the Iwasawa decomposition $KAN$. Thus $g = k(g)a(g)n(g)$, where $$\begin{align*} k(g) &= \begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}\end{align*}\\ a(g) = \begin{pmatrix}e^t & 0\\ 0&e^{-t}\end{pmatrix}\\ n(g) = \begin{pmatrix}1 & u\\ 0&1\end{pmatrix}, $$ where $\theta$, $t$, and $u$ are analytic functions of the coefficients of $g$ (see Keith Conrad's notes).

It follows that $N$ acts on a Gaussian by moving its mean: $$ n(g)\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left( \mu + u\,, \sigma^2\right),$$ and $A$ acts on a Gaussian by dilating the mean and standard deviation: $$ a(g)\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left( e^{2t}\mu\,, e^{4t}\sigma^2\right).$$

$K$'s action is much more interesting:

$$ k(g)\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left(\frac{\cos\theta\sin\theta(\mu^2+\sigma^2) - \cos\theta\sin\theta + (\cos 2\theta)\mu}{((\sin\theta)\mu+\cos\theta)^2+(\sin^2\theta)\sigma^2}, \left(\frac{\sigma}{((\sin\theta)\mu+\cos\theta)^2+(\sin^2\theta)\sigma^2}\right)^2\right).$$

The standard normal $\mathcal{N}(0,1)$ is fixed by $K$. For any other Gaussian $\mathcal{N}(\mu,\sigma^2)$, moving through values of $\theta$ will cause oscillations of the mean and standard deviation, creating a "wobbling" of the Gaussian.

Questions: What are some physical interpretations of this? What are some good resources for exploring similar links among probability and geometry?

Quandle homomorphism does not always induces group homomorphism on inner automorphism groups of quandles

Math Overflow Recent Questions - Tue, 07/24/2018 - 03:39

Let $X$ and $Y$ be two quandles and $f: X \rightarrow Y$ be a quandle homomorphism. Then we can define a map $\bar f: Inn(X) \rightarrow Inn(Y)$ as $\bar f(S_a)=S_{f(a)}$, where $a \in X$. Then $\bar f$ may not be a group homomorphism. But I am not able to construct such example.

I have posted this question on Math Stack Exchange, but have not got any reply.The link is here

Any suggestions or hints would be of great help.

Computing minimum / maximum of strange two variable funcion

Math Overflow Recent Questions - Tue, 07/24/2018 - 03:05

I want to compute $$\max_{\frac{1}{5 \theta }\leq \alpha \leq \frac{1}{2}} \left(\frac{\alpha\log (\alpha)}{1-\alpha} + \log \left( 1 - \alpha\right) + \frac{1}{1-\alpha} \cdot \left( f\left(\frac{1-\alpha}{2\alpha}\right) + 1 + \frac{41}{15\theta} \right)\right)$$ for \begin{align*} f(x) = \begin{cases} - 1.3- \log x + x/2 +1/2 - 1/x&\text{for } x \geq 2\\ -1.3 &\text{otherwise}. \end{cases} \end{align*}

Intuitively, it should be either for $\alpha=1/2$ or for $\alpha = 1/(5\theta)$ (and a plot verifies this). Is there a way how to prove this?

In the end I want to compute the $\theta$ where the whole expression reaches its minimum. Could there be another approach to do this?

Thanks in advance!!

make me idempotent

Math Overflow Recent Questions - Tue, 07/24/2018 - 00:50

$T_n$ be the full transformation semigroup on $X_n= \{1, 2, \cdots , n\}$.

$D_r =\{\alpha \in T_n: |im(\alpha)|=r\}$.

$E(D_r)$ is the set of all idempotents of semigroup $T_n$.

$support(\alpha)=\{x: x\alpha\neq x\}$

We know that $\alpha \in T_n$ is idempotent $\iff$ $x\alpha=x$ for all $x\in im(\alpha)$.

I wonder product of idempotents. For example; $\alpha =\begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 2 & 1 \end{pmatrix}$, $\beta=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 3 & 4 \end{pmatrix} \in E(D_2)$, but $\alpha\beta=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 4 & 3 \end{pmatrix}$ is not idempotent.

$\textbf{Question:}$ Let $\alpha_0, \alpha_1\cdots, \alpha_k\in E(D_r)$ and $support(\alpha_i)\cap support(\alpha_j)=\emptyset, 0\leq i\neq j\leq k $.

Under which condition the product $\alpha_0\alpha_1\cdots \alpha_k$ becomes idempotent ?

Mean of a periodic velocity field and trajectory displacement bound

Math Overflow Recent Questions - Tue, 07/24/2018 - 00:47

Suppose $u(t,x)$ is a smooth velocity field on $[0,\infty)\times \mathbb{R}$ and periodic in space, i.e., $u(t,0)=u(t,1)$ $\forall t$. Assume that $\int_0^1 u(t,x) \,dx = c$, independent of time. Let $x(t)$ be the trajectory with $x(0)=0$. Prove that $\displaystyle \lim_{t\to\infty} \lvert x(t)\rvert/t \le \lvert c\rvert.$ (Alternatively, reduce to the circle and ask about the winding number of a trajectory.)

I'm not sure if this is true, but if it is, it would complete the last step in a theorem I'm trying to prove about the $\mu$-Hunter-Saxton equation. The reason I think it's true is the following elementary argument in the time-independent case: if $u(x)$ has a zero, then some fixed point is attracting and this limit will be zero. Otherwise we can assume $u(x)$ is always positive and compute that the period is $\displaystyle T = \int_0^1 dx/u(x)$. Then by Cauchy-Schwarz we must have $Tc \ge (\int_0^1 \,dx)^2 = 1$, so that $\frac{1}{T}\le c$. But this argument seems cumbersome and doesn't work in the nonautonomous case. Also this seems like a thing that should be well-known.

Embedding a given affine variety as a divisor

Math Overflow Recent Questions - Tue, 07/24/2018 - 00:35

Let $A$ be a finite type algebra over $\mathbb{C}$. Does there exist a finite type $\mathbb{C}$-algebra $B$ and a nonzero divisor $b \in B$ such that $B/b \cong A$ and $B[1/b]$ is Cohen-Macaulay (or, even better, regular)?

EDIT: In the view of the very helpful counterexample of Jason Starr given in the comments below, let us assume in addition that $A$ is (S$_2$).

Problem in analytic number theory involving simultaneous paired product congruences

Math Overflow Recent Questions - Mon, 07/23/2018 - 21:34

Given a large integer $N$, a small $\epsilon>0$ and integers $A,B,C,D,E,F,G,H$ and prime p where

  1. $A,B,C,D\approx N^{\frac14+\epsilon}$ with $A,B,C,D$ mutually coprime

  2. $E,F,G,H$ $\approx$ $N^{\alpha}$

  3. $p$ is of size $N^{\frac12+\alpha+2\epsilon}$

hold what is the minimum $\alpha\geq\frac12$ such that for every pair $(m,m')\in\mathbb Z\times\mathbb Z$ with $p\nmid m$ and $p\nmid m'$ there are $a,b,c,d\in\mathbb Z$ such that $$m\equiv (ACa+BDb)(Gc+Hd)\bmod p$$ $$m'\equiv (ADc+BCd)(Ea+Fb)\bmod p$$ hold?

There are four parameters $a,b,c,d$ and only two targets $m,m'$ and four linear forms which are interleaved in the congruences. That is why I think it might be a pigeonhole type argument.

In principle there is sufficient room to hit $m$ and $m'$ in arithmetic progressions modulo $p$ however how to show this? Is this doable with exponential sums? At least can some $m,m'$ with $1\leq|m|,|m'|\leq N^\epsilon$ be representable? How small can $|mm'|$ be?

Essentially at large enough but fixed $\alpha\geq\frac12$ is there an $N_\alpha$ such that for every integer $N>N_\alpha$ and small enough $\epsilon$ if we choose $A,B,C,D,E,F,G,H,p$ in the way above then for every pair of $(m,m')\in\mathbb Z\times\mathbb Z$ we can find $a,b,c,d$ such that the congruence above works?

Are there an infinite number of circular primes? [on hold]

Math Overflow Recent Questions - Mon, 07/23/2018 - 21:22

A Circular prime example is 193,939. Circular primes are numbers such that regardless of the what way you arrange the digits, the resulting number is a prime.

how to impose a terminal condition in a minimisation problem?

Math Overflow Recent Questions - Mon, 07/23/2018 - 18:39

Consider the problem of minimising $$ J(u(.)) \triangleq \int_0^T l(x(t),u(t)) d t + \phi(x(T)), \: T> 0 $$ over a space of controls $\mathcal{U}$ with the constraint $$ \dot x(t) = f(x(t),u(t)), \: t >0, \: x(0) = x_0. $$ To make it simple, everything is $\mathbb{R}$ valued. We also assume that the functions $l,\phi,f$ and the space of controls $\mathcal{U}$ satisfy appropriate conditions for having a well-posed problem.

Under this form we can employ either Bellman or Pontryagin principles to study the optimal control.

Question. Can we / How to use this framework to impose a terminal value for $x$? say $x(T) = x_\star$.

Shapes defined by points

Math Overflow Recent Questions - Mon, 07/23/2018 - 16:37

Can shapes determined by number of points?

From an amazing theorem in plane curves geometry we know that vertices of triangles similar to arbitrary triangles $T$ is dense on every closed jordan curve in a plane, so if $J$ be such curve and $A,B,C$ be 3 noncollinear points on plane then at least one curve similar to $J$ contains $A,B,C$. If $J$ be a circle then exactly one circle passes through $A,B,C$. From here we reach the problem in its simplest form:

QUESTION: is circle the only shape on a Euclidean plane (not only from closed curves mentioned above) which just one similar to it defined by each set of 3 non-collinear points we consider from $\mathbb{R}^2$ ? (Means not two or more similar of the shape fits $A,B,C$, just one of it).

Note: here the mentioned "shape" can be any subset of $\mathbb{R}^2$. The "trivial triangle" trivially could not be an answer because we can find many similar of it passing through the 3 points creates it as vertices. It seems for an specified set of 3 points with predefined angles of the triangle it creates, also we may find just circle which is unique.

What about generalization to $n$ points in $\mathbb{R}^m$ which exactly define $k$ similar shapes?

Morphisms of isogeny classes of elliptic curves

Math Overflow Recent Questions - Mon, 07/23/2018 - 16:35

As a follow-up to Automorphisms of the L-function associated to an elliptic $\mathbb{Q}$-curve, I would like to know if there is a natural way to define morphisms $ \phi $ of (isogeny classes of) elliptic curves such that :

  • There exists an isogeny $ f $ from $ E_1 $ to $ E_2 $ iff there exists an isogeny $ f_{\phi} $ from $ \phi(E_{1}) $ to $ \phi(E_{2}) $ that is equivalent to $ f $.

  • $ f_{\phi}\circ\phi=\phi\circ f $.

Building from Chris Wuthrich's comment below, it appears that the preservation of an isogeny class makes $ \phi $ an endofunctor of the considered category. As an isogeny class determines the L-function associated to a representant thereof, this endofunctor induces an automorphism of this L-function. The question is thus : does every automorphism of such an L-function arise this way ?

Roots of unity and an extremal problem

Math Overflow Recent Questions - Mon, 07/23/2018 - 14:36

I want to determine the subset of $m$ members ($m < n/2$) of the set $e^{i 2\pi k/n}, \ \ k=0,\dots, n-1$, so that the absolute value of its sum is maximal.

Numerical minimization spectral norm under diagonal similarity

Math Overflow Recent Questions - Mon, 07/23/2018 - 12:16

This question is a follow up.

Let $A$ be a real square matrix of size $n \times n$. How to determine the minimum spectral norm under diagonal similarity, i.e.,

$$ s(A) = \min_{D} \lVert D^{-1} A D\rVert_2, $$ where $D$ is a non-singular, diagonal real matrix. As it is unlikely to find an analytical upper bound, I would like to ask how $s(A)$ could be determined numericallly.

root of identity matrix and lexicographic order

Math Overflow Recent Questions - Mon, 07/23/2018 - 11:12

I asked a question here order of a permutation and lexicographic order but it seems*** that a powerful and rich generalization can be made!

Let $A$ be a ring together with an arbitrary good order $<^*$ and let $L: M_n(A)\to M_n(A)$ be the application that sorts the rows of a matrix according to the lexicographic order (induced by $<^*$) and the columns of the matrice we get, according to the lexicographic order. We then define $L_Q(M):=L(M).Q$ for any $Q\in M_n(A)$.

[edit : I previously just asked $<^*$ to be a total order (see third comment )]

Suppose that $Q^q=Id$ for some $q\in \mathbb N$ and $Q\in GL_n(A)$. Is is true that for any $M\in M_n(A)$, there exists $r\in \mathbb N$ such that for any $i\in \mathbb N$, we have $L_Q^r(M)=L_Q^{r+iq}(M)$ ?

(***the experimentations I made with a computer concern $A=\mathbb F_2$)

[***edit : I also examined the case $A=\mathbb Z$ with $<^*$ defined by

$a<^*b$ iff ($2a^2+a/|a|<2b^2+b/|b|$) the usual order fail in some cases,

where $k\mapsto L_Q^k(M)$ is not almost-periodic, but the usual order it is NOT a good order, and the cases that failed with it are not failing anymore with $<^*$ as defined in this edit !]

A question about the sum of two convex polytopes with the same normals

Math Overflow Recent Questions - Mon, 07/23/2018 - 10:49

Fix throughout a finite set $T \subset S^{d-1}$ intersecting every open hemisphere of $S^{d-1}$. A $T$-polytope is a (necessarily bounded) set $P$ of the form $$ P = \bigcap_{u \in T} \big\{ x \in \mathbb{R}^d : \langle x,u \rangle \leq a_u \big\}. $$ Given a bounded set $X \subset \mathbb{R}^d$, let $\mathrm{cont}(X)$ denote the minimal $T$-polytope containing $X$ (I'm thinking of this as the `container' of $X$).

Now let $A$ and $B$ be two $T$-polytopes such that $\mathbf{0} \in A \cap B$. One can show quite easily that in two dimensions we always have $\mathrm{cont}(A \cup B) \subset A + B$. On the other hand, one can construct examples in three or more dimensions where this relation does not hold. This (hopefully!) motivates the following slightly stranger question. Given a $T$-polytope $P$ as above, let us say that $x$ is a special point of $P$ if there are $d$ linearly independent vectors $u_1,\dots,u_d \in T$ such that $\langle x,u_i \rangle = a_{u_i}$ for all $i$. (In other words, the special points are the intersections of any d extended faces of $P$ in linearly independent directions.) Let $E(P)$ denote the convex hull of the set of all special points of $P$. Thus, as an example, if $P$ is a simplex or a parallelepiped, then $E(P) = P$. Clearly in general, $P \subset E(P)$, since the vertices of $P$ are special. My question is: with $A$ and $B$ as above, is it true that $E(\mathrm{cont}(A \cup B)) \subset E(A) + E(B)$?

Very roughly speaking, the function $E$ is an attempt to circumvent one of the reasons that the relation $\mathrm{cont}(A \cup B) \subset A + B$ fails in three (say) dimensions, which is that $A + B$ need not be a $T$-polytope (by contrast, in two dimensions, $A + B$ is always a $T$-polytope). By this I don't mean that $E(A) + E(B)$ is a $T$-polytope (it clearly isn't), but that the points in $E(A)$ (say) that are not vertices of $A$ seem to be the points that `hurt' us when we try to prove the false relation.

Final question: If the proposed containment is false, could there be another simple function of $T$-polytopes that does obey such a containment relation?

Maximal number of singularities of an algebraic curve

Math Overflow Recent Questions - Mon, 07/23/2018 - 08:23

An algebraic curve defined by a polynomial of degree n can have at most (n−1)(n−2)/2 singularities. What is the maximal number of singularities of a curve in a k-dimensional space in terms of the degree of the k-1 polynomial equations defining the curve?

P.S.: The comment of Mohan posted below gives an upper bound on the number of singularities by using the fact the the geometric genus is non-negative. So, I could reformulate my question as follows. Are there space curves of genus 0 (or 1) such that the upper bound (possibly minus 1) is attained? given for granted that the curve is irreducible.

Differential equation changing sign almost everywhere

Math Overflow Recent Questions - Mon, 07/23/2018 - 03:06

Conjecture: Let $f:\mathbb{R}→\mathbb{R}$ be an everywhere differentiable function and assume that $f(x)+f′(x)∈ \{-1,1\}$ almost everywhere and $f'(0)=0$. Then is $f$ necessarily a constant function?

Can you give me a counter-example? I have already asked the question here on MathSE.

multi-dimensional integral of modified Vandermonde determinant

Math Overflow Recent Questions - Sun, 07/22/2018 - 17:30

I'm looking for suggestions on how one might try to compute the following $(N-1)$-dimensional integral:

$$I_N= \frac{1}{(2\pi)^{N-1}(N-1)!} \int\cdots\int \\ \begin{vmatrix} 1 & 1 & \cdots & 1 & 1 \\ e^{i\alpha_1} & e^{i\alpha_2} & \cdots & e^{i\alpha_{N-1}} & e^{-i(\alpha_1+\dots+\alpha_{N-1})} \\ e^{i2\alpha_1} & e^{i2\alpha_2} & \cdots & e^{i2\alpha_{N-1}} & e^{-i2(\alpha_1+\dots+\alpha_{N-1})} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ e^{i(N-1)\alpha_1} & e^{i(N-1)\alpha_2} & \cdots & e^{i(N-1)\alpha_{N-1}} & e^{-i(N-1)(\alpha_1+\dots+\alpha_{N-1})} \\ \end{vmatrix}^2 {d\alpha_1\cdots d\alpha_{N-1}}$$

over the region given by the following conditions:

$$\alpha_{N-1}\in\left(-\pi, -\pi+\frac{2\pi}{N}\right) \text{ and } \alpha_1,\dots,\alpha_{N-2},-(\alpha_1+\dots+\alpha_{N-1})\in\left(-\pi+\frac{2\pi}{N}, \pi\right)$$

More generally, the integration region can be described as: one of the variables $\alpha_i, 1\leq i\leq N-1$ must be inside the interval $\left(-\pi, -\pi+\frac{2\pi}{N}\right)$, and the rest of the $\alpha_i$, together with $-\alpha_1-\dots-\alpha_{N-1}$ must be outside this interval $\mod 2\pi$. $\alpha_{N-1}$ is just a particular choice and the value of $I_N$ doesn't change if $\alpha_{N-1}$ is replaced by some other $\alpha_i$; it also doesn't change if the interval $\left(-\pi, -\pi+\frac{2\pi}{N}\right)$ gets replaced by $\left(\pi-\frac{2\pi}{N}, \pi\right)$

Regarding the shape of the integration region, in the $N=3$ case it is a hexagon, in the $N=4$ case it's an octahedron, and in general, the region for $I_N$ is an $(N-1)$-dimensional polytope. The normalization factor in front of the integral is just to ensure that $0\leq I_N\leq 1$

The integrand is the square of the absolute value of the determinant, and can also be expressed as $$\prod_{1\leq j<k\leq N-1} |e^{i\alpha_j}-e^{i\alpha_k}|^2 \prod_{1\leq j\leq N-1} |e^{i\alpha_j}-e^{-i(\alpha_1+\dots+\alpha_{N-1})}|^2$$

or equivalently

$$2^{N(N-1)} \prod_{1\leq j<k\leq N-1} \sin^2\frac{\alpha_j-\alpha_k}{2} \prod_{1\leq j\leq N-1} \sin^2\frac{\alpha_j+\alpha_1+\dots+\alpha_{N-1}}{2} $$

So far, I have computed the values for when $N=2,3,4$

\begin{align} I_2 & =1 \\[8pt] I_3 & =\frac{7}{12}+\frac{11}{3\pi^2} \\[8pt] I_4 & =\frac{5}{12}+\frac{1001}{360\pi^2}+\frac{4232}{675\pi^3} \end{align}

but I haven't been able to figure out a method for computing the integral in the general case.

The limits of integration for each variable $\alpha_i$ can be expressed explicitly if we split the integration region into two equal parts, say $R_1$ and $R_2$, for which we have the following limits

\begin{array}{l | l} R_1 & R_2 \\ \hline \alpha_{4k+1}\in \left( -\pi+\frac{2\pi}{N} , \pi-\frac{2\pi}{N}-\sum_{4k+2}^{N-1}\alpha_i \right) & \alpha_{4k+1}\in \left( -\pi-2k\frac{2\pi}{N}-\sum_{4k+2}^{N-1}\alpha_i, \pi \right) \\ \alpha_{4k+2}\in \left( -(2k+1)\frac{2\pi}{N}-\sum_{4k+3}^{N-1}\alpha_i , \pi \right) & \alpha_{4k+2}\in \left( -\pi+\frac{2\pi}{N}, -(2k+1)\frac{2\pi}{N} -\sum_{4k+3}^{N-1}\alpha_i \right) \\ \alpha_{4k+3}\in \left( -\pi-(2k+1)\frac{2\pi}{N}-\sum_{4k+4}^{N-1}\alpha_i , \pi \right) & \alpha_{4k+3}\in \left( -\pi+\frac{2\pi}{N} , \pi-(2k+2)\frac{2\pi}{N} -\sum_{4k+4}^{N-1}\alpha_i \right) \\ \alpha_{4k+4}\in \left( -\pi+\frac{2\pi}{N} , -(2k+2)\frac{2\pi}{N}-\sum_{4k+5}^{N-1}\alpha_i \right) & \alpha_{4k+4}\in \left( -(2k+2)\frac{2\pi}{N} -\sum_{4k+5}^{N-1}\alpha_i , \pi \right) \\ \end{array}

for $4k+1, 4k+2, 4k+3, 4k+4 \in \{1,2,\dots,N-2\}$, and for $N-1$:

\begin{array}{l | l | l} & R_1 & R_2 \\ \hline \text{if } N=2n & \alpha_{N-1}\in \left( -\pi, -\pi+\frac{2\pi}{N} \right) & \alpha_{N-1}\in \left( -\pi, -\pi+\frac{2\pi}{N} \right) \\ \text{if } N=4n+1 & \alpha_{N-1}\in \left( -\pi, -\pi+\frac{\pi}{N} \right) & \alpha_{N-1}\in \left( -\pi+\frac{\pi}{N},-\pi+\frac{2\pi}{N} \right) \\ \text{ if} N=4n+3 & \alpha_{N-1}\in \left( -\pi+\frac{\pi}{N},-\pi+\frac{2\pi}{N} \right) & \alpha_{N-1}\in \left( -\pi, -\pi+\frac{\pi}{N} \right) \\ \end{array}

With these conventions, the integral over $R_1$ is equal to the integral over $R_2$ and equal to one-half of the integral over the total region.

For example, when $N=3$

$\frac{1}{(2\pi)^22!}\int\int_{R_1}|\dots|^2d\alpha_1 d\alpha_2 = \frac{1}{(2\pi)^22!}\int_{-\pi+\frac{\pi}{3}}^{-\pi+\frac{2\pi}{3}}\int_{-\pi+\frac{2\pi}{3}}^{\pi-\frac{2\pi}{3}-\alpha_2} |\dots|^2d\alpha_1 d\alpha_2 = \frac{1}{2}I_3 $

$\frac{1}{(2\pi)^22!}\int\int_{R_2}|\dots|^2d\alpha_1 d\alpha_2 = \frac{1}{(2\pi)^22!}\int_{-\pi}^{-\pi+\frac{\pi}{3}}\int_{-\pi-\alpha_2}^{\pi} |\dots|^2d\alpha_1 d\alpha_2 = \frac{1}{2}I_3 $

and if $N=4$

$\frac{1}{(2\pi)^33!}\int\int\int_{R_1}|\dots|^2d\alpha_1 d\alpha_2 d\alpha_3 = \frac{1}{(2\pi)^33!}\int_{-\pi}^{-\pi+\frac{2\pi}{4}}\int_{-\frac{2\pi}{4}-\alpha_3}^{\pi}\int_{-\pi+\frac{2\pi}{4}}^{\pi-\frac{2\pi}{4}-\alpha_2-\alpha_3} |\dots|^2d\alpha_1 d\alpha_2 d\alpha_3 = \frac{1}{2}I_4 $

$\frac{1}{(2\pi)^33!}\int\int\int_{R_2}|\dots|^2d\alpha_1 d\alpha_2 d\alpha_3 = \frac{1}{(2\pi)^33!}\int_{-\pi}^{-\pi+\frac{2\pi}{4}}\int_{-\pi+\frac{2\pi}{4}}^{-\frac{2\pi}{4}-\alpha_3}\int_{-\pi-\alpha_2-\alpha_3}^{\pi} |\dots|^2d\alpha_1 d\alpha_2 d\alpha_3 = \frac{1}{2}I_4 $

This is how I computed the values for $I_3$ and $I_4$, but for $N\geq 5$ it gets too messy, and I don't think this would be the best approach for the general case.

Absence of eigenfunctions to integral operator

Math Overflow Recent Questions - Sun, 07/22/2018 - 13:01

Consider the unbounded integral operator $T:D(T) \subset L^2((0,\infty),x \ dx ) \rightarrow L^2((0,\infty),x \ dx)$ on its maximal domain $D(T)=\left\{f \in L^2((0,\infty),x \ dx); Tf \in L^2((0,\infty),x \ dx) \right\}$ such that for some $x_0 \in (0,\infty)$ we have

$$(Tf)(y) = \frac{1}{y-x_0} \int_{0}^{\infty} f(x) \frac{xy}{x^2+y^2+1} \ dx.$$

I would like to show that $1$ is not(!) in the point spectrum of this operator (this is what I observed numerically), i.e.

$(Tf)(x)=f(x)$ does not have a solution.


One problem (and perhaps the decisive problem) seems to be the singularity at $y=x_0.$

If such a eigenfunction existed, then

$$f(y) = \frac{1}{y-x_0} \int_{0}^{\infty} f(x) \frac{xy}{x^2+y^2+1} \ dx.$$

For $f$ to be integrable at $y=x_0,$ the integral $\int_{0}^{\infty} f(x) \frac{x}{x^2+y^2+1} \ dx$ needs to have a zero at $y=x_0$ and $f$ changes sign.

I tried to construct a contradiction from this but failed so far.

On the Dirichlet series for $1/\zeta(s)$ at $\Re(s)=1/2$

Math Overflow Recent Questions - Sat, 07/21/2018 - 05:56

Suppose that $1/2+it$ is not a zero of the Riemann zeta function $\zeta$, where $t \in \mathbb{R}$. Can $1/\zeta(1/2+it)$ be expressed as a Dirichlet series ?


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