31

The "Katz-Sarnak philosophy" is just the idea that statistics of various kinds for $L$-functions should, in the large scale limit, match statistics for large random matrices from some particular classical compact group.
First you need to decide what kinds of zeros to look at: the high zeros of an individual $L$-function, the low zeros (near the real axis) ...

27

Edit: According to Dean and Majumdar, the precise value of $c$ in my answer below is $c=\frac{\log 3}{4}$ (and $c=\frac{\log 3}{2}$ for GUE random matrices). I did not read their argument, but I have been told that it can be considered as rigourous. I heard about this result through the recent work of Gayet and Welschinger on the mean Betti number of random ...

answered Jan 10 '13 at 17:46

Mikael de la Salle

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26

this is a limit of a more general result by Majumdar and company, How many eigenvalues of a Gaussian random matrix are positive? (2010), see also their earlier papers from 2006 and 2008.
The coefficient $\sqrt{3}$, or $\frac{1}{2}\log 3$ in the exponent, appears from a saddle point approximation, see Eq. 59 of the 2008 paper, without any particular ...

answered Apr 17 '14 at 13:54

Carlo Beenakker

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22

Very nice problem!
Let me recall you that the determinant of $n \times n$ matrices with entries in $\{0,1\}$ is related to the one of $n+1 \times n+1$ matrices with entries in $\{-1,+1\}$: replace the zeros by $-1$'s and add a row of $-1$'s on the top and a column of ones on the right (you may want to read this arXiv).
I can now tell you that, besides ...

21

Your question appears to be based on a false premise. In fact $AB+BA$ does not tend to be positive definite as $n$ increases, even within the particular distribution you happen to be using.
To demonstrate this, here is a simple piece of Mathematica code that implements precisely the numerical experiment you described:
randMat[n_] := With[{
eigval = ...

19

Here is a general comment. Let $G$ be a compact group and let $V$ be a (finite-dimensional, continuous, complex) representation of $G$. This data determines a locally finite directed graph, the representation graph $\Gamma$ of $G$ and $V$, as follows. The vertices of $\Gamma$ are the irreducible representations of $G$, and the number of edges from an ...

answered Sep 5 '14 at 9:14

Qiaochu Yuan

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17

If you choose the matrix elements of $A$ independently from a Gaussian distribution you have the socalled Ginibre ensemble of random-matrix theory. The statistics of the eigenvalues is known, see for example Eigenvalue statistics of the real Ginibre ensemble. The statistics of the eigenvectors, and the eigenvector-eigenvalue correlations, have been much less ...

answered Sep 18 '14 at 7:16

Carlo Beenakker

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15

My initial answer was wrong. Instead of completely deleting it, I left it at the bottom of the post. I use notation now that slightly differ from my original post.
As before, I give only a partial answer related to the max eigenvalue question and the limsup. Recall that the tail estimates for the max eigenvalue is
$$(P(\lambda_1(n) >2\sqrt{n}+tn^{-1/6})\...

15

Pastur and Vasilchuk have extended the result of Diaconis and Evans for $a_{2k}$ from $2k\leq n/2$ to $2k\leq n-1$:
$$a_{2k}=\pi^{-1/2}2^{k}\Gamma(k+1/2)\;\;\text{for}\;\;2k\leq n-1\quad\quad[*]$$
As suggested by Liviu Nicolaescu, for small $n$ you might directly integrate over the probability distribution of the eigenvalues in $O_n$, which you can ...

answered Sep 4 '14 at 21:16

Carlo Beenakker

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I'm going to give an answer that discusses some things that the other answers don't go into as much detail on. In particular let me try to explain why the results you mention on classical groups having the same $n$-level correlations are in fact relevant to the number theory situation. It will build off KConrad's answer in some ways.
The Katz-Sarnak book ...

14

By the Cauchy-Binet theorem, $\det AA^T=\sum (\det B)^2$, where $B$
ranges over all $m\times m$ submatrices of $A$. The expected value of
$(\det B)^2$ is $(m+1)!/4^m$, so the expected value of $\det AA^T$ is
${n\choose m}(m+1)!/4^m$. This is also the expected value of $|\det
AA^T|$ since $\det AA^T\geq 0$.
Incidentally, it is misleading to say that the ...

answered Jul 2 '15 at 11:01

Richard Stanley

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13

I don't know of a fully intuitive derivation, but there are some informal arguments that give the circular law with a relatively small amount of calculation.
Let $M$ be a matrix where the entries are iid with mean zero and variance one. One can begin with the determinant formula
$$ \log |\det( M - z )| = \sum_{j=1}^n \log |\lambda_j - z|.$$
The circular ...

answered Apr 10 '13 at 15:34

Terry Tao

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13

$\text{tr}(AB+BA) = 2 \text{tr}(A^{1/2} B A^{1/2}) > 0$, so that may produce some bias toward positive eigenvalues. In particular if you generate your "random" matrices in such a way that the eigenvalues of $AB+BA$ will tend to be concentrated very close together, this may produce the results you observed.
But I tried a different experiment: $A = X^T X$...

11

You might be interested in something Jelani Nelson wrote me in an email on Oct. 13, 2011:
"Another notion of derandomizing JL is the following: come up with a
distribution over embeddings that can be sampled using as few random
bits as possible so that, for any vector $x$ in $R^d$, a random vector has
its $\ell_2$ norm preserved up to $1+\epsilon$ with ...

11

I assume your random $M_i$s are from $O(3)$, not $SO(3)$. In fact, I suspect that the answer depends on whether $M_1M_2$ has eigenvalue $1$ or eigenvalue $-1$.
If I did not make a mistake, when you expand $p_{2i}$, you will get essentially
$$\sum_{k=0}^{i-1} U^k x_1+ V\left(\sum_{k=1}^{i-1} U^i\right)^T x_2$$
Here $U=M_2M_1$ and $V$ is also an orthogonal ...

11

This phenomenon is to be expected: the point is to recall the classification of isometries and that indirect isometries almost always have fixed points, while direct ones almost always have no fixed points.
The fact that there are two isometries is not relevant to the boundedness of the dynamic: let us look at $U=T_2\circ T_1$. This is a random isometry, ...

answered Jul 12 '13 at 20:25

Benoît Kloeckner

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11

Involutions $s=s^{-1}$ in $S_n$ are modeled by the Tracy-Widom distribution $F_1$ for real symmetric matrices (GOE):
Take as $S_n^\ast$ the subset of involutions in $S_n$, and let
$M_n$ be the corresponding random variable for the longest increasing subsequence. Then the limit distribution is
$$P\left(\frac{M_n − 2\sqrt n}{n^{1/6}}\leq s\right)\rightarrow ...

answered Sep 9 '13 at 16:29

Carlo Beenakker

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11

A recursion formula for the moments of the Gaussian orthogonal ensemble, M. Ledoux (2009).
The desired recursion formula for the moment $b_p^N\equiv E\,[\,{\rm tr}\,(S_N^{2p})]$ is
I notice a difference in normalization, you'll want to divide $b_p^N$ by $2^p$.

answered Feb 6 '18 at 17:08

Carlo Beenakker

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11

I do not know what is exactly the KS philosophy, or much number theory for that matter, but maybe I can tell you a few things. Take the Riemann zeta function, for instance. It was discovered by Montgomery, with some help from Dyson, that the Riemann zeros have the same correlation function as the eigenvalues of unitary random matrices. Some averages over the ...

11

The answer to the question as stated (maximum of row elements) has been solved in Extreme statistics of complex random and quantum chaotic states, see also this MO posting:
$$\int dU \max_j |U_{1,j}|^2 =\frac{1}{d}\sum_{j=1}^d \frac{1}{j}.$$
For large $d$ this tends to $(1/d)\log d$. The complete probability distribution of the row-maximum is known.
The ...

answered Dec 15 '19 at 8:11

Carlo Beenakker

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11

There is an explicit determinental formula for these numbers due to Gessel in Symmetric functions and P-recursiveness (JCTA, 1990). Asymptotics were known much earlier and appear in a paper by Amitai Regev Asymptotic values for degrees associated with strips of young diagrams (Adv. Math. 1981). The gross asymptotics are that the $k$th root of the number of ...

10

This is really a long comment: I don't know the answers, but I expect that these numbers have been calculated by the geometric probabilists (and physicists) quite some time ago.
If you use the fact that the group $\mathrm{SO}(n)$ acts on the symmetric matrices in the obvious way and parametrize the 'hemisphere' of symmetric matrices of positive trace and ...

answered Jan 9 '13 at 22:57

Robert Bryant

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10

Since you haven't given a distribution, let me make an observation giving the right form of the answer in the case where the $D_{xy}$ are independent uniform $[0,1]$ random variables.
I want to claim that the probability, $p$, that the matrix defines a metric satisfies
$\alpha^{n^2}\le p\le \beta^{n^2}$ for constants $\alpha$ and $\beta$.
First, notice ...

10

### Find a generalized hypergeometric-based function yielding certain ratios of fifth-degree polynomials

Are there any other conditions this function must satisfy?
Otherwise, there is a trivial solution:
$$\frac{(2-2a) \left(64 k^5+128 k^4-340 k^3-1032 k^2-1099 k-384\right)}{k (2 k-1) (2 k+5) (4 k-1) (4 k+1)}+\frac{(2 a-1) \left(8 k^5+36 k^4-82 k^3-681 k^2-1366 k-885\right)}{128 (k+2) (k+3) (k+4) (4 k+5) (4 k+7)}$$

10

I hope I understood the OP correctly, in case not please let me know. And I will discuss the case of eigenvalue instead of singular value without much loss of generality.
In case you are only interested in the distribution of eigenvalues of an i.i.d. sample covariance matrix(which is also a random matrix but underlying distribution is always of dimension ...

10

This next-nearest-neighbor distribution of the Riemann zero's is addressed in Mehta's book on random-matrix theory. It is well reproduced by that of the Gaussian Unitary Ensemble (GUE), compare black curve and black data points:
I added the red curve in the plot, being the convolution of the nearest-spacing distribution in the GUE:
$$p(s)=\int_0^s f(u)f(s-...

answered Feb 1 '16 at 11:35

Carlo Beenakker

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10

I agree with user39115!
I will give a heuristic from random matrix theory because we know the global behaviour of the eigenvalue. First
$$A=p 1 +\sqrt{N(p-p^2)}\frac{B}{\sqrt{N}} $$ where $1$ is the matrix with only 1 entries and $$B_{i,j}=\begin{cases} \frac{1-p}{\sqrt{(p-p^2)}} &\text{with probability } p\\ \frac{-p}{\sqrt{(p-p^2)}} &\text{ with ...

10

The GUE random matrix model predicts that the zeroes should satisfy the local statistics of random matrices. It doesn't predict that the zeroes should satisfy the global statistics of random matrices, because it's not clear what that would even mean unless the zeroes are all contained in some bounded interval. (In fact they get more frequent the further away ...

10

Yes, this follows by the de la Vallée-Poussin necessary and sufficient condition for the uniform integrability. Indeed, suppose that
\begin{equation}
\gamma n^2\to a
\end{equation}
(as $n\to\infty$) for some real $a\ge0$.
Your integral is
$$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma},$$
where the $X_i$'s are independent standard normal random variables. ...

9

This answer is a follow-up to the other answers (particularly to Carlo Beenakker's answer from September 4).
First off, Carlo's conjecture is indeed true. That is:
Theorem. If $n \geq k$ then
$$a_{2k}=\pi^{-1/2}2^{k}\Gamma(k+1/2).\quad [*]$$
Idea of proof: Recall that the Brauer algebra describes the set of invariants of $2k$-fold tensor copies of ...

answered Nov 6 '14 at 18:16

Nathaniel Johnston

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