5
$\begingroup$

Investigations concerning random Morse functions led me to the following problem. Consider the classical GOE of $m\times m$ real symmetric matrices $A$ with independent Gaussian entries with zero means and variances

$$ \boldsymbol{E}(a_{ii}^2)=2 \boldsymbol{E}(a_{ij}^2)= 2 $$

for all $i \neq j$. Consider the function

$$ F_m(x, y) = \boldsymbol{E}_{GOE}\bigl( |\det(y+A)| e^{ -x(tr A)^2 } \bigr), $$

$x,y$ real, $x>0$. What can one say about the behavior of $F_m(x,y)$ as $m\rightarrow \infty$.

Equivalently we can consider the Gaussian ensemble $\mathcal{S}(m,x)$ of symmetric $m\times m$ real matrices with probability density

$$ dP(A)=\frac{1}{Z_{m,x}} e^{-\frac{1}{2}tr(A^2)-x(tr A)^2} \prod_{i\leq j} da_{ij}, $$

$x>0$, and then ask for the bevavior as $m\rightarrow \infty$ of the expectation

$$ \boldsymbol{E}_{\mathcal{S}(m,x)}\left( |\det(A+y)|\right). $$

Observe that

$$GOE= \mathcal{S}(m,x)_{x=0}.$$

The normalizing constant $Z_{m,x}$ can be explicitly computed for any $x$ and thus

$$ F_m(x,y)= \frac{Z_{m,x}}{Z_{m,0}} \boldsymbol{E}_{\mathcal{S}(m,x)}\left( |\det(A+y)|\right). $$

In the geometric problem I am interested $x=\frac{1}{8}$. In this case the ensemble $\mathcal{S}_m:=\mathcal{S}(m, \frac{1}{4})$ can be described as the ensemble of real, symmetric $m\times m$ matrices whose entries are mean zero Gaussian variables satisfying the covariance equalities

$$ \boldsymbol{E}\left( a_{ij} a_{k\ell}\right)=-\frac{2}{2+m}\delta_{ij}\delta_{k\ell} +\left( \delta_{ik}\delta_{j\ell}+ \delta_{i\ell}\delta_{jk}\right).$$

Note that as $m\rightarrow \infty$ this ensemble resembles more and more the classical GOE which satisfies the covariance equalities

$$ \boldsymbol{E}\left( a_{ij} a_{k\ell}\right)= \left(\delta_{ik}\delta_{j\ell}+ \delta_{i\ell}\delta_{jk}\right).$$

Finally, I want to explain how is this related to Morse theory. To put things in perspective observe that if $A$ is a symmetric $m\times m$ matrix, then its spectrum can be identified with the set of critical values of the restriction to the unit sphere in $\mathbb{R}^m$ of the quadratic polynomial

$$\mathbb{R}^m\ni x\mapsto q_A(X)=(Ax,x).$$

To a Morse function $f$ on a compact smooth manifold $M$ of dimension $m$ we can associate two measures.

(a) A measure $K_f$ on $M$ defined as the sum of Dirac delta's concentrated at the critical points of $f$

$$K_f=\sum_{df(p)=0}\delta_p.$$

(b) A measure $\Delta_f$ on $\mathbb{R}$ supported on the set of critical values of $f$ and defined as the pushforward of $K_f$ via $f$,

$$\Delta_f:=f_*(K_f).$$

In other words, $\Delta_f$ counts the critical values with multiplicity. Note that when $f$ is the restriction to the unit sphere of the quadratic form $q_A$ then $\Delta_f$ coincides with the spectral measure of $A$.

Fix a Riemann metric $g$ on $M$ and an orthonormal $(\Psi_k)_{k\geq 0}$ basis of $L^2(M)$ consisting of eigenfunctions of the Laplacian

$$ \Delta \Psi_k=\lambda_k \Psi_k. $$

Fix i.i.d. standard Gaussian random variables $(x_k)_{k\geq 0}$ and for every $L >0$ define the random function

$$f_L=\sum_{\lambda_k\leq L^2}x_k\Psi_k. $$

The function $f_L$ is roughly speaking a random polynomial of large degree. Equivalently one should think of $f_L$ as a random element in the space $U_L$ spanned by the eigenfunctions corresponding to eigenvalues $\leq L^2$ and equipped with the standard Gaussian measure. The large $L$ behavior of $\dim U_L$ is governed by Weyl's asymptotic formula

$$ \dim U_L \sim const. L^m.$$

To $f_L$ we associate two random measures

$$ K_{f_L},\;\; \Delta_{f_L} $$

that have normalized expectations

$$ K_L:=\frac{1}{\dim U_L} \boldsymbol{E}( K_{f_L} ), $$

$$ \Delta_L:=\frac{1}{\dim U_L} \boldsymbol{E}( \Delta_{f_L} ). $$

Above, $K_L$ is a measure on $M$ and $\Delta_L$ is a measure on $\mathbb{R}$. I can show that as $L\to\infty$ the measure $K_L$ converges weakly to $C_m dV_g$, where $dV_g$ denotes the volume measure determined by the metric $g$, and $C_m$ is a certain explicit constant that depends only on $m$ but not on $(M,g)$. Thus, the critical points of a random $f_L$, $L\gg 0$, is uniformly distributed on average.

As $L\to \infty$ the measure a suitable rescaled version of $\Delta_L$ converges to a measure $d\mu_m(y)$ on $\mathbb{R}$ that is absolutely continuous with respect to the Lebesgue measure. More precisely

$$d\mu_m(y)=\rho_m(y) dy= Const_m \times \boldsymbol{E}_{\mathcal{S}(m,1/8)}\left( \;|\det(A-s_my )|\;\right) e^{-\frac{y^2}{2 }} dy,$$

$$s_m=\sqrt{\frac{m+4}{m+2}} $$

Remark. The measure $d\mu_m(y)$ can also be given a description as a conditional expectation. To explain this I need to introduce another Gaussian ensemble of symmetric $m\times m$ matrices.

To describe it observe that to any such matrix $A$ we can associate a quadratic form $q_A$ on $\mathbb{R}^m$,

$$ q_A(x)=(Ax, x).$$

We have a unique, centered Gaussian probability measure on the space of symmetric $m\times m$ matrices with variance

$$V(A)=\int_{\mathbb{R^m}} q_A(x)^2 \frac{e^{-\frac{|x|^2}{2}}}{(2\pi)^{\frac{m}{2}}} dx. $$

Denote by $\mathcal{U}_m$ this Gaussian ensemble of symmetric matrices. (I use the symbol $\mathcal{U}_m$ because this ensemble has a remarkable universality property.)

Now fix a standard (scalar) Gaussian r.v. $Y$ such that the pair $(A,Y)$ is a Gaussian vector satisfying the correlation equalities

$$\boldsymbol{E}(a_{ij} Y)=s_m\delta_{ij} $$.

Then for any Borel subset of $\mathbb{R}$ we have

$$\mu_m(B)=\boldsymbol{E}_{\mathcal{U}_m}\Bigl( |\det A|\;\Bigl|\; Y\in B\Bigr). $$

$\endgroup$
1
  • $\begingroup$ Thanks. You are right. The actual problem I have is a bit more complicated. I'll post a new question. $\endgroup$ Commented Jan 2, 2012 at 14:37

1 Answer 1

1
$\begingroup$

Just an heuristic answer:

The joint eigenvalue distribution of a GOE random matrix is given by $$ \frac{1}{Z_n}\prod_{i<j}|x_i-x_j|\prod_{i=1}^ne^{-x_i^2/2}dx_i. $$ Thus, $E_{GOE}(|\det(y+A)|e^{-x(tr A)^2})$ equals to $$ \frac{1}{Z_n}\int\cdots\int\prod_{i=1}^n|y+x_i|e^{-x\sum_{i,j}x_ix_j} \prod_{i<j}|x_i-x_j|\prod_{i=1}^ne^{-x_i^2/2}dx_i, $$ and, if I introduce the measure $\mu_n=\frac{1}{n}\sum_{i=1}^n\delta_{x_i}$, to \begin{multline} \frac{1}{Z_n}\int\cdots\int\exp\left\{-n^2\left[-\frac{1}{n}\int\log|y+u|\mu_n(du)+x\iint uv\mu_n(du)\mu_n(dv)\\ -\iint_{u<v}\log|u-v|\mu_n(du)\mu_n(dv)+\frac{1}{2n}\int u^2\mu_n(du)\right]\right\}\prod_{i=1}^ndx_i. \end{multline} Similarly, $$ Z_n=\int\cdots\int\exp\left\{-n^2\left[ -\iint_{u<v}\log|u-v|\mu_n(du)\mu_n(dv)+\frac{1}{2n}\int u^2\mu_n(du)\right]\right\}\prod_{i=1}^ndx_i. $$

Now, let me denote by $n^{\alpha}_*\mu$ the push-forwards by $x\mapsto n^{\alpha}x$ of a probability measure $\mu$. It is known that $n^{-1/2}_*\mu_n$ converges weakly towards the semi-circle distribution $\mu_{SC}$ with probability one.

As a consequence, heuristically I would say that as $n\rightarrow\infty$ (which may be made rigorous by large deviation estimates)

\begin{multline} \frac{1}{n^2}\log E_{GOE}(|\det(|y+A|e^{-x(tr A)^2})\\ \simeq \frac{1}{n}\int\log|y+u|\,(n^{1/2}_*\mu_{SC})(du)-x\iint uv \,(n^{1/2}_*\mu_{SC})(du)(n^{1/2}_*\mu_{SC})(dv), \end{multline} the latter being computable explicitly.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .