Random functions on tori

$\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\bC}{\mathbb{C}}$  $\newcommand{\bT}{\mathbb{T}}$ $\newcommand{\eE}{\mathscr{E}}$ $\newcommand{\ve}{{\varepsilon}}$ $\newcommand{\lan}{\langle}$ $\newcommand{\ran}{\rangle}$ $\newcommand{\bu}{\boldsymbol{u}}$ $\newcommand{\eS}{\mathscr{S}}$ $\DeclareMathOperator{\var}{\boldsymbol{var}}$ $\newcommand{\ii}{\boldsymbol{i}}$ $\newcommand{\bsU}{\boldsymbol{U}}$  $\newcommand{\vfi}{\varphi}$ $\newcommand{\bsE}{\boldsymbol{E}}$ $\newcommand{\teE}{\widetilde{\mathscr{E}}} $ $\newcommand{\pa}{\partial}$  $\DeclareMathOperator{\Hess}{\boldsymbol{Hess}}$ $\DeclareMathOperator{\diag}{Diag}$ $\newcommand{\one}{\boldsymbol{1}}$





Consider the $m$-dimensional torus

$$\bT^m:=\bR^m/(2\pi\bZ)^m $$

equipped with the  flat metric

$$g:=\sum_{j=1}^m (d\theta^j)^2. $$

It has volume ${\rm vol}_g(\bT^m) =(2\pi)^m.$   The eigenvalues of the corresponding  Laplacian are

$$|\vec{k}|^2,\;\;\vec{k}=(k_1,\dotsc, k_m)\in\bZ^m. $$

 For $\vec{\theta}=(\theta^1,\dotsc, \theta^m) \in\bR^m$ and $\vec{k}\in\bZ^m$ we set

$$ \lan\vec{k},\vec{\theta}\ran =\sum_jk_k\theta^j. $$

Denote by $\prec$ the lexicographic order on $\bR^m$.  An  orthonormal basis of $L^2(\bT^m)$ is given by the  functions $(\Psi_{\vec{k}})_{\vec{k}\in\bZ^m}$, where


$$ \Psi_{\vec{0}}(\vec{\theta}) =\frac{1}{(2\pi)^{\frac{m}{2}}}$$,


$$\Psi_{\vec{k}}(\vec{\theta})=\frac{\sqrt{2}}{(2\pi)^{m/2}} \sin\lan \vec{k},\vec{\theta}\ran, \;\;\vec{k}\succ\vec{0}, $$


$$ \Psi_{\vec{k}}(\vec{\theta})=\frac{\sqrt{2}}{(2\pi)^{m/2}} \cos\lan\vec{k},\vec{\theta}\ran,\;\;\vec{k}\prec\vec{0}. $$



Fix a  nonnegative Schwartz function $w\in \eS(\bR)$, set $w_\ve(t)=w(\ve t)$ and consider the random  function

$$ \bu_\ve(\vec{\theta})=\sum_{\vec{k}\in\bZ^m} X_{\vec{k}}\Psi_{\vec{k}}(\vec{\theta}), $$

where $X_{\vec{k}}$ are independent  Gaussian random variables with mean $0$ and variances

$$\var(X_{\vec{k}})= w(\ve|\vec{k}|). $$

We denote by $N(\bu_\ve)$ the number of critical points  of $\bu_\ve$ and  by $N_\ve$ its expectation

$$ N_\ve =\bsE\Bigl(\; N(\bu_\ve)\;\Bigr). $$

A simple computation shows that the covariance kernel of this random function is

$$\eE^\ve(\vec{\theta}_1,\vec{\theta}_2)= \frac{1}{(2\pi)^m}\sum_{\vec{k}\in\bZ^m }w(\ve|\vec{k}|)e^{-\ii\lan\vec{k}, \vec{\theta}_2-\vec{\theta_1}\ran}. $$

Set $\vec{\theta}:=\vec{\theta}_2-\vec{\theta}_1$ and define $\phi:\bR^m\to\bC$ by

$$\phi(\vec{x})=e^{-\ii\lan\vec{x},\frac{1}{\ve}\vec{\theta}\ran}  w(|\vec{x}|). $$

We deduce that

$$ \eE^\ve(\vec{\theta}_1,\vec{\theta}_2)=\frac{1}{(2\pi)^m}\sum_{\vec{k}\in\bZ^m}  \phi(\ve\vec{k}). $$
Using Poisson formula we deduce  that for any $a>0$ we have

$$\sum_{\vec{k}\in\bZ^m}\phi\left(\frac{2\pi}{a}\vec{k}\right)=\left(\frac{a}{2\pi}\right)^m \sum_{\vec{\nu}\in\bZ^m}\widehat{\phi}(a\vec{\nu}), $$

where  for any $f\in\eS(\bR^m)$ we denote by $\widehat{f}(\xi)$ its Fourier transform

$$\widehat{f}(\xi)=\int_{\bR^m} e^{-\ii\lan\xi,\vec{x}\ran} f(\vec{x})|d\vec{x}|. $$

If we let  $\frac{2\pi}{a}=\ve$, then we deduce

$$\eE^\ve(\vec{\theta}_1,\vec{\theta}_2)=\frac{1}{(2\pi\ve)^m} \sum_{\vec{\nu}\in\bZ^m}\widehat{\phi}\left(\frac{2\pi}{\ve}\vec{\nu}\right). $$


Let $v:\bR^m\to \bR$,  $v(\vec{x})=w(|\vec{x}|) $. Then

$$\widehat{\phi}(\xi)=\widehat{v}\Bigl(\;\xi+\frac{1}{\ve}\theta\;\Bigr). $$

Hence

$$\eE^\ve(\vec{\theta}_1,\vec{\theta}_2)= \frac{1}{(2\pi\ve)^m}\sum_{\vec{\nu}\in\bZ^m}\widehat{v}\left(\frac{1}{\ve}\vec{\theta}+\frac{2\pi}{\ve}\vec{\nu}\right). $$

Now observe that if $|\theta| \ll 2\pi$, then for $\vec{\nu}\in\bZ^m\setminus 0$ then for any $N>0$ there exists a constant  $C_N>0$ such that

$$ \left|\widehat{v}\left(\frac{1}{\ve}\vec{\theta}+\frac{2\pi}{\ve}\vec{\nu}\right)\right|\leq   C_N\ve^N|\nu|^{-N}. $$

We deduce that

$$ \eE^\ve(\vec{\theta}_1,\vec{\theta}_2) = \frac{1}{(2\pi\ve)^m}\left(\widehat{v}\left(\; \frac{1}{\ve}\vec{\theta}\;\right)+O\bigl(\; \ve^N\;\bigr)\;\right),\;\;\forall N>0. $$

The last asymptotic expansion can be  differentiated with respect to $\vec{\theta}_1$ and $\vec{\theta}_2$.

Now define the random function

$$\bsU_\ve:\bT^m\times \bT^m\to \bR,\;\;\bsU_\ve(\vec{\theta},\vec{\vfi})=\bu_\ve(\vec{\theta})+\bu_\ve(\vec{\vfi}). $$

We denote  by $N(\bsU_\ve)$ the number of critical points of $\bsU_\ve$ situated  outside the diagonal. Note that

$$ N(\bsU_\ve)= N(\bu_\ve)^2-N(\bu_\ve). $$

We would like to  understand the behavior of the expectation of $N(\bsU_\ve)$ as $\ve\searrow 0$.

The covariance kernel  of $\bsU_\ve$ is the function

$$ \widetilde{\eE}^\ve(\vec{\theta}_1,\vec{\vfi}_1; \vec{\theta}_2,\vec{\vfi}_2) =\eE^\ve(\vec{\theta}_1,\vec{\theta}_2)+\eE^\ve(\vec{\theta}_1,\vec{\vfi}_2)+\eE^\ve(\vec{\vfi}_1,\vec{\theta}_2)+\eE^\ve(\vec{\vfi}_1,\vec{\vfi}_2)$$

$$= \frac{1}{(2\pi\ve)^m}\Bigl( \;\widehat{v}(\;\ve^{-1}(\vec{\theta}_2-\vec{\theta}_1)\;)+ \widehat{v}(\ve^{-1}(\vec{\vfi}_2-\vec{\theta}_1)\;)\\
+\widehat{v}(\;\ve^{-1}(\vec{\theta}_2-\vec{\vfi}_1)\;) +\widehat{v}(\;\ve^{-1}(\vec{\vfi}_2-\vec{\vfi}_1)\;)+O(\ve^\infty)\;\Bigr). $$

Let us introduce the notation

$$\Theta:=(\vec{\theta},\vec{\vfi})\in\bT^m\times\bT^m , d(\Theta):=\vec{\vfi}-\vec{\theta}. $$


We  need to understand the quantities

$$\pa^\alpha_{\Theta_1}\pa^\beta_{\Theta_2}\teE^\ve(\Theta_1,\Theta_2)_{\Theta_1=\Theta_2=\Theta}=\bsE\bigl(\;\pa^\alpha_\Theta\bsU_\ve(\Theta)\cdot\pa^\beta_\Theta\bsU_\ve(\Theta)\;\bigr). $$

Note that $\widehat{v}(\xi)$ is radially symmetric, in fact it  can be written as $f(|\xi|^2)$  for some smooth function $f$. Indeed, we have  (see  Michael Taylor's notes;  he uses a different normalization for the Fourier transform.)


$$\widehat{v}(\xi)=\int_{\bR^m} v(|\vec{x}|) e^{-\ii\lan\xi,\vec{x}\ran}  =(2\pi)^{\frac{m}{2}}|\xi|^{1-\frac{m}{2}}\int_0^\infty v(r) J_{\frac{m}{2}-1}(r|\xi|) dr,$$

where $J_\nu$ denotes the Bessel function of first type and order $\nu$. For  any multi-indices $\alpha,\beta$ we have



$$(2\pi)^m\pa^\alpha_{\vec{\theta}_1}\pa^\beta_{\vec{\theta}_2}\teE^\ve(\Theta,\Theta)=  \ve^{-m-|\alpha|-|\beta|} \Bigl(\; (-1)^{|\alpha|}\pa^{\alpha+\beta}_\xi \widehat{v}(0) + O(\ve^\infty)\,\Bigr), \tag{1}$$


$$ (2\pi)^m\pa^\alpha_{\vec{\vfi}_1}\pa^\beta_{\vec{\vfi}_2}\teE^\ve(\Theta,\Theta)= \ve^{-m-|\alpha|-|\beta|}\Bigl( (-1)^{|\alpha|} \pa^{\alpha+\beta}_\xi\widehat{v}(0) +O(\ve^{\infty})\;\Bigr). \tag{2}$$

The  main term of this asymptotics  is trivial if $|\alpha|+|\beta|$ is odd. Next


$$(2\pi)^m\pa^\alpha_{\vec{\theta}_1}\pa^\beta_{\vec{\vfi}_2} \teE^\ve(\Theta,\Theta)= \ve^{-m-|\alpha|-|\beta}\Bigl( (-1)^{|\alpha|}\pa^{\alpha+\beta}_\xi\widehat{v}(\ve^{-1}d(\Theta) ) +O(\ve^\infty)\;\Bigr), \tag{3}$$


$$(2\pi)^m \pa^\alpha_{\vec{\vfi}_1}\pa^\beta_{\vec{\theta}_2}\teE^\ve(\Theta,\Theta) =\ve^{-m-|\alpha|-|\beta|} \Bigl(\;(-1)^{|\alpha|} \pa^{\alpha+\beta}_\xi \widehat{v}(\;-\ve^{-1}d(\Theta)\;)+ O(\ve^\infty)\;\Bigr)\\
=\ve^{-m-|\alpha|-|\beta|}\Bigl(\;(-1)^{|\beta|}\pa^{\alpha+\beta}_\xi\widehat{v}(\;\ve^{-1}d(\Theta)\;)+O(\ve^\infty)\;\Bigr).\tag{4}$$

For example  if $|\alpha|=2$ and $|\beta|=1$ we have

$$ (2\pi)^m\pa^\alpha_{\vec{\theta}_1}\pa^\beta_{\vec{\vfi}_1}\teE^\ve(\Theta,\Theta)= \ve^{-m-3}\Bigl(\;\pa^{\alpha+\beta}_\xi\widehat{v}(\;\ve^{-1}d(\Theta)\;)+O(\ve^\infty)\;\Bigr),\tag{3'} $$

Example 1.    Let us compute $\pa^\alpha_\xi f(|\xi|^2)$,  $|\alpha|\leq 4$.

We have

$$\pa_{\xi_i} f(|\xi|^2) = 2\xi_i f',\;\;\pa^2_{\xi_i\xi_j}f(|\xi|^2)= 2\delta_{ij} f' +4\xi_i\xi_j f'',$$

$$\pa^3_{\xi_i\xi_j\xi_k} f = 4\bigl(\; \delta_{ij}\xi_k+\delta_{ik}\xi_j+\delta_{jk}\xi_i\;\bigr)f''+8\xi_i\xi_j\xi_k f'''. $$

$$\pa^4_{\xi_i\xi_j\xi_k\xi_\ell} f(|\xi|^2)= 4\bigl(\;\delta_{ij}\delta_{k\ell}+\delta_{ik}\delta_{j\ell}+\delta_{jk}\delta_{i\ell}\;\bigr) f'' $$

$$+ 8\bigl(\; \delta_{ij}\xi_k\xi_\ell+\delta_{ik}\xi_j\xi_\ell+\delta_{jk}\xi_i\xi_\ell+\delta_{i\ell}\xi_j\xi_k+\delta_{j\ell}\xi_i\xi_k+\delta_{k\ell}\xi_i\xi_j\;\bigr) f''' +16\xi_i\xi_j\xi_k\xi_\ell f^{(4)}. $$

Example 2. Let's be more specific and  set $w(t)=e^{-t^2/2}$. Then $v(\vec{x})= e^{-|\vec{x}|^2/2}$, $f(s)=e^{-s}$ so that

$$\widehat{v}(\xi) = (2\pi)^{m} e^{-|\xi|^2/2}. $$


We can write

$$\widehat{v}(\xi) =(2\pi)^{m/2}\prod_{j=1}^m e^{-\xi_j^2/2}. $$

For any multi-index $\alpha=(\alpha_1,\dotsc, \alpha_m)$ we have


$$\pa^\alpha_\xi \widehat{v}(\xi) =(-1)^{|\alpha|}\underbrace{\left(\prod_{j=1}^m H_{\alpha_j}(\xi_j)\right) }_{=:H_\alpha(\xi)}\widehat{v}(\xi), $$

where $H_n$ denotes the $n$-th Hermite polynomial defined by

$$ \frac{d^n}{dx^n} e^{-x^2/2}= (-1)^nH_n(x) e^{-x^2/2}. $$


Let us point out that

$$ (-1)^{n+1}H_{n+1}(x) = (-1)^nH_n'(x) +(-1)^{n+1}xH_n(x), $$

$$ H_1(x)=x, \;\; H_2(x)=x^2-1,\;\;H_3(x) =x^3-3x,\;\; H_4(x)=x^4-6x^2 +3. $$



Observe that

$$\Hess\bsU_\ve(\Theta) =\Hess \bu_\ve(\vec{\theta})\oplus \Hess(\bu_\ve(\vec{\vfi}). $$



We need to understand the statistics of the following two random objects.

 $$d\bsU_\ve(\Theta),, $$


$$H_c(\Theta):= \bsE\Bigl(\;\Hess \bsU_\ve(\Theta)\;\bigr|\; d\bsU_\ve(\Theta)=0\;\Bigr),$$

as $d(\Theta)\to 0$, i.e.,  $\Theta$ approaches the diagonal in  $\bT^m\times \bT^m$. The covariance form $V_\ve$ of $d\bsU_\ve(\Theta)$ becomes singular as $d\to 0$.  Fortunately, something miraculous seems to be happening: as $d\to 0$  the Gaussian random variable  $H_c$ ecomes highly concentrated near the trivial matrix and in the limit it becomes the deterministic $0$-matrix. This leads to remarkable compensation is the Kac-Rice formula.