$\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bT}{\mathbb{T}}$ $\newcommand{\eE}{\mathscr{E}}$ $\newcommand{\ve}{{\varepsilon}}$ $\newcommand{\lan}{\langle}$ $\newcommand{\ran}{\rangle}$ $\newcommand{\bu}{\boldsymbol{u}}$ $\newcommand{\eS}{\mathscr{S}}$ $\DeclareMathOperator{\var}{\boldsymbol{var}}$ $\newcommand{\ii}{\boldsymbol{i}}$ $\newcommand{\bsU}{\boldsymbol{U}}$ $\newcommand{\vfi}{\varphi}$ $\newcommand{\bsE}{\boldsymbol{E}}$ $\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.
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