$\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\ve}{{\varepsilon}}$ $\newcommand{\eS}{\mathscr{S}}$ $\newcommand{\ii}{\boldsymbol{i}}$ $\newcommand{\bZ}{\mathbb{Z}}$
Suppose that $w\in \eS(\bR)$ is an even, nonnegative Schwartz function. Assume that $w\not\equiv 0$. $\newcommand{\hw}{\widehat{w}}$ We denote by $\hw(t)$ its Fourier transform
$$ \hw(t)=\int_{\bR}e^{-\ii t x} w(x) dx. $$
For $n\in \bZ$ we set $\newcommand{\be}{\boldsymbol{e}}$
$$ \be_n(\theta) :=\frac{1}{\sqrt{\pi}}\begin{cases}
\frac{1}{\sqrt{2}}, & n=0,\\
\sin n \theta , & n<0,\\
\cos n\theta , & n>0.
\end{cases} $$
Observe that the collection $\lbrace \be_n(\theta)\rbrace_{n\in\bZ}$ is an orthonormal basis of $L^2(\bR/2\pi\bZ)$. $\newcommand{\bT}{\mathbb{T}}$ For any positive integer $N$ we denote by $\bT^N$ the $N$-dimensional torus
$$\bT^N:= (\bR/2\pi\bZ)^N $$.
Consider the random Fourier series
$$ f_\ve(\theta)=\sum_{n\in \bZ} \sqrt{w(\ve n) } c_n \be_n(\theta), $$
where $(c_n)_{n\in\bZ}$ are i.i.d. Gaussian random variables with mean zero and variance $1$. $\newcommand{\eE}{\mathscr{E}}$ The correlation kernel of this random function is $\newcommand{\bsE}{\boldsymbol{E}}$ $\newcommand{\vfi}{\varphi}$
$$\eE^{\ve}:\bT^1\times \bT^1\to \bR,\;\;\eE^\ve (\theta,\vfi)=\bsE\bigl( f_\ve(\theta)\cdot f_\ve(\vfi)) =\sum_{n\in \bZ} w(\ve n) \be_n(\theta)\be_n(\vfi) $$
$$ =\frac{1}{2\pi} w(0)+\frac{1}{\pi}\sum_{n>0}w(\ve n)\cos n(\theta-\vfi)=\frac{1}{2\pi}\sum_{n\in\bZ}w(\ve n) e^{\ii n(\theta-\vfi)}=W_\ve(\theta-\vfi). \tag{1}\label{1}$$
Poisson formula. For any $\phi\in \eS(\bR)$ and any $c\in\bR\setminus 0$ we have
$$ \frac{2\pi}{c}\sum_{n\in\bZ} \phi\Bigl(\frac{2\pi n}{c}\Bigr)= \sum_{\nu\in\bZ} \widehat{\phi}(n c). $$
Suppose $\phi\in\eS(\bR)$ and $c$ are such that
$$ \phi\Bigl(\;\frac{2\pi n}{c}\;\Bigr)= w(\ve n) e^{\ii n(\theta-\vfi)} .$$
If we formally replace $n =\frac{c x}{2\pi}$ we deduce from the above equality that
$$\phi(x)= w\Bigl(\frac{\ve c x}{2\pi}\Bigr) e^{\ii\frac{c(\theta-\vfi)x}{2\pi}}=w(ax)e^{\ii b x}, \;\; a:=\frac{\ve c}{2\pi},\;\;b :=\frac{c(\theta-\vfi)}{2\pi}. $$
Then
$$ \widehat{\phi}(t) =\int_{\bR} e^{-\ii tx} w(ax) e^{\ii bx} dx = \frac{1}{a}\int_{\bR} e^{-\ii \frac{t-b}{a}x} w(y) dy = \frac{1}{a}\hw\Bigl( \frac{t-b}{a}\Bigr). $$
We now set $c:=2\pi$ so that $a=\ve$, $b=(\theta-\vfi)$. Using The Poisson formula in (\ref{1}) we deduce
$$W_\ve(\theta-\vfi)=\eE^\ve(\theta,\vfi) =\frac{1}{2\pi\ve} \sum_{n\in\bZ} \hw\Bigl(\frac{2\pi n-(\theta-\vfi)}{\ve}\Bigr) . \tag{2}\label{2} $$
Now consider the random function
$$ F_\ve:\bT^N\to \bR,\;\; F_\ve(\vec{\theta}) = \sum_{j=1}^n f_\ve(\theta_j). $$
The correlation kernel of this random function is
$$\eE_N^\ve(\vec{\theta},\vec{\vfi}) =\sum_{1\leq j,k\leq N} \eE^\ve(\theta_j-\vfi_k). $$
The differential of $F_\ve$ at a point $\vec{t}\in\bT^N$ is a Gaussian random vector with covariance matrix $\newcommand{\pa}{\partial}$
$$ S^\ve(\vec{t})= \Bigl( S^\ve_{jk}(\vec{t})\;\Bigr)_{1\leq j,k\leq N},\;\; S^\ve_{jk}(\vec{t})= \frac{\pa^2}{\pa\theta_j\pa \vfi_k} \eE_N^\ve\bigl(\;\vec{\theta},\vec{\vfi}\;\bigr)|_{\vec{\theta}=\vec{\vfi}=\vec{t}}=-W_\ve''(t_j-t_k)=\frac{1}{2\pi}\sum_{n\in\bZ} n^2w(\ve n) e^{\ii n(t_j-t_k)}\tag{3}\label{3}. $$
Definition. We say that $\vec{t}\in\bT^n$ is nondegenerate if $t_j-t_k\in\bR\setminus 2\pi\bZ$, $\forall j\neq k$. We denote by $\bT^N_*$ the collection of nondgenerate points in $\bT^N$.
$$\ast\ast\ast $$
We have the following result similar to the one in our previous post.
Proposition 1. There exists $\ve_0=\ve_0(w,N)>0$ such that if $\ve \in (0,\ve_0)$ and $\vec{t}\in \bT^N$ is nondegenerate, then the matrix $S^\ve(\vec{t})$ is positive definite.
Proof. Set
$$ Z_\ve:=\bigl\{ n\in\bZ;\;\;w(\ve n)\neq 0\;\bigr\}. $$
Consider the space $H_\ve$ consisting of functions $\newcommand{\bC}{\mathbb{C}}$
$$ u: Z_\ve \to \bC,\;\;\sum_{n\in Z_\ve} |u(n)|^2 n^2 w(\ve n) <\infty. $$
This is a separable Hilbert space with inner product
$$(u,v)_\ve= \frac{1}{2\pi} \sum_{n\in\bZ} u(n)\cdot \overline{v(n)}\; n^2w(\ve n). $$
We denote by $\Vert-\Vert_\ve$ the associated norm.
For $t\in\bT^1$ consider the truncated character $\chi^\ve_t:Z_\ve\to \bT^1$, $\chi^\ve_t(n)=e^{\ii tn}$. For $\vec{z}\in \bC^N$ $\newcommand{\vez}{{\vec{z}}}$ and $\vec{t}\in \bT^N$ consider $T_{\vez,\vec{t}}\in H_\ve$
$$ T_{\vez,\vec{t}}(n)=\sum_{j=1}^n z_j \chi^\ve _{t_j}(n)=\sum_{j=1}^N z_j e^{\ii t_j n},\;\;n\in Z_\ve. $$
From the equality (\ref{3}) we deduce that
$$ \sum_{j,k=1}^n S^\ve_{jk}(\vec{t}) z_j\bar{z}_k = \Vert T_{\vez,\vec{t}}\Vert_\ve^2. $$
Thus, the matrix $S^\ve(\vec{t})$ has a kernel if and only if the truncated characters $\chi^\ve_{t_1},\dotsc, \chi^\ve_{t_N}$ are linearly dependent. We show that this is not possible if $\vec{t}$ is nondegenerate and $\ve$ is sufficiently small.
Fix $\ve_0=\ve_0(N,w)$ such that if $\ve<\ve_0$ the support of $x\mapsto w(\ve x)$ contains a long interval of the form $[\nu_\ve, \nu_\ve+N-1]$, for some integer $\nu_\ve>0$. (Recall that $w$ is even.) In other words $\nu_\ve,\nu_\ve+1,\cdots,\nu_\ve+N-1\in Z_\ve$.
Let $\ve\in (0,\ve_0)$ and suppose that $\vez\in\bC^N\setminus 0$ and $\vec{t}\in\bT^N$ are such that such that $T_{\vez,\vec{t}}=0$. Thus
$$ (T_\vez, u)_\ve =0,\;\;\forall u\in H_\ve $$
For any $m\in Z_\ve $ consider the Dirac function $\delta_m: Z_\ve\to \bC$, $\delta_m(n)=\delta_{mn}=$ the Kronecker delta.
We deduce that for any $m=\nu_\ve,\nu_\ve+1,\dotsc, \nu_\ve+N-1$ we have
$$ 0 = (T_\vez, \delta_m)_\ve=m^2w(\ve m) \sum_{j=1}^N z_j e^{\theta_j m} . $$
This can happen if and only if
$$0= \det\left[
\begin{array}{cccc}
e^{\ii \nu_\ve t_1} & e^{\ii \nu_\ve t_2} & \cdots & e^{\nu_\ve t_N}\\
e^{\ii(\nu_\ve+1)t_1} & e^{\ii(\nu_\ve+1)t_2} & \cdots & e^{\ii (\nu_\ve+1) t_N}\\
\vdots & \vdots &\vdots &\vdots\\
e^{\ii(\nu_\ve+N-1)t_1} & e^{\ii(\nu_\ve+N-1)t_2} &\cdots & e^{\ii(\nu_\ve+N-1)t_N}
\end{array}
\right] = e^{\ii\nu_\ve(t_1+\cdots +t_N)} \prod_{j<k} \Bigl( e^{\ii t_k}-e^{\ii t_j}\Bigr) .$$
This shows that $S^\ve(\vec{t})$ has a kernel if and only of $\vec{t}$ is degenerate. Q.E.D.
Remark. Here is an alternate proof of Proposition 1 that yields a bit more. The above proof shows that
$$ S^\ve_{jk}(\vec{t})= (\chi_{t_j},\chi_{t_k})_\ve. $$
Suppose for simplicity that $0\in Z_\ve$, i.e., $w(0)>0$. Then for $\ve>0$ sufficiently small we have $1,\dotsc, N\in Z_\ve$. Observe that
$$(\delta_j,\delta_k)_\ve= \frac{k^2w(\ve k)}{2\pi}\delta_{jk},\;\;j,k=1,\dotsc, N. $$
We have a Cauchy-Schwartz inequality
$$\Bigl|\; \bigl( \chi_{t_1}\wedge\cdots \chi_{t_n}, \delta_1\wedge \cdots \wedge \delta_N\;\bigr)_\ve\;\Bigr| \leq \bigl|\; \chi_{t_1}\wedge\cdots \wedge\chi_{t_n}\;\bigr|_\ve\cdot \bigl|\;\delta_1\wedge \cdots \wedge \delta_N\;\bigr|_\ve. $$
This translates to
$$ \Bigl| \det\Bigl(\; (\chi_{t_j},\delta_k)_\ve\;\Bigr)_{1\leq i,j\leq N}\;\Bigr| \leq \sqrt{\det\Bigr( \; (\chi_{t_j},\chi_{t_k})_\ve\;\Bigr)_{1\leq i,j\leq N} } \cdot \sqrt{\det\Bigr( \; (\delta_j,\delta_k)_\ve\;\Bigr)_{1\leq i,j\leq N} }, $$
or, equivalently
$$ \prod_{j<k} \Bigl| e^{\ii t_j}-e^{\ii t_k}\Bigr|^2 \leq \frac{1}{(2\pi)^N}\Bigl(\prod_{j=1}^N j^2w(\ve j)\Bigr) \det S^\ve(\vec{t}). \tag{4} \label{4} $$
$$\ast\ast\ast$$
The basic question that interests me is the following: what happens to $S^\ve(\vec{t})$ as $\ve\to 0$, and $\vec{t}$ is nondegenerate.
Observe that (\ref{2}) implies that
$$W_\ve''(t)=\frac{1}{2\pi\ve^3}\sum_{n\in\bZ}\hw''\Bigl(\frac{2\pi n-t}{\ve}\Bigr). $$
We make the change in variables $t=\ve \tau$, we set
$$ C^\ve(\vec{\tau}) := S^\ve(\ve\vec{\tau})$$
and we deduce
$$C^\ve_{jk}(\vec{\tau})=\frac{1}{2\pi}\sum_{n\in\bZ}n^2w(\ve n) e^{\ii\ve n(\tau_j-\tau_k)} =-\frac{1}{2\pi\ve^3}\sum_{n\in\bZ}\hw''\Bigl(\tau_j-\tau_k-\frac{2\pi n}{\ve}\Bigr),\;\;0\leq \tau_j <\frac{2\pi}{\ve},\;\;j=1,\dotsc, N. \tag{5}\label{5} $$
For $\vec{t}\in\bT^N$ nondegenerate we denote by $X(\vec{t})\subset H_\ve$ the vector space spanned by the characters $\chi_{t_j}$, $j=1,\dotsc, N$. $\newcommand{\bsD}{\boldsymbol{D}}$ Denote by $\bsD_N\subset H_\ve$ the space spanned by $\delta_1,\dotsc, \delta_n$. For $k=1,\dotsc, N$ we set $\newcommand{\bde}{\check{\delta}}$
$$\bde_k=\bde_k^\ve :=\frac{\sqrt{2\pi}}{k\sqrt{w(\ve k)}}\delta_k. $$
By construction, the collection $\bde_1,\dotsc,\bde_N$ is an $(-,-)_\ve$-orthonormal basis of $\bsD_N$.
The $(-,-)_\ve$-orthogonal projection $P_\ve=P_\ve(\vec{t}): X(\vec{t})\to \bsD_n$ is given by
$$P_\ve \chi_{t_j} =\sum_{k=1}^n (\chi_{t_j},\bde_k)_\ve \bde_k = \sum_{k=1}^N e^{\ii kt_j} \delta_k. $$
With respect to the natural bases $\chi_{t_1},\dotsc,\chi_{t_N}$ of $X(\vec{t})$ and $\delta_1,\dotsc, \delta_N$ of $\bsD_N$ the projection is therefore given by the Vandermonde matrix
$$ V= V(\vec{t}),\;\; V_{kj}= e^{\ii k t_j},\;\; V=\left[\begin{array}{cccc}
e^{\ii t_1} & e^{\ii t_2} &\cdots & e^{\ii t_N}\\
e^{2\ii t_1} & e^{2\ii t_2} & \cdots & e^{2\ii t_N}\\
\vdots &\vdots &\vdots &\vdots\\
e^{N\ii t_1} & e^{N\ii t_2} &\cdots & e^{N\ii t_N}
\end{array}
\right]. $$
Hope to add bits of mathematics or mathematical news I find interesting.
Showing posts with label Poisson formula. Show all posts
Showing posts with label Poisson formula. Show all posts
Wednesday, November 21, 2012
Tuesday, July 24, 2012
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.
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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