2. Integral formulas

Proof. (1) Given an ordered partition $\mathbf{S}=(S_1,\ldots,S_{ \nu})$ of [n] into non-empty subsets $S_{\alpha}$ , define $Y_{\alpha}\;:\!=\; \sum_{j\in S_{\alpha}}X_j$ , so that $\sum_{\beta\in [\nu]}Y_{\beta}=\sum_{j\in [n]}X_j$ . $\mathbf{S}$ is perfect if and only if

\begin{equation*} \nu \cdot Y_{\alpha} -\sum_{\beta\in [ \nu]}Y_{\beta}=0\quad \text{for all $ \alpha\in [ \nu]$.}\end{equation*}

Clearly, for a perfect partition to exist, $\sum_{j\in [n]}X_j$ must be divisible by $\nu$ . Let $\mathbb{I}(A)$ denote the indicator of an event A. Then

\begin{align*}\mathbb{I}(\text{$\mathbf{S}$ perfect})&=\prod_{\alpha\in [ \nu]}\mathbb{I}\Biggl( \nu\cdot Y_{\alpha}-\sum_{\beta\in [ \nu]}Y_{\beta}=0\Biggr)\\*&=\prod_{\alpha\in [ \nu]}\dfrac{1}{2\pi}\int_{x\in [\!-\pi,\pi]}\exp\Biggl[{{\mathrm{i}}} x\Biggl( \nu \cdot Y_{\alpha} -\sum_{\beta\in [ \nu]}Y_{\beta}\Biggr)\Biggr]\,{{\mathrm{d}}} x\\*&=\dfrac{1}{(2\pi)^{ \nu}}\int_{\mathbf{x}\in [\!-\pi,\pi]^{ \nu}}\exp\Biggl[{{\mathrm{i}}} \sum_{\alpha\in [\nu]}x_{\alpha}\Biggl( \nu\cdot Y_{\alpha}-\sum_{\beta\in [ \nu]}Y_{\beta}\Biggr)\Biggr]\,{{\mathrm{d}}} \mathbf{x},\quad \mathbf{x}\;:\!=\; \{x_{\alpha}\}_{\alpha\in [\nu]}.\end{align*}

Summing the above equation over all ordered partitions $\mathbf{S}$ with non-empty $S_{\alpha}$ , and interchanging summation and integration on the right-hand side, we obtain

(2.4) \begin{equation}Z_n=\dfrac{1}{(2\pi)^{ \nu}}\int_{\mathbf{x}\in [\!-\pi,\pi]^{ \nu}}\sum_{\mathbf{S}}\exp\Biggl[{{\mathrm{i}}} \sum_{\alpha\in [ \nu]}x_{\alpha}\Biggl( \nu\cdot Y_{\alpha}-\sum_{\beta\in [ \nu]}Y_{\beta}\Biggr)\Biggr]\,{{\mathrm{d}}} \mathbf{x}.\end{equation}

What if $\mathbf{S}$ is such that some $S_{\alpha_0}=\emptyset$ ? In this case $Y_{\alpha_0}=\sum_{j\in S_{\alpha_0}}X_j=0$ ; so the corresponding factor in the exponent sum is $-\sum_{j\in [n]}X_j\neq 0$ . Then the integral of the corresponding exponential function over the (hyper)cube $[\!-\pi,\pi]^{\nu}$ is zero. Therefore we may and will extend the sum in (2.4) to all ordered sequences $\mathbf{S}$ of sets $S_1,S_2,\ldots,S_{ \nu}$ , empty or non-empty, that form a partition of [n]. Regrouping summands in the exponent, we obtain

\begin{gather*}\exp\Biggl[{{\mathrm{i}}} \sum_{\alpha\in [ \nu]}x_{\alpha}\Biggl( \nu\cdot Y_{\alpha}-\sum_{\beta\in [ \nu]}Y_{\beta}\Biggr)\Biggr]=\prod_{j\in [n]}\exp({{\mathrm{i}}} \sigma_jX_j),\\* \sigma_j=\sum_{\alpha\in [ \nu]}x_{\alpha}(\nu\,\delta_{\alpha}(j)-1), \quad \delta_{\alpha}(j)\;:\!=\; \mathbb{I}(j\in S_{\alpha}).\end{gather*}

Such a relaxed partition $\mathbf{S}$ is uniquely determined by values of $\delta_{\alpha}(j)\in\{0,1\}$ , subject to $\sum_{\alpha\in [\nu]}\delta_{\alpha}(j)=1$ , ( $j\in [n]$ ), but not to $\sum_{j\in [n]}\delta_{\alpha}(j)\ge 1$ , ( $\alpha\in [\nu]$ ), because empty $S_{\alpha}$ are allowed now. Thus $\delta_{\alpha}(j)$ vary independently for different j. So, introducing

\[\Sigma(\mathbf{x})\;:\!=\; \Biggl\{\sum_{\alpha\in [ \nu]}x_{\alpha}(\nu\,\delta_{\alpha}-1)\colon \delta_{\gamma}\in \{0,1\},\,\sum_{\gamma\in [\nu]}\delta_{\gamma}=1\Biggr\},\]

for the sum over relaxed partitions we obtain

\begin{align*}\sum_{\mathbf{S}}\exp\Biggl[{{\mathrm{i}}} \sum_{\alpha\in [\nu]}x_{\alpha}\Biggl( \nu\cdot Y_{\alpha}-\sum_{\beta\in [ \nu]}Y_{\beta}\Biggr)\Biggr]&=\prod_{j\in [n]}\sum_{\sigma\in \Sigma(\mathbf{x})}\exp({{\mathrm{i}}} X_j \sigma)\\*&=\prod_{j\in [n]}\sum_{\alpha\in [\nu]}\exp\Biggl[{{\mathrm{i}}} \Biggl(\nu x_{\alpha}-\sum_{\beta\in [\nu]}x_{\beta}\Biggr)X_j\Biggr]\\*&=\prod_{j\in [n]}\sum_{\alpha\in [\nu]}\exp({{\mathrm{i}}} y_{\alpha}(\mathbf{x})X_j).\end{align*}

We conclude that

\begin{equation*}Z_n=\dfrac{1}{(2\pi)^{ \nu}}\int\limits_{\mathbf{x}\in [\!-\pi,\pi]^{ \nu}} \prod_{j\in [n]}\sum_{\alpha\in [\nu]}\exp({{\mathrm{i}}} y_{\alpha}(\mathbf{x})X_j)\,{{\mathrm{d}}} \mathbf{x},\end{equation*}

which is (2.1) in the lemma. Equating the expectations of both sides and using independence of the $X_j$ , we get (2.2).

(2) Equation (2.1) implies that

\[Z_n^2=\dfrac{1}{(2\pi)^{ 2\nu}} \int\limits_{\mathbf{x}, \mathbf{x}'}\prod_{j\in [n]}f(\mathbf{x},X_j)\overline{f(\mathbf{x}',X_j)}\,{{\mathrm{d}}} \mathbf{x} \,{{\mathrm{d}}} \mathbf{x}',\]

so that

\begin{equation*}\mathbb{E}[Z_n^2]=\dfrac{1}{(2\pi)^{ 2\nu}} \int\limits_{\mathbf{x}, \mathbf{x}'}\mathbb{E}^n\bigl[f(\mathbf{x},X)\overline{f(\mathbf{x}',X)}\bigr]\,{{\mathrm{d}}} \mathbf{x} \,{{\mathrm{d}}} \mathbf{x}'.\end{equation*}

Notes. Thus, while there is no hope of dealing with $Z_n$ directly, the above approach delivers exact integral formulas for the first two moments of $Z_n$ . Why do these formulas hold promise? Because the integrands in the equations (2.2) and (2.3) certainly suggest that the dominant contributions to $\mathbb{E}[Z_n]$ and $\mathbb{E}[Z_n^2]$ may come from $\mathbf{x}$ and $\{\mathbf{x},\mathbf{x}'\}$ such that $|\mathbb{E}[f(\mathbf{x},X)]|$ and $|\mathbb{E}[f(\mathbf{x},X),\overline{f(\mathbf{x}',X)}]|$ are close to their respective maxima, $\nu$ and $\nu^2$ respectively. This is definitely similar to what we encounter dealing with the multi-dimensional local limit theorems, based on the observation that the distribution of the sum of n i.i.d. random vectors is the inverse Fourier transform of the individual characteristic function raised to the nth power. That is what made the resulting integrals amenable to a Laplace-type method.

The key difference is that normally, without an exception known to the author, the dominant contributor to the integral is a single tiny ball centered at a certain point, or a set of isolated points. For the integrals in question, though, the dominant contributors turn out to be a collection of wire-thin circular cylinders parallel to the cube’s main diagonal for $\mathbb{E}[Z_n]$ , and Cartesian products of pairs of such cylinders for $\mathbb{E}[Z_n^2]$ . For $\nu=2$ , there was just a single cylinder, one enclosing the main diagonal of the cube $[\!-\pi,\pi]^{2}$ . We will see that for $\nu>2$ there are $\nu^{-1}[\nu^{\nu}-(\nu-1)^{\nu}-(\nu-1)(-1)^{\nu}]$ of those cylinders.

Counting the $\nu$ -dimensional cylinders is combinatorics alone, but next comes an analytical problem of actually showing that, the cylinders aside, the rest of the cube (of the product of two cubes, for $E[Z_n^2]$ ) contributes negligibly to $\mathbb{E}[Z_n]$ ( $\mathbb{E}[Z_n^2]$ ). Through a sequence of progressively sharper estimates, we upper-bound that peripheral contribution when cross-sections of the cylinders are $(\nu-1)$ -dimensional spheres of radius $(M\nu)^{-1}n^{-1/2}\log n$ . These cylinders are thin enough to derive a sharp approximation of their overall contribution. We get definitive results for the asymptotic behavior of $\mathbb{E}[Z_n]$ in the cases when the contribution of the cylinders dwarfs the contribution of the complementary part of $[\!-\pi,\pi]^{\nu}$ .

In the author’s view, the problem and the method of its analysis belong, broadly speaking, to a burgeoning area of analytic combinatorics, that combines enumerative techniques with a sharp asymptotic analysis of the complex-valued integrals; see Flajolet and Sedgewick [Reference Flajolet and Sedgewick6], Pemantle and Wilson [Reference Pemantle and Wilson12], and Pemantle, Wilson, and Melczer [Reference Pemantle, Wilson and Melczer13].

3. Expected number of perfect partitions

We begin with the following result.

Note. We included the case $\nu=2$ , since somehow a rigorous proof of $\mathbb{E}[Z_n]\to\infty$ , letting alone an explicit bound, has not been known.

Proof. Let $n=\nu a +r$ , $r<\nu$ . Introduce the integers $n_1=\cdots=n_{\nu-1}=a$ and $n_{\nu}=a+r$ ; clearly $\sum_{i\in [\nu]}n_i=n$ , that is, $n_1,\ldots,n_{\nu}$ is a composition of n. Furthermore,

\[n_1=\cdots=n_{\nu-1}=\frac{n}{\nu}-\frac{r}{\nu},\quad n_{\nu}=\frac{n}{\nu}+\frac{r(\nu-1)}{\nu}.\]

The number of ways to partition $X_1,\ldots, X_n$ into the subsets of cardinalities $n_1,\ldots, n_{\nu}$ is the multinomial coefficient

\[\frac{n!}{\prod_{i\in [\nu]}n_i!}=(1+{{\mathrm{o}}} (1))c(\nu) n^{-(\nu-1)/2} \nu^n {,}\]

where we used

\[m!=(2\pi m)^{1/2}\biggl(\frac{m}{{{\mathrm{e}}}}\biggr)^m (1+{{\mathrm{O}}} (m^{-1})).\]

Let $S_{n_1},\ldots, S_{n_{\nu}}$ stand for the sums of $X_j$ from those subsets. Introduce the integer

\[s=\biggl\lfloor \frac{n}{\nu}\mathbb{E}[X]\biggr\rfloor.\]

Then

\begin{align*}\mathbb{P}(S_{n_i}=s)&=\mathbb{P}\biggl(\frac{S_{n_i}-\mathbb{E}[S_{n_i}]}{\sqrt{\text{Var}(S_{n_i})}}=\frac{s-n_i\mathbb{E}[X]}{\sqrt{n_i\text{Var}(X)}}\biggr)\\*&=\mathbb{P}\biggl(\frac{S_{n_i}-\mathbb{E}[S_{n_i}]}{\sqrt{\text{Var}(S_{n_i})}}={{\mathrm{O}}} (n^{-1/2})\biggr)\\&=(1+{{\mathrm{o}}} (1))(2\pi\text{Var}(S_{n_i}))^{-1/2}\\*&=(1+{{\mathrm{o}}} (1)) c(\nu)n^{-1/2}.\end{align*}

(For the second line we used the local central limit theorem, applicable since M is fixed.) Consequently

\begin{equation*}\mathbb{E}[Z_n]\ge \frac{n!}{\prod_{i\in [\nu]}n_i!}\prod_{i\in [\nu]}\mathbb{P}(S_{n_i}=s)\ge n^{-\nu}\nu^n\to\infty.\end{equation*}

From now on we focus on $M=M(n)\to\infty$ . Using Lemma 2.1, we prove the following.

Theorem 3.2. Let $\nu\ge 3$ . If

\[\lim\frac{n}{\log M}<\frac{2}{\log \nu},\]

then $\lim\mathbb{E}[Z_n]=0$ .

1. (i) Suppose $\nu=3$ . If

   \[\lim\frac{n}{\log M}\in \biggl(\frac{2}{\log 3},\infty\biggr)\setminus\{2,4\},\]
   then
   (3.1) \begin{equation}\mathbb{E}[Z_n]=(1+{{\mathrm{o}}} (1))\dfrac{\nu^n}{M^{\nu-1}}\cdot\dfrac{\nu^{\nu-3/2}}{(2\pi\nu c_Mn)^{{{(\nu-1)}/{2}}}}\to\infty,\end{equation}
   and $c_{\scriptscriptstyle M}\;:\!=\; M^{-2}\mathbb{E}[X^2]= 1/3+{{\mathrm{O}}} (M^{-1})$ .

2. (ii) Suppose $\nu\ge 4$ . Then (3.1) holds if

   \[\lim\frac{n}{\log M}\in \biggl(\frac{2}{\eta(\nu)},\infty\biggr)\setminus\{2(\nu-1)\},\]
   where
   \[\eta(\nu)\in \biggl(\frac{1}{\nu-1},\frac{2\log \nu}{\nu-1}\biggr)\]
   is the larger of two roots of
   \[\frac{{{\mathrm{e}}}^{(\nu-1)\eta}}{\nu {{\mathrm{e}}} \eta}-1,\]
   the smaller one being below ${{1}/{(\nu-1)}}$ .

Note. It is unclear whether $\mathbb{E}[Z_n]\to\infty$ if $\nu\ge 3$ and $\lim{{n}/{\log M}}=2(\nu-1)$ , or if $\nu=3$ and $\lim{{n}/{\log M}}=2$ . Perhaps $\mathbb{E}[Z_n]$ diverges to infinity in all these cases, albeit at a rate lower then for $\lim{{n}/{\log M}}$ arbitrarily close to but different from $2(\nu-1)$ for $\nu\ge 3$ , and from 2 for $\nu=3$ , respectively. The case $\nu\ge 4$ ,

\[\lim\frac{n}{\log M}\in \biggl(\frac{2}{\log\nu}, \frac{2}{\eta(\nu)}\biggr){,}\]

remains open as well.

Proof. By (2.2), we have

\begin{align*}\mathbb{E}[Z_n]&= \dfrac{1}{(2\pi)^{\nu}} \int\limits_{\mathbf{x}\in [\!-\pi,\pi]^{\nu}} \mathbb{E}^n[f(\mathbf{x},X) ] \,{{\mathrm{d}}} \mathbf{x},\\*\mathbb{E}[f(\mathbf{x},X)]&=\sum_{\alpha\in [\nu]}\mathbb{E}\bigl[{{\mathrm{e}}}^{{{\mathrm{i}}} y_{\alpha}X}\bigr],\quad y_{\alpha}=y_{\alpha}(\mathbf{x})\;:\!=\; \nu x_{\alpha}-\sum_{\beta\in [\nu]}x_{\beta}.\end{align*}

(I) We begin by counting the cylinders that presumably contribute most to the integrals evaluating the first two moments of $Z_n$ . Since X is integer-valued, $|\mathbb{E}[f(\mathbf{x},X) ] |$ certainly attains its maximum $\nu$ at every $\mathbf{x}$ such that for each $\alpha\in [\nu]$ we have $y_{\alpha}=y_{\alpha}(\mathbf{x})=k_{\alpha}\pi$ , and $k_{\alpha}$ is even. Here necessarily $\sum_{\alpha\in [\nu]}k_{\alpha}=0$ , since $\sum_{\alpha\in [\nu]}y_{a}(\mathbf{x})=0$ . Given such $\mathbf{k}=\{k_{\alpha}\}_{\alpha\in [\nu]}$ , the set of $\mathbf{x}$ satisfying

\begin{equation*} \nu x_{\alpha}-\sum_{\beta\in [\nu]} x_{\beta}= k_{\alpha}\pi,\quad {\alpha\in [\nu],}\end{equation*}

is a straight line $\mathcal L_{\mathbf{k}}$ in $\mathbb{R}^{\nu}$ , since the $\nu\times\nu$ matrix on the left-hand side has rank $\nu-1$ . We will prove that $|\mathbb{E}[f(\mathbf{x},X) ] | < \nu$ everywhere else. Let us enumerate the lines $\mathcal L_{\mathbf{k}}$ that contain interior points of the cube $[\!-\pi,\pi]^{\nu}$ . To this end, notice that each line’s parametric equation is $x_{\alpha}=\nu^{-1}(t+k_{\alpha}\pi)$ , $t\in \mathbb{R}$ . This line contains an interior point $\mathbf{x}$ of the cube $[\!-\pi,\pi]^{\nu}$ if and only if, for some t,

\[\nu^{-1}(t+k_{\alpha}\pi)\in (\!-\pi,\pi) \quad\text{for all $\alpha\in [\nu]$,}\]

or equivalently

(3.2) \begin{equation}-\pi(\nu+\min k_{\alpha}) < t < \pi(\nu-\max k_{\alpha}).\end{equation}

In particular, $\max k_{\alpha}-\min k_{\alpha}\le 2(\nu-1)$ , since $k_{\alpha}$ are even. For $\nu=2$ we have $k_1=k_2=0$ since $k_1+k_2=0$ . From now on we focus on $\nu>2$ . We consider these (admissible) lines only, since any other line can share at most one point with the cube. Further, the set of all such touch points is finite.

Indeed, suppose $\mathcal L$ does not contain interior points of the cube. If $\mathcal L$ ’s segment is in the cube, then this segment, i.e. a non-zero multiple of $\mathbf{e}\;:\!=\; (1,\ldots,1)\in {R^{\nu}}$ , belongs to one of $2\nu$ faces, cubes of dimension $\nu-1$ . Hence the segment is orthogonal to a coordinate vector $(0,\ldots,0,1,0,\ldots 0)\in {R^{\nu}}$ – a contradiction. So $\mathcal L$ shares at most one point with the cube. Any such touch point $\mathbf{x}$ satisfies $y_{\alpha}(\mathbf{x})=\pi k_{\alpha}$ , $\alpha\in [\nu]$ , and it has to have a component $x_{\alpha'}=\pi$ , and a component $x_{\alpha''}=-\pi$ , for that matter. Otherwise $\mathcal L$ would have contained interior points of the cube. The linear system $y_a(\mathbf{x})=k_{\alpha}\pi$ , $x_{\alpha'}=\pi$ has at most one solution in the cube, because the associated matrix has rank $\nu$ . So, since $|k_{\alpha}|\le 2(\nu-1)$ , $\alpha\in [\nu]$ , the number of such solutions (i.e. touch points) is at most $\nu(2\nu-1)^{\nu}$ .

By (3.2), $\mathbf{k}=\{k_{\alpha}\}_{\alpha\in [\nu]}$ , which consists of even numbers adding up to 0, is admissible if and only if $d(\mathbf{k})\;:\!=\; 2\nu+\min_{\alpha}k_{\alpha}-\max_{\alpha}k_a\ge 2,$ since the range of the parameter t is $\pi d(\mathbf{k})$ , and the length of the corresponding line segment is

(3.3) \begin{equation}L(\mathbf{k})=\pi d(\mathbf{k})\nu^{-1/2}=\pi \nu^{-1/2} \big(2\nu+\min_{\alpha}k_{\alpha}-\max_{\alpha}k_a\big).\end{equation}

Let even numbers $a\le 0\le b$ be generic values of $\min_{\alpha}k_{\alpha}$ and $\max_{\alpha}k_{\alpha}$ . Let $\mathcal N(a,b)$ be the total number of admissible $\{k_{\alpha}\}_{\alpha\in [\nu]}$ with parameters a, b. Obviously $\mathcal N(a,a)=\mathbb{I}(a=0)$ , and $\mathcal N(a,a+2)=0$ . Consider $r\;:\!=\; b-a\in [4,2(\nu-1)]$ , $2(\nu-1)$ coming from $d(\mathbf{k})\ge 2$ . A generic $\{k_{\alpha}\}$ with given $a=\min_{\alpha}k_{\alpha}$ , $b=\max_{\alpha}k_{\alpha}$ , contains some $\mu_1 > 0$ components equal to a, and $\mu_2 > 0$ components equal to b, with $\mu_1+\mu_2\le \nu$ , and $\mu_3\;:\!=\; \nu-\mu_1-\mu_2$ remaining even components, with values strictly between a and b, such that the total sum of all $\nu$ components is zero. Consequently, if $b\ge a+4$ , then

(3.4) \begin{equation}N(a,b)=[\zeta^0]\sum_{\mu_1,\,\mu_2 > 0}\dfrac{\nu!}{\mu_1!\,\mu_2!\,\mu_3!}\cdot \zeta^{\mu_1 a+\mu_2 b}\Biggl(\sum_{\mathrm{even}\,j=a+2}^{b-2}\zeta^j\Biggr)^{\mu_3},\end{equation}

with the last sum equal to

\begin{equation*} \zeta^{a+2}\sum_{\mathrm{even }j=0}^{r-4}\zeta^j=\frac{\zeta^{a+2}(1-\zeta^{r-2})}{1-\zeta^2},\quad r\;:\!=\; b-a.\end{equation*}

So, by inclusion–exclusion, the function of $\zeta$ in (3.4) is equal to

\begin{align*}&\biggl[\zeta^a+\zeta^b+\dfrac{\zeta^{a+2}(1-\zeta^{r-2})}{1-\zeta^2}\biggr]^{\nu}-\biggl[\zeta^a+\dfrac{\zeta^{a+2}(1-\zeta^{r-2})}{1-\zeta^2}\biggr]^{\nu} \\*&\quad\quad -\biggl[\zeta^b+\dfrac{\zeta^{a+2}(1-\zeta^{r-2})}{1-\zeta^2}\biggr]^{\nu}+\biggl[\dfrac{\zeta^{a+2}(1-\zeta^{r-2})}{1-\zeta^2}\biggr]^{\nu}\\*&\quad =\zeta^{a\nu}\biggl[\biggl(\dfrac{1-\zeta^{r+2}}{1-\zeta^2}\biggr)^{\nu}-\biggl(\dfrac{1-\zeta^r}{1-\zeta^2}\biggr)^{\nu}-\biggl(\dfrac{\zeta^2-\zeta^{r+2}}{1-\zeta^2}\biggr)^{\nu}+\biggl(\dfrac{\zeta^2-\zeta^r}{1-\zeta^2}\biggr)^{\nu}\biggr].\end{align*}

So we have

(3.5) \begin{equation}\begin{aligned}\mathcal N(a,b)&=[\zeta^{-a\nu}] \mathcal F_r(\zeta),\\\mathcal F_r(\zeta)&=\biggl(\dfrac{1-\zeta^{r+2}}{1-\zeta^2}\biggr)^{\nu}-\biggl(\dfrac{1-\zeta^r}{1-\zeta^2}\biggr)^{\nu} -\biggl(\dfrac{\zeta^2-\zeta^{r+2}}{1-\zeta^2}\biggr)^{\nu}+\biggl(\dfrac{\zeta^2-\zeta^r}{1-\zeta^2}\biggr)^{\nu}.\end{aligned}\end{equation}

Since $b=a+r > 0$ , we must have $\mathcal N(a,b)=0$ for $a\le -r$ . As a partial validation, one can check that $[\zeta^{-a\nu}] \mathcal F_r(\zeta)$ is indeed zero for $a\le -r$ , and also for $a=0$ . What we will actually need is

(3.6) \begin{equation}\mathcal N(r)\;:\!=\; \sum_{\mathrm{even }a < 0}\mathcal N(a,a+r),\quad \mathrm{even }r\in [4,2(\nu-1)],\end{equation}

the total number of the relevant $\nu$ -tuples of even integers $\{k_{\alpha}\}_{\alpha\in [\nu]}$ adding up to 0, such that $\max_{\alpha\in [\nu]}k_{\alpha}-\min_{\alpha\in [\nu]}k_{\alpha}=r$ . In fact, ultimately we are after a single number, namely

\[\mathcal M=\mathcal M(\nu)\;:\!=\; 2\nu+\sum_{\mathrm{even }r\in [4,2(\nu-1)]}(2\nu-r)\mathcal N(r),\]

since $\pi\nu^{-1/2}\mathcal M$ is the total length of the in-cube segments of lines for even tuples $\mathbf{k}$ . Combining (3.5) and (3.6) and changing the order of double summation, we obtain

\begin{equation*} \mathcal M=2\nu+\sum_{\mathrm{even }a\le 0} [\zeta^{-a\nu}]\sum_{\mathrm{even }r\in [4,2(\nu-1)]}(2\nu-r)\mathcal F_r(\zeta).\end{equation*}

We included $a=0$ since $\mathcal F_r(0)=0$ . Using the formula for $\mathcal F_r(\zeta)$ , and telescoping, we get, with r even,

\begin{align*} \sum_{r\in [4,2(\nu-1)]} \mathcal F_r(\zeta) &=\biggl(\dfrac{1-\zeta^{2\nu}}{1-\zeta^2} \biggr)^{\nu} -\biggl(\dfrac{1-\zeta^4}{1-\zeta^2} \biggr)^{\nu} -\biggl(\dfrac{\zeta^2-\zeta^{2\nu}}{1-\zeta^2} \biggr)^{\nu} +\biggl(\dfrac{\zeta^2-\zeta^4}{1-\zeta^2} \biggr)^{\nu} \notag \\*&=\biggl(\dfrac{1-\zeta^{2\nu}}{1-\zeta^2} \biggr)^{\nu}-\biggl(\dfrac{\zeta^2-\zeta^{2\nu}}{1-\zeta^2} \biggr)^{\nu}-(1+\zeta^2)^{\nu}+\zeta^{2\nu} \notag \\*&\;:\!=\; S_1(\zeta^2).\end{align*}

Then

\[\sum_{\mathrm{even }a\le 0}[\zeta^{-a\nu}]\sum_{\mathrm{even }r\in [4,2(\nu-1)]}\mathcal F_r(\zeta)=\sum_{\alpha\ge 0}[z^{\alpha\nu}]S_1(z).\]

Now, for $S(z)\;:\!=\; \sum_{r\ge 0}c_rz^r$ and

\[z_j\;:\!=\; \exp\biggl({{\mathrm{i}}} \frac{2\pi j}{\nu}\biggr),\quad j\in [0,\nu-1],\]

we have

\begin{equation*} \frac{1}{\nu}\sum_{j=0}^{\nu-1}S(z_j)=\sum_{r\ge 0}c_r\frac{1}{\nu}\sum_{j=0}^{\nu-1}z_j^r=\sum_{a\ge 0}[z^{a\nu}]S(z).\end{equation*}

Therefore

(3.7) \begin{align}&\sum_{\mathrm{even }a\le 0}[\zeta^{-a\nu}]\sum_{\mathrm{even }r\in [4,2(\nu-1)]}\mathcal F_r(\zeta)\notag \\*&\quad =\sum_{\alpha\ge 0}[z^{\alpha\nu}]S_1(z) \notag \\&\quad =\frac{1}{\nu}\sum_{j=0}^{\nu-1}\biggl[\biggl(\frac{1-z_j^{\nu}}{1-z_j}\biggr)^{\nu}-\biggl(\frac{1-z_j^{\nu-1}}{1-z_j}\biggr)^{\nu}\biggr] -\sum_{a\ge 0}[z^{a\nu}][(1+z)^{\nu}-z^{\nu}]\notag \\*&\quad =\frac{1}{\nu}[\nu^{\nu}-(\nu-1)^{\nu}-(\nu-1)(-1)^{\nu}]-1,\end{align}

since the first sum on the third line is telescoping. Combinatorially, the right-hand side expression is the total number of all tuples associated with lines containing interior points of the cube $[\!-\pi,\pi]^{\nu}$ , distinct from the diagonal $\mathbf{x} =\gamma \mathbf{e}$ , i.e. $k_{\alpha}\equiv 0$ . In particular, it is 0 for $\nu=2$ , and it is $6=3!$ for $\nu=3$ , as it should be: the former because the single admissible 2-tuple is $\{0,0\}$ , and the latter because the admissible tuples $\{k_1,k_2,k_3\}\neq\{0,0,0\}$ are $3!$ permutations of the single tuple $\{-2,0,2\}$ .

Similarly, telescoping again,

(3.8) \begin{align}\sum_{\mathrm{even }r\in [4,2(\nu-1)]} r\,\mathcal F_r(\zeta)& =2(\nu-1)\biggl(\dfrac{1-\zeta^{2\nu}}{1-\zeta^2} \biggr)^{\nu}-2(\nu-1)\zeta^{2\nu}\biggl(\dfrac{1-\zeta^{2(\nu-1)}}{1-\zeta^2} \biggr)^{\nu}\notag \\*&\quad+2 \sum_{\mathrm{even }r\in [6,2(\nu-1)]}\biggl[-\biggl(\dfrac{1-\zeta^r}{1-\zeta^2}\biggr)^{\nu}+\zeta^{2\nu}\biggl(\dfrac{1-\zeta^{r-2}}{1-\zeta^2}\biggr)^{\nu}\biggr]\notag \\*&\quad-4(1+\zeta^2)^{\nu}+4\zeta^{2\nu}.\end{align}

Now, analogously to (3.7), we will replace $\zeta^2$ with z, making

\[\zeta^{2\nu}=z^{\nu}=1\quad\text{for all $z_j=\exp\biggl({{\mathrm{i}}} \frac{2\pi\, j}{\nu}\biggr)$.}\]

Hence we may and will replace the factor $\zeta^{2\nu}$ in (3.8) with 1, and make the middle sum perfectly amenable to another round of telescoping. Thus

\[\sum_{\mathrm{even }a\le 0}[\zeta^{-a\nu}]\sum_{\mathrm{even }r\in [4,2(\nu-1)]}r\mathcal F_r(\zeta)=\sum_{\alpha\ge 0}[z^{\alpha\nu}]S_2(z),\]

where

\begin{align*}S_2(z)&\;:\!=\; 2(\nu-1)\biggl(\dfrac{1-z^{\nu}}{1-z} \biggr)^{\nu} -2(\nu-1)\biggl(\dfrac{1-z^{\nu-1}}{1-z} \biggr)^{\nu} \notag \\*&\quad+2\biggl[(1+z)^{\nu}-\biggl(\frac{1-z^{\nu-1}}{1-z}\biggr)^{\nu}\biggr]-4(1+z)^{\nu}+4\notag \\*&=2(\nu-1)\biggl(\dfrac{1-z^{\nu}}{1-z} \biggr)^{\nu} -2\nu\biggl(\dfrac{1-z^{\nu-1}}{1-z} \biggr)^{\nu}-2(1+z)^{\nu}+4.\end{align*}

Consequently we have

(3.9) \begin{align}\sum_{\alpha\ge 0}[z^{\alpha\nu}]S_2(z)&=\frac{2}{\nu}\sum_{j=0}^{\nu-1}\biggl[(\nu-1)\biggl(\frac{1-z_j^{\nu}}{1-z_j}\biggr)^{\nu}-\nu\biggl(\frac{1-z_j^{\nu-1}}{1-z_j}\biggr)^{\nu}\biggr] \notag \\*&=\frac{2}{\nu}[(\nu-1)\nu^{\nu}-\nu (\nu-1)^{\nu}-\nu (-1)^{\nu} (\nu-1)] \notag \\*&=2[(\nu-1)\nu^{\nu-1}-(\nu-1)^{\nu}-(-1)^{\nu}(\nu-1)].\end{align}

Combining (3.3), (3.7), and (3.9), we obtain

(3.10) \begin{align}\mathcal M &=2\nu\Biggl(1+\sum_{\mathrm{even }a\le 0}[\zeta^{-a\nu}]\sum_{\mathrm{even }r\in [4,2(\nu-1)]}\mathcal F_r(\zeta)\Biggr) \notag \\*&\quad-\sum_{\mathrm{even }a\le 0}[\zeta^{-a\nu}]\sum_{\mathrm{even }r\in [4,2(\nu-1)]}r\mathcal F_r(\zeta)\notag \\*&=2\nu^{\nu-1}.\end{align}

Combining (3.3) and (3.10), we have proved the following.

The reason that asymptotically only $\pi\nu^{-1/2} \mathcal M$ matters is that the dominant contribution to $\mathbb{E}[Z_n]$ comes from thin parallel cylinders in the cube $[\!-\pi,\pi]^{\nu}$ , each enclosing its own line segment, with the integrand asymptotically translation-invariant along the line segment, and dependent only on the Euclidean distance from $\mathbf{x}$ to the point on this segment with the same $\sum_{\alpha\in [\nu]}x_{\alpha}$ .

(II) Now that we have counted the total length of the line segments $\mathcal L_{\mathbf{k}}\cap [\!-\pi,\pi]^{\nu}$ , our next, protracted, step is to upper-bound $|\mathbb{E}[f(\mathbf{x},X)]|$ for $\mathbf{x}$ at a certain sufficiently small distance from the line segments. Start with

\begin{gather*}\mathbb{E}[f(\mathbf{x},X)]\;:\!=\; \sum_{\alpha\in [\nu]}\phi(y_{\alpha}),\quad y_{\alpha}=y_{\alpha}(\mathbf{x}),\\*\phi(y) \;:\!=\; \mathbb{E}\big[{{\mathrm{e}}}^{{{\mathrm{i}}} yX}\big]=\frac{1}{M}\sum_{j\in [M]} {{\mathrm{e}}}^{{{\mathrm{i}}} y\,j}=\dfrac{{{\mathrm{e}}}^{{{\mathrm{i}}} y}({{\mathrm{e}}}^{{{\mathrm{i}}} yM}-1)}{M({{\mathrm{e}}}^{{{\mathrm{i}}} y}-1)},\end{gather*}

so that

(3.11) \begin{equation}|\mathbb{E}[f(\mathbf{x},X)]|\le\sum_{\alpha\in [\nu]}|\phi(y_{\alpha})|,\quad |\phi(y)|=\biggl|\dfrac{\sin (My/2)}{M\sin(y/2)}\biggr|.\end{equation}

To proceed, given $|y|\le 2\pi(\nu-1)$ , define an even integer k(y) by the condition

\[\min_{\mathrm{even }k\colon |k|\le 2(\nu-1)} |y-k\pi|=|y-k(y)\pi|.\]

So $k(y)\pi$ is an even multiple of $\pi$ closest to y among all even multiples $k\pi$ with $|k|\le 2(\nu-1)$ . If y is not an odd multiple of $\pi$ , then k(y) is uniquely defined. So $z(y)\;:\!=\; y-k(y)\pi$ is uniquely defined for almost all y, and for those y we have $\phi(y)=\phi(z(y))$ , since $\phi$ is $2\pi$ -periodic. Furthermore, $|z(y)|$ is the distance from those y to the set of even multiples of $\pi$ , and as such it has a continuous extension to all y. Therefore $|z(y)|< \pi$ for almost all y. Slightly abusing notation, from now on let $[\!-\pi,\pi]^{\nu}$ stand for the original cube minus the set of $\mathbf{x}$ such that at least one $y_{\alpha}(\mathbf{x})$ is an odd multiple of $\pi$ . The discarded subset has zero Lebesgue measure. The reduced cube is a finite disjoint union of subsets, each with its own set $\{k_{\alpha}\}_{\alpha\in [\nu]}$ . Thus, for $\mathbf{x}$ in the reduced cube $[\!-\pi,\pi]^{\nu}$ , we have the following: for each $\alpha\in [\nu]$ , $z(y_{\alpha}(\mathbf{x}))=y_{\alpha}(\mathbf{x})-k(y_{\alpha}(\mathbf{x}))\pi$ is well-defined, and denoting $z_{\alpha}=z(y_{\alpha}(x))$ , $k_{\alpha}=k(y_{\alpha}(x))$ ,

(3.12) \begin{equation}|\mathbb{E}[f(\mathbf{x},X)]|\le\sum_{\alpha\in [\nu]}|\phi(z_{\alpha})|,\quad \sum_{\alpha\in [\nu]}z_{\alpha}=-\pi\sum_{\alpha\in [\nu]}k_{\alpha},\end{equation}

since $\sum_{\alpha}y_{\alpha}(\mathbf{x})=0$ .

So, we need to upper-bound $|\phi(z_{\alpha})|$ , $\alpha\in [\nu]$ , for each of these subsets. Let $z\in [\!-\pi,\pi]$ . Pick $b>1$ . If $|\sin(z/2)|\ge b/M$ , then $|\phi(z)|\le b^{-1}$ . Also, $|\phi(z)|\le \pi/(M|z|)$ , because $\sin\eta\ge 2\eta/\pi$ for $\eta\in (0,\pi/2)$ . Now $|\sin(z/2)|\le b/M$ if and only if $|z|\le b_0(M)/M$ , where $b_0(M)$ is uniquely defined for $M\ge b$ by the condition $\sin(b_0/(2M))=b/M$ with $b_0/(2M)\in (0,\pi/2]$ . A closer look shows that, given b,

(3.13) \begin{equation}b_0(M)=2b [1+{{\mathrm{O}}} (b^2/M^2)],\quad M\to\infty.\end{equation}

Therefore $|\phi(z)|\le b^{-1}$ if $|z|\ge b_0(M)/M$ , implying that

(3.14) \begin{equation}|\phi(z)|\le \min\{\pi/(M|z|);\; b^{-1}\}\quad\text{for $|z|\ge b_0(M)/M$.}\end{equation}

Consider $|z|\le b_0(M)/M$ , and set $z=\xi/M$ , i.e. $|\xi|\le b_0(M)\sim 2b$ for large M. Then

\[|\phi(z)|=\bigg|\dfrac{\sin(\xi/2)}{M\sin(\xi/(2M))}\bigg|=|\psi(\xi)|\cdot |1+{{\mathrm{O}}} (\xi^2/M^2)|,\quad \psi(\xi)\;:\!=\; \dfrac{\sin(\xi/2)}{\xi/2}.\]

Since

\[\psi^2(\xi)=\dfrac{1-\cos(\xi)}{\xi^2/2}=\sum_{j\ge 1}\frac{2(-1)^{j-1}}{(2j)!}\xi^{2(j-1)},\]

which is an alternating series with decreasing terms for $|\xi|\le 3\sqrt{2}$ , we obtain

\[\psi^2(\xi)\le 1-\frac{\xi^2}{30},\quad {|\xi| \le 3\sqrt{2}.}\]

Therefore

\[|\phi(z)|^2\le 1-\frac{\xi^2}{31}\quad\text{for $|\xi|\le 3\sqrt{2}$}\]

and M sufficiently large. Furthermore, $\psi^2(\xi)\le \tfrac{2}{9}$ for $|\xi|\ge 3\sqrt{2}$ . So, given b, there exists $M_1(b)$ such that $|\phi(z)|^2\le \tfrac{1}{3}$ for $|\xi|\in [3\sqrt{2}, b_0(M)]$ and $M\ge M_1(b)$ . Invoking (3.13), we have the following: there exists $M(b)\ge M_1(b)$ such that for $M\ge M(b)$ ,

(3.15) \begin{equation}|\phi(z)|^2\le 1-\lambda(b)\xi^2,\quad \xi\;:\!=\; Mz,\,|\xi|\le b_0(M),\quad \lambda\;:\!=\; \min\biggl\{\frac{1}{31},\frac{1}{7b^2}\biggr\}.\end{equation}

It follows from (3.14) and (3.15) that $|\phi(z)| < 1$ for $0 < |z|\le \pi$ . Consequently, by (3.11) and (3.12), we have $|\mathbb{E}[f(\mathbf{x},X)]| < \nu$ , unless $z_{\alpha}(\mathbf{x})\equiv 0$ , $\alpha\in [\nu]$ . Moreover, using the Cauchy–Schwarz inequality and $1-\gamma\le {{\mathrm{e}}}^{-\gamma}$ , we obtain the following.

Lemma 3.2. Let $\mathbf{x}\in [\!-\pi,\pi]^{\nu}$ . With $z_{\alpha}=z_{\alpha}(\mathbf{x})$ , denote $A(\mathbf{x})=\{\alpha\in [\nu]\colon |z_{\alpha}|\le b_0(M)/M\}$ , $a(\mathbf{x})\;:\!=\; |A(\mathbf{x})|$ , and define $\xi_{\alpha}=\xi_{\alpha}(\mathbf{x})\;:\!=\; Mz_{\alpha}$ , $\alpha\in [\nu]$ . Given $b > 0$ , for $M\ge M(b)$ there exists $\lambda=\lambda(b) > 0$ such that

(3.16) \begin{align} |\mathbb{E}[f(\mathbf{x},X)]|^2 &\le \nu\Biggl(\sum_{\alpha\in A(\mathbf{x})}|\phi(z_{\alpha})|^2+\sum_{\alpha\in A^c(\mathbf{x})} {\min}^2\biggl\{\frac{\pi}{|\xi_{\alpha}|};\; b^{-1}\biggr\}\biggr)\notag \\*&\le\nu\Biggl[ a(\mathbf{x})\exp\biggl(-\frac{\lambda}{a(\mathbf{x})}\sum_{\alpha\in A(\mathbf{x})}\xi_{\alpha}^2\biggr)+ \sum_{\alpha\in A^c(\mathbf{x})} {\min}^2\biggl\{\frac{\pi}{|\xi_{\alpha}|};\; b^{-1}\biggr\}\biggr]\end{align}

if $A(\mathbf{x})\neq\emptyset$ ; otherwise the first term within the brackets is set to zero.

Note. Of course, a similar inequality holds for $|\mathbb{E}[f(\mathbf{x},X)]|$ , but with unsquared summands and without the factor $\nu$ on the right-hand side. The inequality (3.16) has a useful advantage: ${\min}^2\{{{\pi}/{|\xi|}};\; b^{-1}\}$ is integrable on $(\!-\infty,\infty)$ . There is a price to pay, though: to get a bound for $|\mathbb{E}[f(\mathbf{x},X)]|^n$ , and then to integrate it over a peripheral part of the (reduced) cube $[\!-\pi,\pi]^{\nu}$ in question, we will have to raise both sides of the above inequality to the $(n/2)$ th power. So, while working on upper bounds, we assume that n is even, and define $N=n/2$ . For odd n we would raise the right-hand side expression to the integer power $N \;:\!=\; (n-1)/2$ and multiply the result by $\nu\ge|\mathbb{E}[f(\mathbf{x},X)]|$ .

The next step is to use Lemma 3.2 in order to upper-bound the contribution to $\mathbb{E}[Z_n]$ coming from a vast part of the cube $[\!-\pi,\pi]^{\nu}$ , complementary to the union of thin cylinders enclosing the line segments $\mathcal L_{\mathbf{k}}\cap [\!-\pi,\pi]^{\nu}$ .

Lemma 3.3. Define $D=D_1\cup D_2$ ,

\begin{align*}D_1&=\{\mathbf{x}\in [\!-\pi,\pi]^{\nu}\colon a(\mathbf{x})<\nu\},\\*D_2&=\Biggl\{\mathbf{x}\in [\!-\pi,\pi]^{\nu}\colon a(\mathbf{x})=\nu,\,\sum_{\alpha\in [\nu]}\xi^2_{\alpha}\ge \frac{\log ^2 n}{n} \Biggr\}.\end{align*}

where $\xi_{\alpha}=Mz_{\alpha}$ , $z_{\alpha}=y_{\alpha}(\mathbf{x})-k_{\alpha}\pi$ , $\min_{\mathrm{even}k}|y_{\alpha}(\mathbf{x})-k\pi|=|y_{\alpha}(\mathbf{x})-k_{\alpha}\pi|$ . Observe that

\begin{align*}D^c&=\Biggl\{\mathbf{x}\in [\!-\pi,\pi]^{\nu}\colon a(\mathbf{x})=\nu,\sum_{\alpha\in [\nu]}\xi_{\alpha}^2<\frac{\log^2 n}{n}\Biggr\}\\*&=\Biggl\{\mathbf{x}\in [\!-\pi,\pi]^{\nu}\colon \sum_{\alpha\in [\nu]}\xi_{\alpha}^2< \frac{\log^2 n}{n}\Biggr\},\end{align*}

the last equality holding if

\[\frac{\log^2n}{n}\le b_0(M)(={{\mathrm{O}}} (b)).\]

For $M\to\infty$ , and $N=n/2$ , within factor c(b) we have

\begin{align*} I_n&\;:\!=\; \int\limits_{D}\big|\mathbb{E}\bigl[f(\mathbf{x},X)\bigr]\big|^{n}\,{{\mathrm{d}}} \mathbf{x} \notag \\*& \le\frac{\nu^{2N}}{M^{\nu-1}}\exp\biggl(-\frac{\lambda\log^2 N}{4\nu}\biggr)+\begin{cases}\dfrac{(\nu {{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})})^N}{M}, &\lim\dfrac{N}{\log M}<1,\\[7pt]\biggl(\dfrac{\nu N{{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})}}{{{{\mathrm{e}}}\log M}}\biggr)^N,& \lim\dfrac{N}{\log M}\in (1,\nu-1),\\[7pt]\dfrac{(\nu(\nu-1)\,{{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})})^N}{M^{\nu-1}},&\lim\dfrac{N}{\log M}>\nu-1.\end{cases}\end{align*}

Proof. (i) Consider $\mathbf{x}\in D_1$ , so that $|A(\mathbf{x})|\in [0,\nu-1]$ . Let $I_{n,1}$ (resp. $I_{n,1}(a)$ ) denote the contribution of $D_1$ (resp. of $D_1(a)\;:\!=\; \{\mathbf{x}\in D_1\colon |A(\mathbf{x})|=a\}$ ) to $\mathbb{E}[Z_n]$ . For $a > 0$ and $\mathbf{x}\in D_1(a)$ , we use (3.16) and a union-type bound to get

\begin{align*}|\mathbb{E}[f(\mathbf{x},X)]|^{n}&\le\nu^N \sum_{A\subset [\nu]\colon |A|=a}\Biggl[a\exp\Biggl(-\frac{\lambda}{a}\sum_{\alpha\in A}\xi_{\alpha}^2\Biggr)\cdot I\Bigl(\max_{\alpha\in A}|\xi_{\alpha}|\le b_0\Bigr)\\[5pt] &\quad +\sum_{\beta\in A^c}\mathbb{I}(|\xi_{\beta}|\in [b_0,\pi M])\,{\min}^2\biggl\{\frac{\pi}{|\xi_{\beta}|};\; b^{-1}\biggr\} \Biggr]^N,\quad \xi_{\gamma}=\xi_{\gamma}(\mathbf{x}),\ \gamma\in [\nu].\end{align*}

So we bound $I_{n,1}(a)$ by the above right-hand side sum integrated over the full cube $[\!-\pi,\pi]^{\nu}$ . Each of $\binom{\nu}{a}$ terms in the outer sum contributes equally to the resulting integral. Therefore, by the multinomial formula with $\mathbf{r}\;:\!=\; (r_1, r_{a+1},\ldots, r_{\nu})$ ,

(3.17) \begin{align} I_{n,1}(a)&\le\nu^N\binom{\nu}{a}\sum_{\|\mathbf{r}\|=N}\dfrac{N!}{r_1!\,r_{a+1}!\,\cdots r_{\nu}!}\notag \\[5pt] &\quad \times \int\limits_{\mathbf{x}\in [\!-\pi,\pi]^{\nu}}\Biggl[a^{r_1}\exp\Biggl(-\frac{\lambda r_1}{a}\sum_{\alpha\in [a]}\xi_{\alpha}^2\Biggr)\cdot I\Bigl(\max_{\alpha\in [a]}|\xi_{\alpha}|\le b_0\Bigr) \notag \\[5pt] &\quad \times\prod_{\beta=a+1}^{\nu}\mathbb{I}(|\xi_{\beta}|\in[b_0,\pi M])\,{\min}^{2r_{\beta}}\biggl\{\frac{\pi}{|\xi_{\beta}|};\; b^{-1}\biggr\}\Biggr]\,{{\mathrm{d}}} \mathbf{x}.\end{align}

Recall that $\{\xi_{\alpha}\}=\{\xi_{\alpha}(\mathbf{x})\}$ is the piecewise affine function, with the same matrix of rank $\nu-1$ on each of at most $(2\nu-1)^{\nu}$ disjoint subsets forming a partition of $[\!-\pi,\pi]^{\nu}$ . To bound the integral in (3.17), we will switch from $x_{\alpha}$ to $\xi_{\alpha}$ for all but a single component $x_{\mu}$ , choosing $\mu$ dependent on $\mathbf{r}$ . No matter what $\mu$ is, the Jacobian factor is the same, namely $(M^{\nu-1}\nu^{\nu-2})^{-1}$ , on every subset of the resulting partition. Let $\beta\in \{a+1,\ldots,\nu\}$ ; if $\xi_{\beta}$ is among the new variables, then – upon integration of the $\mathbf{r}$ th term – the variable $\xi_{\beta}$ contributes the factor

(3.18) \begin{equation}2\int\limits_{b_0}^{\pi M} \biggl(\min\biggl\{\frac{\pi}{\xi};\; b^{-1} \biggr\} \biggr)^{2r_{\beta}} \,{{\mathrm{d}}} \xi \begin{cases}= 2(\pi M-b_0) \le 2 \pi M,&r_{\beta}=0,\\[5pt]\le \dfrac{2(\pi b-b_0)}{b^{2r_{\beta}}}+\dfrac{2\pi}{b^{2r_{\beta}-1}(2r_{\beta}-1)} \le \dfrac{4\pi}{b^{2r_{\beta}-1}},&r_{\beta}\ge 1.\end{cases}\end{equation}

We will control the factors for $r_{\beta}\ge 1$ by choosing b large. The case $r_{\beta}=0$ is the crux of the matter. Since $a\ge 1$ , the number of zeros in $\{r_{a+1},\ldots, r_{\nu}\}$ can be as high as $\nu-a$ if $\mu\in A$ , thus leading to a factor $M^{\nu-a}$ in the bound for the product of these integrals. However, we may and will select $\mu\in A^c$ if $r_{\mu}$ is one of the zeros in $\{r_{a+1},\ldots, r_{\nu}\}$ . For this choice,

\[2\int_{b_0}^{\pi M} \biggl(\min\biggl\{\frac{\pi}{\xi_{\mu}};\;b^{-1}\biggr\}\biggr)^{2r_{\mu}} \,{{\mathrm{d}}} \xi_{\mu}\approx 2\pi M\]

is replaced by $\int_{-\pi}^{\pi} 1\, {{\mathrm{d}}} x_{\mu}=2\pi$ . Hence the worst-case bound becomes $M^{\nu-a-1}$ instead! Only if there are no zeros in $\{r_{a+1},\ldots, r_{\nu}\}$ do we choose $\mu\in A$ .

Since $\mu\le \nu$ , we obtain

(3.19) \begin{equation}\begin{aligned} I_{n,1}(a)&\le \nu^N\binom{\nu}{a}\frac{(2\nu-1)^{\nu}\nu}{M^{\nu-1}\nu^{\nu-2}}\sum_{\|\mathbf{r}\|=N}\dfrac{N!\, a^{r_1}}{r_1!\,r_{a+1}!\cdots r_{\nu}!} \cdot J(\mathbf{r}),\\[5pt]J(\mathbf{r})&\;:\!=\; \int\limits_{x_{\mu},\,\{\xi_{\gamma}\}_{\gamma \neq \mu}} \exp\Biggl(-\frac{\lambda r_1}{a}\sum_{\alpha\in [a]}\xi_{\alpha}^2\Biggr)\cdot I\Bigl(\max_{\alpha\in [a]}|\xi_{\alpha}|\le b_0\Bigr)\\[5pt]&\quad \times\prod_{\beta=a+1}^{\nu}\mathbb{I}(|\xi_{\beta}|\in[b_0,\pi M])\,{\min}^{2r_{\beta}}\biggl\{\frac{\pi}{|\xi_{\beta}|};\; b^{-1}\biggr\}\,{{\mathrm{d}}} x_{\mu}\prod_{\gamma\neq\mu} \,{{\mathrm{d}}} \xi_{\gamma}.\end{aligned}\end{equation}

Let $\sigma(\mathbf{r})$ stand for the number of zeros in $\{r_{a+1},\ldots, r_{\nu}\}$ . If $\sigma(\mathbf{r}) > 0$ , then we choose $\mu\in \{a+1,\ldots, \nu\}$ such that $r_{\mu}=0$ . Rather crudely, we have

\begin{equation*}\int\limits_{\xi_1,\ldots,\xi_a} \exp\Biggl(-\frac{\lambda r_1}{a}\sum_{\alpha\in [a]}\xi_{\alpha}^2\Biggr)\cdot I\Bigl(\max_{\alpha\in [a]}|\xi_{\alpha}|\le b_0\Bigr) \prod_{\alpha\in [a]}\,{{\mathrm{d}}} \xi_{\alpha}\le (2b_0)^{a}={{\mathrm{O}}} (b^a).\end{equation*}

Furthermore, using (3.18) and

\[\sum_{\beta\in [a+1,\,\nu]\colon r_{\beta} > 0}(2r_{\beta}-1)=2(N-r_1)- [\nu-a-\sigma(\mathbf{r})],\]

we bound

\begin{align*}&\int\limits_{x_{\mu},\,\{\xi_{\gamma}\}_{\mu\neq \gamma>a}}\prod_{\beta=a+1}^{\nu}\mathbb{I}(|\xi_{\beta}|\in[b_0,\pi M])\,{\min}^{2r_{\beta}}\biggl\{\frac{\pi}{|\xi_{\beta}|};\; b^{-1}\biggr\}\,{{\mathrm{d}}} x_{\mu}\prod_{\mu\neq \gamma> a} \,{{\mathrm{d}}} \xi_{\gamma}\\[5pt]&\quad \le 2\pi\cdot 2^{\nu-a-1}\prod_{\mu\neq \gamma>a}\int\limits_{b_0}^{\pi M} \biggl(\min\biggl\{\frac{\pi}{\xi};\; b^{-1} \biggr\} \biggr)^{2r_{\gamma}} \,{{\mathrm{d}}} \xi \\[5pt]&\quad \le 2^{\nu-a}\pi (\pi M)^{\sigma(\mathbf{r})-1} (4\pi)^{\nu-a-\sigma(\mathbf{r})} b^{-2(N-r_1)+(\nu-a-\sigma(\mathbf{r}))}\\[5pt]&\quad \le c_1M^{\sigma(\mathbf{r})-1} b^{-2(N-r_1)}\\*&\quad \le c_1M^{\nu-a-1} b^{-2(N-r_1)}.\end{align*}

Here $c_1$ and $c_j$ below depend on b only. So

\[J(\mathbf{r})\le c_2M^{\nu-a-1} b^{-2(N-r_1)}\quad \text{if $\sigma(\mathbf{r})>0$.}\]

Likewise, if $\sigma(\mathbf{r})=0$ then

\[J(\mathbf{r}) \le c_3 b^{-2(N-r_1)} \le c_3M^{\nu-a-1} b^{-2(N-r_1)},\]

since $\nu > a$ . Therefore

\[J(\mathbf{r})\le c_4M^{\nu-a-1} b^{-2(N-r_1)}.\]

Plugging this bound into (3.19), we have the following: for $a\in [1,\nu-1]$ ,

(3.20) \begin{align}I_{n,1}(a)&\le\frac{c_5\nu^N}{M^{\nu-1}} \sum_{\| \mathbf{r}\|=N}\dfrac{M^{\nu-a-1} N! }{\prod\limits_s r_s!}\cdot a^{r_1} b^{-2(N-r_1)} \notag \\[5pt] &=\frac{c_5\nu^N}{M^{\nu-1}}\cdot M^{\nu-a-1}\sum_{\|\mathbf{r}\|=N}\dfrac{N!}{ \prod\limits_s r_s!}\,a^{r_1}\prod_{s>a} b^{-2r_s} \notag \\[5pt] &\le\frac{c_5\nu^N}{M^{\nu-1}}\cdot M^{\nu-a-1}(a+\nu b^{-2})^N.\end{align}

For $a=0$ , instead of the bound (3.17) we get

\begin{align*}|\mathbb{E}[f(\mathbf{x},X)]|^{n}&\le\nu^N \Biggl(\sum_{\beta=1}^{\nu}\mathbb{I}(|\xi_{\beta}|\in [b_0,\pi M])\,{\min}^2\biggl\{\frac{\pi}{|\xi_{\beta}|};\; b^{-1}\biggr\} \Biggr)^N\\[5pt]&=\nu^N\sum_{\|\mathbf{r}\|=N}\dfrac{N!}{r_1!\cdots r_{\nu}!}\prod_{\beta=1}^{\nu}\mathbb{I}(|\xi_{\beta}|\in[b_0,\pi M])\,{\min}^{2r_{\beta}}\biggl\{\frac{\pi}{|\xi_{\beta}|};\; b^{-1}\biggr\},\end{align*}

$\mathbf{r}=(r_1,\ldots,r_{\nu})$ . Further, analogously to (3.20), we obtain

(3.21) \begin{equation} I_{n,1}(0)\le \frac{c_6\nu^N}{M^{\nu-1}}\cdot M^{\nu-2} b^{-2N}.\end{equation}

Why $M^{\nu-2}$ ? Because $\sigma(\mathbf{r})\le\nu-1$ , and if $\sigma(\mathbf{r})=\nu-1$ , then we select $\mu$ for which $r_{\mu}=0$ .

Adding (3.20) and (3.21), we get

(3.22) \begin{equation}I_{n,1}\le\frac{c_7\nu^N}{M}\Biggl(b^{-2N}+\sum_{a=1}^{\nu-1}M^{-a+1}(a+\nu b^{-2})^N\Biggr).\end{equation}

Let us simplify (3.22). For $n\to\infty$ , the (log-concave) function

\[\psi(a)\;:\!=\; M^{-a+1}\biggl(a+\frac{\nu}{b^2}\biggr)^N,\quad {a\in [1,\nu-1],}\]

attains its unique maximum at

\begin{equation*}a(\nu)=\begin{cases}1,&\lim\dfrac{N}{\log M} < 1+\dfrac{\nu}{b^2},\\[7pt]\dfrac{N}{\log M}-\dfrac{\nu}{b^2},&\lim\dfrac{N}{\log M}\in \biggl(1+\dfrac{\nu}{b^2},\nu-1+\dfrac{\nu}{b^2}\biggr),\\[7pt]\nu-1,&\lim\dfrac{N}{\log M} > \nu-1+\dfrac{\nu}{b^2}.\end{cases}\end{equation*}

Consequently

\begin{equation*}\psi(a(\nu))=\begin{cases}\bigl({{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})}\bigr)^N,&\lim\dfrac{N}{\log M}<1+\dfrac{\nu}{b^2},\\[7pt]M\biggl(\dfrac{N{{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})}}{{{\mathrm{e}}}\log M}\biggr)^N,&\lim\dfrac{N}{\log M}\in \biggl(1+\dfrac{\nu}{b^2},\nu-1+\dfrac{\nu}{b^2}\biggr),\\[7pt]M^{-\nu+2}\bigl((\nu-1)\,{{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})}\bigr)^N,&\lim\dfrac{N}{\log M}>\nu-1+\dfrac{\nu}{b^2}.\end{cases}\end{equation*}

The sum in (3.22) is below $b^{-2N}+\nu \psi(a(\nu))$ , and b can be chosen arbitrarily large. So, for b sufficiently large, we have

(3.23) \begin{equation}I_{n,1}\le c_8\cdot\begin{cases}\dfrac{\bigl(\nu {{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})}\bigr)^N}{M}, &\lim\dfrac{N}{\log M}<1,\\[7pt]\biggl(\dfrac{\nu N{{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})}}{{{\mathrm{e}}}\log M}\biggr)^N,& \lim\dfrac{N}{\log M}\in (1,\nu-1),\\[7pt]\dfrac{\bigl(\nu(\nu-1)\,{{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})}\bigr)^N}{M^{\nu-1}},&\lim\dfrac{N}{\log M}>\nu-1.\end{cases}\end{equation}

(ii) Suppose $\mathbf{x}\in D_2$ . Let $I_{n,2}$ denote the contribution of $D_2$ to $\mathbb{E}[Z_n]$ . We switch from $x_{\alpha}$ to $\xi_{\alpha}$ for all but a single $x_{\mu}$ such that $|\xi_{\mu}|=\min_{\alpha}|\xi_{\alpha}|$ . So, here $D_2$ is a finite disjoint union of subsets, each with its own $\mu\in [\nu]$ , and the choice of the new variables depends on a subset in such a way that the condition

\[\sum_{a\in [\nu]}\xi_{\alpha}^2\ge \frac{\log^2n}{n}\]

implies that

\[\sum_{a\neq \mu}\xi_{\alpha}^2\ge \frac{(\nu-1)\log^2n}{\nu n}\ge \frac{\log^2N}{4N}.\]

Analogously to $I_{n,1}$ , by Lemma 3.2 we have

\begin{equation*}I_{n,2}\le \frac{c_1\nu^{2N}}{M^{\nu-1}} \int\limits_{\sum_{\alpha=1}^{\nu-1}\xi_{\alpha}^2\ge {{\log^2N}/{(4N)}}} \exp\Biggl(-\frac{\lambda N}{\nu}\sum_{\alpha=1}^{\nu-1}\xi_{\alpha}^2\Biggr)\,\prod_{\alpha=1}^{\nu-1} \,{{\mathrm{d}}} \xi_{\alpha}.\end{equation*}

Here $c_1$ and $c_j$ below are independent of b and $\lambda=\lambda(b)$ . Using spherical coordinates, and denoting

\[\zeta_0=\biggl(\frac{\lambda}{4\nu}\biggr)^{1/2}\log N,\]

we bound the last integral by

\begin{align*}c_2 \int\limits_{\rho\ge {{\log N}/{\sqrt{4N}}}} \exp\biggl(-\frac{\lambda N}{\nu}\rho^2\biggr) \rho^{\nu-2}\,{{\mathrm{d}}} \rho&=c_3\,(\lambda N)^{-(\nu-1)/2} \int\limits_{\zeta\ge \zeta_0}{{\mathrm{e}}}^{-\zeta^2/2}\zeta^{\nu-2}\,{{\mathrm{d}}} \zeta\\*&\le c_4\, \lambda^{-1} N^{-(\nu-1)/2} (\log N)^{\nu-3} \exp\biggl(-\frac{\lambda\log^2 N}{4\nu}\biggr).\end{align*}

The last bound follows from integrating the left-hand side integral (call it I) by parts, and bounding the residual integral by $I\cdot(\nu-3)\zeta_0^{-2}$ . We conclude that

(3.24) \begin{equation} I_{n,2}\le \frac{c_5\,\nu^{2N}}{\lambda M^{\nu-1}}\exp\biggl(-\frac{\lambda\log^2 N}{4\nu}\biggr).\end{equation}

Adding (3.23) and (3.24), we complete the proof of Lemma 3.3.

(III) So it remains to sharply evaluate an asymptotic contribution to the integral

\[\mathbb{E}[Z_n]= \dfrac{1}{(2\pi)^{\nu}} \int\limits_{\mathbf{x}\in [\!-\pi,\pi]^{\nu}} \mathbb{E}^n[f(\mathbf{x},X) ] \,{{\mathrm{d}}} \mathbf{x},\]

that comes from $C\;:\!=\; [\!-\pi,\pi]^{\nu}\setminus D$ , or more explicitly from those $\mathbf{x}$ in the (reduced) cube $[\!-\pi,\pi]^{\nu}$ with $\sum_{\alpha}\xi_{\alpha}^2 < n^{-1}\log^2 n$ , where the $\xi_{\alpha}=\xi_{\alpha}(\mathbf{x})$ are uniquely defined by

\[\xi_{\alpha}=Mz_{\alpha},\quad z_{\alpha}=y_{\alpha}(\mathbf{x})-k_{\alpha}(\mathbf{x})\pi,\quad |z_{\alpha}|=\min_{\mathrm{even} k}|y_{\alpha}(\mathbf{x})-k\pi|.\]

That is, $\{k_{\alpha}(\mathbf{x})\}$ is an even tuple. Thus

\begin{align*}&\sum_{\alpha} (y_{\alpha}(\mathbf{x})-k_{\alpha}(\mathbf{x})\pi)^2=M^{-2}\sum_{\alpha}\xi_{\alpha}^2 < M^{-2}n^{-1}\log^2 n\\*& \Longrightarrow \sum_{\alpha}k_{\alpha}(\mathbf{x})=0\Longrightarrow\sum_{\alpha}\xi_{\alpha}=0\end{align*}

if n is large enough. Indeed, $\sum_{\alpha}y_{\alpha}(\mathbf{x})\equiv 0$ . So if $\sum_{\alpha}k_{\alpha}(\mathbf{x})\neq 0$ then $|\sum_{\alpha}k_{\alpha}(\mathbf{x})|\ge 2$ , and for large n we have a contradiction:

\[4\pi^2\le \biggl|\sum_{\alpha}(y_{\alpha}(\mathbf{x})-k_{\alpha}(\mathbf{x})\pi)\biggr|^2\le\nu\sum_{\alpha}(y_{\alpha}(\mathbf{x})-k_{\alpha}(\mathbf{x})\pi)^2\le\frac{\nu\log^2n}{M^2 n}.\]

Conversely, if $\{y_{\alpha}(\mathbf{x})\}$ and an even tuple $\mathbf{k}=\{k_{\alpha}\}$ , $\sum_{\alpha}k_{\alpha}=0$ , satisfy

\[\sum_{\alpha} (y_{\alpha}(\mathbf{x})-k_{\alpha}\pi)^2 < M^{-2}n^{-1}\log^2 n,\]

then $\sum_{\alpha}\xi_{\alpha}^2\le n^{-1}\log^2 n$ . Moreover, $\xi_{\alpha}=M(y_{\alpha}(\mathbf{x})-k_{\alpha}\pi)$ if n is large. Indeed, $\xi_{\alpha}=M(y_{\alpha}(\mathbf{x})-k_{\alpha}'\pi)$ , where $|(y_{\alpha}(\mathbf{x})-k_{\alpha}'\pi)|=\min_{\mathrm{even}\, \kappa}|y_{\alpha}(\mathbf{x})-\kappa\pi|$ , and the conditions

\[\sum_{\alpha} (y_{\alpha}(\mathbf{x})-k_{\alpha}\pi)^2,\quad\sum_{\alpha} (y_{\alpha}(\mathbf{x})-k_{\alpha}'\pi)^2 < M^{-2}n^{-1}\log^2 n,\]

combined with the triangle inequality, imply that

\[\|\mathbf{k}-\mathbf{k}'\|\le \frac{2\log n}{ \pi M n^{1/2}}.\]

Therefore $C=\cup_{\mathbf{k}}C_{\mathbf{k}}$ , where $\mathbf{k}=\{k_{\alpha}\}$ , with even $k_{\alpha}$ adding up to 0, and

\[C_{\mathbf{k}}\;:\!=\; \Biggl\{\mathbf{x}\in [\!-\pi,\pi]^{\nu}\colon \sum_{\alpha} (y_{\alpha}(\mathbf{x})-k_{\alpha}\pi)^2 < M^{-2}n^{-1}\log^2 n\Biggr\}.\]

Further, for large n, by the triangle inequality $C_{\mathbf{k}_1}\cap\, C_{\mathbf{k}_2}=\emptyset$ if $\mathbf{k}_1\neq \mathbf{k}_2$ .

Consider the line $\mathcal L_{\mathbf{k}}$ given by $y_{\alpha}(\mathbf{x})=k_{\alpha}\pi$ , $\alpha\in [\nu]$ , for any such tuple $\mathbf{k}$ . (In part (I) we enumerated all $\mathcal L_{\mathbf{k}}$ that contain interior points of $[\!-\pi,\pi]^{\nu}$ .) A generic $\mathcal L_{\mathbf{k}}$ is given by its parametric equation

(3.25) \begin{equation}x_{\alpha}(t)=\nu^{-1}(t+k_{\alpha}\pi),\quad \alpha\in [\nu].\end{equation}

The lines are parallel to each other, running in the direction of $\mathbf{e}=(1,\ldots,1)$ , and crossing at 90 degrees each of the (hyper)planes $\sum_{\alpha\in [\nu]}x_{\alpha}=t$ , with $\mathbf{x}(t)\;:\!=\; \{x_{\alpha}(t)\}_{\alpha\in [\nu]}$ being the intersection point. Using

\[x_{\alpha}=\nu^{-1}\Biggl(\sum_{\beta}x_{\beta}+y_{\alpha}(\mathbf{x})\Biggr)\]

and (3.25), we obtain the following: if $\mathbf{x}=\{x_{\alpha}\}\in C_{\mathbf{k}}$ , then with $t\;:\!=\; \sum_{\alpha}x_{\alpha}$ ,

\begin{align*} &x_{\alpha}-x_{\alpha}(t)=\frac{y_{\alpha}(\mathbf{x})-k_{\alpha}\pi}{\nu}=\frac{\xi_{\alpha}(\mathbf{x})}{M\nu}\\*&\quad\Longrightarrow \|\mathbf{x}-\mathbf{x}(t)\|^2=\frac{1}{(M\nu)^2}\sum_{\alpha}\xi_{\alpha}^2(\mathbf{x})\le \frac{\log^2 n}{(M\nu)^2n}.\end{align*}

Conversely, if $t=\sum_{\alpha}x_{\alpha}$ , then

\[\|\mathbf{x}-\mathbf{x}(t)\|^2\le \frac{\log^2 n}{(M\nu)^2n}\]

implies that $\mathbf{x}\in C_{\mathbf{k}}$ .

Consider a $\nu$ -dimensional cylinder $\mathcal C_{\mathbf{k}}$ enclosing the line $\mathcal L_{\mathbf{k}}$ , such that each of its cross-sections formed by planes $\sum_{a}x_{\alpha}=t$ orthogonal to $\mathcal L_{\mathbf{k}}$ is a $(\nu-1)$ -dimensional (hyper)sphere in that plane, of radius $r_n\;:\!=\;(M\nu)^{-1}n^{-1/2}\log n$ , which is centered at the common point $\mathbf{x}(t)$ of the plane and the line $\mathcal L_{\mathbf{k}}$ . We denote the sphere by $\mathcal S(\mathbf{x}(t), r_n)$ .

The above discussion means the following. If $\mathcal L_{\mathbf{k}}$ contains interior points of $[\!-\pi,\pi]^{\nu}$ , then $C_{\mathbf{k}}=\mathcal C_{\mathbf{k}}\cap [\!-\pi,\pi]^{\nu}$ . Shortly we will see that the contribution of $C_{\mathbf{k}}$ to $\mathbb{E}[Z_n]$ is of exact order $L(\mathbf{k})\nu^n M^{-\nu+1} n^{-{{(\nu-1)}/{2}}}$ , where $L(\mathbf{k})\ge 2\pi \nu^{-1/2}$ is the length of line segment of $\mathcal L_{\mathbf{k}}$ in $[\!-\pi,\pi]^{\nu}$ ; see (3.3).

In part (I) we proved that if $\mathcal L_{\mathbf{k}}$ contains no interior points of $[\!-\pi,\pi]^{\nu}$ , then $\mathcal L_{\mathbf{k}}$ may only touch $[\!-\pi,\pi]^{\nu}$ at a single point $\mathbf{x}_{\mathbf{k}}=\{x_{\mathbf{k},\alpha}\}$ , such that $x_{\mathbf{k},\alpha_1}=\pi$ and $x_{\mathbf{k},\alpha_2}=-\pi$ for some $\alpha_1,\alpha_2\in [\nu]$ . In this case $C_{\mathbf{k}}$ is the union of the parallel line segments $\mathbf{x}(t)=\tilde {\mathbf{x}}_{\mathbf{k}} +t\mathbf{e} \in [\!-\pi,\pi]^{\nu}$ , where $\sum_{\alpha}\tilde{\mathbf{x}}_{\mathbf{k},\alpha}=\sum_{\alpha}x_{\mathbf{k},\alpha}$ , and $\|\tilde{\mathbf{x}}_{\mathbf{k}}-\mathbf{x}_{\mathbf{k}}\|\le r_n$ . Using $x_{\mathbf{k},\alpha_1}=\pi$ and $x_{\mathbf{k},\alpha_2}=-\pi$ we obtain $t\in [x_{\mathbf{k},\alpha_2}-\tilde x_{\mathbf{k},\alpha_2},x_{\mathbf{k},\alpha_1}-\tilde x_{\mathbf{k}, \alpha_1}]$ . Since $|y|$ is convex, we see then that $|t|\le r_n$ . So $C_{\mathbf{k}}$ is enclosed in a sub-cylinder of $\mathcal C_{\mathbf{k}}$ sandwiched between two parallel cross-sections, for $t_1=\sum_{\alpha} x_{\mathbf{k},\alpha}-r_n$ and $t_2=\sum_{\alpha} x_{\mathbf{k},\alpha}+r_n$ , that are at distance ${{\mathrm{O}}} (r_n)$ from each other. Consequently, contribution of $C_{\mathbf{k}}$ to $\mathbb{E}[Z_n]$ is at most of order $r_n\nu^n M^{-\nu+1} n^{-{{(\nu-1)}/{2}}}$ , and we know that the total number of the points $\mathbf{x}_{\mathbf{k}}$ is $\nu(2\nu-1)^{\nu}$ , at most. So the contribution to $\mathbb{E}[Z_n]$ from the sets $C_{\mathbf{k}}$ with the lines $\mathcal L_{\mathbf{k}}$ merely touching the cube $[\!-\pi,\pi]^{\nu}$ is relatively negligible as $n\to\infty$ . Therefore we focus on $C_{\mathbf{k}}$ with lines $\mathcal L_{\mathbf{k}}$ enumerated in part (I).

For $\mathbf{x}\in \mathcal S(\mathbf{x}(t), r_n)$ , we set $\xi_{\alpha}=M(y_{\alpha}(\mathbf{x})-k_{\alpha}\pi)=M\nu(x_{\alpha}-x_{\alpha}(t))$ , so that $\sum_{\alpha}\xi_{\alpha}^2\le n^{-1}\log ^2n$ and $\sum_{\alpha}\xi_{\alpha}=0$ . Then

\begin{align*}\mathbb{E}[f(\mathbf{x},X) ] &=\sum_{\alpha\in [\nu]} \mathbb{E}\bigl[{{\mathrm{e}}}^{{{\mathrm{i}}} y_{\alpha}(\mathbf{x})X}\bigr]\\*&=\sum_{\alpha\in [\nu]} \mathbb{E}\bigl[{{\mathrm{e}}}^{{{\mathrm{i}}} \xi_{\alpha}M^{-1}X}\bigr]\\&=\sum_{\alpha\in [\nu]}\biggl[1+{{\mathrm{i}}} \frac{\mathbb{E}[X]}{M}\xi_{\alpha}-\frac{\mathbb{E}[X^2]}{2M^2}\xi_{\alpha}^2+{{\mathrm{O}}}\biggl(|\xi_{\alpha}|^3\frac{\mathbb{E}[X^3]}{M^3}\biggr)\biggr]\\&=\nu-\frac{\mathbb{E}[X^2]}{2M^2}\sum_{\alpha\in [\nu]}\xi_{\alpha}^2+{{\mathrm{O}}}\Biggl(\frac{\mathbb{E}[X^3]}{M^3}\sum_{\alpha\in [\nu]}|\xi_{\alpha}|^3\Biggr)\\&=\nu\exp\Biggl(-\frac{c_{\scriptscriptstyle M}}{2\nu}\sum_{\alpha\in [\nu]}\xi_{\alpha}^2+{{\mathrm{O}}} (n^{-3/2}\log^3n)\Biggr)\\*&=\nu\exp\biggl(-\frac{c_{\scriptscriptstyle M}}{2\nu}(M\nu)^2\|\mathbf{x}-\mathbf{x}(t)\|^2+{{\mathrm{O}}} (n^{-3/2}\log^3n)\biggr),\end{align*}

where

\[c_{\scriptscriptstyle M}=\frac{\mathbb{E}[X^2]}{M^2}, \quad t=\sum_{\alpha}x_{\alpha}.\]

So, uniformly for $\mathbf{x}\in \mathcal S(\mathbf{x}(t),r_n)$ , we have

(3.26) \begin{equation}\mathbb{E}^n[f(\mathbf{x},X) ] =\nu^n\exp\biggl(-\frac{c_{\scriptscriptstyle M}n}{2\nu}(M\nu)^2\|\mathbf{x}-\mathbf{x}(t)\|^2+{{\mathrm{O}}} (n^{-1/2}\log^3n)\biggr).\end{equation}

With $\rho_n\;:\!=\; n^{-1/2}\log n\to 0$ , an admissibility issue arises when $|t-t_j|={{\mathrm{O}}} (M^{-1}\rho_n)$ , where $t_1\;:\!=\; -\pi(\nu+\min k_{\alpha})$ and $t_2\;:\!=\; \pi(\nu-\max k_{\alpha})$ are the endpoints of t’s range. Ignoring this lack of complete homogeneity, and using the orthogonality of $\mathcal S(\mathbf{x}(t), r_n)$ and $\mathcal L_{\mathbf{k}}$ , we replace the contribution of $C_{\mathbf{k}}$ to the integral representing $E[Z_n]$ with the integral of the right-hand side expression in (3.26) over the sphere $\mathcal S(\mathbf{x}(t), r_n)$ , times $C_{\mathbf{k}}$ ’s length

\[L(\mathbf{k})\;:\!=\; \|\mathbf{x}(t_2)-\mathbf{x}(t_1)\|=\nu^{-1/2} \pi\Bigl(2\nu+\min_{\alpha}k_{\alpha}-\max_{\alpha}k_\alpha\Bigr)\ge 2\pi\nu^{-1/2};\]

see (3.3). So, introducing $\mathcal S(\mathbf{0},\rho)$ , the $(\nu-1)$ -dimensional sphere of radius $\rho$ , centered at the origin $\mathbf{0}\in E^{\nu-1}$ , and using

\[\text{Volume}\bigl(\mathcal S(\mathbf{0},\rho)\bigr)=\frac{\pi^{{{(\nu-1)}/{2}}}\rho^{\nu-1}}{\Gamma\bigl(\frac{(\nu-1)}{2}+1\bigr)},\]

we obtain the following: within factor $1+{{\mathrm{o}}} (1)$ , the contribution of $C_{\mathbf{k}}$ to $\mathbb{E}[Z_n]$ is

\begin{align*}&\dfrac{\nu^n L(\mathbf{k})}{(2\pi)^{\nu}(M\nu)^{\nu-1}}\int_{\mathcal{S}(\mathbf{0}, \rho_n)}\exp\biggl({-\dfrac{c_M n}{2\nu}\rho^2}\biggr)\,{{\mathrm{d}}} V\\&\quad =\dfrac{\nu^{n} L(\mathbf{k})}{(2\pi)^{\nu}(M\nu)^{\nu-1}}\cdot\dfrac{\pi^{{{(\nu-1)}/{2}}}(\frac{\nu-1}{2})}{\Gamma\bigl(\frac{(\nu-1)}{2}+1\bigr)}\biggl(\frac{2\nu}{c_Mn}\biggr)^{{{(\nu-1)}/{2}}}\int_0^{{{c_{Mn}\rho_{n}^2}/{\nu}}}{{\mathrm{e}}}^{-z} z^{{{(\nu-1)}/{2}}-1}\,{{\mathrm{d}}} z\\ &\quad =(1+{{\mathrm{o}}} (1))\dfrac{\nu^n}{M^{\nu-1}}\cdot\dfrac{L(\mathbf{k})}{2\pi(2\pi\nu c_{\scriptscriptstyle M}n)^{{{(\nu-1)}/{2}}}}.\end{align*}

Indeed, the second line integral is asymptotic to $\Gamma\bigl(\frac{(\nu-1)}{2}\bigr)$ , and for the third line we used $\Gamma(z+1)=z\Gamma(z)$ . (This estimate confirms our earlier claim concerning the contribution of every such $C_{\mathbf{k}}$ to $\mathbb{E}[Z_n]$ .) Combining the estimate with Lemmas 3.1 and 3.3, we have

(3.27) \begin{equation}\mathbb{E}[Z_n]=(1+{{\mathrm{o}}} (1))\dfrac{\nu^n}{M^{\nu-1}}\cdot\dfrac{\nu^{\nu-3/2}}{(2\pi\nu c_{\scriptscriptstyle M} n)^{{{(\nu-1)}/{2}}}}+{{\mathrm{O}}} (\mathcal R_n),\end{equation}

where (with $N=n/2$ )

(3.28) \begin{align}\mathcal R_n=\frac{\nu^{n}}{M^{\nu-1}}\exp\biggl(-\frac{\lambda\log^2 N}{4\nu}\biggr) +\begin{cases}\dfrac{\bigl(\nu {{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})}\bigr)^N}{M}, &\lim\dfrac{N}{\log M}<1,\\[7pt]\biggl(\dfrac{\nu N{{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})}}{{{\mathrm{e}}}\log M}\biggr)^N,& \lim\dfrac{N}{\log M}\in (1,\nu-1),\\[7pt]\dfrac{\bigl(\nu (\nu-1)\,{{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})}\bigr)^N}{M^{\nu-1}},&\lim\dfrac{N}{\log M}>\nu-1.\end{cases}\end{align}

We will refer to this vertical stack as the top, middle, and bottom remainders.

(IV) Let us have a close look at the estimate above.

(a) $\nu=3$ . Suppose that

\[\lim\frac{N}{\log M}\in\biggl(\frac{1}{\log 3},\infty\biggr)\setminus\{1,2\}.\]

Then ${{3^{2N}}/{M^2}}$ grows exponentially fast, hence

\[\frac{3^{2N}}{M^2}\gg\frac{(3{{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})})^N}{M}\]

(the top remainder term in (3.28)), for b chosen sufficiently large. So, if

\[\lim\frac{N}{\log M}\in\biggl(\frac{1}{\log 3}, 1\biggr),\]

then the above formula for $\mathbb{E}[Z_n]$ implies that

(3.29) \begin{equation}\mathbb{E}[Z_n]=(1+{{\mathrm{o}}} (1))\dfrac{\nu^n}{M^{\nu-1}}\cdot\dfrac{\nu^{\nu-3/2}}{(2\pi\nu c_{\scriptscriptstyle M} n)^{{{(\nu-1)}/{2}}}}\Big|_{\nu=3}\to\infty.\end{equation}

Let $\lim{{N}/{\log M}}\in (1,2)$ . Then (3.29) continues to hold for b large enough, provided that

\[\frac{3^{2N}}{M^2}\gg \biggl(\frac{\nu N{{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})}}{{{\mathrm{e}}}\log M}\biggr)^N,\]

the middle remainder in (3.28). This inequality holds if

\[\rho\;:\!=\; \lim \frac{\log M}{N}\in (0.5,1)\]

satisfies

\[(3{{\mathrm{e}}}^{-\rho})^2> \frac{3}{{{\mathrm{e}}}\rho},\]

and it is indeed so, since

\[\min\{9\tau {{\mathrm{e}}}^{-2\tau}\colon \tau\in [1/2,1]\}=\frac{9}{{{\mathrm{e}}}^2}>\frac{3}{{{\mathrm{e}}}}.\]

Let $\lim{{N}/{\log M}}\in (2,\infty)$ . Then (3.29) again holds for b large enough, since $3^n\gg 6^N$ ; see the bottom remainder in (3.28). On the other hand, if

\[\lim\frac{N}{\log M}<\frac{1}{\log 3},\]

then

\[\lim\frac{N}{\log M}<1,\]

and using the top remainder from (3.28), we see that

\[\mathbb{E}[Z_n]={{\mathrm{O}}}\biggl(\frac{(3 {{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})})^N}{M}\biggr)\to 0\quad \text{for}\, \textit{b}\, \text{sufficiently large.}\]

(b) Now let $\nu>3$ . Then (3.29) holds if

\[\lim\frac{N}{\log M}>\nu-1.\]

Indeed, by (3.28),

\begin{gather*}\mathcal R_n =\frac{\nu^{2N}}{M^{\nu-1}}\exp\biggl(-\frac{\lambda\log^2 N}{4\nu}\biggr)+\frac{\bigl(\nu (\nu-1)\,{{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})}\bigr)^N}{M^{\nu-1}},\\*\frac{\nu^{2N}}{M^{\nu-1}} =\exp\biggl[2\log M\cdot\log \nu\cdot\biggl(\frac{N}{\log M}-\frac{\nu-1}{2\log\nu}\biggr)\biggr]\to\infty,\\* \dfrac{{{\nu^{2N}}/{M^{\nu-1}}}}{{{(\nu(\nu-1)\,{{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})})^N}/{M^{\nu-1}}}} =\biggl(\frac{\nu}{\nu-1}{{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})}\biggr)^N\to\infty\end{gather*}

if b is sufficiently large, since

\[\frac{\nu-1}{2\log\nu}<\nu-1.\]

Now consider

\[\frac{\nu-1}{2\log\nu}< \lim\frac{N}{\log M}<\nu-1.\]

Since

\[\min_{\nu>3}\frac{\nu-1}{2\log\nu}=\frac{4-1}{2\log 4}>1\]

(notice that ${{(3-1)}/{(2\log 3)}}<1$ ), we use the middle remainder from (3.28) to obtain

\[\mathcal R_n=\frac{\nu^{2N}}{M^{\nu-1}}\exp\biggl(-\frac{\lambda\log^2 N}{4\nu}\biggr)+\biggl(\frac{\nu N{{\mathrm{e}}}^{{{\mathrm{O}}} (b^{-2})}}{{{\mathrm{e}}}\log M}\biggr)^N.\]

Therefore

(3.30) \begin{equation}\dfrac{\mathcal R_n}{{{\nu^n}/{M^{\nu-1}}}}=\exp\biggl(-\frac{\lambda\log^2 N}{4\nu}\biggr)+\biggl(\frac{{{\mathrm{e}}}^{(\nu-1)\eta_n}+{{\mathrm{O}}} (b^{-2})}{{\nu {{\mathrm{e}}}\eta_n}}\biggr)^N,\quad \eta_n\;:\!=\; \frac{\log M}{N},\end{equation}

that is,

\[\lim\eta_n\in \biggl(\frac{1}{\nu-1},\frac{2\log\nu}{\nu-1}\biggr).\]

The function

\[f_{\nu}(\eta)\;:\!=\; \frac{{{\mathrm{e}}}^{(\nu-1)\eta}}{\nu {{\mathrm{e}}}\eta}\]

increases with $\eta>(\nu-1)^{-1}$ , and

\[f_{\nu}((\nu-1)^{-1})=1-\frac{1}{\nu}<1,\quad f_{\nu}\biggl(\frac{2\log\nu}{\nu-1}\biggr)=\frac{\nu(\nu-1)}{2{{\mathrm{e}}}\log\nu}>1\quad \text{for all $\nu>3$.}\]

So there exists a unique root

\[\eta(\nu)\in \biggl(\frac{1}{\nu-1},\frac{2\log\nu}{\nu-1}\biggr)\]

of $f_{\nu}(\eta)-1$ . Using (3.30), we conclude that for $\nu>3$ the equation (3.29) holds for

\[\frac{1}{\eta(\nu)}<\lim\frac{N}{\log M}<\nu-1\]

if b is chosen large enough.

Finally, like the case $\nu=3$ , $\lim \mathbb{E}[Z_n]=0$ if

\[\lim\frac{N}{\log M}<\frac{1}{\log\nu}.\]

This completes the proof of Theorem 3.2.

4. Second-order moment of the number of perfect partitions

From Theorem 3.2 we know that for

\[\lim\frac{n}{\log M}<\frac{2}{\log \nu}\]

we have $\mathbb{E}[Z_n]\to 0$ , so that $\mathbb{P}(Z_n>0)\le \mathbb{E}[Z_n]\to 0$ . Furthermore, we stated the conditions on $\lim{{n}/{\log M}}$ under which $\lim\mathbb{E}[Z_n]=\infty$ . These conditions make it plausible, but certainly do not imply, that $\mathbb{P}(Z_n>0)$ approaches 1, or even that $\lim \mathbb{P}(Z_n>0)>0$ . In general, if the first-order moment grows very fast, it may well portend that the standard deviation grows even faster. However, our $Z_n$ happens to be a rare exception. (Another such case is the number of Hamilton cycles in a random regular graph; see Robinson and Wormald [Reference Robinson and Wormald14].)

Theorem 4.1. Suppose that

\[\lim\frac{n}{\log M}>\frac{2(\nu-1)}{\log[(1-2\nu^{-2})^{-1}]}.\]

Then, with the limiting probability $\gtrsim (1+\nu^2)^{-1}$ , $Z_n$ is of order $\mathbb{E}[Z_n]$ , that is, the number of perfect partitions is exponentially large.

Note. The limiting probability is certainly below

\[\frac{1}{\nu} = \lim_{n\to\infty}\mathbb{P}\Biggl(\sum_{j\in [n]}X_j \equiv 0 (\mathrm{mod}\, \nu)\Biggr).\]

By Lemma 2.1, we have

\begin{equation*} \mathbb{E}[Z_n^2]=\dfrac{1}{(2\pi)^{ 2\nu}} \int\limits_{\mathbf{x}, \mathbf{x}'}\mathbb{E}^n\bigl[f(\mathbf{x},X)\overline{f(\mathbf{x}',X)}\bigr]\,{{\mathrm{d}}} \mathbf{x} \,{{\mathrm{d}}} \mathbf{x}'.\end{equation*}

Here, with $\phi(y)=\mathbb{E}[{{\mathrm{e}}}^{{{\mathrm{i}}} yX}]$ , we have

(4.1) \begin{align}\mathbb{E}\bigl[f(\mathbf{x},X)\overline{f(\mathbf{x}',X)}\bigr]&=\mathbb{E}\Biggl[\Biggl(\sum_{\alpha\in [\nu]}\exp({{\mathrm{i}}} y_{\alpha}(\mathbf{x})X)\Biggr)\Biggl(\sum_{\beta\in [\nu]}\exp(-{{\mathrm{i}}} y_{\beta}(\mathbf{x}')X)\Biggr)\Biggr] \notag \\*&=\mathbb{E}\Biggl[\sum_{\alpha,\beta \in [\nu]}\exp({{\mathrm{i}}} (y_{\alpha}(\mathbf{x})-y_{\beta}(\mathbf{x}'))X)\Biggr] \notag \\*&=\sum_{\alpha,\beta\in [\nu]}\phi(y_{\alpha}(\mathbf{x})-y_{\beta}(\mathbf{x}')).\end{align}

Let $\varepsilon\in (0, \nu^{-2})$ , and introduce

\[D_{\varepsilon}=\Bigl\{\mathbf{x}, \mathbf{x}'\in [\!-\pi,\pi]^{\nu}\colon \min_{\alpha,\beta\in [\nu]}|\phi(y_{\alpha}(\mathbf{x})-y_{\beta}(\mathbf{x}'))|\le 1-\varepsilon\nu^2\Bigr\}.\]

For $(\mathbf{x}, \mathbf{x}')\in D_{\varepsilon}$ , we have

\[\sum_{\alpha,\beta\in [\nu]}|\phi\bigl(y_{\alpha}(\mathbf{x})-y_{\beta}(\mathbf{x}')\bigr)|\le (1-\varepsilon\nu^2)+\nu^2-1=\nu^2(1-\varepsilon).\]

So $\mathcal E[Z_n^2; D_{\varepsilon}]$ , the contribution of $D_{\varepsilon}$ to $\mathbb{E}[Z_n^2]$ , is at most $\nu^{2n}(1-\varepsilon)^n$ .

Let $(\mathbf{x},\mathbf{x}')\in D_{\varepsilon}^c$ , so that $|\phi(y_{\alpha}(\mathbf{x})-y_{\beta}(\mathbf{x}'))|> 1-\varepsilon\nu^2$ for all $\alpha,\,\beta$ . If $y= y_{\alpha,\beta}\;:\!=\; y_{\alpha}(\mathbf{x})-y_{\beta}(\mathbf{x}')\neq k\pi$ , k even, then

\begin{align*}|\phi(y)|=\bigg|\frac{{{\mathrm{e}}}^{{{\mathrm{i}}} y}({{\mathrm{e}}}^{{{\mathrm{i}}} My}-1)}{M({{\mathrm{e}}}^{{{\mathrm{i}}} y}-1)}\bigg|=\frac{(1-\cos(My))^{1/2}}{M(1-\cos y)^{1/2}}=\bigg|\frac{\sin(My/2)}{M\sin (y/2)}\biggr|> 1-\varepsilon\nu^2.\end{align*}

Let $k_{\alpha,\beta}(\mathbf{x},\mathbf{x}')\pi$ be the even multiple of $\pi$ closest to $y_{\alpha}(\mathbf{x})-y_{\beta}(\mathbf{x}')$ , and set $z_{\alpha,\beta}(\mathbf{x},\mathbf{x}')=y_{\alpha}(\mathbf{x})-y_{\beta}(\mathbf{x}')-k_{\alpha,\beta}(\mathbf{x},\mathbf{x}')\pi$ ; then $|z_{\alpha,\beta}(\mathbf{x},\mathbf{x}')|\le \pi$ , and $\phi(y_{\alpha}(\mathbf{x})-y_{\beta}(\mathbf{x}'))=\phi(z_{\alpha,\beta}(\mathbf{x},\mathbf{x}'))$ . Therefore

\begin{align*}1-\varepsilon\nu^2\le |\phi(y_{\alpha}(\mathbf{x})-y_{\beta}(\mathbf{x}'))|=|\phi(z_{\alpha,\beta}(\mathbf{x},\mathbf{x}'))| \le \dfrac{1}{M|\sin(z_{\alpha,\beta}(\mathbf{x},\mathbf{x}')/2)|}\le\frac{\pi}{M|z_{\alpha,\beta}(\mathbf{x},\mathbf{x}')|},\end{align*}

whence

\[|y_{\alpha}(\mathbf{x})-y_{\beta}(\mathbf{x}')-k_{\alpha,\beta}(\mathbf{x},\mathbf{x}')\pi|= |z_{\alpha,\beta}(\mathbf{x},\mathbf{x}')|\le\frac{\pi}{M(1-\varepsilon\nu^2)}={{\mathrm{O}}} (M^{-1}).\]

Since $\sum_{\alpha}y_{\alpha}(\mathbf{x})=\sum_{\beta}y_{\beta}(\mathbf{x}')=0$ , we then obtain

\begin{align*}&y_{\alpha}(\mathbf{x})-\frac{\pi}{\nu}k_{\alpha}(\mathbf{x},\mathbf{x}')={{\mathrm{O}}} (M^{-1}),\quad k_{\alpha}(\mathbf{x},\mathbf{x}')\;:\!=\; \sum_{\beta}k_{\alpha,\beta}(\mathbf{x},\mathbf{x}'),\\*&y_{\beta}(\mathbf{x}')-\frac{\pi}{\nu}\kappa_{\beta}(\mathbf{x},\mathbf{x}')={{\mathrm{O}}} (M^{-1}),\quad \kappa_{\beta}(\mathbf{x},\mathbf{x}')\;:\!=\; \sum_{\alpha}k_{\alpha,\beta}(\mathbf{x},\mathbf{x}').\end{align*}

Since $\nu$ is fixed and $M\to\infty$ , $({{\pi}/{\nu}}) k_{\alpha}(\mathbf{x},\mathbf{x}')$ (resp. $({{\pi}/{\nu}}) \kappa_{\beta}(\mathbf{x},\mathbf{x}')$ ) is again an even multiple of ${{\pi}/{\nu}}$ closest to $y_{\alpha}(\mathbf{x})$ (resp. $y_{\beta}(\mathbf{x}')$ ), whence $k_{\alpha}(\mathbf{x},\mathbf{x}')=k_{\alpha}(\mathbf{x})$ (resp. $\kappa_{\beta}(\mathbf{x},\mathbf{x}')=k_{\beta}(\mathbf{x}')$ ), i.e. dependent only on $\mathbf{x}$ (resp. on $\mathbf{x}'$ ). Combining the last three equations, and again using $\sum_{\alpha}y_{\alpha}(\mathbf{x})=0$ , $\sum_{\beta}y_{\beta}(\mathbf{x}')=0$ , we obtain

\begin{align*}&\frac{\pi}{\nu}\cdot (k_{\alpha}(\mathbf{x})-\kappa_{\beta}(\mathbf{x}'))=\pi k_{\alpha,\beta}(\mathbf{x},\mathbf{x}')+{{\mathrm{O}}} (M^{-1}),\\*&\sum_{\alpha}k_{\alpha}(\mathbf{x})={{\mathrm{O}}} (M^{-1}),\quad \sum_{\beta}\kappa_{\beta}(\mathbf{x}')={{\mathrm{O}}} (M^{-1}),\end{align*}

implying that for M large

\[k_{\alpha,\beta}(\mathbf{x},\mathbf{x}')=\frac{1}{\nu} (k_{\alpha}(\mathbf{x})-\kappa_{\beta}(\mathbf{x}')),\quad \sum_{\alpha}k_{\alpha}(\mathbf{x})=\sum_{\beta}\kappa_{\beta}(\mathbf{x}')=0.\]

(In particular, all $\nu^2$ differences $k_{\alpha}(\mathbf{x})-\kappa_{\beta}(\mathbf{x}')$ are divisible by $2\nu$ .) The first identity above implies that $z(\mathbf{x},\mathbf{x}')=M^{-1}(\xi_{\alpha}(\mathbf{x})-\eta_{\beta}(\mathbf{x}'))$ , where

\[\xi_{\alpha}(\mathbf{x})\;:\!=\; M\biggl(y_{\alpha}(\mathbf{x})-\frac{\pi}{\nu}k_{\alpha}(\mathbf{x})\biggr),\quad \eta_{\beta}(\mathbf{x}')\;:\!=\;M\biggl(y_{\beta}(\mathbf{x}')-\frac{\pi}{\nu}k_{\beta}(\mathbf{x}')\biggr),\]

and by the second identity, $\sum_{\alpha}\xi_{\alpha}(\mathbf{x})=0$ , $\sum_{\beta}\eta_{\beta}(\mathbf{x}')=0$ . So

\[\phi(z_{\alpha,\beta}(\mathbf{x},\mathbf{x}'))=\phi(M^{-1}(\xi_{\alpha}(\mathbf{x})-\eta_{\beta}(\mathbf{x}'))),\]

and $|\xi_{\alpha}(\mathbf{x})-\eta_{\beta}(\mathbf{x}')|$ is uniformly bounded for $(\mathbf{x},\mathbf{x}')\in D_{\varepsilon}^c$ . Now, for each $b>0$ , there exists $\lambda=\lambda(b)>0$ such that $u^{-1}|\sin u|\in [1-2\lambda u^2, 1-\lambda u^2]$ for $u\in (0,b]$ . Therefore, for $z={{\mathrm{O}}} (M^{-1})$ , there exists sufficiently small $\hat\lambda>0$ such that

\[|\phi(z)|=\biggl|\frac{\sin(Mz/2)}{Mz/2}\biggr| \cdot \biggl|\frac{z/2}{\sin(z/2)}\biggr|\le \frac{1-\hat\lambda (Mz/2)^2}{1-2\hat\lambda(z/2)^2}\le 1-\lambda'(Mz)^2,\]

for some absolute constant $\lambda'>0$ . So, for $(\mathbf{x},\mathbf{x}')\in D_{\varepsilon}^c$ , we have

(4.2) \begin{align}\sum_{\alpha,\beta}|\phi(y_{\alpha}(\mathbf{x})-y_{\beta}(\mathbf{x}'))|&\le \sum_{\alpha,\beta}(1-\lambda' (\xi_{\alpha}(\mathbf{x})-\eta_{\beta}(\mathbf{x}'))^2) \notag \\*&\le \nu^2\exp\Biggl(-\frac{\lambda'}{\nu^2}\sum_{\alpha,\beta}(\xi_{\alpha}(\mathbf{x})-\eta_{\beta}(\mathbf{x}'))^2\Biggr) \notag \\*&=\nu^2\exp\biggl(-\frac{\lambda'}{\nu^2}\bigl(\|\boldsymbol\xi(\mathbf{x})\|^2+\|\boldsymbol \eta(\mathbf{x}')\|^2\bigr)\biggr),\end{align}

since $\sum_{\alpha}\xi_{\alpha}(\mathbf{x})=\sum_{\beta}\eta_{\beta}(\mathbf{x}')=0$ .

Let $\mathcal E[Z_n^2;\, D_{\varepsilon}^c]$ be the contribution to $\mathbb{E}[Z_n^2]$ from $(\mathbf{x},\mathbf{x}')\in D_{\varepsilon}^c$ such that

\[\max\{\|\boldsymbol\xi(\mathbf{x})\|,\,\|\boldsymbol \eta(\mathbf{x}')\|\}\ge \frac{\log n}{n^{1/2}}.\]

Using (4.2), similarly to part (ii) of the proof of Lemma 3.3, we obtain

\begin{equation*}\mathcal E[Z_n^2;\, D_{\varepsilon}^c]\le c_1(\nu)\frac{\nu^{2n}}{\lambda' M^{2(\nu-1)}}\exp\biggl(-\frac{\lambda'\log^2 n}{\nu^2}\biggr).\end{equation*}

Consequently

(4.3) \begin{align}\mathcal E[Z_n^2;\, D_{\varepsilon}]+\mathcal E[Z_n^2;\, D_{\varepsilon}^c] &={{\mathrm{O}}}\biggl(\nu^{2n}(1-\varepsilon)^n+\frac{\nu^{2n}}{\lambda' M^{2(\nu-1)}}\exp\biggl(-\frac{\lambda'\log^2 n}{\nu^2}\biggr)\biggr)\notag \\*&={{\mathrm{O}}}\biggl(\frac{\nu^{2n}}{\lambda' M^{2(\nu-1)}}\exp\biggl(-\frac{\lambda'\log^2 n}{\nu^2}\biggr)\biggr),\end{align}

provided that

\begin{equation*} \lim\frac{n}{\log M}>\dfrac{2(\nu-1)}{\log\frac{1}{1-\nu^{-2}}}.\end{equation*}

It remains to sharply estimate $\mathcal E[Z_n^2;\,\mathcal { C^*}]$ , the contribution to $\mathbb{E}[Z_n^2]$ from

\[\mathcal C^*\;:\!=\; \biggl\{(\mathbf{x},\mathbf{x}')\in D_{\varepsilon}^c\colon \max\{\|\boldsymbol\xi(\mathbf{x})\|,\,\|\boldsymbol \eta(\mathbf{x}')\|\}\le \frac{\log n}{n^{1/2}}\biggr\}.\]

Analogously to $\mathbb{E}[Z_n]$ (see part (III) of the proof of Theorem 3.2), $\mathcal C$ is the disjoint union of $\mathcal C^*_{\mathbf{k},\boldsymbol\kappa}$ , over all pairs $\{\mathbf{k},\boldsymbol\kappa\}$ of $\nu$ -long even tuples $\mathbf{k}=\{k_{\alpha}\}$ , $\boldsymbol\kappa=\{\kappa_{\beta}\}$ , with $\sum_{\alpha}k_{\alpha}=\sum_{\beta}\kappa_{\beta}=0$ , $k_{\alpha}-\kappa_{\beta}\equiv 0(\mathrm{mod} 2\nu)$ , and

\begin{align*}\mathcal C^*_{\mathbf{k},\boldsymbol\kappa}&=\biggl\{(\mathbf{x},\mathbf{x}')\in D_{\varepsilon}^c\colon \max\{\|\boldsymbol\xi(\mathbf{x})\|,\|\boldsymbol \eta(\mathbf{x}')\|\} \le \frac{\log n}{n^{1/2}},\, k_{\alpha}(\mathbf{x}) =k_{\alpha},\, \kappa_{\beta}(\mathbf{x}') = \kappa_{\beta}\biggr\}.\end{align*}

Geometrically, $\mathcal C^*_{\mathbf{k},\boldsymbol\kappa}$ is the Cartesian product of two thin $\nu$ -dimensional cylinders, enclosing respectively the line $\mathcal L^*_{\mathbf{k}}$ and the line $\mathcal L^*_{\boldsymbol\kappa}$ , given by the parametric equations

(4.4) \begin{equation}x_{\alpha}(t)=\nu^{-1}\biggl(t+\frac{\pi}{\nu}k_{\alpha}\biggr),\quad \alpha\in [\nu],\quad x'_{\beta}(t')=\nu^{-1}\biggl(t'+\frac{\pi}{\nu}\kappa_{\beta}\biggr),\quad \beta\in [\nu].\end{equation}

As in the case of $\mathbb{E}[Z_n]$ , the dominant contributors to $\mathbb{E}[Z_n^2]$ are the pairs $\{\mathcal L^*_{\mathbf{k}},\mathcal L^*_{\boldsymbol\kappa}\}$ which contain interior points of their respective cubes $[\!-\pi,\pi]^{\nu}$ . However, this time we need to consider only the pairs for which the differences $k_{\alpha}-\kappa_{\beta}$ are all divisible by $2\nu$ . Each of the cross-sections of the two cylinders formed by planes $\sum_{\alpha}x_{\alpha}=t$ , and $\sum_{\beta}x'_{\beta}=t'$ , respectively, is the $(\nu-1)$ -dimensional sphere of radius $r_n\;:\!=\; (M\nu)^{-1}n^{-1/2}\log n$ , coming from

\begin{align*}\|\mathbf{x}-\mathbf{x}(t)\|^2&=\frac{1}{(M\nu)^2}\sum_{\alpha}\xi_{\alpha}^2(\mathbf{x})\le \frac{\log^2n}{(M\nu)^2n},\\*\|\mathbf{x}'-\mathbf{x}'(t')\|^2&=\frac{1}{(M\nu)^2}\sum_{\beta}\eta_{\beta}^2(\mathbf{x}')\le \frac{\log^2n}{(M\nu)^2n}.\end{align*}

For $(\mathbf{x},\mathbf{x}')\in \mathcal C^*_{\mathbf{k},\boldsymbol\kappa}$ , we have

\begin{align*}&\phi(y_{\alpha}(\mathbf{x})-y_{\beta}(\mathbf{x}'))\\*&\quad =\phi(z_{\alpha,\beta}(\mathbf{x},\mathbf{x}'))\\&\quad =\phi(M^{-1}(\xi_{\alpha}(\mathbf{x})-\eta_{\beta}(\mathbf{x}')))\\*&\quad =1+{{\mathrm{i}}} \frac{\mathbb{E}[X]}{M}(\xi_{\alpha}(\mathbf{x})-\eta_{\beta}(\mathbf{x}'))-\frac{\mathbb{E}[X^2]}{2M^2}(\xi_{\alpha}(\mathbf{x})-\eta_{\beta}(\mathbf{x}'))^2+{{\mathrm{O}}}\bigl(|\xi_{\alpha}(\mathbf{x})-\eta_{\beta}(\mathbf{x}')|^3\bigr).\end{align*}

Summing this equation over $\alpha,\beta \in [\nu]$ , and using $\sum_{\alpha}\xi_{\alpha}(\mathbf{x})=0$ , $\sum_{\beta}\eta_{\beta}(\mathbf{x}')=0$ , we transform (4.1) into

\begin{align*}&\mathbb{E}\bigl[f(\mathbf{x},X)\overline{f(\mathbf{x}',X)}\bigr]\\*&\quad =\nu^2-\frac{\mathbb{E}[X^2]}{2M^2}\bigl(\|\boldsymbol\xi(\mathbf{x})\|^2+\|\boldsymbol \eta(\mathbf{x}')\|^2\bigr)+{{\mathrm{O}}}\bigl(\|\boldsymbol\xi(\mathbf{x})\|^3+\|\boldsymbol \eta(\mathbf{x}')\|^3\bigr) \\&\quad =\nu^2\biggl(1-\frac{M^2c_M}{2}\bigl(\|\mathbf{x}-\mathbf{x}(t)\|^2+\|\mathbf{x}'-\mathbf{x}'(t')\|^2\bigr)+{{\mathrm{O}}}\bigl(M^3\bigl(\|\mathbf{x}-\mathbf{x}(t)\|^3+\|\mathbf{x}'-\mathbf{x}'(t')\|^3\bigr)\biggr)\\*&\quad =\nu^2\exp\biggl(-\frac{M^2c_M}{2}\bigl(\|\mathbf{x}-\mathbf{x}(t)\|^2+\|\mathbf{x}'-\mathbf{x}'(t')\|^2\bigr)+{{\mathrm{O}}} (n^{-3/2}\log^3 n)\biggr).\end{align*}

So

\begin{align*}&\mathbb{E}^n\bigl[f(\mathbf{x},X)\overline{f(\mathbf{x}',X)}\bigr] =\nu^{2n}\exp\biggl(-\frac{M^2nc_M}{2}\bigl(\|\mathbf{x}-\mathbf{x}(t)\|^2+\|\mathbf{x}'-\mathbf{x}'(t')\|^2\bigr)+{{\mathrm{O}}} (n^{-1/2}\log^3 n)\biggr).\end{align*}

So $\mathcal E\bigl[Z_n^2;\,\mathcal C^*_{\mathbf{k},\boldsymbol{\kappa}}\bigr]$ , the contribution of $\mathcal C^*_{\mathbf{k},\boldsymbol{\kappa}}$ to $\mathbb{E}[Z_n^2]$ , is given by

\[\mathcal E\bigl[Z_n^2;\,\mathcal C^*_{\mathbf{k},\boldsymbol{\kappa}}\bigr]=(1+{{\mathrm{o}}} (1))\frac{\nu^{2n}L^*(\mathbf{k})L^*(\boldsymbol\kappa)}{(2\pi)^{2\nu}(M\nu)^{2(\nu-1)}}\biggl(\int_{\rho\ge 0}\exp\biggl({-\frac{c_Mn}{2\nu}\rho^2}\biggr)\,{{\mathrm{d}}} V(\rho)\biggr)^2,\]

$V(\rho)$ being the volume of the $(\nu-1)$ -dimensional sphere of radius $\rho$ . Here $L^*(\mathbf{k})$ and $L^*(\boldsymbol\kappa)$ are the lengths of segments of $\mathcal L^*(\mathbf{k})$ and $\mathcal L^*(\boldsymbol\kappa)$ within the respective cubes $[\!-\pi,\pi]^{\nu}$ . Summing over all admissible pairs $\{\mathbf{k},\boldsymbol\kappa\}$ , we have

\[\mathcal E\bigl[Z_n^2;\;\mathcal C^*\bigr] = (1+{{\mathrm{o}}} (1))\biggl(\frac{\nu^n}{(2\pi)^{\nu}(M\nu)^{\nu-1}}\int_{\rho\ge 0}\exp\biggl({-\frac{c_Mn}{2\nu}\rho^2}\biggr)\,{{\mathrm{d}}} V(\rho) \biggr)^2 \sum_{\{\mathbf{k},\boldsymbol\kappa\}} L^*(\mathbf{k}) L^*(\boldsymbol\kappa).\]

We upper-bound the last product by $(L^*)^2$ , $L^*\;:\!=\; \sum_{\mathbf{k}}L^*(\mathbf{k})$ , thus replacing the condition that all $k_{\alpha}-\kappa_{\beta}$ are divisible by $2\nu$ with the weaker condition that marginally all the differences $k_{\alpha}-k_{\alpha'}$ and all the differences $\kappa_{\beta}-\kappa_{\beta'}$ are divisible by $2\nu$ .

By (4.4), for the interior segment of $\mathcal L^*(\mathbf{k})$ we have

\[\!-\pi\biggl(\nu+\frac{\min k_{\alpha}}{\nu}\biggr) < t < \pi\biggl(\nu-\frac{\max k_{\alpha}}{\nu}\biggr).\]

So

\[L^{*}(\mathbf{k})=\pi d^*(\mathbf{k})\nu^{-1/2},\quad d^*(\mathbf{k})\;:\!=\; 2\nu-\frac{r}{\nu}, \quad r\;:\!=\; \max k_{\alpha}-\min k_{\alpha},\]

and $d^*(\mathbf{k})>0$ if either $k_{\alpha}\equiv 0$ , or ${{r}/{(2\nu)}}\in [\nu-1]$ . Consequently

\begin{gather*}\sum_{\mathbf{k}}d^*(\mathbf{k}) =2\nu(1+\mathcal M_1^*)-\frac{1}{\nu}\mathcal M_2^*,\\*\mathcal M_1^* =\sum_{{{r}/{(2\nu)}}\in [\nu-1]}\mathcal N^*(r),\quad \mathcal M_2^*=\sum_{{{r}/{(2\nu)}}\in [\nu-1]}r\mathcal N^*(r),\end{gather*}

Here, closely following the line enumeration in part (I) of the proof of Theorem 3.2, we have

\begin{align*}\mathcal N^*(r) & =\sum_{\mathrm{even}\, a\le 0}[\zeta^{-a\nu}]\mathcal F^*_r(\zeta),\quad\frac{r}{2\nu}\in [\nu-1], \\*\mathcal F^*_r(\zeta) &=\biggl(\frac{1-\zeta^{r+2\nu}}{1-\zeta^{2\nu}}\biggr)^{\nu}-\biggl(\frac{1-\zeta^r}{1-\zeta^{2\nu}}\biggr)^{\nu}-\biggl(\frac{\zeta^{2\nu}-\zeta^{r+2\nu}}{1-\zeta^{2\nu}}\biggr)^{\nu}+\biggl(\frac{\zeta^{2\nu}-\zeta^r}{1-\zeta^{2\nu}}\biggr)^{\nu}.\end{align*}

So

\begin{align*}\mathcal M_1^*&=\sum_{\mathrm{even}\, a\le 0}[\zeta^{-a\nu}]\sum_{{{r}/{(2\nu)}}\in [\nu-1]}\mathcal F^*_r(\zeta)\\*&=\sum_{\mathrm{even}\, a\le 0}[\zeta^{-a\nu}]\biggl(\biggl(\frac{1-\zeta^{2\nu^2}}{1-\zeta^{2\nu}}\biggr)^{\nu}-\biggl(\frac{\zeta^{2\nu}-\zeta^{2\nu^2}}{1-\zeta^{2\nu}}\biggr)^{\nu}-1\biggr)\\&=\sum_{\alpha\ge 0}[t^{\alpha}]\biggl[\biggl(\frac{1-t^{\nu}}{1-t}\biggr)^{\nu}-\biggl(\frac{t-t^{\nu}}{1-t}\biggr)^{\nu}-1\biggr]\\&=\lim_{t\to 1}\biggl[\biggl(\frac{1-t^{\nu}}{1-t}\biggr)^{\nu}-\biggl(\frac{t-t^{\nu}}{1-t}\biggr)^{\nu}-1\biggr]\\*&= \nu^{\nu}-(\nu-1)^{\nu}-1.\end{align*}

Next, with a bit of telescoping again,

\begin{align*}\sum_{{{r}/{(2\nu)}}\in [\nu-1]}r\mathcal F^*_r(\zeta)&=2\nu(\nu-1)\biggl(\frac{1-\zeta^{2\nu^2}}{1-\zeta^{2\nu}}\biggr)^{\nu}\\*&\quad -(2\nu(\nu-1)\zeta^{2\nu^2}+2\nu)\biggl(\frac{1-\zeta^{2\nu(\nu-1)}}{1-\zeta^{2\nu}}\biggr)^{\nu}\\*&\quad +2\nu(\zeta^{2\nu^2}-1)\sum_{j=1}^{\nu-2}\biggl(\frac{1-\zeta^{2\nu j}}{1-\zeta^{2\nu}}\biggr)^{\nu}\,=\!:\, S^*(\zeta^{2\nu}).\end{align*}

Consequently,

\begin{align*}\mathcal M_2^*&=\sum_{\mathrm{even}\, a\le 0}[\zeta^{-a\nu}]\sum_{{{r}/{(2\nu)}}\in [\nu-1]}r\mathcal F^*_r(\zeta)\\*&=\sum_{\alpha\ge 0}[t^{\alpha}] S^*(t)\\*&=\lim_{t\to 1}S^*(t)=2\nu(\nu-1)\nu^{\nu}-2\nu^2(\nu-1)^{\nu}.\end{align*}

Therefore

\begin{align*}\sum_{\mathbf{k}}d^*(\mathbf{k})&=2\nu(1+\mathcal M_1^*)-\frac{1}{\nu}\mathcal M_2^*\\*&=2\nu(\nu^{\nu}-(\nu-1)^{\nu})-2(\nu-1)\nu^{\nu}+2\nu(\nu-1)^{\nu}\\*&=2\nu^{\nu},\end{align*}

so that $L^*=2\pi\nu^{\nu-1/2}$ . Hence

\begin{align*}\mathcal E\bigl[Z_n^2;\,\mathcal C^*\bigr] &\le (1+{{\mathrm{o}}} (1))\biggl(\frac{\nu^n \cdot2\pi\nu^{\nu-1/2}}{(2\pi)^{\nu}(M\nu)^{\nu-1}}\int_{\rho\ge 0}\exp\biggl({-{\frac{c_Mn}{2\nu}}\rho^2}\biggr)\,{{\mathrm{d}}} V(\rho) \biggr)^2\\ &=(1+{{\mathrm{o}}} (1))\biggl(\frac{\nu^n}{M^{\nu}-1}\cdot\dfrac{\nu^{\nu-1/2}}{(2\pi\nu c_{M}n)^{{{(\nu-1)}/{2}}}}\biggr)^2.\end{align*}

In combination with (4.3), and (3.27), this leads to

\begin{align*}\mathbb{E}\bigl[Z_n^2\bigr]&\le (1+{{\mathrm{o}}} (1))\biggl(\frac{\nu^n}{M^{\nu}-1}\cdot\dfrac{\nu^{\nu-1/2}}{(2\pi\nu c_{M}n)^{{{(\nu-1)}/{2}}}}\biggr)^2\\*&=(1+{{\mathrm{o}}} (1)) \nu^2 \mathbb{E}^2[Z_n],\end{align*}

provided that

\[\lim\frac{n}{\log M}>\frac{2(\nu-1)}{\log[(1-2\nu^{-2})^{-1}]}.\]

By Cantelli’s inequality (Billingsley [Reference Billingsley2]),

\begin{align*}\liminf\mathbb{P}(Z_n\ge \delta\mathbb{E}[Z_n])\ge \lim \frac{(1-\delta)^2}{(1-\delta)^2+{{\mathbb{E}[Z^2_n]}/{\mathbb{E}^2[Z_n]}}}\ge \frac{(1-\delta)^2}{(1-\delta)^2+\nu^2}\quad\text{for all $\delta\in (0,1)$.}\end{align*}