2. Preliminaries

2.1. Notation

For the sake of clarity, we fix some standard notation and conventions. This repeats some discussion from the introduction. Let $q \geq 2$ be a prime power, and let $\mathbb{F}_q[t]$ be the polynomial ring over the finite field $\mathbb{F}_q$ with q elements. Let $\mathcal{M}$ denote the set of all monic polynomials in $\mathbb{F}_q[t]$ and let $\mathcal{M}_N \subseteq \mathcal{M}$ be the subset of degree N polynomials, so $|\mathcal{M}_N| = q^N$ . The letters F and G will always denote monic polynomials, and the letters P and Q will always denote monic irreducible polynomials. The zeta function $\zeta_q$ for function fields is defined as

\begin{align*} \zeta_q(z) &= \sum_{N=0}^{\infty} |\mathcal{M}_N| z^{{\kern1pt}N} = \prod_{P}\left(1-z^{\deg P}\right)^{-1} = \frac{1}{1-qz},\end{align*}

where products over P will always run over all irreducible monic polynomials.

A magic square of size k with magic constant N is a $k\times k$ matrix whose entries are nonnegative integers such that the entries of each row and column sum to N. The number of such matrices will be denoted by $\mathcal{S}_k(N)$ .

For variables a and b, we write $a \ll b$ or $a = O(b)$ to say that there exists an absolute positive constant C such that $|a| \leq C b$ . If the constant C depends on a parameter, say k, we shall write $a \ll_k b$ or $a = O_k(b)$ . If a and b depend on a positive parameter x, then we say that $a = o(b)$ as $x \to \infty$ if the ratio ${a}/{b}$ converges to 0 as $x \to \infty$ .

Sums of the form

\[\mathop{\sum\cdots\sum}_{m_1+\dots + m_k = m_{k+1} + \dots + m_{2k}}\]

will be taken over non-negative integers $m_1, \dots, m_{2k} \geq 0$ .

2.2. Setup

For the entirety of the paper, let $f \!\colon \mathcal{M} \to \mathbb{C}$ be a Steinhaus random multiplicative function over $\mathbb{F}_q[t]$ . Our goal is to calculate the 2kth moments of (1·5) for integers k. All of our proofs begin by reducing to a natural counting problem. By complete multiplicativity and independence of f on distinct irreducibles P, it follows for any $F,G \in \mathcal{M}$ that

\[\mathbb{E}[f(F)\overline{f(G)}] = \unicode{x1D7D9}_{\{F = G\}},\]

where $\unicode{x1D7D9}_{\mathcal{A}}$ denotes the indicator function of a set $\mathcal{A}$ . For the case $k=1$ , this implies that

\begin{align*}\mathbb{E}\left| \sum_{F\in \mathcal{M}_N} f(F)\right|^{2} = \mathop{\sum\sum}_{F, G\in \mathcal{M}_N}\mathbb{E}[f(F)\overline{f(G)}] = \mathop{\sum\sum}_{F, G\in \mathcal{M}_N}\unicode{x1D7D9}_{\{F=G\}} = |\mathcal{M}_N|=q^N.\end{align*}

Similarly, the 2kth moments are given by

\begin{align*}\mathop{\sum\cdots\sum}_{F_1, \dots, F_{2k}\in \mathcal{M}_N} \mathbb{E}[f(F_1)\dots f(F_k) \overline{f(F_{k+1})\dots f(F_{2k})}] = \mathop{\sum\cdots\sum}_{F_1, \dots, F_{2k}\in \mathcal{M}_N} \unicode{x1D7D9}_{\{F_1\dots F_k = F_{k+1}\dots F_{2k}\}},\end{align*}

leading to the key initial identity:

(2·1) \begin{equation} \mathbb{E}\left| \sum_{F \in \mathcal{M}_N} f(F) \right|^{2k} = \#\Big\{(F_1, \dots, F_{2k})\in \mathcal{M}_N^{2k}\ \Big|\ F_1F_2\cdots F_k = F_{k+1}F_{k+2}\cdots F_{2k}\Big\}.\end{equation}

2.3. Remarks on the integer setting

The moment problem over the integers can be reduced to a counting problem similar to (2·1). More precisely, the left-hand side of (1·2) is equal to

(2·2) \begin{equation} \# \Big\{ (n_1,\dots,n_{2k}) \in \{1,2,\dots,\lfloor x \rfloor \}^{2k} \ \Big|\ n_1\cdots n_k = n_{k+1} \cdots n_{2k} \Big\}.\end{equation}

Estimating this quantity takes several approaches in the literature. Granville–Soundararajan [ Reference Granville and SoundararajanGS01 ] give upper and lower bounds for (2·2) via their GCD matrix construction. Our proof of Theorem 1·2 applies this method over $\mathbb{F}_q[t]$ to produce an asymptotic in the q-limit; see also Section 3 for the case $k=2$ .

Harper–Nikeghbali–Radziwiłł [ Reference Harper, Nikeghbali and RadziwiHNR15 ] and Heap–Lindqvist [ Reference Heap and LindqvistHL16 ] count (2·2) using the analytic theory of multivariable Dirichlet series. They enforce the condition $1 \leq n_i \leq N$ with a suitable bump function, which is written as 2k line integrals via Mellin inversion. This essentially rewrites (2·2) as an iterated integral over the multivariable Dirichlet series

(2·3) \begin{equation} \mathop{\sum\cdots\sum}_{\substack{n_1, n_2, \dots, n_{2k} \\ n_1 \cdots n_k = n_{k+1} \cdots n_{2k}}} \frac{1}{n_1^{s_1} \cdots n_{2k}^{s_{2k}}}.\end{equation}

By expressing this Dirichlet series as a product of Riemann zeta functions and an Euler product with an expanded region of absolute convergence, they can extract the integral’s asymptotic size. Roughly speaking, after a change of variables of the form $s_j \mapsto {1}/{2}+it_j$ , they show that the main contribution is concentrated when $t_j \in \mathbb{R}$ is very small for many j. This truncation requires a careful case analysis by subdviding the remaining large region of integration, where the complicated integrand decays rapidly as $t_j$ grows for some fixed j.

Our proof of Theorem 1·1 in Section 6 adapts this strategy to the function field setting. The multivariable Dirichlet series (2·3) over the integers corresponds to a multivariable power series (6·1) over $\mathbb{F}_q[t]$ . As established in Section 7, this power series will also factor as a product of zeta functions and another series with expanded radius of convergence. By construction, the quantity (2·1) is a specific coefficient of our multivariable power series, expressed as an iterated integral via Cauchy’s integral formula. The case $k=2$ for Theorem 1·3 yields a sufficiently simple series factorisation and integral expression, which we compute directly in Section 5. For $k \geq 3$ , we parameterise the resulting integrals, and perform a truncation like Heap–Lindqvist. After an analogous change of variables in Section 8, the main contribution near the origin is extracted in Section 9. The proof concludes in Section 10 with a technical analysis of the integral over the complement of the truncated region.

3. Fourth moment computation using a combinatorial approach

We provide a short proof of Theorem 1·3 by counting coprime pairs of monic polynomials of fixed degree. We require the following lemma.

Proof of Theorem 1·3. By (2·1), the fourth moment is given by

\begin{align*}\mathop{\sum\sum\sum\sum}_{\substack{F_1, F_2, G_1, G_2 \in \mathcal{M}_N \\ F_1 F_2=G_1 G_2}}1.\end{align*}

We note that $F_1F_2 = G_1 G_2$ if and only if

\begin{align*} \frac{F_1}{(F_1, G_2)}\frac{F_2}{(F_2, G_1)} = \frac{G_1}{(F_2, G_1)}\frac{G_2}{(F_1, G_2)},\end{align*}

where (F,G) denotes the GCD of F and G, which is equivalent to

\begin{align*}\frac{F_1}{(F_1, G_2)}=\frac{G_1}{(F_2, G_1)} \text{ and }\frac{F_2}{(F_2, G_1)} = \frac{G_2}{(F_1, G_2)}.\end{align*}

Letting $A \,{:\!=}\, {F_1}/{(F_1, G_2)}$ and $B \,{:\!=}\, {G_2}/{(F_1, G_2)}$ , we observe that A and B can be any monic, coprime polynomials such that

\begin{align*}\deg A = \deg B = N-\deg(F_1, G_2) = N-\deg(F_2, G_1),\end{align*}

while $(F_1, G_2)$ and $(F_2, G_1)$ can be any monic polynomials of degree $N-\deg A = N-\deg B$ . Thus, counting the possible values of $(F_1, G_2)$ and $(F_2, G_1)$ for each possible value of A and B, we have

\begin{align*} \mathop{\sum\sum\sum\sum}_{\substack{F_1, F_2, G_1, G_2 \in \mathcal{M}_N \\ F_1 F_2=G_1 G_2}}1&= \mathop{\sum\sum}_{\substack{(A,B)=1\\ \deg A = \deg B \leq N}}\left(\sum_{F \in \mathcal{M}_{N-\deg A}} 1\right)^2\\ &= \mathop{\sum\sum}_{\substack{(A,B)=1\\ \deg A = \deg B \leq N}}\left(q^{N-\deg A}\right)^2\\ &= \sum_{m=0}^N q^{2(N-m)} \mathop{\sum\sum}_{\substack{(A,B)=1\\ \deg A=\deg B=m}}1\\ &= N q^{2N}\left(1-\frac{1}{q}\right)+q^{2N},\end{align*}

where the last line follows from Lemma 3·1.

4. Combinatorial method in the large q limit: proof of Theorem 1·2

This section is dedicated to the proof of Theorem 1·2 by generalising the combinatorial approach of Section 3. As in the $k=2$ case, we can divide the equation

\begin{align*}F_1 \cdots F_k = G_1 \cdots G_k\end{align*}

by certain greatest common divisors to reduce the problem of computing the 2kth moment to a problem of counting sets of polynomials with certain coprimality conditions. For $k\gt 2$ , these coprimality conditions are more complex and, to the best of our knowledge, there are no known exact formulas for counting the number of polynomials satisfying these conditions. For simplicity, rather than describing equivalent conditions to $F_1\cdots F_k = G_1 \cdots G_k$ , we define an injection from

(4·1) \begin{equation}\{(F_1, \dots, F_k, G_1, \dots, G_k) \in \mathcal{M}_N^{2k}\ \big|\ F_1\cdots F_k = G_1 \cdots G_k\}\end{equation}

to $\mathcal{M}^{k^2}$ and then apply simple bounds to the image of (4·1).

We first present the argument for $k=3$ as an example that will illustrate the general case. We consider

\begin{align*}F_1\,F_2\,F_3 = G_1\, G_2\, G_3,\end{align*}

where $F_i, G_i \in \mathcal{M}_N$ . Due to the additional factor compared to the $k=2$ case, we will require multiple recursive steps to decompose the products. We begin by letting

\begin{align*} A_{12} &= (F_1, G_2)\\ A_{23} &= (F_2, G_3)\\ A_{31} &= (F_3, G_1).\end{align*}

Here the indices are determined by the permutation $\sigma = (1\,2\,3)$ , so that $A_{i\sigma(i)} = (F_i, G_{\sigma(i)})$ . For the next set of GCDs, we repeat this shift in the indices by letting

\begin{align*}A_{i\,\sigma^2(i)} = \left(\frac{F_i}{A_{i\,\sigma(i)}}, \frac{G_{\sigma^2(i)}}{A_{\sigma(i)\, \sigma^2(i)}}\right),\end{align*}

so that

\begin{align*} A_{13} &= \left(\frac{F_1}{A_{12}}, \frac{G_3}{A_{23}}\right),\\ A_{21} &= \left(\frac{F_2}{A_{23}}, \frac{G_1}{A_{31}}\right),\\ A_{32} &= \left(\frac{F_3}{A_{31}}, \frac{G_2}{A_{12}}\right).\end{align*}

Repeating the shifts a third time, we have

\begin{align*} A_{i\sigma(i)^3} = A_{ii} = \frac{F_i}{A_{i\sigma(i)} A_{i \sigma^2(i)}}= \frac{G_i}{A_{\sigma^{-1}(i) i} A_{ \sigma^{-2}(i) i}},\end{align*}

with the last equality following from the fact that for $i\neq j$ ,

\begin{align*}\Big(\frac{F_i}{A_{i\sigma(i)} A_{i \sigma^2(i)}}, \frac{G_j}{A_{\sigma^{-1}(j) j} A_{ \sigma^{-2}(j) j}}\Big) = 1.\end{align*}

We then have that

\begin{align*} F_i = A_{i \sigma(i)} A_{i \sigma^2(i)} A_{ii} \qquad \text{and}\qquad G_j = A_{\sigma^{-1}(j) j} A_{ \sigma^{-2}(j) j} A_{jj}.\end{align*}

Thus, we have mapped $(F_1, F_2, F_3, G_1, G_2, G_3)$ to a unique matrix

\begin{align*}\begin{bmatrix}A_{11} &A_{12} & A_{13}\\ A_{21} &A_{22} &A_{23}\\ A_{31} &A_{32} &A_{33}\end{bmatrix}\end{align*}

in which the product of the entries in row i is $F_i$ and the product of the entries in column j is $G_j$ . We will call this matrix the GCD matrix of $(F_1, F_2, F_3, G_1, G_2, G_3)$ .

Now, letting $\deg A_{ij} = a_{ij}$ , we note that the matrix containing the degrees of the polynomials in the GCD matrix,

\begin{align*}\begin{bmatrix}a_{11} &a_{12} & a_{13}\\ a_{21} &a_{22} &a_{23}\\ a_{31} &a_{32} &a_{33}\end{bmatrix},\end{align*}

is a magic square since row i must sum to $\deg F_i = N$ and column j must sum to $\deg G_j = N$ . We will call this the GCD degree matrix of $(F_1, F_2, F_3, G_1, G_2, G_3)$ .

For S a magic square of size 3 for the number N, we let R(S) count the number of tuples in

\begin{align*}\{(F_1, F_2, F_3, G_1, G_2, G_3) \in \mathcal{M}_N^6\ \big|\ F_1 F_2 F_3 = G_1 G_2 G_3 \}\end{align*}

with GCD degree matrix given by S. We then have that

\begin{align*}\mathbb{E}\Big|\sum_{F\in \mathcal{M}_N}f(F)\Big|^{6} = \sum_{S}R(S),\end{align*}

where the sum is taken over all magic squares of size 3 for the number N. We would now like to bound R(S). We first note that R(S) is bounded above by the total number of $3\times3$ matrices with entries in $\mathcal{M}$ and with the degrees of the entries given by S. Thus $R(S) \leq q^{a_{11}+a_{12}+\cdots+a_{32}+a_{33}} = q^{3N}$ . To bound R(S) from below, we consider a $3\times 3$ matrix with entries $A_{ij}$ in $\mathcal{M}$ such that the entries are pairwise coprime and the degrees of the entries are given by S. We have that $(A_{i1} A_{i2} A_{i3},\, A_{1j}A_{2j}A_{3j}) = A_{ij}$ , so this matrix is the GCD matrix of $(F_1, F_2, F_3, G_1, G_2, G_3)$ , where

\begin{align*}F_i = A_{i1}A_{i2} A_{i3}\end{align*}

and

\begin{align*}G_j = A_{1j}A_{2j}A_{3j}.\end{align*}

It follows that R(S) is bounded below by the number of pairwise coprime monic polynomials with degrees equal to the entries of S.

Before establishing an asymptotic for this lower bound, we return for the moment to the general case. For $k\geq 3$ , given $\sigma = (1\, 2\, \dots\, k)$ and $s\leq k$ , we let

\begin{align*}A_{i\,\sigma^s(i)} = \Big(\frac{F_i}{\prod_{r\lt s}A_{i\,\sigma^r(i)}}, \frac{G_{\sigma^s(i)}}{\prod_{r\lt s}A_{\sigma^{-r}(i)\,\sigma^s(i)}}\Big).\end{align*}

As we saw for $k = 3$ , we have

\begin{align*}A_{ii} = A_{i\sigma^k(i)} = \frac{F_i}{\prod_{r\lt k}A_{i\,\sigma^r(i)}} = \frac{G_i}{\prod_{r\lt k}A_{\sigma^{-r}(i)\,i}},\end{align*}

meaning that

\begin{align*} F_i = \prod_{h = 1}^k A_{i\, h} \qquad \text{and}\qquad G_j = \prod_{h = 1}^k A_{h\,j}.\end{align*}

In this way, we map $(F_1, \dots, F_k, G_1, \dots, G_k)$ to the GCD matrix

\begin{align*}\begin{bmatrix}A_{11} &\cdots & A_{1k}\\ \vdots &\ddots &\vdots\\ A_{k1} &\cdots &A_{kk}\end{bmatrix}\end{align*}

in which the product of the entries in row i is $F_i$ and the product of the entries in column j is $G_j$ . Letting $\deg A_{ij} = a_{ij}$ , we note again that the GCD degree matrix

\begin{align*}\begin{bmatrix}a_{11} &\cdots& a_{1k}\\ \vdots &\ddots &\vdots\\ a_{k1} &\cdots &a_{kk}\end{bmatrix}\end{align*}

is a magic square. We remark that a similar GCD matrix construction was explored by Granville and Soundararajan in [ Reference Granville and SoundararajanGS01 ] in their study of 2kth moments in the integer setting.

For S a magic square of size k for the number N, let R(S) count the number of tuples in

\begin{align*}\{(F_1, \dots, F_k, G_1, \dots, G_k) \in \mathcal{M}_N^{2k}\ \big|\ F_1 \cdots F_k = G_1 \cdots G_k \}\end{align*}

with GCD degree matrix S, so that

\begin{align*}\mathbb{E}\Big|\sum_{F\in \mathcal{M}_N}f(F)\Big|^{2k} = \sum_{S}R(S).\end{align*}

As before, the sum is taken over all magic squares of size k for the number N. We observe that R(S) is bounded above by the total number of $k\times k$ matrices with entries in $\mathcal{M}$ and with the degrees of the entries given by S. Thus $R(S) \leq q^{\sum_{i=1}^k \sum_{j=1}^k a_{ij}} = q^{kN}$ . To bound R(S) from below, we consider a $k\times k$ matrix with pairwise coprime entries $A_{ij}$ in $\mathcal{M}$ such that the degrees of the entries are given by S. We have that

\begin{align*}\Big(\prod_{h=1}^k A_{ih},\, \prod_{h=1}^k A_{hj}\Big) = A_{ij},\end{align*}

so this matrix is the GCD matrix of $(F_1, \dots, F_k, G_1, \dots, G_k)$ , where

\begin{align*} F_i = \prod_{h = 1}^k A_{i\, h} \qquad \text{and}\qquad G_j = \prod_{h = 1}^k A_{h\,j}.\end{align*}

It follows that R(S) is bounded below by the number of pairwise coprime monic polynomials with degrees equal to the entries of S.

To obtain the q-limit asymptotic and complete the proof of Theorem 1·2, we will need the following result on the number of pairwise coprime polynomials of fixed degrees.

Lemma 4·1 ([ Reference Helmke, Jordan and LiebHJL16 , corollary 3]). For $(d_1, \dots, d_s)\in \mathbb{N}^s$ , the number of monic pairwise coprime polynomials $f_i \in \mathbb{F}_q[t]$ with $\deg(f_i)= d_i$ for $i=1, \dots, s$ is

\begin{align*} q^{d_1+\cdots+d_s}\left( 1-\frac{s(s-1)}{2}\cdot\frac{1}{q}+O_{s,d_1,\dots,d_s}\left(\frac{1}{q^2}\right)\right).\end{align*}

Applying Lemma 4·1 to our lower bound for R(S), we see that

\begin{align*}R(S) \geq q^{kN}\left( 1-\frac{k^2(k^2-1)}{2}\cdot\frac{1}{q}+O_{k,N}\left(\frac{1}{q^2}\right)\right).\end{align*}

Thus $R(S) = q^{kN}\big(1 + O_{k,N}(q^{-1}) \big)$ , and Theorem 1·2 follows.

5. Fourth moment using an analytic approach

We provide another proof of Theorem 1·3 by studying a power series in 4 variables indexed over all 4-tuples of monic polynomials:

(5·1) \begin{equation}{}A(z_1,z_2,z_3,z_4) \,{:\!=}\, \mathop{\sum\sum\sum\sum}_{F_1, \dots, F_{4}\in \mathcal{M}} \unicode{x1D7D9}_{\{F_1F_2 = F_{3}F_{4}\}}z_1^{\deg F_1}\cdots z_{4}^{\deg F_{4}}.\end{equation}

We will check at the beginning of Section 6 in a more general case that A converges uniformly on compact subsets of $\{ (z_1,z_2,z_3,z_4) \in \mathbb{C}^{4} : |z_i| \lt {1}/{q} \text{ for } 1 \leq i \leq 4\}$ . By (2·1), calculating the fourth moment $\mathbb{E}| \sum_{F \in \mathcal{M}_N} f(F) |^4$ amounts to computing the coefficient of $z_1^N\cdots z_{4}^N$ in the above power series.

Observing that the indicator function $\unicode{x1D7D9}_{\{F_1F_2 = F_{3}F_{4}\}}$ is multiplicative (see the Section 6 footnote for a definition of multiplicative), we find that

(5·2) \begin{equation}{} A(z_1,z_2,z_3,z_4) = \prod_{P} \left(\mathop{\sum\sum\sum\sum}_{m_1+m_2 = m_{3}+m_4} (z_1^{m_1}z_2^{m_2}z_3^{m_3}z_{4}^{m_{4}})^{\deg P}\right)\!.\end{equation}

The product runs over all irreducible monic polynomials and the internal sum iterates over the possible ways to distribute a given irreducible factor P among $F_1, \dots, F_{4}$ such that the number of factors of P in $F_1F_2$ and $F_{3}F_{4}$ are equal. As in [ Reference Heap and LindqvistHL16 ], we factor this product as

(5·3) \begin{equation}A(z_1, \dots, z_{4}) = B(z_1, \dots, z_{4}) \prod_{i=1}^2 \prod_{j=3}^{4} \zeta_q(z_iz_j).\end{equation}

Here, B is defined implicitly by (5·3), namely

(5·4) \begin{equation}B(z_1, z_2, z_3, z_4) = \prod_{P}\Big(\mathop{\sum\sum\sum\sum}_{\substack{m_1+m_2=m_3+m_4}}\Big(z_1^{m_1}z_2^{m_2}z_3^{m_3}z_4^{m_4}\Big)^{\deg P}\Big)\prod_{i=1}^2\prod_{j=3}^4 \Big(1-(z_iz_j)^{\deg P}\Big).\end{equation}

We note that absolute and uniform convergence of this expression for $|z_i| \lt {1.1}/{\sqrt{q}}$ will follow by applying Proposition 6·1, which we prove in Section 7, for $k =2$ .

Recall that by the multivariate version of Cauchy’s integral formula, obtaining the desired coefficient of A can be done by computing

\begin{align*}{} &\frac{1}{(2\pi i)^{4}}\int_{|z_{4}|=\varepsilon_{4}}\cdots \int_{|z_{1}|=\varepsilon_1} \frac{A(z_1, \dots, z_{4})}{z_1^{N+1}\cdots z_{4}^{N+1}}dz_1\cdots dz_{4}\\ &\quad= \frac{1}{(2\pi i)^{4}}\int_{|z_{4}|=\varepsilon_{4}}\cdots \int_{|z_{1}|=\varepsilon_1} \frac{B(z_1, \dots, z_{4}) \prod_{i=1}^2 \prod_{j=3}^{4} \zeta_q(z_iz_j)}{z_1^{N+1}\cdots z_{4}^{N+1}}dz_1\cdots dz_{4},\end{align*}

where we integrate over a polydisc with small radii $0 \lt \varepsilon_i \lt q^{-1}$ around the origin. We proceed to find a simple formula for (5·4) which will enable us to explicitly compute the above integral. To do so, we study the local factors of the B-term in the following lemma.

Lemma 5·1. If $|z_iz_j| \lt 1$ for $i\in \{1,2\}$ and $j\in \{3,4\}$ , then the following identity holds:

\begin{align*}\Big(\mathop{\sum\sum\sum\sum}_{m_1+m_2=m_3+m_4}z_1^{m_1}z_2^{m_2}z_3^{m_3}z_4^{m_4}\Big)\prod_{i=1}^2\prod_{j=3}^4 \left(1-z_iz_j\right) = 1-z_1z_2z_3z_4.\end{align*}

Proof. Note that if $|z_iz_j| \lt 1$ for $i\in \{1,2\}$ and $j\in \{3,4\}$ , then

(5·5) \begin{equation}\prod_{i=1}^2 \prod_{j = 3}^4 \frac{1}{1-z_i z_j} = \mathop{\sum\sum\sum\sum}_{k_{13},k_{14},k_{23},k_{24}\geq 0} z_1^{k_{13}+k_{14}} z_2^{k_{23}+k_{24}} z_3^{k_{13}+k_{23}} z_4^{k_{14}+k_{24}}.\end{equation}

We can thus write the product of (5·5) times $z_1z_2z_3z_4$ as

(5·6) \begin{equation}\mathop{\sum\sum\sum\sum}_{k_{13},k_{14},k_{23},k_{24}\geq 0} z_1^{(k_{13}+1)+k_{14}} z_2^{k_{23}+(k_{24}+1)} z_3^{(k_{13}+1)+k_{23}} z_4^{k_{14}+(k_{24}+1)}.\end{equation}

Subtracting this from (5·5) amounts to multiplying (5·5) by $(1-z_1 z_2 z_3 z_4)$ . Now, notice that the monomial in (5·6) associated to the index $(k_{13}-1,k_{14},k_{23},k_{24}-1)$ can be paired with the monomial associated to the index $(k_{13},k_{14},k_{23},k_{24})$ in (5·5) to cancel each other out. This way of pairing terms is not unique, but with this choice the only remaining terms come from monomials in (5·5) where either $k_{13} = 0$ or $k_{24} = 0$ .

Now, it suffices to show that given a tuple $(m_1,m_2,m_3,m_4)$ such that

\begin{align*}m_1 + m_2 = m_3 + m_4,\end{align*}

there is a unique tuple $(k_{13},k_{14},k_{23},k_{24})$ with either $k_{13} = 0$ or $k_{24} = 0$ such that

\begin{align*}(m_1,m_2,m_3,m_4) = (k_{13}+k_{14},k_{23}+k_{24},k_{13}+k_{23},k_{14}+k_{24}).\end{align*}

To see this, assume without loss of generality that $m_1$ is maximal among the $m_i$ . Then $m_1-m_4=k_{13}-k_{24}\geq 0$ so we must have $k_{24}=0$ . It then follows that $k_{23}=m_2$ , $k_{14}=m_4$ , and $k_{13}=m_3-m_2$ .

We then observe

\begin{align*} B(z_1, \dots, z_4) &= \prod_{P} \Big( 1-(z_1z_2z_3z_4)^{\deg P} \Big)\\ &= \zeta_q(z_1z_2z_3z_4)^{-1}\\ &= 1-qz_1z_2z_3z_4.\end{align*}

Substituting our formula for $B(z_1, \dots, z_4)$ , the precise moment for $k=2$ is therefore

\begin{align*}\frac{1}{(2\pi i)^{4}}\int_{|z_{4}|=\varepsilon_{4}}\cdots \int_{|z_{1}|=\varepsilon_1} \frac{(1-qz_1z_2z_3z_4) \prod_{i=1}^2 \prod_{j=3}^{4} \zeta_q(z_iz_j)}{z_1^{N+1}\cdots z_{4}^{N+1}}dz_1\cdots dz_{4}, \end{align*}

which we may split using linearity as

\begin{align*}\frac{1}{(2\pi i)^{4}}\int_{|z_{4}|=\varepsilon_{4}}\cdots \int_{|z_{1}|=\varepsilon_1} &\frac{\prod_{i=1}^2 \prod_{j=3}^{4} \zeta_q(z_iz_j)}{z_1^{N+1}\cdots z_{4}^{N+1}}dz_1\cdots dz_{4} \\-&q \frac{1}{(2\pi i)^{4}}\int_{|z_{4}|=\varepsilon_{4}}\cdots \int_{|z_{1}|=\varepsilon_1} \frac{\prod_{i=1}^2 \prod_{j=3}^{4} \zeta_q(z_iz_j)}{z_1^{N}\cdots z_{4}^{N}}dz_1\cdots dz_{4}.\end{align*}

The following proposition will allow us to conclude.

Proposition 5·2. Fix $0 \lt \varepsilon \lt q^{-1}$ . For each integer $k\geq 1$ ,

(5·7) \begin{equation}{} \frac{1}{(2\pi i)^{2k}}\int_{|z_{2k}|=\varepsilon}\cdots \int_{|z_{1}|=\varepsilon} \frac{\prod_{i=1}^k\prod_{j=k+1}^{2k} \zeta_q(z_iz_j)}{z_1^{N+1}\cdots z_{2k}^{N+1}}dz_1\cdots dz_{2k} = q^{Nk}\mathcal{S}_k(N). \end{equation}

Indeed, noting that $\mathcal{S}_2(N) = N+1$ , it then follows that the fourth moment equals

\begin{align*}(N+1)q^{2N} - q\cdot Nq^{2(N-1)} = Nq^{2N}\Big(1-\frac{1}{q}\Big) + q^{2N},\end{align*}

as desired.

Proof of Proposition 5·2. Fix $j\in \{k+1,\dots, 2k\}$ . Note that we can compute the integral over $z_j$ by computing the coefficient of $z_j^N$ in the product

(5·8) \begin{equation}\prod_{i=1}^{k} \zeta_q(z_iz_j).\end{equation}

Recall that $\zeta_q(z) = \sum_{n\geq 0}q^n\cdot z^n$ . The coefficient of $z_j^N$ in (5·8) is then just

\begin{align*}q^N \mathop{\sum\cdots\sum}_{m_{1}+\cdots +m_{k} = N} z_{1}^{m_{1}}\cdots z_{k}^{m_{k}}.\end{align*}

Observe that this product is independent of our choice of j, and since the j’s are independent of each other in the product, we can integrate out $z_{k+1}, \dots, z_{2k}$ . In this way we can transform the integral expression in (5·7) into

\begin{align*}\frac{1}{(2\pi i)^k}\int_{|z_{k}|=\varepsilon}\dots \int_{|z_{1}|=\varepsilon} \frac{q^{Nk}\Big(\displaystyle\sum_{m_{1}+\cdots +m_{k} = N} z_{1}^{m_{1}}\cdots z_{k}^{m_{k}}\Big)^k}{z_{1}^{N+1} \cdots z_{k}^{N+1}}dz_{1}\cdots dz_{k}.\end{align*}

At this point we can directly compute the coefficient of $z_{1}^N\cdots z_{k}^N$ in the numerator.

Reindexing the sums, we expand it as

\begin{align*}q^{Nk} \Big(\mathop{\sum\cdots\sum}_{m_{1, 1}+\cdots +m_{k, 1} = N} z_{1}^{m_{1, 1}}\cdots z_{k}^{m_{k, 1}}\Big) \cdots\Big(\mathop{\sum\cdots\sum}_{m_{1, k}+\cdots +m_{k, k} = N} z_{1}^{m_{1, k}}\cdots z_{k}^{m_{k, k}}\Big).\end{align*}

This product is precisely

\begin{align*}\mathop{\sum\cdots\sum}_{\substack{m_{1, 1}+\cdots +m_{k, 1} = N\\ \vdots\\ m_{1, k}+\cdots +m_{k, k} = N}} z_{1}^{\sum_{j=1}^k m_{1, j}}\cdots z_{k}^{\sum_{j=1}^k m_{k, j}}.\end{align*}

We observe that the coefficient of $z_1^N\cdots z_k^N$ is given by the number of sets of $k^2$ integers $m_{i,j}$ such that in the square

\begin{align*}\begin{bmatrix}m_{1,1} &\cdots &m_{k,1}\\ \vdots &\ddots &\vdots\\ m_{1,k} &\cdots &m_{k,k}\end{bmatrix},\end{align*}

each row sums to N (enforcing the condition over the summation) and each column sums to N (ensuring that all exponents are raised to the Nth power). Thus the integral is precisely equal to $q^{Nk} \mathcal{S}_k(N)$ .

6. Analytic computation of Steinhaus moments: proof of theorem 1·1

The remainder of the paper will be dedicated to the proof of Theorem 1·1. We provide a brief outline. This section establishes the theorem assuming some key propositions whose proofs will be postponed to subsequent sections; the strategy closely parallels Section 5. First, we form the formal power series $A(z_1,\dots,z_{2k})$ for counting solutions to (2·1). We express this series as a product of zeta functions and another power series, denoted $B(z_1,\dots,z_{2k})$ , and then form the contour integral representation of (2·1). Second, using analytic properties of B (Proposition 6·1 is proved in Section 7), we express (2·1) as an iterated integral over an expanded polydisk after a suitable change of variables (Proposition 6·2 is proved in Section 8). Third, we extract the main contribution of this integral in terms of the arithmetic constant $b_k(q)$ and the number of magic squares $\mathcal{S}_k(N)$ (Proposition 6·3 is proved in Section 9). Finally, we compute the size of the residual error term via a delicate analysis of an oscillatory integral (Proposition 6·4 is proved in Section 10), thus establishing the main theorem.

Proof of Theorem 1·1. One can check by hand that the explicit formula for $k=1$ matches the asymptotics given in Theorem 1·1. We may thus assume $k \geq 2$ . Furthermore, note that if N eventually remains bounded by 20 as $q^N \to \infty$ , then Theorem 1·1 is already implied by Theorem 1·2. We may therefore also assume that N is at least 20. Both of these assumptions will be convenient for technical estimates later in the proof.

We proceed as in Section 5. Namely, we want to recover the coefficient of $z_1^N\cdots z_{2k}^N$ in the power series

(6·1) \begin{equation} A(z_1,\dots, z_{2k})\,{:\!=}\,\mathop{\sum\cdots\sum}_{F_1, \dots, F_{2k}\in \mathcal{M}} \unicode{x1D7D9}_{\{F_1\cdots F_k = F_{k+1}\cdots F_{2k}\}}z_1^{\deg F_1}\cdots z_{2k}^{\deg F_{2k}}\end{equation}

in 2k variables, which is precisely the 2kth moment by (2·1). Note that this function is dominated by

\begin{align*}\mathop{\sum\cdots\sum}_{F_1,\dots,F_{2k} \in \mathcal{M}} \left|z_1^{\deg F_1} \cdots z_{2k}^{\deg F_{2k}}\right| = \prod_{i = 1}^{2k} \zeta_q(|z_i|),\end{align*}

which converges uniformly on compact subsets of $\big\{ (z_1,\dots,z_{2k}) \in \mathbb{C}^{2k} \;:\; |z_i| \lt {1}/{q} \text{for } 1 \leq i \leq 2k\big\}$ .

Since the indicator function is multiplicativeFootnote ¹ , we have the product expansion

\begin{equation*} A(z_1,\dots, z_{2k}) = \prod_{P} \Bigg(\mathop{\sum\cdots\sum}_{\substack{m_1+\cdots+m_k\\ = m_{k+1}+\cdots+m_{2k}}} (z_1^{m_1}\cdots z_{2k}^{m_{2k}})^{\deg P}\Bigg).\end{equation*}

Proposition 6·1. Let

\begin{align*}B(z_1,\dots,z_{2k}) \,{:\!=}\, \prod_{P} \Bigg(\mathop{\sum\cdots\sum}_{\substack{m_1+\cdots+m_k\\ = m_{k+1}+\cdots+m_{2k}}} (z_1^{m_1}\cdots z_{2k}^{m_{2k}})^{\deg P}\Bigg) \prod_{i = 1}^k \prod_{j = k+1}^{2k} (1-z_i^{\deg P}z_j^{\deg P}).\end{align*}

The infinite product B(z) converges absolutely and uniformly on

\begin{align*}\mathcal{R}\,{:\!=}\, \Big\{ (z_1,\dots,z_{2k}) \in \mathbb{C}^{2k} : |z_i| \lt \frac{1.1}{\sqrt{q}} \text{ for } 1 \leq i \leq 2k \Big\},\end{align*}

and hence B(z) is holomorphic on $\mathcal{R}$ . Furthermore, if $z,w\in \mathcal{R}$ , then

\begin{align*}|B(z)| \ll_k 1 \quad \text{ and } \quad |B(z)-B(w)|\ll_{k} \frac{1}{\sqrt{q}} \|z-w\|_{\infty}.\end{align*}

We may thus factor $A(z_1,\dots, z_{2k})$ as

\begin{align*}A(z_1, \dots, z_{2k}) = B(z_1, \dots, z_{2k}) \prod_{i=1}^k \prod_{j=k+1}^{2k} \zeta_q(z_iz_j),\end{align*}

where this equation holds when both functions are defined.

By the multivariate version of Cauchy’s integral formula, we have

\begin{align*} \mathbb{E} \Big | \sum_{F \in \mathcal{M}_N} f(F) \Big|^{2k} = \frac{1}{(2\pi i)^{2k}}\int_{|z_{2k}|=\varepsilon_{2k}}\cdots \int_{|z_{1}|=\varepsilon_1} \frac{A(z_1, \dots, z_{2k})}{z_1^{N+1}\cdots z_{2k}^{N+1}} \, dz_1\cdots dz_{2k},\end{align*}

where $0 \lt \varepsilon_i \lt q^{-1}$ , so that A converges on the closed polydisk with radii $\varepsilon_i$ .

Proposition 6·2. Let $c_i \in (1/4,3/4)$ for $i \in \{2,\dots,2k\}$ . We have the following equality:

(6·2) \begin{align} \mathbb{E} \Big | \sum_{F \in \mathcal{M}_N} f(F) \Big|^{2k} &= \frac{N^{(k-1)^2} }{(2\pi)^{2k-1}}q^{kN}\int_{-\pi N}^{\pi N}\cdots \int_{-\pi N}^{\pi N} \frac{e^{-(c_2+i\theta_2 + \cdots + c_{2k}+i\theta_{2k})}}{\prod_{i=2}^k\prod_{j=k+1}^{2k} N\left(1-e^{(c_i+i\theta_i + c_j+i\theta_j)/N}\right)}\\ &\quad\times \prod_{j=k+1}^{2k}\frac{d\theta_j}{N\left(1-e^{(c_j+i\theta_j)/N}\right)} \nonumber \\&\quad\times B\left(q^{-1/2}, q^{-1/2}e^{(c_2 + i \theta_2)/N}, \dots, q^{-1/2}e^{(c_{2k}+i\theta_{2k})/N}\right) \nonumber \\&\quad\times \ d\theta_2\cdots d\theta_k. \nonumber\end{align}

Now, we let $b_k(q) \,{:\!=}\, B(q^{-1/2},\dots,q^{-1/2})$ so that $0 \lt b_k(q) \ll_k 1$ by Proposition 6·1. Rewrite

\begin{align*}\mathbb{E} \Big | \sum_{F \in \mathcal{M}_N} f(F) \Big|^{2k}=b_k(q) q^{kN}(I_k+J_k),\end{align*}

where

\begin{multline}I_{k}= \frac{N^{(k-1)^2}}{(2\pi)^{2k-1}}\int_{-\pi N}^{\pi N} \cdots \int_{-\pi N}^{\pi N} \frac{e^{-(c_2+i\theta_2 + \cdots + c_{2k}+i\theta_{2k})}}{\prod_{i=2}^k\prod_{j=k+1}^{2k} N \left(1-e^{(c_i+i\theta_i + c_j+i\theta_j)/N}\right)} \\\times \prod_{j=k+1}^{2k}\frac{d\theta_j}{N \left(1-e^{(c_j+i\theta_j)/N}\right)}\, d\theta_2\cdots d\theta_{k} \nonumber\end{multline}

and

\begin{align*}\displaystyle J_k= \frac{1}{b_k(q) q^{kN}} \mathbb{E} \Big | \sum_{F \in \mathcal{M}_N} f(F) \Big|^{2k}-I_{k}.\end{align*}

We claim that $b_k(q) q^{kN} I_k$ is our main term.

It remains to show that the error term $J_k$ is $o(\mathcal{S}_k(N))$ as $qN\to\infty$ . Since ${\mathcal{S}_k(N)/ N^{(k-1)^2} \ll_k 1}$ (see the remarks after Theorem 1·1), it suffices to estimate the quantity

(6·3) \begin{align} \widetilde{J_k} \,{:\!=}\, \frac{1}{(2\pi)^{2k-1}}\int_{-\pi N}^{\pi N}&\cdots \int_{-\pi N}^{\pi N} \frac{e^{-(c_2+i\theta_2 + \cdots + c_{2k}+i\theta_{2k})}}{\prod_{i=2}^k\prod_{j=k+1}^{2k} N\left(1-e^{(c_i+i\theta_i + c_j+i\theta_j)/N}\right)}\\\times &\prod_{j=k+1}^{2k}\frac{d\theta_j}{N\left(1-e^{(c_j+i\theta_j)/N}\right)} \nonumber \\\times &\left( B(q^{-1/2}, q^{-1/2}e^{(c_2 + i \theta_2)/N}, \dots, q^{-1/2}e^{(c_{2k}+i\theta_{2k})/N}) - b_k(q) \right) \nonumber \\\times &\ d\theta_2\cdots d\theta_{k} \nonumber,\end{align}

and show that $\widetilde{J_k}$ converges to 0 as $qN\to \infty$ .

Proposition 6·4. Fix $k \in \mathbb{N}$ with $k \geq 2$ . For $2 \leq X \leq N/10$ and $10X \leq Y \leq N$ (recall the assumption $N \geq 20$ ),

\begin{align*}|\widetilde{J_k}| &\ll_k \frac{1}{q} \frac{(\!\log N)^{2k-1}}{X^{1-1/k}} + \frac{1}{q} (\!\log N)^{2k-2} \frac{X}{N} + \frac{1}{q} \Big( \frac{X}{Y} \Big)^{k-1} + \frac{1}{q} \frac{Y}{N} \left( \log N \right)^k.\end{align*}

Choosing

\begin{align*}X = (N \log N)^{\frac{k}{2k-1}}\quad \text{ and } \quad Y = \frac{N^{1-1/k}}{X} = \frac{N^{\frac{k^2+k-1}{(2k-1)k}}}{(\!\log N)^{\frac{k}{2k-1}}},\end{align*}

we obtain the final error estimate

\begin{align*}\frac{1}{q} \Big( \frac{(\!\log N)^{4k^2-5k+2}}{N^{k-1}}\Big)^{\frac{1}{2k-1}} + \frac{1}{q} \Big( \frac{(\!\log N)^{2k^2-2k}}{N^{\frac{(k-1)^2}{k}}}\Big)^{\frac{1}{2k-1}} \ll_k \frac{1}{q}\frac{(\!\log N)^{k}}{N^{\frac{1}{2} - \frac{3k-2}{4k^2-2k}} }.\end{align*}

A short computation shows that

\begin{align*} b_k(q) &= B(q^{-1/2}, \dots, q^{-1/2})\\ &= \prod_{P} \sum_{m=0}^\infty \Big(\sum_{m_1+\cdots+m_k=m} \Big(\frac{1}{\sqrt{q}}\Big)^{m\deg P} \Big)^2 \prod_{i=1}^k \prod_{j=k+1}^{2k} \Big( 1-\Big(\frac{1}{q}\Big)^{\deg P}\Big).\end{align*}

Using stars and bars for the inner sum (cf. [ Reference Montgomery and VaughanMV07 , lemma 7·7]), and noting that the second double product no longer depends on i,j, this reduces to

\begin{align*}\prod_{P}\sum_{m=0}^\infty \left(\binom{m+k-1}{k-1}^2 \Big(\frac{1}{\sqrt{q}}\Big)^{2m\deg P} \right) \left( 1-\left(\frac{1}{q}\right)^{\deg P}\right)^{k^2}.\end{align*}

Thus,

\begin{align*}b_k(q) = \prod_{P \in \mathcal{M}} \Big( 1-\Big(\frac{1}{q}\Big)^{\deg P}\Big)^{k^2}\sum_{m=0}^\infty \binom{m+k-1}{k-1}^2 \Big(\frac{1}{q}\Big)^{m\deg P} .\end{align*}

This completes the proof of Theorem 1·1 assuming Propositions 6·1 to 6·4.

The last four sections will establish the remaining propositions.

7. Proof of Proposition 6·1: convergence of B series

Recall sums over $m_1,\dots,m_{2k}$ will always be over non-negative integers and

\begin{align*}B(z_1,\dots,z_{2k}) \,{:\!=}\, \prod_{P} \Big(\mathop{\sum\cdots\sum}_{\substack{m_1+\cdots+m_k\\ = m_{k+1}+\cdots+m_{2k}}} (z_1^{m_1}\cdots z_{2k}^{m_{2k}})^{\deg P}\Big) \prod_{i = 1}^k \prod_{j = k+1}^{2k} (1-z_i^{\deg P}z_j^{\deg P}).\end{align*}

Our proof proceeds by first looking at the local factors (i.e. for fixed P). Write $\ell \,{:\!=}\, \deg P$ and let

\begin{align*}F_{\ell}(z_1,\dots,z_{2k}) \,{:\!=}\, \mathop{\sum\cdots\sum}_{\substack{m_1+\cdots+m_{k} \\ =m_{k+1}+\cdots+m_{2k}}}z_1^{m_1 \ell}\cdots z_{2k}^{m_{2k}\ell}\end{align*}

and

\begin{align*}G_{\ell}(z_1,\dots,z_{2k}) \,{:\!=}\,\prod_{i=1}^k\prod_{j=k+1}^{2k}(1-z_i^{\ell} z_j^{\ell}).\end{align*}

Expanding the sums and products, we get

\begin{align*}F_{\ell}(z_1,\dots,z_{2k})=1+\sum_{i=1}^k\sum_{j=k+1}^{2k}z_i^{\ell} z_j^{\ell} + \sum_{M=2}^\infty \left(\mathop{\sum\cdots\sum}_{\substack{m_1+\cdots+m_k = M \\ m_{k+1}+\cdots+m_{2k} =M}} (z_1^{m_1}\cdots z_{2k}^{m_{2k}})^{\ell}\right)\end{align*}

and

\begin{align*}G_{\ell}(z_1,\dots,z_{2k})=1-\sum_{i=1}^k\sum_{j=k+1}^{2k}z_i^{\ell} z_j^{\ell}+ \sum_{L=2}^{k^2}\left(\mathop{\sum\cdots\sum}_{\substack{n_1+\cdots+n_k =L\\n_{k+1}+\cdots+n_{2k} =L \\ 0 \leq n_1,\dots,n_{2k}\leq k}} c(n_1,\dots, n_{2k})(z_1^{n_1}\cdots z_{2k}^{n_{2k}})^{\ell}\right)\end{align*}

for some constants $c(n_1,\dots,n_{2k})$ depending on $n_1,\dots,n_{2k}$ . Consequently,

(7·1) \begin{multline} (F_{\ell}\cdot G_{\ell})(z_1,\dots,z_{2k}) =1+\sum_{M=1}^\infty \left(\mathop{\sum\cdots\sum}_{\substack{m_1+\cdots+m_k=M \\ m_{k+1}+\cdots+m_{2k} =M}} (z_1^{m_1}\cdots z_{2k}^{m_{2k}})^{\ell}\right)\\ -\sum_{M=0}^\infty\left(\mathop{\sum\cdots\sum}_{\substack{m_1+\cdots+m_k=m_{k+1}+\cdots+m_{2k}=M\\n_1+\cdots+n_k =n_{k+1}+\cdots+n_{2k} =1}}(z_1^{m_1+n_1}\cdots z_{2k}^{m_{2k}+n_{2k}})^{\ell}\right)\\ +\sum_{M=0}^\infty\sum_{L=2}^{k^2}\left(\mathop{\sum\cdots\sum}_{\substack{0\leq n_1,\dots,n_{2k}\leq k\\m_1+\cdots+m_k=m_{k+1}+\cdots+m_{2k}=M\\n_1+\cdots+n_k=n_{k+1}+\cdots+n_{2k}=L}}c(n_1,\dots,n_{2k})(z_1^{m_1+n_1}\cdots z_{2k}^{m_{2k}+n_{2k}})^{\ell}\right).\end{multline}

Collecting terms, we see that the $M=1$ term in the first series cancels with the $M=0$ term in the second series and that when $m_1'+\cdots+m_k'=m_{k+1}'+\cdots+m_{2k}'\geq 2$ , the coefficient of $(z_1^{m_1'}\cdots z_{2k}^{m_{2k}'})^\ell$ is given by

\[c'(m_1',\cdots,m_{2k}')\,{:\!=}\,1-\mathop{\sum\cdots\sum}_{\substack{0 \leq n_1,\dots,n_{2k}\leq k\\n_1+\cdots+n_k =n_{k+1}+\cdots+n_{2k} =1\\m_i'-n_i\geq 0}}1+\mathop{\sum\cdots\sum}_{\substack{0\leq n_1, \dots, n_{2k}\leq k\\n_1+\cdots+n_k =n_{k+1}+\cdots+n_{2k} \geq 2\\m_i'-n_i\geq0}}c(n_1,\dots,n_{2k}).\]

We observe that

\begin{align*}c'(m_1',\dots,m_{2k}')\leq 1+k^2+\mathop{\sum\cdots\sum}_{0\leq n_1, \dots, n_{2k}\leq k}|c(n_1,\dots,n_{2k})|,\end{align*}

so these constants are uniformly bounded in k. Thus, (7·1) can be rewritten as $1+C_P(z)$ , where

(7·2) \begin{equation} C_P(z)= \sum_{M=2}^\infty \mathop{\sum\cdots\sum}_{\substack{m_1+\cdots+m_k=M \\ m_{k+1}+\cdots+m_{2k} =M}} c'(m_1,\dots,m_{2k}) z_1^{m_1 \ell}\cdots z_{2k}^{m_{2k} \ell}.\end{equation}

For simplicity, fix $D = 1.21 e^{0.1} \gt 1$ (this choice is somewhat arbitrary). We now give bounds on the local factor.

Lemma 7·1. For $z, w\in\mathcal{R} = \{ (z_1,\dots,z_{2k}) \in \mathbb{C}^{2k} : |z_i| \lt 1.1 q^{-1/2} \text{ for } 1 \leq i \leq 2k \}$ , we have

\begin{align*}|C_P(z)| \ll_k {\Big( \frac{D}{q} \Big)}^{2 \ell}\end{align*}

and

\begin{align*}|C_P(z)- C_P(w)| \ll_k \ell ||z-w||_{\infty} \sqrt{q} {\Big( \frac{D}{q} \Big)}^{2\ell}.\end{align*}

Proof. For $z\in\mathcal{R}$ , we have

(7·3) \begin{align} |C_P(z)|&\ll_k \sum_{M=2}^\infty \Big(\mathop{\sum\cdots\sum}_{\substack{m_1+\cdots+m_k =M\\ m_{k+1}+\cdots+m_{2k} =M}}1\Big)\Big(\frac{1.1}{\sqrt{q}}\Big)^{2M\ell} \nonumber \\ &= \sum_{M=2}^\infty \binom{M+k-1}{k-1}^2 \Big(\frac{(1.1)^2}{q}\Big)^{M\ell}\nonumber \\ &\ll_k \sum_{M=2}^{\infty} M^{2(k-1)} \Big(\frac{1.21}{q}\Big)^{M\ell}, \end{align}

where the last inequality follows since $\binom{M+k-1}{k-1}^2$ is a polynomial in M of degree at most $2(k-1)$ . Observe that $\big({1.21}/{q}\big)^{M\ell} = e^{-0.1M\ell}\big({D}/{q}\big)^{M\ell}$ , and further for $M \geq 2$ ,

\begin{align*}M^{2(k-1)}e^{-0.1M\ell} \leq M^{2(k-1)}e^{-0.1M} \ll_k 1.\end{align*}

We conclude that

(7·4) \begin{align}{} \sum_{M=2}^{\infty} M^{2(k-1)} \Big(\frac{1.21}{q}\Big)^{M\ell} &\ll_k \sum_{M=2}^{\infty} \Big(\frac{D}{q}\Big)^{M\ell} \ll_k \Big( \frac{D}{q}\Big)^{2\ell},\end{align}

where the last estimate comes from bounding the convergent geometric series by its first term. This proves the first estimate in the lemma.

Now, we prove the upper bound for the difference $|C_P(z)-C_P(w)|$ . From the definition in (7·2) and using that $|z_i|,|w_i|\lt 1.1 q^{-1/2}$ , we find that

(7·5) \begin{align} |C_P(z)-C_P(w)|&\ll_k\sum_{M=2}^\infty \mathop{\sum\cdots\sum}_{\substack{m_1+\cdots+m_k=M \\ m_{k+1}+\cdots+m_{2k} =M}} \big|(z_1^{m_1 }\cdots z_{2k}^{m_{2k}})^{\ell} - (w_1^{m_1 }\cdots w_{2k}^{m_{2k}})^{\ell} \big|\nonumber \\&\leq \ell \sum_{M=2}^\infty \Big(\frac{1.1}{\sqrt{q}} \Big)^{2M(\ell-1)} \mathop{\sum\cdots\sum}_{\substack{m_1+\cdots+m_k=M \\ m_{k+1}+\cdots+m_{2k}=M}} \big|z_1^{m_1 }\cdots z_{2k}^{m_{2k}} - w_1^{m_1}\cdots w_{2k}^{m_{2k}}\big|, \end{align}

where we implicitly used the identity $A^n-B^n=(A-B)\sum_{k=0}^{n-1}A^{n-1-k}B^k$ .

Now, we claim

(7·6) \begin{equation} |z_1^{m_1 }\cdots z_{2k}^{m_{2k}} - w_1^{m_1}\cdots w_{2k}^{m_{2k}}| \leq 2M \Big(\frac{1.1}{\sqrt{q}}\Big)^{2M-1}\|z-w\|_\infty,\end{equation}

where $m_1+\cdots +m_k=m_{k+1}+\cdots+m_{2k}=M$ . Observe that when $k=1$ , we can write

\begin{align*}|z_1^{m_1}z_2^{m_2}-w_1^{m_1}w_2^{m_2}| &= |z_1^{m_1}z_2^{m_2}-z_1^{m_1}w_2^{m_2}+z_1^{m_1}w_2^{m_2}-w_1^{m_1}w_2^{m_2}| \\&\leq |z_1|^{m_1}|z_2^{m_2}-w_2^{m_2}| + |w_2^{m_2}||z_1^{m_1}-w_1^{m_1}| \\& \leq \Big(\frac{1.1}{\sqrt{q}}\Big)^{m_1}|z_2-w_2|m_2\Big(\frac{1.1}{\sqrt{q}}\Big)^{m_2-1} + \Big(\frac{1.1}{\sqrt{q}}\Big)^{m_2}|z_1-w_1|m_1\Big(\frac{1.1}{\sqrt{q}}\Big)^{m_1-1} \\&\leq \|z-w\|_{\infty}(m_1+m_2) \Big(\frac{1.1}{\sqrt{q}}\Big)^{m_1+m_2-1}.\end{align*}

For any other $k\geq 2$ , the claim follows by induction after rewriting

\begin{align*} |z_1^{m_1 }\cdots z_{2k}^{m_{2k}} - w_1^{m_1}\cdots w_{2k}^{m_{2k}}| & = |z_1^{m_1 }z_2^{m_2}(z_3^{m_3}\cdots z_{2k}^{m_{2k}} - w_3^{m_3}\cdots w_{2k}^{m_{2k}}) \\ & \qquad +w_3^{m_3}\cdots w_{2k}^{m_{2k}}(z_1^{m_1}z_2^{m_2}-w_1^{m_1}w_2^{m_2})| \\ & \leq |z_1|^{m_2} |z_2|^{m_2} \cdot (m_3+\cdots+m_{2k}) \Big(\frac{1.1}{\sqrt{q}}\Big)^{m_3+\cdots+m_{2k}-1} ||z-w||_{\infty} \, \\ & \qquad + |w_3|^{m_3} \cdots |w_{2k}|^{m_{2k}} \cdot (m_1+m_2) \Big(\frac{1.1}{\sqrt{q}}\Big)^{m_1+m_2-1} ||z-w||_{\infty} \\ & \leq (m_1+\cdots+m_{2k}) \Big(\frac{1.1}{\sqrt{q}}\Big)^{m_1+\cdots+m_{2k}-1} ||z-w||_{\infty},\end{align*}

as required. Thus, the estimate (7·6) implies that (7·5) is

\begin{align*}\ll \ell\, \|z-w\|_\infty \sqrt{q}\sum_{M=2}^\infty M \binom{M+k-1}{k-1}^2 \Big(\frac{1.21}{q}\Big)^{M\ell}.\end{align*}

At this point, we use the same manipulation as we did in going from (7·3) to (7·4) to recover the estimate

\begin{align*}|C_P(z)-C_P(w)| \ll_k \ell ||z-w||_{\infty} \sqrt{q}\left(\frac{D}{q}\right)^{2\ell}. \unicode{x8718}\end{align*}

Recall that by definition, $B(z) = \prod_{P } (1+C_P(z))$ . By Lemma 7·1 and the observation ${D \lt 2 \leq q}$ , if $\ell_0 \geq 1$ is sufficiently large depending only on k, then

(7·7) \begin{equation} |C_P(z)| \leq \frac{1}{2} \quad \text{whenever } \deg P \geq \ell_0.\end{equation}

For $\ell_0 \geq 1$ , define

\begin{align*}J(z) \,{:\!=}\, \prod_{\substack{P \\ \deg P \lt \ell_0}}\big(1+C_P(z)\big), \qquad K(z) \,{:\!=}\, \prod_{\substack{P \\ \deg P \geq \ell_0}}\big(1+C_P(z)\big)\end{align*}

so that $B(z) = J(z)K(z)$ .

Lemma 7·2. Assume $\ell_0 = \ell_0(k)$ depends only on k and satisfies (7·7). For $z \in \mathcal{R}$ ,

\begin{align*} J(z) \ll_{k} 1, \qquad K(z) \ll_{k} 1. \end{align*}

In particular, $|B(z)| \ll_k 1$ .

Proof. Recall the prime polynomial theorem implies that

(7·8) \begin{equation} \pi_q(\ell) \,{:\!=}\, \#\{ P \in \mathcal{M}_{\ell}\, :\, P\ \text{ irreducible} \} \ll \frac{q^{\ell}}{\ell},\end{equation}

so we have

\begin{align*} |J(z)| &= \Big|\prod_{\substack{P \\ \deg P \lt \ell_0}} (1 + C_P(z)) \Big| \\ &\ll_k \prod_{\substack{P \\ \deg P \lt \ell_0}} \Big(1 + \frac{D^{2 \deg P}}{q^{2 \deg P}}\Big) = \prod_{1 \leq \ell \lt \ell_0} \Big(1 + \frac{D^{2 \ell}}{q^{2l}}\Big)^{\pi_q(\ell)} \\ &\leq \prod_{1 \leq \ell \lt \ell_0} \exp\Big( O(q^{-\ell}D^{2\ell}/\ell) \Big) \ll_k 1. \end{align*}

Here, we have used that $(1+1/x)^x \lt e$ for all $x \gt 0$ and that $D=1.21 e^{0.1}\lt \sqrt{2}$ .

By our assumption on $\ell_0$ , we have $|C_P(z)| \leq 1/2$ , so we may use the expansion ${|\log(1+w)| = O(|w|)}$ for $|w| \leq 1/2$ . Thus, the prime polynomial theorem (7·8), Lemma 7·1, and the fact that $D\lt \sqrt{2}$ together imply that

\begin{align*}|\log(K(z))| &\ll \sum_{\substack{P \\ \deg P \geq \ell_0}} |C_P(z)| \ll_k \sum_{\substack{P \\ \deg P \geq \ell_0}} \frac{D^{2 \ell}}{q^{2 \ell}} \ll_k \sum_{\ell \geq \ell_0} \frac{D^{2\ell}}{q^{2\ell}} \frac{q^{\ell}}{\ell} \ll_k q^{-\ell_0} \ll_k 1.\end{align*}

In particular, $|K(z)| \ll_{k} 1$ .

Lemma 7·2 shows in particular that $|B(z)| \ll_k 1$ as claimed in Proposition 6·1 The remainder of Proposition 6·1 rests on the following consequence of the triangle inequality: for $z,w \in \mathcal{R}$ ,

(7·9) \begin{align} |B(z)-B(w)| &= |J(z)K(z) - J(w)K(w)| \nonumber\\ &\leq |J(z)-J(w)| \cdot |K(z)| + | K(z) - K(w) | \cdot | J(w)|.\end{align}

Therefore, showing that the differences $|J(z)-J(w)|$ and $|K(w)-K(z)|$ are bounded in terms of k and explicitly in terms of q will allow us to complete the proof of Proposition 6·1.

We now choose a suitable value for $\ell_0$ by means of the next lemma.

Lemma 7·3. There exists $\ell_0 = \ell_0(k)$ sufficiently large such that (7·7) holds and for $z,w \in \mathcal{R}$ ,

\[ \Big|\log \frac{K(z)}{K(w)} \Big| \leq \dfrac{1}{2} \quad \text{ and } \quad \Big|\log \frac{K(z)}{K(w)} \Big| \ll_k \sqrt{q} ||z-w||_\infty \Big(\frac{D^2}{q}\Big)^{\ell_0}. \]

Proof. By Lemma 7·1, K(z) is an absolutely convergent infinite product inside $\mathcal{R}$ . So, it is never zero in that region and we can take its logarithm. Using the Taylor expansion of $\log(1+z)$ for $|z| \leq 1/2$ , we find that

\begin{align*} \Big|\log \frac{K(z)}{K(w)}\Big| &= \Big|\sum_{\substack{P \\ \deg P \geq \ell_0}} \big(\!\log(1+C_P(z))-\log(1+C_P(w))\big)\Big| \\ &= \Big| \sum_{\substack{P \\ \deg P \geq \ell_0}} \sum_{n = 1}^\infty ({-}1)^{n+1} \frac{C_P(z)^n-C_P(w)^n}{n} \Big| \\ &\leq \sum_{\substack{P \\ \deg P \geq \ell_0}} \sum_{n = 1}^\infty \frac{|C_P(z) - C_p(w)|}{n} \sum_{i = 0}^{n-1} |C_P(z)^i C_P(w)^{n-1-i}| \\ &\leq \sum_{\substack{P \\ \deg P \geq \ell_0}} |C_P(z) - C_p(w)|. \end{align*}

Applying Lemma 7·1, as well as the prime polynomial theorem (7·8), we see that

\begin{align*} \sum_{\substack{P \\ \deg P \geq \ell_0}} \big(C_P(z)-C_P(w)\big) &\ll_k \sum_{\substack{\ell \geq \ell_0}} \ell ||z-w||_{\infty} \sqrt{q} \Big( \frac{D}{q}\Big)^{2\ell}\frac{q^{\ell}}{\ell}\\ &= \sqrt{q} ||z-w||_{\infty}\sum_{\ell \geq \ell_0} \Big(\frac{D^2}{q}\Big)^{\ell}\\ &\ll_k \sqrt{q}||z-w||_\infty \Big(\frac{D^2}{q}\Big)^{\ell_0}. \end{align*}

This is the second estimate that we claimed. Since $||z-w||_\infty \ll 1/\sqrt{q}$ and $D^2/q \lt 1$ for any $q \geq 2$ , the first estimate also holds provided $\ell_0 = \ell_0(k)$ is taken to be sufficiently large.

We proceed to bound the differences in (7·9), starting with $|K(z)- K(w)|$ .

Lemma 7·4. If $z,w \in \mathcal{R}$ and $\ell_0 = \ell_0(k)$ is sufficiently large according to Lemma 7·3, then

\begin{align*} |K(z)- K(w)| \ll_k \sqrt{q} ||z-w||_{\infty} {\Big(\frac{D^2}{q}\Big)}^{\ell_0}. \end{align*}

Proof. We simply observe that

\begin{align*}|K(z)-K(w)| \ll_k \Big|\frac{K(z)}{K(w)}-1\Big| \ll \Big|\log \frac{K(z)}{K(w)}\Big|,\end{align*}

where we used Lemma 7·2 for the first bound, and $e^z = 1 + O(|z|)$ for $|z| \leq 1/2$ and Lemma 7·3 for the second bound.

The last estimate is an upper bound for the difference $|J(z)-J(w)|$ .

Lemma 7·5. For $z,w \in \mathcal{R}$ , we have that

\begin{align*} |J(z) - J(w)| \ll_{k} \ell_0 ||z-w||_\infty \frac{D^2}{\sqrt{q}}. \end{align*}

Proof. Let $\mathcal{N}$ be some subset of the set of all monic irreducible polynomials P of degree less than $\ell_0$ . We will show by induction on $|\mathcal{N}|$ that

\begin{align*}\Big|\prod_{P \in \mathcal{N}} (1+C_P(z)) - \prod_{P \in \mathcal{N}} (1+C_P(w))\Big| \ll_{k} \ell_0 \sqrt{q} \sum_{P \in \mathcal{N}} ||z-w||_\infty \Big(\frac{D}{q}\Big)^{2 \deg P}.\end{align*}

If $|\mathcal{N}| = 1$ , this statement is just Lemma 7·1 and the fact that $\deg P \lt \ell_0$ . In general, we have

\begin{align*} \Big|\prod_{P \in \mathcal{N}}& (1+C_P(z)) - \prod_{P \in \mathcal{N}} (1+C_P(w))\Big| \\ &\leq |C_{P_1}(z) - C_{P_1}(w)| \prod_{\substack{P \in \mathcal{N} \setminus \{P_1\}}} \left| 1 + C_P(z) \right| \\ &\quad + \Big| (1 + C_{P_1}(w)) \Big( \prod_{\substack{P \in \mathcal{N} \setminus \{P_1\}}} (1 + C_P(z)) - \prod_{\substack{P \in \mathcal{N} \setminus \{P_1\}}} (1 + C_P(w))\Big) \Big|\end{align*}

for any $P_1 \in \mathcal{N}$ . Now, we can trivially bound $|1+C_{P_1}(w)| \ll_{k} 1$ , and by essentially the same argument as in Lemma 7·2, $\left|\prod_{\substack{P \in \mathcal{N} \setminus \{P_1\}}} (1 + C_P(z))\right| \ll_{k} 1$ . Applying the induction hypothesis and Lemma 7·1, we find that the above is bounded by

\begin{align*}\ell_0\sqrt{q}\,||z-w||_\infty \Big(\frac{D}{q}\Big)^{2 \deg P_1} + \ell_0\sum_{P \in \mathcal{N} \setminus \{P_1\}} \sqrt{q}\,||z-w||_\infty \Big(\frac{D}{q}\Big)^{2 \deg P},\end{align*}

up to constants depending only on k, as claimed. In particular, we now get

\begin{align*} |J(z) - J(w)| &\ll_{k} \ell_0 \sqrt{q} \sum_{\substack{P \\ \deg P \lt \ell_0}} ||z-w||_\infty {\Big(\frac{D}{q}\Big)}^{2 \deg P} \\ &\ll \ell_0\sqrt{q} \sum_{\ell \lt \ell_0} ||z-w||_\infty {\Big(\frac{D}{q}\Big)}^{2 \ell} \frac{q^{\ell}}{\ell} \ll \ell_0 ||z-w||_\infty \frac{D^2}{\sqrt{q}},\end{align*}

by the prime polynomial theorem (7·8). This proves the lemma.

Inserting Lemmas 7·2, 7·4 and 7·5 in (7·9), we obtain

\begin{align*} |B(z)-B(w)| &\ll_k ||z-w||_{\infty}\Big( \frac{2}{\sqrt{q}}\Big) + ||z-w||_{\infty}\sqrt{q}\Big(\frac{2}{q}\Big)^{\ell_0} \ll_k \frac{1}{\sqrt{q}} ||z-w||_{\infty},\end{align*}

as desired. This completes the proof of Proposition 6·1.

8. Proof of Proposition 6·2: integral representation

Recall that

\begin{align*} \mathbb{E} \Big | \sum_{F \in \mathcal{M}_N} f(F) \Big|^{2k} = \frac{1}{(2\pi i)^{2k}}\int_{|z_{2k}|=\varepsilon_{2k}}\cdots \int_{|z_{1}|=\varepsilon_1} \frac{A(z_1, \dots, z_{2k})}{z_1^{N+1}\cdots z_{2k}^{N+1}} \, dz_1\cdots dz_{2k}.\end{align*}

Recalling that $\zeta_q(z)= {1}/({1-qz})$ , we can further rewrite this as

(8·1) \begin{equation}{} \frac{1}{(2\pi i)^{2k}} \int_{|z_{2k}| = \varepsilon_{2k}} \cdots\int_{|z_1|=\varepsilon_1} \frac{(z_1\cdots z_{2k})^{-(N+1)}}{\prod_{i=1}^k\prod_{j=k+1}^{2k}(1-qz_iz_j)}\cdot B(z_1, \dots, z_{2k}) \, dz_1\cdots dz_{2k}.\end{equation}

We make several changes of variables as a means of centering the integral around $z_1$ . In each case the contour integrals can still be taken along circles of small radii around 0. First, we modify $z_{k+1}$ through $z_{2k}$ by sending

\begin{align*}z_j \mapsto z_1^{-1}q^{-1}z_j \quad \text{ for }\quad k+1\leq j\leq 2k.\end{align*}

This yields the expression

\begin{align*}\frac{q^{kN}}{(2\pi i)^{2k}} \int \cdots\int \frac{(z_1^{1-k} z_2 \cdots z_{2k})^{-(N+1)}}{\prod_{i=2}^k\prod_{j=k+1}^{2k}(1-z_iz_j/z_1)}\cdot \frac{B\left(z_1, \dots, z_1^{-1}q^{-1}z_{2k}\right)}{z_1^k\prod_{j=k+1}^{2k} (1-z_j)} \, dz_1\cdots dz_{2k}.\end{align*}

Here, the second product in the denominator of the integrand corresponds to the contribution from the double product in (8·1) in the case where $z_i = z_1$ . We may also factor out a $q^{kN}$ by considering the contributions from the numerator of the integrand in (8·1) as well as the changes in the increment.

We make the second substitution

\begin{align*}z_i \mapsto z_1z_i \quad \text{ for }\quad 2\leq i\leq k,\end{align*}

and our integral becomes

(8·2) \begin{equation}{} \frac{q^{kN}}{(2\pi i)^{2k}} \int \cdots\int \frac{1}{z_1}\frac{(z_2 \cdots z_{2k})^{-(N+1)}}{\prod_{i=2}^k\prod_{j=k+1}^{2k}(1-z_iz_j)}\cdot \frac{B\left(z_1, z_1z_2,\dots, z_1^{-1}q^{-1}z_{2k}\right)}{\prod_{j=k+1}^{2k} (1-z_j)} \, dz_1\cdots dz_{2k}.\end{equation}

Before proceeding with our analysis of (8·2), we observe that along this contour, where $|z_1|\gt 0$ , the term $B(z_1, z_1z_2, \dots, z_1^{-1}q^{-1}z_{2k})$ is independent of the value of $z_1$ . Indeed,

\begin{align*} B(z_1, z_1z_2, \dots, z_1^{-1}q^{-1}z_{2k}) = \prod_{P} & \mathop{\sum\cdots\sum}_{\substack{m_1+\cdots+m_k=\\ m_{k+1}+\cdots+m_{2k}}} \Big(\frac{z_1^{m_1+\cdots+m_k}}{z_1^{m_{k+1}+\cdots+m_{2k}}}\Big)^{\deg P}(z_2^{m_2}\cdots q^{-m_{2k}}z_{2k}^{m_{2k}})^{\deg P}\\ \times &\prod_{i=2}^k \prod_{j=k+1}^{2k} \Big( 1-(q^{-1}z_iz_j)^{\deg P}\Big)\prod_{j=k+1}^{2k} \Big( 1-(q^{-1}z_j)^{\deg P} \Big).\end{align*}

Since $m_1+\cdots +m_k = m_{k+1} + \cdots +m_{2k}$ , the $z_1$ term vanishes and we are left with

\begin{align*}\prod_{P} \mathop{\sum\cdots\sum}_{\substack{m_1+\cdots+m_k\\ = m_{k+1}+\cdots+m_{2k}}}(z_2^{m_2}\cdots q^{-m_{2k}}z_{2k}^{m_{2k}})^{\deg P}\prod_{i=2}^k \prod_{j=k+1}^{2k} (1-(q^{-1}z_iz_j)^{\deg P}) \prod_{j=k+1}^{2k} (1-(q^{-1}z_j)^{\deg P}),\end{align*}

which is clearly independent of $z_1$ . We further note that in (8·2), the only term that depends on $z_1$ is the $1/z_1$ factor. Thus, we can integrate out $z_1$ multiplying our expression by $2\pi i$ .

In the input of B, we can thus pick any value for $z_1$ , so without loss of generality take $z_1=q^{-1/2}$ . Our integral now becomes

(8·3) \begin{equation}{} \frac{q^{kN}}{(2\pi i)^{2k-1}}\int\cdots \int \frac{(z_2 \cdots z_{2k})^{-(N+1)}}{\prod_{i=2}^k\prod_{j=k+1}^{2k}(1-z_iz_j)}\cdot \frac{B\left(q^{-1/2}, q^{-1/2}z_2\cdots, q^{-1/2}z_{2k}\right)}{\prod_{j=k+1}^{2k} (1-z_j)} \, dz_2\cdots dz_{2k}.\end{equation}

Despite the changes of variables, the ith integral is still taken around some disk of positive radius $\varepsilon_i$ which now only have to satisfy $0 \lt \varepsilon_i \lt 1.1$ due to Proposition 6·1 and our choice $z_1=q^{-1/2}$ . For each integral, we can parametrise its contour in terms of its argument $\theta_i$ by

\begin{align*}\gamma(\theta_i) = e^{c_i/N} e^{i\theta_i}.\end{align*}

Here, the $c_i \in (1/2,3/4)$ are some constants, chosen to be independent of N and q. Note the radius satisfies $e^{c_i/N} \leq e^{3/80} \lt 1.1$ as $N \geq 20$ by assumption. Explicitly substituting the contours, we rewrite (8·3) as

(8·4) \begin{align}{} \frac{q^{kN}}{(2\pi)^{2k-1}} &\int_{-\pi}^\pi \cdots\int_{-\pi}^\pi \frac{e^{-c_2-\cdots-c_{2k}-iN(\theta_2+\cdots + \theta_{2k})}}{\prod_{i=2}^k\prod_{j=k+1}^{2k}\left(1-e^{(c_i+c_j)/N+i\theta_i+i\theta_j}\right)}\cdot \prod_{j=k+1}^{2k} \frac{1}{1-e^{c_j/N+i\theta_j}}\nonumber\\ &\times B\left(q^{-1/2}, q^{-1/2}e^{c_2/N + i\theta_2}, \dots, q^{-1/2}e^{c_{2k}/N+i\theta_{2k}}\right)d\theta_2\cdots d{\theta_{2k}}. \end{align}

Here, we see some cancellations arising from the derivative term and the original numerator of the integrand from (8·3). Note that the factor of $i^{2k-1}$ cancels with the constant term preceding the integral. At this point, we make the change of variables

\begin{align*}&\theta_j\mapsto \theta_j/N \quad \text{ for }\quad j=2,3,\dots,2k.\end{align*}

We will also multiply the numerator and denominator by $N^{(k-1)^2}$ , preceding each term in each product of the denominator in (8·4) by N. This ultimately yields

\begin{align*}\frac{N^{(k-1)^2} }{(2\pi)^{2k-1}}q^{kN}\int_{-\pi N}^{\pi N}\cdots \int_{-\pi N}^{\pi N} &\frac{e^{-(c_2+i\theta_2 + \cdots + c_{2k}+i\theta_{2k})}}{\prod_{i=2}^k\prod_{j=k+1}^{2k} N\left(1-e^{(c_i+i\theta_i + c_j+i\theta_j)/N}\right)}\\\times &\prod_{j=k+1}^{2k}\frac{d\theta_j}{N\left(1-e^{(c_j+i\theta_j)/N}\right)} \nonumber \\\times &B\left(q^{-1/2}, q^{-1/2}e^{(c_2 + i \theta_2)/N}, \dots, q^{-1/2}e^{(c_{2k}+i\theta_{2k})/N}\right) \nonumber \\\times &\ d\theta_2\cdots d\theta_k, \nonumber\end{align*}

completing the proof of Proposition 6·2.

9. Proof of Proposition 6·3: magic squares

This section is nearly identical to [ Reference Conrey and GamburdCG06 , section 3], but we have included the argument for the sake of completeness. Making the change of variables $\theta_j \mapsto -\theta_j$ for $j=2,\dots,2k$ , we turn each integral into a complex line integral along some vertical strip of length $2\pi N$ with real part $-c_j$ . Letting $s_j = -c_j + i \theta_j$ , our integral becomes

\begin{align*}I_k = \frac{N^{1-2k}}{(2\pi i)^{2k-1}}\int_{-c_{2k} - i \pi N }^{ -c_{2k} + i \pi N}\cdots \int_{-c_{2} - i\pi N}^{-c_2 + i \pi N} &\frac{e^{s_2+\cdots+s_{2k}}}{\prod_{i=2}^k\prod_{j=k+1}^{2k} \left(1-e^{-(s_i+s_j)/N}\right)}\\\times &\prod_{j=k+1}^{2k}\frac{ds_j}{\left(1-e^{-s_j/N}\right)} \nonumber \\\times &\ ds_2\cdots ds_{2k}. \nonumber\end{align*}

Let $u \,{:\!=}\, (s_2,\dots,s_k)$ and $v \,{:\!=}\, (s_{k+1}, \dots, s_{2k})$ . Then, using the geometric series formula, the integrand becomes

\begin{align*} &e^{s_2+\cdots+s_{2k}} \prod_{i=2}^k\prod_{j=k+1}^{2k} \frac{1}{\Big(1-e^{-(s_i+s_j)/N}\Big)} \times \prod_{j=k+1}^{2k}\frac{1}{\Big(1-e^{-s_j/N}\Big)} \\ &\quad = e^{s_2+\cdots+s_{2k}} \prod_{i=2}^k\prod_{j=k+1}^{2k} \Big( \sum_{a_{ij} \geq 0} (e^{({-}s_i-s_j)/N})^{a_{ij}} \Big) \prod_{j = k+1}^{2k}\Big( \sum_{b_j \geq 0} (e^{-s_j/N})^{b_j} \Big) \\ &\quad = e^{s_2+\cdots+s_{2k}} \sum_{\alpha \in \mathbb{N}_0^{k-1},\; \beta \in \mathbb{N}_0^{k}} M_{\alpha, \beta} e^{-\alpha \cdot u / N} e^{-\beta \cdot v / N},\end{align*}

where $\cdot$ denotes the usual dot product and, letting $\alpha=(\alpha_2,\ldots,\alpha_k)$ and $\beta=(\beta_{k+1},\ldots,\beta_{2k})$ , the coefficient $M_{\alpha, \beta} \geq 0$ counts the possible choices for $a_{ij}$ and $b_j$ such that

\begin{equation*} \sum_{j = k+1}^{2k} a_{ij} = \alpha_i \quad \text{ for }\quad i=2,\dots,k, \qquad \text{and} \qquad b_j + \sum_{i = 2}^k a_{ij} = \beta_j \quad \text{ for }\quad j=k+1,\dots,2k.\end{equation*}

The sum is convergent on the compact region of integration. Applying Fubini’s theorem, we may thus interchange sum and integral to obtain

\begin{align*}I_k &{=} \frac{N^{1{-}2k}}{(2\pi i)^{2k{-}1}} \sum_{\substack{\alpha \in \mathbb{N}_0^{k{-}1},\\ \beta \in \mathbb{N}_0^{k}}} M_{\alpha, \beta} \int_{{-}c_{2k} {-} i \pi N }^{ {-}c_{2k} + i \pi N}\cdots \int_{{-}c_{2} {-} i\pi N}^{{-}c_2 + i \pi N} e^{{-}\alpha \cdot u / N} e^{{-}\beta \cdot v / N} e^{s_2+\cdots+s_{2k}} ds_2 \cdots ds_{2k} \\&{=} \frac{N^{1{-}2k}}{(2\pi i)^{2k{-}1}} \sum_{\substack{\alpha \in \mathbb{N}_0^{k{-}1},\\ \beta \in \mathbb{N}_0^{k}}}\!\! M_{\alpha, \beta} \Big(\int_{{-}c_{2k} {-} i\pi N}^{{-}c_{2k} + i \pi N} e^{{-}(\beta_{2k}{-}N) s_{2k} / N} ds_{2k}\Big)\cdots\Big(\int_{{-}c_{2} {-} i\pi N}^{{-}c_2 + i \pi N} e^{{-}(\alpha_2{-}N) s_2 / N } ds_2\Big) \\&{=} \frac{N^{1{-}2k}}{(2\pi i)^{2k{-}1}}\! \sum_{\substack{\alpha \in \mathbb{N}_0^{k{-}1},\\ \beta \in \mathbb{N}_0^{k}}}\! M_{\alpha, \beta} \prod_{\substack{i{=}2 \\ \alpha_i \neq N}}^k\! \frac{N\left( e^{{-}(\alpha_i{-}N)({-}c_i + i\pi N)/N} {-} e^{{-}(\alpha_i {-} N)({-}c_i {-} i \pi N)/N} \right)}{N {{-}} \alpha_i} \prod_{\substack{i{=}2 \\ \alpha_i {=} N}}^k \!(2\pi i N) \\&\hspace{3.5cm} \times \prod_{\substack{j {=} k+1 \\ \beta_j \neq N}}^{2k}\! \frac{N\left( e^{{-}(\beta_j {-} N)({-}c_j + i \pi N)/N} {-} e^{{-}(\beta_j {-} N)({-}c_j {-} i \pi N) / N} \right)}{N {-} \beta_j} \prod_{\substack{j {=} k{+}1 \\ \beta_j {=} N}}^{2k}\! ( 2 \pi i N).\end{align*}

Now, observe that

\begin{align*}e^{-(\alpha_i-N)({-}c_i + i\pi N)/N} - e^{-(\alpha_i - N)({-}c_i - i \pi N)/N} = e^{(\alpha_i/N - 1)c_i} 2i\sin(\left({-}\alpha_i+N)\pi\right)),\end{align*}

and as $\alpha_i\in \mathbb{Z}$ , this (and the corresponding expression for $\beta_j$ ) evaluates to 0 for all i,j. Thus, the only term in the summation that contributes to $I_k$ is $M_{(N, \dots, N), (N,\dots, N)}$ , which is just the number of possible choices for $a_{ij}$ and $b_j$ such that

\begin{equation*} \sum_{j = k+1}^{2k} a_{ij} = N \quad \text{ for }\quad i=2,\dots,k, \qquad \text{and} \qquad b_j + \sum_{i = 2}^k a_{ij} = N \quad \text{ for }\quad j=k+1,\dots,2k.\end{equation*}

It follows that we also have

\begin{equation*}\sum_{j = k+1}^{2k} b_j = \sum_{j = k+1}^{2k} \Big (b_j + \sum_{i = 2}^k a_{ij} \Big) - \sum_{i = 2}^k \sum_{j = k+1}^{2k} a_{ij} = kN - (k-1)N = N.\end{equation*}

This gives the desired equality $I_k = \mathcal{S}_k(N)$ , establishing Proposition 6·3.

10. Proof of Proposition 6·4: estimating the error term

To facilitate notation, we define

\begin{align*}F(w) \,{:\!=}\, F(w, N) \,{:\!=}\, \frac{1}{N(1-e^{-w/N})}\end{align*}

for $w \in \mathbb{C}$ . For $i = 2,\dots,2k$ , denote

\begin{align*}z_i = c_i + i \theta_i, \quad c_i \in \left( \frac{1}{4}, \frac{3}{4}\right), \ \theta_i \in \mathbb{R}.\end{align*}

For any Lebesgue-measurable region $R \subset [{-}\pi N, \pi N]^{2k-1}$ , we further define

\begin{align*}I(R) \,{:\!=}\, \underset{R}{\int \cdots \int} \prod_{i = 2}^{k} \prod_{j = k+1}^{2k} |F(z_i + z_j)| \prod_{j = k+1}^{2k} |F(z_j)| \ d\theta_2 \cdots d\theta_{2k}\end{align*}

and

(10·1) \begin{equation}{} \Delta(R) \,{:\!=}\, \sup_{(\theta_2, ..., \theta_{2k})\in R}\left|B\left(q^{-1/2}, q^{-1/2}e^{(c_2+i\theta_2)/N}, \dots, q^{-1/2}e^{(c_{2k}+i\theta_{2k})/N}\right) - b_k(q)\right|\!.\end{equation}

To give a brief outline, the proof of Proposition 6·4 will involve splitting the region of integration, $[{-}\pi N, \pi N]^{2k-1}$ , into three regions:

\[R_1 \,{:\!=}\, [{-}\pi N, \pi N]^{2k-1}\setminus \mathfrak{S}(X),\qquad R_2 \,{:\!=}\, \mathfrak{S}(X)\setminus \mathfrak{S}(X, Y),\qquad R_3 \,{:\!=}\, \mathfrak{S}(X, Y),\]

with $\mathfrak{S}(X)$ and $\mathfrak{S}(X,Y)\subseteq\mathfrak{S}(X)$ specified later by (10·2) and (10·4) and dependent on the parameters X and Y given in the proposition. We observe that the integral $\widetilde{J_k}$ from (6·3) can be bounded by

\begin{align*}I(R_1)\Delta(R_1) + I(R_2)\Delta(R_2) + I(R_3)\Delta(R_3).\end{align*}

In Lemmas 10·3 and 10·4, we will give estimates for $I(R_1)$ and $I(R_2)$ , respectively. Here we take the trivial estimates for $\Delta(R_1)$ and $ \Delta(R_2)$ coming from Proposition 6·1. In $R_3$ , we loosely estimate $I(R_3)$ in Lemma 10·2 and carefully bound $\Delta(R_3)$ in Lemma 10·5. Combining these estimates yields the desired error term in Proposition 6·4.

We start with some simple upper bounds for F.

Lemma 10·1. For $w \in \mathbb{C}$ with $0 \lt \mathrm{Re}(w) \leq N$ , we have

\begin{align*} |F(w)| \ll \begin{cases} \frac{\textstyle 1}{\textstyle |w|} & \text{if $\ |\mathrm{Im}(w)| \leq \pi N$,} \\ \frac{\textstyle 1}{\textstyle |w-2\pi i N|} & \text{if $\ \pi N \leq \mathrm{Im}(w) \leq 2 \pi N$,} \\ \frac{\textstyle 1}{\textstyle |w+2\pi i N|} & \text{if $\ -2 \pi N \leq \mathrm{Im}(w) \leq -\pi N$}. \end{cases} \end{align*}

Proof. Note that the function

\begin{align*}\frac{w}{N(1-e^{-w/N})}\end{align*}

is holomorphic on an open set containing $R = \{ w \in \mathbb{C} \, | \, |\mathrm{Im}(w)| \leq \pi N, \, 0 \leq \mathrm{Re}(w) \leq N\}$ since the singularity at 0 is removable. Making the substitution $w \mapsto N \cdot w$ , we see that it suffices to show that ${w}/{(1-e^{-w})}$ is bounded on $\tilde{R} = \{ w \in \mathbb{C}\, | \, |\mathrm{Im}(w)| \leq \pi, \, 0 \leq \mathrm{Re}(w) \leq 1 \}$ . This follows immediately since the above expression is holomorphic on an open set containing $\tilde{R}$ . One uses the same argument in the other cases.

We will also need the following estimate:

Lemma 10·2.

\begin{align*} I([{-}\pi N, \pi N]^{2k-1}) \ll_k (\!\log N)^k. \end{align*}

Proof. For fixed $i \in \{2,\dots,k\}$ , we use Hölder’s inequality and the previous lemma to obtain

\begin{align*}\int_{-\pi N}^{\pi N} \prod_{j = k+1}^{2k} |F(z_i + z_j)|d\theta_i &\leq \prod_{j = k+1}^{2k} \Big( \int_{-\pi N}^{\pi N} |F(z_i + z_j)|^k d\theta_i \Big)^{1/k} \\&\ll_k \prod_{j = k+1}^{2k} \Big( \int_{|\theta_i + \theta_j| \leq \pi N} \frac{1}{|z_i+z_j|^k} d\theta_i \Big. \\ &\hspace{1.1cm} + \int_{-2\pi N \leq \theta_i + \theta_j \leq -\pi N} \frac{1}{|z_i + z_j + 2\pi i N |^k} d \theta_i \\ &\hspace{1.1cm} + \Big. \int_{ \pi N \leq \theta_i + \theta_j \leq 2 \pi N} \frac{1}{|z_i + z_j - 2\pi i N|^k} d \theta_i \Big)^{1/k}. \end{align*}

Upon applying the inequality $1/|w| \leq \sqrt{2}/(\mathrm{Re}(w)+|\mathrm{Im}(w)|)$ with $\mathrm{Re}(w) \gt 0$ , we have that the above is

\begin{align*}&\ll \prod_{j = k+1}^{2k} \Big( \int_{\mathbb{R}} \frac{1}{(c_i+c_j+|\theta_i+\theta_j|)^k} d\theta_i \Big.+ \int_{\mathbb{R}} \frac{1}{(c_i+c_j+|\theta_i + \theta_j + 2\pi N |)^k} d \theta_i\\ &\hspace{1.5cm} + \Big. \int_{\mathbb{R}} \frac{1}{(c_i+c_j + |\theta_i + \theta_j - 2\pi N|)^k} d \theta_i \Big)^{1/k}, \end{align*}

and this is clearly bounded by a constant independent of N for $k \geq 2$ . (Recall that we made this assumption at the very beginning of Section 6.) Thus,

\begin{align*} I([{-}\pi N, \pi N]^{2k-1}) &\ll_k \int_{-\pi N}^{\pi N} \cdots \int_{- \pi N}^{\pi N} \prod_{j = k+1}^{2k} |F(z_j)| d \theta_{k+1} \cdots d \theta_{2k} \\ &= \prod_{j = k+1}^{2k} \int_{-\pi N}^{\pi N} |F(z_j)| d \theta_j \\ &\ll \prod_{j = k+1}^{2k} \int_{0}^{\pi N} \frac{1}{c_j+\theta_j} d \theta_j \\ &\ll \prod_{j = k+1}^{2k} \log N = (\!\log N)^k. \unicode{x8718}\end{align*}

The next few lemmas estimate different regions of the integral I. Let $2 \leq X \leq N/10$ be a parameter. Define the set $\mathfrak{S}(X) \subseteq [{-}\pi N, \pi N]^{2k-1}$ by

(10·2) \begin{equation}\{ (\theta_2,\dots,\theta_{2k}) \in [{-}\pi N, \pi N]^{2k-1}\ \big|\ |z_i + z_j| \leq X \text{ for } 2 \leq i \leq k, \ k+1 \leq j \leq 2k\}.\end{equation}

We begin by estimating the contribution outside $\mathfrak{S}(X)$ .

Lemma 10·3.

\begin{align*} I([{-}\pi N, \pi N]^{2k-1} \setminus \mathfrak{S}(X)) \ll_k \frac{(\!\log N)^{2k-1}}{X^{1-1/k}} + (\!\log N)^{2k-2} \frac{X}{N}. \end{align*}

Proof. Let $\Omega \,{:\!=}\, \{2,\dots,k\} \times \{k+1,\dots,2k\}$ with $|\Omega| = k(k-1)$ . We say a k-partition $\Lambda_1 \sqcup \Lambda_2 \sqcup \cdots \sqcup \Lambda_k$ of $\Omega$ is weak if we allow $\Lambda_i = \varnothing$ . For every weak 3-partition $\Lambda = (\Lambda_1,\Lambda_2,\Lambda_3)$ of $\Omega$ , define the region $R_{\Lambda}(X)$ to be the set

\begin{align*} \left\{(\theta_2, \dots, \theta_{2k}) \in [{-}\pi N, \pi N]^{2k{-}1} \Bigg| \begin{array}{cc} |z_i {+} z_j| \leq X & \forall\ (i,j) \in \Lambda_1, \\ |z_i {+} z_j {-} 2\pi i N| \leq X \text{ or } |z_i {+} z_j {+} 2\pi i N| \leq X & \forall\ (i,j) \in \Lambda_2, \\ |z_i {+} z_j| \gt X \text{ and } |z_i {+} z_j \pm 2\pi i N| \gt X & \forall\ (i,j) \in \Lambda_3 \end{array}\right\}\!. \end{align*}

It is easy to see that $[{-}\pi N, \pi N]^{2k-1}$ is partitioned by $\{R_\Lambda\ |\ \Lambda \text{ a weak 3-partition of }\Omega\}$ and that $\mathfrak{S}(X) = R_{(\Omega, \varnothing, \varnothing)}(X)$ . We further define two projection maps $\pi_i, \ i=1,2$ which respectively return the first and second components of a pair of integers, that is, $\pi_1(a,b) = a$ and $\pi_2(a,b) = b$ for $a,b \in \mathbb{Z}$ .

It now suffices to prove that

(10·3) \begin{equation} I(R_\Lambda(X)) \ll_k \frac{(\!\log N)^{2k-1-|\pi_2(\Lambda_2)|}}{X^{|\Lambda_3|(1-1/k)}} \Big(\frac{X}{N}\Big)^{|\pi_2(\Lambda_2)|}, \end{equation}

because if $\Lambda_2 \neq \varnothing$ , then the RHS of the above expression is $\ll_k (\!\log N)^{2k-2} X/N$ . If, on the other hand, $\Lambda_3 \neq \varnothing$ , then the RHS is $\ll_k {(\!\log N)^{2k-1}}/{X^{1-1/k}}$ , and, since $\mathfrak{S}(X)=R_{(\Omega, \varnothing, \varnothing)}(X)$ , at least one of these two cases must occur. Now, fix a weak partition $\Lambda$ of $\Omega$ and write $R_\Lambda \,{:\!=}\, R_\Lambda(X)$ . For $(\theta_2,\dots,\theta_{2k}) \in R_\Lambda$ , notice that

\begin{align*} \prod_{i = 2}^{k} \prod_{j = k+1}^{2k} |F(z_i + z_j)| \ll_k \frac{1}{X^{|\Lambda_3|}} \end{align*}

by assumption and Lemma 10·1. Hence,

\begin{align*} I(R_\Lambda) \ll_k \frac{1}{X^{|\Lambda_3|(1-1/k)}} \underset{R_\Lambda}{\int \cdots \int} \prod_{i = 2}^{k} \prod_{j = k+1}^{2k} |F(z_i + z_j)|^{\frac{1}{k}}\ \prod_{j = k+1}^{2k} |F(z_j)|\ d\theta_2 \cdots d\theta_{2k}. \end{align*}

Further note that $R_\Lambda \subset [{-}\pi N, \pi N]^{k-1} \times \tilde{R}_\Lambda$ where

\begin{align*} \tilde{R}_\Lambda \,{:\!=}\, \{ (\theta_{k+1},\dots,\theta_{2k}) \in [{-}\pi N, \pi N]^{k}\ \big|\ \pi N - X \leq |\theta_j| \leq \pi N\;\quad \forall\ j \in \pi_2(\Lambda_2) \}. \end{align*}

This inclusion holds because if we had $|\theta_j| \lt \pi N - X$ for $(i,j) \in \Lambda_2$ , then

\begin{align*} |z_i + z_j \pm 2 \pi i N| \geq \left| |\theta_i + \theta_j| - 2 \pi N \right| \geq 2\pi N - |\theta_i| - |\theta_j| \gt 2 \pi N - \pi N + X - \pi N = X. \end{align*}

Consequently, the above gives

\begin{align*} I(R_\Lambda) \ll_k \frac{1}{X^{|\Lambda_3|(1-1/k)}} \underset{[{-}\pi N, \pi N]^{k-1} \times \tilde{R}_\Lambda}{\int \cdots \int} \prod_{i = 2}^{k} \prod_{j = k+1}^{2k} |F(z_i + z_j)|^{\frac{1}{k}}\ \prod_{j = k+1}^{2k} |F(z_j)|\ d\theta_2 \cdots d\theta_{2k}. \end{align*}

By Hölder’s inequality,

\begin{align*} \prod_{i = 2}^{k} \int_{- \pi N}^{\pi N} \prod_{j = k+1}^{2k} |F(z_i + z_j)|^{\frac{1}{k}} d\theta_i \leq \prod_{i = 2}^{k} \prod_{j = k+1}^{2k} \Big(\underbrace{\int_{- \pi N}^{\pi N} |F(z_i + z_j)| d\theta_i}_{\ll \log N} \Big)^{\frac{1}{k}} \ll (\!\log N)^{k-1}. \end{align*}

Finally, we obtain

\begin{align*} I(R_\Lambda) &\ll_k \frac{(\!\log N)^{k-1}}{X^{|\Lambda_3|(1-1/k)}} \underset{\tilde{R}_\Lambda}{\int \cdots \int} \prod_{j = k+1}^{2k} |F(z_j)| d \theta_{k+1} \cdots d \theta_{2k} \\ &\ll_k \frac{(\!\log N)^{k-1}}{X^{|\Lambda_3|(1-1/k)}} \prod_{j \notin \pi_2(\Lambda_2)} \underbrace{\int_{- \pi N}^{\pi N} |F(z_j)| d\theta_j}_{\ll \log N} \prod_{j \in \pi_2(\Lambda_2)} \Big(\int_{-\pi N}^{-\pi N + X} + \int_{\pi N - X}^{\pi N}\Big) \underbrace{|F(z_j)|}_{\ll \frac{1}{N}} d \theta_j \\ &\ll \frac{(\!\log N)^{2k-1-|\pi_2(\Lambda_2)|}}{X^{|\Lambda_3|(1-1/k)}} \Big( \frac{X}{N}\Big)^{|\pi_2(\Lambda_2)|}. \end{align*}

Thus, we have shown (10·3).

Let $10X \leq Y \leq N$ be a parameter. Define the region $\mathfrak{S}(X,Y) \subseteq \mathfrak{S}(X)$ to be the set

(10·4) \begin{equation}\left\{ (\theta_2,\dots,\theta_{2k}) \in [{-}\pi N, \pi N]^{2k-1}\ \bigg|\ \begin{array}{cc} |z_i + z_j| \leq X & \ \ \forall\ 2 \leq i \leq k, \ k+1 \leq j \leq 2k, \\ |z_j| \leq Y & \forall\ k+1 \leq j \leq 2k\end{array} \right\}\!.\end{equation}

Lemma 10·4. We have

\begin{align*} I(\mathfrak{S}(X) \setminus \mathfrak{S}(X,Y)) \ll_k \Big( \frac{X}{Y} \Big)^{k-1}. \end{align*}

Proof. For $l \in \{ k+1,\dots,2k\}$ , we define

\begin{align*}\mathfrak{T}_l(X,Y) \,{:\!=}\, \left\{(\theta_2,\dots,\theta_{2k}) \in [{-}\pi N, \pi N]^{2k-1}\ \bigg| \begin{array}{cc} |z_i + z_j| \leq X & \ \ \forall\ 2 \leq i \leq k, \ k+1 \leq j \leq 2k, \\ |z_l | \gt Y &\end{array}\right\}\!.\end{align*}

We then have $\mathfrak{S}(X) \setminus \mathfrak{S}(X,Y) \subset \bigcup_{l = k+1}^{2k} \mathfrak{T}_l(X,Y)$ , and it suffices to prove

\begin{align*}I(\mathfrak{T}_l(X,Y)) \ll_k \Big( \frac{X}{Y} \Big)^{k-1} \quad \forall\ l = k+1,\dots,2k.\end{align*}

We claim that

\begin{align*}\mathfrak{T}_l(X,Y) \subset [{-}\pi N, \pi N]^{k-1} \times \tilde{\mathfrak{T}_l}(X,Y),\end{align*}

where

\begin{align*}\tilde{\mathfrak{T}}_l(X,Y) \,{:\!=}\, \left\{(\theta_2,\dots,\theta_{2k}) \in [{-}\pi N, \pi N]^{2k-1}\ \bigg|\begin{array}{cc} |\theta_j - \theta_l| \leq 2X & \ \ \forall\ k+1 \leq j \leq 2k, \\ |\theta_l| \geq Y/2 &\end{array}\right\}\!.\end{align*}

To prove this, let $(\theta_2,\dots,\theta_{2k}) \in \mathfrak{T}_l(X,Y)$ . As $|z_l| \gt Y$ , it follows that

\begin{align*}|\theta_l| \geq Y - |c_l| \geq Y - \frac{3}{4} \geq Y - \frac{X}{2} \geq \frac{Y}{2}.\end{align*}

By the triangle inequality, we also have

\begin{align*}|\theta_j - \theta_l| \leq |z_j - z_l| \leq |z_j + z_2| + |z_2 + z_l| \leq 2X.\end{align*}

This proves the claim. Therefore,

\begin{align*} I(\mathfrak{T}_l(X,Y)) &\leq I([{-}\pi N, \pi N]^{k-1}\times \tilde{\mathfrak{T}}_l(X,Y)) \\&= \underset{\tilde{\mathfrak{T}}_l(X,Y)}{\int \cdots \int} \Big( \underbrace{\int_{-\pi N}^{\pi N} \cdots \int_{-\pi N}^{\pi N} \prod_{i = 2}^k \prod_{j = k+1}^{2k} |F(z_i + z_j)| d \theta_i}_{\ll_k 1 \text{ by H\"{o}lder as in (Lemma 10$\cdot$2)}} \Big) \prod_{j = k+1}^{2k} |F(z_j)|\ d \theta_{k+1} \cdots d \theta_{2k} \\&\ll_k \underset{\tilde{\mathfrak{T}}_l(X,Y)}{\int \cdots \int} \prod_{j = k+1}^{2k} |F(z_j)|\ d \theta_{k+1} \cdots d \theta_{2k} \\&\ll_k \int_{\frac{Y}{2} \leq |\theta_l| \leq \pi N} \bigg( \prod_{\substack{j = k+1 \\ j \neq l}}^{2k} \int_{|\theta_j - \theta_l| \leq 2X} |F(z_j)|\ d \theta_j \bigg) |F(z_l)|\ d \theta_l.\end{align*}

Note that for $j \neq l$ , $|F(z_j)| \ll |\theta_j|^{-1} \ll |\theta_l|^{-1}$ where the first bound follows from Lemma 10·1 and the second from

\begin{align*}|\theta_j| \geq |\theta_l| - |\theta_j - \theta_l|\end{align*}

and

\begin{align*}|\theta_j - \theta_l| \leq 2X \leq \frac{Y}{5} \leq \frac{|\theta_l|}{2}.\end{align*}

Furthermore, the length of the interval of integration for each j is O(X). Thus,

\begin{align*} I(\mathfrak{T}_l(X,Y)) &\ll_k \int_{|\theta_l| \geq \frac{Y}{2}} \bigg( \underbrace{|F(z_l)|}_{\ll \frac{1}{|\theta_l|}} \prod_{\substack{j = k+1 \\ j \neq l}}^{2k} \frac{X}{|\theta_l|} d \theta_l \bigg) \\ &\ll_k \int_{|\theta_l| \geq \frac{Y}{2}} \frac{X^{k-1}}{|\theta_l|^{k}} d\theta_l \ll_k \Big( \frac{X}{Y} \Big)^{k-1}.\end{align*}

This proves the lemma.

For the remaining region $\mathfrak{S}(X,Y)$ , we estimate trivially

\begin{align*}I(\mathfrak{S}(X,Y)) \ll_k (\!\log N)^k.\end{align*}

Here, our savings will come from the fact that the difference

\begin{align*}B\left(q^{-1/2}, q^{-1/2}e^{(c_2+i\theta_2)/N}, \dots, q^{-1/2}e^{(c_{2k}+i\theta_{2k})/N}\right) - b_k(q) \end{align*}

is small on this region, where all of the exponents $z_i+z_j$ and $z_j$ are small. The next lemma makes this precise. Recall the definition of $\Delta(R)$ from equation (10·1) for any Lebesgue-measurable region R of $[{-}\pi N,\pi N]^{2k-1}$ :

\[ \Delta(R) \,{:\!=}\, \sup_{(\theta_2, ..., \theta_{2k})\in R}\left|B\left(q^{-1/2}, q^{-1/2}e^{(c_2+i\theta_2)/N}, \dots, q^{-1/2}e^{(c_{2k}+i\theta_{2k})/N}\right) - b_k(q)\right|\!.\]

Lemma 10·5. We have

\begin{align*}\Delta([{-}\pi N, \pi N]^{2k-1}) \ll_{k} \frac{1}{q},\end{align*}

and

\begin{align*}\Delta(\mathfrak{S}(X,Y)) \ll_{k} \frac{Y}{qN}.\end{align*}

Proof. Note that the first bound is an immediate consequence of Proposition 6·1 and

\begin{align*}\max_{-\pi N \leq \theta_i \leq N \pi}|e^{c_i/N + i \theta_i/N}-1| \ll 1.\end{align*}

If we have $(\theta_2,\dots,\theta_{2k}) \in \mathfrak{S}(X,Y)$ , then $|\theta_i| \leq 2Y$ for all $i=2,\dots,2k$ . Thus, by Proposition 6·1, it again suffices to show that

\begin{align*}\max_{(\theta_2,\dots,\theta_{2k}) \in \mathfrak{S}(X,Y)} ||(q^{-1/2}e^{c_2/N + i\theta_2/N}, \dots, q^{-1/2}e^{c_{2k}/N + i\theta_{2k}/N})-(q^{-1/2},\dots,q^{-1/2})||_\infty \ll \frac{Y}{\sqrt{q} N}.\end{align*}

Using $|\theta_i| \leq 2Y$ , we can simplify this statement to

(10·5) \begin{equation}\max_{-2Y \leq \theta_i \leq 2Y}|e^{c_i/N + i \theta_i/N}-1| \ll \frac{Y}{N}\end{equation}

for each i. Recalling that for bounded $z\in \mathbb{C}$ , we have $\lvert{e^z-1}\ll |z|$ , we conclude

\begin{align*}|e^{c_i/N + i\theta_i/N}-1| &= |e^{c_i/N + i\theta_i/N}-e^{i\theta_i/N}+e^{i\theta_i/N}-1|\\&\leq |e^{c_i/N}-1| + |e^{i\theta_i/N}-1| \\ &\ll_{c_i} \frac{1}{N} ( 1 + |\theta|).\end{align*}

We now take the maximum over $-2Y \leq \theta \leq 2Y$ to establish (10·5).

Thus, we obtain the desired bound on the integral $\widetilde{J_k}$ :

\begin{align*}|\widetilde{J_k}| &\ll_k \frac{1}{q} \frac{(\!\log N)^{2k-1}}{X^{1-1/k}} + \frac{1}{q} (\!\log N)^{2k-2} \frac{X}{N} + \frac{1}{q} \left( \frac{X}{Y} \right)^{k-1} + \frac{Y}{q N} \left( \log N \right)^k.\end{align*}

The first two terms are a consequence of estimating $I(R_1)$ using Lemma 10·3 and taking the trivial $1/q$ estimate for $\Delta(R_1)$ . The third term makes use of Lemma 10·4 to bound $I(R_2)$ and uses the same estimate for $\Delta(R_2)$ . Lastly, the fourth term comes from upper bounding $I(R_3)$ by $I([{-}\pi N, \pi N]^{2k-1})$ and using Lemma 10·5 to bound $\Delta(R_3)$ . This completes the proof of Proposition 6·4 and thus the proof of Theorem 1·1.