# Sign changes in statistics for plane partitions

Walter Bridges, Johann Franke, and Joshua Males

###### Abstract.

Recent work of Cesana, Craig and the third author shows that the trace of
plane partitions is asymptotically equidistributed in residue classes mod $b$.
Applying a technique of the first two authors and Garnowski, we prove
asymptotic formulas for the secondary terms in this equidistribution, which
are controlled by certain complex numbers generated by a twisted MacMahon-type
product. We further carry out a similar analysis for a statistic related to
plane overpartitions.

###### Key words and phrases:

plane partition, trace, equidistribution, secondary term, plane overpartition,
circle method

##  1\. Introduction and statement of results

Partitions are ubiquitous in number theory, and in the last century have
become one of the most well-studied combinatorial objects. For example, if
$p(n)$ denotes the number of partitions of a positive integer $n$, Hardy and
Ramanujan [22] introduced their famed Circle Method in order to prove that as
$n\to\infty$ we have

$\displaystyle
p(n)\sim\frac{1}{4n\sqrt{3}}\exp\left(\pi\sqrt{\frac{2n}{3}}\right),$

a technique which was perfected by Rademacher [29], with further versions
introduced by Wright [36, 37] among many others. Partitions also appear in
many areas outside of number theory and combinatorics, such as in algebraic
topology [8] and representation theory [19, 20].

One of the most natural generalizations of partitions are the so-called plane
partitions. A plane partition of $n\in\mathbb{N}$ (see e.g. [1]) is a two-
dimensional array $\pi=\\{\pi_{j,k}\\}_{j,k\geq 1}$ of non-negative integers
that is non-increasing in both indices, i.e $\pi_{j,k}\geq\pi_{j+1,k}$,
$\pi_{j,k}\geq\pi_{j,k+1}$ for all $j$ and $k$, and fulfills
$|\pi|:=\sum_{j,k}\pi_{j,k}=n$. Plane partitions can be visualized in the
plane as in Figure 1.

$4\qquad\begin{matrix}3&1\end{matrix}\qquad\begin{matrix}3\\\
1\end{matrix}\qquad\begin{matrix}2&2\end{matrix}\qquad\begin{matrix}2\\\
2\end{matrix}\qquad\begin{matrix}2&1&1\end{matrix}\qquad\begin{matrix}2&1\\\
1&\end{matrix}\qquad\begin{matrix}2\\\ 1\\\
1\end{matrix}\qquad\begin{matrix}1&1&1&1\end{matrix}\qquad\begin{matrix}1&1&1\\\
1&&\end{matrix}\qquad\begin{matrix}1&1\\\
1&1\end{matrix}\qquad\begin{matrix}1&1\\\ 1&\\\
1&\end{matrix}\qquad\begin{matrix}1\\\ 1\\\ 1\\\ 1\end{matrix}$ Figure 1.  All
plane partitions of size 4.

Within number theory plane partitions have many applications, with an
excellent overview given in the classical reference [31]. They also appear in
areas outside of number theory and combinatorics, and have seen a surge in
interest in recent years. For example, they are intricately related to certain
aspects of superconformal field theory [26] and in connection to counting
small black holes in string theory [15]. In the present paper, we determine
refined information on certain statistics for plane partitions, complementing
the growing literature in the area.

Let $pp(n)$ count the number of plane partitions of $n$ (with $pp(0):=1$). A
classical result of MacMahon [25] gives the product generating function

$\sum_{\pi}q^{|\pi|}=\sum_{n\geq 0}pp(n)q^{n}=\prod\limits_{n\geq
1}\dfrac{1}{(1-q^{n})^{n}}.$ (1.1)

One well-studied statistic associated a plane partition $\pi$ is its trace
$t(\pi)$, which is the sum of the diagonal entries

$t(\pi)=\sum_{j\geq 1}\pi_{j,j}.$

Let $pp(m,n)$ denote the number of plane partitions with trace $m$. Stanley
[30] applied a bijection of Bender and Knuth [2] to show that the trace is
generated by a simple refinement of (1.1),

$PP(\zeta;q):=\sum_{\pi}\zeta^{t(\pi)}q^{|\pi|}=\sum_{n,m\geq
0}pp(m,n)\zeta^{m}q^{n}=\prod\limits_{n\geq 1}\dfrac{1}{(1-\zeta q^{n})^{n}}.$
(1.2)

The coefficient of $q^{n}$ above is a polynomial in $\zeta$, which we denote
by $T_{n}(\zeta)$.

To enumerate plane partitions with trace in residue classes, we can substitute
roots of unity for $\zeta$ and use orthogonality. Let $pp(a,b,n)$ denote those
plane partitions of $n$ with trace congruent to $a\
\,\left(\mathrm{mod}\,b\right)$ and let $\zeta_{k}^{h}:=e^{2\pi
i\frac{h}{k}}$. Then

$pp(a,b,n)=\frac{pp(n)}{b}+\frac{1}{b}\sum_{1\leq\nu\leq
b-1}\zeta_{b}^{-a\nu}T_{n}(\zeta_{b}^{\nu}).$ (1.3)

Work of Cesana, Craig and the third author shows that the term $pp(n)$
dominates the right-hand side as $n\to\infty$; that is, we have asymptotic
equidistribution of the trace in residue classes mod $b$.

######  Theorem (Theorem 1.4, [10]).

For all $a,b\in\mathbb{N}$, we have, as $n\to\infty$,

$\frac{pp(a,b,n)}{pp(n)}\sim\frac{1}{b}.$

Secondary terms in the asymptotic behavior of $pp(a,b,n)$ are controlled by
the polynomials $T_{n}(\zeta_{b}^{\nu})$. In particular, consider two distinct
residue classes $a_{1}\not\equiv a_{2}\ \,\left(\mathrm{mod}\,b\right)$. Then
in the difference
$pp(a_{1},b,n)-pp(a_{2},b,n)$, the term $pp(n)$ in (1.3) cancels and to
predict the oscillation in this difference as $n\to\infty$, we require the
asymptotic behavior of the complex numbers $T_{n}(\zeta_{b}^{\nu})$.

To state our main result, recall the trilogarithm which is defined for
$|z|\leq 1$ by the series $\mathrm{Li}_{3}(z)=\sum_{n\geq
1}\frac{z^{n}}{n^{3}}.$ Let $\theta_{12}=0.47585\dots$ be the solution to

$\mathrm{Re}\left(\sqrt[3]{\mathrm{Li}_{3}\left(e^{2\pi
i\theta}\right)}\right)=\frac{\mathrm{Re}\left(\sqrt[3]{\mathrm{Li}_{3}\left(e^{4\pi
i\theta}\right)}\right)}{2},\qquad 0\leq\theta\leq\frac{1}{2}.$

(See Corollary 3.5.) Then we have the following asymptotic formulas. As
$T_{n}\left(\overline{\zeta}\right)=\overline{T_{n}(\zeta)},$ we state them
only for Im$(\zeta)\geq 0$.

######  Theorem 1.1.

Let $a,b$ be integers with $1\leq a\leq b$ and $\gcd(a,b)=1$.

  1. (1)

If $0<\frac{a}{b}<\theta_{12}$, then

$T_{n}\left(\zeta_{b}^{a}\right)\sim\frac{\left(1-\zeta_{b}^{a}\right)^{\frac{1}{12}}\mathrm{Li}_{3}\left(\zeta_{b}^{a}\right)^{\frac{1}{6}}}{2^{\frac{1}{3}}\sqrt{3\pi}n^{\frac{2}{3}}}\exp\left(\frac{3}{2^{\frac{2}{3}}}\mathrm{Li}_{3}\left(\zeta_{b}^{a}\right)^{\frac{1}{3}}n^{\frac{2}{3}}\right).$

  2. (2)

If $\theta_{12}<\frac{a}{b}<\frac{1}{2}$, then

$T_{n}\left(\zeta_{b}^{a}\right)\sim(-1)^{n}\frac{\left(1-\zeta_{b}^{a}\right)^{\frac{1}{6}}\mathrm{Li}_{3}\left(\zeta_{b}^{2a}\right)^{\frac{1}{6}}}{\left(1+\zeta_{b}^{a}\right)^{\frac{1}{12}}2^{\frac{5}{6}}\sqrt{3\pi}n^{\frac{2}{3}}}\exp\left(\frac{3}{2^{\frac{5}{3}}}\mathrm{Li}_{3}\left(\zeta_{b}^{2a}\right)^{\frac{1}{3}}n^{\frac{2}{3}}\right).$

  3. (3)

We have

$T_{n}(-1)\sim(-1)^{n}\frac{e^{-\zeta^{\prime}(-1)}\zeta(3)^{\frac{5}{36}}}{2^{\frac{3}{4}}\sqrt{3\pi}n^{\frac{23}{36}}}\mathrm{exp}\left(\frac{3}{2^{\frac{5}{3}}}\zeta(3)^{\frac{1}{3}}n^{\frac{2}{3}}\right).$

######  Remark.

The case $\zeta=1$ is the classical asymptotic formula for plane partitions
due to Wright [35],

$T_{n}(1)=pp(n)\sim\frac{\zeta(3)^{\frac{7}{36}}e^{\zeta^{\prime}(-1)}}{2^{\frac{11}{36}}\sqrt{3\pi}n^{\frac{25}{36}}}\exp\left(\frac{3\zeta(3)^{\frac{1}{3}}}{2^{\frac{2}{3}}}n^{\frac{2}{3}}\right).$

When $\zeta=-1$, we have $T_{n}(-1)=pp(0,2,n)-pp(1,2,n),$ the difference of
plane partitions with even and odd trace.

From our asymptotic formulas, it follows that the differences
$pp(a_{1},b,n)-pp(a_{2},b,n)$, when rescaled, oscillate like a cosine. (See
Figure 2.)

Figure 2. The blue dots depict the function
$\frac{pp(1,5,n)-pp(4,5,n)}{Bn^{-\frac{3}{4}}e^{3\cdot
2^{-\frac{2}{3}}\lambda_{1}n^{\frac{2}{3}}}}$, and the red line is the
asymptotic prediction $\cos\left(\alpha+3\cdot
2^{-\frac{2}{3}}\lambda_{2}n^{\frac{2}{3}}\right)$. The approximate values of
the constants are $B\approx 0.19971$, $\alpha\approx-1.41897$,
$\lambda_{1}\approx 0.89873$, and $\lambda_{2}\approx 0.44610$.

######  Corollary 1.2.

Let $b\geq 3$, and $a_{1}\not\equiv a_{2}$ be two classes modulo $b$. Then we
have

$\displaystyle\frac{pp(a_{1},b,n)-pp(a_{2},b,n)}{B_{a_{1},a_{2},b}n^{-\frac{2}{3}}\exp\left(\frac{3}{2^{\frac{2}{3}}}\lambda_{1}n^{\frac{2}{3}}\right)}=\cos\left(\alpha_{a_{1},a_{2},b}+\frac{3}{2^{\frac{2}{3}}}\lambda_{2}n^{\frac{2}{3}}\right)+o(1),$

where $\lambda_{1}+i\lambda_{2}:=\mathrm{Li}_{3}(\zeta_{b})^{\frac{1}{3}}$,
and $B_{a_{1},a_{2},b}>0$ and $\alpha_{a_{1},a_{2},b}\in[0,2\pi)$ are defined
by

$\displaystyle
B_{a_{1},a_{2},b}e^{i\alpha_{a_{1},a_{2},b}}=\frac{2^{\frac{2}{3}}}{b\sqrt{3\pi}}\left(\zeta_{b}^{-a_{1}}-\zeta_{b}^{-a_{2}}\right)\left(1-\zeta_{b}\right)^{\frac{1}{12}}(\lambda_{1}+i\lambda_{2})^{\frac{1}{2}}.$

######  Remark.

Corollary 1.2 can also be generalized to the case of higher differences, i.e.
terms $\sum_{a\ \,\left(\mathrm{mod}\,b\right)}v_{a}pp(a,b,n)$ with $\sum_{a\
\,\left(\mathrm{mod}\,b\right)}v_{a}=0$, completely analogously to the case of
ordinary partitions in [7].

We prove Theorem 1.1 using an adaptation of the Hardy–Ramanujan Circle Method
due to the first two authors and Garnowski [7]. Asymptotics for $T_{n}(\zeta)$
in most cases with $|\zeta|\neq 1$ were proved by Boyer and Parry [3, 4, 5] in
connection with the zero attractors of these polynomials. The case of generic
products was considered by Parry in [27], again for $|\zeta|\neq 1.$ As in
[7], the primary difficulty is the bounding of minor arcs, as classical
techniques do not apply when $\zeta$ is complex and has absolute value 1. The
present work demonstrates the wide applicability of the techniques in [7] to
twisted infinite product generating functions.

We also consider the following MacMahon-type product,

$\overline{PP}(\zeta;q)=\sum_{n\geq
0}\mathcal{A}_{n}(\zeta)q^{n}:=\prod_{n\geq 1}\left(\frac{1-\zeta
q^{n}}{1-q^{n}}\right)^{n}.$

For an explicit description of the statistics for plane partitions generated
by $\mathcal{A}_{n}(\zeta)$, we refer the readers to the work of Vuletić [33,
34]. If one substitutes $\zeta=-1$, then $\overline{PP}(-1;q)$ counts plane
overpartitions (see [14]). Again, the work of Cesana, Craig and the third
author [10] implies asymptotic equidistribution in residue classes, and we
extend this to secondary terms by proving the following asymptotic formulas.

######  Theorem 1.3.

For $\frac{a}{b}\neq 1$, we have

$\mathcal{A}_{n}\left(\zeta_{b}^{a}\right)\sim\left(1-\zeta_{b}^{a}\right)^{-\frac{1}{12}}\frac{e^{\zeta^{\prime}(-1)}\left(\zeta(3)-\mathrm{Li}_{3}\left(\zeta_{b}^{a}\right)\right)^{\frac{7}{36}}}{2^{\frac{11}{36}}\sqrt{3\pi}n^{\frac{25}{36}}}\exp\left(\frac{3}{2^{\frac{2}{3}}}\left(\zeta(3)-\mathrm{Li}_{3}\left(\zeta_{b}^{a}\right)\right)^{\frac{1}{3}}n^{\frac{2}{3}}\right).$

In particular, we have the following asymptotic formula for the number of
plane overpartitions of size $n$, $\overline{pp}(n):=\mathcal{A}_{n}(-1)$.

######  Corollary 1.4.

As $n\to\infty,$ we have

$\overline{pp}(n)\sim\frac{e^{\zeta^{\prime}(-1)}(7\zeta(3))^{\frac{7}{36}}}{2^{\frac{7}{9}}\sqrt{3\pi}n^{\frac{25}{36}}}\mathrm{exp}\left(\frac{3}{2^{\frac{4}{3}}}(7\zeta(3))^{\frac{1}{3}}n^{\frac{2}{3}}\right).$

We structure the article as follows. Section 2 recalls a few classical results
useful for asymptotic analysis. In Section 3, we make a careful study of the
trilogarithm on the unit circle; this is needed to identify the major arcs in
the circle method. In Section 4, we use the techniques of the first two
authors and Garnowski [7] to prove Theorems 1.1 and 1.3. Concluding remarks
regarding further extensions to generic product generating functions are given
in Section 5.

## Acknowledgements

We are grateful to Daniel Parry for his commentary on the proof of Lemma 3.6.
We thank Giulia Cesana, William Craig and Taylor Garnowski for helpful
feedback that greatly improved the exposition.

The first author is partially supported by the SFB/TRR 191 “Symplectic
Structures in Geometry, Algebra and Dynamics”, funded by the DFG
(Projektnummer 281071066 TRR 191). The second author is partially supported by
the Alfried Krupp prize. The research of the third author conducted for this
paper is supported by the Pacific Institute for the Mathematical Sciences
(PIMS). The research and findings may not reflect those of the Institute

##  2\. Preliminaries

###  2.1. Major arcs

This section contains results we use in the course of evaluating the major
arcs in the circle method. We first recall Laplace’s method.

######  Theorem 2.1 (Section 1.1.5 of [28]).

Let $A,B:[a,b]\to\mathbb{C}$ be continuous functions. Suppose $x\neq
x_{0}\in[a,b]$ such that $\mathrm{Re}(B(x))<\mathrm{Re}(B(x_{0})),$ and that

$\displaystyle\lim_{x\to
x_{0}}\frac{B(x)-B(x_{0})}{(x-x_{0})^{2}}=-k\in\mathbb{C},$

with $\mathrm{Re}(k)>0.$ Then as $t\to\infty$

$\displaystyle\int^{b}_{a}A(x)e^{tB(x)}dx=e^{tB(x_{0})}\left(A(x_{0})\sqrt{\frac{\pi}{tk}}+o\left(\frac{1}{\sqrt{t}}\right)\right).$

We recall classical Euler–Maclaurin summation.

######  Theorem 2.2 (p. 66 of [23]).

Let $\\{x\\}:=x-\lfloor x\rfloor$ denote the fractional part of $x$. For
$N\in\mathbb{N}$ and
$f:[1,\infty)\to\mathbb{C}$ a continuously differentiable function, we have

$\sum_{1\leq n\leq
N}f(n)=\int_{1}^{N}f(x)dx+\frac{1}{2}(f(N)+f(1))+\int_{1}^{N}f^{\prime}(x)\left(\\{x\\}-\frac{1}{2}\right)dx.$

We also need the following lemma that may be proved with elementary calculus.

######  Lemma 2.3.

Let $a\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. Then $x=\frac{a}{3}$ is
the unique maximum of the function $\cos(a-2x)\cos^{2}(x)$ for
$x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

###  2.2. Minor arcs

This section contains lemmas we use to bound the minor arcs in the circle
method. The following is known as Abel-partial summation.

######  Proposition 2.4 (p. 3 of [32]).

Let $N\in\mathbb{N}_{0}$ and $M\in\mathbb{N}$. For sequences
$\\{a_{n}\\}_{n\geq N}$, $\\{b_{n}\\}_{n\geq N}$ of complex numbers, if
$A_{n}:=\sum_{N<m\leq n}a_{m},$

$\sum_{N<n\leq
N+M}a_{n}b_{n}=A_{N+M}b_{N+M}+\sum_{N<n<N+M}A_{n}(b_{n}-b_{n+1}).$

We use a version of Euler-Maclaurin summation that applies in cones when the
summand has a simple pole at $0$. Recall that the Bernoulli polynomials
$B_{n}(x)$ may be defined through the power series

$\sum_{n\geq 0}\frac{B_{n}(x)t^{n}}{n!}=\frac{te^{tx}}{e^{t}-1}.$

######  Theorem 2.5 (Theorem 1.3 of [9]).

Let $D_{\vartheta}:=\\{re^{i\phi}:r\geq
0\;\textnormal{and}\;|\phi|\leq\vartheta\\}$ and suppose that
$0\leq\vartheta<\frac{\pi}{2}$. Let $f:\mathbb{C}\to\mathbb{C}$ be holomorphic
in a domain containing $D_{\vartheta}$ with the exception of a simple pole at
the origin, and assume that $f$ and all of its derivatives are of sufficient
decay in $D_{\vartheta}$. If $f(w)=\sum_{n\geq-1}b_{n}w^{n}$ near $0$, then
for $a\in\mathbb{R}\setminus\left(-\mathbb{N}_{0}\right)$, and
$N\in\mathbb{N}_{0}$, uniformly as $w\to 0$ in $D_{\vartheta}$,

$\displaystyle\sum_{m\geq 0}f(w(m+a))=$
$\displaystyle\frac{b_{-1}\operatorname{Log}(w)}{w}+\frac{b_{-1}C_{a}}{w}+\frac{1}{w}\int^{\infty}_{0}\left(f(x)-\frac{b_{-1}e^{-x}}{x}\right)$
$\displaystyle-\sum^{N-1}_{n=0}\frac{B_{n+1}(a)b_{n}}{n+1}w^{n}+O_{N}\left(w^{N}\right),$

where

$C_{a}:=(1-a)\sum_{m\geq 0}\frac{1}{(m+a)(m+1)}.$

The following bound for holomorphic functions is a consequence of the mean
value theorem.

######  Lemma 2.6.

Let $f:U\to\mathbb{C}$ be a holomorphic function and
$\overline{B_{r}(c)}\subset U$ a compact disk. Then, for all $a,b\in B_{r}(c)$
with $a\not=b$, we have

$\displaystyle\left|f(b)-f(a)\right|\leq\sup_{|z-c|=r}|f^{\prime}(z)||b-a|.$

We apply this lemma to the function

$\displaystyle\phi_{a}(w):=\frac{e^{-w-aw}}{(1-e^{-w})^{2}}+\frac{ae^{-aw}}{1-e^{-w}}-\frac{1}{w^{2}},$
(2.1)

where $0\leq a\leq 1$ is a real parameter. Note that $\phi_{a}$ is holomorphic
at 0 and in the cone $|\mathrm{Arg}(w)|\leq\frac{\pi}{2}-\eta,$ for any
$\eta>0.$

######  Lemma 2.7.

Let $x$ be a complex number with positive imaginary part and $|x|\leq 1$. Then
there is a constant $c>0$ independent from $a$ and $x$, such that for all
$m\leq\frac{1}{|x|}$ we have

$\displaystyle\left|\phi_{a}(xm)-\phi_{a}(x(m+1))\right|\leq c|x|.$

###### Proof.

One uses Lemma 2.6 and the maximum modulus principle. The details are
analogous to Lemma 2.12 in [7]. ∎

######  Lemma 2.8.

Let $x$ be a complex number with positive imaginary part. Then all
$m>\frac{1}{|x|}$ we have

$\displaystyle\left|\phi_{a}(xm)-\phi_{a}(x(m+1))\right|\ll\frac{1}{|x|^{2}}\left(\frac{1}{m^{2}}-\frac{1}{(m+1)^{2}}\right)+|x|e^{-m\mathrm{Re}(x)}+|x|e^{-m\mathrm{Re}(x)(1+a)}+a^{2}|x|e^{-am\mathrm{Re}(x)}.$

###### Proof.

We estimate the different parts separately. First we have

$\displaystyle\left|\frac{1}{x^{2}m^{2}}-\frac{1}{x^{2}(m+1)^{2}}\right|=\frac{1}{|x|^{2}}\left(\frac{1}{m^{2}}-\frac{1}{(m+1)^{2}}\right).$

For the second term in (2.1) we obtain

$\displaystyle\left|\frac{ae^{-amx}}{1-e^{-mx}}-\frac{ae^{-a(m+1)x}}{1-e^{-(m+1)x}}\right|\leq
ae^{-am\mathrm{Re}(x)}\left(\left|\frac{1}{1-e^{-mx}}-\frac{1}{1-e^{-(m+1)x}}\right|+\left|\frac{e^{-ax}-1}{1-e^{-(m+1)x}}\right|\right)$

and since $m>\frac{1}{|x|}$, we obtain

$\displaystyle\ll a|x|e^{-m\mathrm{Re}(x)}+a^{2}|x|e^{-a\mathrm{Re}(x)m}.$

For the first term, we obtain

$\displaystyle\left|\frac{e^{-mx(a+1)}}{(1-e^{-mx})^{2}}-\frac{e^{-(m+1)x(a+1)}}{(1-e^{-(m+1)x})^{2}}\right|$
$\displaystyle=e^{-m\mathrm{Re}(x)(a+1)}\left|\frac{1}{(1-e^{-mx})^{2}}-\frac{1}{(1-e^{-(m+1)x})^{2}}+O(x)\right|$
$\displaystyle\ll|x|e^{-m\mathrm{Re}(x)(a+1)},$

where we have used $|xm|>1$ and the mean value theorem. ∎

Finally, we need some bounds from [7] on partial sums of the twisted harmonic
series,

$\displaystyle G_{M}(\theta):=\sum_{1\leq m\leq M}\frac{e^{2\pi i\theta
m}}{m}.$

######  Lemma 2.9 (Lemma 2.16 of [7]).

We have, uniformly for $0<\theta<1$,

$\displaystyle\left|G_{M}(\theta)\right|\ll\log\left(\frac{1}{\theta}\right)+\log\left(\frac{1}{1-\theta}\right),\qquad
M\to\infty.$

######  Lemma 2.10 (Lemma 2.17 of [7]).

Let $a$, $b$, $h$ and $k$ be positive integers, such that
$\gcd(a,b)=\gcd(h,k)=1$. We assume that $a$ and $b$ are fixed. Then we have,
as $k\to\infty$

$\displaystyle\sum_{\begin{subarray}{c}1\leq j\leq k\\\ ak+bjh\not\equiv 0\
\,\left(\mathrm{mod}\,bk\right)\end{subarray}}\max_{m\geq
1}\left|G_{M}\left(\frac{a}{b}+\frac{hj}{k}\right)\right|=O(k)$

uniformly in $h$.

##  3\. Trilogarithms on the unit circle

We recall that for complex numbers $s$ and $z$ with $|z|<1$ the polylogarithm
$\mathrm{Li}_{s}(z)$ is defined by the series

$\displaystyle\mathrm{Li}_{s}(z):=\sum_{n\geq 1}\frac{z^{n}}{n^{s}}.$

We are especially interested in the case $s=3$, where $\mathrm{Li}_{3}(z)$ is
called the trilogarithm. In this section, we derive some tools concerning the
trilogarithm that are crucial for the choice of major arcs in the circle
method. In doing so, we follow ideas of Boyer and Parry, who have already made
similar considerations in detail for the dilogarithm $\mathrm{Li}_{2}(z)$ in
the appendix Analysis of the Root Dilogarithm of [6]. Some of the following
results regarding the trilogarithm have already been proved by Boyer and Parry
[5]. To determine major arcs in the circle method for $T_{n}(\zeta)$, we need
to study the function

$\displaystyle\theta\mapsto\mathrm{Re}\left(\sqrt[3]{\mathrm{Li}_{3}(e^{i\theta})}\right),$
(3.1)

for $\theta\in(0,\pi).$ Here we are primarily interested in monotonicity
questions, since these will later answer which parts of our asymptotic
formulas dominate others. In general, the mapping behavior of power series is
not an easy problem. However, under assumptions on the coefficients, results
can be obtained in this direction. This concerns, for instance, questions
whether holomorphic mappings map the unit circle onto convex sets, or
monotonicity problems. One possibility is to consider higher differences of
the coefficients. This plays into our hands above all in the theory of
polylogarithms, where the coefficients behave very uniformly. For a sequence
of real numbers $a:=a_{n}$, we define

$\displaystyle\Delta^{r}(a)_{n}:=\sum_{j=0}^{r}\binom{r}{j}(-1)^{j}a_{n+j}.$

We call $a_{n}$ strictly $r$-fold monotone, if $\Delta^{m}(a)_{n}>0$ for all
$n\in\mathbb{N}$ and $1\leq m\leq r$. We need the following theorem by Fejer.

######  Theorem 3.1 (see [17]).

Let $(a_{n})_{n\in\mathbb{N}}$ be a sequence of real numbers that is strictly
4-fold monotone and satisfies $\lim_{n\to\infty}a_{n}=0$. Then the function
$F$, that is defined by the Fourier series

$\displaystyle F(\theta):=\sum_{n=1}^{\infty}a_{n}\cos(n\theta)$

on the interval $(0,2\pi)$, is strictly decreasing on the interval
$(0,\pi)$.111In fact, Fejer shows something slightly different: He only
assumes that the generating sequence is 4-fold monotone in the sense
$\Delta^{m}(a)\geq 0$ and concludes that the function $\theta\mapsto
P(e^{i\theta})$ is monotone in $(0,\pi)$. However, as he does exclude the case
$a_{j}=0$ for all $j$, he argues that there is a $\mu$ such that
$\Delta^{4}(a)_{\mu}>0$. With this, one can actually show that the derivative
is negative and the function is strictly decreasing.

The proof uses multiple Abel-partial summation. We can use this result to
obtain the following statement about the boundary behavior of power series.

######  Theorem 3.2.

Let $(a_{n})_{n\in\mathbb{N}}$ be a sequence of real numbers that is strictly
4-fold monotone. Assume that the series

$\displaystyle P(z):=\sum_{n=1}^{\infty}a_{n}z^{n}$

converges absolutely on $|z|\leq 1$. Then the function $t\to|P(e^{it})|$ is
strictly decreasing on the interval $(0,\pi)$.

###### Proof.

We follow Fejer [18]. First, one has

$\displaystyle\left|P\left(e^{it}\right)\right|^{2}=\sum_{n=1}^{\infty}A_{n}\cos(nt)$

where

$\displaystyle A_{n}:=\sum_{j=1}^{\infty}a_{j}a_{n+j}.$

As the sequence $(a_{n})_{n\in\mathbb{N}}$ is strictly 4-fold monotone, so is
$(A_{n})_{n\in\mathbb{N}}$. The theorem now follows from Theorem 3.1. ∎

The concatenation of real part and root in (3.1) is not easy to handle in
monotonicity questions at first. Therefore we use a trick and split the
problem into argument and absolute value. The idea of proof of the following
Lemma is adapted from Boyer and Parry [6] (Appendix A, Lemma 7).

######  Lemma 3.3.

Let $k\geq 2$. Then the function $\theta\mapsto|\mathrm{Li}_{k}(e^{i\theta})|$
is decreasing on $(0,\pi)$.

###### Proof.

By Theorem 3.2, it is sufficient to show that the sequence
$a_{n}:=\frac{1}{n^{k}}$ is strictly 4-fold monotone. To verify this, use the
identity

$\displaystyle\frac{1}{n^{k}}=\frac{1}{\Gamma(k)}\int_{0}^{1}x^{n-1}(-\log(x))^{k-1}dx$

to find that for $n\geq 1$,

$\displaystyle\Delta^{m}\left(\frac{1}{n^{k}}\right)$
$\displaystyle=\sum_{j=0}^{m}{m\choose
j}\frac{(-1)^{j}}{(n+j)^{k}}=\frac{1}{\Gamma(k)}\int_{0}^{1}\sum_{j=0}^{m}{m\choose
j}(-1)^{j}x^{n+j-1}(-\log(x))^{k-1}dx$
$\displaystyle=\frac{1}{\Gamma(k)}\int_{0}^{1}x^{n-1}\sum_{j=0}^{m}{m\choose
j}(-1)^{j}x^{j}(-\log(x))^{k-1}dx=\frac{1}{\Gamma(k)}\int_{0}^{1}x^{n-1}(1-x)^{m}(-\log(x))^{k-1}dx>0.$

∎

For the convenience of the reader we provide detailed proofs for Proposition
3.4 and Corollary 3.5. Note that these results have been proved by Boyer and
Parry [5].

######  Proposition 3.4.

The function $\theta\mapsto\mathrm{Re}\left(\mathrm{Li}_{3}(e^{2\pi
i\theta})^{\frac{1}{3}}\right)$ is decreasing on $\left(0,\frac{1}{2}\right).$

###### Proof.

First we argue that the function
$t\mapsto\mathrm{Arg}(\mathrm{Li}_{3}(e^{it}))$ is increasing on the interval
$(0,\pi)$. By a result of Lewis [24], any polylogarithm $\mathrm{Li}_{3}(z)$
is star-like and univalent on the unit disk. Now, by Theorem 2.10 of [16] on
p. 41, this is equivalent to

$\displaystyle\frac{d}{dt}\left(\mathrm{Arg}(\mathrm{Li}_{3}(\rho
e^{it}))\right)=\mathrm{Re}\left({\frac{\rho
e^{it}\mathrm{Li}_{3}^{\prime}(\rho e^{it})}{\mathrm{Li}_{3}(\rho
e^{it})}}\right)>0,\qquad 0<\rho<1.$

By taking limits, we conclude

$\displaystyle\mathrm{Re}\left({\frac{e^{it}\mathrm{Li}_{3}^{\prime}(e^{it})}{\mathrm{Li}_{3}(e^{it})}}\right)\geq
0.$

By Lemma 3.3, we know that the function $t\mapsto|\mathrm{Li}_{3}(e^{it})|$ is
decreasing on the interval $(0,\pi)$. Finally, as we have

$\displaystyle\mathrm{Re}\left(\sqrt[3]{\mathrm{Li}_{3}(e^{it})}\right)=\left|\mathrm{Li}_{3}(e^{it})\right|^{\frac{1}{3}}\cos\left(\frac{1}{3}\mathrm{Arg}(\mathrm{Li}_{3}(e^{it}))\right),$

the function on the left hand side is strictly decreasing as the product of a
strictly decreasing function on the interval $(0,\pi)$. ∎

The following corollary determines the major arcs for the proof of Theorem
1.1.

######  Corollary 3.5.

We have

$\displaystyle L(\theta):=\max_{k\geq
1}\frac{\mathrm{Re}\left(\sqrt[3]{\mathrm{Li}_{3}\left(e^{2\pi
ik\theta}\right)}\right)}{k}=\begin{cases}\mathrm{Re}\left(\sqrt[3]{\mathrm{Li}_{3}(e^{2\pi
i\theta})}\right),&\qquad 0\leq\theta\leq\theta_{12},\\\
\frac{\mathrm{Re}\left(\sqrt[3]{\mathrm{Li}_{3}(e^{4\pi
i\theta})}\right)}{2},&\qquad\theta_{12}<\theta\leq\frac{1}{2},\end{cases}$

where $\theta_{12}=0.47585\ldots$ is the unique solution of
$\mathrm{Re}\left(\sqrt[3]{\mathrm{Li}_{3}(e^{2\pi
i\theta})}\right)=\frac{\mathrm{Re}\left(\sqrt[3]{\mathrm{Li}_{3}(e^{4\pi
i\theta})}\right)}{2}$ in the interval $[0,\frac{1}{2}]$. Additionally, we
have $L(\frac{1}{3})>L(\theta)$ for all $\frac{1}{3}<\theta\leq\frac{1}{2}$.

###### Proof.

Put $f_{k}(\theta):=\frac{\mathrm{Re}(\sqrt[3]{\mathrm{Li}_{3}(e^{2\pi
ik\theta})})}{k}$. By Proposition 3.4 we see with

$\displaystyle
f_{1}\left(\frac{1}{4}\right)=0.8391145\ldots>f_{2}(0)=\frac{\sqrt[3]{\zeta(3)}}{2}=0.531632\ldots\geq
f_{k}(0)$

that the corollary holds for $0\leq\theta\leq\frac{1}{4}$. Now we consider the
case $\frac{1}{4}\leq\theta\leq\frac{1}{2}$. By symmetry we argue that
$f_{2}(\theta)$ is increasing in the interval $[\frac{1}{4},\frac{1}{2}]$, so
the equation $f_{1}(\theta)=f_{2}(\theta)$ only has one solution
$\frac{1}{4}<\theta_{12}$ in the interval $[0,\frac{1}{2}]$. It is clear that
$f_{2}(\theta)>f_{1}(\theta)$ for $\theta_{12}<\theta\leq\frac{1}{2}$ and
$f_{2}(\theta)<f_{1}(\theta)$ for $\frac{1}{4}\leq\theta\leq\theta_{12}$.
Hence, we are done after showing $f_{1}(\theta_{12})>f_{k}(0)$ for all $k\geq
3$. This can be checked numerically, we have
$f_{1}(\theta_{12})=0.5212\ldots$, but on the other hand

$\displaystyle f_{3}(0)=\frac{\sqrt[3]{\zeta(3)}}{3}=0.3544\ldots.$

Finally, note that $L(\frac{1}{3})=0.7304\dots$. For
$\theta\in(\frac{1}{3},\theta_{12})$ the assertion $L(\frac{1}{3})>L(\theta)$
is clear by monotonicity. For $\theta\in[\theta_{12},\frac{1}{2}]$ it follows
by monotonicity and $L(\frac{1}{2})=0.5316\cdots<L(\frac{1}{3})$. ∎

We now make the same considerations as above for the case of genuine regular
plane partitions. The decisive function here is of type

$\theta\mapsto\mathrm{Re}\left(\left(\zeta(3)-\mathrm{Li}_{3}\left(e^{i\theta}\right)\right)^{\frac{1}{3}}\right).$

######  Lemma 3.6.

The function

$\theta\mapsto\mathrm{Arg}\left(\zeta(3)-\mathrm{Li}_{3}\left(e^{i\theta}\right)\right)$

is increasing on $(0,\pi)$ and surjects onto $(-\frac{\pi}{2},0)$.

###### Proof.

As $\mathrm{Li}_{3}(z)$ is a convex mapping, the curve
$(0,2\pi)\to\mathbb{C}$, $\theta\mapsto\mathrm{Li}_{3}(e^{i\theta})$ bounds a
convex set. Starting at $\zeta(3)$, the curve runs with monotonically
increasing argument over the upper half plane to the point
$-\frac{3}{4}\zeta(3)$ (halfway at $\theta=\pi$). As a result, the curve
$(0,\pi)\to\mathbb{C}$, $\theta\mapsto\zeta(3)-\mathrm{Li}_{3}(e^{i\theta})$
starts at the origin222Note that the fact that the origin is contained in the
closure of the image is necessary. and runs with monotonically increasing
argument over the lower half plane to the point $\frac{7}{4}\zeta(3)$.
Finally, we need to show that the map surjects. It is clear that

$\displaystyle\lim_{\theta\to\pi^{-}}\mathrm{Arg}\left(\left(\zeta(3)-\mathrm{Li}_{3}(e^{i\theta})\right)\right)=0.$

It remains to show that

$\lim_{\theta\to
0^{+}}\mathrm{Arg}\left(\left(\zeta(3)-\mathrm{Li}_{3}(e^{i\theta})\right)\right)=-\frac{\pi}{2}.$

We compute for $\theta$ near 0,

$\mathrm{Arg}\left(\left(\zeta(3)-\mathrm{Li}_{3}(e^{i\theta})\right)\right)=\arctan\left(\frac{-\sum_{n\geq
1}\frac{\sin(n\theta)}{n^{3}}}{\zeta(3)-\sum_{n\geq
1}\frac{\cos(n\theta)}{n^{3}}}\right).$

As $\theta\to 0$, we have $\sum_{n\geq
1}\frac{\sin(n\theta)}{n^{3}}\sim\zeta(2)\theta$ and from Theorem 2.5
(choosing $\vartheta=0$), it follows that

$\sum_{n\geq
1}\frac{\cos(n\theta)}{n^{3}}\sim\zeta(3)-\frac{\theta^{2}\log\left(\frac{1}{\theta}\right)}{2},$

whence

$\lim_{\theta\to
0^{+}}\mathrm{Arg}\left(\left(\zeta(3)-\mathrm{Li}_{3}(e^{i\theta})\right)\right)=\lim_{\theta\to
0^{+}}\arctan\left(-\frac{2\zeta(2)}{\theta\log\left(\frac{1}{\theta}\right)}\right)=-\frac{\pi}{2},$

proving the lemma.

∎

######  Proposition 3.7.

The function

$\theta\mapsto\mathrm{Re}\left(\left(\zeta(3)-\mathrm{Li}_{3}\left(e^{i\theta}\right)\right)^{\frac{1}{3}}\right)$

is strictly increasing on $(0,\pi)$.

###### Proof.

First we show that the sequence

$\displaystyle
A_{n}:=\frac{\zeta(3)}{n^{3}}-\sum_{k=1}^{\infty}\frac{1}{k^{3}(k+n)^{3}}=\sum_{k=1}^{\infty}\frac{1}{k^{3}}\left(\frac{1}{n^{3}}-\frac{1}{(k+n)^{3}}\right)$

is 4-fold monotone for $n\in\mathbb{N}$. We find

$\displaystyle\frac{1}{n^{3}}-\frac{1}{(k+n)^{3}}=\frac{1}{2}\int_{0}^{1}\left(x^{n-1}-x^{k+n-1}\right)\log(x)^{2}dx.$

Hence, for all $m\in\mathbb{N}_{0}$

$\displaystyle\Delta^{m}\left(\frac{1}{n^{3}}-\frac{1}{(k+n)^{3}}\right)$
$\displaystyle=\frac{1}{2}\int_{0}^{1}\sum_{j=0}^{m}\binom{m}{j}(-1)^{j}\left(x^{n+j-1}-x^{k+n+j-1}\right)\log(x)^{2}dx$
$\displaystyle=\frac{1}{2}\int_{0}^{1}\left(x^{n-1}-x^{k+n-1}\right)\sum_{j=0}^{m}\binom{m}{j}(-1)^{j}x^{j}\log(x)^{2}dx$
$\displaystyle=\frac{1}{2}\int_{0}^{1}x^{n-1}\left(1-x^{k}\right)\left(1-x\right)^{m}\log(x)^{2}dx>0.$

We conclude, that the $A_{n}$ are 4-fold monotone. Now, we have

$\displaystyle\left|\zeta(3)-\mathrm{Li}_{3}\left(e^{i\theta}\right)\right|^{2}=\zeta(3)^{2}+\zeta(6)-2\sum_{n=1}^{\infty}A_{n}\cos(n\theta).$

To show that this function is increasing on $(0,\pi)$ it suffices to show that
$\theta\mapsto\sum_{n=1}^{\infty}A_{n}\cos(n\theta)$ is decreasing on
$(0,\pi)$, which follows from Theorem 3.2 and the fact that the sequence
$A_{n}$ is 4-fold monotone. Now we have

$\displaystyle\mathrm{Re}\left(\sqrt[3]{\zeta(3)-\mathrm{Li}_{3}\left(e^{i\theta}\right)}\right)=\left|\zeta(3)-\mathrm{Li}_{3}\left(e^{i\theta}\right)\right|^{\frac{1}{3}}\cos\left(\frac{1}{3}\mathrm{Arg}\left(\zeta(3)-\mathrm{Li}_{3}\left(e^{i\theta}\right)\right)\right).$

By Lemma 3.6, the cosine term is increasing, as the inner function is
increasing in the interval $(-\frac{\pi}{6},0)$, so the left hand side is a
product of two increasing functions. This proves the proposition. ∎

The following proposition determines the major arc in the proof of Theorem
1.3.

######  Proposition 3.8.

For any root of unity $\zeta\neq 1$ and all integers $k\geq 2$,

$\mathrm{Re}\left(\frac{\left(\zeta(3)-\mathrm{Li}_{3}\left(\zeta^{k}\right)\right)^{\frac{1}{3}}}{k}\right)<\mathrm{Re}\left(\left(\zeta(3)-\mathrm{Li}_{3}(\zeta)\right)^{\frac{1}{3}}\right).$

###### Proof.

Let $\theta_{1}=0.23792\dots$ be the unique solution in $(0,\pi)$ to

$\frac{(7\zeta(3))^{\frac{1}{3}}}{2^{\frac{5}{3}}}=\mathrm{Re}\left(\left(\zeta(3)-\mathrm{Li}_{3}\left(e^{i\theta}\right)\right)^{\frac{1}{3}}\right),$

which exists by Proposition 3.7. Then for $\theta\in(\theta_{1},\pi]$, we have

$\mathrm{Re}\left(\frac{\left(\zeta(3)-\mathrm{Li}_{3}\left(e^{ik\theta}\right)\right)^{\frac{1}{3}}}{k}\right)\leq\mathrm{Re}\left(\frac{\left(\zeta(3)-\mathrm{Li}_{3}\left((-1)\right)\right)^{\frac{1}{3}}}{k}\right)=\frac{(7\zeta(3))^{\frac{1}{3}}}{2^{\frac{2}{3}}k}<\mathrm{Re}\left(\left(\zeta(3)-\mathrm{Li}_{3}\left(e^{i\theta}\right)\right)^{\frac{1}{3}}\right).$

Now let $\theta\in(0,\theta_{1}]$ and write

$r_{\theta}e^{3i\psi_{\theta}}:=\zeta(3)-\mathrm{Li}_{3}(e^{i\theta}).$

We want to show

$\frac{r_{k\theta}^{\frac{1}{3}}\cos(\psi_{k\theta})}{k}<r_{\theta}^{\frac{1}{3}}\cos(\psi_{\theta}),$

or equivalently,

$\frac{\cos(\psi_{k\theta})}{\cos(\psi_{\theta})}<k\left(\frac{r_{\theta}}{r_{k\theta}}\right)^{\frac{1}{3}}.$
(3.2)

To show the above, we bound the right-hand side from below and the left-hand
side from above. Beginning with the right-hand side, we use the mean value
theorem to find $\xi_{\theta}\in(0,\theta)$ so that

$\zeta(3)-\mathrm{Li}_{3}(e^{i\theta})=\theta\frac{d}{dx}\left(\zeta(3)-\mathrm{Li}_{3}(e^{ix})\right)\rvert_{x=\theta}=-i\theta\mathrm{Li}_{2}(e^{i\xi_{\theta}}).$

Using the fact that $x\mapsto|\mathrm{Li}_{2}(e^{ix})|$ is decreasing (see
Lemma 3.3), we have for $k\geq 2,$

$k\left(\frac{r_{\theta}}{r_{k\theta}}\right)^{\frac{1}{3}}=k\left|\frac{\mathrm{Li}_{2}\left(e^{i\xi_{\theta}}\right)}{k\mathrm{Li}_{2}\left(e^{i\xi_{k\theta}}\right)}\right|^{\frac{1}{3}}\geq
k^{\frac{2}{3}}\left(\frac{|\mathrm{Li}_{2}\left(e^{i\theta_{1}}\right)|}{\frac{\pi^{2}}{6}}\right)^{\frac{1}{3}}\geq
1.5.$ (3.3)

For the left-hand side in (3.2), we first note that by Lemma 3.6,
$\psi_{\theta}$ is increasing and negative on $(0,\theta_{1}].$ To find
$\lim_{\theta\to 0^{+}}\psi_{\theta}$, we first observe

$\lim_{\theta\to
0^{+}}\frac{\zeta(3)-\mathrm{Li}_{3}\left(e^{i\theta}\right)}{\theta}=\lim_{\theta\to
0^{+}}\left(-i\mathrm{Li}_{2}\left(e^{i\theta}\right)\right)=-i\frac{\pi^{2}}{6},$

from which it follows that, as $\theta\to 0^{+}$

$3\psi_{\theta}\sim\arg\left(-i\frac{\pi^{2}}{6}\theta\right)=-\frac{\pi}{2}\implies\lim_{\theta\to
0^{+}}\psi_{\theta}=-\frac{\pi}{6}.$

Thus, for $\theta\in(0,\theta_{1}],$

$\frac{\cos(\psi_{k\theta})}{\cos(\psi_{\theta})}\leq\frac{1}{\cos\left(-\frac{\pi}{6}\right)}=\frac{2}{\sqrt{3}}<1.2,$

which with (3.3) proves (3.2). The proposition follows. ∎

##  4\. Proof of Theorems 1.1 and 1.3

We now apply the circle method in the form of [7] and [27] (see also [1, Ch.
5]). Choose $N=\lfloor\delta n^{\frac{1}{3}}\rfloor$ where $\delta$ will be
taken fixed and sufficiently small in the course of the proof. Let
$\mathcal{F}_{N}$ be the Farey sequence of order $N$ and let
$\theta_{h,k}^{\prime}$ and $\theta_{h,k}^{\prime\prime}$ be the differences
between $\frac{h}{k}$ and the mediants; i.e., if
$\frac{h^{\prime}}{k^{\prime}},\frac{h}{k},\frac{h^{\prime\prime}}{k^{\prime\prime}}$
are three consecutive members of $\mathcal{F}_{N},$ then

$\theta_{h,k}^{\prime}:=\frac{h}{k}-\frac{h^{\prime}+h}{k^{\prime}+k},\qquad\theta_{h,k}^{\prime\prime}:=\frac{h}{k}-\frac{h+h^{\prime\prime}}{k+k^{\prime\prime}}.$

Let

$t_{n}:=\begin{cases}\frac{2^{\frac{1}{3}}\mathrm{Li}_{3}\left(\zeta_{b}^{a}\right)^{\frac{1}{3}}}{n^{\frac{1}{3}}}&\text{if
$\frac{a}{b}<\theta_{12}$,}\\\
\frac{\mathrm{Li}_{3}\left(\zeta_{b}^{2a}\right)^{\frac{1}{3}}}{2^{\frac{2}{3}}n^{\frac{1}{3}}}&\text{if
$\theta_{12}<\frac{a}{b}\leq\frac{1}{2}$,}\end{cases}$

and note that $t_{n}$ is complex. We apply Cauchy’s integral theorem and add
and subtract the main term to obtain

$\displaystyle T_{n}\left(\zeta_{b}^{a}\right)$
$\displaystyle=\sum_{\begin{subarray}{c}\gcd(h,k)=1\\\
\frac{h}{k}\in\mathcal{F}_{N}\end{subarray}}\zeta_{k}^{nh}\int_{-\theta_{h,k}^{\prime}}^{\theta_{h,k}^{\prime\prime}}\exp\left(\log
PP\left(\zeta_{b}^{a};e^{-t_{n}+2\pi i\theta}\right)+nt_{n}-2\pi
in\theta\right)d\theta$ $\displaystyle=\sum_{\begin{subarray}{c}\gcd(h,k)=1\\\
\frac{h}{k}\in\mathcal{F}_{N}\end{subarray}}\zeta_{k}^{nh}\int_{-\theta_{h,k}^{\prime}}^{\theta_{h,k}^{\prime\prime}}\exp\left(\frac{\mathrm{Li}_{3}\left(\zeta_{b}^{ka}\right)}{k^{3}(t_{n}-2\pi
i\theta)^{2}}+nt_{n}-2\pi in\theta+E_{h,k}(\zeta_{b}^{a};t_{n}-2\pi
i\theta)\right)d\theta,$

where we define the error terms,

$E_{h,k}(z;t):=\log
PP\left(z;e^{-t}\right)-\frac{\mathrm{Li}_{3}\left(z^{k}\right)}{k^{3}t^{2}}.$

Corollary 3.5 will imply that the major arc will correspond either to
$\frac{h}{k}=\frac{0}{1}$ or $\frac{h}{k}=\frac{1}{2}$ depending on whether
$\frac{a}{b}<\theta_{12}$ or $\theta_{12}<\frac{a}{b}<\frac{1}{2}$,
respectively. We need to estimate $E_{h,k}$ up to $o(1)$ on the major arcs and
bound $E_{h,k}$ by $O(N^{2})$ on the minor arcs. The former is a
straightforward application of Euler–Maclaurin summation, while the latter is
one of the core difficulties we overcome in this article following the
techniques of [7].

###  4.1. Bounds for the error terms

######  Lemma 4.1.

  1. (1)

If $\frac{a}{b}<\theta_{12},$ then uniformly for
$\theta\in[-\theta_{0,1}^{\prime},\theta_{0,1}^{\prime\prime}]$ we have

$E_{0,1}\left(\zeta_{b}^{a};t_{n}-2\pi
i\theta\right)=\frac{1}{12}\log\left(1-\zeta_{b}^{a}\right)+O\left(\frac{1}{n^{\frac{2}{3}}}\right).$

  2. (2)

Uniformly for $\theta\in[-\theta_{0,1}^{\prime},\theta_{0,1}^{\prime\prime}]$
we have

$E_{0,1}\left(1;t_{n}-2\pi i\theta\right)=\frac{1}{12}\log\left(t_{n}-2\pi
i\theta\right)+\zeta^{\prime}(-1)+O\left(\frac{1}{n^{\frac{2}{3}}}\right).$

  3. (3)

If $\theta_{12}<\frac{a}{b}<\frac{1}{2},$ then uniformly for
$\theta\in[-\theta_{1,2}^{\prime},\theta_{1,2}^{\prime\prime}]$ we have

$E_{1,2}\left(\zeta_{b}^{a};t_{n}-2\pi
i\theta\right)=\frac{1}{6}\log\left(1-\zeta_{b}^{a}\right)-\frac{1}{12}\log\left(1+\zeta_{b}^{a}\right)+O\left(\frac{1}{n^{\frac{2}{3}}}\right).$

  4. (4)

Uniformly for $\theta\in[-\theta_{1,2}^{\prime},\theta_{1,2}^{\prime\prime}]$,
we have

$E_{1,2}(-1;t_{n}-2\pi i\theta)=-\frac{1}{12}\log(t_{n}-2\pi
i\theta)-\zeta^{\prime}(-1)++O\left(\frac{1}{n^{\frac{2}{3}}}\right).$

  5. (5)

Uniformly for $k\leq N$ and
$\theta\in[-\theta_{h,k}^{\prime},\theta_{h,k}^{\prime\prime}]$, we have

$E_{h,k}\left(\zeta_{b}^{a};t_{n}-2\pi i\theta\right)=O\left(N^{2}\right).$

We begin by rewriting $E_{h,k}$.

######  Lemma 4.2.

For $\mathrm{Re}(t)>0$

$\displaystyle
E_{h,k}\left(\zeta_{b}^{a};t\right)=\sum_{\begin{subarray}{c}1\leq j\leq k\\\
1\leq m\leq bk\end{subarray}}\zeta_{b}^{ma}\zeta_{k}^{mjh}k^{2}t\sum_{\ell\geq
0}g_{j,k}\left(t\left(bk^{2}\ell+km\right)\right),$

where

$\displaystyle
g_{j,k}(w):=\frac{e^{-w-\frac{j}{k}w}}{w(1-e^{-w})^{2}}+\frac{j}{k}\frac{e^{-\frac{j}{k}w}}{w(1-e^{-w})}-\frac{1}{w^{3}}.$

###### Proof.

Taking logarithms in (1.2) gives

$\displaystyle\operatorname{Log}\left(PP(\zeta_{b}^{a};\zeta_{k}^{h}e^{-t})\right)$
$\displaystyle=-\sum_{\nu\geq
1}\nu\operatorname{Log}\left(1-\zeta_{b}^{a}\zeta_{k}^{h\nu}e^{-\nu
t}\right)=\sum_{\nu\geq 1}\sum_{\ell\geq
1}\nu\frac{\zeta_{b}^{a\ell}\zeta_{k}^{h\ell\nu}e^{-\nu\ell t}}{\ell}.$

Letting $\ell\mapsto bk\ell+m$ and $\nu\mapsto\nu k+j$ we obtain

$\displaystyle\operatorname{Log}\left(PP(\zeta_{b}^{a};\zeta_{k}^{h}e^{-t})\right)=$
$\displaystyle\sum_{\begin{subarray}{c}1\leq j\leq k\\\ 1\leq m\leq
bk\end{subarray}}\zeta_{b}^{ma}\zeta_{k}^{mjh}\sum_{\ell\geq
0}\frac{e^{-j(bk\ell+m)t}}{bk\ell+m}\sum_{\nu\geq 0}(\nu k+j)e^{-\nu
k(bk\ell+m)t}$ $\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}1\leq
j\leq k\\\ 1\leq m\leq
bk\end{subarray}}\zeta_{b}^{ma}\zeta_{k}^{mjh}k^{2}t\sum_{\ell\geq
0}\frac{e^{-j(bk\ell+m)t}}{k(bk\ell+m)t}\left(\frac{e^{-k(bk\ell+m)t}}{\left(1-e^{-k(bk\ell+m)t}\right)^{2}}+\frac{j}{k}\frac{1}{1-e^{-k(bk\ell+m)t}}\right).$

For the right term in $g_{j,k}$, we compute (using $\gcd(h,k)=1$)

$\displaystyle-\sum_{\begin{subarray}{c}1\leq j\leq k\\\ 1\leq m\leq
bk\end{subarray}}\zeta_{b}^{ma}\zeta_{k}^{mjh}k^{2}t\sum_{\ell\geq
0}\frac{1}{t^{3}(bk^{2}\ell+mk)^{3}}$
$\displaystyle=-\frac{1}{t^{2}}\sum_{\begin{subarray}{c}1\leq m\leq bk\\\
m\equiv 0\ \,\left(\mathrm{mod}\,k\right)\end{subarray}}\sum_{\ell\geq
0}\frac{\zeta_{b}^{ma}}{(bk\ell+m)^{3}}$
$\displaystyle=-\frac{1}{t^{2}}\sum_{\ell\geq
0}\sum_{\nu=1}^{b}\frac{\zeta_{b}^{\nu ka}}{(bk\ell+k\nu)^{3}}$
$\displaystyle=-\frac{1}{k^{3}t^{2}}\mathrm{Li}_{3}\left(\zeta_{b}^{ak}\right).\qed$

###### Proof of Lemma 4.1.

Throughout this section we write $t_{\theta}:=t_{n}-2\pi i\theta.$ Recall (see
[1, Ch. 5]) that
$\theta_{h,k}^{\prime},\theta_{h,k}^{\prime\prime}\asymp\frac{1}{kN}\ll\frac{1}{n^{\frac{1}{3}}}$.
It follows that for some cone $D_{\psi}$ described in Theorem 2.5, we have
$t_{\theta}\in D_{\psi}$ for all
$\theta\in[\theta_{h,k}^{\prime},\theta_{h,k}^{\prime\prime}]$. Furthermore,
as $w\to 0$, we have

$g_{j,k}(w)=\left(-\frac{j^{2}}{2k^{2}}-\frac{1}{12}+\frac{j}{2k}\right)w^{-1}+O(1),$

and thus Theorem 2.5 applies to those $\theta$ in Lemma 4.1. Hence with

$b_{-1}=b_{-1}\left(\frac{j}{k}\right):=-\frac{j^{2}}{2k^{2}}-\frac{1}{12}+\frac{j}{2k},$

we have

$\displaystyle\sum_{\ell\geq
0}g_{j,k}(t_{\theta}(bk^{2}\ell+km))=\frac{b_{-1}\operatorname{Log}\left(\frac{1}{bk^{2}t_{\theta}}\right)}{bk^{2}t_{\theta}}+\frac{b_{-1}C_{\frac{m}{bk}}}{bk^{2}t_{\theta}}+\frac{1}{bk^{2}t_{\theta}}\int^{\infty}_{0}\left(g_{j,k}(x)-\frac{b_{-1}e^{-x}}{x}\right)dx+O(t_{\theta}k^{2}),$
(4.1)

with

$C_{\frac{m}{bk}}=\left(1-\frac{m}{bk}\right)\sum_{r\geq
0}\frac{1}{(r+\frac{m}{bk})(r+1)}=O(1).$

Here, $b_{-1}(1)=-\frac{1}{12}$, so for part (1) we have

$\displaystyle E_{0,1}\left(\zeta_{b}^{a};t_{\theta}\right)$
$\displaystyle=\sum_{1\leq m\leq
b}\zeta_{b}^{ma}t_{\theta}\left(-\frac{\operatorname{Log}\left(\frac{1}{bt_{\theta}}\right)}{12bt_{\theta}}-\frac{C_{\frac{m}{b}}}{12bt_{\theta}}+\frac{1}{bt_{\theta}}\int_{0}^{\infty}\left(g_{1,1}(x)+\frac{e^{-x}}{12x}\right)dx+O\left(t_{\theta}\right)\right)$
$\displaystyle=-\frac{1}{12b}\sum_{1\leq m\leq
b}\zeta_{b}^{ma}C_{\frac{m}{b}}+O\left(t_{\theta}^{2}\right)$
$\displaystyle=\frac{1}{12}\log\left(1-\zeta_{b}^{a}\right)+O\left(\frac{1}{n^{\frac{2}{3}}}\right),$

as claimed. The proof of parts (2)-(4) are similar.

For part (5), we follow [7] and separate the cases for small and large $k$.
First let $k\leq n^{\frac{1}{6}}$ and define

$S_{a,b}:=\\{(h,k)\in\mathbb{N}^{2}:ak+bjh\equiv 0\
\,\left(\mathrm{mod}\,bk\right),\ \text{for some $1\leq j\leq k$}\\}.$

It is easy to see that if $(h,k)\in S_{a,b}$ then there is only one
corresponding $j$; let this be $j_{0}=j_{0}(h,k).$ Thus, plugging (4.1) into
Lemma 4.2, we obtain

$\displaystyle E_{h,k}(\zeta_{b}^{a};t_{\theta})$
$\displaystyle=\sum_{\begin{subarray}{c}1\leq j\leq k\\\ j\neq j_{0}\\\ 1\leq
m\leq
bk\end{subarray}}\frac{\zeta_{b}^{ma}\zeta_{k}^{mjk}}{b}\left(b_{-1}\left(\frac{j}{k}\right)C_{\frac{m}{bk}}+O(t_{\theta}k^{2})\right)+1_{(h,k)\in
S_{a,b}}\left(kb_{-1}\left(\frac{j_{0}}{k}\right)\operatorname{Log}\left(\frac{1}{bk^{2}t_{\theta}}\right)\right.$
$\displaystyle\qquad\left.+kb_{-1}\left(\frac{j_{0}}{k}\right)+k\int^{\infty}_{0}\left(g_{j_{0},k}(x)-\frac{b_{-1}\left(\frac{j_{0}}{k}\right)e^{-x}}{x}\right)dx+O(t_{\theta}k^{3})\right)$
$\displaystyle=O\left(k^{2}+t_{\theta}k^{4}\right),$

and this is certainly $O(N^{2})$.

Now suppose that $n^{\frac{1}{6}}\leq k\leq N$. We consider two separate cases
for the sums in Lemma 4.2, $\ell=0$ and $\ell\geq 1$. First let $\ell=0$. Note
that we have $\phi_{\frac{j}{k}}(z)=zg_{j,k}(z)$, where $\phi_{a}(z)$ was
defined in (2.1). Hence, we find

$\displaystyle\sum_{\begin{subarray}{c}1\leq j\leq k\\\ 1\leq m\leq
bk\end{subarray}}\zeta_{b}^{ma}\zeta_{k}^{mjh}k^{2}tg_{j,k}\left(ktm\right)$
$\displaystyle=k\sum_{\begin{subarray}{c}1\leq j\leq k\\\ 1\leq m\leq
bk\end{subarray}}\zeta_{b}^{ma}\zeta_{k}^{mjh}\frac{\phi_{\frac{j}{k}}(ktm)}{m}.$

We now adapt the methodology of [7] and prove the following.

$\displaystyle k\sum_{\begin{subarray}{c}1\leq j\leq k\\\ 1\leq m\leq
bk\end{subarray}}\frac{\zeta_{b}^{ma}\zeta_{k}^{mjh}}{m}\phi_{\frac{j}{k}}(kt_{\theta}m)=O\left(k^{2}\right).$
(4.2)

Put $x:=kt_{\theta}$ and $a:=\frac{j}{k}$. Note that we have $0<a\leq 1$. We
have $\mathrm{Re}(x)>0$ and $\frac{|x|}{\mathrm{Re}(x)}\ll 1$ uniformly in
$k$. We use partial summation and split the sum into two parts.

$\displaystyle\sum_{\begin{subarray}{c}1\leq m\leq bk\\\ 1\leq j\leq
k\end{subarray}}=\sum_{1\leq j\leq
k}\left(\sum_{0<m\leq\min\left\\{bk,\frac{1}{|x|}\right\\}}+\sum_{\min\left\\{bk,\frac{1}{|x|}\right\\}<m\leq
bk}\right).$

In the case $|x|>1$, the first sum is empty, so we can assume $|x|\leq 1$. We
first find with Proposition 2.4

$\displaystyle\sum_{0<m\leq\min\left\\{bk,\frac{1}{|x|}\right\\}}\zeta_{bk}^{m(ak+hjb)}\frac{1}{m}\phi_{a}(xm)=G_{\min\left\\{bk,\frac{1}{|x|}\right\\}}\left(\frac{a}{b}+\frac{hj}{k}\right)\phi_{a}\left(x\min\left\\{bk,\Big{\lfloor}\frac{1}{|x|}\Big{\rfloor}\right\\}\right)$
$\displaystyle+\sum_{m\leq\min\left\\{bk,\frac{1}{|x|}\right\\}-1}G_{m}\left(\frac{a}{b}+\frac{hj}{k}\right)\left(\phi_{a}(mx)-\phi_{a}((m+1)x)\right).$

It follows with Lemma 2.7

$\displaystyle k\left|\sum_{1\leq j\leq
k}\sum_{m\leq\min\left\\{bk,\frac{1}{|x|}\right\\}}\zeta_{b}^{ma}\zeta_{k}^{mhj}\cdot\frac{1}{m}\phi_{a}(mx)\right|$
$\displaystyle\leq k\sum_{1\leq j\leq
k}\left|G_{\min\left\\{bk,\frac{1}{|x|}\right\\}}\left(\frac{a}{b}+\frac{hj}{k}\right)\right|\left|\phi_{a}\left(x\min\left\\{bk,\Big{\lfloor}\frac{1}{|x|}\Big{\rfloor}\right\\}\right)\right|$
$\displaystyle\hskip 85.35826pt+k\left|\sum_{1\leq j\leq
k}\sum_{0<m\leq\min\left\\{bk,\frac{1}{|x|}\right\\}}G_{m}\left(\frac{a}{b}+\frac{hj}{k}\right)\left(\phi_{a}(mx)-\phi_{a}((m+1)x)\right)\right|$
$\displaystyle\ll k\sum_{1\leq j\leq
k}\left|G_{\min\left\\{bk,\frac{1}{|x|}\right\\}}\left(\frac{a}{b}+\frac{hj}{k}\right)\right|+k\sum_{1\leq
j\leq k}\max_{1\leq
m\leq\min\left\\{bk,\frac{1}{|x|}\right\\}}\left|G_{m}\left(\frac{a}{b}+\frac{hj}{k}\right)\right|\sum_{0<m\leq\frac{1}{|x|}}|x|=O\left(k^{2}\right),$

where we have used Lemma 2.10 and $G_{bk}(1)=H_{bk}=O(\log(k))$ in the last
step. Similarly, we find with Lemma 2.8 (without loss of generality we assume
$\frac{1}{|x|}<bk$)

$\displaystyle\left|\sum_{1\leq j\leq k}\sum_{\frac{1}{|x|}<m\leq
bk}\zeta_{b}^{ma}\zeta_{k}^{mhj}\cdot\frac{1}{m}\phi_{a}(mx)\right|$
$\displaystyle\ll\sum_{1\leq j\leq
k}\left|G_{bk}\left(\frac{a}{b}+\frac{hj}{k}\right)\right|\left|\phi_{a}\left(xbk\right)\right|+\left|\sum_{1\leq
j\leq k}\sum_{\frac{1}{|x|}<m\leq
bk}G_{m}\left(\frac{a}{b}+\frac{hj}{k}\right)\left(\phi_{a}(mx)-\phi_{a}((m+1)x)\right)\right|$
$\displaystyle\ll O(k)+\sum_{1\leq j\leq k}\max_{\frac{1}{|x|}\leq m\leq
bk}\left|G_{m}\left(\frac{a}{b}+\frac{hj}{k}\right)\right|$
$\displaystyle\hskip 28.45274pt\times\sum_{\frac{1}{|x|}<m\leq
bk}\left(\frac{1}{|x|^{2}}\left(\frac{1}{m^{2}}-\frac{1}{(m+1)^{2}}\right)+|x|e^{-m\mathrm{Re}(x)}+|x|e^{-m\mathrm{Re}(x)(1+a)}+a^{2}|x|e^{-am\mathrm{Re}(x)}\right).$

Note that we uniformly have $\phi_{a}(xbk)\ll 1$ (as $1\ll|xbk|$ and $x$ is
part of a fixed cone $|\mathrm{Arg}(x)|\leq\frac{\pi}{2}-\eta$) as well as

$\displaystyle\sum_{\frac{1}{|x|}<m\leq
bk}\frac{1}{|x|^{2}}\left(\frac{1}{m^{2}}-\frac{1}{(m+1)^{2}}\right)$
$\displaystyle\leq\sum_{\frac{1}{|x|}<m<\infty}\frac{1}{|x|^{2}}\left(\frac{1}{m^{2}}-\frac{1}{(m+1)^{2}}\right)\leq\frac{1}{|x|^{2}\frac{1}{|x|^{2}}}=1,$
$\displaystyle\sum_{\frac{1}{|x|}<m\leq bk}|x|e^{-m\mathrm{Re}(x)}$
$\displaystyle\leq\frac{|x|}{1-e^{-\mathrm{Re}(x)}}\ll
1,\qquad\text{as}\quad\frac{|x|}{\mathrm{Re}(x)}\ll 1,|x|\ll 1,$ and similarly
$\displaystyle\sum_{\frac{1}{|x|}<m\leq bk}|x|e^{-m\mathrm{Re}(x)(a+1)}$
$\displaystyle\leq\frac{|x|}{1-e^{-\mathrm{Re}(x)(a+1)}}\ll 1.$

As a result, using Lemma 2.10 (again up to at most one summand in
$O(\log(k))$),

$\displaystyle\left|k\sum_{1\leq j\leq k}\sum_{\frac{1}{|x|}<m\leq
bk}\zeta_{b}^{ma}\zeta_{k}^{mhj}\cdot\frac{1}{m}\phi_{a}(mx)\right|$
$\displaystyle\ll O\left(k^{2}\right)+k\sum_{1\leq j\leq
k}\max_{\frac{1}{|x|}\leq m\leq
bk}\left|G_{m}\left(\frac{a}{b}+\frac{hj}{k}\right)\right|=O\left(k^{2}\right),$

which yields (4.2).

For the summands $\ell\geq 1$ in Lemma 4.2, we first focus on the $j$ sum,

$\displaystyle\sum_{1\leq j\leq k}\zeta_{k}^{mjh}g_{j,k}\left(w\right)$
$\displaystyle=\sum_{1\leq j\leq
k}\zeta_{k}^{mjh}\left(\frac{e^{-w-\frac{jw}{k}}}{w(1-e^{-w})^{2}}+\frac{j}{k}\frac{e^{-\frac{jw}{k}}}{w(1-e^{-w})}-\frac{1}{w^{3}}\right)$
$\displaystyle=-\frac{\mathrm{1}_{k|mh}}{w^{3}}+\frac{\zeta_{k}^{mh}e^{-w-\frac{w}{k}}}{w(1-e^{-w})^{2}}\left(\frac{1-e^{-w}}{1-\zeta_{k}^{mh}e^{-\frac{w}{k}}}\right)$
$\displaystyle\hskip
28.45274pt+\frac{\zeta_{k}^{mh}e^{-\frac{w}{k}}}{kw(1-e^{-w})}\left(\frac{1+k\zeta_{k}^{mh}e^{-\frac{w}{k}-w}-(k+1)e^{-w}}{(1-\zeta_{k}^{mh}e^{-\frac{w}{k}})^{2}}\right),$

where we used the formulas

$\sum^{k}_{j=1}a^{j}=\frac{1-a^{k+1}}{1-a}-1,\qquad\sum^{k}_{j=1}ja^{j}=a\frac{\left(ka^{k+1}+1-(k+1)a^{k}\right)}{1-a^{2}}.$

When $k$ divides $m$, we have

$\displaystyle\sum_{\begin{subarray}{c}1\leq j\leq
k\end{subarray}}\zeta_{k}^{mjh}g_{j,k}\left(w\right)$
$\displaystyle=-\frac{1}{w^{3}}+\frac{e^{-w-\frac{w}{k}}}{w(1-e^{-w})(1-e^{-\frac{w}{k}})}+\frac{e^{-\frac{w}{k}}\left(1-e^{-w}+ke^{-w}(-1+e^{-\frac{w}{k}})\right)}{kw(1-e^{-w})(1-e^{-\frac{w}{k}})^{2}}$
$\displaystyle=-\frac{1}{w^{3}}+\frac{e^{-\frac{w}{k}}}{kw(1-e^{-\frac{w}{k}})^{2}}=:f_{1}(w),$

and when $k$ does not divide $m$ we have

$\displaystyle\sum_{\begin{subarray}{c}1\leq j\leq
k\end{subarray}}\zeta_{k}^{mjh}g_{j,k}\left(w\right)$
$\displaystyle=\frac{\zeta_{k}^{mh}e^{-w-\frac{w}{k}}}{w(1-e^{-w})(1-\zeta^{mh}_{k}e^{-\frac{w}{k}})}+\frac{\zeta_{k}^{mh}e^{-\frac{w}{k}}}{kw(1-e^{-w})}\left(\frac{1-e^{-w}+k\zeta_{k}^{mh}e^{-w-\frac{w}{k}}-ke^{-w}}{(1-\zeta_{k}^{mh}e^{-\frac{w}{k}})^{2}}\right)$
$\displaystyle=\frac{\zeta^{mh}_{k}e^{-\frac{w}{k}}}{kw(1-\zeta_{k}^{mh}e^{-\frac{w}{k}})^{2}}=:f_{2}(w).$

We will finish the proof of Lemma 4.1 by showing

$k^{2}t\sum_{\begin{subarray}{c}m\leq bk\\\ k\mid
m\end{subarray}}\sum_{\ell\geq
1}\zeta_{b}^{ma}f_{1}\left(t\left(bk\ell+m\right)\right)=O\left(k^{2}\right),$
(4.3)

and

$k^{2}t\sum_{\begin{subarray}{c}m\leq bk\\\ k\nmid
m\end{subarray}}\sum_{\ell\geq
1}\zeta_{b}^{ma}f_{2}\left(tk\left(bk\ell+m\right)\right)=O\left(k^{2}\right).$
(4.4)

First, we apply Theorem 2.2 to the sum on $f_{1}$. Computing the derivative of
$f_{1}(w)$, we find

$\displaystyle f^{{}^{\prime}}_{1}(w)=3\,{w}^{-4}-{\frac{1}{{k}^{2}w}{{\rm
e}^{-{\frac{w}{k}}}}\left(1-{e}^{-{\frac{w}{k}}}\right)^{-2}}-{\frac{1}{k{w}^{2}}{{\rm
e}^{-{\frac{w}{k}}}}\left(1-{e}^{-{\frac{w}{k}}}\right)^{-2}}-2\,{\frac{1}{{k}^{2}w}{{\rm
e}^{-{\frac{w}{k}}}}{e}^{-{\frac{w}{k}}}\left(1-{e}^{-{\frac{w}{k}}}\right)^{-3}}.$

Thus,

$\displaystyle\sum_{\ell\geq 1}f_{1}\left(t\left(bk\ell+m\right)\right)$
$\displaystyle=\int^{\infty}_{1}f_{1}(t_{\theta}(bkw+m))dw+\frac{\displaystyle{\lim_{\ell\to\infty}f_{1}(t_{\theta}(bk\ell+m))}+f_{1}(t_{\theta}(bk+m))}{2}$
$\displaystyle\hskip
28.45274pt+t_{\theta}bk\int^{\infty}_{1}f_{1}^{{}^{\prime}}(t_{\theta}(bkw+m))\left(\\{w\\}-\frac{1}{2}\right)dw$
$\displaystyle=\frac{1}{t_{\theta}bk}\int^{\infty}_{t_{\theta}(bk+m)}f_{1}(x))dx+\frac{f_{1}(t_{\theta}(bk+m))}{2}$
$\displaystyle\hskip
28.45274pt+\int^{\infty}_{t_{\theta}(bk+m)}f_{1}^{{}^{\prime}}(x)\left(\left\\{\frac{x-t_{\theta}m}{t_{\theta}bk}\right\\}-\frac{1}{2}\right)dw.$

We note that $f_{1}(w)\ll\frac{k}{w^{3}}$ as $w\to 0$ in a cone, which comes
from the Laurent expansion

$f_{1}(w)=\left(-1+k\right){w}^{-3}+-\frac{7}{12}w^{-1}+O(1).$

Since $|t_{\theta}|\asymp\frac{1}{n^{\frac{1}{3}}}$ and $k\ll
n^{\frac{1}{3}}$, there exists a fixed complex number $c$ on the line of
integration such that $|t_{\theta}(bk+m)|<|c|.$ Then
$\int^{\infty}_{c}f_{1}(w)dw=O_{c}(1)$. Near $0$, we have

$\int^{c}_{t_{\theta}(bk+m)}f_{1}(w)dw\ll\int^{c}_{t_{\theta}(bk+m)}\frac{1}{w^{3}}dw\ll
O\left(\frac{1}{t_{\theta}^{2}(bk+m)^{2}}\right)=O\left(\frac{1}{t_{\theta}^{2}k^{2}}\right),$

where the last step follows from the fact that $k|m$. Therefore,

$k^{2}t_{\theta}\sum_{k|m\leq
bk}\zeta_{b}^{ma}\frac{1}{t_{\theta}kb}\int^{\infty}_{1}f_{1}(t_{\theta}(bkw+m))dw=O\left(\frac{1}{t_{\theta}^{2}k}\right)\ll
N^{\frac{3}{2}}.$

We have $f_{1}(t_{\theta}(bk+m))\ll\frac{1}{t_{\theta}^{3}k^{3}}$ which
implies that

$k^{2}t_{\theta}\sum_{k|m\leq bk}\zeta_{b}^{ma}f_{1}(t_{\theta}(bk+m))\ll
n^{1/2}\ll N^{\frac{3}{2}}.$

It remains to contend with the integral involving the derivative. The
derivative has the Laurent expansion

$\displaystyle
f^{{}^{\prime}}_{1}(w)=3(1-k)w^{-4}+O(w^{-2})\ll\frac{k}{w^{4}}.$

Thus, breaking the integral as before into two parts (with $c$ fixed) we find

$k^{2}t_{\theta}\sum_{k|m\leq
bk}\zeta_{b}^{ma}\int^{\infty}_{1}f^{{}^{\prime}}_{1}(t_{\theta}(bkw+m))dw=k^{3}t_{\theta}\frac{1}{t^{2}_{\theta}k^{2}}\ll
N^{\frac{3}{2}}.$

This completes the proof of (4.3).

Now turning to $\eqref{E:f2bound},$ it is clear that in the right half plane

$\left|f_{2}(w)\right|\leq\frac{e^{-\frac{\mathrm{Re}(w)}{k}}}{k\mathrm{Re}(w)(1-e^{-\frac{\mathrm{Re}(w)}{k}})^{2}}:=\tilde{f}_{2}(\mathrm{Re}(w)),$

where $\tilde{f}_{2}(\mathrm{Re}(w))$ is a decreasing function. Approximating
by a Riemann sum, we have

$\displaystyle k^{2}t_{\theta}\sum_{k\not|m\leq bk}\sum_{\ell\geq
1}\zeta_{b}^{ma}f_{2}\left(t_{\theta}k\left(bk\ell+m\right)\right)\leq
k^{3}t_{\theta}\sum_{\ell\geq
1}\tilde{f}_{2}\left(\mathrm{Re}(t_{\theta}k^{2}b\ell\right)$
$\displaystyle\leq\frac{k|t_{\theta}|}{b\mathrm{Re}(t_{\theta})}\left(\frac{1}{\mathrm{Re}(t_{\theta})k^{2}b}\int^{\infty}_{\mathrm{Re}(t_{\theta})k^{2}b}\tilde{f}_{2}(x)dx+bk^{2}\mathrm{Re}(t_{\theta})\tilde{f}_{2}\left(\mathrm{Re}(t_{\theta}k^{2}b\right)\right).$

Since $k\gg n^{1/6}$, the lower bound of the integral is bounded away from
$0$. Therefore,

$\displaystyle k^{2}t_{\theta}\sum_{k\not|m\leq bk}\sum_{\ell\geq
1}\zeta_{b}^{ma}f_{2}\left(t_{\theta}k\left(bk\ell+m\right)\right)$
$\displaystyle\ll
O(n^{1/6})+k^{3}|t_{\theta}|\tilde{f}_{2}\left(\mathrm{t_{\theta}}k^{2}b\right).$

Furthermore,

$\tilde{f}_{2}\left(\mathrm{t_{\theta}}k^{2}b\right)\ll\frac{1}{k^{4}|t_{\theta}|^{2}}\ll
1,$

and (4.4) follows, completing the proof of Lemma 4.1

∎

###  4.2. Proof of Theorem 1.1

In this section we follow Parry [27]. We begin with the case
$\frac{a}{b}<\theta_{12}.$ Write

$\lambda_{1}+i\lambda_{2}:=\mathrm{Li}_{3}\left(\zeta_{b}^{a}\right)^{\frac{1}{3}}.$

Then by Lemma 4.1,

$\displaystyle
e^{-\frac{3}{2^{\frac{2}{3}}}\lambda_{1}n^{\frac{2}{3}}}T_{n}\left(\zeta_{b}^{a}\right)$
$\displaystyle=\left(1-\zeta_{b}^{a}\right)^{\frac{1}{12}}\int_{-\theta_{0,1}^{\prime}}^{\theta_{0,1}^{\prime\prime}}\exp\left(-\frac{3}{2^{\frac{2}{3}}}\lambda_{1}n^{\frac{2}{3}}+\frac{(\lambda_{1}+i\lambda_{2})^{3}}{(t_{n}-2\pi
i\theta)^{2}}+nt_{n}-2\pi in\theta+o(1)\right)d\theta$
$\displaystyle+\sum_{\begin{subarray}{c}k\geq 2\\\
\frac{h}{k}\in\mathcal{F}_{N}\end{subarray}}\zeta_{k}^{nh}\int_{-\theta_{h,k}^{\prime}}^{\theta_{h,k}^{\prime\prime}}\exp\left(-\frac{3}{2^{\frac{2}{3}}}\lambda_{1}n^{\frac{2}{3}}+\frac{\mathrm{Li}_{3}\left(\zeta_{b}^{ka}\right)}{k^{3}(t_{n}-2\pi
i\theta)^{2}}+nt_{n}-2\pi in\theta+\delta O(n^{\frac{2}{3}})\right)d\theta,$
(4.5)

where the constants are independent of $k$. For the major arc integral above,
we rewrite the exponent as

$\displaystyle\frac{\lambda_{1}+i\lambda_{2}}{2^{\frac{2}{3}}\left(1-\frac{2\pi
in^{\frac{1}{3}}\theta}{2^{\frac{1}{3}}(\lambda_{1}+i\lambda_{2})}\right)^{2}}n^{\frac{2}{3}}-\frac{1}{2^{\frac{2}{3}}}\lambda_{1}n^{\frac{2}{3}}+i2^{\frac{1}{3}}\lambda_{2}n^{\frac{2}{3}}-2\pi
in\theta+o(1).$

Setting $\theta\mapsto n^{-\frac{1}{3}}\theta$ and using the fact that
$\theta_{0,1}^{\prime}=\theta_{0,1}^{\prime\prime}=\frac{1}{N+1}\asymp\frac{1}{n^{\frac{1}{3}}},$
the major arc integral is asymptotic to

$\frac{1}{n^{\frac{1}{3}}}\int_{-c}^{c}e^{n^{\frac{2}{3}}B(\theta)}d\theta,$

for some $c>0$, where

$B(\theta):=\frac{\lambda_{1}+i\lambda_{2}}{2^{\frac{2}{3}}\left(1-\frac{2\pi
i\theta}{2^{\frac{1}{3}}(\lambda_{1}+i\lambda_{2})}\right)^{2}}-\frac{1}{2^{\frac{2}{3}}}\lambda_{1}+i2^{\frac{1}{3}}\lambda_{2}-2\pi
i\theta.$

Here, $B(0)=i\frac{3}{2^{\frac{2}{3}}}\lambda_{2}$ and using
$\frac{1}{(1-x)^{2}}=1+2x+3x^{2}+O(x^{3}),$ we have

$\lim_{\theta\to 0}\frac{B(\theta)-B(0)}{\theta^{2}}=3\frac{(2\pi
i)^{2}}{2^{4/3}(\lambda_{1}+i\lambda_{2})}=-\frac{2^{\frac{2}{3}}3\pi^{2}}{\lambda_{1}+i\lambda_{2}}.$

Now we show that $\mathrm{Re}(B(\theta))\leq\mathrm{Re}(B(0))=0$ with equality
if and only if $\theta=0.$ Following [27, Lemma 5.2], we write

$\displaystyle\mathrm{Re}(B(\theta))$
$\displaystyle=\mathrm{Re}\left(\frac{\lambda_{1}+i\lambda_{2}}{2^{\frac{2}{3}}\left(1-\frac{2\pi
i\theta}{2^{\frac{1}{3}}(\lambda_{1}+i\lambda_{2})}\right)^{2}}\right)-\frac{\lambda_{1}}{2^{\frac{2}{3}}}$
$\displaystyle=\frac{|\lambda_{1}+i\lambda_{2}|^{3}}{2^{\frac{2}{3}}\lambda_{1}^{2}}\mathrm{Re}\left(\frac{e^{i3r}}{\left(1+i\left(\frac{\lambda_{2}}{\lambda_{1}}-\frac{2^{\frac{2}{3}}\pi\theta}{\lambda_{1}}\right)\right)^{2}}\right)-\frac{\lambda_{1}}{2^{\frac{2}{3}}}$
$\displaystyle\text{($r:=\arg(\lambda_{1}+i\lambda_{2})$)}.$

Next, we write

$\left|1+i\left(\frac{\lambda_{2}}{\lambda_{1}}-\frac{2^{\frac{2}{3}}\pi\theta}{\lambda_{1}}\right)\right|e^{i\psi_{\theta}}:=1+i\left(\frac{\lambda_{2}}{\lambda_{1}}-\frac{2^{\frac{2}{3}}\pi\theta}{\lambda_{1}}\right).$

Note that $\theta\mapsto\psi_{\theta}$ is a bijection from
$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ to $\mathbb{R}$, and further by
matching real and imaginary parts of both sides, we see that

$\frac{1}{\left|1+i\left(\frac{\lambda_{2}}{\lambda_{1}}-\frac{2^{\frac{2}{3}}\pi\theta}{\lambda_{1}}\right)\right|}=\cos\left(\psi_{\theta}\right).$

Thus,

$\mathrm{Re}(B(\theta))=\frac{|\lambda_{1}+i\lambda_{2}|^{3}}{2^{\frac{2}{3}}\lambda_{1}^{2}}\cos(3r-2\psi_{\theta})\cos^{2}(\psi_{\theta})-\frac{\lambda_{1}}{2^{\frac{2}{3}}}.$

By Lemma 2.3, the maximum is achieved when $3r-2\psi_{\theta}=\psi_{\theta}$,
i.e. $\psi_{\theta}=r$, which gives

$\mathrm{Re}(B(\theta))\leq\frac{|\lambda_{1}+i\lambda_{2}|^{3}}{2^{\frac{2}{3}}\lambda_{1}^{2}}\cos(r)^{3}-\frac{\lambda_{1}}{2^{\frac{2}{3}}}=\frac{\lambda_{1}^{3}}{2^{\frac{2}{3}}\lambda_{1}^{2}}-\frac{\lambda_{1}}{2^{\frac{2}{3}}}=0,$

with equality if and only if $\theta=0$. Thus by Theorem 2.1, we have

$\frac{1}{n^{\frac{1}{3}}}\int_{-c}^{c}e^{n^{\frac{2}{3}}B(\theta)}d\theta\sim\frac{1}{n^{\frac{1}{3}}}\sqrt{\frac{\pi}{n^{\frac{2}{3}}\frac{2^{\frac{2}{3}}3\pi^{2}}{\lambda_{1}+i\lambda_{2}}}}=\frac{\sqrt{\lambda_{1}+i\lambda_{2}}}{2^{\frac{1}{3}}\sqrt{3\pi}n^{\frac{2}{3}}}e^{i\frac{3}{2^{\frac{2}{3}}}\lambda_{2}n^{\frac{2}{3}}},$

and this is the main term in case (1) of Theorem 1.1 after multiplying both
sides again by $e^{\frac{3}{2^{\frac{2}{3}}}\lambda_{1}n^{\frac{2}{3}}}$ and
the constant $\left(1-\zeta_{b}^{a}\right)^{1/12}$ in (4.5).

We now show that the minor arcs in (4.5) tend to 0; taking the real part of
the exponent, we get

$\displaystyle\frac{n^{\frac{2}{3}}\lambda_{1}}{2^{\frac{2}{3}}}\left(-1+\frac{\left|\mathrm{Li}_{3}\left(\zeta_{b}^{ka}\right)\right|}{k^{3}\lambda_{1}^{3}}\mathrm{Re}\left(\frac{e^{i3r_{k}}}{\left(1+i\left(\frac{\lambda_{2}}{\lambda_{1}}-\frac{2^{\frac{2}{3}}\pi\theta
n^{\frac{1}{3}}}{\lambda_{1}}\right)\right)^{2}}\right)+\delta
O(1)\right)\qquad\text{$\left(r_{k}:=\arg\mathrm{Li}_{3}\left(\zeta_{b}^{ak}\right)^{\frac{1}{3}}\right)$}.$

As before, the maximum of the expression Re$(\cdot)$ is $\cos(r_{k})^{3}$, so
we get that the above is at most

$\frac{n^{\frac{2}{3}}\lambda_{1}}{2^{\frac{2}{3}}}\left(-1+\left(\frac{\mathrm{Re}\left(\mathrm{Li}_{3}\left(\zeta_{b}^{ka}\right)\right)}{k\lambda_{1}}\right)^{3}+\delta
O(1)\right)\leq\frac{n^{\frac{2}{3}}\lambda_{1}}{2^{\frac{2}{3}}}\left(-\Delta+\delta
O(1)\right),$

for some $\Delta>0$ uniformly in $k$ by Corollary 3.5. Choosing $\delta$ small
enough so that the constant above is still negative, we get that the minor
arcs decay exponentially, finishing the proof in case (1).

In the case that $\theta_{12}<\frac{a}{b}<\frac{1}{2}$, we set
$\lambda_{1}+i\lambda_{2}:=\frac{\mathrm{Li}_{3}\left(\zeta_{b}^{2a}\right)^{\frac{1}{3}}}{2},$
and again by Lemma 4.1 we have

$\displaystyle
e^{-\frac{3}{2^{\frac{2}{3}}}\lambda_{1}n^{\frac{2}{3}}}T_{n}\left(\zeta_{b}^{a}\right)$
$\displaystyle=(-1)^{n}\frac{\left(1-\zeta_{b}^{a}\right)^{\frac{1}{6}}}{\left(1+\zeta_{b}^{a}\right)^{\frac{1}{12}}}\int_{-\theta_{1,2}^{\prime}}^{\theta_{1,2}^{\prime\prime}}\exp\left(-\frac{3}{2^{\frac{2}{3}}}\lambda_{1}n^{\frac{2}{3}}+\frac{(\lambda_{1}+i\lambda_{2})^{3}}{(t_{n}-2\pi
i\theta)^{2}}+nt_{n}-2\pi in\theta+o(1)\right)d\theta$
$\displaystyle+\sum_{\begin{subarray}{c}k\neq 2\\\
\frac{h}{k}\in\mathcal{F}_{N}\end{subarray}}\zeta_{k}^{nh}\int_{-\theta_{h,k}^{\prime}}^{\theta_{h,k}^{\prime\prime}}\exp\left(-\frac{3}{2^{\frac{2}{3}}}\lambda_{1}n^{\frac{2}{3}}+\frac{\mathrm{Li}_{3}\left(\zeta_{b}^{ka}\right)}{k^{3}(t_{n}-2\pi
i\theta)^{2}}+nt_{n}-2\pi in\theta+\delta O(n^{\frac{2}{3}})\right)d\theta.$
(4.6)

Whereas for $\frac{a}{b}=\frac{1}{2}$, we set
$\lambda_{1}:=\frac{\zeta(3)^{\frac{1}{3}}}{2},$ and Lemma 4.1 implies

$\displaystyle
e^{-\frac{3}{2^{\frac{2}{3}}}\lambda_{1}n^{\frac{2}{3}}}T_{n}\left(-1\right)$
$\displaystyle\hskip
28.45274pt=(-1)^{n}e^{-\zeta(-1)}\int_{-\theta_{1,2}^{\prime}}^{\theta_{1,2}^{\prime\prime}}(t_{n}-2\pi
i\theta)^{-\frac{1}{12}}\exp\left(-\frac{3}{2^{\frac{2}{3}}}\lambda_{1}n^{\frac{2}{3}}+\frac{\lambda_{1}^{3}}{(t_{n}-2\pi
i\theta)^{2}}+nt_{n}-2\pi in\theta+o(1)\right)d\theta$ $\displaystyle\hskip
56.9055pt+\sum_{\begin{subarray}{c}k\neq 2\\\
\frac{h}{k}\in\mathcal{F}_{N}\end{subarray}}\zeta_{k}^{nh}\int_{-\theta_{h,k}^{\prime}}^{\theta_{h,k}^{\prime\prime}}\exp\left(-\frac{3}{2^{\frac{2}{3}}}\lambda_{1}n^{\frac{2}{3}}+\frac{\mathrm{Li}_{3}\left(\zeta_{b}^{ka}\right)}{k^{3}(t_{n}-2\pi
i\theta)^{2}}+nt_{n}-2\pi in\theta+\delta O(n^{\frac{2}{3}})\right)d\theta.$
(4.7)

The analysis of (4.6) and (4.7) is carried out as before.

###  4.3. Proof of Corollary 1.2

By (1.3), we have

$\displaystyle
pp(a_{1},b,n)-pp(a_{2},b,n)=\frac{2}{b}\mathrm{Re}\left(\left(\zeta_{b}^{-a_{1}}-\zeta_{b}^{-a_{2}}\right)T_{n}(\zeta_{b})\right)+\frac{1}{b}\sum_{2\leq
j\leq
b-2}\left(\zeta_{b}^{-a_{1}j}-\zeta_{b}^{-a_{2}j}\right)T_{n}\left(\zeta_{b}^{j}\right).$
(4.8)

Note that $0<\frac{1}{b}\leq\frac{1}{3}<\theta_{12}$. By definition and
Theorem 1.1, we find

$\displaystyle\frac{2}{b}\left(\zeta_{b}^{-a_{1}}-\zeta_{b}^{-a_{2}}\right)T_{n}(\zeta_{b})$
$\displaystyle\sim\frac{2}{b}\left(\zeta_{b}^{-a_{1}}-\zeta_{b}^{-a_{2}}\right)\frac{\left(1-\zeta_{b}^{a}\right)^{\frac{1}{12}}\mathrm{Li}_{3}\left(\zeta_{b}^{a}\right)^{\frac{1}{6}}}{2^{\frac{1}{3}}\sqrt{3\pi}n^{\frac{2}{3}}}\exp\left(\frac{3}{2^{\frac{2}{3}}}\mathrm{Li}_{3}\left(\zeta_{b}^{a}\right)^{\frac{1}{3}}n^{\frac{2}{3}}\right)$
$\displaystyle=\frac{B_{a_{1},a_{2},b}e^{i\alpha_{a_{1},a_{2},b}}}{n^{\frac{2}{3}}}\exp\left(\frac{3}{2^{\frac{2}{3}}}\lambda_{1}n^{\frac{2}{3}}\right)\exp\left(\frac{3}{2^{\frac{2}{3}}}\lambda_{2}n^{\frac{2}{3}}i\right)$
$\displaystyle=\frac{B_{a_{1},a_{2},b}}{n^{\frac{2}{3}}}\exp\left(\frac{3}{2^{\frac{2}{3}}}\lambda_{1}n^{\frac{2}{3}}\right)\exp\left(\left(\alpha_{a_{1},a_{2},b}+\frac{3}{2^{\frac{2}{3}}}\lambda_{2}n^{\frac{2}{3}}\right)i\right).$

Hence,

$\displaystyle\frac{\frac{2}{b}\left(\zeta_{b}^{-a_{1}}-\zeta_{b}^{-a_{2}}\right)T_{n}(\zeta_{b})}{B_{a_{1},a_{2},b}n^{-\frac{2}{3}}\exp\left(\frac{3}{2^{\frac{2}{3}}}\lambda_{1}n^{\frac{2}{3}}\right)}=\exp\left(\left(\alpha_{a_{1},a_{2},b}+\frac{3}{2^{\frac{2}{3}}}\lambda_{2}n^{\frac{2}{3}}\right)i\right)+o(1).$
(4.9)

When taking real parts in (4.9), we find with $\mathbb{R}$-linearity

$\displaystyle\frac{\frac{2}{b}\mathrm{Re}\left(\left(\zeta_{b}^{-a_{1}}-\zeta_{b}^{-a_{2}}\right)T_{n}(\zeta_{b})\right)}{B_{a_{1},a_{2},b}n^{-\frac{2}{3}}\exp\left(\frac{3}{2^{\frac{2}{3}}}\lambda_{1}n^{\frac{2}{3}}\right)}=\cos\left(\alpha_{a_{1},a_{2},b}+\frac{3}{2^{\frac{2}{3}}}\lambda_{2}n^{\frac{2}{3}}\right)+o(1).$
(4.10)

Now, again with Theorem 1.1 and Corollary 3.5, we have

$\displaystyle\frac{\frac{1}{b}\sum_{2\leq j\leq
b-2}\left(\zeta_{b}^{-a_{1}j}-\zeta_{b}^{-a_{2}j}\right)T_{n}\left(\zeta_{b}^{j}\right)}{B_{a_{1},a_{2},b}n^{-\frac{2}{3}}\exp\left(\frac{3}{2^{\frac{2}{3}}}\lambda_{1}n^{\frac{2}{3}}\right)}=O\left(e^{-c_{b}n^{\frac{2}{3}}}\right),$
(4.11)

for some $c_{b}>0$. The claim follows with (4.8), (4.10) and (4.11).

###  4.4. Proof of Theorem 1.3

In fact there is very little to change between the proofs of Theorems 1.1 and
1.3, as the bounds for error terms in Section 4.1, together with Proposition
3.7 to identify the major arc, can readily be applied. With $N=\lfloor\delta
n^{\frac{1}{3}}\rfloor$ as before, we write

$\displaystyle\mathcal{A}_{n}\left(\zeta_{b}^{a}\right)$
$\displaystyle=\sum_{\begin{subarray}{c}\gcd(h,k)=1\\\
\frac{h}{k}\in\mathcal{F}_{N}\end{subarray}}\zeta_{k}^{nh}\int_{-\theta_{h,k}^{\prime}}^{\theta_{h,k}^{\prime\prime}}\exp\left(\log\overline{PP}\left(\zeta_{b}^{a};e^{-t_{n}+2\pi
i\theta}\right)+nt_{n}-2\pi in\theta\right)d\theta$
$\displaystyle=\sum_{\begin{subarray}{c}\gcd(h,k)=1\\\
\frac{h}{k}\in\mathcal{F}_{N}\end{subarray}}\zeta_{k}^{nh}\int_{-\theta_{h,k}^{\prime}}^{\theta_{h,k}^{\prime\prime}}\exp\left(\frac{\zeta(3)-\mathrm{Li}_{3}\left(\zeta_{b}^{ka}\right)}{k^{3}(t_{n}-2\pi
i\theta)^{2}}+nt_{n}-2\pi in\theta+\overline{E}_{h,k}(\zeta_{b}^{a};t_{n}-2\pi
i\theta)\right)d\theta,$

where

$\overline{E}_{h,k}(z,t):=\log\overline{PP}\left(z;e^{-t}\right)-\frac{\zeta(3)-\mathrm{Li}_{3}\left(z^{k}\right)}{k^{3}t^{2}},\qquad
t_{n}:=\frac{2^{\frac{1}{3}}\left(\zeta(3)-\mathrm{Li}_{3}\left(\zeta_{b}^{a}\right)\right)^{\frac{1}{3}}}{n^{\frac{1}{3}}}.$

Note that

$\operatorname{Log}\left(\overline{PP}(\zeta;q)\right)=\operatorname{Log}\left(PP(1;q)\right)-\operatorname{Log}\left(PP(\zeta;q)\right),$

so

$\overline{E}_{h,k}(z;t)=E_{0,1}(1;t)-E_{h,k}(z;t),$

and we can show as in Lemma 4.1 that uniformly for
$\theta\in[-\theta_{h,k}^{\prime},\theta_{h,k}^{\prime\prime}]$ we have

$\overline{E}_{h,k}(\zeta_{b}^{a};t_{n}-2\pi i\theta)=O(N^{2}).$ (4.12)

The only reason we cannot directly quote Lemma 4.1 is that our $t_{n}$ is a
different constant times $n^{-\frac{1}{3}}.$ But this constant does not affect
the proofs of Lemmas 4.1, so we obtain (4.12). The proof of Theorem 1.3 then
follows Section 4.2 mutatis mutandis, this time using Proposition 3.7 to show
that the major arc occurs at $\frac{h}{k}=\frac{0}{1}$ for all $\frac{a}{b}$.

##  5\. Outlook

In general, one could ask for asymptotic formulae for the polynomials
generated by products of the shape

$\prod_{n\geq 1}\frac{1}{(1-\zeta q^{n})^{a_{n}}},\qquad(a_{n}\geq 0),$

for $\zeta$ any root of unity. This would generalize Meinardus’ classical
theorem [1, Ch. 6] which is the case $\zeta=1$ and extend a theorem of Parry
who proved asymptotic formulas when $|\zeta|<1.$ For Meinardus’ theorem, one
requires certain technical hypotheses on the sequence $\\{a_{n}\\}_{n\geq 1}$,
and an extension of our method appears to require an excessive number of such
hypotheses. We thus leave the general case as an open problem.

Such an extension of Meinardus’ theorem would immediately yield information on
the sign changes for many partition-theoretic objects. One notable example
comes from partitions into $\ell$-th powers, i.e.

$\sum_{n\geq 0}Q_{n}^{(\ell)}(\zeta)q^{n}=\prod_{n\geq 1}\frac{1}{1-\zeta
q^{n^{\ell}}}.$

When $\zeta=-1$ the polynomials $Q_{n}^{(\ell)}(-1)$ count the difference of
partitions into an even and odd number of $\ell$-th powers. Ciolan [11, 12]
overcame significant technical obstacles to prove asymptotic formulae for
$Q^{(\ell)}_{n}(-1)$ and additionally gave a combinatorial proof of the
following.

######  Theorem (Theorem 2 of [12]).

If $\ell\geq 2$, then

$Q^{(\ell)}_{n}(-1)>0\quad\text{(if $n$ is even),}\qquad\text{and}\qquad
Q^{(\ell)}_{n}(-1)<0\quad\text{(if $n$ is odd).}$

It would thus be of great interest to see asymptotic formulae for
$Q_{n}^{(\ell)}(\zeta)$ for arbitrary roots of unity $\zeta$.

In work of Corteel–Safelief–Vuletić, there are also analogues for cylindric
partitions of the statistics generated by $\mathcal{A}_{n}(\zeta)$, namely the
polynomials generated by

$\prod_{\begin{subarray}{c}n\geq 1\\\ j\in S\end{subarray}}\frac{1-\zeta
q^{j+nT}}{1-q^{j+nT}},$

for some finite set $S$ (see [14, Theorem 8]). We expect our techniques apply
to these polynomials as well.

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The fifth term in $E_{s,2}$ reads

$\begin{split}&\mathrm{i}f_{2}(h)\exp\left(\operatorname{ad}_{i\frac{h}{2}f_{1}(h/2)H_{1}}\right)H_{2}\\\
&=\mathrm{i}f_{2}(0)H_{2}+\mathrm{i}hf_{2}^{\prime}(0)H_{2}-\frac{h}{2}f_{2}(0)f_{1}(0)[H_{1},H_{2}]\\\
&+\mathrm{i}\int_{0}^{h}ds(h-s)f_{2}^{\prime\prime}(s)\exp\left(\operatorname{ad}_{i\frac{s}{2}f_{1}(s/2)H_{1}}\right)H_{2}(h-s)\\\
&+\mathrm{i}\int_{0}^{h}ds(h-s)f_{2}(s)\left(\frac{\mathrm{i}}{2}f_{1}(s/2)+\frac{\mathrm{i}s}{4}f_{1}^{\prime}(s/2)\right)^{2}\exp\left(\operatorname{ad}_{i\frac{s}{2}f_{1}(s/2)H_{1}}\right)[H_{1},[H_{1},H_{2}]]\Big{]}(h-s)\\\
&+\mathrm{i}\int_{0}^{h}ds(h-s)\left(\frac{\mathrm{i}}{2}f_{2}^{\prime}(s)f_{1}(s/2)+\frac{\mathrm{i}}{2}f_{2}(s)f_{1}^{\prime}(s/2)+\frac{\mathrm{i}s}{4}f_{2}^{\prime}(s)f_{1}^{\prime}(s/2)+\frac{\mathrm{i}s}{8}f_{2}(s)f_{1}^{\prime\prime}(s/2)\right)\\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\exp\left(\operatorname{ad}_{i\frac{s}{2}f_{1}(s/2)H_{1}}\right)[H_{1},H_{2}](h-s),\end{split}$
(96)

The last term in $E_{s,2}$ is the sum of

$-\mathrm{i}f_{2}(h/2)H_{2}=-\mathrm{i}f_{2}(0)H_{2}-\mathrm{i}\frac{h}{2}f_{2}^{\prime}(0)H_{2}-\mathrm{i}\int_{0}^{h}\frac{1}{4}f_{2}^{\prime\prime}(s/2)(h-s)ds,$
(97)

and

$-\frac{\mathrm{i}h}{2}f_{2}^{\prime}(h/2)H_{2}=-\frac{\mathrm{i}h}{2}f_{2}^{\prime}(0)H_{2}-\frac{\mathrm{i}h}{4}\int_{0}^{h}f_{2}^{\prime\prime}(s/2)H_{2}ds.$
(98)

Notice that, if we add the above eight equations together, all the zeroth-
order and first-order terms of $h$ cancel, then the desired expression of
$E_{s,2}(h)$ is achieved.

∎

###### Proof of Lemma 6.

The strategy for proving Lemma 6 is the same as that for Lemma 5. By taking
derivatives with respect to $h$ in both $U(h,0)$ and $U_{g,2}(h,0)$, we have
Eq. 76 and

$\begin{split}\partial_{h}U_{g,2}(h,0)&=\left(-\mathrm{i}f_{1}(h)+\frac{\mathrm{i}}{2}f_{1}(h/2)\right)H_{1}U_{g,2}(h,0)\\\
&\quad+\exp\left(-\mathrm{i}\int_{h/2}^{h}f_{1}(s)dsH_{1}\right)(-\mathrm{i}f_{2}(h)H_{2})\exp\left(-\mathrm{i}\int_{0}^{h}f_{2}(s)dsH_{2}\right)\exp\left(-\mathrm{i}\int_{0}^{h/2}f_{1}(s)dsH_{1}\right)\\\
&\quad+\exp\left(-\mathrm{i}\int_{h/2}^{h}f_{1}(s)dsH_{1}\right)\exp\left(-\mathrm{i}\int_{0}^{h}f_{2}(s)dsH_{2}\right)\\\
&\quad\quad\times\left(-\frac{\mathrm{i}}{2}f_{1}(h/2)/H_{1}\right)\exp\left(-\mathrm{i}\int_{0}^{h/2}f_{1}(s)dsH_{1}\right)\\\
&=-(\mathrm{i}f_{1}(h)H_{1}+\mathrm{i}f_{2}(h)H_{2})U_{g,2}(h,0)\\\
&\quad\quad+\exp\left(-\mathrm{i}\int_{h/2}^{h}f_{1}(s)dsH_{1}\right)E_{g,2}(h)\exp\left(-\mathrm{i}\int_{0}^{h}f_{2}(s)dsH_{2}\right)\exp\left(-\mathrm{i}\int_{0}^{h/2}f_{1}(s)dsH_{1}\right),\end{split}$
(99)

where $E_{g,2}(h)$ denotes

$\begin{split}E_{g,2}(h)&=if_{2}(h)\exp(\operatorname{ad}_{\mathrm{i}\int_{h/2}^{h}f_{1}(s)dsH_{1}})H_{2}+\frac{\mathrm{i}}{2}f_{1}(h/2)H_{1}\\\
&\quad-\mathrm{i}f_{2}(h)H_{2}-\frac{\mathrm{i}}{2}f_{1}(h/2)\exp\left(\operatorname{ad}_{-\mathrm{i}\int_{0}^{h}f_{2}(s)dsH_{2}}\right)H_{1}\\\
&=\mathrm{i}f_{2}(h)\left[\exp\left(\operatorname{ad}_{\mathrm{i}\int_{h/2}^{h}f_{1}(s)dsH_{1}}\right)H_{2}-H_{2}\right]\\\
&\quad-\frac{\mathrm{i}}{2}f_{1}(h/2)\left[\exp\left(\operatorname{ad}_{-\mathrm{i}\int_{0}^{h}f_{2}(s)dsH_{2}}\right)H_{1}-H_{1}\right].\end{split}$
(100)

By applying Lemma 2, we have

$\begin{split}U_{g,2}(h,0)=U(h,0)+&\int_{0}^{h}U(h,s)\exp\left(-\mathrm{i}\int_{s/2}^{s}f_{1}(s^{\prime})ds^{\prime}H_{1}\right)E_{g,2}(s)\\\
&\quad\times\exp\left(-\mathrm{i}\int_{0}^{s}f_{2}(s^{\prime})ds^{\prime}H_{2}\right)\exp\left(-\mathrm{i}\int_{0}^{s/2}f_{1}(s^{\prime})ds^{\prime}H_{1}\right)ds.\end{split}$
(101)

It remains to derive the representation of $E_{g,2}$. It follows from the
Taylor’s theorem (Lemma 1) that

$\displaystyle\quad\exp\left(\operatorname{ad}_{\mathrm{i}\int_{h/2}^{h}f_{1}(s)dsH_{1}}\right)H_{2}-H_{2}$
$\displaystyle=\frac{\mathrm{i}h}{2}f_{1}(0)[H_{1},H_{2}]+\int_{0}^{h}\left(\mathrm{i}f_{1}^{\prime}(s)-\frac{1}{4}\mathrm{i}f_{1}^{\prime}(s/2)\right)\left(\exp\left(\operatorname{ad}_{\mathrm{i}\int_{s/2}^{s}f_{1}(s^{\prime})ds^{\prime}H_{1}}\right)[H_{1},H_{2}]\right)(h-s)ds$
$\displaystyle\quad\quad+\int_{0}^{h}\left(\mathrm{i}f_{1}(s)-\frac{1}{2}\mathrm{i}f_{1}(s/2)\right)^{2}\left(\exp\left(\operatorname{ad}_{\mathrm{i}\int_{s/2}^{s}f_{1}(s^{\prime})ds^{\prime}H_{1}}\right)[H_{1},[H_{1},H_{2}]]\right)(h-s)ds,$

and

$\displaystyle\quad\exp\left(\operatorname{ad}_{-\mathrm{i}\int_{0}^{h}f_{2}(s)dsH_{2}}\right)H_{1}-H_{1}$
$\displaystyle=-\mathrm{i}hf_{2}(0)[H_{2},H_{1}]-\int_{0}^{h}\mathrm{i}f_{2}^{\prime}(s)\left(\exp\left(\operatorname{ad}_{-\mathrm{i}\int_{0}^{s}f_{2}(s^{\prime})ds^{\prime}H_{2}}\right)[H_{2},H_{1}]\right)(h-s)ds$
$\displaystyle\quad\quad+\int_{0}^{h}(\mathrm{i}f_{2}(s))^{2}\left(\exp\left(\operatorname{ad}_{-\mathrm{i}\int_{0}^{s}f_{2}(s^{\prime})ds^{\prime}H_{2}}\right)[H_{2},[H_{2},H_{1}]]\right)(h-s)ds.$

Thus we have

$\displaystyle E_{g,2}(h)$
$\displaystyle=\mathrm{i}f_{2}(h)\frac{\mathrm{i}h}{2}f_{1}(0)[H_{1},H_{2}]+\frac{\mathrm{i}}{2}f_{1}(h/2)\mathrm{i}hf_{2}(0)[H_{2},H_{1}]$
$\displaystyle\quad+\mathrm{i}f_{2}(h)\int_{0}^{h}\left(\mathrm{i}f_{1}^{\prime}(s)-\frac{\mathrm{i}}{4}f_{1}^{\prime}(s/2)\right)\left(\exp\left(\operatorname{ad}_{\mathrm{i}\int_{s/2}^{s}f_{1}(s^{\prime})ds^{\prime}H_{1}}\right)[H_{1},H_{2}]\right)(h-s)ds$
$\displaystyle\quad+\mathrm{i}f_{2}(h)\int_{0}^{h}\left(\mathrm{i}f_{1}(s)-\frac{\mathrm{i}}{2}f_{1}(s/2)\right)^{2}\left(\exp\left(\operatorname{ad}_{\mathrm{i}\int_{s/2}^{s}f_{1}(s^{\prime})ds^{\prime}H_{1}}\right)[H_{1},[H_{1},H_{2}]]\right)(h-s)ds$
$\displaystyle\quad+\frac{\mathrm{i}}{2}f_{1}(h/2)\int_{0}^{h}\mathrm{i}f_{2}^{\prime}(s)\left(\exp\left(\operatorname{ad}_{-\mathrm{i}\int_{0}^{s}f_{2}(s^{\prime})ds^{\prime}H_{2}}\right)[H_{2},H_{1}]\right)(h-s)ds$
$\displaystyle\quad-\frac{\mathrm{i}}{2}f_{1}(h/2)\int_{0}^{h}(\mathrm{i}f_{2}(s))^{2}\left(\exp\left(\operatorname{ad}_{-\mathrm{i}\int_{0}^{s}f_{2}(s^{\prime})ds^{\prime}H_{2}}\right)[H_{2},[H_{2},H_{1}]]\right)(h-s)ds$
$\displaystyle=-\frac{h}{2}f_{1}(0)\int_{0}^{h}f_{2}^{\prime}(s)ds[H_{1},H_{2}]+\frac{h}{4}f_{2}(0)\int_{0}^{h}f_{1}^{\prime}(s/2)ds[H_{1},H_{2}]$
$\displaystyle\quad-
f_{2}(h)\int_{0}^{h}(f_{1}^{\prime}(s)-\frac{1}{4}f_{1}^{\prime}(s/2))\left(\exp\left(\operatorname{ad}_{\mathrm{i}\int_{s/2}^{s}f_{1}(s^{\prime})ds^{\prime}H_{1}}\right)[H_{1},H_{2}]\right)(h-s)ds$
$\displaystyle\quad+\frac{1}{2}f_{1}(h/2)\int_{0}^{h}f_{2}^{\prime}(s)\left(\exp\left(\operatorname{ad}_{-\mathrm{i}\int_{0}^{s}f_{2}(s^{\prime})ds^{\prime}H_{2}}\right)[H_{1},H_{2}]\right)(h-s)ds$
$\displaystyle\quad-\mathrm{i}f_{2}(h)\int_{0}^{h}\left(f_{1}(s)-\frac{1}{2}f_{1}(s/2)\right)^{2}\left(\exp\left(\operatorname{ad}_{\mathrm{i}\int_{s/2}^{s}f_{1}(s^{\prime})ds^{\prime}H_{1}}\right)[H_{1},[H_{1},H_{2}]]\right)(h-s)ds$
$\displaystyle\quad+\frac{\mathrm{i}}{2}f_{1}(h/2)\int_{0}^{h}f_{2}^{2}(s)\left(\exp\left(\operatorname{ad}_{-\mathrm{i}\int_{0}^{s}f_{2}(s^{\prime})ds^{\prime}H_{2}}\right)[H_{2},[H_{2},H_{1}]]\right)(h-s)ds.$

∎

##  Appendix C Proof of Lemma 8

###### Proof.

Let $V_{k}^{(0)}=V_{k}=V(x_{k})$ for $0\leq k\leq n-1$ and
$V_{k+n}=V_{k}=V_{k-n}$ defined in a cyclic manner. Recursively we define
$V_{k}^{(j+1)}=n(V_{k+1}^{(j)}-V_{k}^{(j)})$. Notice that $V_{k}^{(j)}$ is an
approximation of the $j$-th order derivative of $V(x)$ evaluated at $x=x_{k}$.
By Taylor’s theorem and the assumption that $V(x)$ has bounded derivatives up
to fourth order, we obtain that $V_{k}^{(j)}$ is bounded for any $k$ and
$0\leq j\leq 4$. The equality $H_{1}=D_{1}^{\dagger}D_{1}$ directly follows
from the definition. We focus on the proof of the commutator bounds.

We start with the calculation of an explicit expression of $[H_{1},H_{2}]$,

$H_{1}H_{2}=n^{2}\left(\begin{array}[]{ccccc}2V_{0}&-V_{1}&&&-V_{n-1}\\\
-V_{0}&2V_{1}&-V_{2}&&\\\ &\ddots&\ddots&\ddots&\\\
&&-V_{n-3}&2V_{n-2}&-V_{n-1}\\\ -V_{0}&&&-V_{n-2}&2V_{n-1}\\\
\end{array}\right),$ (102)
$H_{2}H_{1}=n^{2}\left(\begin{array}[]{ccccc}2V_{0}&-V_{0}&&&-V_{0}\\\
-V_{1}&2V_{1}&-V_{1}&&\\\ &\ddots&\ddots&\ddots&\\\
&&-V_{n-2}&2V_{n-2}&-V_{n-2}\\\ -V_{n-1}&&&-V_{n-1}&2V_{n-1}\\\
\end{array}\right).$ (103)

Then

$\begin{split}[H_{1},H_{2}]&=n^{2}\left(\begin{array}[]{ccccc}0&V_{0}-V_{1}&&&V_{0}-V_{n-1}\\\
V_{1}-V_{0}&0&V_{1}-V_{2}&&\\\ &\ddots&\ddots&\ddots&\\\
&&V_{n-2}-V_{n-3}&0&V_{n-2}-V_{n-1}\\\ V_{n-1}-V_{0}&&&V_{n-1}-V_{n-2}&0\\\
\end{array}\right)\\\
&=n\left(\begin{array}[]{ccccc}0&-V_{0}^{(1)}&&&V_{n-1}^{(1)}\\\
V_{0}^{(1)}&0&-V_{1}^{(1)}&&\\\ &\ddots&\ddots&\ddots&\\\
&&V_{n-3}^{(1)}&0&-V_{n-2}^{(1)}\\\ -V_{n-1}^{(1)}&&&V_{n-2}^{(1)}&0\\\
\end{array}\right).\end{split}$ (104)

We further split $[H_{1},H_{2}]=D_{L}+D_{R}+S$ where

$D_{L}=n\left(\begin{array}[]{ccccc}-V_{n-1}^{(1)}&&&&V_{n-1}^{(1)}\\\
V_{0}^{(1)}&-V_{0}^{(1)}&&&\\\ &\ddots&\ddots&&\\\
&&V_{n-3}^{(1)}&-V_{n-3}^{(1)}&\\\ &&&V_{n-2}^{(1)}&-V_{n-2}^{(1)}\\\
\end{array}\right),$ (105)
$D_{R}=n\left(\begin{array}[]{ccccc}V_{0}^{(1)}&-V_{0}^{(1)}&&&\\\
&V_{1}^{(1)}&-V_{1}^{(1)}&&\\\ &&\ddots&\ddots&\\\
&&&V_{n-2}^{(1)}&-V_{n-2}^{(1)}\\\ -V_{n-1}^{(1)}&&&&V_{n-1}^{(1)}\\\
\end{array}\right).$ (106)

and
$S=-\text{diag}(V_{n-1}^{(2)},V_{0}^{(2)},V_{1}^{(2)},\cdots,V_{n-2}^{(2)})$.
Notice that for any vector $\vec{v}=(v_{k})_{k=0}^{n-1}$,

$\begin{split}\|D_{L}\vec{v}\|_{\star}^{2}&=n\sum_{k=0}^{n-1}|V_{k-1}^{(1)}|^{2}|v_{k-1}-v_{k}|^{2}\\\
&\leq
n\sup|V_{k}^{(1)}|^{2}\sum_{k=0}^{n-1}|v_{k-1}-v_{k}|^{2}=n\sup|V_{k}^{(1)}|^{2}\left(\vec{v}^{\dagger}\frac{H_{1}}{n^{2}}\vec{v}\right)\\\
&=\frac{1}{n}\sup|V_{k}^{(1)}|^{2}\left(\vec{v}^{\dagger}H_{1}\vec{v}\right)=\sup|V_{k}^{(1)}|^{2}\|D_{1}\vec{v}\|_{\star}^{2},\end{split}$
(107)

and similarly
$\|D_{R}\vec{v}\|_{\star}^{2}\leq\sup|V_{k}^{(1)}|^{2}\|D_{1}\vec{v}\|_{\star}^{2}$.
Furthermore we have
$\|S\vec{v}\|_{\star}\leq\sup|V_{k}^{(2)}|\|\vec{v}\|_{\star}$, thus there
exists $\widetilde{C}$ such that

$\|[H_{1},H_{2}]\vec{v}\|_{\star}\leq\|D_{L}\vec{v}\|_{\star}+\|D_{R}\vec{v}\|_{\star}+\|S\vec{v}\|_{\star}\leq\widetilde{C}(\|D_{1}\vec{v}\|_{\star}+\|\vec{v}\|_{\star}).$
(108)

To bound $\|[H_{1},[H_{1},H_{2}]]\vec{v}\|_{\star}$, we first compute
$[H_{1},[H_{1},H_{2}]]$ and it gives

$[H_{1},[H_{1},H_{2}]]=n^{2}\left(\begin{array}[]{ccccccc}-2V_{n-1}^{(2)}&&V_{0}^{(2)}&&.&V_{n-2}^{(2)}&\\\
&-2V_{0}^{(2)}&&V_{1}^{(2)}&&&V_{n-1}^{(2)}\\\
V_{0}^{(2)}&&-2V_{2}^{(2)}&&&&\\\ &V_{1}^{(2)}&&-2V_{3}^{(2)}&&&\\\
&&\ddots&\ddots&\ddots&\ddots&V_{n-3}^{(2)}\\\
V_{n-2}^{(2)}&&&&&-2V_{n-3}^{(2)}&\\\
&V_{n-1}^{(2)}&&&V_{n-3}^{(2)}&&-2V_{n-2}^{(2)}\end{array}\right),$ (109)

_i.e._ the only non-zero entries are

$[H_{1},[H_{1},H_{2}]]_{k+1,k+1}=-2V_{k}^{(2)},\quad[H_{1},[H_{1},H_{2}]]_{k,k+2}=[H_{1},[H_{1},H_{2}]]_{k+2,k}=V_{k}^{(2)}.$

Then we split $[H_{1},[H_{1},H_{2}]]=H_{L}+H_{R}+2H_{C}+2D_{DL}+2D_{DR}+W$
where

$H_{L}=n^{2}\left(\begin{array}[]{ccccc}V_{n-2}^{(2)}&&&V_{n-2}^{(2)}&-2V_{n-2}^{(2)}\\\
-2V_{n-1}^{(2)}&V_{n-1}^{(2)}&&&V_{n-1}^{(2)}\\\ &&\ddots&\ddots&\\\
&&&\ddots&\\\ V_{n-3}^{(2)}&&&-2V_{n-3}^{(2)}&V_{n-3}^{(2)}\\\
\end{array}\right),$ (110)
$H_{R}=n^{2}\left(\begin{array}[]{ccccc}V_{0}^{(2)}&-2V_{0}^{(2)}&V_{0}^{(2)}&&\\\
&V_{1}^{(2)}&-2V_{1}^{(2)}&V_{1}^{(2)}&\\\ &&\ddots&\ddots&\\\ &&&\ddots&\\\
-2V_{n-1}^{(2)}&V_{n-1}^{(2)}&&&V_{n-1}^{(2)}\\\ \end{array}\right),$ (111)
$H_{C}=n^{2}\left(\begin{array}[]{ccccc}-2V_{n-1}^{(2)}&V_{n-1}^{(2)}&&&V_{n-1}^{(2)}\\\
V_{0}^{(2)}&-2V_{0}^{(2)}&V_{0}^{(2)}&&\\\ &&\ddots&\ddots&\\\ &&&\ddots&\\\
V_{n-2}^{(2)}&&&V_{n-2}^{(2)}&-2V_{n-2}^{(2)}\\\ \end{array}\right),$ (112)
$D_{DL}=n\left(\begin{array}[]{ccccc}V_{n-2}^{(3)}&&&&-V_{n-2}^{(3)}\\\
-V_{n-1}^{(3)}&V_{n-1}^{(3)}&&&\\\ &\ddots&\ddots&&\\\
&&-V_{n-4}^{(3)}&V_{n-4}^{(3)}&\\\ &&&-V_{n-3}^{(3)}&V_{n-3}^{(3)}\\\
\end{array}\right),$ (113)
$D_{DR}=n\left(\begin{array}[]{ccccc}-V_{n-1}^{(3)}&V_{n-1}^{(3)}&&&\\\
&-V_{0}^{(3)}&V_{0}^{(3)}&&\\\ &&\ddots&\ddots&\\\
&&&-V_{n-3}^{(3)}&V_{n-3}^{(3)}\\\ V_{n-2}^{(3)}&&&&-V_{n-2}^{(3)}\\\
\end{array}\right),$ (114)

and
$W=n^{2}\text{diag}(V_{k}^{(2)}+V_{k-2}^{(2)}-2V_{k-1}^{(2)})_{k=0}^{n-1}$.
Notice that $H_{L},H_{R},H_{C}$ are modifications from $H_{1}$, with the
center of the central formula to be on the diagonal or subdiagonal and
multiplying each row by different bounded parameters. $D_{DL}$ and $D_{DR}$
are very similar to $D_{L}$ and $D_{R}$, with higher order potential.
Furthermore, we have

$\begin{split}\|H_{L}\vec{v}\|_{\star}^{2}&=n^{3}\sum_{k=0}^{n-1}|V_{k-2}^{(2)}|^{2}|v_{k}-2v_{k-1}+v_{k-2}|^{2}\\\
&\leq
n^{3}\sup|V_{k}^{(2)}|^{2}\sum_{k=0}^{n-1}|v_{k}-2v_{k-1}+v_{k-2}|^{2}\\\
&=\sup|V_{k}^{(2)}|^{2}\|H_{1}\vec{v}\|_{\star}^{2},\end{split}$ (115)

and similarly

$\|H_{R}\vec{v}\|_{\star}\leq\widetilde{C}\|H_{1}\vec{v}\|_{\star},\quad\|H_{C}\vec{v}\|_{\star}\leq\widetilde{C}\|H_{1}\vec{v}\|_{\star}.$

For $D_{DL}$ and $D_{DR}$, they can be bounded by the same way we bound
$D_{L}$ and $D_{R}$ before, and thus

$\|D_{DL}\vec{v}\|_{\star}\leq\widetilde{C}\|D_{1}\vec{v}\|_{\star}\leq\widetilde{C}(\|H_{1}\vec{v}\|_{\star}+\|\vec{v}\|_{\star})$

and

$\|D_{DR}\vec{v}\|_{\star}\leq\widetilde{C}(\|H_{1}\vec{v}\|_{\star}+\|\vec{v}\|_{\star}).$

Finally since $V$ has bounded fourth order derivative, the term
$n^{2}(V_{k}^{(2)}+V_{k-2}^{(2)}-2V_{k-1}^{(2)})$ is bounded. Therefore

$\|W\vec{v}\|_{\star}\leq\widetilde{C}\|\vec{v}\|_{\star}.$

Combining all the estimates together, we have

$\begin{split}\|[H_{1},[H_{1},H_{2}]]\vec{v}\|_{\star}&\leq\|H_{L}\vec{v}\|_{\star}+\|H_{R}\vec{v}\|_{\star}+2\|H_{C}\vec{v}\|_{\star}+2\|D_{DL}\vec{v}\|_{\star}+2\|D_{DR}\vec{v}\|_{\star}+\|W\vec{v}\|_{\star}\\\
&\leq\widetilde{C}(\|H_{1}\vec{v}\|_{\star}+\|\vec{v}\|_{\star}).\end{split}$
(116)

∎

##  Appendix D Proof of Lemma 10

###### Proof.

1\. From Taylor’s theorem,

$\begin{split}&\quad
n^{2}(\phi(t,x+1/n)-2\phi(t,x)+\phi(t,x-1/n))-\Delta\phi(t,x)\\\
&=\frac{n^{2}}{6}\left[\int_{x-1/n}^{x}(y-(x-1/n))^{3}\frac{\partial^{4}}{\partial
x^{4}}\phi(t,y)dy+\int_{x}^{x+1/n}(x+1/n-y)^{3}\frac{\partial^{4}}{\partial
x^{4}}\phi(t,y)dy\right].\end{split}$ (117)

Therefore

$\begin{split}&\quad|n^{2}(\phi(t,x+1/n)-2\phi(t,x)+\phi(t,x-1/n))-\Delta\phi(t,x)|\\\
&\leq\frac{1}{6n}\left[\int_{x-1/n}^{x}\left|\frac{\partial^{4}}{\partial
x^{4}}\phi(t,y)\right|dy+\int_{x}^{x+1/n}\left|\frac{\partial^{4}}{\partial
x^{4}}\phi(t,y)\right|dy\right]\\\
&\leq\frac{1}{3n^{2}}\sup_{y\in[0,1]}\left|\frac{\partial^{4}}{\partial
x^{4}}\phi(t,y)\right|.\end{split}$ (118)

2\. Since $\phi(t,x)$ satisfies the equation

$\begin{split}\mathrm{i}\partial_{t}\phi(t,x)&=H(t)\phi(t,x)\\\
&=-f_{1}(t)n^{2}(\phi(t,x+1/n)-2\phi(t,x)+\phi(t,x-1/n))+f_{2}(t)V(x)\phi(t,x)+f_{1}(t)r(t,x)\end{split}$
(119)

where $r(t,x)=n^{2}(\phi(t,x+1/n)-2\phi(t,x)+\phi(t,x-1/n))-\Delta\phi(t,x)$,
the vector $\vec{\phi}(t)=(\phi(t,k/n))_{k=0}^{n-1}$ satisfies the ordinary
differential equation

$\mathrm{i}\partial_{t}\vec{\phi}(t)=(f_{1}(t)H_{1}+f_{2}(t)H_{2})\vec{\phi}(t)+f_{1}(t)\vec{R}(t)$
(120)

where $\vec{R}(t)=(r(t,k/n))_{k=0}^{n-1}$. Same as our previous notations, let
$U(t,s)$ denote the evolution operator from time $s$ to $t$ of the dynamics
Eq. 1 with Hamiltonian Eq. 3. By the variation of parameters formula (Lemma
2),

$\vec{\phi}(t)=\vec{\psi}(t)+\int_{0}^{t}U(t,s)f_{1}(s)\vec{R}(s)ds,$ (121)

and thus

$\|\vec{\phi}(t)-\vec{\psi}(t)\|_{\star}\leq
t\|f_{1}\|_{\infty}\sup_{s\in[0,t]}\|\vec{R}(s)\|_{\star}.$ (122)

It remains to bound $\|\vec{R}(s)\|_{\star}$.

By the definition of $\vec{R}$ and the first part of this lemma, for any $s$,

$\begin{split}\|\vec{R}(s)\|_{\star}^{2}&=\frac{1}{n}\sum_{k=0}^{n-1}|r(s,k/n)|^{2}\\\
&\leq\frac{1}{n}\sum_{k=0}^{n-1}\left(\frac{1}{3n^{2}}\sup_{y\in[0,1]}\left|\frac{\partial^{4}}{\partial
x^{4}}\phi(s,y)\right|\right)^{2}\\\
&\leq\frac{1}{9n^{4}}\left(\sup_{y\in[0,1]}\left|\frac{\partial^{4}}{\partial
x^{4}}\phi(s,y)\right|\right)^{2}.\end{split}$ (123)

Plug this estimate back to Eq. 122, we complete the proof.

∎