# Bounds for discrete moments of Weyl sums and applications

Karin Halupczok  Karin Halupczok, Mathematisches Institut der HHU Düsseldorf,
Universitätsstraße 1, D-40225 Düsseldorf, Germany  karin.halupczok@uni-
duesseldorf.de

###### Abstract.

We prove two bounds for discrete moments of Weyl sums. The first one can be
obtained using a standard approach. The second one involves an observation how
this method can be improved, which leads to a sharper bound in certain ranges.
The proofs both build on the recently proved main conjecture for Vinogradov’s
mean value theorem.

We present two selected applications: First, we prove a new $k$-th derivative
test for the number of integer points close to a curve by an exponential sum
approach. This yields a stronger bound than existing results obtained via
geometric methods, but it is only applicable for specific functions. As second
application we prove a new improvement of the polynomial large sieve
inequality for one-variable polynomials of degree $k\geq 4$.

###### Key words and phrases:

Weyl sums, discrete moments, Vinogradov’s Mean Value Theorem, integer points
close to a curve, polynomial large sieve inequality

###### 1991 Mathematics Subject Classification:

Primary 11L15, Secondary 11J54

##  1\. Introduction

The recent breakthroughs of Bourgain, Demeter and Guth in [7] and Wooley in
[19, 20] has led to a full proof of the main conjecture in Vinogradov’s Mean
Value Theorem (VMVT for short). As one consequence among many, new estimates
for Weyl sums are available. With a standard approach, in this article, we
show that these already lead to strong estimates for moments of Weyl sums (see
Theorem 9).

In this context we record an observation that for moments of Weyl sums, a
small extra-improvement can be made using Montgomery’s so-called alternative
derivation from [17, §4] incorporating VMVT (see Theorem 8). This additional
gain can be exploited in a certain range for the approximating denominator
(see (8)) assuming the length of summation is large enough. We formulate a
conjecture stating where this gain might lead to, if further refinements were
available (see Conjecture 10).

Then, in Section 4, the Weyl sum moment estimates are used to prove $k$-th
derivative tests for discrete moments of exponential sums with smooth
functions (Theorems 12 and 13, the mentioned extra-improvement is incorporated
in Theorem 12).

The achieved bounds for moments of Weyl sums and exponential sums with smooth
functions lead to improvements in some number-theoretic applications, and we
present two such applications.

The first application, discussed in Section 5, is the problem of counting
integer points close to smooth curves. For this, we use a new approach
involving exponential sums such that strong bounds for the counting quantity
$\mathcal{R}(f,N,\delta)$, see Definition 15, can be obtained. The novelty is
to perform an efficient Weyl shift step over a set of indices known to be
$H^{\prime}$-spaced so that the cluster structure of the indices is respected.
This allows a saving of an extra factor $H^{-1}$ in the proof of Theorem 17,
see the arguments before (29).

Compared to some existing bounds the resulting bound in Theorem 17 is
stronger, but is valid only for certain appropriate functions. This is
discussed at the end of the section.

The second application, discussed in Section 6, concerns the polynomial large
sieve inequality from [11, 12]. In the one-dimensional case we obtain a new
improvement of the bound. That new bound comes from the extra-improvement in
Theorem 8.

###  1.1. Notations and conventions

Let $k$ denote a fixed positive integer and let $\varepsilon$ be an arbitrary
small positive real number that may change its value during calculations. By
$s,s_{0},s_{1}\geq 1$ we denote integers that depend on $k$. In this article,
we suppress the dependence of the implicit constants on $k$, $s$ or
$\varepsilon$ in our notation, simply writing $\ll$ for
$\ll_{k,s,\varepsilon}$. Moreover, we write $f\lll g$ if $f(x)=o(g(x))$, that
is if $f(x)/g(x)\to 0$ for $x\to\infty$.

For $\alpha\in\mathbb{R}$ we write $\mathrm{e}(\alpha):=\exp(2\pi i\alpha)$
for the complex exponential function and $\|\alpha\|$ denotes the distance
from $\alpha$ to the nearest integer.

For integers $k\geq 1$, $s\geq 1$ and a real number $x>0$ we use the notation
$J_{k}(x,s)$ for Vinogradov’s integral, that is the number of solutions to
Vinogradov’s system

$\displaystyle m_{1}+\dots+m_{s}$ $\displaystyle=n_{1}+\dots+n_{s}$
$\displaystyle m_{1}^{2}+\dots+m_{s}^{2}$
$\displaystyle=n_{1}^{2}+\dots+n_{s}^{2}$ $\displaystyle\cdots$ $\displaystyle
m_{1}^{k}+\dots+m_{s}^{k}$ $\displaystyle=n_{1}^{k}+\dots+n_{s}^{k}$

with $1\leq m_{1},\dots,m_{s},n_{1},\dots,n_{s}\leq x$. In this work, although
no use of the integral representation of $J_{k}(x,s)$ is made, the
$\ell_{2}$-norm of the counting function $r_{s}(\boldsymbol{\lambda})$ is used
in the proof of Theorem 8.

Given a positive integer $n$ we write $\tau(n)$ for the number of divisors of
$n$, and $\tau_{3}(n)$ denotes the number of ways one can write $n$ as a
product of $3$ factors. We will use the well-known estimates $\tau(n)\ll
n^{\varepsilon}$ and $\tau_{3}(n)\ll n^{\varepsilon}$.

The set of real functions with continuous derivatives of order up to $k$ on an
interval $I$ is denoted by $C^{k}(I)$.

###  1.2. Auxiliaries

We collect some auxiliary results needed as tools in this article.

The following is the well-known sum lemma, see e.g. [18, Lemma 4C] for a
proof.

######  Lemma 1 (Sum lemma).

For $\alpha\in\mathbb{R}$ let $u,q$ be integers with $(u,q)=1$, $0\leq u\leq
q-1$ and $|\alpha-u/q|<q^{-2}$. Let $\beta\in\mathbb{R}$. Then

$\sum_{Z\leq h\leq Y}\min(X,\|\alpha
h+\beta\|^{-1})\ll(X+q\log(q))((Y-Z)q^{-1}+1).$

For further improvements, we will also make use of the following result from
[18, Lemma 9C].

######  Lemma 2 (Variant of sum lemma).

For $\alpha\in\mathbb{R}$ let $u,q$ be integers with $(u,q)=1$, $0\leq u\leq
q-1$ and $|\alpha-u/q|<q^{-1}X^{-1}$. Let $\beta\in\mathbb{R}$. Then

$\sum_{1\leq j\leq q}\min(X,\|\alpha j+\beta\|^{-1})\ll\min(X,q\|\beta
q\|^{-1})+q\log q.$

Next, we need the following simple bound for the number of curve points close
to integer points. This is Lemma 2 in [14], see also [6, Thm. 5.6], where a
proof is provided.

######  Lemma 3 (Curve points close to integer points).

Let $N$ be a positive integer, and suppose that $g(x):[0,N]\to\mathbb{R}$ has
a continuous derivative on $(0,N)$. Suppose further that $0<\lambda\leq
g^{\prime}(x)\leq A\lambda$ for all $x\in(0,N)$. Then

$\\#\\{n\leq N:\|g(n)\|\leq\delta\\}\ll(1+A\lambda N)(1+\delta/\lambda).$

We also need the following simple assertion.

######  Lemma 4.

Consider positive real functions $S,f$ and $A$. Assume that $S(x)\ll
f(x)S(x)+A(x)$. If $f(x)$ tends to zero for $x\to\infty$, then $S(x)\ll A(x)$.

###### Proof.

If $C$ denotes the implicit constant, then $S(x)(1-Cf(x))\leq CA(x)$. With
$f(x)\leq C/2$ for all large $x$ we deduce $S(x)\leq 2CA(x)$. ∎

Another important ingredient is Vinogradov’s Mean Value Theorem. The theorem
is elementary for $k=1$ and $k=2$. For the highly nontrivial cases $k\geq 3$
it has been proved in [7] by Bourgain, Demeter and Guth for $k\geq 4$ and in
[19] by Wooley for $k=3$, and again in [20] by Wooley for $k\geq 3$. In our
analysis, we will make use of this deep estimate.

######  Theorem 5 (VMVT).

Let $s\geq 1$, $k\geq 1$ be integers and $\varepsilon>0$. Then
$J_{k}(x,s)\ll(x^{s}+x^{2s-k(k+1)/2})x^{\varepsilon}$.

##  2\. Discrete moments of Weyl sums

It is known that VMVT has the following impact on Weyl sum estimates:

######  Theorem 6 (Weyl sum estimate).

Let $P\in\mathbb{R}[X]$ be a polynomial of degree $k\geq 2$, and for the
leading coefficient $\alpha_{k}$ of $P$ let $u,q$ be integers with $(u,q)=1$,
$q\geq 1$ and $|\alpha_{k}-u/q|<q^{-2}$. Then we have

$\sum_{n\leq x}\mathrm{e}(P(n))\ll
x\Big{(}\frac{1}{q}+\frac{1}{x}+\frac{q}{x^{k}}\Big{)}^{1/k(k-1)}x^{\varepsilon}.$

A proof can easily be found using Montgomery’s exposition [17, §4]. Our
analysis of this proof yields a generalization of this estimate for discrete
moments of Weyl sums. By changing a small aspect, it comes with an extra-
improvement, stated below as Theorem 8. Compared to this, Theorem 9 below is
just a straight forward generalization of Theorem 6 that stems from
Montgomery’s original approach presented in [17, §4].

We give the definition of the discrete moments we look at.

######  Definition 7.

Let $k\geq 2$ be a fixed integer and consider a fixed polynomial
$P_{\boldsymbol{\alpha}}\in\mathbb{R}[X]$ of degree $k$ with
$P_{\boldsymbol{\alpha}}(0)=0$, say

$P_{\boldsymbol{\alpha}}(X)=\alpha_{k}X^{k}+\alpha_{k-1}X^{k-1}+\dots+\alpha_{1}X$

with $\alpha_{1},\dots,\alpha_{k}\in\mathbb{R}$. Let $x>1$ be a sufficiently
large real number and for $a\in\mathbb{N}$ let

$S_{a}(\boldsymbol{\alpha}):=\sum_{m\leq
x}\mathrm{e}(aP_{\boldsymbol{\alpha}}(m))$

be a corresponding Weyl sum of $P_{\boldsymbol{\alpha}}$. The twist with $a$
allows us to consider discrete moments of the form

$\sum_{a\leq T}|S_{az}(\boldsymbol{\alpha})|^{2s}$ (1)

with large real $T>1$ and with fixed numbers $z,s\in\mathbb{N}$.

The role of $z$ is to control a possible dependence of a further factor in the
argument of the exponential. We might think of a small $z$, or even $z=1$.

Sums of the shape (1) occur in numerous applications, like in Dirichlet’s
divisor problem, counting integer points close to curves, or, as we will see
below, in the polynomial large sieve inequality (for one variable
polynomials). We will restrict on presenting just the latter two applications
which work well.

Our first goal is to give good estimates for the expression in (1) depending
on $x$ and $T$.

Note that bounds for other moments can then easily be derived by Hölder’s
inequality

$\sum_{a\leq T}|S_{az}(\boldsymbol{\alpha})|^{\ell}\leq
T^{1-\ell/2s}\Big{(}\sum_{a\leq
T}|S_{az}(\boldsymbol{\alpha})|^{2s}\Big{)}^{\ell/2s},\ 0<\ell<2s.$ (2)

Since our results use different values of $s$, it is convenient to state
bounds for the first moment, which makes the statements easy to compare.
Therefore, the results Theorem 8 and Theorem 9 below are stated for the first
moment.

##  3\. Improved moment estimate

We are following the estimate of Weyl sums along the lines of Montgomery’s so-
called alternative derivation in [17, §4.4] and carry it over to the situation
of discrete moments. This approach yields the following result for Weyl sums
as given in Definition 7. We call it the improved moment estimate. The direct
approach leading to Theorem 9 yields a bound that is weaker in certain ranges.
This is discussed in Subsection 3.1.

######  Theorem 8 (Improved moment estimate).

Let $k\geq 3$, $s_{0}=(k-1)(k-2)/2+1$ and $u,q$ be integers with $q\geq 1$,
$(u,q)=1$ and $|\alpha_{k}-u/q|<q^{-2}$. Then

$\sum_{a\leq T}|S_{az}(\boldsymbol{\alpha})|\ll
Tx\Big{(}\frac{zx^{k-1}}{q}+\frac{zx^{k-1}\log(q)}{T}+\frac{1}{x}+\frac{q\log(q)}{Tx}\Big{)}^{1/2s_{0}}x^{\varepsilon}.$

In the bound, we ordered the factors on the right hand side: it starts with
the trivial estimate $Tx$, then we give the improvement factor and then a
small additional factor $x^{\varepsilon}$.

###### Proof.

We need to introduce some of the notations from [17, §4.4], but writing $x$
instead of $N$.

Thus for $j\in\mathbb{N}$, the $j$-th power sum of a tuple
$\mathbf{m}=(m_{1},\dots,m_{s})\in\mathbb{N}^{s}$ is written as
$s_{j}(\mathbf{m}):=m_{1}^{j}+\dots+m_{s}^{j}$, and the difference of two
power sums as
$d_{j}=d_{j}(\mathbf{u},\mathbf{v}):=s_{j}(\mathbf{u})-s_{j}(\mathbf{v})$,
where $u_{i},v_{i}\in\\{-x,\dots,x\\}$.

Multiplying $S_{az}(\boldsymbol{\alpha})^{s}$ out, sorting the summands
according to the value of the power sums with power $j=1,\dots,k-2$, and an
application of Cauchy–Schwarz’s inequality yields

$|S_{az}(\boldsymbol{\alpha})|^{2s}\leq s^{k-2}x^{(k-1)(k-2)/2}\mathcal{T}(a)$

with

$\displaystyle\mathcal{T}(a)$
$\displaystyle:=\sum_{\begin{subarray}{c}\mathbf{m},\mathbf{n}\\\
s_{j}(\mathbf{m})=s_{j}(\mathbf{n}),\
j=1,\dots,k-2\end{subarray}}\mathrm{e}(azP(m_{1}))\cdots\mathrm{e}(azP(m_{s}))$
$\displaystyle\hskip
142.26378pt\cdot\mathrm{e}(-azP(n_{1}))\cdots\mathrm{e}(-azP(n_{s}))$
$\displaystyle=\sum_{\begin{subarray}{c}\mathbf{m},\mathbf{n}\\\
s_{j}(\mathbf{m})=s_{j}(\mathbf{n}),\
j=1,\dots,k-2\end{subarray}}\mathrm{e}\big{(}(s_{k}(\mathbf{m})-s_{k}(\mathbf{n}))az\alpha_{k}$
$\displaystyle\hskip
142.26378pt+(s_{k-1}(\mathbf{m})-s_{k-1}(\mathbf{n}))az\alpha_{k-1}\big{)}$
$\displaystyle=\sum_{\begin{subarray}{c}\mathbf{u},\mathbf{v}\\\
d_{j}(\mathbf{u},\mathbf{v})=0,\
j=1,\dots,k-2\end{subarray}}\mathrm{e}\big{(}d_{k}az\alpha_{k}+d_{k-1}az\alpha_{k-1}\big{)}\sum_{m\in
I}\mathrm{e}\big{(}kd_{k-1}maz\alpha_{k}\big{)},$

compare [17, Eq. (34)–(38) in §4.4]. Here, $m$ runs through an interval
$I=I(\mathbf{u},\mathbf{v},x)$ that contains at most $x$ many successive
integers, and we have put $m=m_{1}$, $m_{i}=m+u_{i}$ for $2\leq i\leq s$, and
$n_{i}=m+v_{i}$ for $1\leq i\leq s$. Note that the vector $\mathbf{u}$
consists of one variable less, it has $s-1$ components. In this step, all
variables $m_{i},n_{i}$ in the Vinogradov System
$s_{j}(\mathbf{m})=s_{j}(\mathbf{n})$, $j=1,\dots,k-2$, have been translated
by $m=m_{1}$, thus we make use of the translation invariance of the Vinogradov
system.

Now let $h=d_{k-1}\leq 2skx^{k-1}$ and sort the tuples $\mathbf{u},\mathbf{v}$
by their value for $d_{k-1}=h$.

Then the summation of $\mathcal{T}(a)$ over $a\leq T$ yields

$\sum_{a\leq T}\mathcal{T}(a)=\sum_{h\ll
x^{k-1}}\sum_{\begin{subarray}{c}\mathbf{u},\mathbf{v}\\\ d_{k-1}=h\\\
d_{j}=0,\ j=1,\dots,k-2\end{subarray}}\sum_{m\in I}\sum_{a\leq
T}\mathrm{e}(az(\alpha_{k}d_{k}+\alpha_{k}khm+h\alpha_{k-1})),$ (3)

where the last geometric sum can be estimated by

$\ll\min(T,\|\alpha_{k}zd_{k}+\alpha_{k}zkhm+zh\alpha_{k-1}\|^{-1}).$

In the following, the notation $\sum^{\prime}$ at the sum over
$\mathbf{u},\mathbf{v}$ abbreviates the condition that $d_{j}=0$ holds for
$j=1,\dots,k-2$.

Using this, we obtain

$\displaystyle\sum_{a\leq T}\mathcal{T}(a)$ $\displaystyle\ll\sum_{h\ll
x^{k-1}}\sideset{}{{}^{\prime}}{\sum}_{\begin{subarray}{c}\mathbf{u},\mathbf{v}\\\
d_{k-1}=h\end{subarray}}\sum_{m\in
I}\min(T,\|\alpha_{k}zd_{k}+\alpha_{k}zkhm+zh\alpha_{k-1}\|^{-1})$
$\displaystyle\ll\sum_{h\ll x^{k-1}}\sum_{m\in
I^{\prime}}\sideset{}{{}^{\prime}}{\sum}_{\begin{subarray}{c}\mathbf{u},\mathbf{v}\\\
d_{k-1}=h\end{subarray}}\min(T,\|\alpha_{k}zd_{k}+\alpha_{k}zkhm+zh\alpha_{k-1}\|^{-1}),$

where we extended the interval $I$ of length at most $x$ to an interval
$I^{\prime}$ of length at most $3x$, in order to remove the dependence on the
variables $\mathbf{u},\mathbf{v}$ except on $h$, which makes the separation of
summation possible. We continue with

$\sum_{a\leq T}\mathcal{T}(a)\ll\sum_{d\ll x^{k}}\sum_{h\ll x^{k-1}}\sum_{m\in
I^{\prime}}\min(T,\|\alpha_{k}zd+\alpha_{k}zkhm+zh\alpha_{k-1}\|^{-1})\\\
\cdot\sideset{}{{}^{\prime}}{\sum}_{\begin{subarray}{c}\mathbf{u},\mathbf{v}\\\
d_{k-1}=h,\ d_{k}=d\end{subarray}}1,$

and changing $hm$ to $w$ yields

$\displaystyle\sum_{a\leq T}\mathcal{T}(a)$ $\displaystyle\ll\sum_{d\ll
x^{k}}\sum_{w\ll x^{k}}\sum_{\begin{subarray}{c}h\mid w\\\ h\ll
x^{k-1}\end{subarray}}\min(T,\|\alpha_{k}z(d+kw)+zh\alpha_{k-1}\|^{-1})$
$\displaystyle\hskip
199.16928pt\cdot\sideset{}{{}^{\prime}}{\sum}_{\begin{subarray}{c}\mathbf{u},\mathbf{v}\\\
d_{k-1}=h,\ d_{k}=d\end{subarray}}1$ $\displaystyle\ll\sum_{j\ll
zx^{k}}\sum_{\begin{subarray}{c}d\ll x^{k}\\\ dz\equiv
j(zk)\end{subarray}}\sum_{\begin{subarray}{c}h\ll x^{k-1}\\\
h\mid(j-zd)/zk\end{subarray}}\min(T,\|\alpha_{k}j+zh\alpha_{k-1}\|^{-1})\sideset{}{{}^{\prime}}{\sum}_{\begin{subarray}{c}\mathbf{u},\mathbf{v}\\\
d_{k-1}=h,\ d_{k}=d\end{subarray}}1$ $\displaystyle\ll\sum_{d\ll
x^{k}}\sum_{h\ll x^{k-1}}\Big{(}\sum_{j\ll
zx^{k}}\min(T,\|\alpha_{k}j+zh\alpha_{k-1}\|^{-1})\Big{)}\sideset{}{{}^{\prime}}{\sum}_{\begin{subarray}{c}\mathbf{u},\mathbf{v}\\\
d_{k-1}=h,\ d_{k}=d\end{subarray}}1,$ (4)

and an application of Lemma 1 to the sum in large brackets yields

$\displaystyle\sum_{a\leq T}\mathcal{T}(a)$ $\displaystyle\ll\sum_{d\ll
x^{k}}\sum_{h\ll
x^{k-1}}(T+q\log(q))(zx^{k}/q+1)\sideset{}{{}^{\prime}}{\sum}_{\begin{subarray}{c}\mathbf{u},\mathbf{v}\\\
d_{k-1}=h,\ d_{k}=d\end{subarray}}1$
$\displaystyle=(T+q\log(q))(zx^{k}/q+1)\sideset{}{{}^{\prime}}{\sum}_{\mathbf{u},\mathbf{v}}1,$

assuming that the integer $q\geq 1$ is such that there exists an integer $u$
with $(u,q)=1$ and $|\alpha_{k}-u/q|<q^{-2}$.

Now we shall give an estimate for the last sum. For
$\boldsymbol{\lambda}\in\mathbb{Z}^{k-2}$ let

$r_{s-1}(\boldsymbol{\lambda}):=\\#\\{\mathbf{u};\
u_{2}+\dots+u_{s}=\lambda_{1},\dots,\
u_{2}^{k-2}+\dots+u_{s}^{k-2}=\lambda_{k-2}\\}$

and similarly

$r_{s}(\boldsymbol{\lambda}):=\\#\\{\mathbf{v};\
v_{1}+\dots+v_{s}=\lambda_{1},\dots,\
v_{1}^{k-2}+\dots+v_{s}^{k-2}=\lambda_{k-2}\\}.$

Then the Cauchy–Schwarz inequality yields

$\displaystyle\sideset{}{{}^{\prime}}{\sum}_{\mathbf{u},\mathbf{v}}1$
$\displaystyle=\sum_{\begin{subarray}{c}\mathbf{u},\mathbf{v}\\\ d_{j}=0,\
j=1,\dots,k-2\end{subarray}}1=\sum_{\boldsymbol{\lambda}\in\mathbb{Z}^{k-2}}r_{s-1}(\boldsymbol{\lambda})r_{s}(\boldsymbol{\lambda})$
$\displaystyle\leq\Big{(}\sum_{\boldsymbol{\lambda}}r_{s-1}(\boldsymbol{\lambda})^{2}\Big{)}^{1/2}\Big{(}\sum_{\boldsymbol{\lambda}}r_{s}(\boldsymbol{\lambda})^{2}\Big{)}^{1/2}=(J_{k-2}(x,s-1)J_{k-2}(x,s))^{1/2},$

and so we obtain

$\displaystyle\sum_{a\leq T}\mathcal{T}(a)$
$\displaystyle\ll(J_{k-2}(x,s-1)J_{k-2}(x,s))^{1/2}\Big{(}\frac{Tzx^{k}}{q}+zx^{k}\log(q)+T+q\log(q)\Big{)}$
$\displaystyle=Tx^{k}(J_{k-2}(x,s-1)J_{k-2}(x,s))^{1/2}\Big{(}\frac{z}{q}+\frac{z\log(q)}{T}+\frac{1}{x^{k}}+\frac{q\log(q)}{Tx^{k}}\Big{)}.$

For the desired moment of Weyl sums this yields

$\displaystyle\sum_{a\leq T}$
$\displaystyle|S_{az}(\boldsymbol{\alpha})|^{2s}$ $\displaystyle\ll
Tx^{(k-1)(k-2)/2+k}(J_{k-2}(x,s-1)J_{k-2}(x,s))^{1/2}$ $\displaystyle\hskip
170.71652pt\cdot\Big{(}\frac{z}{q}+\frac{z\log(q)}{T}+\frac{1}{x^{k}}+\frac{q\log(q)}{Tx^{k}}\Big{)}$
$\displaystyle=Tx^{2s}\Big{(}\frac{J_{k-2}(x,s-1)J_{k-2}(x,s)}{x^{4s-(k-1)(k-2)-2k}}\Big{)}^{1/2}$
$\displaystyle\hskip
170.71652pt\cdot\Big{(}\frac{z}{q}+\frac{z\log(q)}{T}+\frac{1}{x^{k}}+\frac{q\log(q)}{Tx^{k}}\Big{)}.$

Now that we have VMVT, Theorem 5, at hand, we can apply the best possible
bound for the term in big brackets that includes the Vinogradov integrals.
Choosing $s=s_{0}$ with $s_{0}=(k-1)(k-2)/2+1$, we have

$\frac{J_{k-2}(x,s-1)J_{k-2}(x,s)}{x^{4s-(k-1)(k-2)-2k}}\\\ \ll
x^{s-1}x^{2s-(k-1)(k-2)/2}x^{-4s+(k-1)(k-2)+2k}x^{\varepsilon}=x^{2k-2+\varepsilon}$

for this value of $s$. (Note that we have $\ll x^{2k-1+\varepsilon}$ when
choosing $(k-1)(k-2)/2$ for $s$ instead, so the choice $s=s_{0}$ is optimal.)
We arrive at the following estimate.

$\sum_{a\leq T}|S_{az}(\boldsymbol{\alpha})|^{2s_{0}}\ll
Tx^{2s_{0}}x^{k-1}\Big{(}\frac{z}{q}+\frac{z\log(q)}{T}+\frac{1}{x^{k}}+\frac{q\log(q)}{Tx^{k}}\Big{)}x^{\varepsilon}.$
(5)

Using Hölder’s inequality (2), we obtain an estimate for the first moment. In
this way, we obtain the asserted bound from equation (5). ∎

Theorem 8 has to be compared with the result obtained by the straight-forward
approach, that is, the following bound.

######  Theorem 9 (Standard approach estimate).

Let $k\geq 2$, $s_{1}=k(k-1)/2$ and $u,q$ be integers with $q\geq 1$,
$(u,q)=1$ and $|\alpha_{k}-u/q|<q^{-2}$. Then

$\sum_{a\leq T}|S_{az}(\boldsymbol{\alpha})|\ll
Tx\Big{(}\frac{z}{q}+\frac{z}{x}+\frac{q}{Tx^{k}}\Big{)}^{1/2s_{1}}(Txz)^{\varepsilon}.$

Note that with $T=1$ and $z=1$, we get back Theorem 6 above as a special case.

###### Proof.

Montgomery’s original approach in [17, §4.4,p.81,l.15] yields

$\sum_{a\leq T}|S_{az}(\boldsymbol{\alpha})|^{2s}\\\ \ll
x^{(k-1)(k-2)/2}x^{-1}J_{k-1}(3x,s)\sum_{a\leq T}\sum_{0\leq h\leq
2sx^{k-1}}\min(x,\|akhz\alpha_{k}\|^{-1}),$

where the last double sum can be estimated using Lemma 1. Together with the
substitution $w=akhz$ this yields

$\ll\sum_{w\leq
2sTzkx^{k-1}}\tau_{3}(w)\min(x,\|w\alpha_{k}\|^{-1})\ll\Big{(}\frac{zTx^{k}}{q}+zTx^{k-1}+q\Big{)}(Txz)^{\varepsilon},$

where there exist integers $u,q$ with $q\geq 1$, $(u,q)=1$ and
$|\alpha_{k}-u/q|<q^{-2}$. We proceed with

$\sum_{a\leq T}|S_{az}(\boldsymbol{\alpha})|^{2s}\ll
x^{2s}T\Big{(}\frac{J_{k-1}(3x,s)}{x^{2s-k(k-1)/2}}\Big{)}\Big{(}\frac{z}{q}+\frac{z}{x}+\frac{q}{Tx^{k}}\Big{)}(Txz)^{\varepsilon}.$

Next, using VMVT (Theorem 5) with the optimal $s=s_{1}=k(k-1)/2$ leads to the
estimate

$\sum_{a\leq T}|S_{az}(\boldsymbol{\alpha})|^{2s_{1}}\ll
x^{2s_{1}}T\Big{(}\frac{z}{q}+\frac{z}{x}+\frac{q}{Tx^{k}}\Big{)}(Txz)^{\varepsilon}.$
(6)

Applying Hölder’s inequality (2) to equation (6) yields the desired first
moment as given in the assertion of the theorem. ∎

###  3.1. Comparison and conjectural considerations

The expressions in large brackets in Theorems 8 and 9 show the improvements
compared to the trivial estimate $Tx$. They lead to a nontrivial assertion if
$zx^{k-1}\lll q\lll Tx$ in Theorem 8 respectively if $z\lll q\lll Tx^{k}$ in
Theorem 9. Let $s_{0}=(k-1)(k-2)/2+1$ and $s_{1}=k(k-1)/2$.

We compare these improvement expressions (supposing $z$ is small in this
comparison) and obtain the following assertions.

  1. 1.)

In these expressions, we compare the typical dominant terms, $x^{-1/2s_{0}}$
(for $zx^{k}\ll q\ll T$) with $(z/x)^{1/2s_{1}}$ (for $x\ll q\ll zTx^{k-1}$),
we immediately see that Theorem 8 yields a sharper estimate in the
intersection range $zx^{k}\ll q\ll T$.

  2. 2.)

The dominant term $(zx^{k-1}/q)^{1/2s_{0}}$ for $q\ll\min(zx^{k},T)$ in
Theorem 8 is sharper than $(z/x)^{1/2s_{1}}$ for $x\ll q\ll T$, if $q\gg
z^{\sigma}x^{k-\sigma}$ with

$\sigma=\sigma_{k}:=1-s_{0}/s_{1}=2/k-2/k(k-1).$ (7)

To summarize, with Theorem 8 we obtain an improvement in the range
$z^{\sigma}x^{k-\sigma}\ll q\ll\min(zx^{k},T)$.

  3. 3.)

The dominant term $(zx^{k-1}/T)^{1/2s_{0}}$ for $T\ll q\ll zx^{k}$ is sharper
than $(z/x)^{1/2s_{1}}$ for $x\ll q\ll zTx^{k-1}$ if $T\gg
z^{\sigma}x^{k-\sigma}$. Thus in the intersection range $T\ll q\ll zx^{k}$ we
obtain an improvement.

  4. 4.)

The dominant term $(q/Tx)^{1/2s_{0}}$ for $q\gg\max(T,zx^{k})$ is sharper than
$(z/x)^{1/2s_{1}}$ for $x\ll q\ll zTx^{k-1}$ if $q\ll
Tx^{\sigma}z^{1-\sigma}$. Thus in the intersection range $\max(T,zx^{k})\ll
q\ll Tx^{\sigma}z^{1-\sigma}$, for which $T\gg z^{\sigma}x^{k-\sigma}$ has to
hold necessarily, we obtain an improvement.

To summarize, Theorem 8 yields an improvement only if $T\gg
z^{\sigma}x^{k-\sigma}$, so this term $z^{\sigma}x^{k-\sigma}$ turns out to be
a critical value for $T$ from which on we obtain improvements. Moreover, above
conditions on $q$ have to hold, that is the range

$z^{\sigma}x^{k-\sigma}\ll q\ll Tx^{\sigma}z^{1-\sigma}.$ (8)

For any other $q$, Theorem 9 gives a sharper bound.

An observation is that in (4) we made a very coarse estimate. Heuristically,
one would expect that it could be doable with the mean value over $h$. This
would provide a gain of an extra factor $x^{k-1}$ in the estimate. In this
way, we would save it also in (5) and arrive at the following conjectural
bound.

######  Conjecture 10.

For $k\geq 3$, $s_{0}=(k-1)(k-2)/2+1$, and $|\alpha_{k}-a/q|<q^{-2}$ for
$(a,q)=1$, the estimate

$\sum_{a\leq T}|S_{az}(\boldsymbol{\alpha})|\ll
Tx\Big{(}\frac{z}{q}+\frac{z\log(q)}{T}+\frac{1}{x^{k}}+\frac{q\log(q)}{Tx^{k}}\Big{)}^{1/2s_{0}}(Txz)^{\varepsilon}$

is conjectured to hold true.

Compared to Theorem 8, this would lead to an improvement factor
$x^{-k/2s_{0}}$ (around $x^{-1/k}$) instead of $x^{-1/2s_{0}}$, provided that
the secondary terms do not matter. It is interesting to see that we can indeed
improve further towards Conjecture 10 if we assume suitable rational
approximations to $\alpha_{k}$ and $\alpha_{k-1}$ as follows.

######  Theorem 11 (Second improved moment estimate).

Let $k\geq 2$, $s_{2}=k(k-1)/2+1$ and $u,q$ be integers with $q\geq 1$,
$(u,q)=1$ and $|\alpha_{k}-u/q|<q^{-1}T^{-1}$. Further, let $v,w$ be integers
with $1\leq w\leq x^{k-1}z$ and $|zq\alpha_{k-1}-v/w|<w^{-2}$. Then

$\sum_{a\leq T}|S_{az}(\boldsymbol{\alpha})|\ll
Tx\Big{(}\frac{x^{k-1}}{qw}+\frac{x^{k-1}}{T}+\frac{1}{xw}+\frac{q}{Tx}\Big{)}^{1/2s_{2}}x^{\varepsilon}.$

###### Proof.

We start as in the proof of Theorem 8, but continue (4) with

$\displaystyle\sum_{a\leq T}\mathcal{T}(a)$ $\displaystyle\ll\sum_{d\ll
x^{k}}\sum_{h\ll x^{k-1}}\Big{(}\sum_{j\ll
zx^{k}}\min(T,\|\alpha_{k}j+hz\alpha_{k-1}\|^{-1})\Big{)}\sideset{}{{}^{\prime}}{\sum}_{\begin{subarray}{c}\mathbf{u},\mathbf{v}\\\
d_{k-1}=h,\ d_{k}=d\end{subarray}}1$ $\displaystyle\ll\Big{(}\sum_{h\ll
x^{k-1}}\sum_{j\ll
zx^{k}}\min(T,\|\alpha_{k}j+zh\alpha_{k-1}\|^{-1})\Big{)}\max_{h_{0}}\sideset{}{{}^{\prime}}{\sum}_{\begin{subarray}{c}\mathbf{u},\mathbf{v}\\\
d_{k-1}=h_{0}\end{subarray}}1.$ (9)

To handle the last sum, let

$\tilde{r}_{s}(\boldsymbol{\lambda},h_{0}):=\\#\\{\mathbf{v};\
v_{1}+\dots+v_{s}=\lambda_{1},\dots,\
v_{1}^{k-2}+\dots+v_{s}^{k-2}=\lambda_{k-2},\\\
v_{1}^{k-2}+\dots+v_{s}^{k-2}=\lambda_{k-2}+h_{0}\\},$

we obtain

$\displaystyle\sideset{}{{}^{\prime}}{\sum}_{\begin{subarray}{c}\mathbf{u},\mathbf{v}\\\
d_{k-1}=h_{0}\end{subarray}}1$
$\displaystyle=\sum_{\boldsymbol{\lambda}\in\mathbb{Z}^{k-1}}\tilde{r}_{s-1}(\boldsymbol{\lambda},0)\tilde{r}_{s}(\boldsymbol{\lambda},h_{0})$
$\displaystyle\leq\Big{(}\sum_{\boldsymbol{\lambda}}\tilde{r}_{s-1}(\boldsymbol{\lambda},0)^{2}\Big{)}^{1/2}\Big{(}\sum_{\boldsymbol{\lambda}}\tilde{r}_{s}(\boldsymbol{\lambda},h_{0})^{2}\Big{)}^{1/2}$
$\displaystyle=\Big{(}J_{k-1}(x,s-1)J_{k-1}(x,s)\Big{)}^{1/2},$

uniformly in $h_{0}$. Now we turn to the sum over $j,h$ in (3.1). For each
block $B=[1+bq,\dots,q-1+bq]$ of consecutive positive integers with $b\geq 1$,
we have

$\sum_{j\in B}\min(T,\|j\alpha_{k}+hz\alpha_{k-1}\|^{-1})\ll\sum_{1\leq j\leq
q}\min(T,\|j\alpha_{k}+hz\alpha_{k-1}\|^{-1}),$

since
$\min(T,\|j\alpha_{k}+q\alpha_{k}+hz\alpha_{k-1}\|^{-1})\ll\min(T,\|j\alpha_{k}+hz\alpha_{k-1}\|^{-1})$
holds true under the assumption $\|q\alpha_{k}\|<T^{-1}$. Therefore

$\displaystyle\sum_{h\ll x^{k-1}}$ $\displaystyle\sum_{j\ll
zx^{k}}\min(T,\|\alpha_{k}j+hz\alpha_{k-1}\|^{-1})$
$\displaystyle\ll(zx^{k}/q+1)\sum_{h\ll x^{k-1}}\sum_{1\leq j\leq
q}\min(T,\|\alpha_{k}j+hz\alpha_{k-1}\|^{-1})$
$\displaystyle\ll(zx^{k}/q+1)\sum_{h\ll
x^{k-1}}(\min(T,q\|hzq\alpha_{k-1}\|^{-1})+q\log q),$

where we applied Lemma 2 in the last step. Writing $m=hz$ and noting that the
number of divisors of $m$ is $\ll x^{\varepsilon}$, we continue using Lemma 1
by

$\ll(zx^{k}/q+1)x^{\varepsilon}q(T/q+w\log
w)(x^{k-1}z/w+1)+zx^{2k-1+\varepsilon}+x^{k-1+\varepsilon}q\\\ \ll
x^{\varepsilon}z^{2}(x^{k}+q)(T/q+w)x^{k-1}/w+zx^{2k-1+\varepsilon}+x^{k-1+\varepsilon}q\\\
\ll
x^{\varepsilon}z^{2}Tx^{k}\Big{(}\frac{x^{k-1}}{qw}+\frac{x^{k-1}}{T}+\frac{1}{xw}+\frac{q}{Tx}\Big{)},$

supposing $w\leq x^{k-1}z$. Then together with (3.1), we arrive at

$\sum_{a\leq T}\mathcal{T}(a)\ll
x^{\varepsilon}x^{k}T\Big{(}\frac{x^{k-1}}{qw}+\frac{x^{k-1}}{T}+\frac{1}{xw}+\frac{q}{Tx}\Big{)}(J_{k-1}(x,s-1)J_{k-1}(x,s))^{1/2}.$

For the desired moment of Weyl sums this yields now

$\displaystyle\sum_{a\leq T}$
$\displaystyle|S_{az}(\boldsymbol{\alpha})|^{2s}$ $\displaystyle\ll
Tx^{(k-1)(k-2)/2+k+\varepsilon}(J_{k-1}(x,s-1)J_{k-1}(x,s))^{1/2}$
$\displaystyle\hskip
28.45274pt\cdot\Big{(}\frac{x^{k-1}}{qw}+\frac{x^{k-1}}{T}+\frac{1}{xw}+\frac{q}{Tx}\Big{)}$
$\displaystyle=Tx^{2s+\varepsilon}\Big{(}\frac{J_{k-1}(x,s-1)J_{k-1}(x,s)}{x^{4s-(k-1)(k-2)-2k}}\Big{)}^{1/2}\cdot\Big{(}\frac{x^{k-1}}{qw}+\frac{x^{k-1}}{T}+\frac{1}{xw}+\frac{q}{Tx}\Big{)}.$

Again, we need to choose the optimal parameter $s$ which fits best with the
Vinogradov integrals. This is provided by the choice $s_{2}=k(k-1)/2+1$, an
application of VMVT (Theorem 5) yields

$\frac{J_{k-1}(x,s-1)J_{k-1}(x,s)}{x^{4s-(k-1)(k-2)-2k}}\ll
x^{s-1}x^{2s-k(k-1)/2}x^{-4s+(k-1)(k-2)+2k}\ll 1.$

Thus applying Hölder’s inequality, we arrive at the assertion

$\sum_{a\leq T}|S_{az}(\boldsymbol{\alpha})|\ll
Tx\Big{(}\frac{x^{k-1}}{qw}+\frac{x^{k-1}}{T}+\frac{1}{xw}+\frac{q}{Tx}\Big{)}^{1/2s_{2}}.$

∎

We see that in the setting of Theorem 11 we improved the term $1/x$ by $1/xw$,
where $w$ may be taken as large as $x^{k-1}$. This would allow a saving of up
to $x^{-1/k}$ in the estimate, namely

$x^{-k/2s_{2}}=x^{-k/(k(k-1)+2)}=x^{-1/(k-1+2/k))}\ll x^{-1/k},$

assuming best parameter choices for $q$ and $T$ (say $x^{k}\leq q$ and $T\geq
qw$). Like this, we come close to Conjecture 10, but the assumptions on
$\alpha_{k}$ and $\alpha_{k-1}$ are more restrictive.

Theorem 11 also suggests that there might be limitations to Conjecture 10,
such as if $\alpha_{k-1}$ is close to $0$ mod $1$ or has good approximation to
a rational $v/w$ with small denominator $w$. In cases like these it seems that
we may not estimate the sum in (4) much better than the way we proceed.

##  4\. Discrete moments of exponential sums

We turn now to discrete moments of general exponential sums with smooth
functions $f$. The main idea is to approximate $f$ with a polynomial using
Taylor’s theorem and apply the bounds of the previous sections.

We proceed similar as in Bordellès’ book [6, §6.6.7], or in Heath-Brown’s
recent article [14]. The first result is as follows.

######  Theorem 12.

Let $N$ be a large positive integer, and let $f\in\mathbb{C}^{k}((0,3N))$,
$k\geq 3$. Suppose that there exists real numbers $\lambda,A$ such that
$0<\lambda\leq f^{(k)}(x)\leq A\lambda$ holds for all $x\in(0,3N)$. Let
$\rho=1/((k-2)(k-3)+2)$ and $\mu=1+A\lambda N$. Let $z$ be a positive integer
that is considered to be small, and let $T$ be a real number with
$N^{-k}(zA\lambda)^{-1}\leq T\leq(zA\lambda)^{-1}$. Then

$\sum_{a\leq T}\Big{|}\sum_{N<m<2N}\mathrm{e}(azf(m))\Big{|}\ll NT(zA\lambda
T)^{\rho/k+\varepsilon}+T(zA\lambda T)^{-1/k}\\\ +T\mu z^{2}(zA\lambda
T)^{2/k-2}+\mu z(zA\lambda T)^{1/k-1}\lambda^{-1}.$ (10)

We note that $\lambda$ as well as $A$ and $z$ may depend on $N$ and $T$. In
the case if $A$ and $z$ is depending on $k$ only, we may hide $A$ and $z$ in
the implicit constant leading to a slightly easier expression. Additionally
assuming $\mu=1$, the upper bound simplifies to

$NT(\lambda T)^{\rho/k}+T(\lambda T)^{-1/k}+T(\lambda T)^{2/k-2}+(\lambda
T)^{1/k-1}\lambda^{-1}.$ (11)

The proof uses an adapted circle method. The first term in the bound (10)
respectively (11) comes from the the minor arc contribution, the second gives
a trivial contribution from a Weyl-shift, and the last two terms come from the
major arc contribution.

###### Proof.

Let $\mathcal{L}_{f}$ denote the left hand side of (10).

We start with a Weyl-shift with $1\leq H\leq N$. For this, let
$\beta_{m}=\mathrm{e}(azf(m))$ if $N<m<2N$, and $\beta_{m}=0$ otherwise. Then
for each $h^{\prime}\leq H$,

$\sum_{N<m<2N}\mathrm{e}(azf(m))=\sum_{m\in\mathbb{Z}}\beta_{m+h^{\prime}}=\frac{1}{H}\sum_{h\leq
H}\ \sum_{m\in\mathbb{Z}}\beta_{m+h}\\\ =\frac{1}{H}\sum_{m=N-H+1}^{2N-1}\
\sum_{h\leq H}\beta_{m+h}=\frac{1}{H}\sum_{m=N+1}^{2N-1}\ \sum_{h\leq
H}\mathrm{e}(azf(m+h))+O(H).$

We obtain

$\mathcal{L}_{f}=\frac{1}{H}\sum_{a\leq T}\ \sum_{N<m<2N}\Big{|}\sum_{h\leq
H}\mathrm{e}(azf(m+h))\Big{|}+O(TH).$ (12)

An application of Taylor’s theorem provides the expansion
$f(m+h)=Q_{m}(h)+u_{m}(h)$ with

$Q_{m}(h)=hf^{\prime}(m)+h^{2}f^{\prime\prime}(m)/2!+\dots+h^{k-1}f^{(k-1)}(m)/(k-1)!.$

Note that $f^{(k-1)}(m)/(k-1)!$ is the leading coefficient of this polynomial
of degree $k-1$ in $h$, and that

$u_{m}(h)=f(m)+\frac{1}{(k-1)!}\int_{0}^{h}(h-v)^{k-1}f^{(k)}(m+v)dv,$

so that $\mathrm{e}(azf(m+h))=\mathrm{e}(azQ_{m}(h))\mathrm{e}(azu_{m}(h))$.

We separate the exponential expressions containing $Q_{m}$ and $u_{m}$ by
partial summation, this yields

$\mathcal{L}_{f}\leq\mathcal{S}_{1}+\mathcal{S}_{2}+O(TH)$

with

$\displaystyle\mathcal{S}_{1}$
$\displaystyle\leq\frac{1}{H}\sum_{N<m<2N}\sum_{a\leq T}\Big{|}\sum_{h\leq
H}\mathrm{e}(azQ_{m}(h))\Big{|}$

and

$\displaystyle\mathcal{S}_{2}$ $\displaystyle\leq\sum_{a\leq T}\frac{2\pi
az}{H}\sum_{N<m<2N}\int_{1}^{H}\Big{|}\sum_{h\leq
x}\mathrm{e}(azQ_{m}(h))\Big{|}\cdot|u^{\prime}_{m}(x)|\,\mathrm{d}x$
$\displaystyle\ll zH^{k-2}\sum_{N<m<2N}\int_{1}^{H}\sum_{a\leq
T}a\Big{|}\sum_{h\leq
x}\mathrm{e}(azQ_{m}(h))\Big{|}\,\mathrm{d}x\cdot\sup_{v\in(0,H)}|f^{(k)}(m+v)|$
$\displaystyle\ll zA\lambda H^{k-2}\int_{1}^{H}\sum_{N<m<2N}\sum_{a\leq
T}a\Big{|}\sum_{h\leq x}\mathrm{e}(azQ_{m}(h))\Big{|}\,\mathrm{d}x.$

Next, we abbreviate

$S_{a,m}(x):=\sum_{h\leq x}\mathrm{e}(azQ_{m}(h)),$

summarize the bounds for $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$ and arrive at

$\mathcal{L}_{f}\ll TH+\frac{1}{H}\sum_{N<m<2N}\sum_{a\leq T}|S_{a,m}(H)|\\\
+zA\lambda H^{k-2}T\int_{1}^{H}\sum_{N<m<2N}\sum_{a\leq
T}|S_{a,m}(x)|\,\mathrm{d}x.$ (13)

For the next argument, fix $x$ with $x\leq H\leq N$ and let
$\Delta_{0}:=z^{-1}T^{-1}H^{1-k}$. Consider $m\in(N,2N)\cap\mathbb{Z}$ and let

$\mathcal{A}_{m}:=\Big{\\{}\alpha\in[0,1];\
\Big{\|}\frac{f^{(k-1)}(m)}{(k-1)!}-\alpha\Big{\|}\leq\Delta_{0}\Big{\\}}.$

Fix an $\alpha\in\mathcal{A}_{m}$.

We replace the leading coefficient in $Q_{m}(h)$ by $b_{k-1}\in\mathbb{R}$
such that $\frac{f^{(k-1)}(m)}{(k-1)!}-b_{k-1}\in\mathbb{Z}$, so that
$|b_{k-1}-\alpha|\leq\Delta_{0}$. Like this, we look at

$f_{m}^{*}(h):=hf^{\prime}(m)+\dots+h^{k-2}f^{(k-2)}(m)/(k-2)!+b_{k-1}h^{k-1}.$

Let $S_{a,m}^{*}(x):=\sum_{h\leq x}\mathrm{e}(azf_{m}^{*}(h))$, so that
$|S_{a,m}^{*}(x)|=|S_{a,m}(x)|$ and we are able to work with $S_{a,m}^{*}(x)$
instead of $S_{a,m}(x)$ in (13). Moreover, let

$\tilde{f}_{m,\alpha}(h):=hf^{\prime}(m)+\dots+h^{k-2}f^{(k-2)}(m)/(k-2)!+\alpha
h^{k-1}$

and

$\tilde{S}_{a,m}(\alpha,x):=\sum_{h\leq
x}\mathrm{e}(az\tilde{f}_{m,\alpha}(h)).$

Then we have

$\frac{d}{dx}(f_{m}^{*}(x)-\tilde{f}_{m,\alpha}(x))\ll|b_{k-1}-\alpha|x^{k-2}\ll\Delta_{0}x^{k-2},$

and we conclude by a partial summation, that

$S_{a,m}^{*}(x)\ll|\tilde{S}_{a,m}(\alpha,x)|+az\int_{1}^{x}y^{k-2}\Delta_{0}|\tilde{S}_{a,m}(\alpha,y)|\,\mathrm{d}y.$
(14)

Our task is reduced to prove good upper bounds for the term

$\mathcal{T}_{x}:=\sum_{N<m<2N}\sum_{a\leq T}|\tilde{S}_{a,m}(\alpha,x)|$

with $x\leq H$. For each $m$ in the sum there is a chosen
$\alpha\in\mathcal{A}_{m}$. We intend to apply Theorem 8. We expect a good
result if we assume $T$ to be much bigger than $zx^{k-2}$. (Note that
$\deg\tilde{S}_{a,m}(\alpha,x)=k-1$.)

For this purpose, introduce appropriate major and minor arcs. Let

$\mathfrak{M}=\bigcup_{q\leq
zx^{k-1}}\bigcup_{(u,q)=1}\Big{[}\frac{u}{q}-\frac{1}{qT},\frac{u}{q}+\frac{1}{qT}\Big{]}$

denote the set of major arcs, and $\mathfrak{m}=[0,1]\setminus\mathfrak{M}$.

Now we distinguish two cases:

Say case ($\mathfrak{m}$) occurs if $m$ is such that there exists a real
number $\alpha\in\mathcal{A}_{m}\cap\mathfrak{m}$. We choose then such an
$\alpha$ for each such $m$. By Dirichlet’s approximation theorem, there exists
coprime integers $u$ and $q$ with $1\leq q\leq T$ such that

$\Big{|}\alpha-\frac{u}{q}\Big{|}\leq\frac{1}{qT}.$

Since $\alpha$ is contained in $\mathfrak{m}$, we conclude that even $q\geq
zx^{k-1}$ holds true.

A closer look at the improvement expression in Theorem 8 yields

$\Big{(}\frac{zx^{k-2}}{q}+\frac{zx^{k-2}}{T}+\frac{1}{x}+\frac{q}{Tx}\Big{)}^{\rho}\ll
x^{-\rho}$

with $\rho=1/((k-2)(k-3)+2)$. Therefore by Theorem 8,

$\sum_{a\leq T}|\tilde{S}_{a,m}(\alpha,x)|\ll Tx^{1-\rho+\varepsilon},$

hence, summing up over these $m$,

$\mathcal{T}_{x,\text{($\mathfrak{m}$)}}:=\sideset{}{{}^{(\mathfrak{m})}}{\sum}_{N<m<2N}\
\sum_{a\leq T}|\tilde{S}_{a,m}(\alpha,x)|\ll NTx^{1-\rho+\varepsilon}.$ (15)

In the major arc case, $\mathcal{A}_{m}$ is contained completely in a major
arc interval. Then we conclude for $m$ with $N<m<2N$ in case ($\mathfrak{M}$),
that there exist $q\leq zx^{k-1}$ and $(u,q)=1$ such that
$\|f^{(k-1)}(m)/(k-1)!-u/q\|<1/qT$. Summing up over these $m$, we obtain

$\mathcal{T}_{x,\text{($\mathfrak{M}$)}}:=\sideset{}{{}^{(\mathfrak{M})}}{\sum}_{N<m<2N}\
\sum_{a\leq T}|\tilde{S}_{a,m}(\alpha,x)|\\\ \ll Tx\sum_{q\leq
zx^{k-1}}\sum_{(u,q)=1}\\#\\{m\in(N,2N);\ \\\ \hskip
142.26378pt\|f^{(k-1)}(m)/(k-1)!-u/q\|<1/qT\\}\\\ \ll Tx\sum_{q\leq
zx^{k-1}}\varphi(q)(1+A\lambda N)(1+1/\lambda qT),$

where in the last step we applied Lemma 3 with
$g(x):=f^{(k-1)}(x)/(k-1)!-u/q$. From now on we make use of the abbreviation
$\mu=1+A\lambda N$. This yields the bound

$\mathcal{T}_{x,\text{($\mathfrak{M}$)}}\ll Tx(\mu z^{2}x^{2k-2}+\mu
zx^{k-1}\lambda^{-1}T^{-1}),$

hence, in the major arc case,

$\mathcal{T}_{x,\text{($\mathfrak{M}$)}}\ll T\mu z^{2}x^{2k-1}+\mu
zx^{k}\lambda^{-1}.$ (16)

Then joining the estimates (15) and (16) together yields

$\mathcal{T}_{x}\ll\mathcal{T}_{x,\text{($\mathfrak{m}$)}}+\mathcal{T}_{x,\text{($\mathfrak{M}$)}}\ll
NTx^{1-\rho+\varepsilon}+T\mu z^{2}x^{2k-1}+\mu zx^{k}\lambda^{-1}.$ (17)

Next, from estimate (14) together with (17) we obtain

$\sum_{N<m<2N}\sum_{a\leq
T}|S_{a,m}^{*}(x)|\ll\mathcal{T}_{x}+Tz\Delta_{0}\int_{1}^{x}y^{k-2}\mathcal{T}_{y}\,\mathrm{d}y\\\
\ll
NTx^{1-\rho+\varepsilon}+Tz\Delta_{0}\int_{1}^{x}y^{k-2}NTy^{1-\rho+\varepsilon}\,\mathrm{d}y\\\
+T\mu z^{2}x^{2k-1}+Tz\Delta_{0}\int_{1}^{x}y^{k-2}T\mu
z^{2}y^{2k-1}\,\mathrm{d}y\\\ +\mu
zx^{k}\lambda^{-1}+Tz\Delta_{0}\int_{1}^{x}y^{k-2}\mu
zy^{k}\lambda^{-1}\,\mathrm{d}y\\\ \ll NTx^{1-\rho+\varepsilon}+T\mu
z^{2}x^{2k-1}+\mu zx^{k}\lambda^{-1},$

where only in the last step we used $\Delta_{0}=z^{-1}T^{-1}H^{1-k}$ and
$x\leq H$.

Therefore, by (13), we arrive at

$\displaystyle\mathcal{L}_{f}$ $\displaystyle\ll TH+H^{-1}\sum_{m}\sum_{a\leq
T}|S_{a,m}^{*}(H)|$ $\displaystyle\qquad+zA\lambda
H^{k-2}T\int_{1}^{H}\sum_{m}\sum_{a\leq T}|S_{a,m}^{*}(x)|\,\mathrm{d}x$
$\displaystyle\ll TH+H^{-1}NTH^{1-\rho+\varepsilon}+zA\lambda
H^{k-2}T\int_{1}^{H}NTx^{1-\rho+\varepsilon}\,\mathrm{d}x$
$\displaystyle+H^{-1}T\mu z^{2}H^{2k-1}+zA\lambda H^{k-2}T\int_{1}^{H}T\mu
z^{2}x^{2k-1}\,\mathrm{d}x$ $\displaystyle+H^{-1}\mu
zH^{k}\lambda^{-1}+zA\lambda H^{k-2}T\int_{1}^{H}\mu
zx^{k-1}\lambda^{-1}\,\mathrm{d}x$ $\displaystyle\ll
NTH^{-\rho+\varepsilon}+TH+T\mu z^{2}H^{2k-2}+\mu zH^{k-1}\lambda^{-1},$ (18)

where we have chosen $H=[(zA\lambda T)^{-1/k}]$ in (18). This gives the bound

$\mathcal{L}_{f}\ll NT(zA\lambda T)^{\rho/k+\varepsilon}+T(zA\lambda
T)^{-1/k}\\\ +T\mu z^{2}(zA\lambda T)^{2/k-2}+\mu z(zA\lambda
T)^{1/k-1}\lambda^{-1}.$

As necessary constraint for $T$ we get $N^{-k}\leq zA\lambda T\leq 1$, since
we need $1\leq H=[(zA\lambda T)^{-1/k}]\leq N$. ∎

Remark. We have to discuss in which range for $T$ Theorem 12 provides a
nontrivial upper bound for $\mathcal{L}_{F}$.

The first two terms of the bound 10 clearly give a nontrivial upper bound
$\lll TN$, and also the third term is $\lll TN$ provided that $T\mu
z^{2}H^{2k-2}\lll TN$ which means

$T\ggg\mu^{k/(2k-2)}z^{1/(k-1)}(A\lambda)^{-1}N^{k/(2-2k)}.$ (19)

And also the fourth term is $\lll TN$ provided that

$T\ggg\mu^{k/(2k-1)}z^{1/(2k-1)}A^{(1-k)/(2k-1)}\lambda^{-1}N^{k/(1-2k)}.$
(20)

Note that this means $T\ggg\mu(\lambda N)^{-1}zH^{k-1}$, which is stronger
than just $T\ggg zH^{k-1}$ which was expected in the proof to lead to
nontrivial results.

A short calculation shows that these lower bounds (19) and (20) for $T$ are
admissible with the constraint $T\leq(zA\lambda)^{-1}$ provided that
$z^{2}\mu\ll N$ and $z^{2}\mu A\ll N^{k}$. We conclude that then, for small
$z$, there exists a range for $T$ where a nontrivial bound is achieved.

The lower bounds (19) and (20) for $T$ are quite restrictive, but realize the
advantage of Theorem 8 compared to Theorem 9. Using Theorem 9 in the proof
instead will lead to the following slightly weaker bound (21) since
$\tau<\rho$, but provides a larger range for $T$.

######  Theorem 13.

Let $N$ be a large positive integer and let $f\in\mathbb{C}^{k}((0,3N))$,
$k\geq 3$. Suppose that there exists real numbers $\lambda,A$ such that
$0<\lambda\leq f^{(k)}(x)\leq A\lambda$ holds for all $x\in(0,3N)$. Let
$\tau=1/(k-1)(k-2)$ and $\mu=1+A\lambda N$. Let $z$ be a positive integer that
is considered to be small and let $T$ be a positive real number with
$N^{-k}(zA\lambda)^{-1}\leq T\leq(zA\lambda)^{-1}$. Then

$\sum_{a\leq T}\Big{|}\sum_{N<m<2N}\mathrm{e}(azf(m))\Big{|}\ll NT(zA\lambda
T)^{\tau/k+\varepsilon}+T(zA\lambda T)^{-1/k}\\\ +A\mu T(zA\lambda T)^{-2/k}.$
(21)

###### Proof.

We proceed as before in Theorem 12, but choose now the major arc set to be

$\mathfrak{M}=\bigcup_{q\leq
x}\bigcup_{(u,q)=1}\Big{[}\frac{u}{q}-\frac{1}{qTzx^{k-1}},\frac{u}{q}+\frac{1}{qTzx^{k-1}}\Big{]}.$

In the minor arc case, we treat $m$ with $x\leq q\leq zx^{k-1}T$ and we are in
the situation to use Theorem 9 instead, leading to the slightly weaker
estimate

$\mathcal{T}_{x,\text{($\mathfrak{m}$)}}\ll NTx^{1-\tau+\varepsilon},$

since $\tau<\rho$, where $\tau=1/(k-1)(k-2)$. Like this, we estimate the major
arc contribution in a better way, namely

$\mathcal{T}_{x,\text{($\mathfrak{M}$)}}:=\sideset{}{{}^{(\mathfrak{M})}}{\sum}_{N-H<m<2N}\
\sum_{a\leq T}|\tilde{S}_{a,m}(\alpha,x)|\\\ \ll Tx\sum_{q\leq
x}\sum_{(u,q)=1}\\#\\{m\in(N-H,2N);\ \\\ \hskip
142.26378pt\|f^{(k-1)}(m)/(k-1)!-u/q\|<1/qTzx^{k-1}\\}\\\ \ll Tx\sum_{q\leq
x}\varphi(q)(1+A\lambda N)(1+1/\lambda qTzx^{k-1})\\\ \ll Tx(\mu x^{2}+\mu
x/\lambda Tzx^{k-1})\ll\mu Tx^{3}+\mu(\lambda z)^{-1}x^{3-k},$

with $\mu=1+A\lambda N$, again by using Lemma 3. We similarly arrive at

$\mathcal{L}_{f}\ll NTH^{-\tau+\varepsilon}+TH+\mu TH^{2}+\mu(\lambda
z)^{-1}H^{2-k},$

if we choose $H=[(zA\lambda T)^{-1/k}]$ again. Since

$\max\\{1,(T\lambda z)^{-1}H^{-k}\\}=\max\\{1,A\\}=A,$

the last two terms are $\ll A\mu TH^{2}$. Again noting that $1\leq H\leq N$
provides the assertion of Theorem 13. ∎

Remark. Again, we give the range for $T$ where Theorem 13 provides a
nontrivial bound for $\mathcal{L}_{f}$.

We need to inspect the third term in this bound, it is $\lll TN$ provided that
$A\mu TH^{2}\lll TN$ which means

$T\ggg\mu^{k/2}A^{k/2-1}(z\lambda)^{-1}N^{-k/2}.$ (22)

A short calculations shows that this lower bound (22) for $T$ is admissible
with the constraint $T\leq(zA\lambda)^{-1}$ provided that $\mu A\ll N$.

Compared to (19) and (20), the range for $T$ due to (22) will be much bigger
in most cases.

We compare our theorems with the direct application of the following recent
result of Heath-Brown in [14, Thm. 1].

######  Theorem 14 (Heath-Brown).

Let $k\geq 3$, let $f:[0,N]\to\mathbb{R}$ denote a function in $C^{k}((0,N))$,
and suppose that $0<\lambda\leq f^{(k)}(x)\leq A\lambda$ for all $x\in(0,N)$.
Then

$\sum_{n\leq
N}\mathrm{e}(f(n))\ll_{A,k,\varepsilon}N^{1+\varepsilon}(\lambda^{1/k(k-1)}+N^{-1/k(k-1)}+N^{-2/k(k-1)}\lambda^{-2/k^{2}(k-1)}).$

In principle, $\mathcal{L}_{f}$ can be estimated by using Theorem 14, but one
needs then the dependence of the implicit constant on $A$ explicitly since the
term $az$ occurs in the argument of the complex exponential function, so that
$A$ in Theorem 14 contains this factor $az\leq zT$.

Writing down the dependence on $A$ from the proof in [14, Thm. 1] explicitly,
we will have a factor $A^{4}$ occurring in the quantity $\mathcal{N}$ there.
The resulting bound for $\mathcal{L}_{f}$ will then contain the factor
$A^{4/2s}=A^{4/k(k-1)}$. Thus the main term from this method will provide the
extra factor $T^{4/k(k-1)}$ which is much larger than the factors $T^{\rho/k}$
or $T^{\tau/k}$ from our Theorems here.

So compared to this, Theorems 12 and 13 give sharper estimates for long Weyl
sum averages. When no or short averages are considered, Heath-Brown’s bound is
sharper.

Note that the potential improvements depend also on the type of functions
considered. For example, if $f(n)=t\log n$ and $k$ is large with $t=N^{k/2}$,
then the minor arc contribution from Theorem 12 is around
$tN^{1-1/k(k-1)(k-2)}$, whereas Heath-Brown’s bound is around
$tN^{1-1/k^{2}}$. It may be that Theorem 12, as it stands, can _not_ be
improved if uniformity for _all_ functions is desired, but further
improvements may be feasible if a suitable type of function is assumed.

A careful combination of the methods from Theorem 8 and Theorem 12 together
with the idea from the proof of Theorem 11 could lead to further improvements,
which is not discussed in this article.

##  5\. First application: Integer points close to smooth curves

In this application, we will use Theorem 13. We introduce the following
quantity.

######  Definition 15.

Let $N$ be a large positive integer, let $f\in\mathbb{C}^{k}((0,3N))$, $k\geq
3$ and $0<\delta<1/4$. Define

$\mathcal{R}(f,N,\delta):=\\#\\{n\in[N,2N]\cap\mathbb{Z};\
\|f(n)\|<\delta\\}.$

Like this, we count lattice points in $\mathbb{Z}^{2}$ close to the graph of
$f$. Bordellès gives in [6, Ch. 5] an overview of several known nontrivial
bounds for this quantity and their applications. In Lemma 3 we gave already a
bound for $\mathcal{R}(f,N,\delta)$ in the case $k=1$, it is also known as the
first derivative test for $\mathcal{R}(f,N,\delta)$.

In what follows, we use a property of the set which is counted by
$\mathcal{R}(f,N,\delta)$. It is proved in [6, Thm. 5.11].

######  Lemma 16.

Suppose that there exists real numbers $\lambda,A>0$ such that $\lambda\leq
f^{(k)}(x)\leq A\lambda\leq 1/4$ holds for all $x\in(0,3N)$. In the set
$\\{n\in[N,2N]\cap\mathbb{Z};\ \|f(n)\|<\delta\\}$ there exists a
$H^{\prime}$-spaced subset $\mathcal{R}$ such that
$\mathcal{R}(f,N,\delta)\leq(k+1)(1+\\#\mathcal{R})$, where
$H^{\prime}=(A\lambda)^{-2/k(k+1)}$.

A set is called $H^{\prime}$-spaced if any two elements differ by more than
$H^{\prime}>0$.

Theorem 13 of Section 4 allows us to prove a strong bound for
$\mathcal{R}(f,N,\delta)$ which is the following.

######  Theorem 17.

Suppose that there exists real numbers $\lambda,A>0$ such that $\lambda\leq
f^{(k)}(x)\leq A\lambda\leq c_{0}$ holds for all $x\in(0,3N)$ and some small
constant $c_{0}<1/4$, and assume $(A\lambda)^{-2/k(k+1)}\leq N$. Let
$\lambda_{1}>0$ be such that $|f^{\prime}(x)|\leq\lambda_{1}$ holds for all
$x\in(0,3N)$. Assume that $\delta>0$ is such that

$A\lambda\ll\delta+\lambda_{1}\ll(A\lambda)^{1-2(k-1)/k(k+1)},$ (23)

then we have the bound

$\mathcal{R}(f,N,\delta)\ll 1+A(1+A\lambda
N)((\delta+\lambda_{1})/A\lambda)^{1/(k-1)}.$ (24)

###### Proof.

We begin the proof as indicated in [6, Ex. 6.7.4]. Let $m\in\mathbb{Z}$ with
$N\leq m\leq 2N$ and $\|f(m)\|<\delta$, then

$\Big{|}\sum_{a=0}^{T-1}\mathrm{e}(af(m))\Big{|}^{2}=\sum_{a_{1},a_{2}}\mathrm{e}((a_{1}-a_{2})f(m))\\\
=\sum_{a_{1},a_{2}}\Re(\mathrm{e}((a_{1}-a_{2})f(m)))\geq T^{2}/2,$ (25)

since we have $\Re(\mathrm{e}(af(m)))\geq\sqrt{2}/2$ for all $a\in\mathbb{Z}$
with $|a|<T$, provided that $1\leq T\leq[1/8\delta]+1$.

From Lemma 16 we know that there is a $H^{\prime}$-spaced subset $\mathcal{R}$
of $\\{m;\ N\leq m\leq 2N,\|f(m)\|<\delta\\}$ with
$\mathcal{R}(f,N,\delta)\leq(k+1)(1+\\#\mathcal{R})$, where
$H^{\prime}=(A\lambda)^{-2/k(k+1)}$.

Hence

$\mathcal{R}(f,N,\delta)\ll 1+\sum_{m\in\mathcal{R}}1\leq
1+\frac{2}{T^{2}}\sum_{m\in\mathcal{R}}\Big{|}\sum_{a=0}^{T-1}\mathrm{e}(af(m))\Big{|}^{2}.$

Now opening the square and separating the summand for $a=0$ shows

$\Big{|}\sum_{a=0}^{T-1}\mathrm{e}(af(m))\Big{|}^{2}=\sum_{|a|\leq
T-1}(T-|a|)\mathrm{e}(af(m))\\\ =T+\sum_{0<|a|\leq
T-1}(T-|a|)\mathrm{e}(af(m))=:T+a_{m}$ (26)

say. Clearly $a_{m}>0$ for large $T$ and for $m\in\mathcal{R}$ due to (25). We
proceed with

$\mathcal{R}(f,N,\delta)\ll
1+\mathcal{R}(f,N,\delta)T^{-1}+T^{-2}\sum_{m\in\mathcal{R}}a_{m}\ll
1+T^{-2}\sum_{m\in\mathcal{R}}a_{m},$ (27)

by an application of Lemma 4 if we take $T\gg 1$ (in fact $T\geq 4(k+1)$
suffices).

Let $m\in\mathcal{R}$, then there exists an integer $n$ with
$|f(m)-n|<\delta$. Thus, by an application of the mean value theorem,
$|f(m-h)-f(m)|=|f^{\prime}(t)|h\leq\lambda_{1}H$ for some $t\in(-h,0)$. We
conclude $|f(m-h)-n|\leq|f(m-h)-f(m)|+|f(m)-n|\leq\lambda_{1}H+\delta$, so
$\|f(m-h)\|\leq\lambda_{1}H+\delta$ for all $h\leq H$.

This argument shows that for $m\in\mathcal{R}$ we have $a_{m-h}\geq
T^{2}/2-T\gg T^{2}$ by (25) assuming

$T\leq\frac{1}{8(\lambda_{1}H+\delta)}.$ (28)

Since $T^{2}\geq a_{m}$ we conclude that

$a_{m}\ll\frac{1}{H}\sum_{h\leq H}a_{m-h}.$

This implies

$\sum_{m\in\mathcal{R}}a_{m}\ll\sum_{m\in\mathcal{R}}\frac{1}{H}\sum_{h\leq
H}a_{m-h}.$

Now we have

$\sum_{m\in\mathcal{R}}\frac{1}{H}\sum_{h\leq
H}a_{m-h}=\frac{1}{H}\sum_{\begin{subarray}{c}n\in\mathbb{Z}\\\ \exists h\leq
H:n+h\in\mathcal{R}\end{subarray}}a_{n},$

since for each $m\in\mathcal{R}$, the summands $a_{m-1},\dots,a_{m-h}$ occur
in the sum on the right hand side. Assuming $H\leq H^{\prime}$, the elements
of $\mathcal{R}$ are $H$-spaced, so each such summand occurs exactly once on
both sides of this equality.

This improves upon the Weyl step in the proofs above, namely

$\sum_{m\in\mathcal{R}}a_{m}\ll
H^{-1}\sum_{\begin{subarray}{c}n\in\mathbb{Z}\\\ \exists h^{\prime}\leq
H:n+h^{\prime}\in\mathcal{R}\end{subarray}}a_{n}=H^{-2}\sum_{h\leq
H}\sum_{\begin{subarray}{c}n\in\mathbb{Z}\\\ \exists h^{\prime}\leq
H:n+h+h^{\prime}\in\mathcal{R}\end{subarray}}a_{n+h}\\\
=H^{-2}\sum_{\begin{subarray}{c}n\in\mathbb{Z}\\\ \exists
h_{0}\in[1,2H]:n+h_{0}\in\mathcal{R}\end{subarray}}\
\sum_{\begin{subarray}{c}h\leq H\\\
h\in[h_{0}-H,h_{0}-1]\end{subarray}}a_{n+h}\\\
=H^{-2}\sum_{\begin{subarray}{c}n\in\mathbb{Z}\cap[N-2H,2N]\\\ \exists
h_{0}\in[1,2H]:n+h_{0}\in\mathcal{R}\end{subarray}}\
\sum_{\begin{subarray}{c}h\leq H\\\
h\in[h_{0}-H,h_{0}-1]\end{subarray}}a_{n+h}.$

Note that the number of $n\in\mathbb{Z}$ in this sum is
$\ll\min\\{H\\#\mathcal{R},N\\}=:R$, which gives an improvement for a sparse
set $\mathcal{R}$ since then $H\\#\mathcal{R}\ll N$. Further, $h$ lies in the
intersection $\mathcal{H}:=[1,H]\cap[h_{0}-H,h_{0}-1]$ which is an interval of
length at most $H$. The extra factor $H^{-1}$ due to the improved Weyl step
will be an advantage in the minor arc analysis.

With above definition (26) of $a_{m}$, we derive

$\sum_{m\in\mathcal{R}}a_{m}\ll TH^{-2}\sum_{a\leq
T}\sum_{\begin{subarray}{c}m\in[N-2H,2N]\\\ \exists h_{0}\leq
2H:m+h_{0}\in\mathcal{R}\end{subarray}}\Big{|}\sum_{h\in\mathcal{H}}\mathrm{e}(af(m+h))\Big{|}.$
(29)

The right hand side in (29) can now be handled like $\mathcal{L}_{f}$ in the
proofs of the Theorems 12 and 13 above. For this, let us denote in analogy to
$\mathcal{L}_{f}$,

$\mathcal{L}_{f}^{\prime}:=H^{-1}\sum_{a\leq
T}\sum_{\begin{subarray}{c}m\in[N-2H,2N]\\\ \exists h_{0}\leq
2H:m+h_{0}\in\mathcal{R}\end{subarray}}\Big{|}\sum_{h\in\mathcal{H}}\mathrm{e}(af(m+h))\Big{|}.$

Thus

$\mathcal{R}(f,N,\delta)\ll 1+T^{-1}H^{-1}\mathcal{L}_{f}^{\prime}$ (30)

by (27) and (29). Note that $\mathcal{L}_{f}$ differs from
$\mathcal{L}_{f}^{\prime}$ only by the summation over $m$ and that $h$ runs
through an interval $\mathcal{H}$ of length at most $H$, which boundary points
depend on $m$.

The estimation of $\mathcal{L}_{f}^{\prime}$ follows now that of
$\mathcal{L}_{f}$. The only small change in above proof of Theorem 13 lies in
the minor arc estimate (15), where we are able to replace $N$ by $R$ so that

$\mathcal{T}^{\prime}_{x,(\mathfrak{m})}\ll RTx^{1-\tau+\varepsilon}.$

To verify this, note that

$\mathcal{T}_{x}^{\prime}:=\sum_{\begin{subarray}{c}N-2H<m<2N\\\ \exists
h_{0}\leq 2H:m+h_{0}\in\mathcal{R}\end{subarray}}\sum_{a\leq
T}|\tilde{S}_{a,m}^{\prime}(\alpha,x)|$

involves a restriction of $h$ in the sum

$\tilde{S}_{a,m}^{\prime}(\alpha,x)=\sum_{h\in\mathcal{H},h\leq
x}\mathrm{e}(a\tilde{f}_{m,\alpha}(h))$

to $h\in\mathcal{H}$. Still Theorem 9 can be applied since the condition
$h\in\mathcal{H}$ restricts the summation down to a set which is an interval.
Since this interval has an upper bound of at most $x$, this provides the
stated bound for $\mathcal{T}^{\prime}_{x,(\mathfrak{m})}$.

Next, for appropriate $H$, choose $T=[\frac{1}{H^{k}A\lambda}]$. Especially,
if $H^{k}A\lambda-8\lambda_{1}H\geq 8\delta$ then (28) is true. We obtain in
the same way as in Theorem 13 that

$\mathcal{L}_{f}^{\prime}\ll RTH^{-\tau+\varepsilon}+A\mu TH^{2}.$

We put this bound inside (30) above and obtain

$\mathcal{R}(f,N,\delta)\ll 1+T^{-1}H^{-1}(RTH^{-\tau+\varepsilon}+A\mu
TH^{2})\\\ \ll 1+\mathcal{R}(f,N,\delta)H^{-\tau+\varepsilon}+A\mu H.$

By an application of Lemma 4 we leave out the term on the right hand side
containing $\mathcal{R}(f,N,\delta)$ assuming that $H$ is large enough in
terms of the implicit constant. This works since $H^{-\tau+\varepsilon}$ gets
arbitrary small if $H$ increases.

We arrive at the bound

$\mathcal{R}(f,N,\delta)\ll 1+A(1+A\lambda N)H.$

Now we collect all the assumptions made on $H$. Due to Lemma 16 we need
$H\leq(A\lambda)^{-2/k(k+1)}\leq N$.

Moreover, due to (28) we need $H^{k}A\lambda-8\lambda_{1}H\geq 8\delta$, that
is if

$8\delta\leq H(H^{k-1}A\lambda-8\lambda_{1}),$ (31)

for which necessarily $H^{k-1}A\lambda>8\lambda_{1}$ has to be true. Let
$H=((8\delta+8\lambda_{1})/A\lambda)^{1/(k-1)}$, so that (31) holds true. Now
if

$((\delta+\lambda_{1})/A\lambda)^{1/(k-1)}\ll(A\lambda)^{-2/k(k+1)},$

that is, if

$\delta+\lambda_{1}\ll(A\lambda)^{1-2(k-1)/k(k+1)}$ (32)

holds true, we conclude that $H$ is appropriate for the asserted bound in the
theorem to be valid for all small $\delta$ such that (32) holds. We further
need to assume that $H$ is bigger than some constant which makes above step
with Lemma 4 work, so we shall assume also $A\lambda\ll\delta+\lambda_{1}$.

This yields the assertion. ∎

We shall compare the bound in Theorem 17 with the well-known theorem of Huxley
and Sargos from [15, Thm. 1], it states the following bound for
$\mathcal{R}(f,N,\delta)$. The given version here is explicit in $A$ and has
been taken from [6, Thm. 5.12] where a proof is provided. The known proofs are
geometric and do not depend on any exponential sum technique.

######  Theorem 18 (Huxley and Sargos, explicit in $A$).

Let $k\geq 3$ be an integer and $f\in C^{k}([N,2N])$ be such that there exist
$\lambda,A>0$ with $\lambda\leq|f^{(k)}(x)|\leq A\lambda$ for all
$x\in[N,2N]$. Let $0<\delta<1/4$. Then

$\mathcal{R}(f,N,\delta)\ll
N(A\lambda)^{2/k(k+1)}+N(A\delta)^{2/k(k-1)}+(\delta/\lambda)^{1/k}+1.$

In Theorem 18, the first term $N(A\lambda)^{2/k(k+1)}$ dominates if
$\delta\leq\lambda^{1-2/(k+1)}$ $A^{-2/(k+1)}$ and $\delta\ll
N^{k}A^{2/(k+1)}\lambda^{1+2/(k+1)}$, especially when examining very small
$\delta$. The main term $N(A\lambda)^{2/k(k+1)}$ is commonly called the
smoothness term and being regarded as very difficult to improve, compare also
[6, p.275].

A comparison of the term
$NA^{2}\lambda((\delta+\lambda_{1})/A\lambda)^{1/(k-1)}$ in the bound of
Theorem 17 with the smoothness term shows that under the stated assumptions,
Theorem 17 gives a sharper bound. Also note that if we let $\delta\to 0$ in
Theorem 17, we obtain the improved bound $\mathcal{R}(f,N,0)\ll 1+A(1+A\lambda
N)(\lambda_{1}/A\lambda)^{1/(k-1)}$, compared to Theorem 17 which yields the
smoothness term as upper bound for $\mathcal{R}(f,N,0)$.

The bound of Theorem 17 is quite satisfying, the only obstruction lies in the
restrictive assumption

$\lambda_{1}\ll(A\lambda)^{1-2(k-1)/k(k+1)}$ (33)

needed for $\lambda_{1}$ and $A\lambda$. This shows that Theorem 17 is in fact
of limited use in applications since this condition is true only for certain
appropriate functions. Unfortunately, functions like $f(x)=B/x^{r}$ or
$f(x)=(B/x)^{1/r}$ for integers $r\geq 2$, which are occurring in interesting
applications, are not of this kind when $A$ is taken as constant. And if $A$
is so big such that (33) holds, the bound of Theorem 17 can be quite weak.

In this context, we state the following theorem of Gorny [10].

######  Theorem 19 (Gorny).

Let $k\geq 2$, $f\in C^{k}([1,1+3N])$. Let $M,A,\lambda\in\mathbb{R}$ such
that $|f(x)|\leq M$ and $|f^{(k)}(x)|\leq A\lambda$ holds for all
$x\in[1,1+3N]$. Then for all $x\in[1,1+3N]$,

$|f^{\prime}(x)|\ll MN^{-1}+M^{1-1/k}(A\lambda)^{1/k},$

with implicit constant that depends on $k$ only.

By this theorem, we conclude that (33) is true for all sufficiently large $N$
provided that $M\leq(A\lambda)^{1-2/(k+1)}$. Therefore, functions on
$[1,1+3N]$ that are much smaller in absolute value compared to the maximum of
the absolute value of the $k$-th derivative are admissible for Theorem 17.
This gives a nice criterion for Theorem 17 to hold, but it seems to be hard to
find easy examples.

Remark. Applying Theorem 12 instead of Theorem 13 in the proof of Theorem 17
would also lead to a vanishing of the minor arc contribution in the bound, but
the resulting bound for $\mathcal{R}(f,N,\delta)$ would be weaker due to the
bigger major arc contribution. Instead, the presented vanishing trick may be
used with even bigger minor arc contributions probably leading to further
refinements.

##  6\. Second application: The polynomial large sieve inequality (LSI) in the
one-dimensional case

In this section we present an application of Theorem 8, namely to the
polynomial large sieve inequality. This generalization of the classical large
sieve inequality to sparse moduli sets, usually given as values of some fixed
polynomial, has been studied intensely in past research and has already
influenced some other areas of Number Theory, especially the version with
$k=2$ from [2]. To name a few such topics, it has been found useful to find
variants of Bombieri–Vinogradov’s theorem [4, 13], new results on primes of
polynomial shape [9] or primes in APs to spaced moduli [4], divisibility
questions with Fermat quotients [8], and mean value estimates for character
sums with applications [1, 3, 5, 16]. Furthermore, the multidimensional
polynomial LSI can be used for sieving with high powers like seen in [12] and
reaches questions related to the abc-conjecture.

We start by giving the setting and basic assumptions in the polynomial LSI in
the one-dimensional case.

Setting. Let $P\in\mathbb{R}[x]$ be a fixed monic polynomial of degree $k\geq
2$ with $P(0)=0$. Assume that $P$ has only positive values in $[Q,2Q]$ for
each real $Q\geq 1$ and let $M_{Q}:=\max\\{P(q);\ q\in[Q,2Q]\\}$ be the
maximal value for integers $q\in[Q,2Q]$. Clearly $M_{Q}\ll Q^{k}$, assume also
that $P(q)\gg Q^{k}$ holds true for all integers $q\in[Q,2Q]$ and some
implicit constant that may depend only on $k$. Let $N,M$ be integers and
$(v_{n})_{n\in\mathbb{N}}$ be a sequence of complex numbers.

In the theory of the large polynomial LSI, see [2, 11, 21], we aim to give
upper bounds for the quantity

$\Sigma_{P}:=\sum_{q\leq Q}\sum_{\begin{subarray}{c}1\leq a\leq P(q)\\\
(a,P(q))=1\end{subarray}}\Big{|}\sum_{M<n\leq
M+N}v_{n}\mathrm{e}\Big{(}\frac{an}{P(q)}\Big{)}\Big{|}^{2}.$

When we put the current form for Weyl’s inequality, Theorem 6, in the
machinery of [11, 12], the bound $\Sigma_{P}\ll
Q^{\varepsilon}\|v\|^{2}(Q^{k+1}+A_{k}(Q,N))$ is easily derived with

$A_{k}(Q,N):=NQ^{1-1/k(k-1)}+N^{1-1/k(k-1)}Q^{1+1/(k-1)}.$

The interesting range for $N$ is $Q^{k}\ll N\ll Q^{2k}$, since outside that,
it is already known by an application of the standard large sieve inequality
that the sharp bound $\Sigma_{P}\ll Q^{\varepsilon}\|v\|^{2}(Q^{k+1}+N)$ holds
true. So we assume without loss of generality that $N$ lies in this range.

This result already offers an improvement compared to [3] in the range
$Q^{k}\ll N\ll Q^{2k-2+2/k(k-1)}$ when $k\geq 3$. We will now improve on that.

###  6.1. The connection of the Polynomial LSI with Weyl sums

From [11, Lemma 1], we know that

$\Sigma_{P}\ll Q^{\varepsilon}\|v\|^{2}\bigg{(}\sum_{Q<q\leq
2Q}P(q)+\max_{Q<r\leq 2Q}\max_{\begin{subarray}{c}1\leq b<P(r)\\\
\operatorname{gcd}(b,P(r))=1\end{subarray}}\int_{1/N}^{1/2}\\#\mathcal{F}_{b,P(r)}(x)\frac{dx}{x^{2}}\bigg{)}.$
(34)

The first term in large brackets can be estimated as $\ll QM_{Q}$ and is
admissible. It is not necessary to repeat the definition of
$\\#\mathcal{F}_{b,P(r)}(x)$ since we will just make use of the upper bound

$\\#\mathcal{F}_{b,P(r)}(x)\ll B^{-1}Q+B^{-1}\sum_{1\leq a\leq B}|S_{a}|$

with $B^{-1}=2M_{Q}x$ and

$S_{a}:=\sum_{q\leq 2Q}\mathrm{e}\Big{(}\frac{ab}{P(r)}P(q)\Big{)},$

which has been shown in the deduction of [11, (8)].

To consider the integral expression in (34), fix a pair $b,r$ with
$r\in[Q,2Q]$, $1\leq b<P(r)$ and $\operatorname{gcd}(b,P(r))=1$. We substitute
$B^{-1}=2M_{Q}x$ and estimate as follows.

$\begin{aligned}
\int_{1/N}^{1/2}&\\#\mathcal{F}_{b,P(r)}(x)\frac{dx}{x^{2}}\\\
&\ll\int_{1/4M_{Q}}^{N/2M_{Q}}\bigg{(}B^{-1}Q+B^{-1}\sum_{a\leq
B}|S_{a}|\bigg{)}M_{Q}\,dB\\\ &\ll QM_{Q}\log
N+M_{Q}\int_{1/4M_{Q}}^{N/2M_{Q}}B^{-1}\sum_{a\leq B}|S_{a}|\,dB\\\ &\ll
QM_{Q}\log N+M_{Q}\sum_{a\leq N/2M_{Q}}|S_{a}|\int_{a}^{N/2M_{Q}}B^{-1}\,dB\\\
&\ll QM_{Q}\log N+M_{Q}\log N\sum_{a\leq N/2M_{Q}}|S_{a}|.\end{aligned}$

Here the last sum is a discrete moment of a Weyl sum with the polynomial
$bP(x)/P(r)$ and leading term $b/P(r)$ since $P$ is monic. We are able to
apply Theorem 8 directly with $P(r)$ as approximating denominator (when $k\geq
3$). By this, we have

$\sum_{a\leq
N/2M_{Q}}|S_{a}|\ll\frac{N}{M_{Q}}Q^{1+\varepsilon}\Big{(}\frac{Q^{k-1}}{P(r)}+\frac{Q^{k-1}}{N/M_{Q}}+\frac{1}{Q}+\frac{P(r)}{QN/M_{Q}}\Big{)}^{1/2s_{0}}$

with

$s_{0}=(k-1)(k-2)/2+1,\text{ so that }2s_{0}=k(k-1)-2k+4.$

Let $\omega:=1/2s_{0}$.

In the big bracket expression, the last summand $P(r)M_{Q}/NQ$ dominates since
$P(r)\gg Q^{k}$ and $1/Q\ll P(r)M_{Q}/NQ$ for $N\ll Q^{2k}$.

So we continue with

$\displaystyle\int_{1/N}^{1/2}$
$\displaystyle\\#\mathcal{F}_{b,P(r)}(x)\frac{dx}{x^{2}}$ $\displaystyle\ll
QM_{Q}\log Q+N^{1-\omega}M_{Q}^{\omega}Q^{1-\omega+\varepsilon}P(r)^{\omega}$
$\displaystyle\ll QM_{Q}\log
Q+N^{1-\omega}M_{Q}^{2\omega}Q^{1-\omega+\varepsilon}$ $\displaystyle\ll
Q^{k+1}\log Q+N^{1-\omega}Q^{1+2k\omega-\omega+\varepsilon},$

where we used $M_{Q}\ll Q^{k}$ in the last step.

Compared to the dominating term $NQ^{1-1/k(k-1)}$ in the former bound
$A_{k}(Q,N)$, we get an advantage if $N^{1-\omega}Q^{1+(2k-1)\omega}\leq
NQ^{1-1/k(k-1)}$, which is the case if $N\geq Q^{2k-2/(k-1)+4/k(k-1)}$, so
when $N$ is close to $Q^{2k}$, but still in the interesting range $Q^{k}\ll
N\ll Q^{2k}$. We have therefore shown the following new improved bound for the
polynomial LSI.

######  Theorem 20.

In the setting of Section 6 when $k\geq 3$,

$\Sigma_{P}\ll
Q^{\varepsilon}\|v\|^{2}(Q^{k+1}+\min\\{A_{k}(Q,N),N^{1-\omega}Q^{1+(2k-1)\omega}\\})$
(35)

with $\omega=1/((k-1)(k-2)+2)$.

Theorem 20 offers an improvement compared to known previous results when
$k\geq 4$. This is since for $k=3$, the additional bound from [3] is still
stronger in this case.

It is interesting what we would obtain having Conjecture 10. In this case, we
would be able to gain a factor $Q^{(1-k)\omega}$. Then, we would arrive at the
following result.

######  Conjecture 21.

In the setting of Section 6 when $k\geq 3$,

$\Sigma_{P}\ll
Q^{\varepsilon}\|v\|^{2}(Q^{k+1}+\min\\{A_{k}(Q,N),N^{1-\omega}Q^{1+k\omega}\\})$

with $\omega=1/((k-1)(k-2)+2)$.

Note that if we could take even $1/k(k-1)$ at the place of $\omega$, the
expression $N^{1-1/k(k-1)}Q^{1+1/(k-1)}$ coincides with the second summand in
$A_{k}(Q,N)$.

Still, these conjectural bounds are far from Zhao’s conjecture in [21] stating

$\Sigma_{P}\ll Q^{\varepsilon}\|v\|^{2}(Q^{k+1}+N).$

Conjecture 21 might be rather within reach of further refinements of the
methods presented in this article.

We remark that an attempt to use Theorem 11 will not give more if more on the
coefficient of $q^{k-1}$ in $P(q)$, say $\alpha_{k-1}$, is known. This is
since the coefficient of $q^{k-1}$ in the polynomial $bP(q)/P(r)$ is then
$\alpha_{k-1}b/P(r)$. By Theorem 11, one needs to look then at the rational
approximations to $\alpha_{k-1}b$. Since $b$ is supposed to be _any_ coprime
residue mod $P(r)$, there will always be one with small denominator which
offers no advantage in Theorem 11.

##  7\. Acknowledgements

The author thanks the organizers of the Workshop on Efficient Congruencing in
March 2017 at the Fields institute in Toronto during the Thematic Program on
Unlikely Intersections, Heights, and Efficient Congruencing for an inspiring
stay. The author also thanks the referee for many useful suggestions and
comments on the manuscript which led to an improvement of some of the
material.

## References

  * [1] S. Baier, M.P. Young, _Mean values with cubic characters_ , J. Number Theory 130 (2010), no. 4, 879–903.
  * [2] S. Baier and L. Zhao, _An improvement for the large sieve for square moduli_ , J. Number Theory 128 (2008), no. 1, 154–174.
  * [3] S. Baier and L. Zhao, _Large sieve inequality with characters for powerful moduli_ , Int. J. Number Theory 1 (2005), no. 2, 265–279.
  * [4] R. Baker, _Primes in arithmetic progressions to spaced moduli. III_ , Acta Arith. 179 (2017), no. 2, 125–132.
  * [5] V. Blomer, L. Goldmakher, B. Louvel, _L-functions with n-th-order twists_ , Int. Math. Res. Not. IMRN 2014, no. 7, 1925–1955.
  * [6] O. Bordellès, _Arithmetic Tales_ , Universitext. Springer, London, 2012. xxii+556 pp. ISBN: 978-1-4471-4095-5
  * [7] J. Bourgain, C. Demeter and L. Guth, _Proof of the main conjecture in Vinogradov’s mean value theorem for degrees higher than three_ , Ann. of Math. (2) 184 (2016), no. 2, 633–682.
  * [8] J. Bourgain, K. Ford, S.V. Konyagin, I.E. Shparlinski _On the divisibility of Fermat quotients_ , Michigan Math. J., 59 (2) (2010), 313–328.
  * [9] T. Foo, L. Zhao, _On primes represented by cubic polynomials_ , Math. Z. 274 (2013), no. 1-2, 323–340.
  * [10] A. Gorny, _Contribution à l’étude des fonctions dérivables d’une variable réelle_ , Acta Math. 71 (1939), 317–358.
  * [11] K. Halupczok, _A new bound for the large sieve inequality with power moduli_ , Int. J. Number Theory 8 (2012), no. 3, 689–695.
  * [12] K. Halupczok, _Large sieve inequalities with general polynomial moduli_ , Q. J. Math. 66 (2015), no. 2, 529–545.
  * [13] K. Halupczok, _A Bombieri-Vinogradov theorem with products of Gaussian primes as moduli_ , Funct. Approx. Comment. Math. 57 (2017), no. 1, 77–91.
  * [14] D. R. Heath-Brown, _A New k-th Derivative Estimate for Exponential Sums via Vinogradov’s Mean Value_ , arXiv:1601.04493v3 [math.NT]
  * [15] M. N. Huxley and P. Sargos, _Points entiers au voisinage d’une courbe plane de classe Cn_. (French) [Integral points near a plane curve of class Cn] Acta Arith. 69 (1995), no. 4, 359–366.
  * [16] Y. Lamzouri, _Large values of $L(1,\chi)$ for kth order characters $\chi$ and applications to character sums_, Mathematika 63 (2017), no. 1, 53–71.
  * [17] H. L. Montgomery, _Ten Lectures on the Interface Between Analytic Number Theory and Harmonic Analysis_ , CBMS Number 84.
  * [18] W. Schmidt, Small fractional parts of polynomials, CBMS Number 32.
  * [19] T. Wooley, _Vinogradov’s mean value theorem via efficient congruencing_ , Ann. of Math. (2) 175 (2012), no. 3, 1575–1627.
  * [20] T. Wooley, _Nested efficient congruencing and relatives of Vinogradov’s mean value theorem_ , arXiv:1708.01220v2 [math.NT]
  * [21] L. Zhao, _Large sieve inequality with characters to square moduli_ , Acta Arith. 112 (2004), no. 3, 297–308.


$\Lambda^{ab}_{3,ij}(\tilde{q})$ as a function of $\tilde{q}$,
$\Lambda(\tilde{q})=\Lambda(0)+(\partial_{\bf
q}\Lambda)_{\tilde{q}=0}\cdot{\bf
q}+(\partial_{\omega_{n}}\Lambda)_{\tilde{q}=0}\omega_{n}+...$. At the zero
order term, we have

$\displaystyle
W^{(0)}=\frac{1}{2}\sum_{\tilde{q}}\left(\bar{u}(\tilde{q})\Lambda_{0}(0)\bar{u}(-\tilde{q})+\sum_{i,a}\bar{u}(\tilde{q})\Lambda^{a}_{2,i}(0)u_{ia}(-\tilde{q})+\sum_{ij,ab}u_{ia}(\tilde{q})\Lambda^{ab}_{3,ij}(0)u_{jb}(-\tilde{q})\right)$
$\displaystyle=\frac{1}{2}\int_{0}^{\beta}d\tau\int d^{3}r\left(\bar{u}({\bf
r},\tau)\Lambda_{0}(0)\bar{u}({\bf r},\tau)+\sum_{i,a}\bar{u}({\bf
r},\tau)\Lambda^{a}_{2,i}(0)u_{ia}({\bf r},\tau)+\sum_{ij,ab}u_{ia}({\bf
r},\tau)\Lambda^{ab}_{3,ij}(0)u_{jb}({\bf r},\tau)\right).$ (166)

Here all the terms are quadratic in the strain tensor $u_{ij}$ and let’s
consider the general form of the effective action for acoustic phonons
(elastic wave), which is defined asBarkeshli  _et al._ (2012); Landau  _et
al._ (1986)

$S_{ph,0}=\frac{1}{2}\int
dtd^{3}r\left(\rho\sum_{j}\partial_{t}u_{j}\partial_{t}u_{j}-\sum_{ijkl}\lambda_{ijkl}\partial_{i}u_{j}\partial_{k}u_{l}\right),$
(167)

where the rank-4 tensor $\lambda_{ijkl}$ is the elastic modulus. It is clear
that $W^{(0)}$ just provides the correction of the elastic modulus
$\lambda_{ijkl}$.

Next we focus on the first order term with the linear ${\bf q}$ dependence,
which takes the form

$\displaystyle
W^{(1)}=\frac{1}{2}\sum_{\tilde{q}}\left(\sum_{i}(\partial_{i}\Lambda_{0})\bar{u}(\tilde{q})q_{i}\bar{u}(-\tilde{q})+\sum_{ij,a}(\partial_{j}\Lambda^{a}_{2,i})\bar{u}(\tilde{q})q_{j}u_{ia}(-\tilde{q})+\sum_{ijl,ab}(\partial_{l}\Lambda^{ab}_{3,ij})u_{ia}(\tilde{q})q_{l}u_{jb}(-\tilde{q})\right)$
$\displaystyle=\frac{1}{2}\int_{0}^{\beta}d\tau\int
d^{3}r\left(\sum_{j}(\partial_{j}\Lambda_{0})\bar{u}({\bf
r},\tau)\left(i\frac{\partial}{\partial r_{j}}\bar{u}({\bf
r},\tau)\right)+\sum_{ij,a}(\partial_{j}\Lambda^{a}_{2,i})\bar{u}({\bf
r},\tau)\left(i\frac{\partial}{\partial r_{j}}u_{ia}({\bf
r},\tau)\right)\right.$
$\displaystyle\left.+\sum_{ijk,ab}(\partial_{k}\Lambda^{ab}_{3,ij})u_{ia}({\bf
r},\tau)\left(i\frac{\partial}{\partial r_{k}}u_{jb}({\bf
r},\tau)\right)\right),$ (168)

where $(\partial_{j}\Lambda)=\left(\frac{\partial\Lambda}{\partial
q_{j}}\right)_{\tilde{q}=0}$ is just a number. From Eq. (165), one can see
that the last term is just the Nieh-Yan term. Although the above expression
seems quite complex, the symmetry gives a strong constraint on the form of the
allowed effective action. It is clear that the above action contributes to the
following general form of the effective action of acoustic phonons

$S_{ph,1}=\int dtd^{3}r\sum_{ijklm}\xi_{ijklm}(\partial_{i}u_{jk})u_{lm},$
(169)

where $\partial_{i}=\frac{\partial}{\partial r_{i}}$. Now the question is what
is the most general form of the coefficient $\xi_{ijklm}$, a rank-5 tensor,
for a uniform isotropic system. Since the strain tensor $u_{ij}$ is symmetric
with respect to the indices $i$ and $j$, the coefficient $\xi_{ijklm}$ should
satisfy the relations $\xi_{ijklm}=\xi_{ikjlm}=\xi_{ijkml}$. Furthermore, it
also satisfies the anti-symmetric relation $\xi_{ijklm}=-\xi_{ilmjk}$. These
relations together with the full rotation symmetry will fix the form of
$\xi_{ijklm}$.

It is also equivalent to ask how to construct an invariant term with the form
of $(\partial_{i}u_{jk})u_{lm}$. We analyze this problem based on the
irreducible representation of the angular momentum. As discussed in Sec. B,
the symmetric rank-2 tensor $u_{ij}$ can be decomposed to two parts, $\bar{u}$
with the angular momentum 0 and $u_{J=2,M}$ with the angular momentum 2. Since
the expression involves two strain tensors, the corresponding angular momentum
decomposition can be given by

$\displaystyle{\bf 2\otimes 2=4\oplus 3\oplus 2\oplus 1\oplus 0},$ (170)
$\displaystyle{\bf 2\otimes 0=2},$ (171) $\displaystyle{\bf 0\otimes 0=0},$
(172)

where the bold number represents the total angular momentum. Furthermore, the
expression $(\partial_{i}u_{jk})u_{lm}$ involves one $\partial_{i}$ in
addition to two strain tensors, and $\partial_{i}$ possesses the angular
momentum ${\bf 1}$. To make the term $(\partial_{i}u_{jk})u_{lm}$ invariant
(with the angular momentum ${\bf 0}$), only the irreducible representation
with the angular momentum ${\bf 1}$ in ${\bf 2\otimes 2}$ can lead to ${\bf
1\otimes 1=2\oplus 1\oplus 0}$. From such angular momentum combination, only a
single term is allowed for $\xi_{ijklm}$ in a uniform isotropic system. This
term can be explicitly constructed as

$S_{ph,1}=\xi_{0}\int
dtd^{3}r\sum_{ijklm}\delta_{jl}\epsilon_{ikm}(\partial_{i}u_{jk})u_{lm},$
(173)

so that the parameter $\xi_{ijklm}$ given by

$\xi_{ijklm}=\frac{\xi_{0}}{4}(\delta_{jl}\epsilon_{ikm}+\delta_{kl}\epsilon_{ijm}+\delta_{jm}\epsilon_{ikl}+\delta_{km}\epsilon_{ijl}).$
(174)

We can apply the above analysis to the action $W^{(1)}$, and it is clear that
it is impossible to construct an invariant term from the first two terms in
Eq. (G), and thus the coefficient $\partial_{j}\Lambda_{0}$ and
$\partial_{j}\Lambda^{a}_{2,i}$ must vanish. The effective action then takes
the form

$W^{(1)}=\frac{1}{2}\int_{0}^{\beta}d\tau\int
d^{3}r\left(\sum_{ijk,ab}(i\partial_{k}\Lambda^{ab}_{3,ij})u_{ia}({\bf
r},\tau)\left(\frac{\partial}{\partial r_{k}}u_{jb}({\bf
r},\tau)\right)\right),$ (175)

with the coefficient

$(i\partial_{k}\Lambda^{ab}_{3,ij})=i\frac{g_{1}^{2}}{(\hbar
v_{f})^{2}}\left(\partial_{k}\Phi^{ab}_{ij}(\tilde{q})\right)_{\tilde{q}=0}=i\frac{g_{1}^{2}}{(\hbar
v_{f})^{2}}\Phi^{ab}_{ij,k}=\frac{g_{1}^{2}}{(\hbar
v_{f})^{2}}\epsilon^{abk}\delta_{ij}\chi\mu\mathcal{F}_{0}.$ (176)

Here we have used Eq. 104. Substituting the coefficient and also changing from
imaginary time to real time, we have

$\displaystyle W^{(1)}=-\frac{g_{1}^{2}\chi\mu\mathcal{F}_{0}}{2(\hbar
v_{f})^{2}}\sum_{ijk,ab}\int dtd^{3}r\epsilon^{abk}\delta_{ij}u_{ia}({\bf
r},\tau)\left(\frac{\partial}{\partial r_{k}}u_{jb}({\bf r},\tau)\right)$
$\displaystyle=\frac{g_{1}^{2}\chi\mu\mathcal{F}_{0}}{2(\hbar
v_{f})^{2}}\sum_{ijk,ab}\int
dtd^{3}r\epsilon^{kab}\delta_{ij}\left(\frac{\partial}{\partial
r_{k}}u_{ia}({\bf r},\tau)\right)u_{jb}({\bf r},\tau).$ (177)

Comparing with $S_{ph,1}$ in Eq. (173), one can see that the Nieh-Yan term can
contribute to a new term in the phonon effective action and the corresponding
coefficient is given by

$\xi_{0}=\frac{g_{1}^{2}\chi\mu\mathcal{F}_{0}}{2(\hbar
v_{f})^{2}}=\frac{g_{1}^{2}\mu_{5}\mathcal{F}_{0}}{(\hbar v_{f})^{2}}.$ (178)

In the above discussion, we have dropped the term with the form of
$u_{ij}\omega_{n}u_{kl}$ and in the time-domain, this term should have the
form of $u_{ij}(\partial_{\tau}u_{kl})$. In the elastic theory, this term is
nothing but the viscosity, generally defined as Barkeshli  _et al._ (2012);
Avron  _et al._ (1995); Landau  _et al._ (1986)

$S_{ph,2}=\int dtd^{3}r\sum_{ijkl}\eta_{ijkl}(\partial_{t}u_{jk})u_{lm}.$
(179)

However, the existence of the viscosity requires the breaking of TR symmetry,
while TR symmetry exists in our system. Thus, we will expect the viscosity
coefficient $\eta_{ijkl}$ extracted from our model should vanish.

Now we hope to solve the acoustic phonon dispersion (or elastic wave) in our
system based on the actions (167) and (173). For the uniform isotropic system,
there are two independent parameters for the elastic moduli $\lambda_{ijkl}$
and the corresponding equation of motion for the displacement field ${\bf u}$
is given by

$\frac{d^{2}}{dt^{2}}{\bf u}=c_{t}^{2}\nabla^{2}{\bf
u}+(c_{l}^{2}-c_{t}^{2})\nabla(\nabla\cdot{\bf u}),$ (180)

where $c_{l}$ and $c_{t}$ are the velocities of longitudinal and transverse
waves. The Nieh-Yan term $S_{ph,1}$ can add an additional term into the above
elastic wave equation, as given by

$\displaystyle\frac{\delta S_{ph,1}}{\delta
u_{n}}=\frac{\xi_{0}}{2}\epsilon_{ikm}\left(\delta_{kn}\partial_{i}\partial_{j}u_{jm}+\delta_{jn}\partial_{i}\partial_{k}u_{jm}-\delta_{mn}\partial_{i}\partial_{j}u_{jk}-\delta_{jn}\partial_{i}\partial_{m}u_{jk}\right)$
$\displaystyle=\frac{\xi_{0}}{2}\left(\epsilon_{inm}\partial_{i}\partial_{j}u_{jm}+\epsilon_{ikm}\partial_{i}\partial_{k}u_{nm}-\epsilon_{ikn}\partial_{i}\partial_{j}u_{jk}-\epsilon_{ikm}\partial_{i}\partial_{m}u_{nk}\right)$
$\displaystyle=\frac{\xi_{0}}{2}\left(\epsilon_{inm}\partial_{i}\partial_{j}u_{jm}-\epsilon_{ikn}\partial_{i}\partial_{j}u_{jk}\right)=\frac{\xi_{0}}{2}\epsilon_{inm}\partial_{i}\partial_{j}(\partial_{j}u_{m}+\partial_{m}u_{j})$
$\displaystyle=\frac{\xi_{0}}{2}\epsilon_{inm}\partial_{i}(\nabla^{2}u_{m})=-\frac{\xi_{0}}{2}(\nabla\times(\nabla^{2}{\bf
u}))_{n}$ (181)

By including the Nieh-Yan term, the equation of motion for the displacement
field is now written as

$\frac{d^{2}}{dt^{2}}{\bf u}=c_{t}^{2}\nabla^{2}{\bf
u}+(c_{l}^{2}-c_{t}^{2})\nabla(\nabla\cdot{\bf
u})+\frac{\xi_{0}}{2}\nabla\times(\nabla^{2}{\bf u}).$ (182)

We can decompose the displacement field into the longitudinal and transverse
components, ${\bf u=u_{l}+u_{t}}$. One can show that the longitudinal and
transverse components are still decoupled from each other. The longitudinal
part does not have any correction from the Nieh-Yan term, while the equation
of motion for the transverse part is revised as

$\frac{d^{2}}{dt^{2}}{\bf u}_{t}=c_{t}^{2}\nabla^{2}{\bf
u}_{t}+\frac{\xi_{0}}{2}\nabla\times(\nabla^{2}{\bf u}_{t}).$ (183)

Let’s take ${\bf u}_{t}={\bf u}_{0}e^{i{\bf q\cdot r}-i\omega t}$ and the
equation of motion is changed to

$\omega^{2}{\bf u}_{0}=c_{t}^{2}q^{2}{\bf u}_{0}+\frac{\xi_{0}q^{2}}{2}i{\bf
q}\times{\bf u}_{0}$ (184)

Now let’s choose ${\bf q}=q{\bf\hat{e}_{z}}$ and ${\bf
u}_{0}=(u_{0x},u_{0y},0)^{T}$ (the superscript $T$ here represents transpose)
and then the equation of motion (184) can be written as

$\omega^{2}\left(\begin{array}[]{c}u_{0x}\\\
u_{0y}\end{array}\right)=\left(\begin{array}[]{cc}c_{t}^{2}q^{2}&-i\frac{\xi_{0}}{2}q^{3}\\\
i\frac{\xi_{0}}{2}q^{3}&c_{t}^{2}q^{2}\end{array}\right)\left(\begin{array}[]{c}u_{0x}\\\
u_{0y}\end{array}\right).$ (185)

This eigen equation can be easily solved. Two branches of dispersion relations
for the transverse modes are given by
$\omega^{t}_{\pm}=\sqrt{c_{t}^{2}q^{2}\pm\frac{|\xi_{0}|}{2}q^{3}}=q\sqrt{c_{t}^{2}\pm\frac{|\xi_{0}|}{2}q}$,
from which one can see that the Nieh-Yan term mainly corrects the phonon
velocity and gives rise to the different velocities for two transverse phonon
modes. The corresponding normalized eigen vectors are given by

${\bf u}_{0,\pm}=\frac{1}{\sqrt{2}}\left(\begin{array}[]{c}1\\\ \pm i\\\
0\end{array}\right)$ (186)

so that ${\bf u}^{{\dagger}}_{0,s}({\bf k}){\bf u}_{0,s}({\bf k})=1$. The
angular momentum of the phonon modes is defined as Hamada  _et al._ (2018);
Zhang and Niu (2014); Hamada and Murakami (2020); Zhu  _et al._ (2018); Zhang
and Niu (2015)

$l_{s,i}({\bf q})=\hbar{\bf u}^{{\dagger}}_{0,s}({\bf q})M_{i}{\bf
u}_{0,s}({\bf q})$ (187)

where $s=\pm$, $i=x,y,z$, and $(M_{i})_{jk}=(-i)\epsilon_{ijk}$. One can
easily show that only $l_{s,z}$ is non-zero and $l_{\pm,z}=\pm\hbar$ for the
momentum ${\bf q}=q{\bf\hat{e}_{z}}$. Due to the full rotation symmetry of our
model, the general form of angular momentum for the $s$-mode is given by ${\bf
l}_{s}=s\hbar{\bf\hat{q}}$ with ${\bf\hat{q}}=\frac{\bf q}{q}$.

Next let’s consider the phonon total angular momentum induced by a temperature
gradient. The total phonon angular momentum per volume can be related to the
temperature gradient by

$I^{ph}_{i}=\alpha_{ij}\frac{\partial T}{\partial r_{j}}$ (188)

where $i,j=x,y,z$ and ${\bf r}$ is the spatial coordinate. From Ref. Hamada
_et al._ , 2018, the response coefficient $\alpha_{ij}$ is derived as

$\alpha_{ij}=-\frac{\tau}{V}\sum_{{\bf q},s}l_{s,i}v^{ph}_{s,j}\frac{\partial
f_{0}(\omega^{t}_{s})}{\partial T}$ (189)

in the linear response regime, where $\tau$ is the phonon relaxation time,
${\bf v}^{ph}_{s}$ is the group velocity of $s$-phonon mode and
$f_{0}(\omega^{t}_{s})$ is the Bose distribution function for phonons. The
group velocity of the transverse mode should be given by

$\displaystyle{\bf v}^{ph}_{s}=\frac{\partial\omega^{t}_{s}}{\partial{\bf
q}}=\frac{\partial\omega^{t}_{s}}{\partial q}\frac{\partial q}{\partial{\bf
q}}=\frac{c_{t}^{2}+\frac{3s}{4}|\xi_{0}|q}{\sqrt{c_{t}^{2}+\frac{s}{2}|\xi_{0}|q}}{\bf\hat{q}}.$
(190)

For our isotropic system, there is only a single non-zero independent
component $\alpha_{xx}=\alpha_{yy}=\alpha_{zz}$, so let’s evaluate
$\alpha_{zz}$, which is given by

$\alpha_{zz}=-\tau\hbar\int\frac{d^{3}q}{(2\pi)^{3}}\sum_{s}\frac{sq_{z}^{2}}{q^{2}}\frac{c_{t}^{2}+\frac{3s}{4}|\xi_{0}|q}{\sqrt{c_{t}^{2}+\frac{s}{2}|\xi_{0}|q}}\frac{\partial
f_{0}}{\partial T}=-\frac{\tau\hbar}{6\pi^{2}}\int
q^{2}dq\sum_{s}s\frac{c_{t}^{2}+\frac{3s}{4}|\xi_{0}|q}{\sqrt{c_{t}^{2}+\frac{s}{2}|\xi_{0}|q}}\frac{\partial
f_{0}}{\partial T},$ (191)

where

$\frac{\partial f_{0}({\omega})}{\partial
T}=k_{B}\beta^{2}\omega\frac{e^{\beta\omega}}{(e^{\beta\omega}-1)^{2}}.$ (192)

The angular part of the phonon momentum in the above expression can be
integrated out since $\frac{\partial f_{0}({\omega})}{\partial T}$ only
depends on the magnitude of ${\bf q}$. This expression can be evaluated
numerically and to get some analytical understanding, we can make the further
approximations. According to the Bose distribution $f_{0}({\omega})$, we know
that the phonons are mainly excited in the energy range determined by the
temperature $T$, $\omega\ll k_{B}T$. This means the contribution in the above
expression mainly comes from the momentum range $q<\frac{k_{B}T}{c_{t}}$. Now
let’s consider the momentum range, in which the Nieh-Yan term contribution to
the velocity $|\xi_{0}|q$ is much smaller than $c_{t}^{2}$. This is naturally
satisfied when the temperature $k_{B}T\ll\frac{c_{t}^{2}}{|\xi_{0}|}$ and in
this temperature range, we can treat $|\xi_{0}|q$ as a perturbation and expand
the integrand in Eq. (191). First, we have

$\frac{c_{t}^{2}+\frac{3s}{4}|\xi_{0}|q}{\sqrt{c_{t}^{2}+\frac{s}{2}|\xi_{0}|q}}\approx
c_{t}\left(1+\frac{3s}{4c_{t}^{2}}|\xi_{0}|q\right)\left(1-\frac{s}{4c_{t}^{2}}|\xi_{0}|q\right)\approx
c_{t}\left(1+\frac{s}{2c_{t}^{2}}|\xi_{0}|q\right)$ (193)

and

$\frac{\partial f_{0}}{\partial T}=\frac{\partial f_{0}(\omega_{0})}{\partial
T}+\frac{\partial^{2}f_{0}}{\partial T\partial\omega}\delta\omega_{s},$ (194)

where $\omega_{0}=c_{t}q$ and
$\delta\omega_{s}=\frac{s|\xi_{0}|}{4c_{t}}q^{2}$. Let’s define
$x=\beta\omega_{0}=\frac{c_{t}q}{k_{B}T}$ and then we obtain

$\frac{\partial f_{0}(\omega_{0})}{\partial
T}=k_{B}\beta^{2}\omega_{0}\frac{e^{\beta\omega_{0}}}{(e^{\beta\omega_{0}}-1)^{2}}=\frac{x}{T}\frac{e^{x}}{(e^{x}-1)^{2}},$
(195)

and

$\frac{\partial^{2}f_{0}(\omega_{0})}{\partial
T\partial\omega}=k_{B}\beta^{2}\frac{e^{\beta\omega_{0}}}{(e^{\beta\omega_{0}}-1)^{3}}\left(e^{\beta\omega_{0}}(1-\beta\omega_{0})-\beta\omega_{0}-1\right)=\frac{1}{k_{B}T^{2}}\frac{e^{x}}{(e^{x}-1)^{3}}\left(e^{x}(1-x)-x-1\right).$
(196)

Thus, we have

$\displaystyle\alpha_{zz}\approx-\frac{\tau\hbar}{6\pi^{2}}\int
q^{2}dq\sum_{s}sc_{t}\left(1+\frac{s}{2c_{t}^{2}}|\xi_{0}|q\right)\left(\frac{\partial
f_{0}(\omega_{0})}{\partial T}+\frac{\partial^{2}f_{0}}{\partial
T\partial\omega}\frac{s|\xi_{0}|}{4c_{t}}q^{2}\right)$
$\displaystyle=-\frac{\tau\hbar}{6\pi^{2}}\int
q^{2}dqc_{t}\left(\frac{\partial f_{0}(\omega_{0})}{\partial
T}\frac{|\xi_{0}|}{c_{t}^{2}}q+\frac{\partial^{2}f_{0}}{\partial
T\partial\omega}\frac{|\xi_{0}|}{2c_{t}}q^{2}\right)$
$\displaystyle=-\frac{\tau\hbar}{6\pi^{2}}\left(\frac{k_{B}T}{c_{t}}\right)^{3}\frac{|\xi_{0}|}{c_{t}^{2}}\int
x^{2}dx\left(\frac{\partial f_{0}(\omega_{0})}{\partial
T}xk_{B}T+\frac{\partial^{2}f_{0}}{\partial
T\partial\omega}\frac{x^{2}}{2}\left(k_{B}T\right)^{2}\right)$
$\displaystyle=-\frac{\tau\hbar|\xi_{0}|k_{B}}{6\pi^{2}c_{t}^{5}}\left(k_{B}T\right)^{3}\int
x^{4}dx\left(\frac{e^{x}}{(e^{x}-1)^{2}}+\frac{1}{2}\frac{e^{x}}{(e^{x}-1)^{3}}\left(e^{x}(1-x)-x-1\right)\right)$

In the above expression, the integral of $x$ is convergent, and thus just
gives a number. Therefore, all the temperature dependence is given by the
coefficient before the $x$-integral. Since
$\xi_{0}=g_{1}^{2}\mu_{5}\mathcal{F}_{0}$ with
$\mathcal{F}_{0}=F_{0}+F_{1}(k_{B}T)^{2}$, we expect $\alpha_{zz}$ has $T^{3}$
dependence for the standard Nieh-Yan term and $T^{5}$ dependence for the
thermal Nieh-Yan term. Let’s denote $a_{0}=\int
x^{4}dx\left(\frac{e^{x}}{(e^{x}-1)^{2}}+\frac{1}{2}\frac{e^{x}}{(e^{x}-1)^{3}}\left(e^{x}(1-x)-x-1\right)\right)=-4\pi^{4}/15$,
and then we have

$\alpha_{zz}\approx-\frac{\tau\hbar
a_{0}|\xi_{0}|k_{B}}{6\pi^{2}c_{t}^{5}}\left(k_{B}T\right)^{3}=-\frac{\tau\hbar
a_{0}k_{B}g_{1}^{2}}{6\pi^{2}c_{t}^{5}}\mu_{5}\left(F_{0}\left(k_{B}T\right)^{3}+F_{1}\left(k_{B}T\right)^{5}\right)=\frac{2\tau\hbar\pi^{2}k_{B}g_{1}^{2}}{45c_{t}^{5}}\mu_{5}\left(F_{0}\left(k_{B}T\right)^{3}+F_{1}\left(k_{B}T\right)^{5}\right).$
(198)

##  Appendix H Electron-electron Interaction

Since the electron-phonon interaction may be influenced by electron-electron
interaction, one may wonder how electron-electron interaction affects the
proposed phenomena here. In this section, we will demonstrate that electron-
electron interaction is actually decoupled from the phonon modes discussed in
this work and thus all the results remain valid even taking into account
electron-electron interaction.

To see that, let’s start from the effective action $S_{eff}$ in Eq. (65) with
an additional electron-electron Coulomb interaction. We consider the
Stratonovich-Hubbard transformation by introducing an additional $\varphi$
field with the effective action

$S_{C}=\int d\tau d^{3}r\left(\frac{1}{2}(\nabla\varphi({\bf
r},\tau))^{2}+ie\varphi({\bf
r},\tau)\hat{\psi}^{{\dagger}}_{\bf\Gamma_{i}}({\bf
r},\tau)\hat{\psi}_{\bf\Gamma_{i}}({\bf r},\tau)\right).$ (199)

With the Fourier transform

$\displaystyle\hat{\psi}_{\bf\Gamma_{i}}({\bf
r},\tau)=\sum_{\tilde{k}}e^{i{\bf k\cdot
r}-i\omega_{n}\tau}\hat{\psi}_{\bf\Gamma_{i}}(\tilde{k}),$ (200)
$\displaystyle\varphi({\bf r},\tau)=\sum_{\tilde{q}}e^{i{\bf q\cdot
r}-i\nu_{m}\tau}\varphi(\tilde{q}),$ (201)

we have

$S_{C}=\sum_{\tilde{q}}\frac{q^{2}}{2}\varphi(\tilde{q})\varphi(-\tilde{q})+ie\sum_{\tilde{k},\tilde{q}}\varphi(\tilde{q})\hat{\psi}^{{\dagger}}_{\bf\Gamma_{i}}(\tilde{k})\hat{\psi}_{\bf\Gamma_{i}}(\tilde{k}-\tilde{q}),$
(202)

where $\sum_{\tilde{k}}=\frac{1}{\beta V}\sum_{{\bf k},i\omega_{n}}$. Now
let’s perform the transformation

$\displaystyle\tilde{\varphi}(-\tilde{q})=\varphi(-\tilde{q})+i\frac{e}{q^{2}}\sum_{\tilde{k}}\hat{\psi}^{{\dagger}}_{\bf\Gamma_{i}}(\tilde{k})\hat{\psi}_{\bf\Gamma_{i}}(\tilde{k}-\tilde{q})$
(203)
$\displaystyle\tilde{\varphi}(\tilde{q})=\varphi(\tilde{q})+i\frac{e}{q^{2}}\sum_{\tilde{k}}\hat{\psi}^{{\dagger}}_{\bf\Gamma_{i}}(\tilde{k})\hat{\psi}_{\bf\Gamma_{i}}(\tilde{k}+\tilde{q})$
(204)

and

$\displaystyle\sum_{\tilde{q}}\frac{q^{2}}{2}\tilde{\varphi}(\tilde{q})\tilde{\varphi}(-\tilde{q})=\sum_{\tilde{q}}\frac{q^{2}}{2}\left(\varphi(\tilde{q})+i\frac{e}{q^{2}}\sum_{\tilde{k}}\hat{\psi}^{{\dagger}}_{\bf\Gamma_{i}}(\tilde{k})\hat{\psi}_{\bf\Gamma_{i}}(\tilde{k}+\tilde{q})\right)\left(\varphi(-\tilde{q})+i\frac{e}{q^{2}}\sum_{\tilde{k}}\hat{\psi}^{{\dagger}}_{\bf\Gamma_{i}}(\tilde{k})\hat{\psi}_{\bf\Gamma_{i}}(\tilde{k}-\tilde{q})\right)$
$\displaystyle=\sum_{\tilde{q}}\left(\frac{q^{2}}{2}\varphi(\tilde{q})\varphi(-\tilde{q})+ie\varphi(\tilde{q})\sum_{\tilde{k}}\hat{\psi}^{{\dagger}}_{\bf\Gamma_{i}}(\tilde{k})\hat{\psi}_{\bf\Gamma_{i}}(\tilde{k}-\tilde{q})-\frac{e^{2}}{2q^{2}}\sum_{\tilde{k}\tilde{k}^{\prime}}\hat{\psi}^{{\dagger}}_{\bf\Gamma_{i}}(\tilde{k})\hat{\psi}_{\bf\Gamma_{i}}(\tilde{k}+\tilde{q})\hat{\psi}^{{\dagger}}_{\bf\Gamma_{i}}(\tilde{k}^{\prime})\hat{\psi}_{\bf\Gamma_{i}}(\tilde{k}^{\prime}-\tilde{q})\right).$
(205)

From the above quality, we have

$S_{C}=\sum_{\tilde{q}}\frac{q^{2}}{2}\tilde{\varphi}(\tilde{q})\tilde{\varphi}(-\tilde{q})+\frac{e^{2}}{2q^{2}}\sum_{\tilde{q}\tilde{k}\tilde{k}^{\prime}}\hat{\psi}^{{\dagger}}_{\bf\Gamma_{i}}(\tilde{k})\hat{\psi}_{\bf\Gamma_{i}}(\tilde{k}+\tilde{q})\hat{\psi}^{{\dagger}}_{\bf\Gamma_{i}}(\tilde{k}^{\prime})\hat{\psi}_{\bf\Gamma_{i}}(\tilde{k}^{\prime}-\tilde{q}),$
(206)

the latter term of which is indeed four-fermion Coulomb interaction with the
coefficient $V_{q}=\frac{e^{2}}{q^{2}}$ the Fourier transform of Coulomb
interaction in the momentum space (we have set $\epsilon_{0}=1$).

Now let’s compare the second term in $S_{C}$ (Eq. (199)) with the first term
in $S_{1}$ (Eq. (67)), and one can see that the $A_{0}$ field is identical to
the $-ie\varphi$ field. Similar to the derivation for the effective action
$W[A_{0},\Delta]$ in Eq. (C), we consider the effective action for the
$\varphi$ and ${\bf\Delta}$ fields (we drop the $A_{0}$ field here) by
integrating out the fermion operators, and the resulting effective action is
given by

$\displaystyle
W_{1}[\varphi,{\bf\Delta}]=\frac{1}{2}\sum_{\tilde{q}}\left(q^{2}\varphi(\tilde{q})\varphi(-\tilde{q})-e^{2}\Pi_{0}(\tilde{q})\varphi(\tilde{q})\varphi(-\tilde{q})+\Phi^{ab}_{ij}(\tilde{q})\Delta^{i}_{a}(\tilde{q})\Delta^{j}_{b}(-\tilde{q})+ie\Theta^{a}_{i}(\tilde{q})\varphi(\tilde{q})\Delta^{i}_{a}(-\tilde{q})+ie\Theta^{a}_{i}(-\tilde{q})\Delta^{i}_{a}(\tilde{q})\varphi(-\tilde{q})\right)$
$\displaystyle=\frac{1}{2}\sum_{\tilde{q}}\left(D^{-1}(\tilde{q})\varphi(\tilde{q})\varphi(-\tilde{q})+\Phi^{ab}_{ij}(\tilde{q})\Delta^{i}_{a}(\tilde{q})\Delta^{j}_{b}(-\tilde{q})+ie\Theta^{a}_{i}(\tilde{q})\varphi(\tilde{q})\Delta^{i}_{a}(-\tilde{q})+ie\Theta^{a}_{i}(-\tilde{q})\Delta^{i}_{a}(\tilde{q})\varphi(-\tilde{q})\right),$
(207)

where the duplicated indices should be summed over. Here
$D^{-1}(\tilde{q})=q^{2}-e^{2}\Pi_{0}(\tilde{q})=D_{0}^{-1}-e^{2}\Pi_{0}(\tilde{q})$
defines the full Green’s function for the $\varphi$ field.

Next let’s consider the path integral

$Z_{1}[{\bf\Delta}]=\int\mathcal{D}\varphi e^{-W_{1}[\varphi,{\bf\Delta}]}$
(208)

and we can perform the transformation

$\displaystyle\tilde{\varphi}(\tilde{q})=\varphi(\tilde{q})+ie\Theta^{a}_{i}(-\tilde{q})\Delta^{i}_{a}(\tilde{q})D(\tilde{q})$
(209)
$\displaystyle\tilde{\varphi}(-\tilde{q})=\varphi(-\tilde{q})+ie\Theta^{a}_{i}(\tilde{q})\Delta^{i}_{a}(-\tilde{q})D(-\tilde{q}).$
(210)

Correspondingly,

$\displaystyle\frac{1}{2}\sum_{\tilde{q}}D^{-1}(\tilde{q})\tilde{\varphi}(\tilde{q})\tilde{\varphi}(-\tilde{q})=\frac{1}{2}\sum_{\tilde{q}}D^{-1}(\tilde{q})(\varphi(\tilde{q})+ie\Theta^{a}_{i}(-\tilde{q})\Delta^{i}_{a}(\tilde{q})D(\tilde{q}))(\varphi(-\tilde{q})+ie\Theta^{b}_{j}(\tilde{q})\Delta^{j}_{b}(-\tilde{q})D(-\tilde{q}))$
$\displaystyle=\frac{1}{2}\sum_{\tilde{q}}\left(D^{-1}(\tilde{q})\varphi(\tilde{q})\varphi(-\tilde{q})+ie\varphi(\tilde{q})\Theta^{b}_{j}(\tilde{q})\Delta^{j}_{b}(-\tilde{q})+ie\Theta^{a}_{i}(-\tilde{q})\Delta^{i}_{a}(\tilde{q})\varphi(-\tilde{q})-e^{2}\Theta^{a}_{i}(-\tilde{q})\Delta^{i}_{a}(\tilde{q})D(\tilde{q})\Theta^{b}_{j}(\tilde{q})\Delta^{j}_{b}(-\tilde{q})\right)$

and

$\displaystyle
W_{1}[\varphi,{\bf\Delta}]=\frac{1}{2}\sum_{\tilde{q}}\left(D^{-1}(\tilde{q})\tilde{\varphi}(\tilde{q})\tilde{\varphi}(-\tilde{q})+\left(\Phi^{ab}_{ij}(\tilde{q})+e^{2}\Theta^{a}_{i}(-\tilde{q})D(\tilde{q})\Theta^{b}_{j}(\tilde{q})\right)\Delta^{i}_{a}(\tilde{q})\Delta^{j}_{b}(-\tilde{q})\right).$
(212)

Now one can see that the $\tilde{\varphi}$ and ${\bf\Delta}$ fields are
decoupled from each other. Let’s write $Z_{1}[{\bf\Delta}]\sim
e^{-W_{2}[{\bf\Delta}]}$ and then

$W_{2}[{\bf\Delta}]=\frac{1}{2}\sum_{\tilde{q}}\left(\Phi^{ab}_{ij}(\tilde{q})+e^{2}\Theta^{a}_{i}(-\tilde{q})D(\tilde{q})\Theta^{b}_{j}(\tilde{q})\right)\Delta^{i}_{a}(\tilde{q})\Delta^{j}_{b}(-\tilde{q}).$
(213)

From this expression, we conclude that the full stress-stress correlation
function, denoted as $\tilde{\Phi}^{ab}_{ij}(\tilde{q})$, should generally
acquire a correction from the electron-electron interaction through

$\tilde{\Phi}^{ab}_{ij}(\tilde{q})=\Phi^{ab}_{ij}(\tilde{q})+e^{2}\Theta^{a}_{i}(-\tilde{q})D(\tilde{q})\Theta^{b}_{j}(\tilde{q}),$
(214)

where the stress-density correlation function $\Theta^{a}_{i}(\tilde{q})$ is
defined in Eq. (82).

One can see that the $\Theta^{a}_{i}(\tilde{q})$ function couples electrons to
acoustic phonons and this function involves one density vertex and one stress
tensor vertex from Eq. (82). Next we will show that for the phonon modes that
we are interested in, the $\Theta^{a}_{i}(\tilde{q})$ function is always zero
if the system has full rotation symmetry. In the above expressions,
$a,b,i,j=x,y,z$ which is not convenient for the symmetry analysis. It is more
convenient to relabel the strain and stress tensor with the angular momentum,
as shown in Eq. (52)-(61). The rotationally symmetric form of the Hamiltonian
(62) is useful for our symmetry analysis below.

To make the problem more concrete, let’s consider the phonon momentum along
the z-direction, ${\bf q}=q_{z}\hat{e}_{z}$, and we first figure out which
strain tensors, as well as the stress tensors, are involved in the Nieh-Yan
term. To see that, let’s write down explicitly the terms in Eq. (173) along
the $q_{z}\hat{e}_{z}$ direction, which involves

$\displaystyle\delta_{jl}\epsilon_{zkm}(\partial_{z}u_{jk})u_{lm}=\epsilon_{zxy}(\partial_{z}u_{jx})u_{jy}+\epsilon_{zyx}(\partial_{z}u_{jy})u_{jx}$
$\displaystyle=\left((\partial_{z}u_{xx})u_{xy}+(\partial_{z}u_{yx})u_{yy}+(\partial_{z}u_{zx})u_{zy}\right)-\left((\partial_{z}u_{xy})u_{xx}+(\partial_{z}u_{yy})u_{yx}+(\partial_{z}u_{zy})u_{zx}\right)$
$\displaystyle=2c\partial_{z}(u_{xx}-u_{yy})+2(\partial_{z}u_{zx})u_{zy},$
(215)

in which all the total derivative term has been dropped since they will not
contribute to the effective action. Now we notice that only the components
$u_{xx}-u_{yy}$, $u_{xy}$, $u_{zx}$ and $u_{zy}$ of the strain tensor are
involved for the acoustic phonons along the z direction that we are interested
in. From the Eq. (52)-(56), one can see that only $u_{J=2,M}$ with $M=\pm
1,\pm 2$ are involved, which means the the z-direction angular momentum $M$
only takes $\pm 1$ and $\pm 2$, but not $0$. On the other hand, the electron
density operator carries z-direction angular momentum $0$. Since the
z-direction rotation symmetry is preserved for the momentum ${\bf q}$ along
the z direction, we expect elections cannot directly couple to the phonon
modes with higher angular momentum.

To make this argument more explicitly, let’s define $\mathcal{T}^{J}_{M}$ as

$\displaystyle\mathcal{T}^{2}_{2}=\frac{1}{2}\left(\mathcal{T}^{x}_{x}-\mathcal{T}^{y}_{y}+i(\mathcal{T}^{y}_{x}+\mathcal{T}^{x}_{y}\right)$
(216)
$\displaystyle\mathcal{T}^{2}_{1}=\left(-\frac{1}{2}\right)\left(\mathcal{T}^{z}_{x}+\mathcal{T}^{x}_{z}+i(\mathcal{T}^{z}_{y}+\mathcal{T}^{y}_{z})\right)$
(217)
$\displaystyle\mathcal{T}^{2}_{0}=\sqrt{\frac{1}{6}}\left(2\mathcal{T}^{z}_{z}-\mathcal{T}^{x}_{x}-\mathcal{T}^{y}_{y}\right)$
(218)
$\displaystyle\mathcal{T}^{2}_{-1}=\frac{1}{2}\left(\mathcal{T}^{z}_{x}+\mathcal{T}^{x}_{z}-i(\mathcal{T}^{z}_{y}+\mathcal{T}^{y}_{z})\right)$
(219)
$\displaystyle\mathcal{T}^{2}_{-2}=\frac{1}{2}\left(\mathcal{T}^{x}_{x}-\mathcal{T}^{y}_{y}-i(\mathcal{T}^{y}_{x}+\mathcal{T}^{x}_{y})\right)$
(220)

and the stress-density correlation function can also be re-constructed as

$\Theta^{J}_{M}({\bf q},i\nu_{m})=\frac{1}{\beta V}\sum_{{\bf
k},i\omega_{n}}Tr_{\sigma}\left(\mathcal{G}_{0}({\bf
k-q},i\omega_{n}-i\nu_{m})\mathcal{T}^{J}_{M}({\bf k-q},{\bf
k})\mathcal{G}_{0}({\bf k},i\omega_{n})\right)$ (221)

for the angular momentum $J,M$ of the stress operator. Here we also add the
density operator $n$ which is an identity. Let’s consider a symmetry operator
$\mathcal{R}$, which transforms the stress tensor as
$\mathcal{R}\mathcal{T}^{J}_{M}({\bf k-q},{\bf
k})\mathcal{R}^{-1}=\sum_{M^{\prime}}D^{J}_{MM^{\prime}}(\mathcal{R})\mathcal{T}^{J}_{M^{\prime}}(\mathcal{R}^{-1}({\bf
k-q}),\mathcal{R}^{-1}{\bf k})$. The Green function should be invariant under
the symmetry operation $\mathcal{R}\mathcal{G}_{0}({\bf
k},i\omega_{n})\mathcal{R}^{-1}=\mathcal{G}_{0}(\mathcal{R}^{-1}{\bf
k},i\omega_{n})$. Thus, one can insert symmetry operation into the definition
of stress-density correlation function and find

$\displaystyle\Theta^{J}_{M}({\bf q},i\nu_{m})=\frac{1}{\beta V}\sum_{{\bf
k},i\omega_{n}}Tr_{\sigma}\left(\mathcal{R}\mathcal{G}_{0}({\bf
k-q},i\omega_{n}-i\nu_{m})\mathcal{R}^{-1}\mathcal{R}\mathcal{T}^{J}_{M}({\bf
k-q},{\bf k})\mathcal{R}^{-1}\mathcal{R}\mathcal{G}_{0}({\bf
k},i\omega_{n})\mathcal{R}^{-1}\right)$ $\displaystyle=\frac{1}{\beta
V}\sum_{{\bf
k},i\omega_{n}}Tr_{\sigma}\left(\mathcal{G}_{0}(\mathcal{R}^{-1}({\bf
k-q}),i\omega_{n}-i\nu_{m})\sum_{M^{\prime}}D^{J}_{MM^{\prime}}(\mathcal{R})\mathcal{T}^{J}_{M^{\prime}}(\mathcal{R}^{-1}({\bf
k-q}),{\bf k})\mathcal{G}_{0}(\mathcal{R}^{-1}{\bf k},i\omega_{n})\right)$
$\displaystyle=\frac{1}{\beta V}\sum_{{\bf
k}^{\prime},i\omega_{n},M^{\prime}}D^{J}_{MM^{\prime}}(\mathcal{R})Tr_{\sigma}\left(\mathcal{G}_{0}({\bf
k}^{\prime}-\mathcal{R}^{-1}{\bf
q}),i\omega_{n}-i\nu_{m})\mathcal{T}^{J}_{M^{\prime}}({\bf
k}^{\prime}-\mathcal{R}^{-1}{\bf q},{\bf k}^{\prime})\mathcal{G}_{0}({\bf
k}^{\prime},i\omega_{n})\right)$
$\displaystyle=\sum_{M^{\prime}}D^{J}_{MM^{\prime}}(\mathcal{R})\Theta^{J}_{M^{\prime}}(\mathcal{R}^{-1}{\bf
q},i\nu_{m}).$ (222)

In the above derivation, we have used ${\bf k}^{\prime}=\mathcal{R}^{-1}{\bf
k}$. The above equation gives rise to the symmetry constraint on the form of
correlation functions. Along ${\bf q}=q_{z}\hat{e}_{z}$, we have
$\mathcal{R}^{-1}{\bf q}={\bf q}$ and

$\Theta^{J}_{M}(q_{z},i\nu_{m})=\sum_{M^{\prime}}D^{J}_{MM^{\prime}}(\mathcal{R})\Theta^{J}_{M^{\prime}}(q_{z},i\nu_{m})$
(223)

Now let’s consider the orthonormality relation for irreducible representations
of a symmetry group, given by
$\sum_{\mathcal{R}}D^{\Gamma_{j}}_{MM^{\prime}}(\mathcal{R})[D^{\Gamma_{j}^{\prime}}_{NN^{\prime}}(\mathcal{R})]^{*}=\frac{h}{l_{j}}\delta_{jj^{\prime}}\delta_{MN}\delta_{M^{\prime}N^{\prime}}$.
We should consider the symmetry group formed by the rotation along the z
directions and thus only the z-directional angular momentum $M$ is a good
quantum number that characterizes the rotation group, while the total angular
momentum $J$ is not. Consequently, we expect the orthonormality relation is
then given by
$\sum_{\mathcal{R}}D^{J}_{MM^{\prime}}(\mathcal{R})[D^{J^{\prime}}_{NN^{\prime}}(\mathcal{R})]^{*}\propto\delta_{MN}\delta_{M^{\prime}N^{\prime}}$.
Now let’s choose $D^{J^{\prime}}_{NN^{\prime}}$ to be $J^{\prime}=0$ (identity
representation), so $N=N^{\prime}=0$ and $D^{0}_{00}=1$ for any $\mathcal{R}$.
Then the orthonormality relation takes the form
$\sum_{\mathcal{R}}D^{J}_{MM^{\prime}}(\mathcal{R})\propto\delta_{M0}\delta_{M^{\prime}0}$,
which leads to

$\displaystyle\sum_{\mathcal{R}}\Theta^{J}_{M}(q_{z},i\nu_{m})=\sum_{M^{\prime}}\sum_{\mathcal{R}}D^{J}_{MM^{\prime}}(\mathcal{R})\Theta^{J}_{M^{\prime}}(q_{z},i\nu_{m})\propto\delta_{M0}$
$\displaystyle\rightarrow\Theta^{J}_{M}(q_{z},i\nu_{m})\propto\delta_{M0}.$
(224)

So the correlation function $\Theta^{J}_{M}$ can only be nonzero for the
z-directional angular momentum $M=0$. On the other hand, as we have shown
above, the Nieh-Yan anomaly term only involves the strain tensor with $M=\pm
1,\pm 2$, and this concludes that the electron-electron Coulomb interaction
will not contribute to the acoustic phonon modes, whose dynamics is influenced
by the Nieh-Yan anomaly.