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Ed. by Blomer, Valentin / Cohen, Frederick R. / Droste, Manfred / Duzaar, Frank / Echterhoff, Siegfried / Frahm, Jan / Gordina, Maria / Shahidi, Freydoon / Sogge, Christopher D. / Takayama, Shigeharu / Wienhard, Anna

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Volume 31, Issue 1

# Expansion for the product of matrices in groups

Doowon Koh
/ Thang Pham
/ Chun-Yen Shen
/ Anh Vinh Le
Published Online: 2018-08-18 | DOI: https://doi.org/10.1515/forum-2018-0063

## Abstract

In this paper, we give strong lower bounds on the size of the sets of products of matrices in some certain groups. More precisely, we prove an analogue of a result due to Chapman and Iosevich for matrices in ${\mathrm{SL}}_{2}\left({𝔽}_{p}\right)$ with restricted entries on a small set. We also provide extensions of some recent results on expansion for cubes in Heisenberg group due to Hegyvári and Hennecart.

Keywords: Matrix; finite fields; expanders

MSC 2010: 11B75; 20G40

## 1 Introduction

Let ${𝔽}_{p}$ be a prime field. We denote by ${\mathrm{SL}}_{2}\left({𝔽}_{p}\right)$ the set of $2×2$ matrices with determinant one over ${𝔽}_{p}$. Given $A\subset {𝔽}_{p}$, we define

$R\left(A\right):=\left\{\left(\begin{array}{cc}\hfill {a}_{11}\hfill & \hfill {a}_{12}\hfill \\ \hfill {a}_{21}\hfill & \hfill {a}_{22}\hfill \end{array}\right)\in {\mathrm{SL}}_{2}\left({𝔽}_{p}\right):{a}_{11},{a}_{12},{a}_{21}\in A\right\}.$

For any two sets X and Y of matrices, we define $XY:=\left\{x\cdot y:x\in X,y\in Y\right\}$. It was proved by Chapman and Iosevich [1] by Fourier analytic methods that if $|A|\gg {p}^{\frac{5}{6}}$, then

$|R\left(A\right)R\left(A\right)|\gg {p}^{3}.$

Throughout this paper, the notation $U\ll V$ means $U\le cV$ for some absolute constant $c>0$, and $U\gtrsim V$ means $U\gg {\left(\mathrm{log}U\right)}^{-c}V$ for some absolute constant $c>0$. It has been extensively studied about the size of the products of $R\left(A\right)$. In particular, the breakthrough work of H. A. Helfgott [5] asserts that if E is a subset of ${\mathrm{SL}}_{2}\left({𝔽}_{p}\right)$ and is not contained in any proper subgroup with $|E|<{p}^{3-\delta }$, then $|EEE|>c{|E|}^{1+ϵ}$ for some $ϵ=ϵ\left(\delta \right)>0$. The result mentioned above, by Chapman and Iosevich, is to give a quantitative estimate when the size of the set A is large. However, it is considered to be a difficult problem to obtain some quantitative estimate for the same problem when the size of the set A is not large. It is basically because the Fourier analytic methods are effective only when the size of the set A is large. In this paper, we address the case of small sets, and give a lower bound on the size of $R\left(A\right)R\left(A\right)$. Our first result is as follows.

#### Theorem 1.1.

Let $A\mathrm{\subset }{\mathrm{F}}_{p}$. If $\mathrm{|}A\mathrm{|}\mathrm{\le }c\mathit{}{p}^{\frac{\mathrm{12}}{\mathrm{19}}}$ for some small constant $c\mathrm{>}\mathrm{0}$, then

$|R\left(A\right)R\left(A\right)|\gg {|A|}^{\frac{7}{2}+\frac{1}{12}}={|R\left(A\right)|}^{\frac{7}{6}+\frac{1}{36}}.$

#### Remark 1.1.

We note that the exponent $\frac{7}{6}$ in Theorem 1.1 can be easily obtained by using the estimate $|A\cdot A+A\cdot A|\gg {|A|}^{\frac{3}{2}}$ given in [10]. Indeed, one can check that

$|R\left(A\right)R\left(A\right)|\gg |AA+AA|{|A|}^{2}\ge {|A|}^{2+\frac{3}{2}}={|A|}^{\frac{7}{2}}={|R\left(A\right)|}^{\frac{7}{6}}.$

However, in order to improve the exponent $\frac{7}{2}$, we need to prove some more results on sum-product type problems (see Section 2 for more details).

Let ${𝔽}_{p}$ be a prime field. For an integer $n\ge 1$, the Heisenberg group of degree n, denoted by ${𝐇}_{n}\left({𝔽}_{p}\right)$, is defined by a set of the following matrices:

$\left[𝐱,𝐲,z\right]:=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 𝐱\hfill & \hfill z\hfill \\ \hfill 𝟎\hfill & \hfill {I}_{n}\hfill & \hfill {𝐲}^{𝐭}\hfill \\ \hfill 0\hfill & \hfill 𝟎\hfill & \hfill 1\hfill \end{array}\right],$

where $𝐱,𝐲\in {𝔽}_{p}^{n}$, $z\in {𝔽}_{p}$, ${𝐲}^{𝐭}$ denotes the column vector of $𝐲$, and ${I}_{n}$ is the $n×n$ identity matrix. For $A,B,C\subset {𝔽}_{p}$, we define

$\left[{A}^{n},{B}^{n},C\right]:=\left\{\left[𝐱,𝐲,z\right]:𝐱\in {A}^{n},𝐲\in {B}^{n},z\in C\right\}.$

A similar question in the setting of the Heisenberg group over prime fields has been recently investigated by Hegyvári and Hennecart in [3], namely, they proved the following theorem.

#### Theorem 1.2 (Hegyvári–Hennecart, [3]).

For every $ϵ\mathrm{>}\mathrm{0}$, there exists a positive integer ${n}_{\mathrm{0}}\mathrm{=}{n}_{\mathrm{0}}\mathit{}\mathrm{\left(}ϵ\mathrm{\right)}$ such that if $n\mathrm{\ge }{n}_{\mathrm{0}}$, and $\mathrm{\left[}{A}^{n}\mathrm{,}{B}^{n}\mathrm{,}C\mathrm{\right]}\mathrm{\subseteq }{\mathrm{H}}_{n}\mathit{}\mathrm{\left(}{\mathrm{F}}_{p}\mathrm{\right)}$ with

$|\left[{A}^{n},{B}^{n},C\right]|>{|{𝐇}_{n}\left(\left({𝔽}_{p}\right)\right)|}^{\frac{3}{4}+ϵ},$

then there exists a nontrivial subgroup G of ${\mathrm{H}}_{n}\mathit{}\mathrm{\left(}{\mathrm{F}}_{p}\mathrm{\right)}$ such that $\mathrm{\left[}{A}^{n}\mathrm{,}{B}^{n}\mathrm{,}C\mathrm{\right]}\mathit{}\mathrm{\left[}{A}^{n}\mathrm{,}{B}^{n}\mathrm{,}C\mathrm{\right]}$ contains at least

$\frac{|\left[{A}^{n},{B}^{n},C\right]|}{p}$

cosets of G.

In a very recent paper, using results on sum-product estimates, Hegyvári and Hennecart [4] established some results in the case $n=1$. In particular, they proved that if $A\subset {𝔽}_{p}$ with $|A|\ge {p}^{\frac{1}{2}}$, then

$|\left[A,A,0\right]\left[A,A,0\right]|\gg \mathrm{min}\left\{{p}^{\frac{1}{2}}{|\left[A,A,0\right]|}^{\frac{5}{4}},{p}^{-\frac{1}{2}}{|\left[A,A,0\right]|}^{2}\right\}.$

In the case when $|A|\le {p}^{\frac{2}{3}}$, they also showed that

$|\left[A,A,0\right]\left[A,A,0\right]|\gg {|\left[A,A,0\right]|}^{\frac{7}{4}}.$

In this paper, we also extend this result to the setting of Heisenberg group of degree two. For simplicity, we write ${\left[{A}^{2},{A}^{2},0\right]}^{2}$ and ${\left[{A}^{2},{A}^{2},A\right]}^{2}$ for the products $\left[{A}^{2},{A}^{2},0\right]\left[{A}^{2},{A}^{2},0\right]$ and $\left[{A}^{2},{A}^{2},A\right]\left[{A}^{2},{A}^{2},A\right]$, respectively. We have the following theorems.

#### Theorem 1.3.

If $A\mathrm{\subset }{\mathrm{F}}_{p}$ with $\mathrm{|}A\mathrm{|}\mathrm{\le }{p}^{\frac{\mathrm{1}}{\mathrm{2}}}$, then we have

$|{\left[{A}^{2},{A}^{2},0\right]}^{2}|\gtrsim {|A|}^{\frac{11}{2}+\frac{25}{262}}={|\left[{A}^{2},{A}^{2},0\right]|}^{\frac{11}{8}+\frac{25}{1048}}.$

#### Theorem 1.4.

Let $A\mathrm{\subset }{\mathrm{F}}_{p}$ with $\mathrm{|}A\mathrm{|}\mathrm{\le }{p}^{\frac{\mathrm{9}}{\mathrm{16}}}$. Then we have

$|{\left[{A}^{2},{A}^{2},A\right]}^{2}|\gtrsim {|A|}^{\frac{11}{2}+\frac{23}{90}}={|\left[{A}^{2},{A}^{2},A\right]|}^{\frac{11}{10}+\frac{23}{450}}.$

The rest of this paper is organized to provide the complete proofs of our main theorems. More precisely, in Section 2 we give the proof of Theorem 1.1, and in Section 3 we complete the proofs of Theorems 1.3 and 1.4.

## 2 Proof of Theorem 1.1

In this section, without loss of generality, we assume that $0\notin A$. To prove Theorem 1.1, we need the following lemmas.

#### Lemma 2.1 ([9, Corollary 3.1]).

Let $X\mathrm{,}A\mathrm{\subset }{\mathrm{F}}_{p}$ with $\mathrm{|}X\mathrm{|}\mathrm{\ge }\mathrm{|}A\mathrm{|}$. Then we have

$|X+A\cdot A|\gg \mathrm{min}\left\{{|X|}^{\frac{1}{2}}|A|,p\right\}.$

#### Lemma 2.2.

Let $A\mathrm{\subset }{\mathrm{F}}_{p}$ with $\mathrm{|}A\mathrm{|}\mathrm{\le }c\mathit{}{p}^{\frac{\mathrm{2}}{\mathrm{3}}}$ for a sufficiently small $c\mathrm{>}\mathrm{0}$. Then the number of tuples $\mathrm{\left(}{a}_{\mathrm{1}}\mathrm{,}{a}_{\mathrm{2}}\mathrm{,}{a}_{\mathrm{3}}\mathrm{,}{a}_{\mathrm{4}}\mathrm{,}{a}_{\mathrm{1}}^{\mathrm{\prime }}\mathrm{,}{a}_{\mathrm{2}}^{\mathrm{\prime }}\mathrm{,}{a}_{\mathrm{3}}^{\mathrm{\prime }}\mathrm{,}{a}_{\mathrm{4}}^{\mathrm{\prime }}\mathrm{\right)}\mathrm{\in }{A}^{\mathrm{8}}$ satisfying

${a}_{1}{a}_{2}+{a}_{3}{a}_{4}={a}_{1}^{\prime }{a}_{2}^{\prime }+{a}_{3}^{\prime }{a}_{4}^{\prime }$

is $\mathrm{\ll }{\mathrm{|}A\mathrm{|}}^{\frac{\mathrm{13}}{\mathrm{2}}}$.

#### Proof.

For $\lambda ,\beta \in {𝔽}_{p}\setminus \left\{0\right\}$, one can follow the proof of [10, Theorem 3] to prove that the number of tuples $\left({a}_{1},{a}_{2},{a}_{3},{a}_{1}^{\prime },{a}_{2}^{\prime },{a}_{3}^{\prime }\right)\in {A}^{6}$ such that ${a}_{1}{a}_{2}+\lambda {a}_{3}={a}_{1}^{\prime }{a}_{2}^{\prime }+\beta {a}_{3}^{\prime }$ is $\ll {|A|}^{\frac{9}{2}}$. Thus, we see that for each fixed pair $\left({a}_{4},{a}_{4}^{\prime }\right)\in {A}^{2}$ the number of tuples $\left({a}_{1},{a}_{2},{a}_{3},{a}_{4},{a}_{1}^{\prime },{a}_{2}^{\prime },{a}_{3}^{\prime },{a}_{4}^{\prime }\right)\in {A}^{8}$ satisfying

${a}_{1}{a}_{2}+{a}_{3}{a}_{4}={a}_{1}^{\prime }{a}_{2}^{\prime }+{a}_{3}^{\prime }{a}_{4}^{\prime }$

is $\ll {|A|}^{\frac{9}{2}}$. Taking the sum over all pairs $\left({a}_{4},{a}_{4}^{\prime }\right)\in {A}^{2}$, the lemma follows. ∎

#### Lemma 2.3 ([15, Theorem 4]).

Let $A\mathrm{,}B\mathrm{\subset }{\mathrm{F}}_{p}$ with $\mathrm{|}A\mathrm{|}\mathrm{\le }\mathrm{|}B\mathrm{|}$, and let L be a finite set of lines in ${\mathrm{F}}_{p}^{\mathrm{2}}$. Suppose that $\mathrm{|}A\mathrm{|}\mathit{}{\mathrm{|}B\mathrm{|}}^{\mathrm{2}}\mathrm{\le }{\mathrm{|}L\mathrm{|}}^{\mathrm{3}}$ and $\mathrm{|}A\mathrm{|}\mathit{}\mathrm{|}L\mathrm{|}\mathrm{\ll }{p}^{\mathrm{2}}$. Then the number of incidences between $A\mathrm{×}B$ and lines in L, denoted by $I\mathit{}\mathrm{\left(}A\mathrm{×}B\mathrm{,}L\mathrm{\right)}$, satisfies

$I\left(A×B,L\right)\ll {|A|}^{\frac{3}{4}}{|B|}^{\frac{1}{2}}{|L|}^{\frac{3}{4}}+|L|.$

The following is an improvement of [13, Lemma 23].

#### Lemma 2.4.

Let $A\mathrm{,}B\mathrm{\subset }{\mathrm{F}}_{p}$. Then if $\mathrm{|}A\mathrm{|}\mathrm{=}\mathrm{|}B\mathrm{|}$, and ${\mathrm{|}A\mathrm{|}}^{\mathrm{2}}\mathit{}\mathrm{|}A\mathit{}B\mathrm{|}\mathrm{\ll }{p}^{\mathrm{2}}$, we have

$|A\cap \left(B+x\right)|\ll {|A|}^{-\frac{1}{2}}{|AB|}^{\frac{5}{4}}$

for any $x\mathrm{\ne }\mathrm{0}$.

#### Proof.

It is clear that

$|A\cap \left(B+x\right)|\ll \frac{1}{|A||B|}|\left\{\left(u,{u}_{*},a,b\right)\in AB×AB×A×B:u{b}^{-1}-{u}_{*}{a}^{-1}=x\right\}|.$

The number of such tuples $\left(u,{u}_{*},a,b\right)$ is bounded by the number of incidences between points in ${A}^{-1}×AB$ and a set L of lines of the form ${b}^{-1}Y-{u}_{*}X=x$ with $b\in B$, ${u}_{*}\in AB$. Notice that $|A|=|{A}^{-1}|$ and $|L|=|B||AB|$. Thus, if $|A|=|B|$ and ${|A|}^{2}|AB|\ll {p}^{2}$, Lemma 2.3 implies that

$I\left({A}^{-1}×AB,L\right)\ll {|A|}^{\frac{3}{2}}{|AB|}^{\frac{5}{4}},$

which completes proof of the theorem. ∎

#### Lemma 2.5 ([8, Theorem 2]).

If $A\mathrm{\subset }{\mathrm{F}}_{p}$ with $\mathrm{|}A\mathrm{|}\mathrm{\le }{p}^{\frac{\mathrm{9}}{\mathrm{16}}}$, then we have

${|A±A|}^{18}{|AA|}^{9}\gtrsim {|A|}^{32}.$

We are now ready to prove Theorem 1.1.

#### Proof of Theorem 1.1.

Without loss of generality, we may assume that $0\notin A$. Let ${M}_{1}$ and ${M}_{2}$ be matrices in $R\left(A\right)$ presented as follows:

${M}_{1}:=\left(\begin{array}{cc}\hfill {a}_{11}\hfill & \hfill {a}_{12}\hfill \\ \hfill {a}_{21}\hfill & \hfill \frac{1+{a}_{12}{a}_{21}}{{a}_{11}}\hfill \end{array}\right),{M}_{2}:=\left(\begin{array}{cc}\hfill {b}_{11}\hfill & \hfill {b}_{12}\hfill \\ \hfill {b}_{21}\hfill & \hfill \frac{1+{b}_{12}{b}_{21}}{{b}_{11}}\hfill \end{array}\right).$

Suppose that

${M}_{1}\cdot {M}_{2}=\left(\begin{array}{cc}\hfill t\hfill & \hfill \alpha \hfill \\ \hfill \beta \hfill & \hfill \frac{1+\alpha \beta }{t}\hfill \end{array}\right),$

where $t\ne 0$ and $\alpha ,\beta \in {𝔽}_{p}$. Then we have the following system

${a}_{11}{b}_{11}+{a}_{12}{b}_{21}=t,\frac{{b}_{12}t}{{b}_{11}}+\frac{{a}_{12}}{{b}_{11}}=\alpha ,\frac{{a}_{21}t}{{a}_{11}}+\frac{{b}_{21}}{{a}_{11}}=\beta .$(2.1)

Let us identify the matrix ${M}_{1}\cdot {M}_{2}$ with

$\left(t,\alpha ,\beta \right)\in {𝔽}_{p}^{*}×{𝔽}_{p}^{2}.$

Notice that $R\left(A\right)R\left(A\right)$ contains each $\left(t,\alpha ,\beta \right)\in {𝔽}_{p}^{*}×{𝔽}_{p}^{2}$ satisfying system (2.1) for some $\left({a}_{11},{a}_{12},{a}_{21},{b}_{11},{b}_{12},{b}_{21}\right)$ in ${A}^{6}$. Therefore, we aim to estimate a lower bound of the number of $\left(t,\alpha ,\beta \right)\in {𝔽}_{p}^{*}×{𝔽}_{p}^{2}$ such that system (2.1) holds for some $\left({a}_{11},{a}_{12},{a}_{21},{b}_{11},{b}_{12},{b}_{21}\right)\in {A}^{6}$. To this end, let $ϵ>0$ be a parameter chosen later. We now consider the following two cases:

Case 1. If $|AA|\ge {|A|}^{1+ϵ}$, then it follows from Lemma 2.1 that

$|AA+AA|\gg \mathrm{min}\left\{{|A|}^{\frac{3}{2}+\frac{ϵ}{2}},p\right\}={|A|}^{\frac{3}{2}+\frac{ϵ}{2}},$

where we assume that $|A|\le {p}^{\frac{2}{3+ϵ}}$. From system (2.1) and the above fact, we obtain that if $|A|\le {p}^{\frac{2}{3+ϵ}}$ and $|AA|\ge {|A|}^{1+ϵ}$, then

$|R\left(A\right)R\left(A\right)|\gg |AA+AA|{|A|}^{2}\gg {|A|}^{\frac{7}{2}+\frac{ϵ}{2}},$(2.2)

where the first $\gg$ follows, because in system (2.1), for each nonzero $t\in AA+AA$, if we fix a quadruple $\left({a}_{11},{b}_{11},{a}_{12},{b}_{21}\right)\in {A}^{4}$ with ${a}_{11}{b}_{11}+{a}_{12}{b}_{21}=t$, then $\alpha ,\beta$ are determined in terms of ${b}_{12}\in A$ and ${a}_{21}\in A$, respectively.

Case 2. If $|AA|\le {|A|}^{1+ϵ}$, then we consider as follows. For $t,\alpha ,\beta \in {𝔽}_{p}$, let $\nu \left(t,\alpha ,\beta \right)$ be the number of solutions $\left({a}_{11},{a}_{12},{a}_{21},{b}_{11},{b}_{12},{b}_{21}\right)$ of system (2.1). For the case $t=0$, we have

$\sum _{\alpha ,\beta }\nu \left(0,\alpha ,\beta \right)\le {|A|}^{5}.$

Indeed, for each choice of $\left({b}_{11},{a}_{12},{b}_{21}\right)\in {A}^{3}$, ${a}_{11}$ is determined uniquely, and $\alpha ,\beta$ are determined. In addition, ${a}_{21}$ and ${b}_{12}$ can be taken as arbitrary elements of A.

By the Cauchy–Schwarz inequality, we have

${\left({|A|}^{6}-{|A|}^{5}\right)}^{2}\le {\left(\sum _{t\ne 0,\alpha ,\beta }\nu \left(t,\alpha ,\beta \right)\right)}^{2}\le |R\left(A\right)R\left(A\right)|\sum _{t\ne 0,\alpha ,\beta }{\nu }^{2}\left(t,\alpha ,\beta \right).$

This implies that

$|R\left(A\right)R\left(A\right)|\gg \frac{{|A|}^{12}}{T},$(2.3)

where $T:={\sum }_{t\ne 0,\alpha ,\beta }{\nu }^{2}\left(t,\alpha ,\beta \right)$.

In the next step, we are going to show that

$T\ll {|A|}^{8+\frac{5ϵ}{2}}.$

To see this, observe by definition of $\nu \left(t,\alpha ,\beta \right)$ that for each $\left(t,\alpha ,\beta \right)\in {𝔽}_{p}^{*}×{𝔽}_{p}^{2}$, the value ${\nu }^{2}\left(t,\alpha ,\beta \right)$ is the number of 12-tuples $\left({a}_{11},{a}_{12},{a}_{21},{b}_{11},{b}_{12},{b}_{21},{a}_{11}^{\prime },{a}_{12}^{\prime },{a}_{21}^{\prime },{b}_{11}^{\prime },{b}_{12}^{\prime },{b}_{21}^{\prime }\right)\in {A}^{12}$ satisfying the following:

${a}_{11}{b}_{11}+{a}_{12}{b}_{21}=t={a}_{11}^{\prime }{b}_{11}^{\prime }+{a}_{12}^{\prime }{b}_{21}^{\prime },$$\frac{{b}_{12}t}{{b}_{11}}+\frac{{a}_{12}}{{b}_{11}}=\alpha =\frac{{b}_{12}^{\prime }t}{{b}_{11}^{\prime }}+\frac{{a}_{12}^{\prime }}{{b}_{11}^{\prime }},$$\frac{{a}_{21}t}{{a}_{11}}+\frac{{b}_{21}}{{a}_{11}}=\beta =\frac{{a}_{21}^{\prime }t}{{a}_{11}^{\prime }}+\frac{{b}_{21}^{\prime }}{{a}_{11}^{\prime }}.$

Thus the value of $T={\sum }_{t\ne 0,\alpha ,\beta }{\nu }^{2}\left(t,\alpha ,\beta \right)$ can be written as ${\sum }_{t\ne 0}\mathrm{\Omega }\left(t\right)$, where $\mathrm{\Omega }\left(t\right)$ denotes the number of 12-tuples $\left({a}_{11},{a}_{12},{a}_{21},{b}_{11},{b}_{12},{b}_{21},{a}_{11}^{\prime },{a}_{12}^{\prime },{a}_{21}^{\prime },{b}_{11}^{\prime },{b}_{12}^{\prime },{b}_{21}^{\prime }\right)\in {A}^{12}$ satisfying the following:

${a}_{11}{b}_{11}+{a}_{12}{b}_{21}=t={a}_{11}^{\prime }{b}_{11}^{\prime }+{a}_{12}^{\prime }{b}_{21}^{\prime },$(2.4)$\frac{{b}_{12}t}{{b}_{11}}+\frac{{a}_{12}}{{b}_{11}}=\frac{{b}_{12}^{\prime }t}{{b}_{11}^{\prime }}+\frac{{a}_{12}^{\prime }}{{b}_{11}^{\prime }},$(2.5)$\frac{{a}_{21}t}{{a}_{11}}+\frac{{b}_{21}}{{a}_{11}}=\frac{{a}_{21}^{\prime }t}{{a}_{11}^{\prime }}+\frac{{b}_{21}^{\prime }}{{a}_{11}^{\prime }}.$(2.6)

Now notice that Lemma 2.2 implies that if $|A|\ll {p}^{\frac{2}{3}}$, then there are at most ${|A|}^{\frac{13}{2}}$ 8-tuples $\left({a}_{11},{b}_{11},{a}_{12},{b}_{21},{a}_{11}^{\prime },{b}_{11}^{\prime },{a}_{12}^{\prime },{b}_{21}^{\prime }\right)$ in ${A}^{8}$ satisfying equations (2.4) for some $t\ne 0$. One can also check that among these tuples, there are at most ${|A|}^{6}$ ($\le \frac{1}{2}{|A|}^{\frac{13}{2}}$) tuples with ${a}_{12}^{\prime }{b}_{11}^{\prime ⁣-1}-{a}_{12}{b}_{11}^{-1}=0$. Hence, without loss of generality, we may assume that all tuples satisfy ${a}_{12}^{\prime }{b}_{11}^{\prime ⁣-1}-{a}_{12}{b}_{11}^{-1}\ne 0$.

For such a fixed 8-tuple $\left({a}_{11},{b}_{11},{a}_{12},{b}_{21},{a}_{11}^{\prime },{b}_{11}^{\prime },{a}_{12}^{\prime },{b}_{21}^{\prime }\right)\in {A}^{8}$, we now deal with equation (2.5) which can be rewritten as

$\frac{{b}_{12}}{{t}^{-1}{b}_{11}}+{a}_{12}{b}_{11}^{-1}=\frac{{b}_{12}^{\prime }}{{t}^{-1}{b}_{11}^{\prime }}+{a}_{12}^{\prime }{b}_{11}^{\prime ⁣-1}.$(2.7)

Set

$Q=\frac{t}{{b}_{11}}\cdot A,{Q}^{\prime }=\frac{t}{{b}_{11}^{\prime }}\cdot A\mathit{ }\text{and}\mathit{ }x={a}_{12}{b}_{11}^{-1}-{a}_{12}^{\prime }{b}_{11}^{\prime ⁣-1}\ne 0.$

Then the number of solutions $\left({b}_{12},{b}_{12}^{\prime }\right)\in {A}^{2}$ of (2.7) is the size of $Q\cap \left({Q}^{\prime }-x\right)$. It is clear that $|Q|=|{Q}^{\prime }|=|A|$, because $t\ne 0$, and we have assumed that $0\notin A$ so that $\frac{t}{{b}_{11}},\frac{t}{{b}_{11}^{\prime }}\ne 0$. We also see that

${|Q|}^{2}|Q\cdot {Q}^{\prime }|={|A|}^{2}|AA|\le {|A|}^{3+ϵ},$

where we used the assumption that $|AA|\le {|A|}^{1+ϵ}$. Applying Lemma 2.4, we obtain that if $|A|\ll {p}^{\frac{2}{3+ϵ}}$, then

$|Q\cap \left({Q}^{\prime }-x\right)|\ll {|A|}^{\frac{3}{4}+\frac{5ϵ}{4}}.$

The same argument works identically for equation (2.6) which can be restated as

$\frac{{a}_{21}}{{t}^{-1}{a}_{11}}+{b}_{21}{a}_{11}^{-1}=\frac{{a}_{21}^{\prime }}{{t}^{-1}{a}_{11}^{\prime }}+{b}_{21}^{\prime }{a}_{11}^{\prime ⁣-1}.$

In short, we have proved that if $|A|\ll {p}^{\frac{2}{3+ϵ}}$ and $|AA|\le {|A|}^{1+ϵ}$, then

$T\ll {|A|}^{\frac{13}{2}}{|A|}^{\frac{3}{4}+\frac{5ϵ}{4}}{|A|}^{\frac{3}{4}+\frac{5ϵ}{4}}={|A|}^{8+\frac{5ϵ}{2}}.$

Therefore, combining (2.3) with this estimate yields that if $|A|\ll {p}^{\frac{2}{3+ϵ}}$ and $|AA|\le {|A|}^{1+ϵ}$, then

$|R\left(A\right)R\left(A\right)|\gg {|A|}^{4-\frac{5ϵ}{2}}.$(2.8)

Finally, if we choose $ϵ=\frac{1}{6}$, then it follows from (2.2) and (2.8) that if $|A|\ll {p}^{\frac{12}{19}}$, then

$|R\left(A\right)R\left(A\right)|\gg {|A|}^{\frac{7}{2}+\frac{1}{12}},$

which completes the proof of Theorem 1.1. ∎

In the case of arbitrary finite fields, we have the following result.

#### Theorem 2.6.

Let $q\mathrm{=}{p}^{n}$ and let A be a subset of ${\mathrm{F}}_{q}^{\mathrm{*}}$. If $\mathrm{|}A\mathrm{\cap }\lambda \mathit{}F\mathrm{|}\mathrm{\le }{\mathrm{|}F\mathrm{|}}^{\frac{\mathrm{1}}{\mathrm{2}}}$ for any proper subfield F of ${\mathrm{F}}_{q}$ and any $\lambda \mathrm{\in }{\mathrm{F}}_{q}$, then we have

$|R\left(A\right)R\left(A\right)|\gtrsim {|A|}^{3+\frac{1}{5}}={|R\left(A\right)|}^{1+\frac{1}{15}}.$

To prove Theorem 2.6, we make use of the following results.

#### Theorem 2.7 ([7]).

With the assumptions of Theorem 2.6, we have

$\mathrm{max}\left\{|A+A|,|AA|\right\}\gtrsim {|A|}^{\frac{12}{11}}.$

#### Theorem 2.8.

For $A\mathrm{,}B\mathrm{\subset }{\mathrm{F}}_{q}$, suppose that $\mathrm{|}A\mathrm{\cap }\lambda \mathit{}F\mathrm{|}\mathrm{,}\mathrm{|}B\mathrm{\cap }\lambda \mathit{}F\mathrm{|}\mathrm{\le }{\mathrm{|}F\mathrm{|}}^{\frac{\mathrm{1}}{\mathrm{2}}}$ for any subfield F of ${\mathrm{F}}_{q}$ and any $\lambda \mathrm{\in }{\mathrm{F}}_{q}$. Then we have

$|A+AB|\gg \mathrm{min}\left\{|A|{|B|}^{\frac{1}{5}},{|A|}^{\frac{3}{4}}{|B|}^{\frac{2}{4}}\right\}.$

#### Corollary 2.9.

For $A\mathrm{\subset }{\mathrm{F}}_{q}$, suppose that $\mathrm{|}A\mathrm{\cap }\lambda \mathit{}F\mathrm{|}\mathrm{\le }{\mathrm{|}F\mathrm{|}}^{\frac{\mathrm{1}}{\mathrm{2}}}$ for any subfield F of ${\mathrm{F}}_{q}$ and any $\lambda \mathrm{\in }{\mathrm{F}}_{q}$. Then we have

$|AA+AA|\gg {|A|}^{\frac{6}{5}}.$

#### Proof.

Given a nonzero $x\in A$, we have

$|AA+AA|=|\frac{A}{x}\frac{A}{x}+\frac{A}{x}\frac{A}{x}|\ge |\frac{A}{x}\frac{A}{x}+\frac{A}{x}|\gg {|A|}^{\frac{6}{5}}$

by Theorem 2.8. ∎

We are now ready to prove Theorem 2.6.

#### Proof of Theorem 2.6.

Recall from (2.2) that

$|R\left(A\right)R\left(A\right)|\gg |AA+AA|{|A|}^{2}.$

Thus the theorem follows directly from Corollary 2.9. ∎

In the rest of this section, we present the proof of Theorem 2.8, for which the authors communicated with Oliver Roche-Newton.

## 2.1 Proof of Theorem 2.8

To prove Theorem 2.8, we make use of the following lemmas.

The first lemma is the Plünnecke–Ruzsa inequality.

#### Lemma 2.10 ([11, Theorems 1.6.1 and 1.81]).

Let $X\mathrm{,}{B}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{B}_{k}$ be subsets of ${\mathrm{F}}_{q}$. Then we have

$|{B}_{1}+\mathrm{\dots }+{B}_{k}|\le \frac{|X+{B}_{1}|\mathrm{\cdots }|X+{B}_{k}|}{{|X|}^{k-1}}\mathit{ }\text{𝑎𝑛𝑑}\mathit{ }|{B}_{1}-{B}_{2}|\le \frac{|X+{B}_{1}||X+{B}_{2}|}{|X|}.$

One can modify the proof of [6, Corollary 1.5] due to Katz and Shen to obtain the following:

#### Lemma 2.11.

Let $X\mathrm{,}{B}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{B}_{k}$ be subsets in ${\mathrm{F}}_{q}$. Then for any $\mathrm{0}\mathrm{<}ϵ\mathrm{<}\mathrm{1}$, there exists a subset ${X}^{\mathrm{\prime }}\mathrm{\subset }X$ such that $\mathrm{|}{X}^{\mathrm{\prime }}\mathrm{|}\mathrm{\ge }\mathrm{\left(}\mathrm{1}\mathrm{-}ϵ\mathrm{\right)}\mathit{}\mathrm{|}X\mathrm{|}$ and

$|{X}^{\prime }+{B}_{1}+\mathrm{\dots }+{B}_{k}|\le c\cdot \frac{|X+{B}_{1}|\mathrm{\cdots }|X+{B}_{k}|}{{|X|}^{k-1}}$

for some positive constant $c\mathrm{=}c\mathit{}\mathrm{\left(}ϵ\mathrm{\right)}$.

We also have the following lemma from [7].

#### Lemma 2.12.

Let B be a subset of ${\mathrm{F}}_{q}$ with at least two elements, and define ${\mathrm{F}}_{B}$ as the subfield generated by B. Then there exists a polynomial $P\mathit{}\mathrm{\left(}{x}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{x}_{n}\mathrm{\right)}$ of several variables with coefficients belonging to the prime field ${\mathrm{F}}_{p}$ such that

$P\left(B,\mathrm{\dots },B\right)={𝔽}_{B}.$

We are now ready to prove Theorem 2.8.

#### Proof of Theorem 2.8.

Without loss of generality, we may assume $1\in A$ by considering $\frac{1}{x}A$ for some $x\in A$. We first define the ratio set

$R\left(A,B\right):=\left\{\frac{{a}_{1}-{a}_{2}}{{b}_{1}-{b}_{2}}:{a}_{1},{a}_{2}\in A,{b}_{1},{b}_{2}\in B,{b}_{1}\ne {b}_{2}\right\}.$

We now consider the following cases.

Case 1: $\mathrm{1}\mathrm{+}R\mathit{}\mathrm{\left(}A\mathrm{,}B\mathrm{\right)}\mathrm{\not\subset }R\mathit{}\mathrm{\left(}A\mathrm{,}B\mathrm{\right)}$. In this case, there exist ${a}_{1},{a}_{2}\in A$ and ${b}_{1}\ne {b}_{2}\in B$ such that

$r:=1+\frac{{a}_{1}-{a}_{2}}{{b}_{1}-{b}_{2}}\notin R\left(A,B\right).$

First, we apply Lemma 2.11 so that there exists a subset ${A}^{\prime }\subset A$ such that $|{A}^{\prime }|\gg |A|$ and

$|\left({b}_{1}-{b}_{2}\right){A}^{\prime }+\left({b}_{1}-{b}_{2}\right)B+\left({a}_{1}-{a}_{2}\right)B|\ll \frac{|A+B||\left({b}_{1}-{b}_{2}\right)A+\left({a}_{1}-{a}_{2}\right)B|}{|A|}.$(2.9)

On the other hand, we have

$|\left({b}_{1}-{b}_{2}\right){A}^{\prime }+\left({b}_{1}-{b}_{2}\right)B+\left({a}_{1}-{a}_{2}\right)B|\ge |{A}^{\prime }+rB|.$(2.10)

Since $r\notin R\left(A,B\right)$, the equation ${a}_{1}+r{b}_{1}={a}_{2}+r{b}_{2}$ has no nontrivial solutions, i.e. solutions $\left({a}_{1},{b}_{1},{a}_{2},{b}_{2}\right)$ with ${b}_{1}\ne {b}_{2}$. This implies that

$|{A}^{\prime }+rB|=|{A}^{\prime }||B|.$(2.11)

We now give an upper bound for $\left({b}_{1}-{b}_{2}\right)A+\left({a}_{1}-{a}_{2}\right)B={b}_{1}A+{a}_{1}B-{b}_{2}A-{a}_{2}B$ which will be used in the rest of the proof.

Lemma 2.11 tells us that there exists a subset $X\subset A$ such that $|X|\gg |A|$ and

$|X+{b}_{1}A+{a}_{1}B|\ll \frac{|A+{b}_{1}A||A+{a}_{1}B|}{|A|}\ll \frac{{|A+AB|}^{2}}{|A|},$

and there exists a subset ${X}^{\prime }\subset X$ with $|{X}^{\prime }|\gg |X|$ such that

$|{X}^{\prime }+{b}_{2}A+{a}_{2}B|\le \frac{|X+{b}_{2}A||X+{a}_{2}B|}{|X|}\ll \frac{{|A+AB|}^{2}}{|A|}.$

Applying Lemma 2.10, we have

$\begin{array}{cc}\hfill |{b}_{1}A+{a}_{1}B-{b}_{2}A-{a}_{2}B|& \le \frac{|{X}^{\prime }+{b}_{1}A+{a}_{1}B||{X}^{\prime }+{b}_{2}A+{a}_{2}B|}{|{X}^{\prime }|}\hfill \\ & \ll \frac{|X+{b}_{1}A+{a}_{1}B||{X}^{\prime }+{b}_{2}A+{a}_{2}B|}{|A|}\hfill \\ & \le \frac{{|A+BA|}^{4}}{{|A|}^{3}}.\hfill \end{array}$(2.12)

Putting (2.9)–(2.12) together, we obtain

${|A+AB|}^{5}\gg {|A|}^{5}|B|,$

and we are done.

Case 2: $B\mathrm{\cdot }R\mathit{}\mathrm{\left(}A\mathrm{,}B\mathrm{\right)}\mathrm{\not\subset }R\mathit{}\mathrm{\left(}A\mathrm{,}B\mathrm{\right)}$. Similarly, in this case, there exist ${a}_{1},{a}_{2}\in A$ and $b,{b}_{1},{b}_{2}\in B$ such that

$r:=b\cdot \frac{{a}_{1}-{a}_{2}}{{b}_{1}-{b}_{2}}\notin R\left(A,B\right).$

Since $0\in R\left(A,B\right)$, we have $b\ne 0$ and ${a}_{1}\ne {a}_{2}$. This gives us that ${r}^{-1}$ exists.

Using the same argument as above, we have

$\begin{array}{cc}\hfill |A||B|=|A+rB|& =|{r}^{-1}A+B|\le \frac{|{b}^{-1}A+A||\left({a}_{1}-{a}_{2}\right)B+\left({b}_{1}-{b}_{2}\right)A|}{|A|}\hfill \\ & \le \frac{|A+AB|{|A+AB|}^{4}}{{|A|}^{4}}.\hfill \end{array}$

Thus, we obtain

${|A+AB|}^{5}\gg {|A|}^{5}|B|,$

and we are done.

Case 3: ${B}^{\mathrm{-}\mathrm{1}}\mathrm{\cdot }R\mathit{}\mathrm{\left(}A\mathrm{,}B\mathrm{\right)}\mathrm{\not\subset }R\mathit{}\mathrm{\left(}A\mathrm{,}B\mathrm{\right)}$. As above, in this case, there exist ${a}_{1},{a}_{2}\in A$ and $b\ne 0$, ${b}_{1}\ne {b}_{2}\in B$ such that

$r:={b}^{-1}\cdot \frac{{a}_{1}-{a}_{2}}{{b}_{1}-{b}_{2}}\notin R\left(A,B\right).$

Since $0\in R\left(A,B\right)$, we have ${a}_{1}\ne {a}_{2}$. This gives us that ${r}^{-1}$ exists. The rest is the same as Case 2.

Case 4. We consider the case when

$1+R\left(A,B\right)\subset R\left(A,B\right),$$B\cdot R\left(A,B\right)\subset R\left(A,B\right),$${B}^{-1}\cdot R\left(A,B\right)\subset R\left(A,B\right).$

Now we are going to show that for any polynomial $P\left({x}_{1},\mathrm{\dots },{x}_{n}\right)$ in n variables, for some positive integer n, and coefficients belonging in ${𝔽}_{p}$ such that

$P\left(B,\mathrm{\dots },B\right)+R\left(A,B\right)\subset R\left(A,B\right).$

Indeed, it is enough to show that

$1+R\left(A,B\right)\subset R\left(A,B\right),{B}^{d}+R\left(A,B\right)\subset R\left(A,B\right)$

for any integer $d\ge 1$ and ${B}^{d}=B\mathrm{\cdots }B$ (d times).

It follows from the assumption that the first condition $1+R\left(A,B\right)\subset R\left(A,B\right)$ is satisfied. For the second condition, it is sufficient to prove it for $d=2$, since we can use inductive arguments.

Let $b,{b}^{\prime }$ be arbitrary elements in B. We now show that

$b{b}^{\prime }+R\left(A,B\right)\subset R\left(A,B\right).$

If either $b=0$ or ${b}^{\prime }=0$, then we are done. Thus, we may assume that $b\ne 0$ and ${b}^{\prime }\ne 0$.

First, we have

$b+R\left(A,B\right)=b\left(1+{b}^{-1}R\left(A,B\right)\right)\subset b\left(1+R\left(A,B\right)\right)\right)\subset R\left(A,B\right)$

and

$b{b}^{\prime }+R\left(A,B\right)=b\left({b}^{\prime }+{b}^{-1}R\left(A,B\right)\right)\subset b\left({b}^{\prime }+R\left(A,B\right)\right)\subset bR\left(A,B\right)\subset R\left(A,B\right).$

In other words, for any polynomial $P\left({x}_{1},{x}_{2},\mathrm{\dots },{x}_{n}\right)\in {𝔽}_{p}\left[{x}_{1},\mathrm{\dots },{x}_{n}\right]$, we have

$P\left(B,\mathrm{\dots },B\right)+R\left(A,B\right)\subset R\left(A,B\right).$

Furthermore, Lemma 2.12 tells us that there exists a polynomial P such that $P\left(B,\mathrm{\dots },B\right)={𝔽}_{B}$. This implies that

${𝔽}_{B}+R\left(A,B\right)\subset R\left(A,B\right).$

It follows from our assumption of the theorem that

$|B|=|B\cap {𝔽}_{B}|\le {|{𝔽}_{B}|}^{\frac{1}{2}}.$

Hence, $|R\left(A,B\right)|\ge |{𝔽}_{B}|\ge {|B|}^{2}$.

Next, we shall show that there exists $r\in R\left(A,B\right)$ such that either

$|A+rB|\gg |A||B|\mathit{ }\text{or}\mathit{ }|A+rB|\gg {|B|}^{2}.$

Indeed, let ${E}^{+}\left(X,Y\right)$ be the number of tuples $\left({x}_{1},{x}_{2},{y}_{1},{y}_{2}\right)\in {X}^{2}×{Y}^{2}$ such that

${x}_{1}+{y}_{1}={x}_{2}+{y}_{2}.$

We have that the sum ${\sum }_{r\in R\left(A,B\right)}{E}^{+}\left(A,rB\right)$ is the number of tuples $\left({a}_{1},{a}_{2},{b}_{1},{b}_{2}\right)\in {A}^{2}×{B}^{2}$ such that

${a}_{1}+r{b}_{1}={a}_{2}+r{b}_{2}$

with ${a}_{1},{a}_{2}\in A$, ${b}_{1},{b}_{2}\in B$ and $r\in R\left(A,B\right)$. It is easy to see that there are at most $|R\left(A,B\right)||A||B|$ tuples with ${a}_{1}={a}_{2}$, ${b}_{1}={b}_{2}$, and at most ${|A|}^{2}{|B|}^{2}$ tuples with ${b}_{1}\ne {b}_{2}$. Therefore, we get

$\sum _{r\in R\left(A,B\right)}{E}^{+}\left(A,rB\right)\le |R\left(A,B\right)||A||B|+{|A|}^{2}{|B|}^{2}\le |R\left(A,B\right)|\left(|A||B|+{|A|}^{2}\right).$

By the pigeon-hole principle, there exists $r:=\frac{{a}_{1}-{a}_{2}}{{b}_{1}-{b}_{2}}\in R\left(A,B\right)$ such that

$E\left(A,rB\right)\le |A||B|+{|A|}^{2}.$

So, either

$|A+rB|\gg |A||B|\mathit{ }\text{or}\mathit{ }|A+rB|\gg {|B|}^{2}.$

We now fall into two small cases:

• (1)

If $|A+rB|\gg |A||B|$, then, applying Lemma 2.10, we have

$|A||B|=|A+rB|=|\left({b}_{1}-{b}_{2}\right)A+\left({a}_{1}-{a}_{2}\right)B|\le \frac{{|AB+A|}^{4}}{{|A|}^{3}},$

which gives us

$|A+AB|\gg |A|{|B|}^{\frac{1}{2}}.$

• (2)

If $|A+rB|\gg {|B|}^{2}$, then we have

${|B|}^{2}\ll |A+rB|=|\left({b}_{1}-{b}_{2}\right)A+\left({a}_{1}-{a}_{2}\right)B|\le \frac{{|AB+A|}^{4}}{{|A|}^{3}},$

which gives us

$|A+AB|\gg {|A|}^{\frac{3}{4}}{|B|}^{\frac{2}{4}}.$

This completes the proof of the theorem. ∎

## 3 Proofs of Theorems 1.3 and 1.4

In the proof of Theorem 1.3, we make use of the following version of the Balog–Szemerédi–Gowers theorem due to Schoen [12].

#### Theorem 3.1 ([12, Theorem 1.1]).

Let G be an Abelian group. Suppose that A is a subset of G, and ${E}^{\mathrm{+}}\mathit{}\mathrm{\left(}A\mathrm{\right)}$ denotes the additive energy which is the number of solutions $\mathrm{\left(}a\mathrm{,}b\mathrm{,}c\mathrm{,}d\mathrm{\right)}\mathrm{\in }{A}^{\mathrm{4}}$ to the equation $a\mathrm{+}b\mathrm{=}c\mathrm{+}d$. If ${E}^{\mathrm{+}}\mathit{}\mathrm{\left(}A\mathrm{\right)}$ is equal to $k\mathit{}{\mathrm{|}A\mathrm{|}}^{\mathrm{3}}$, then there exists ${A}^{\mathrm{\prime }}\mathrm{\subset }A$ with $\mathrm{|}{A}^{\mathrm{\prime }}\mathrm{|}\mathrm{\gg }k\mathit{}\mathrm{|}A\mathrm{|}$ such that

$|{A}^{\prime }-{A}^{\prime }|\ll {k}^{-4}|{A}^{\prime }|.$

We will also need the following results.

#### Theorem 3.2.

For $A\mathrm{,}B\mathrm{,}C\mathrm{,}D\mathrm{\subset }{\mathrm{F}}_{p}$, let $Q\mathit{}\mathrm{\left(}A\mathrm{,}B\mathrm{,}C\mathrm{,}D\mathrm{\right)}$ be the number of 8-tuples

$\left({a}_{1},{b}_{1},{c}_{1},{d}_{1},{a}_{2},{b}_{2},{c}_{2},{d}_{2}\right)\in {\left(A×B×C×D\right)}^{2}$

such that

${a}_{1}{b}_{1}+{c}_{1}{d}_{1}={a}_{2}{b}_{2}+{c}_{2}{d}_{2}.$

We have

$Q\left(A,B,C,D\right)\lesssim \frac{{|A|}^{2}{|B|}^{2}{|C|}^{2}{|D|}^{2}}{p}+{|C|}^{2}|B|{|D|}^{\frac{3}{2}}{|A|}^{\frac{1}{2}}{E}^{×}{\left(A,B\right)}^{\frac{1}{2}}+|A|{|D|}^{3}|B||C|+{|A|}^{3}|D||B||C|,$

where

${E}^{×}\left(A,B\right)=\mathrm{#}\left\{\left({a}_{1},{a}_{2},{b}_{1},{b}_{2}\right)\in {A}^{2}×{B}^{2}:{a}_{1}{b}_{1}={a}_{2}{b}_{2}\right\}.$

To prove this theorem, we need the following version of the point-plane incidence bound due to Rudnev in [8].

#### Theorem 3.3 ([8]).

Let P be a set of points in ${\mathrm{F}}_{p}^{\mathrm{3}}$ and let Π be a set of planes in ${\mathrm{F}}_{p}^{\mathrm{3}}$. Suppose that $\mathrm{|}P\mathrm{|}\mathrm{\le }\mathrm{|}\mathrm{\Pi }\mathrm{|}$, and there are at most k collinear points in P for some k. Then the number of incidences between P and Π is bounded by

$I\left(P,\mathrm{\Pi }\right)\le \frac{|P||\mathrm{\Pi }|}{p}+{|P|}^{\frac{1}{2}}|\mathrm{\Pi }|+k|P|.$

We are now ready to prove Theorem 3.2. We will follow the ideas of [14, proof of Theorem 32].

#### Proof of Theorem 3.2.

We have

$Q\left(A,B,C,D\right)=\sum _{\lambda ,\mu }{r}_{CD}\left(\lambda \right){r}_{AB}\left(\mu \right)n\left(\lambda ,\mu \right),$

where ${r}_{CD}\left(\lambda \right)$ is the number of pairs $\left(c,d\right)\in C×D$ such that $cd=\lambda$, ${r}_{AB}\left(\mu \right)$ is the number of pairs $\left(a,b\right)\in A×B$ such that $ab=\mu$, and $n\left(\lambda ,\mu \right)={\sum }_{x}{r}_{AB+\lambda }\left(x\right){r}_{CD+\mu }\left(x\right)$. If we split the sum $Q\left(A,B,C,D\right)$ into intervals, we get

$Q\left(A,B,C,D\right)\ll \sum _{i=1}^{{L}_{1}}\sum _{j=1}^{{L}_{2}}\sum _{\lambda ,\mu }n\left(\lambda ,\mu \right){r}_{CD}^{\left(i\right)}\left(\lambda \right){r}_{AB}^{\left(j\right)}\left(\mu \right),$

where ${L}_{1}\le \mathrm{log}\left(|C||D|\right)$, ${L}_{2}\le \mathrm{log}\left(|A||B|\right)$, ${r}_{AB}^{\left(i\right)}\left(\mu \right)$ is the restriction of the function ${r}_{AB}\left(x\right)$ on the set

${P}_{i}:=\left\{\mu :{\mathrm{\Delta }}_{i}\le {r}_{AB}\left(\mu \right)<2{\mathrm{\Delta }}_{i}\right\},$

and ${r}_{CD}^{\left(i\right)}\left(\lambda \right)$ is the restriction of the function ${r}_{CD}\left(x\right)$ on the set

${P}_{i}:=\left\{\lambda :{\mathrm{\Delta }}_{i}\le {r}_{CD}\left(\lambda \right)<2{\mathrm{\Delta }}_{i}\right\}.$

Applying the pigeon-hole principle twice, there exist sets ${P}_{i}$ and ${P}_{j}$ such that

$Q\left(A,B,C,D\right)\lesssim \sum _{\lambda ,\mu }n\left(\lambda ,\mu \right){r}_{CD}^{\left(i\right)}\left(\lambda \right){r}_{AB}^{\left(j\right)}\left(\mu \right)\lesssim {\mathrm{\Delta }}_{i}{\mathrm{\Delta }}_{j}\sum _{\lambda ,\mu }n\left(\lambda ,\mu \right){P}_{i}\left(\mu \right){P}_{j}\left(\lambda \right),$

where ${P}_{i}\left(x\right)$ is the indicator function of the set ${P}_{i}$. For simplicity, we suppose that $i=1$ and $j=2$.

One can check that the sum

$\sum _{\lambda ,\mu }n\left(\lambda ,\mu \right){P}_{1}\left(\lambda \right){P}_{2}\left(\mu \right)$

is the number of incidences between points $\left(a,d,\lambda \right)\in A×D×{P}_{1}\subset {𝔽}_{p}^{3}$ and planes in ${𝔽}_{p}^{3}$ defined by

$bX-cY+Z=\mu ,$

where $b\in B$, $c\in C$, $\mu \in {P}_{2}$.

With the way we define the plane set, it follows from [2] that we can apply Theorem 3.3 with

$k=\mathrm{max}\left\{|A|,|D|\right\}.$

Thus, we obtain

$\begin{array}{cc}\hfill Q\left(A,B,C,D\right)& \lesssim {\mathrm{\Delta }}_{1}{\mathrm{\Delta }}_{2}\left(\frac{|A||B||C||D||{P}_{1}||{P}_{2}|}{p}\right)\hfill \\ & +{\mathrm{\Delta }}_{1}{\mathrm{\Delta }}_{2}\left({|A|}^{\frac{1}{2}}|B||C|{|D|}^{\frac{1}{2}}{|{P}_{1}|}^{\frac{1}{2}}|{P}_{2}|+\mathrm{max}\left\{|A|,|D|\right\}|A||D||{P}_{1}|\right).\hfill \end{array}$(3.1)

It is clear in our argument that we can switch the point set and the plane set, we also can do the same thing for ${P}_{1}$ and ${P}_{2}$ in the definition of the point set and the plane set. So, without loss of generality, we can assume that $|{P}_{1}|\le |{P}_{2}|$, $|A||D|\le |B||C|$. We now consider the following cases:

• If the second term dominates, then we have

$Q\left(A,B,C,D\right)\lesssim {|C|}^{2}|B|{|D|}^{\frac{3}{2}}{|A|}^{\frac{1}{2}}{E}^{×}{\left(A,B\right)}^{\frac{1}{2}},$

since

${\mathrm{\Delta }}_{2}|{P}_{2}|\le |C||D|,{\mathrm{\Delta }}_{1}{|{P}_{1}|}^{\frac{1}{2}}\le {E}^{×}{\left(A,B\right)}^{\frac{1}{2}}.$

• If the first term dominates, then we have

$Q\left(A,B,C,D\right)\lesssim \frac{{|A|}^{2}{|B|}^{2}{|C|}^{2}{|D|}^{2}}{p},$

since

${\mathrm{\Delta }}_{2}|{P}_{2}|\le |C||D|,{\mathrm{\Delta }}_{1}|{P}_{1}|\le |A||B|.$

• If the last term dominates, then we study the following:

• (1)

Suppose $|A|\le |D|$. If $|D|\le |{P}_{2}|$, then it is easy to check that the second term in (3.1) will be bigger than the last term. Thus, we can suppose that $|D|\ge |{P}_{2}|$. Since $|{P}_{1}|\le |{P}_{2}|$, we have

$|A|{|D|}^{2}|{P}_{1}|\le |A|{|D|}^{3}.$

On the other hand, it is clear that ${\mathrm{\Delta }}_{1}{\mathrm{\Delta }}_{2}\le |B||C|$. This means

$Q\left(A,B,C,D\right)\lesssim |A|{|D|}^{3}|B||C|.$

• (2)

Suppose $|A|\ge |D|$. By repeating the same argument, we obtain

$Q\left(A,B,C,D\right)\lesssim {|A|}^{3}|D||B||C|.$

This completes the proof of the theorem. ∎

#### Proof of Theorem 1.3.

Let N be the number of tuples

$\left({x}_{1},{y}_{1},{z}_{1},{t}_{1},{x}_{2},{y}_{2},{z}_{2},{t}_{2},{x}_{1}^{\prime },{y}_{1}^{\prime },{z}_{1}^{\prime },{t}_{1}^{\prime },{x}_{2}^{\prime },{y}_{2}^{\prime },{z}_{2}^{\prime },{t}_{2}^{\prime }\right)\in {A}^{16}$

such that $\left[𝐱,𝐲,0\right]\cdot \left[𝐳,𝐭,0\right]=\left[{𝐱}^{\prime },{𝐲}^{\prime },0\right]\cdot \left[{𝐳}^{\prime },{𝐭}^{\prime },0\right]$. This can be expressed as follows:

$\left(\begin{array}{cccc}\hfill 1\hfill & \hfill {x}_{1}\hfill & \hfill {x}_{2}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill {y}_{1}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill {y}_{2}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\cdot \left(\begin{array}{cccc}\hfill 1\hfill & \hfill {z}_{1}\hfill & \hfill {z}_{2}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill {t}_{1}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill {t}_{2}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)=\left(\begin{array}{cccc}\hfill 1\hfill & \hfill {x}_{1}^{\prime }\hfill & \hfill {x}_{2}^{\prime }\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill {y}_{1}^{\prime }\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill {y}_{2}^{\prime }\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\cdot \left(\begin{array}{cccc}\hfill 1\hfill & \hfill {z}_{1}^{\prime }\hfill & \hfill {z}_{2}^{\prime }\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill {t}_{1}^{\prime }\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill {t}_{2}^{\prime }\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right).$(3.2)

Thus, by the Cauchy–Schwarz inequality, we have

$|{\left[{A}^{2},{A}^{2},0\right]}^{2}|\ge \frac{{|A|}^{16}}{N}.$(3.3)

From (3.2), observe that N is the number of tuples

$\left({x}_{1},{y}_{1},{z}_{1},{t}_{1},{x}_{2},{y}_{2},{z}_{2},{t}_{2},{x}_{1}^{\prime },{y}_{1}^{\prime },{z}_{1}^{\prime },{t}_{1}^{\prime },{x}_{2}^{\prime },{y}_{2}^{\prime },{z}_{2}^{\prime },{t}_{2}^{\prime }\right)\in {A}^{16}$

satisfying the following system:

${x}_{1}+{z}_{1}={x}_{1}^{\prime }+{z}_{1}^{\prime },{x}_{2}+{z}_{2}={x}_{2}^{\prime }+{z}_{2}^{\prime },$(3.4)${y}_{1}+{t}_{1}={y}_{1}^{\prime }+{t}_{1}^{\prime },{y}_{2}+{t}_{2}={y}_{2}^{\prime }+{t}_{2}^{\prime },$(3.5)${x}_{1}{t}_{1}+{x}_{2}{t}_{2}={x}_{1}^{\prime }{t}_{1}^{\prime }+{x}_{2}^{\prime }{t}_{2}^{\prime }.$(3.6)

We now insert new variables ${s}_{1},{s}_{2},{s}_{3},{s}_{4}\in A+A$ in (3.4) and (3.5) as follows:

${x}_{1}+{z}_{1}={s}_{1}={x}_{1}^{\prime }+{z}_{1}^{\prime },$${x}_{2}+{z}_{2}={s}_{2}={x}_{2}^{\prime }+{z}_{2}^{\prime },$${y}_{1}+{t}_{1}={s}_{3}={y}_{1}^{\prime }+{t}_{1}^{\prime },$${y}_{2}+{t}_{2}={s}_{4}={y}_{2}^{\prime }+{t}_{2}^{\prime }.$

Set ${A}_{{s}_{i}}=A\cap \left({s}_{i}-A\right)$, then we have ${x}_{1},{x}_{1}^{\prime }\in {A}_{{s}_{1}}$, ${x}_{2},{x}_{2}^{\prime }\in {A}_{{s}_{1}}$, ${y}_{1},{y}_{1}^{\prime }\in {A}_{{s}_{3}}$, ${y}_{2},{y}_{2}^{\prime }\in {A}_{{s}_{4}}$.

In this setting, we have

$N=\sum _{{s}_{1},{s}_{2},{s}_{3},{s}_{4}\in A+A}Q\left({A}_{{s}_{1}},{A}_{{s}_{2}},{A}_{{s}_{3}},{A}_{{s}_{4}}\right).$

Applying Theorem 3.2, and assuming the second and third terms are larger than the first term, we have

$\begin{array}{cc}\hfill N& \lesssim \sum _{{s}_{1},{s}_{2},{s}_{3},{s}_{4}}{|{A}_{{s}_{1}}|}^{\frac{1}{2}}|{A}_{{s}_{2}}|{|{A}_{{s}_{3}}|}^{2}{|{A}_{{s}_{4}}|}^{\frac{3}{2}}{E}^{×}{\left({A}_{{s}_{1}},{A}_{{s}_{2}}\right)}^{\frac{1}{2}}+\sum _{{s}_{1},{s}_{2},{s}_{3},{s}_{4}}|{A}_{{s}_{1}}||{A}_{{s}_{2}}||{A}_{{s}_{3}}|{|{A}_{{s}_{4}}|}^{3}\hfill \\ & +\sum _{{s}_{1},{s}_{2},{s}_{3},{s}_{4}}{|{A}_{{s}_{1}}|}^{3}|{A}_{{s}_{2}}||{A}_{{s}_{3}}||{A}_{{s}_{4}}|\hfill \\ & \lesssim \left(\sum _{{s}_{1},{s}_{2}}{|{A}_{{s}_{1}}|}^{\frac{1}{2}}|{A}_{{s}_{2}}|{E}^{×}{\left({A}_{{s}_{1}},{A}_{{s}_{2}}\right)}^{\frac{1}{2}}\right)\left(\sum _{{s}_{3}}{|{A}_{{s}_{3}}|}^{2}\right)\left(\sum _{{s}_{4}}{|{A}_{{s}_{4}}|}^{\frac{3}{2}}\right)+{|A|}^{10}\hfill \\ & \le |A|{E}^{+}{\left(A\right)}^{\frac{3}{2}}\left(\sum _{{s}_{1},{s}_{2}}{|{A}_{{s}_{1}}|}^{\frac{1}{2}}|{A}_{{s}_{2}}|{E}^{×}{\left({A}_{{s}_{1}},{A}_{{s}_{2}}\right)}^{\frac{1}{2}}\right)+{|A|}^{10},\hfill \end{array}$

where we have used the fact that

$\sum _{s}{|{A}_{s}|}^{\frac{3}{2}}\le {\left(\sum _{s}|{A}_{s}|\right)}^{\frac{1}{2}}{\left(\sum _{s}{|{A}_{s}|}^{2}\right)}^{\frac{1}{2}},\sum _{s}{|{A}_{s}|}^{2}\le {E}^{+}\left(A\right).$

Moreover, using the fact ${E}^{×}\left(A,B\right)\le {|A|}^{2}|B|$, we have

$\sum _{{s}_{1},{s}_{2}}{|{A}_{{s}_{1}}|}^{\frac{1}{2}}|{A}_{{s}_{2}}|{E}^{×}{\left({A}_{{s}_{1}},{A}_{{s}_{2}}\right)}^{\frac{1}{2}}\le \sum _{{s}_{1},{s}_{2}}{|{A}_{{s}_{1}}|}^{\frac{3}{2}}{|{A}_{{s}_{2}}|}^{\frac{3}{2}}\le {|A|}^{2}{E}^{+}\left(A,A\right).$

In other words, we have proved that

$N\lesssim {|A|}^{3}{E}^{+}{\left(A\right)}^{\frac{5}{2}}+{|A|}^{10}.$

If $N\lesssim {|A|}^{10}$, then the theorem follows from (3.3). Therefore, we can assume that

$N\lesssim {|A|}^{3}{E}^{+}{\left(A\right)}^{\frac{5}{2}}.$

Let ϵ be a parameter chosen later. We now consider two cases:

Case 1. Suppose that ${E}^{+}\left(A\right)<{|A|}^{3-ϵ}$. Then we have

$N\le {|A|}^{\frac{9}{2}+6-\frac{5ϵ}{2}}.$

From (3.3), this implies that

$|{\left[{A}^{2},{A}^{2},0\right]}^{2}|\ge {|A|}^{\frac{11}{2}+\frac{5ϵ}{2}}.$

Case 2. Suppose that ${E}^{+}\left(A\right)\ge {|A|}^{3-ϵ}$. Then we can write ${E}^{+}\left(A\right)={|A|}^{3-{ϵ}^{\prime }}$ for some ${ϵ}^{\prime }<ϵ<1$. Notice that Theorem 3.1 implies that there exists a subset ${A}^{\prime }\subset A$ such that $|{A}^{\prime }|\gg {|A|}^{1-ϵ}$ and

$|{A}^{\prime }-{A}^{\prime }|\ll {|A|}^{4ϵ}|{A}^{\prime }|\ll {|{A}^{\prime }|}^{1+\frac{4ϵ}{1-ϵ}}.$

Since $|A|\le {p}^{\frac{9}{16}}$ by our assumption, using Lemma 2.5 with the above inequality gives

$|{A}^{\prime }\cdot {A}^{\prime }|\gtrsim {|{A}^{\prime }|}^{\frac{14}{9}-\frac{8ϵ}{1-ϵ}}.$(3.7)

Moreover, one can easily check that

$|{\left[{A}^{2},{A}^{2},0\right]}^{2}|\ge {|A|}^{4}|\left\{{x}_{1}{t}_{1}+{x}_{2}{t}_{2}:{x}_{1},{t}_{1},{x}_{2},{t}_{2}\in A\right\}|.$

Therefore,

$|{\left[{A}^{2},{A}^{2},0\right]}^{2}|\ge {|A|}^{4}|\left\{{x}_{1}{t}_{1}+{x}_{2}{t}_{2}:{x}_{1},{t}_{1}\in A,{x}_{2},{t}_{2}\in {A}^{\prime }\right\}|.$

Moreover, Lemma 2.1 gives us

$\begin{array}{cc}\hfill |\left\{{x}_{1}{t}_{1}+{x}_{2}{t}_{2}:{x}_{1},{t}_{1}\in A,{x}_{2},{t}_{2}\in {A}^{\prime }\right\}|& \gg \mathrm{min}\left\{|A|{|{A}^{\prime }\cdot {A}^{\prime }|}^{\frac{1}{2}},p\right\}\hfill \\ & \gtrsim \mathrm{min}\left\{{|A|}^{1+\left(1-ϵ\right)\left(\frac{7}{9}-\frac{4ϵ}{1-ϵ}\right)},p\right\},\hfill \end{array}$

where we also utilized inequality (3.7) and the fact that $|{A}^{\prime }|\gg {|A|}^{1-ϵ}$.

Thus, we obtain that if $|A|\le {p}^{\frac{9}{16}}$, then

$\begin{array}{cc}\hfill |{\left[{A}^{2},{A}^{2},0\right]}^{2}|& \gtrsim \mathrm{min}\left\{{|A|}^{5+\left(1-ϵ\right)\left(\frac{7}{9}-\frac{4ϵ}{1-ϵ}\right)},p{|A|}^{4}\right\}\hfill \\ & \ge {|A|}^{5+\left(1-ϵ\right)\left(\frac{7}{9}-\frac{4ϵ}{1-ϵ}\right)}\hfill \end{array}$

provided that ${|A|}^{1+\left(1-ϵ\right)\left(\frac{7}{9}-\frac{4ϵ}{1-ϵ}\right)}\le p$. It is clear that with $ϵ=\frac{5}{131}$, this estimate is satisfied since $|A|\le {p}^{\frac{1}{2}}$. We also obtain

$|{\left[{A}^{2},{A}^{2},0\right]}^{2}|\gtrsim {|A|}^{\frac{11}{2}+\frac{25}{262}}.$

Finally, suppose the first term is the largest term. Then we have $N\le \frac{{E}^{+}{\left(A,A\right)}^{4}}{p}$ which is smaller than $\frac{{|A|}^{12}}{p}$. Thus (3.3) shows that

$|{\left[{A}^{2},{A}^{2},0\right]}^{2}|\gtrsim p{|A|}^{4}\ge {|A|}^{6}.$

This is more than what we want and completes the proof of the theorem. ∎

Over finite arbitrary fields ${𝔽}_{q}$, we have the following result.

#### Theorem 3.4.

Let $q\mathrm{=}{p}^{n}$ and let A be a subset of ${\mathrm{F}}_{q}^{\mathrm{*}}$. If $\mathrm{|}A\mathrm{\cap }\lambda \mathit{}F\mathrm{|}\mathrm{\le }{\mathrm{|}F\mathrm{|}}^{\frac{\mathrm{1}}{\mathrm{2}}}$ for any proper subfield F of ${\mathrm{F}}_{q}$ and any $\lambda \mathrm{\in }{\mathrm{F}}_{q}$, then we have

$|{\left[A,A,0\right]}^{2}|\gtrsim {|A|}^{3+\frac{1}{11}},\text{𝑎𝑛𝑑}\mathit{ }|{\left[{A}^{2},{A}^{2},0\right]}^{2}|\gtrsim {|A|}^{5+\frac{1}{5}}.$

#### Proof.

We first observe that

$|{\left[A,A,0\right]}^{2}|\ge {|A|}^{2}\cdot \mathrm{max}\left\{|A+A|,|A\cdot A|\right\}$

and

$|{\left[{A}^{2},{A}^{2},0\right]}^{2}|\ge {|A|}^{4}\cdot |AA+AA|.$

It follows from Theorems 2.7 and 2.8 that

$|AA+AA|\gg {|A|}^{\frac{6}{5}}\mathit{ }\text{and}\mathit{ }\mathrm{max}\left\{|A+A|,|A\cdot A|\right\}\gg {|A|}^{\frac{12}{11}}.$

Therefore, we obtain

$|{\left[A,A,0\right]}^{2}|\gtrsim {|A|}^{3+\frac{1}{11}}\mathit{ }\text{and}\mathit{ }|{\left[{A}^{2},{A}^{2},0\right]}^{2}|\gtrsim {|A|}^{5+\frac{1}{5}},$

which completes the proof of the theorem. ∎

#### Proof of Theorem 1.4.

One can observe that

$|{\left[{A}^{2},{A}^{2},A\right]}^{2}|\ge {|A|}^{4}|AA+AA+A+A|.$(3.8)

We now prove that if $|A|\le {p}^{\frac{9}{16}}$, then

$|AA+AA+A+A|\gg {|A|}^{\frac{79}{45}}.$

Indeed, we first prove that if $|A|\le {p}^{\frac{9}{16}}$, then

$|AA+A+A|\gtrsim {|A|}^{\frac{3}{2}+\frac{1}{90}}.$(3.9)

To prove this inequality, we consider the following cases:

Case 1. If $|A+A|\ge {|A|}^{1+ϵ}$, then it follows from Lemma 2.1 that

$|AA+A+A|\ge \mathrm{min}\left\{{|A|}^{\frac{3}{2}+\frac{ϵ}{2}},p\right\}={|A|}^{\frac{3}{2}+\frac{ϵ}{2}},$(3.10)

whenever $|A|\le {p}^{\frac{2}{3+ϵ}}$.

Case 1. If $|A+A|\le {|A|}^{1+ϵ}$, then Lemma 2.5 gives us that $|AA|\gtrsim {|A|}^{\frac{14}{9}-2ϵ}$ under the condition $|A|\le {p}^{\frac{9}{16}}$. Hence, if $|A|\le {p}^{\frac{9}{16}}$, then

$|AA+A+A|\ge |AA|\gtrsim {|A|}^{\frac{14}{9}-2ϵ}.$(3.11)

Choosing $ϵ=\frac{1}{45}$, we see from (3.10) and (3.11) that if $|A|\le {p}^{\frac{45}{68}}$ and $|A|\le {p}^{\frac{9}{16}}$, then

$|AA+A+A|\gtrsim {|A|}^{\frac{3}{2}+\frac{1}{90}}.$

Since ${p}^{\frac{45}{68}}\ge {p}^{\frac{9}{16}}$, we establish inequality (3.9).

By Lemma 2.1 and inequality (3.9), we see that if $|A|\le {p}^{\frac{9}{16}}$, then

$|AA+\left(AA+A+A\right)|\gg \mathrm{min}\left\{|A|{|AA+A+A|}^{\frac{1}{2}},p\right\}\gtrsim \mathrm{min}\left\{{|A|}^{\frac{7}{4}+\frac{1}{180}},p\right\}={|A|}^{\frac{79}{45}}.$

Finally, combining (3.8) and this estimate, we conclude that if $|A|\le {p}^{\frac{9}{16}}$, then

$|{\left[{A}^{2},{A}^{2},A\right]}^{2}|\gtrsim {|A|}^{\frac{11}{2}+\frac{23}{90}},$

which completes the proof of Theorem 1.4. ∎

## Acknowledgements

The authors would like to deeply thank Oliver Roche-Newton and Ilya Shkredov for many helpful discussions that make nice improvement for our Theorem 1.3. The authors would like to thank the referee for valuable suggestions.

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Revised: 2018-07-05

Published Online: 2018-08-18

Published in Print: 2019-01-01

Funding Source: National Research Foundation of Korea

Award identifier / Grant number: NRF-2015R1A1A1A05001374

Award identifier / Grant number: P2ELP2175050

Funding Source: Ministry of Science and Technology, Taiwan

Award identifier / Grant number: 104-2628-M-002-015-MY4

D. Koh was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (NRF-2015R1A1A1A05001374). T. Pham was supported by Swiss National Science Foundation grant P2ELP2175050. C.-Y. Shen was supported in part by MOST, through grant 104-2628-M-002-015-MY4.

Citation Information: Forum Mathematicum, Volume 31, Issue 1, Pages 35–48, ISSN (Online) 1435-5337, ISSN (Print) 0933-7741,

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