1 Introduction
We begin with the following question.
Question 1.1 Let p be a prime, and
$A, \ B$
be subsets in
$\mathbb {F}_p^d$
. Under what conditions on A and B do we have
$|gA-B|\gg p^d$
for a positive proportion of
$g\in SO_d(\mathbb {F}_p)$
?
The original version of this question was posed in the Euclidean setting and has a well-known history, which we will not detail here. It is necessary to mention that for
$A, B\subset \mathbb {R}^d$
with
$\dim _H(A)+\dim _H(B)>d$
, Mattila [Reference Mattila8] proved that if
$$ \begin{align*}\frac{(d-1)}{d}\dim_H(A)+\dim_H(B)>d ~~\mbox{or}~~\dim_H(B)>\frac{d+1}{2},\end{align*} $$
then
$\mathcal {L}^d(gA-B)>0$
for almost all orthogonal matrices g in
$\mathbb {R}^d$
. Here, by
$\dim _H$
and
$\mathcal {L}(\cdot )$
we mean the Hausdorff dimension and the Lebesgue measure in
$\mathbb {R}^d$
, respectively. In two dimensions with
$\dim _H(A)=\dim _H(B)=t$
, he asked in the survey paper [Reference Mattila9] whether or not the condition
$t>5/4$
would be enough to guarantee that
${\mathcal {L}(gA-B)>0}$
by using the techniques due to Guth, Iosevich, Ou, and Wang in [Reference Guth, Iosevich, Ou and Wang2] on the Falconer distance problem.
In a recent paper, Pham and Yoo [Reference Pham and Yoo12] answered this question affirmatively in the prime field setting by showing that if
$|A|=|B|\gg p^{5/4}$
and
$p\equiv 3\ \pmod 4$
, then there exists a set
$E\subset SO_2(\mathbb {F}_p)$
with
$|E|<p/2$
such that for each
$g\in SO_2(\mathbb {F}_p)\setminus E$
there are at least
$\gg p^2$
elements
$z\in \mathbb {F}_p^2$
satisfying
$|(gA+z)\cap B|\sim \frac {|A||B|}{p^2}$
. When
$p\equiv 1\ \pmod 4$
, their method implies a weaker exponent, namely,
$3/2$
instead of
$5/4$
. Notice that these exponents cannot be reduced to a number less than
$1$
, to see this, one takes two sets A and B being on a circle centered at the origin of radius
$1$
, then it is clear that
${|gA-B|\sim |A||B|}$
.
In this note, we are interested in studying this topic in the plane over a finite p-adic ring. In particular, we consider the following question. Write
$\delta _A=|A|/p^{2r}$
for any set
$A\subset ({\mathbb Z}/p^r{\mathbb Z})^2$
.
Question 1.2 Let
$A\ and\ B$
be sets in
$(\mathbb {Z}/p^r\mathbb {Z})^2$
. Under what conditions on
$\delta _A$
and
$\delta _B$
do we have that
$|gA-B|\gg p^{2r}$
for a positive proportion of
$g\in SO_2(\mathbb {Z}/p^r\mathbb {Z})$
?
The following examples show that in order to have
$|gA-B|\gg p^{2r}$
, the sets A and B cannot both be small.
Example 1.3 Let
$X=\{\mathbf {x} \in ({\mathbb Z}/p^r{\mathbb Z})^2:\, \mathbf {x}\equiv \mathbf {0}\ \pmod p\}$
. Let A (and B) be disjoint unions of m (and n, respectively,) cosets of X, where
$1\leq m,n\leq p^2$
. So,
$|A|=mp^{2r-2}$
and
$|B|=np^{2r-2}$
. One sees that
$|g A|\leq mp^{2r-2}$
and
$|gA-B|\leq mnp^{2r-2}$
.
Our main theorem reads as follows.
Theorem 1.4 Let p be a prime with
$p\equiv 3\ \pmod 4$
, and r be a positive integer. Let
$A\ and\ B \subset ({\mathbb Z} /p^r {\mathbb Z})^2$
be such that
$\delta _A^{1/2}\cdot \delta _B\ge 2p^{-1}$
. Then for a positive proportion of
$g \in SO_2({\mathbb Z} /p^r {\mathbb Z})$
, we have
$$\begin{align*}\left\vert gA -B\right\vert \gg p^{2r}. \end{align*}$$
Based on Example 1.3, the condition
$\delta _A^{1/2}\cdot \delta _B\ge 2p^{-1}$
is sharp up to constant factors. Indeed, if we take
$m=1$
and
$n=\gamma (p)p^2$
with
$\gamma (p)$
tending to
$0$
arbitrarily slowly as p tends to infinity, then
$\delta _A^{1/2}\delta _B=\gamma (p)p^{-1}$
and
$|gA-B|\leq \gamma (p)p^{2r}$
. Example 1.3 is most interesting when
$r>1$
since it shows the sharpness of the exponent
$1/2$
on the density condition of A.
To prove this result, we make use of tools from discrete Fourier analysis and results in restriction/extension theory. As in the finite field setting [Reference Pham and Yoo12], the key estimate in our proof is the following sum:
$$\begin{align*}\sum_{\mathbf{m} \ne \mathbf{0}}\sum_{ \mathbf{m}^{\prime} \in V_{\mathbf{m}}}\left\vert \widehat{A}(\mathbf{m})\right\vert^2 \left\vert \widehat{B}(\mathbf{m} ^{\prime})\right\vert^2,\end{align*}$$
where
$V_{\mathbf {m}}:=\{g \mathbf {m}\colon g\in SO_2(\mathbb {Z}/p^r\mathbb {Z})\}$
. There are several approaches one can use to bound this sum:
-
1. Using the Plancherel formula is a standard argument and often yields a weak bound.
-
2. A more advanced approach employs restriction/extension estimates associated with circles. While such estimates are well understood in the finite field setting, see [Reference Chapman, Erdogan, Hart, Iosevich and Koh1, Reference Koh and Sun3], for example, they have only recently been developed in the p-adic setting [Reference Pham and Xue11]. This method will be used to prove Theorem 1.4.
-
3. When
$r=1$
, a novel method, combining the
$L^2$
norm of the distance function from [Reference Murphy, Petridis, Pham, Rudnev and Stevens7] with a double-counting argument, is introduced in [Reference Pham and Yoo12]. Extending this approach to the p-adic setting, however, presents challenges due to the limited understanding of incidence structures, particularly those involving points and planes in
$(\mathbb {Z}/p^r\mathbb {Z})^3$
. We hope to address this issue in the near future.
Although Theorem 1.4 is not as strong as Pham and Yoo’s result in [Reference Pham and Yoo12], it establishes that if
$\delta _A^{1/2}\cdot \delta _B \ge 2p^{-1}$
, then
$|gA-B|\gg p^2$
for a positive proportion of
${g\in SO_2(\mathbb {Z}/p\mathbb {Z})}$
. This aligns directly with Mattila’s theorem in [Reference Mattila8] with
$d=2$
, as mentioned above.
Notice that when
$p\equiv 1\ \pmod 4$
, the restriction/extension estimates are weaker, so the proof of Theorem 1.4 implies the condition of
$\delta _A^{1/2}\cdot \delta _B\ge 2p^{-1/2}$
. This is worse than the next theorem, which will be proved by using the Plancherel formula directly.
Theorem 1.5 Let p be a prime with
$p\equiv 1\ \pmod 4$
, and r be a positive integer. Let
$A\ and\ B \subset ({\mathbb Z} /p^r {\mathbb Z})^2$
be such that
$\delta _A\cdot \delta _B \ge 2p^{-1}$
. Then for a positive proportion of
${g\in SO_2({\mathbb Z} /p^r {\mathbb Z})}$
, we have
$$\begin{align*}\left\vert gA -B\right\vert \gg p^{2r}. \end{align*}$$
In Example 1.3, by taking
$A=B=X+Y$
, where
$Y\subset \{(x_1, x_2)\in (\mathbb {Z}/{p}\mathbb {Z})\times (\mathbb {Z}/{p}\mathbb {Z})\colon x_1^2+x_2^2\equiv 1\ \pmod p \}$
, then, for any
$g\in SO_2(\mathbb {Z}/p^r\mathbb {Z})$
, we have
$|A|=|B|=|Y|\cdot p^{2r-2}$
and
$|gA-B|\ll |Y|^2\cdot p^{2r-2}$
. Therefore, it is reasonable to make the following conjecture on the balanced case.
Conjecture 1.6 Let p be an odd prime, and r be a positive integer. Let
$A\ and B \subset ({\mathbb Z} /p^r {\mathbb Z})^2$
be such that
$\delta _A=\delta _B\gg p^{-1}$
. Then for a positive proportion of
${g \in SO_2({\mathbb Z} /p^r {\mathbb Z})}$
, we have
$$\begin{align*}\left\vert gA -B\right\vert \gg p^{2r}. \end{align*}$$
This conjecture appears highly challenging and could be as difficult as the Erdős-Falconer distance problem. Regarding further open questions, a natural direction is to explore similar problems in other group settings or in higher dimensions. To generalize Theorem 1.4 to higher dimensions, the first step is to extend the results in Section 2, but we found these extensions too complicated in this setting. This suggests exploring alternative approaches to address the problem. We refer the reader to [Reference Lichtin4, Reference Lichtin5, Reference Lichtin6, Reference Pham and Xue11] for recent progress on the distance problem and extensions in the finite p-adic setting.
Notations 1 In this article, by
$X\gg Y$
we mean
$X\ge CY$
for some absolute positive constant C, and
$X\sim Y$
means
$X\gg Y$
and
$Y\gg X$
.
2 Preliminaries
Throughout this article, the letter p always denotes an odd prime, and r a positive integer.
Let
$$\begin{align*}G_r:=SO_2(\mathbb{Z}/p^r\mathbb{Z})=\left\{\begin{bmatrix} a &-b\\ b & a \end{bmatrix}\,(\textsf{mod } p^r):\,\, a^2+b^2\equiv 1\,(\textsf{mod } p^r)\right\}. \end{align*}$$
For
$\mathbf {x}=(x_1,x_2)\in (\mathbb {Z}/p^r\mathbb {Z})^2$
, the distance function is denoted by
$\|\mathbf {x}\|:=x_1^2+x_2^2\, (\textsf {mod } p^r)$
. Then
$\|g\mathbf {x}\|\equiv \|\mathbf {x}\|\, (\textsf {mod } p^r)$
for any
$g\in G_r$
and
$\mathbf {x}\in (\mathbb {Z}/p^r\mathbb {Z})^2$
. Write the orbit of
$\mathbf {x}$
by
$$\begin{align*}\textsf{orb}_r(\mathbf{x}) = \big\{\theta \mathbf{x}:\, \theta \in G_r\big\}, \end{align*}$$
and the stabilizer of
$\mathbf {x}$
by
$$\begin{align*}\textsf{stab}_r(\mathbf{x}) = \{\theta\in G_r:\, \theta \mathbf{x} \equiv \mathbf{x}\, (\textsf{mod } p^r)\}. \end{align*}$$
For any
$j\in \mathbb {Z}/p^r\mathbb {Z}$
, let
$C_{j,r}$
be the circle centered at the origin of radius j, i.e.,
$$\begin{align*}C_{j,r}=\{\mathbf{x}\in (\mathbb{Z}/p^r\mathbb{Z})^2:\, \|\mathbf{x}\|\equiv j\,(\textsf{mod } p^r)\}. \end{align*}$$
For
$x\in \mathbb {Z}/p^r\mathbb {Z}$
and
$u\in \{0,1,\ldots ,r-1\}$
, the expression
$v_p(x)=u$
means that
${x\equiv 0\,(\textsf {mod } p^u)}$
and
$x\not \equiv 0\,(\textsf {mod } p^{u+1})$
. We also denote
$v_p(0) = r$
for
$0\in \mathbb {Z}/p^r\mathbb {Z}$
. When
$\mathbf {x}=(x_1,x_2)$
, we write
$v_{\mathbf {x}}=\min \{v_p(x_1),\, v_p(x_2)\}$
. Then, the vector
$\mathbf {x}$
can be expressed as
$\mathbf {x}=p^{v_{\mathbf {x}}}\tilde {\mathbf {x}}$
, where
$\tilde {\mathbf {x}}$
is a vector in
$(\mathbb {Z}/p^{r-v_{\mathbf {z}}}\mathbb {Z})^n$
such that
$v_{\tilde {\mathbf {x}}}=0$
.
We recall the following facts from [Reference Pham and Xue11].
Lemma 2.1 ([Reference Pham and Xue11], Lemma 4.5)
Let
$p\equiv 3 \, ({\sf mod }\ 4)$
. For any
$\mathbf {0}\neq \mathbf {x}\in (\mathbb {Z}/p^r\mathbb {Z})^2$
, let
${\mathbf {x}=p^{v_{\mathbf {x}}}\tilde {\mathbf {x}}}$
for some
$\tilde {\mathbf {x}}\in (\mathbb {Z}/p^{r-v_{\mathbf {x}}}\mathbb {Z})^2$
with
$v_{\tilde {\mathbf {x}}}=0$
. Then
$\textsf {orb}_r(\mathbf {x}) = p^{v_{\mathbf {x}}}C_{\|\tilde {\mathbf {x}}\|,r-v_{\mathbf {x}}}$
, and
$$\begin{align*}|\textsf{stab}_r(\mathbf{x})| = p^{v_{\mathbf{x}}},\qquad |\textsf{orb}_r(\mathbf{x})| = p^{r-v_{\mathbf{x}}}\left(1+\frac{1}{p} \right). \end{align*}$$
Lemma 2.2 ([Reference Pham and Xue11], Lemma 4.9)
Let
$p\equiv 1 \, ({\sf mod }\ 4)$
. For any
$\mathbf {0}\neq \mathbf {x}\in (\mathbb {Z}/p^r\mathbb {Z})^2$
, let
${\mathbf {x}=p^{v_{\mathbf {x}}}\tilde {\mathbf {x}}}$
for some
$\tilde {\mathbf {x}}\in (\mathbb {Z}/p^{r-v_{\mathbf {x}}}\mathbb {Z})^2$
with
$v_{\tilde {\mathbf {x}}}=0$
. Then
$\textsf {orb}_r(\mathbf {x}) = p^{v_{\mathbf {x}}}\textsf {orb}_{r-v_{\mathbf {x}}}(\tilde {\mathbf {x}})$
, and
$$\begin{align*}|\textsf{stab}_r({\mathbf{x}})| = p^{v_{\mathbf{x}}},\qquad |\textsf{orb}_r(\mathbf{x})| = p^{r-v_{\mathbf{x}}}\left( 1-\frac{1}{p}\right). \end{align*}$$
For a function
$f:\, ({\mathbb Z} /p^r {\mathbb Z})^2 \rightarrow {\mathbb C}$
, the Fourier transform
$\widehat {f}:\,({\mathbb Z} /p^r {\mathbb Z})^2 \rightarrow {\mathbb C}$
is defined by
$$\begin{align*}\widehat{f}(\mathbf{m}):=p^{-2r}\sum_{\mathbf{x}\in ({\mathbb Z} /p^r {\mathbb Z})^2} \chi_r (-\mathbf{m}\cdot \mathbf{x})f(\mathbf{x}), \end{align*}$$
where
$\chi _r (x) = e^{\frac {2\pi ix}{p^r}} (x\, \, \textsf {mod } p^r)$
.
We have the following basic properties of
$\chi $
and
$\widehat {f}$
.
-
• The orthogonality property
$$\begin{align*}\sum_{\boldsymbol{\alpha } \in ({\mathbb Z} /p^r{\mathbb Z})^2}\chi_r(\boldsymbol{\beta }\cdot \boldsymbol{\alpha }) = \begin{cases} p^{2r} ,& \text{if}\quad \boldsymbol{\beta } \equiv \mathbf{0} \, (\textsf{mod } p^r) ,\\ 0 ,& \text{if}\quad \boldsymbol{\beta } \not \equiv \mathbf{0} \,(\textsf{mod } p^r). \end{cases} \end{align*}$$
-
• The Fourier inversion formula
$$\begin{align*}f(\mathbf{x})= \sum_{\mathbf{m}\in ({\mathbb Z} /p^r{\mathbb Z})^2} \widehat{f}(\mathbf{m})\chi_r(\mathbf{m}\cdot \mathbf{x}). \end{align*}$$
-
• The Plancherel formula
$$\begin{align*}\sum_{\mathbf{m}\in ({\mathbb Z} /p^r {\mathbb Z})^2} \left\vert \widehat{f} (\mathbf{m})\right\vert^2 =p^{-2r} \sum_{\mathbf{x}\in ({\mathbb Z} /p^r {\mathbb Z})^2} \left\vert f(\mathbf{x}) \right\vert^2. \end{align*}$$
For a set A, by abuse of notation, we also denote its characteristic function by
$A(x)$
, i.e.
$A(x)=1$
if
$x\in A$
and
$A(x)=0$
if
$x\not \in A$
.
Now we state a more general restriction problem,.
Let
$\mathbf {m}\in (\mathbb {Z}/p^r\mathbb {Z})^2$
. When r is given, we also use the simplified notation
$V_{\mathbf {m}}$
for
$\textsf {orb}_r(\mathbf {m})$
. Let
$d\sigma _{V_{\mathbf {m}}}$
be the corresponding surface measure on
$V_{\mathbf {m}}$
and for all
${f: V_{\mathbf {m}} \rightarrow {\mathbb C}}$
, define
$$\begin{align*}(fd\sigma_{V_{\mathbf{m}}} )^\vee (\mathbf{y}) := \frac{1}{\left\vert V_{\mathbf{m}}\right\vert} \sum_{\mathbf{x} \in V_{\mathbf{m}}} f(\mathbf{x} ) \chi_r(\mathbf{y} \cdot \mathbf{x}) .\end{align*}$$
As computed in [Reference Pham and Xue11], we have
$$ \begin{align*} &\sum_{\mathbf{m}\,(\textsf{mod } p^r)}|(fd\sigma_{V_{\mathbf{m}}})^\vee(\mathbf{m})|^4=\frac{p^{2r}}{|V_{\mathbf{m}}|^4}\sum_{\xi,\xi^{\prime},\eta,\eta^{\prime}\in V_{\mathbf{m}} \atop \xi-\eta\equiv \xi^{\prime}-\eta^{\prime}\,(\textsf{mod } p^r)}f(\xi)f(\xi^{\prime})\overline{f(\eta)f(\eta^{\prime})}. \end{align*} $$
If f is the characteristic function of a set
$A\subset V_{\mathbf {m}}$
, then
$|V_{\mathbf {m}}|^4p^{-2r}\sum _{\mathbf {m}\,(\textsf {mod } p^r)}| (fd\sigma _{V_{\mathbf {m}}})^\vee (\mathbf {m})|^4$
counts the number of energy quadruples in A.
The following theorems give upper bounds of that sum for general functions f.
Theorem 2.3 ([Reference Pham and Xue11], Theorem 4.1)
Let
$p\equiv 3\, ({\sf mod }\ 4)$
be a prime and
$r \ge 1$
be an integer. Let
$\mathbf {m}\in (\mathbb {Z}/p^r\mathbb {Z})^2$
be such that
$\mathbf {m}\neq \mathbf {0}$
. Then
$$\begin{align*}\left(\sum_{\mathbf{x}\in (\mathbb{Z}/p^r\mathbb{Z})^2}|(fd\sigma_{V_{\mathbf{m}}})^\vee(\mathbf{x})|^4\right)^{\frac{1}{2}}\ll p^{-\frac{r-3v_{\mathbf{m}}+1}{2}}\sum_{\mathbf{x}\in V_{\mathbf{m}}}|f(\mathbf{x})|^2.\end{align*}$$
Theorem 2.4 ([Reference Pham and Xue11], Theorem 4.2)
Let
$p\equiv 1 \, ({\sf mod }\ 4)$
be a prime and
$r \ge 1$
be an integer. Let
$\mathbf {m}\in (\mathbb {Z}/p^r\mathbb {Z})^2$
be such that
$\mathbf {m}\neq \mathbf {0}$
. Suppose that
$\mathbf {m}=p^{v_m}\tilde {\mathbf {m}}$
with
${\tilde {\mathbf {m}}\in (\mathbb {Z}/p^{r-v_m}\mathbb {Z})^2}$
.
If
$\|\tilde {\mathbf {m}}\|\not \equiv 0\,(\textsf {mod } p)$
, then
$$\begin{align*}\left( \sum_{\mathbf{x}\in (\mathbb{Z} /p^r\mathbb{Z})^2} \vert (fd\sigma_{V_{\mathbf{m}}})^{\vee}(\mathbf{x}) \vert^4\right)^{\frac{1}{2}} \ll p^{-\frac{ r-3v_{\mathbf{m}}+1}{2}}\sum_{\mathbf{x}\in V_{\mathbf{m}}} \vert f(\mathbf{x})\vert^2 .\end{align*}$$
If
$\Vert \tilde {\mathbf {m}} \Vert \equiv 0 \, (\textsf {mod } p)$
,we have
$$\begin{align*}\left( \sum_{\mathbf{x}\in (\mathbb{Z} /p^r\mathbb{Z})^2} \vert (fd\sigma_{V_{\mathbf{m}}})^{\vee}(\mathbf{x}) \vert^4\right)^{\frac{1}{2}} \ll p^{-\frac{r-3v_{\mathbf{m}}}{2}}\sum_{\mathbf{x}\in V_{\mathbf{m}}} \vert f(\mathbf{x})\vert^2 .\end{align*}$$
Remark 2.5 In comparison between Theorem 2.3 and Theorem 2.4, the latter is weaker, which comes from elements
$\mathbf {m}$
with
$\Vert \tilde {\mathbf {m}} \Vert \equiv 0 \, (\textsf {mod } p)$
.
The next two lemmas are duality versions of Theorem 2.3 and Theorem 2.4.
Lemma 2.6 Let
$p\equiv 3\ \pmod 4$
. Let
$E\subset ({\mathbb Z}/p^r{\mathbb Z})^2$
and
$\mathbf {m}\in ({\mathbb Z}/p^r{\mathbb Z})^2$
be such that
$\mathbf {m}\neq \mathbf {0}$
. Then
$$\begin{align*}\sum_{\mathbf{x} \in V_{\mathbf{m}}} \left\vert \widehat{E}(\mathbf{x})\right\vert^2 \ll p^{-\frac{5r+v_{\mathbf{m}}+1}{2}}\left\vert E\right\vert^{3/2}. \end{align*}$$
Proof By Hölder’s inequality and Theorem 2.3, we have
$$ \begin{align*} &\frac{p^{2r}}{\left\vert V_{\mathbf{m}}\right\vert}\sum_{\mathbf{x} \in V_{\mathbf{m}}} \left\vert \widehat{E}(\mathbf{x}) \right\vert^2 = \frac{p^{2r}}{\left\vert V_{\mathbf{m}}\right\vert}\sum_{\mathbf{x} \in V_{\mathbf{m}}} \widehat{E}(\mathbf{x}) \overline{\widehat{E}(\mathbf{x})} = \sum_{\mathbf{y} \in ({\mathbb Z}/p^r {\mathbb Z})^2} E(\mathbf{y}) \overline{\frac{1}{\left\vert V_{\mathbf{m}}\right\vert}\sum_{\mathbf{x} \in V_{\mathbf{m}}} \chi_r(\mathbf{y} \cdot \mathbf{x}) \widehat{E}(\mathbf{x})} \\ &= \sum_{\mathbf{y} \in ({\mathbb Z} /p^r {\mathbb Z})^2}E(\mathbf{y}) (\widehat{E}d\sigma_{V_{\mathbf{m}}})^\vee (\mathbf{y}) \le \left( \sum_{\mathbf{y} \in ({\mathbb Z}/p^r {\mathbb Z})^2} \left\vert E(\mathbf{y})\right\vert^{\frac{4}{3}} \right)^{\frac{3}{4}} \left( \sum_{\mathbf{y} \in ({\mathbb Z}/p^r {\mathbb Z})^2}\left\vert (\widehat{E}d\sigma_{V_{\mathbf{m}}})\right\vert^4 \right)^{\frac{1}{4}}\\ & \ll |E|^{\frac{3}{4}} \left( p^{-\frac{r-3v_{\mathbf{m}}+1}{2}}\sum_{\mathbf{x} \in V_{\mathbf{m}}}\left\vert \widehat{E}(\mathbf{x} )\right\vert^2 \right)^{\frac{1}{2}}. \end{align*} $$
Recall from Lemma 2.1 that
$\left \vert V_{\mathbf {m}}\right \vert \sim p^{r-v_{\mathbf {m}}}$
, the previous inequality implies
$$ \begin{align*} p^{r+v_{\mathbf{m}}} \sum_{\mathbf{x} \in V_{\mathbf{m}}} \left\vert \widehat{E}(\mathbf{x})\right\vert^2 \ll |E|^{\frac{3}{4}} \left(p^{\frac{-r+3v_{\mathbf{m}} -1}{2}} \sum_{\mathbf{x} \in V_{\mathbf{m}}} \left\vert \widehat{E}(\mathbf{x})\right\vert^2\right)^{\frac{1}{2}}. \end{align*} $$
It follows that:
$$\begin{align*}\sum_{\mathbf{x} \in V_{\mathbf{m}}} \left\vert \widehat{E}(\mathbf{x})\right\vert^2 \ll p^{-\frac{5r+v_{\mathbf{m}}+1}{2}}\left\vert E\right\vert^{\frac{3}{2}}.\\[-38pt]\end{align*}$$
Lemma 2.7 Let
$p\equiv 1\, ({\sf mod }\ 4)$
be a prime and
$r\ge 1$
be an integer. Let
$E\subset (\mathbb {Z}/p^r\mathbb {Z})^2$
and
$\mathbf {m}\in (\mathbb {Z}/p^r\mathbb {Z})^2$
be such that
$\mathbf {m}\neq \mathbf {0}$
. Suppose that
$\mathbf {m} = p^{v_{\mathbf {m}}}\tilde {\mathbf {m}}$
, where
$v_{\tilde {\mathbf {m}}} =0$
.
If
$\left \Vert \tilde {\mathbf {m}}\right \Vert \not \equiv 0\ \pmod p$
, then
$$\begin{align*}\sum_{\mathbf{x} \in V_{\mathbf{m}}} \left\vert \widehat{E}(\mathbf{x})\right\vert^2 \ll p^{-\frac{5r+v_{\mathbf{m}}+1}{2}}|E|^{\frac{3}{2}}; \end{align*}$$
If
$\left \Vert \tilde {\mathbf {m}}\right \Vert \equiv 0\ \pmod p$
, then
$$\begin{align*}\sum_{\mathbf{x} \in V_{\mathbf{m}}} \left\vert \widehat{E}(\mathbf{x})\right\vert^2 \ll p^{-\frac{5r+v_{\mathbf{m}}}{2}}|E|^{\frac{3}{2}}. \end{align*}$$
Corollary 2.8 Let p be a prime, and Let
$p\equiv 3\ \pmod 4$
. Let
$A, B$
be sets in
$({\mathbb Z} /p^r {\mathbb Z})^2$
. Then we have
$$\begin{align*}\sum_{\mathbf{m} \ne \mathbf{0}}\sum_{ \mathbf{m}^{\prime} \in V_{\mathbf{m}}}p^{v_{\mathbf{m}}} \left\vert \widehat{A}(\mathbf{m})\right\vert^2 \left\vert \widehat{B}(\mathbf{m} ^{\prime})\right\vert^2\ll p^{-4r-1}|A||B|^{\frac{3}{2}}.\end{align*}$$
Proof Applying Lemma 2.6 and Plancherel formula, as above, one has
$$ \begin{align*} & \sum_{\substack{\mathbf{m} \ne \mathbf{0},\\ \mathbf{m}' \in V_{\mathbf{m}}}}p^{v_{\mathbf{m}}}\left\vert \widehat{A}(\mathbf{m})\right\vert ^2\left\vert \widehat{B}(\mathbf{m}^{\prime})\right\vert^2 \le \sum_{\mathbf{m}}|\widehat{A}(\mathbf{m})|^2 \cdot \max_{\mathbf{z} \ne \mathbf{0}} \left( p^{v_{\mathbf{z}}} \sum_{\mathbf{m} ^{\prime}\in V_{\mathbf{z}}} |\widehat{B}(\mathbf{m}^{\prime})|^2 \right) \\ &\qquad\qquad \ll p^{-2r}|A| \cdot \max_{\mathbf{z} \ne \mathbf{0}} \left( p^{v_{\mathbf{z}}} \cdot p^{-\frac{5r+v_{\mathbf{z}} +1}{2}}|B|^{\frac{3}{2}} \right) \ll p^{-4r-1}|A||B|^{\frac{3}{2}}, \end{align*} $$
by noting that
$\max v_{\mathbf {z}} = r-1$
for
$\mathbf {z} \ne \mathbf {0}$
. Thus, the theorem follows.
Corollary 2.9 Let
$p\equiv 1\ \pmod 4$
be a prime, let
$A, B$
be sets in
$({\mathbb Z} /p^r {\mathbb Z})^2$
. Then we have
$$\begin{align*}\sum_{\mathbf{m} \ne \mathbf{0}} \sum_{\mathbf{m}^{\prime} \in V_{\mathbf{m}} } p^{v_{\mathbf{m}}} \left\vert \widehat{A}(\mathbf{m} ) \right\vert \left\vert \widehat{B} (\mathbf{m}^{\prime}) \right\vert \ll p^{-4r-1/2}|A||B|^{3/2}.\end{align*}$$
3 Incidences between points and rigid-motions
Recall that
$G_r=SO_2({\mathbb Z} /p^r {\mathbb Z} )$
. We denote the set of rigid motion over
${\mathbb Z}/p^r{\mathbb Z}$
by
$$\begin{align*}\mathcal{R}_r :=\{(g,\mathbf{z}):\, g\in G_r,\,\mathbf{z}\in ({\mathbb Z} /p^r {\mathbb Z})^2\}. \end{align*}$$
Indeed, it is a semi-direct product of groups (see [Reference Pham and Xue10] for example). As a set, one has
$\mathcal {R}_r = G_r\times ({\mathbb Z} /p^r {\mathbb Z})^2$
and
$|\mathcal {R}_r|\sim p^{3r}$
.
We say a pair of points
$(\mathbf {x},\mathbf {y})\in ({\mathbb Z} /p^r {\mathbb Z} )^2 \times ({\mathbb Z} /p^r {\mathbb Z} )^2$
is incident to a rigid-motion
$(g, \mathbf {z})$
if
$g\mathbf {y}+\mathbf {z}=\mathbf {x}$
. For
$P = A \times B \subset ({\mathbb Z} /p^r {\mathbb Z} )^2 \times ({\mathbb Z} /p^r {\mathbb Z} )^2$
and
$R \subset \mathcal {R}_r $
, we define
$$\begin{align*}\mathcal{I}(P,R) := \Big\{ \big((g,\mathbf{z}),(\mathbf{x},\mathbf{y})\big) \in R\times P:\, g\mathbf{y} +\mathbf{z} =\mathbf{x} \Big\}\end{align*}$$
as the number of incidences between P and R. This section is devoted to bounding
$\mathcal {I}(P, R)$
from above and below.
We first state a universal bound that will be proved by using a standard discrete Fourier analysis argument, followed by an improvement obtained via Corollary 2.8.
Theorem 3.1 Let p be an odd prime, let
$P=A\times B \subset ({\mathbb Z} /p^r{\mathbb Z})^2 \times ({\mathbb Z} /p^r {\mathbb Z})^2$
and
$ {R \subset \mathcal {R}_r }$
. Then
$$\begin{align*}\left\vert \mathcal{I}(P,R) - \frac{|P||R|}{p^{2r}} \right\vert \ll p^{\frac{3r-1}{2}}|P|^{\frac{1}{2}}|R|^{\frac{1}{2}}.\nonumber\end{align*}$$
Proof We have
$$ \begin{align*} \mathcal{I}(P, R)&=\sum_{\substack{(\mathbf{x}, \mathbf{y})\in P, \\ (g, \mathbf{z})\in R}}1_{\mathbf{x}=g\mathbf{y}+\mathbf{z}}=\frac{1}{p^{2r}}\sum_{\mathbf{m} \in ({\mathbb Z} /p^r {\mathbb Z})^2}\sum_{\substack{(\mathbf{x}, \mathbf{y})\in P,\\ (g, \mathbf{z})\in R}}\chi_r\left(\mathbf{m}\cdot (\mathbf{x}-g\mathbf{y}-\mathbf{z} )\right)\\ &=\frac{|P||R|}{p^{2r}}+\frac{1}{p^{2r}}\sum_{\mathbf{m} \in ({\mathbb Z} /p^r {\mathbb Z})^2\setminus \{ \mathbf{0} \}}\sum_{\substack{(\mathbf{x}, \mathbf{y})\in P,\\ (g, \mathbf{z})\in R}}\chi_r(\mathbf{m} \cdot(\mathbf{x}-g\mathbf{y}-\mathbf{z}))\\ &=\frac{|P||R|}{p^{2r}}+p^{2r}\sum_{\mathbf{m}\ne \mathbf{0} }\sum_{(g, \mathbf{z})\in R}\widehat{P}(-\mathbf{m}, g\mathbf{m})\chi_r(-\mathbf{m} \mathbf{z})=:I+II, \end{align*} $$
where,
$$\begin{align*}\widehat{P}(\mathbf{u}, \mathbf{v})=p^{-4r}\sum_{(\mathbf{x}, \mathbf{y}) \in P} \chi_r(-\mathbf{x} \cdot \mathbf{u}-\mathbf{y} \cdot \mathbf{v}). \end{align*}$$
We next bound the second term. By the Cauchy-Schwarz inequality, we have
$$ \begin{align*} II&= p^{2r}\sum_{(g, \mathbf{z})\in R}\sum_{\mathbf{m}\ne \mathbf{0} }\widehat{P}(-\mathbf{m}, g\mathbf{m})\chi_r(-\mathbf{m} \mathbf{z})\\ &\le p^{2r} |R|^{1/2}\left(\sum_{(g, \mathbf{z})\in \mathcal{R}_r}\sum_{\mathbf{m}_1, \mathbf{m}_2\ne \mathbf{0}} \widehat{P}(-\mathbf{m}_1, g\mathbf{m}_1)\overline{\widehat{P}(-\mathbf{m}_2, g\mathbf{m}_2)}\chi_r(\mathbf{z} \cdot (-\mathbf{m}_1+\mathbf{m}_2))\right)^{1/2}\\ &= p^{2r} |R|^{1/2}\left(\sum_{g\in G_r}\sum_{\mathbf{m}_1, \mathbf{m}_2\ne \mathbf{0}} \widehat{P}(-\mathbf{m}_1, g\mathbf{m}_1)\overline{\widehat{P}(-\mathbf{m}_2, g\mathbf{m}_2)}\sum\limits_{\mathbf{z}\in (\mathbb{Z}/p^r\mathbb{Z})^2}\chi_r(\mathbf{z} \cdot (-\mathbf{m}_1+\mathbf{m}_2))\right)^{1/2}\\ &=p^{2r} |R|^{1/2}\left(p^{2r}\sum_{g\in G_r}\sum_{\mathbf{m} \ne \mathbf{0}}|\widehat{P}(\mathbf{m}, -g\mathbf{m})|^2\right)^{1/2}. \end{align*} $$
Since
$P=A\times B$
, we have
$$\begin{align*}\widehat{P}(\mathbf{m}, -g\mathbf{m})=\widehat{A}(\mathbf{m})\widehat{B}(-g\mathbf{m}).\end{align*}$$
Thus, following Lemma 2.2 and Plancherel fomula, we have
$$ \begin{align*} &\sum_{g\in G_r}\sum_{\mathbf{m} \ne \mathbf{0}}|\widehat{P}(\mathbf{m}, -g\mathbf{m})|^2 =\sum_{\mathbf{m} \ne \mathbf{0}}|\widehat{A}(\mathbf{m})|^2\sum_{g\in G_r}|\widehat{B}(-g\mathbf{m})|^2\\ & \ll \sum_{\mathbf{m} \ne \mathbf{0}}|\widehat{A} (\mathbf{m})|^2 \sum_{\mathbf{m}^{\prime} \in V_{\mathbf{m}}}p^{v_{\mathbf{m}}} |\widehat{B}(\mathbf{m}^{\prime})|^2 \le \sum_{\mathbf{m} \ne \mathbf{0}}p^{v_{\mathbf{m}}}|\widehat{A} (\mathbf{m})|^2 \sum_{\mathbf{m}^{\prime}} |\widehat{B}(\mathbf{m}^{\prime})|^2 \\ &\le p^{r-1}\sum_{\mathbf{m}} |\widehat{A}(\mathbf{m})|^2 \sum_{\mathbf{m}^{\prime}} |\widehat{B}(\mathbf{m}^{\prime})|^2 = p^{r-1}\cdot \frac{|A|}{p^{2r}}\cdot \frac{|B|}{p^{2r}} = \frac{|A||B|}{p^{3r+1}}. \end{align*} $$
Here, we have used the fact that the stabilizer of a nonzero element
$\mathbf {m}$
in
$SO_2({\mathbb Z} /p^r {\mathbb Z})$
is
$\sim p^{v_{\mathbf {m}}}\le p^{r-1}$
.
Finally, we obtain that
$$ \begin{align*} II & \ll p^{3r}|R|^{\frac{1}{2}}\left( \frac{|A||B|}{p^{3r+1}} \right)^{\frac{1}{2}} = p^{\frac{3r-1}{2}}|R|^{\frac{1}{2}}|P|^{\frac{1}{2}}. \end{align*} $$
This completes the proof.
Next, we estimate the number of incidences in another way.
Theorem 3.2 Let p be an odd prime. Let
$P=A\times B \subset ({\mathbb Z} /p^r{\mathbb Z})^2 \times ({\mathbb Z} /p^r {\mathbb Z})^2$
and
$ R \subset \mathcal {R}_r$
.
If
$p\equiv 3\ \pmod 4$
, then
$$\begin{align*}\left\vert \mathcal{I}(P,R) - \frac{|P||R|}{p^{2r}} \right\vert \ll p^{r-\frac{1}{2}}|P|^{\frac{1}{2}}|R|^{\frac{1}{2}}|B|^{\frac{1}{4}} .\end{align*}$$
If
$p\equiv 1\ \pmod 4$
, then
$$\begin{align*}\left\vert \mathcal{I}(P,R) - \frac{|P||R|}{p^{2r}} \right\vert \ll p^{r-\frac{1}{4}}|P|^{\frac{1}{2}}|R|^{\frac{1}{2}}|B|^{\frac{1}{4}} .\end{align*}$$
Proof As previous, we have
$$ \begin{align*} &\mathcal{I}(P, R)= \frac{|P||R|}{p^{2r}}+p^{2r}\sum_{\mathbf{m}\ne \mathbf{0} }\sum_{(g, \mathbf{z})\in R}\widehat{P}(-\mathbf{m}, g\mathbf{m})\chi_r(-\mathbf{m} \cdot \mathbf{z})=:I+II, \end{align*} $$
and
$$ \begin{align*} II& \leq p^{3r}|R|^{\frac{1}{2}}\left(\sum_{g\in G_r}\sum_{\mathbf{m} \ne \mathbf{0}}|\widehat{P}(\mathbf{m}, -g\mathbf{m})|^2\right)^{\frac{1}{2}}\\ &\ll p^{3r}|R|^{\frac{1}{2}}\left(\sum_{\substack{\mathbf{m} \ne \mathbf{0},\\ \mathbf{m}^{\prime} \in V_{\mathbf{m}}}} p^{v_{\mathbf{m}}} |\widehat{A}(\mathbf{m})|^2|\widehat{B}(\mathbf{m}^{\prime})|^2\right)^{\frac{1}{2}}. \end{align*} $$
To prove the first assertion, we use Corollary 2.8 and obtain
$$ \begin{align*} II & \ll p^{3r}|R|^{\frac{1}{2}}\left( p^{-4r-1}|A||B|^{\frac{3}{2}}\right)^{\frac{1}{2}}= p^{2r-\frac{1}{2}}|R|^{\frac{1}{2}}|P|^{\frac{1}{2}} |B|^{\frac{1}{4}}. \end{align*} $$
To prove the second assertion, we use Corollary 2.9 and obtain
$$ \begin{align*} II & \ll p^{3r}|R|^{\frac{1}{2}}\left( p^{-4r-\frac{1}{2}}|A||B|^{\frac{3}{2}} \right)^{\frac{1}{2}}= p^{r-\frac{1}{4}}|R|^{\frac{1}{2}}|P|^{\frac{1}{2}} |B|^{\frac{1}{4}}. \end{align*} $$
This completes the proof.
4 Proof of main theorems
Proof of Theorem 1.4
Since
$\delta _A^{1/2}\cdot \delta _B\ge 2p^{-1}$
, we have
$\delta _A\ge 4p^{-2}$
. Thus,
${|A|\ge 4p^{2r-2}}$
. The number of elements
$\mathbf {x}$
in
$({\mathbb Z}/p^r\mathbb {Z})^2$
with
$v_{\mathbf {x}}\neq 0$
does not exceed
$p^{2r-2}$
. So, without loss of generality, we may assume that
$|A|\geq 3p^{2r-2}$
and
$v_{\mathbf {x}} =0$
for any
$\mathbf {x} \in A$
.
Let
$$\begin{align*}\widetilde{G} = \left\{ g \in G_r\colon \left\vert \left\{ \mathbf{z} \in ({\mathbb Z} /p^r {\mathbb Z})^2 \colon B \cap (g A +\mathbf{z} ) =\emptyset \right\} \right\vert \ge \frac{p^{2r}}{2} \right\} .\end{align*}$$
It is sufficient to show that
$|\widetilde {G}| \ll |G_r|$
.
Let
$\widetilde {R}$
be the set of pairs
$(g ,\mathbf {z} ) \in \mathcal {R}_r$
such that
$g \in \widetilde {G}$
and
$ B \cap (g A +\mathbf {z} ) = \emptyset $
. It is clear that
$\mathcal {I}(B \times A ,\widetilde {R}) = 0$
. Alternatively, from Theorem 3.2 (1), we have
$$ \begin{align} \left\vert \mathcal{I}(B\times A ,\widetilde{R}) -\frac{|A||B||\widetilde{R}|}{p^{2r}}\right\vert \ll p^{r-\frac {1}{2}}|A|^{\frac{3}{4}}|B|^{\frac{1}{2}}|\widetilde{R}|^{\frac{1}{2} }. \end{align} $$
Thus, we deduce that
$$\begin{align*}|\widetilde{R}| \ll p^{6r-1} |A|^{-\frac{1}{2}}|B|^{-1}. \end{align*}$$
On the other hand, one sees by the construction of
$\widetilde {G}$
that
$| \widetilde {R}| \ge |\widetilde {G}| \cdot p^{2r}/2$
, we have
$$ \begin{align} |\widetilde{G} | \ll p^{4r-1} |A|^{-\frac{1}{2}}|B|^{-1}. \end{align} $$
It leads to
$|\widetilde {G} |\ll p^r$
, in view of the condition
$|A|^{1/2}|B| \ge 2p^{3r-1}.$
In above arguments, if we use Theorem 3.2 (2), then the condition
$|A|^{1/2}|B|\ge 2p^{3r-\frac {1}{2}}$
is required. By a direct computation, we can see that this condition is worse than those of Theorem 1.5.
Proof of Theorem 1.5
The proof is same as that of Theorem 2.6. In this case, one deduces from
$\delta _A\cdot \delta _B\ge 2p^{-1}$
that
$\delta _A\ge 2p^{-1}$
. Then the cardinality of A is at least
$2p^{2r-1}$
, which is much larger than the number of elements
$\mathbf {x}$
with
$v_{\mathbf {x}}\neq 0$
. From Theorem 3.1, the bound on the right-hand side of (1) is replaced by
$p^{\frac {3r-1}{2}}|A|^{\frac {1}{2}}|B|^{\frac {1}{2}}|\widetilde {R}|^{\frac {1}{2}}$
. And (2) becomes
$$\begin{align*}|\widetilde{G}| \ll p^{-2r}|\widetilde{R}| \ll p^{5r-1}|A|^{-1}|B|^{-1}, \end{align*}$$
which gives
$|\widetilde {G}| \ll p^r$
provided that
$|A||B|\geq 2p^{4r-1}$
.
Appendix: Proof of Theorem 2.4
As shown in [Reference Pham and Xue11], the proof of Theorem 2.4 is almost identical with that of Theorem 2.3 for the case
$p\equiv 3\ \pmod 4$
. For clarity, we provide a detailed proof here.
In this section, the letter p is always a prime with
$p \equiv 1\ (\textsf {mod } 4)$
, and r is a positive integer. Recall that
$G_r=SO_2({\mathbb Z}/p^r{\mathbb Z})$
.
Lemma 5.1 ([Reference Pham and Xue11], Lemma 2.1)
For any prime p, let
$$\begin{align*}\mathbf{G}(\mathbf{x}) = \big(\widetilde{G}(x_1,\ldots,x_n),\ldots,G_m(x_1,\ldots,x_n)\big) \end{align*}$$
be a map from
$\mathbb {Z}^n$
to
$\mathbb {Z}^m$
, with
$G_i$
polynomials with integer coefficients. Let l be a positive integer and
$\mathbf {y}\in \mathbb {Z}^n$
. Suppose that
$\mathbf {G}(\mathbf {y})\equiv \mathbf {0}\ (\textsf {mod } p^l)$
. Let
$R=\text {rank}\ J_{\mathbf {G}}(\mathbf {y})$
, where
$J_{\mathbf {G}}(\mathbf {y})$
is the Jocobi matrix modulo p, i.e.,
$$\begin{align*}J_{\mathbf{G}}(\mathbf{y})=\begin{bmatrix} \frac{\partial \widetilde{G}}{\partial x_1}(\mathbf{y}) &\frac{\partial \widetilde{G}}{\partial x_2}(\mathbf{y}) &\ldots &\frac{\partial \widetilde{G}}{\partial x_n}(\mathbf{y}) \\ \frac{\partial G_2}{\partial x_1}(\mathbf{y}) &\frac{\partial G_2}{\partial x_2}(\mathbf{y}) &\ldots &\frac{\partial G_2}{\partial x_n}(\mathbf{y}) \\ \ldots &\ldots &\ldots &\ldots \\ \frac{\partial G_m}{\partial x_1}(\mathbf{y}) &\frac{\partial G_m}{\partial x_2}(\mathbf{y}) &\ldots &\frac{\partial G_m}{\partial x_n}(\mathbf{y}) \end{bmatrix}\qquad (\textsf{mod } p). \end{align*}$$
Then
$$ \begin{align} \#\big\{\mathbf{z}\,(\textsf{mod } p^k):\, \mathbf{G}(\mathbf{y}+p^l\mathbf{z})\equiv \mathbf{0}\,(\textsf{mod } p^{l+k})\big\} \leq p^{k(n-R)} \end{align} $$
for any integer
$k\geq 1$
. When
$R=m$
, the “
$\leq $
” can be replaced by “
$=$
”.
Lemma 5.2 We have
$\left \vert G_r \right \vert =p^r (1-1/p)$
.
Proof For
$r=1$
, we have
$|G_1|=p-1$
. Now consider the circumstances that
$r\geq 2$
. For any
$\theta =\begin {bmatrix}a & -b\\b &a\end {bmatrix} \in G_r$
, there is some
$\theta _0 = \begin {bmatrix}a_0 & -b_0\\b_0 &a_0\end {bmatrix}\in G_1$
such that
$\theta \equiv \theta _0\,(\textsf {mod } p)$
. Applying Lemma 5.1 to
$(a_0,b_0)$
and the polynomial
$F(x,y)=x^2+y^2-1$
, one obtains that
$(\nabla F)(a_0,b_0) = (2a_0, 2b_0) \not \equiv (0,0) \,(\textsf {mod } p)$
, since
$a_0^2+b_0^2\equiv 1\ \pmod p$
. Thus,
$$\begin{align*}\#\left\{(z_1,z_2)\,(\textsf{mod } p^{r-1}):\, (a_0+pz_1)^2+(b_0+pz_2)^2 \equiv 1\ (\textsf{mod } p^r)\right\} = p^{r-1}. \end{align*}$$
It follows that
$$\begin{align*}|G_r|=p^{r-1}|G_1|=p^{r}(1-1/p).\\[-34pt] \end{align*}$$
Proof of Lemma 2.2
Here
$0 \le v_{\mathbf {x}} \le r-1$
. The equation
$\theta \mathbf {x} \equiv \mathbf {x} (\textsf {mod } p^r)$
is equivalent to
$$ \begin{align} \begin{bmatrix} a-1 & -b \\ b & a-1 \end{bmatrix} \begin{bmatrix} p^{v_{\mathbf{x}}} \tilde{x_1} \\ p^{v_{\mathbf{x}}} \tilde{x_2} \end{bmatrix} \equiv \begin{bmatrix} 0 \\ 0 \end{bmatrix} \quad (\textsf{mod } p^r) , \end{align} $$
or equivalently,
$$ \begin{align} \begin{bmatrix} \tilde{x_1} &-\tilde{x_2}\\ \tilde{x_2} & \tilde{x_1} \end{bmatrix} \begin{bmatrix} a-1\\ b \end{bmatrix}\equiv \begin{bmatrix} 0\\ 0 \end{bmatrix}\quad (\textsf{mod } p^{r-v_{\mathbf{x}}}). \end{align} $$
If
$\tilde {x_1}^2+\tilde {x_2}^2 \not \equiv 0 \, (\textsf {mod } p) $
, then the coefficient matrix is invertible modulo
$p^r$
, and
$$\begin{align*}\begin{bmatrix} a\\ b \end{bmatrix}\equiv \begin{bmatrix} 1\\ 0 \end{bmatrix},\quad (\textsf{mod } p^{r-v_{\mathbf{x}}}). \end{align*}$$
By Lemma 5.1, the number of
$(a,b)\in (\mathbb {Z}/p^r \mathbb {Z})^2$
satisfying
$a^2+b^2\equiv 1\ \pmod {p^r}$
and the above equivalence is exactly
$p^{v_{\mathbf {x}}}$
.
If
$\tilde {x_1}^2+\tilde {x_2}^2 \equiv 0 \, (\textsf {mod } p)$
, then one deduce from
$(\tilde {x_1},\tilde {x_2}) \not \equiv (0,0)\ \pmod p$
that
$\tilde {x_1}, \tilde {x_2}\not \equiv 0\ \pmod p$
. Therefore, let
$u=a-1, v=b$
, we have a system of linear equation
$$\begin{align*}\begin{cases} \tilde{x_1}u -\tilde{x_2}v \equiv 0 \\ \tilde{x_2}u +\tilde{x_1}v \equiv 0 \\ (u+1)^2 +v^2 \equiv 1 \end{cases} \quad (\textsf{mod } p^{r-v_{\mathbf{x}}}). \end{align*}$$
It follows that
$$\begin{align*}\begin{cases} u \equiv \tilde{x_1}^{-1}\tilde{x_2}v \\ 2\tilde{x_1}\tilde{x_2}v \equiv 0 \end{cases} \quad (\textsf{mod } p^{r-v_{\mathbf{x}}}).\end{align*}$$
Hence,
$(u,v)\equiv (0,0)\ \pmod {p^{r-v_{\mathbf {x}}}}$
, or equivalently,
$(a,b)\equiv (1,0)\ \pmod {p^{r-v_{\mathbf {x}}}}$
. By Lemma 5.1 again, the number of such
$(a,b)$
modulo
$p^r$
is exactly
$p^{v_{\mathbf {x}}}$
. So
$$\begin{align*}\big|\textsf{stab}_{\mathbf{x}}\big| = p^{v_{\mathbf{x}}}. \end{align*}$$
It then follows that
$|\textsf {orb}(\mathbf {x})| = |G_r|/|\textsf {stab}_{\mathbf {x}}| = p^{r-v_{\mathbf {x}}}(1-1/p)$
.
Moreover, for any
$\theta \in G_r$
, there is some
$\theta _0\in G_{r-v_{\mathbf {x}}}$
such that
$\theta \equiv \theta _0\,(\textsf {mod } p^{r-v_{\mathbf {x}}})$
. It can be verified that
$\theta \tilde {\mathbf {x}} \equiv \theta _0 \tilde {\mathbf {x}}\, (\textsf {mod } p^{r-v_{\mathbf {x}}})$
. So
$$\begin{align*}\textsf{orb}_r(\mathbf{x}) = \big\{\theta (p^{v_{\mathbf{x}}} \tilde{\mathbf{x}}):\, \theta \in G_r\big\} = \big\{p^{v_{\mathbf{x}}} \theta_0\tilde{\mathbf{x}}:\, \theta_0 \in G_{r-v_{\mathbf{x}}}\big\}. \end{align*}$$
Note that
$|G_{r-v_{\mathbf {x}}}|=p^{r-v_{\mathbf {x}}}(1-1/p)$
by Lemma 5.2, the elements on the right-hand side of the above formula give different members of the orbit. The proof is completed.
Lemma 5.3 Let
$p\equiv 1\, ({\sf mod }\ 4)$
. Suppose that
$ \mathbf {m} =p^{v_{\mathbf {m}}}\tilde {\mathbf {m}} \in (\mathbb {Z} /p^r \mathbb {Z})^2$
with
$\tilde {\mathbf {m}} \in \left ( (\mathbb {Z} /p^{r-v_{\mathbf {m}}} \mathbb {Z})^\ast \right )^2$
. Then for any
$\mathbf {z} \in (\mathbb {Z} /p^r\mathbb {Z})^2$
, we have
$$ \begin{align} \# \{ (\mathbf{x} ,\mathbf{y} )\in (\textsf{orb}_r (\mathbf{m}))^2 \colon \mathbf{x} -\mathbf{y} \equiv \mathbf{z} (\textsf{mod } p^r) \} \ll \begin{cases} p^{r-v_{\mathbf{m}}-1}, & \text{if } \Vert \tilde{\mathbf{m}}\Vert \not\equiv 0\, (\textsf{mod } p),\\ p^{r-v_{\mathbf{m}}}, & \text{if } \Vert \tilde{\mathbf{m}} \Vert \equiv 0 \, (\textsf{mod } p). \end{cases} \end{align} $$
Proof When
$\mathbf {m}=\mathbf {0}$
, the conclusion holds automatically. In the following, we assume that
$\mathbf {m}\neq \mathbf {0}$
. By Lemma 2.2, one has
$\textsf {orb}_r(\mathbf {m})=p^{v_{\mathbf {m}}}\textsf {orb}_{r-v_{\mathbf {m}}}(\tilde {m})$
. Let us write
$\mathbf {x}=p^{v_{\mathbf {m}}}\tilde {\mathbf {x}}$
and
$\mathbf {y}=p^{v_{\mathbf {m}}}\tilde {\mathbf {y}}$
, with
$\tilde {\mathbf {x}},\,\tilde {\mathbf {y}} \in \textsf {orb}_{r-v_{\mathbf {m}}}(\tilde {\mathbf {m}})$
. When
$v_{\mathbf {z}}<v_{\mathbf {m}}$
, the equation
$\mathbf {x}-\mathbf {y}\equiv \mathbf {z}\ \pmod {p^r}$
has no solution. When
$v_{\mathbf {z}}\geq v_{\mathbf {m}}$
, we denote
$\mathbf {z}=p^{v_{\mathbf {m}}}\tilde {\mathbf {z}}$
and obtain that
$\tilde {\mathbf {x}}-\tilde {\mathbf {y}}\equiv \tilde {\mathbf {z}}\ \pmod {p^{r-v_{\mathbf {m}}}}$
. Noting that
$v_{\tilde {\mathbf {m}}}=0$
, it is sufficient to show that
$$\begin{align*}\# \{ (\tilde{\mathbf{x}} ,\tilde{\mathbf{y}} )\in (\textsf{orb}_r (\tilde{\mathbf{m}}))^2 \colon \tilde{\mathbf{x}} -\tilde{\mathbf{y}} \equiv \tilde{\mathbf{z}} (\textsf{mod } p^{r-v_{\mathbf{m}}}) \} \ll \begin{cases} p^{(r-v_{\mathbf{m}})-v_{\tilde{\mathbf{m}}}-1}, & \text{if } \Vert \tilde{\mathbf{m}}\Vert \not\equiv 0\, (\textsf{mod } p),\\ p^{(r-v_{\mathbf{m}})-v_{\tilde{v_m}}}, & \text{if }\Vert \tilde{\mathbf{m}} \Vert \equiv 0 \, (\textsf{mod } p). \end{cases} \end{align*}$$
Therefore, we may assume at the beginning of the proof that
$v_{\mathbf {m}}=0$
.
Since
$\mathbf {y}$
is determined by
$\mathbf {x}$
modulo
$p^r$
, the cardinality on the left-hand-side of (6) does not exceed
$|\textsf {orb}_r (\tilde {\mathbf {m}})|$
, which is
$\sim p^r$
by Lemma 2.2. The second upper bound follows.
Next, we consider the case
$\|\mathbf {m}\|\not \equiv 0\ \pmod p$
. We have assumed that
$v_{\mathbf {m}}=0$
, so
$|\textsf {stab}_r(\mathbf {m})|=1$
by Lemma 2.2. For
$\mathbf {x} ,\mathbf {y} \in \textsf {orb}_r(\mathbf {m})$
, there exist unique
$a,b,a^{\prime },b^{\prime }\in \mathbb {Z} /p^r \mathbb {Z}$
such that
$a^2+b^2 \equiv a^{\prime 2} +b^{\prime 2} \equiv 1\, (\textsf {mod } p^r)$
and
$$\begin{align*}\begin{bmatrix} a & -b \\ b & a \end{bmatrix}\mathbf{m} \equiv \mathbf{x},\quad \begin{bmatrix} a^{\prime} & -b^{\prime}\\ b^{\prime} & a^{\prime} \end{bmatrix}\mathbf{m} \equiv \mathbf{y}, \quad \pmod {p^r}.\end{align*}$$
The set on the left-hand side of (6) involves a system of congruences
$$ \begin{align} \begin{cases} (a-a^{\prime})m_1 -(b-b^{\prime})m_2 \equiv z_1,\\ (b-b^{\prime})m_1 + (a-a^{\prime})m_2 \equiv z_2,\\ a^2 + b^2 \equiv 1,\\ a^{\prime 2} +b^{\prime 2} \equiv 1. \end{cases} \end{align} $$
We first consider (7) modulo p.
Since
$m_1^2+m_2^2\not \equiv 0\ \pmod p$
, either
$m_2$
or
$m_1 \not \equiv 0\ \pmod p$
. Without loss of generality, we assume that
$m_1 \not \equiv 0\, (\textsf {mod } p)$
, so one sees that
$$\begin{align*}\begin{cases} a-a^{\prime} \equiv m_1^{-1}(z_1 +(b-b^{\prime})m_2),\\ (b-b^{\prime})m_1 + m_1^{-1}(z_1+(b-b^{\prime})m_2)m_2 \equiv z_2, \end{cases} \quad (\textsf{mod } p), \end{align*}$$
or equivalently,
$$\begin{align*}\begin{cases} a-a^{\prime} \equiv m_1^{-1}(z_1 +(b-b^{\prime})m_2), \\ (b-b^{\prime})(m_1^2+m_2^2) \equiv z_2m_1-z_1m_2. \end{cases} \quad (\textsf{mod } p). \end{align*}$$
Now
$(a,b)$
is determined by
$(a^{\prime },b^{\prime })$
and system (7) has at most
$2$
solutions modulo p.
In the following, we will apply Hensel’s lemma. The Jacobi matrix of
$G(a,b,a^{\prime },b^{\prime }) \kern1pt{=} {((a\kern1.5pt{-}\kern1.5pt a^{\prime })m_1\kern1.5pt{-}\kern1.5pt (b-b^{\prime })m_2\kern1.5pt{-}\kern1.5pt z_1 ,(b\kern1.5pt{-}\kern1.5pt b^{\prime })m_1\kern1.5pt{+}\kern1.5pt(a\kern1.5pt{-}\kern1.5pt a^{\prime })m_2\kern1.5pt{-}\kern1.5pt z_2,a^2+b^2\kern1.5pt{-}\kern1.5pt 1,a^{\prime 2}+b^{\prime 2}\kern1.5pt{-}\kern1.5pt 1)}$
in
$(a,b,a^{\prime },b^{\prime })$
is given by
$$\begin{align*}\begin{bmatrix} m_1 & -m_2 & -m_1 & m_2\\ m_2 & m_1 & -m_2 & -m_1\\ 2a & 2b & 0 & 0\\ 0 & 0 & 2a^{\prime} & 2b^{\prime} \end{bmatrix}.\end{align*}$$
By elementary operations, we obtain
$$\begin{align*}\begin{bmatrix} 0 & 0 & -m_1 & m_2\\ 0 & 0 & -m_2 & -m_1\\ 2a & 2b & 0 & 0\\ 2a^{\prime} & 2b^{\prime} & 2a^{\prime} & 2b^{\prime} \end{bmatrix}. \end{align*}$$
In view of the fact that
$m_1^2+m_2^2 \not \equiv 0 \, (\textsf {mod } p)$
, the rank of the above Jacobi matrix, modulo p, is at least
$3$
. By Lemma 5.1, the number of solutions of
$(a,b,a^{\prime },b^{\prime })$
to (7) modulo
$p^r$
is at most
$p^{r-1}$
. The first upper bound then follows.
Next, we complete the proof of Theorem 2.4.
Proof of Theorem 2.4
We have
$$ \begin{align*} \sum_{\mathbf{m}\in (\mathbb{Z} / p^r{\mathbb Z})^2}|(fd\sigma_{V_{\mathbf{m}}})^\vee(\mathbf{m})|^4&=\sum_{\mathbf{m}}\left\vert \frac{1}{|V_{\mathbf{m}}|}\sum_{\mathbf{x}\in V_{\mathbf{m}}}\chi_r(\mathbf{m}\cdot \mathbf{x})f(\mathbf{x})\right\vert^4\\ &=\frac{p^{2r}}{|V_{\mathbf{m}}|^4}\sum_{\xi,\xi^{\prime},\eta,\eta^{\prime}\in V_{\mathbf{m}}\colon \xi-\eta=\xi^{\prime}-\eta^{\prime}}f(\xi)f(\xi^{\prime})\overline{f(\eta)f(\eta^{\prime}).} \end{align*} $$
Moreover,
$$\begin{align*}\sum_{\xi,\xi^{\prime},\eta,\eta^{\prime}\in V\colon \xi-\eta=\xi^{\prime}-\eta^{\prime}}f(\xi)f(\xi^{\prime})\overline{f(\eta)f(\eta^{\prime})} = \sum_{\zeta} \left|\sum_{\xi-\eta=\zeta}f(\xi)\overline{f(\eta)}V_{\mathbf{m}}(\xi)V_{\mathbf{m}}(\eta) \right|^2. \end{align*}$$
For
$\zeta \equiv \mathbf {0}\,(\textsf {mod } p^r)$
, we have
$$\begin{align*}\sum_{\zeta\equiv \mathbf{0}\,(\textsf{mod } p^r)} \left| \sum_{\xi-\eta=\zeta}f(\xi)\overline{f(\eta)}V_{\mathbf{m}}(\xi)V_{\mathbf{m}}(\eta)\right|^2\ll \left(\sum_{\xi\in V_{\mathbf{m}}}|f(\xi)|^2\right)^2.\end{align*}$$
For
$\zeta \not \equiv \mathbf {0}\,(\textsf {mod } p^r)$
, the Cauchy-Schwarz inequality implies
$$ \begin{align*} \phantom{-}& \sum_{\zeta\not\equiv \mathbf{0}\,(\textsf{mod } p^r)} \left| \sum_{\xi-\eta=\zeta}f(\xi)\overline{f(\eta)}V_{\mathbf{m}}(\xi)V_{\mathbf{m}}(\eta)\right|^2 \\ &\leq \sum_{\zeta\not\equiv \mathbf{0}\,(\textsf{mod } p^r)} \left( \sum_{\xi-\eta=\zeta}V_{\mathbf{m}}(\xi)V_{\mathbf{m}}(\eta) \right) \sum_{\xi-\eta=\zeta}|f(\xi)|^2|f(\eta)|^2V_{\mathbf{m}}(\xi)V_{\mathbf{m}}(\eta).\\ \end{align*} $$
Now, to bound the sum
$ \sum _{\xi -\eta =\zeta }V_{\mathbf {m}}(\xi )V_{\mathbf {m}}(\eta ) $
, we use Lemma 5.3 and get
$$ \begin{align*} & \sum_{\zeta\not\equiv \mathbf{0}\,(\textsf{mod } p^r)} \left| \sum_{\xi-\eta=\zeta}f(\xi)\overline{f(\eta)}V_{\mathbf{m}}(\xi)V_{\mathbf{m}}(\eta)\right|^2 \\ \ll & \begin{cases} p^{r-v_{\mathbf{m}} -1} \left( \sum_{\xi \in V_{\mathbf{m}}} |f(\xi)|^2 \right)^2 & \text{if } \left\Vert \tilde{\mathbf{m}} \right\Vert \not\equiv 0\quad \pmod p , \\ p^{r-v_{\mathbf{m}} } \left( \sum_{\xi \in V_{\mathbf{m}}} |f(\xi)|^2 \right)^2 & \text{if }\left\Vert \tilde{\mathbf{m}} \right\Vert \equiv 0\quad \pmod p. \end{cases} \end{align*} $$
So, one has
$$ \begin{align*} \left( \sum_{x\in (\mathbb{Z} /p^r\mathbb{Z})^2} \vert (fd\sigma_{V_{\mathbf{m}}})^{\vee}(x) \vert^4\right)^{1/2} \ll \frac{p^{r}}{\left\vert V_{\mathbf{m}}\right\vert^2 } \begin{cases} p^{\frac{r-v_{\mathbf{m}} -1}{2}} \sum_{\xi \in V_{\mathbf{m}} } |f(\xi)|^2 & \text{if }\left\Vert \tilde{\mathbf{m}} \right\Vert \not\equiv 0\quad \pmod p , \\ p^{\frac{r-v_{\mathbf{m}}}{2} } \sum_{\xi \in V_{\mathbf{m}}} |f(\xi)|^2 & \text{if }\left\Vert \tilde{\mathbf{m}} \right\Vert \equiv 0\quad \pmod p. \end{cases} \end{align*} $$
Next, Lemma 2.2 gives that
$|V_{\mathbf {m}} | \sim p^{r-v_{\mathbf {m}}}$
. Hence, the theorem follows.
Acknowledgements
Thang Pham would like to thank the Vietnam Institute for Advanced Study in Mathematics (VIASM) for the hospitality and for the excellent working condition. Thang Pham, Nguyen Duy Phuong, and Le Quang Ham were supported by Vietnam National Foundation for Science and Technology Development (NAFOSTED) under grant number 101.99–2021.09.
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