# Spherical configurations over finite fields

Neil Lyall, Ákos Magyar, Hans Parshall  Department of Mathematics, The
University of Georgia, Athens, GA 30602, USA<EMAIL_ADDRESS>Department of
Mathematics, The University of Georgia, Athens, GA 30602, USA
<EMAIL_ADDRESS>Department of Mathematics, The Ohio State University,
Columbus, OH 43210, USA<EMAIL_ADDRESS>
###### Abstract.

We establish that if $d\geq 2k+6$ and $q$ is odd and sufficiently large with
respect to $\alpha\in(0,1)$, then every set $A\subseteq\mathbf{F}_{q}^{d}$ of
size $|A|\geq\alpha q^{d}$ will contain an isometric copy of every spherical
$(k+2)$-point configuration that spans $k$ dimensions.

###### 2010 Mathematics Subject Classification:

11B30

The first author was partially supported by NSF-DMS grant 1702411 and Simons
Foundation Collaboration Grant for Mathematicians 245792. The second author
was partially supported by Grants NSF-DMS 1600840 and ERC-AdG 321104.

##  1\. Introduction

Geometric Ramsey theory has its origins in series of papers by Erdős et al.
[6, 7, 8], where they studied geometric configurations which cannot be
destroyed by partitioning Euclidean space into finitely many classes. The
fundamental problem is to classify those finite sets $X$ which are _Ramsey_ ,
in the sense that for every number of colors $r\in\mathbf{N}$ there is a
dimension $d=d(r,X)$ for which every $r$-coloring of $\mathbf{R}^{d}$ contains
a monochromatic, congruent copy of $X$.

The simplest example of a Ramsey set is a regular $k$-simplex; that is, $k+1$
equidistant points. Indeed, for any dimension $d\geq kr$, any $r$-coloring of
a regular $d$-simplex contains $k+1$ points of the same color, forming a
monochromatic regular $k$-simplex. On the other hand, a simple construction
using the geometry of the Euclidean metric shows that any set of three
collinear points is not Ramsey. In fact, Erdős et al. [6] showed that every
Ramsey set must be _spherical_ ; that is, contained in some sphere. This has
led to the conjecture by Graham [13] that a finite set $X$ is Ramsey if and
only if it is spherical.

This conjecture is far from settled. Examples of sets known to be Ramsey
include vertices of “bricks” ($k$-dimensional rectangles) [6], non-degenerate
simplices [9], trapezoids [18], regular polygons and regular polyhedra [17].
Common to many of these results was the exploitation of additional symmetries
of the configuration. It was observed by Leader, Russell and Walters [20] that
all known examples of Ramsey sets are _subtransitive_ in the sense that they
can be embedded in a higher dimensional set on which the rotation group acts
transitively. They introduced a rival conjecture that a finite set $X$ is
Ramsey if and only if it is subtransitive, and further showed [19] that almost
all 4-point subsets of a circle are not subtransitive. This was later extended
by Eberhard [5] to show that almost all $(k+2)$-point sets on the
$(k-1)$-sphere are not subtransitive. It remains an open question whether or
not such configurations are Ramsey.

The aim of this article is to show that an analogue of Graham’s conjecture
holds in finite field geometries for 4-point spherical configurations spanning
two dimensions, and more generally for spherical $(k+2)$-point configurations
spanning $k$ dimensions. We in fact prove a stronger _density_ version; namely
that if $d\geq 2k+6$ and $q$ is taken to be odd and sufficiently large with
respect to $\alpha\in(0,1)$, then every set $A\subseteq\mathbf{F}_{q}^{d}$ of
size $|A|\geq\alpha q^{d}$ contains an isometric copy of every such
configuration $X$. To be clear, here we say that two sets $X$ and $X^{\prime}$
are _isometric_ if there is a bijection $\phi:X\to X^{\prime}$ such that
$|\phi(x)-\phi(x^{\prime})|^{2}=|x-x^{\prime}|^{2}$ for all $x\in X$ and
$x^{\prime}\in X^{\prime}$, where $|x|^{2}=x\cdot x$ is the usual dot product
of the vector $x\in\mathbf{F}_{q}^{d}$ with itself.

Our approach takes the point of view modern arithmetic combinatorics which has
been very successful in the study of linear patterns in subsets of
$\mathbf{Z}$ of positive density [10, 11]. In fact, one of the main purpose of
this article is to extend these techniques to the setting of geometric Ramsey
theory, where one counts configurations determined by both linear as well as
certain non-linear relations, i.e. isometries.

The setting of vector spaces over finite fields provides a useful model to
study many problems in arithmetic combinatorics; see especially the surveys
[14, 25, 28]. In the context of geometric Ramsey theory over finite fields,
notable results have been obtained by a number of authors [3, 15, 16, 24].
However, those results concern patterns consisting of points in general
position, with no linear relations between them, and hence are fundamentally
different.

Analogous results for simplices in Euclidean spaces and the integer lattice
have been given in [1, 22], and it reasonable to expect that our approach here
may be successfully adapted to these settings. In the context of geometric
(density) Ramsey theory in $\mathbf{R}^{d}$, some results using this approach
were recently obtained in [4, 21]. We hope to address further adaptation in
the near future.

###  1.1. Outline of paper

The main results of the paper are stated is Section 2 below, and some
preliminaries and reductions are presented in Sections 3 and 4.

A key observation of the paper, see 6 in Section 5.1, is that the count of
isometric copies of a fixed configuration $X$ along bounded functions is
controlled by a certain uniformity norm. This norm, which we introduce in
Section 5.1, measures the uniformity or randomness of a function along
geometric rectangles, and it should be compared with the so-called
$U^{2}$-uniformity norm of Gowers [10] which measures uniformity along
combinatorial rectangles. If a set $A$ is sufficiently uniform with respect to
this norm, then it quickly follows from 6 that $A$ contains many, in fact the
statistically correct number of, isometric copies of $X$. The proof of 6 is
presented in Section 6.

In order to handle arbitrary sets $A$ we prove an _inverse theorem_ , see 10
in Section 7.1, to establish that functions with large uniformity norm
correlate with structured sets. Given such an inverse theorem there are then
various standard iterative procedures that one may hope to adapt to this
setting to complete the argument. We follow an _energy increment_ route which
to leads a so-called _arithmetic regularity lemma_ , namely 7, that allows us
to decompose the indicator function of $A$ as $1_{A}=f_{1}+f_{2}+f_{3}$, where
$f_{1}$ is highly structured, $f_{2}$ has small $L^{2}$ norm and $f_{3}$ has
small uniformity norm. The proof of 7 is presented in Section 7.

In Sections 5.2 and 5.3 we demonstrate how 7 leads to a proof of our main
results. This consists of counting the isometric copies of $X$ along the main
term $f_{1}$ and showing the contribution of the functions $f_{2}$ and $f_{3}$
are negligible. The setback of this approach is that it leads to very weak
bounds, in fact the dependence of $q$ on $\alpha$ is tower-exponential. It
seems quite possible that one could instead proceed via a density increment
argument and obtain better, exponential type bounds, but we do not pursue this
here.

We conclude the paper with an appendix in which we discuss the necessity of
the spherical condition in the statement of our main result.

##  2\. Main results

We will always work with a finite field $\mathbf{F}_{q}$ of odd
characteristic. For vectors $v,w\in\mathbf{F}_{q}^{d}$, we define their dot
product $v\cdot w:=\sum_{j=1}^{d}v_{j}w_{j}$ as usual and we will work with
the isotropic measurements of _length_ $|v|^{2}:=v\cdot v$ and _distance_
$|v-w|^{2}$. For any $u\in\mathbf{F}_{q}^{d}$ and $\lambda\in\mathbf{F}_{q}$,
we define the _sphere_
$S_{\lambda}(u)=\\{x\in\mathbf{F}_{q}^{d}:|x-u|^{2}=\lambda\\}$, and we will
simply write $S_{\lambda}$ when $u=0$. For $k\in\mathbf{N}$, we will say that
$X\subseteq\mathbf{F}_{q}^{d}$ _spans $k$ dimensions_ when
$\dim(\operatorname{Span}(X-X))=k$, and we will call $X$ _spherical_ provided
exists a sphere $S_{\lambda}(u)\subseteq\mathbf{F}_{q}^{d}$ with $X\subseteq
S_{\lambda}(u)$. Note the $(k+1)$-point spherical configurations spanning $k$
dimensions are exactly the $k$-simplices, which were shown in [24] to appear
as isometric copies in sufficiently dense subsets of $\mathbf{F}_{q}^{d}$
provided $d>k$.

In Euclidean spaces it is easy to see that if the finite sets $X$ and
$X^{\prime}$ are isometric, then $X^{\prime}=z+U(X)$ for some vector $z$ and
orthogonal transformation $U\subseteq O(d)$, i.e. $X^{\prime}$ can be obtained
from $X$ by a rigid motion. The same may not hold in finite field geometries
due to the presence of self-orthogonal vectors $x$ for which $|x|^{2}=0$.
However, it follows from Witt’s extension theorem [27] that if the subspace
$V:=\operatorname{Span}(X-X)$ is non-degenerate in the sense that $V\cap
V^{\perp}=\\{0\\}$, then isometric copies of $X$ are indeed obtained by rigid
motions. We use this fact in the appendix to construct dense subsets avoiding
isometric copies of non-spherical sets, establishing the necessity of
restricting our attention to spherical configurations $X$.

The main result of this paper is the following.

######  Theorem 1.

Let $d,k\in\mathbf{N}$ with $d\geq 2k+6$, $\alpha\in(0,1)$, and $q\geq
q(\alpha,k)$. If $A\subseteq\mathbf{F}_{q}^{d}$ with $|A|\geq\alpha q^{d}$,
then $A$ contains at least $c(\alpha,k)q^{(k+1)d-k(k+1)/2}$ isometric copies
of every $(k+2)$-point spherical configuration spanning $k$ dimensions.

Here we write $c(\alpha,k)$ to stand in for some positive constant depending
only on $\alpha$ and $k$, and $q\geq q(\alpha,k)$ to indicate $q$ is taken
sufficiently large with respect to $\alpha$ and $k$. We will use similar
notation to indicate the dependency of constants $c,C>0$ that may change
between occurrences. It is helpful to think of $\alpha$ and $k$ as fixed with
$q$ allowed to tend toward infinity, and implicit constants in our big-O
notation may depend on $k$.

Note that if $A$ would be a random subset of $\mathbf{F}_{q}^{d}$ of density
$\alpha$ then it would contain $\alpha^{k+2}q^{(k+1)d-k(k+1)/2}$ isometric
copies of a $(k+2)$-point spherical configuration $X$ up to an error. This is
because there are $k(k+1)/2$ quadratic relations, given by the length of the
edges, between the points of the configuration $X$ and each vertex is
contained in $A$ with probability $\alpha$.

It immediately follows from 1 that for any fixed number of colors $r$, if we
take $q$ sufficiently large with respect to $r$, then any $r$-coloring of
$\mathbf{F}_{q}^{10}$ contains monochromatic, isometric copies of all 4-point
spherical sets spanning 2 dimensions; this establishes a finite field version
of Graham’s conjecture that all cyclic quadrilaterals are Ramsey. In fact, we
prove a stronger statement relative to spheres; see [12, 23] for some so-
called “sphere Ramsey” results in the Euclidean setting.

######  Theorem 2.

Let $d,k\in\mathbf{N}$ with $d\geq 2k+6$, $\alpha\in(0,1)$, and $q\geq
q(\alpha,k)$. If $\lambda\in\mathbf{F}_{q}^{*}$ and $A\subseteq S_{\lambda}$
with $|A|\geq\alpha q^{d-1}$, then $A$ contains at least
$c(\alpha,k)q^{(k+1)d-(k+1)(k+2)/2}$ isometric copies of every $(k+2)$-point
spherical configuration spanning $k$ dimensions.

A straightforward counting argument reveals that 2 quickly implies 1, this
argument is presented in Section 4 below.

We remark here that the relationship between $d$ and $k$ in both theorems
could be improved if one were only interested in “high rank” configurations;
see the comments following 3. Our methods are further able to prove a version
of 2 when $\lambda=0$ provided $d>2k+6$, but we do not pursue this since it
does not impact our proof of 1.

##  3\. Preliminaries

Here we record notation and ingredients that we will require for the proof of
2. Given any function $f:\mathbf{F}_{q}^{d}\rightarrow\mathbf{C}$ and
$B\subseteq\mathbf{F}_{q}^{d}$, we write

$\mathbf{E}_{x\in B}f(x):=\frac{1}{|B|}\sum_{x\in B}f(x)$

for the average of $f$ over $B$, and we will understand the average
$\mathbf{E}_{x}f(x)$ is taken over $\mathbf{F}_{q}^{d}$. We will condense
multiple averages
$\mathbf{E}_{y_{1}}\mathbf{E}_{y_{2}}\cdots\mathbf{E}_{y_{k}}$ as
$\mathbf{E}_{y_{1},\ldots,y_{k}}$, and to indicate linear independence we will
use the notation

$\mathbf{E}_{y_{1},\ldots,y_{k}}^{*}:=\frac{1}{q^{kd}}\sum_{\begin{subarray}{c}y_{1},\ldots,y_{k}\in\mathbf{F}_{q}^{d}\\\
\text{linearly independent}\end{subarray}}$

Letting $\chi$ denote the canonical additive character of $\mathbf{F}_{q}$, we
define the Fourier transform
$\widehat{f}:\mathbf{F}_{q}^{d}\rightarrow\mathbf{C}$ by

$\widehat{f}(\xi):=\mathbf{E}_{x}f(x)\chi(-\xi\cdot x)$

and we recall the Fourier inversion formula
$f(x)=\sum_{\xi\in\mathbf{F}_{q}^{d}}\widehat{f}(\xi)\chi(\xi\cdot x).$ Given
two functions $f,g:\mathbf{F}_{q}^{d}\rightarrow\mathbf{C}$, we recall
Plancherel’s identity

$\mathbf{E}_{x}f(x)\overline{g(x)}=\sum_{\xi\in\mathbf{F}_{q}^{d}}\widehat{f}(\xi)\overline{\widehat{g}(\xi)}$

and, defining the convolution $f*g(x)=\mathbf{E}_{y}f(y)g(x-y)$, we also
recall $\widehat{f*g}(\xi)=\widehat{f}(\xi)\widehat{g}(\xi)$.

We will write $\sigma_{\lambda}=q1_{S_{\lambda}}$ for a normalized indicator
function of $S_{\lambda}$, where we have the asymptotic

(1) $\widehat{\sigma}_{\lambda}(\xi)=\begin{cases}1+O(q^{-1/2})&\text{if
}\xi=0\\\ O(q^{-1/2})&\text{otherwise}\end{cases}$

valid for $d\geq 2$ and $\lambda\in\mathbf{F}_{q}^{*}$; see, for instance,
[16, Lemma 2.2] exploiting Weil’s bounds on Kloosterman sums.

To vectors $y_{1},\ldots,y_{j-1}\in\mathbf{F}_{q}^{d}$ and constants
$c_{1},\ldots,c_{j}\in\mathbf{F}_{q}$, we will associate the _spherical
measure_

(2)
$\sigma_{y_{1},\ldots,y_{j-1}}^{c_{1},\ldots,c_{j}}(y_{j})=\begin{cases}q^{j}&\text{if
}y_{i}\cdot y_{j}=c_{i}\text{ for all }1\leq i\leq j\\\
0&\text{otherwise}\end{cases}.$

This is essentially an $L^{1}$-normalized indicator function for the
intersection of the sphere $S_{c_{j}}$ with $j-1$ hyperplanes, so one should
expect Fourier decay in appropriate directions. To import the corresponding
Fourier asymptotics, we set

$\delta_{y_{1},\ldots,y_{j-1}}(\xi):=\begin{cases}1&\text{if
}\xi\in\operatorname{Span}(y_{1},\ldots,y_{j-1})\\\
0&\text{otherwise,}\end{cases}$

and record the simplest case of [24, Lemma 7].

######  Lemma 3.

Let $c_{1},\ldots,c_{j}\in\mathbf{F}_{q}$ with $c_{j}\neq 0$ and
$y_{1},\ldots,y_{j-1}\in\mathbf{F}_{q}^{d}$ linearly independent. If $d\geq
2j$, then

$|\widehat{\sigma}_{y_{1},\ldots,y_{j-1}}^{c_{1},\ldots,c_{j}}(\xi)|=\delta_{y_{1},\ldots,y_{j-1}}(\xi)+O(q^{-1/2}).$

It is worth mentioning if one is willing to impose technical conditions on the
defining vectors and scalars, then results within [24] include stronger
asymptotics valid in the range $d>j$. These would allow us to improve the
required relationship of $d\geq 2k+6$ in 2 for configurations avoiding self-
orthogonal subspaces, but we opt instead for a uniform result valid for all
$(k+2)$-point spherical configurations spanning $k$ dimensions. For clarity of
presentation, we will often suppress the scalars $c_{1},\ldots,c_{j}$ in the
notation (2) since we will always restrict ourselves to $c_{j}\neq 0$ and 3
does not depend on the other scalars.

We will also require the notion of Bohr sets, which provide a substitute for
fine subgroup structure. We define the Bohr set of spectrum
$\Gamma\subseteq\mathbf{F}_{q}^{d}$ and radius $\rho\in(0,1]$ by

(3) $B(\Gamma,\rho):=\\{x\in\mathbf{F}_{q}^{d}:|\chi(\xi\cdot
x)-1|\leq\rho\text{ for all }\xi\in\Gamma\\}.$

Setting $\beta=|B(\Gamma,\rho)|/q^{d}$, we record the standard bound

(4) $\beta\geq\Big{(}\frac{\rho}{2\pi}\Big{)}^{|\Gamma|}$

which can be found in [26]. Setting $B=B(\Gamma,\rho)$, we will use the
$L^{1}$ normalized indicator function $\mu_{B}=\beta^{-1}1_{B}$ and its
repeated convolution $\nu_{B}=\mu_{B}*\mu_{B}*\mu_{B}*\mu_{B}$. The structure
provided by repeated convolution will frequently be useful. For example,
provided $d\geq 2$, we can apply Fourier inversion, (1) and Plancherel to see

(5)
$\mu_{B}*\mu_{B}*\sigma(x)=\sum_{\xi\in\mathbf{F}_{q}^{d}}\widehat{\mu}_{B}(\xi)^{2}\widehat{\sigma}(\xi)\chi(\xi\cdot
x)=1+O\Big{(}q^{-1/2}\sum_{\xi\in\mathbf{F}_{q}^{d}}|\widehat{\mu}_{B}(\xi)|^{2}\Big{)}=1+O(\beta^{-1}q^{-1/2}),$

from which it immediately follows that

(6) $\nu_{B}*\sigma(x)=1+O(\beta^{-1}q^{-1/2}).$

In other words, provided $q$ sufficiently large with respect to $\beta$,
repeated convolutions of $\mu_{B}$ with $\sigma$ can be considered essentially
constant.

We call the Bohr set $B(\Gamma,\rho)$ regular if for every $\epsilon>0$, we
have both

$\displaystyle|B(\Gamma,(1+\epsilon)\rho)|$
$\displaystyle\leq(1+100\epsilon|\Gamma|)|B(\Gamma,\rho)|$
$\displaystyle|B(\Gamma,(1-\epsilon)\rho)|$
$\displaystyle\geq(1-100\epsilon|\Gamma|)|B(\Gamma,\rho)|;$

This definition, due to Bourgain [2], ensures that all Bohr sets are only a
small dilation away from being regular (see [26, Lemma 4.25]), and regular
Bohr sets are essentially closed under addition by elements of their interior.
Given a regular Bohr set $B=B(\Gamma,\rho)$, we will write
$B^{\prime}\prec_{\epsilon}B$ in the case that
$B^{\prime}=B(\Gamma^{\prime},\rho^{\prime})$ is another regular Bohr set with
$\Gamma^{\prime}\supseteq\Gamma$ and
$\rho^{\prime}\leq\epsilon\rho/(200\lvert\Gamma\rvert)$. The following
standard lemma provides the main consequence of regularity for our purposes;
we include the proof for completeness.

######  Lemma 4.

Let $\epsilon\in(0,1)$, $B=B(\Gamma,\rho)$ regular with
$B^{\prime}\prec_{\epsilon}B$, and $f:\mathbf{F}_{q}^{d}\rightarrow\mathbf{C}$
with $|f|\leq 1$. For any $y\in B^{\prime}$,

$\lvert\mathbf{E}_{x\in B}f(x)-\mathbf{E}_{x\in B}f(x+y)\rvert\leq\epsilon.$

###### Proof.

We have

$\lvert\mathbf{E}_{x\in B}f(x)-\mathbf{E}_{x\in
B}f(x+y)\rvert=\frac{1}{|B|}\Big{|}\sum_{x\in\mathbf{F}_{q}^{d}}f(x)(1_{B}(x)-1_{B}(x-y))\Big{|}\leq\frac{|B\triangle(y+B)|}{|B|}$

Since $B^{\prime}\prec_{\epsilon}B$ and $y\in B^{\prime}$ we have the
relationship

$B\triangle(y+B)\subseteq B(\Gamma,\rho+\rho^{\prime})\setminus
B(\Gamma,\rho-\rho^{\prime})$

and our claim follows from the bound $|B(\Gamma,\rho+\rho^{\prime})\setminus
B(\Gamma,\rho-\rho^{\prime})|\leq\epsilon|B|$ resulting from regularity. ∎

##  4\. Reduction to Dense Spherical Sets

In this section we present the straightforward counting argument that
estblishes 1 as a consequence of 2. The main observation is that the
collection of spheres of a fixed radius $\lambda\in\mathbf{F}_{q}^{*}$
provides a uniform cover of $\mathbf{F}_{q}^{d}$. Hence, any subset
$A\subseteq\mathbf{F}_{q}^{d}$ with density $\alpha$ has density nearly
$\alpha$ on a large number, in fact a positive proportion, of these spheres. 2
then implies that within each of these spheres, $A$ contains many of the
sought after configurations. By counting the contribution of each of these
spheres, it is easy to see that $A$ contains a positive proportion of the
count of all such configurations within $\mathbf{F}_{q}^{d}$, as each fixed
configuration is contained in approximately the same number of spheres.

We first record the characterization of spherical configurations that will be
most useful as we proceed. This follows in a straightforward way from 16 in
the Appendix.

######  Lemma 5.

Let $X\subseteq\mathbf{F}_{q}^{d}$ be any $(k+2)$-point spherical
configuration spanning $k$ dimensions. If any $(k+1)$-point subset of $X$ that
spans $k$ dimensions is contained in a sphere $S_{\lambda}(u)$, then
$X\subseteq S_{\lambda}(u)$ as well.

Let us now see how 2 implies 1.

###### Proof of 1.

Fixing $\lambda\in\mathbf{F}_{q}^{*}$, we first establish that for many
centers $u\in\mathbf{F}_{q}^{d}$, $|A\cap S_{\lambda}(u)|$ is large. For
$d\geq 2$ and $x\in\mathbf{F}_{q}^{d}$, (1) implies
$\mathbf{E}_{u}\sigma_{\lambda}(u-x)=1+O(q^{-1/2}),$ in which case we can
ensure $q$ is sufficiently large for

$\displaystyle(\alpha/2)q^{2d-1}$
$\displaystyle\leq\sum_{u\in\mathbf{F}_{q}^{d}}|A\cap S_{\lambda}(u)|$
$\displaystyle=\sum_{\begin{subarray}{c}u\in\mathbf{F}_{q}^{d}\\\ |A\cap
S_{\lambda}(u)|\geq(\alpha/4)q^{d-1}\end{subarray}}|A\cap
S_{\lambda}(u)|+\sum_{\begin{subarray}{c}u\in\mathbf{F}_{q}^{d}\\\ |A\cap
S_{\lambda}(u)|<(\alpha/4)q^{d-1}\end{subarray}}|A\cap S_{\lambda}(u)|$
$\displaystyle\leq|S_{\lambda}||\\{u\in\mathbf{F}_{q}^{d}:|A\cap
S_{\lambda}(u)|\geq(\alpha/4)q^{d-1}\\}|+(\alpha/4)q^{2d}.$

Again using (1) and ensuring $q$ sufficiently large, we can conclude

$|\\{u\in\mathbf{F}_{q}^{d}:|(A-u)\cap
S_{\lambda}|\geq(\alpha/4)q^{d-1}\\}|\geq(\alpha/8)q^{d}.$

Fixing now some $(k+2)$-point spherical configuration $X$ spanning $k$
dimensions, we see that for each of these good centers $u$, 2 guarantees that
$A\cap S_{\lambda}(u)$ contains at least
$c(\alpha,k)q^{(k+1)d-\binom{k+2}{2}}$ isometric copies of $X$. We need to
account for how many spheres we count each isometric copy within. By
translating $X$ if necessary, we can parametrize it as

$X=\\{0,v_{1},\ldots,v_{k},a_{1}v_{1}+\cdots+a_{k}v_{k}\\}$

for linearly independent $v_{1},\ldots,v_{k}\in\mathbf{F}_{q}^{d}$ and
$a_{1},\ldots,a_{k}\in\mathbf{F}_{q}$. Then for a translation
$x\in\mathbf{F}_{q}^{d}$ and linearly independent vectors
$y_{1},\ldots,y_{k}\in\mathbf{F}_{q}^{d}$ with $y_{i}\cdot y_{j}=v_{i}\cdot
v_{j}$ for all $1\leq i\leq j\leq k$, we wish to count how many spheres
$S_{\lambda}(u)$ contain the configuration

$X^{\prime}=\\{x,x+y_{1},\ldots,x+y_{k},x+a_{1}y_{1}+\cdots+a_{k}y_{k}\\}.$

Since $y_{1},\ldots,y_{k}$ are linearly independent, 3 applies to the
spherical measure

$\sigma_{y_{1},\ldots,y_{k}}(u):=\sigma_{\lambda}(u)\prod_{j=1}^{k}\sigma_{\lambda}(y_{j}-u)$

so that, for $d\geq 2k+2$,

$\mathbf{E}_{u\in\mathbf{F}_{q}^{d}}\sigma_{y_{1},\ldots,y_{k}}(x-u)=1+O(q^{-1/2}).$

5 ensures that $\\{x,x+y_{1},\ldots,x+y_{k}\\}\subseteq S_{\lambda}(u)$
implies $X^{\prime}\subseteq S_{\lambda}(u)$ as well, so we have shown

$|\\{u\in\mathbf{F}_{q}^{d}:X^{\prime}\subseteq
S_{\lambda}(u)\\}|=(1+O(q^{-1/2}))q^{d-k-1}.$

Hence, within each of the $(\alpha/8)q^{d}$ good spheres $S_{\lambda}(u)$,
$A\cap S_{\lambda}(u)$ contains $c(\alpha,k)q^{(k+1)d-\binom{k+2}{2}}$
isometric copies of $X$, and each of these copies is contained in roughly
$q^{d-k-1}$ spheres. In total this yields that $A$ contains
$c(\alpha,k)q^{(k+1)d-\binom{k+1}{2}}$ isometric copies of $X$ as claimed. ∎

##  5\. Proof of 2

In this section we reduce the task of proving 2 to that of establishing
Propositions 6 and 7 below.

###  5.1. Counting Configurations, a Uniformity Norm, and two key Propositions

For the remainder, we fix $\lambda\in\mathbf{F}_{q}^{*}$ and aim to establish
2; there is no harm in assuming $\lambda=1$. For brevity, we set
$S=S_{\lambda}$ and $\sigma=\sigma_{\lambda}$. We will be considering
$(k+2)$-point spherical configurations spanning $k$ dimensions typically
parameterized as

$X:=\\{0,v_{1},\ldots,v_{k},a_{1}v_{1}+\cdots+a_{k}v_{k}\\}\subseteq\mathbf{F}_{q}^{d},$

with $v_{1},\ldots,v_{k}\in\mathbf{F}_{q}^{d}$ linearly independent and
coefficients $a_{1},\ldots,a_{k}\in\mathbf{F}_{q}$. We will consider other
sets of $k+2$ points of the form

$X^{\prime}=\\{x_{0},x_{0}+x_{1},\ldots,x_{0}+x_{k},x_{0}+a_{1}x_{1}+\cdots+a_{k}x_{k}\\}\subseteq\mathbf{F}_{q}^{d}$

and check whether these are indeed isometric copies of $X$ contained within
$S$ by checking whether $x_{1},\ldots,x_{k}$ are linearly independent and
further satisfy the conditions

$|x_{0}|^{2}=|x_{0}+x_{1}|^{2}=\cdots=|x_{0}+x_{k}|^{2}=\lambda\text{ and
}x_{i}\cdot x_{j}=v_{i}\cdot v_{j}\text{ for each }1\leq i\leq j\leq k;$

when all of this is true, we write $X^{\prime}\simeq X$. We remark that this
notation only explicitly insists that $k+1$ points of $X^{\prime}$ lie on $S$.
However, since we are working with the same coefficients $a_{1},\ldots,a_{k}$,
16 ensures $X^{\prime}$ is spherical and 5 ensures $X^{\prime}\subset S$ as
well. That is, if $X^{\prime}\simeq X$, then it is also the case that
$X^{\prime}\subseteq S$.

To count copies of $X$ parameterized as $X^{\prime}$ above, we define the
weight

$\mathcal{S}_{X}(x_{0},\ldots,x_{k}):=\begin{cases}q^{(k+1)(k+2)/2}&\text{if
}X^{\prime}\simeq X\\\ 0&\text{otherwise}\end{cases}$

and a normalized counting operator on functions
$f_{0},f_{1},\ldots,f_{k+1}:\mathbf{F}_{q}^{d}\rightarrow\mathbf{C}$ by

(7)
$\mathcal{N}_{X}(f_{0},\ldots,f_{k+1}):=\mathbf{E}_{x_{0}}f_{0}(x_{0})\mathbf{E}_{x_{1},\ldots,x_{k}}^{*}\Big{(}\prod_{j=1}^{k}f_{j}(x_{0}+x_{j})\Big{)}f_{k+1}(x_{0}+a_{1}x_{1}+\cdots+a_{k}x_{k})\mathcal{S}_{X}(x_{0},\ldots,x_{k}).$

Note that as long as we restrict our attention to linearly independent
$x_{1},\ldots,x_{k}$, we can write out

$\mathcal{S}_{X}(x_{0},\ldots,x_{k})=\sigma(x_{0})\prod_{j=1}^{k}\sigma_{x_{0},\ldots,x_{j-1}}(x_{j})$

for appropriate spherical measures with implicit scalars determined by our
sphere $S$ and the dot products between the defining vectors
$v_{1},\ldots,v_{k}$ of $X$. Moreover, the contribution of linearly dependent
$x_{1},\ldots,x_{k}$ is negligible, in the sense that we trivially have

$\frac{1}{q^{d}}\sum_{x_{j}\in\operatorname{Span}(x_{1},\ldots,x_{j-1})}\sigma_{x_{1},\ldots,x_{j-1}}(x_{j})\leq
q^{-1}$

whenever $d\geq 2j$. Together with 3, this allows us to freely add in linearly
dependent $x_{1},\ldots,x_{k}$ to our count (7) at the cost of an acceptable
$O(q^{-1})$ error, provided each $f_{j}$ is bounded as will typically be the
case.

The starting point for our argument is to show that the counting operator (7)
is controlled by what we call the $U^{2}_{\perp}(S)$ norm, defined for
$f:\mathbf{F}_{q}^{d}\rightarrow\mathbf{C}$ by

$\lVert
f\rVert_{U^{2}_{\perp}(S)}:=\Big{(}\mathbf{E}_{x,h,h^{\prime}}f\sigma(x)\overline{f}\sigma(x+h)\overline{f}\sigma(x+h^{\prime})f\sigma(x+h+h^{\prime})\Big{)}^{1/4},$

which is the usual Gowers $U^{2}(\mathbf{F}_{q}^{d})$ norm of the function
$f\sigma$. While the $U^{2}(\mathbf{F}_{q}^{d})$ norm averages over
combinatorial rectangles, the $U^{2}_{\perp}(S)$ norm averages instead over
geometric rectangles contained within our sphere $S$. Note that if

$|x|^{2}=|x+h|^{2}=|x+h^{\prime}|^{2}=|x+h+h^{\prime}|^{2}=\lambda,$

then it is also the case that $h\cdot h^{\prime}=0$.

We will show that the operator (7) is controlled by the $U^{2}_{\perp}(S)$
norm in the following sense.

######  Proposition 6.

Let $f_{0},\ldots,f_{k+1}:\mathbf{F}_{q}^{d}\rightarrow\mathbf{C}$ with
$|f_{j}|\leq 1$. If $d\geq 2k+6$, then

$|\mathcal{N}_{X}(f_{0},\ldots,f_{k+1})|\leq\min_{0\leq j\leq k+1}\lVert
f_{j}\rVert_{U^{2}_{\perp}(S)}+O(q^{-1/8})$

Results of this type are often called _generalized von-Neumann inequalities_
in the arithmetic combinatorics literature. The proof of 6 is presented in
Section 6 below.

To see the utility of such a result, consider a set $A\subseteq S$ with
$|A|\geq\alpha q^{d-1}$ and $f_{A}=1_{A}-\alpha$. If $A$ is sufficiently
uniform, in the sense that $\lVert f_{A}\rVert_{U^{2}_{\perp}(S)}$ is
sufficiently small with respect to $\alpha$ and $k$, then the decomposition
$1_{A}=\alpha+f_{A}$ along with 6 provides

$\mathcal{N}_{X}(1_{A},\ldots,1_{A})\gtrsim\alpha^{k+2}$

provided $q$ is taken sufficiently large with respect to $\alpha$ and $k$. Of
course, not all sets must be uniform in this sense, and we will require a more
sophisticated decomposition. Defining the $L^{2}(S)$ norm by

$\lVert
f\rVert_{L^{2}(S)}:=\Big{(}\mathbf{E}_{x}|f(x)|^{2}\sigma(x)\Big{)}^{1/2},$

we will use the following decomposition.

######  Proposition 7.

Let $\eta\in(0,1)$, $f:\mathbf{F}_{q}^{d}\rightarrow[-1,1]$ and
$\varphi:(0,1]^{2}\rightarrow(0,1)$ increasing in both coordinates. If $d\geq
2$ and $q\geq q(\eta,f)$, then there exists $B=B(\Gamma,\rho)$ regular with
$|\Gamma|\leq C(\eta,f)$ and $\rho\geq c(\eta,f)$, and there exist functions
$f_{1},f_{2},f_{3}:\mathbf{F}_{q}^{d}\rightarrow[-2,2]$ with

$\displaystyle f$ $\displaystyle=f_{1}+f_{2}+f_{3}$ $\displaystyle f_{1}$
$\displaystyle=\nu_{B}*(f\sigma)$ $\displaystyle\lVert
f_{2}\rVert_{U^{2}_{\perp}(S)}$ $\displaystyle\leq\varphi(|\Gamma|^{-1},\rho)$
$\displaystyle\lVert f_{3}\rVert_{L^{2}(S)}$ $\displaystyle\leq\eta.$

Results of this type are often called _arithmetic regularity lemmas_ in the
arithmetic combinatorics literature. In light of (6), it should at least be
clear that ensuring $f_{1}$ is bounded amounts to ensuring $q$ is sufficiently
large with respect to the other parameters. The proof of 7 is presented in
Section 7 below.

We close this section by demonstrating, as promised, how Propositions 6 and 7
can be applied to give a proof of 2. We initially specialize, in Section 5.2
below, to the case of spherical quadrilaterals, that is when $k=2$. The proof
of the general case follows along similar lines and is presented in Section
5.3.

###  5.2. Proof that Propositions 6 and 7 imply the $k=2$ case of 2

It clearly suffices to establish the following

######  Theorem 8.

Let $\alpha\in(0,1)$ and $A\subseteq S$ with $|A|\geq\alpha q^{d-1}$. If
$d\geq 10$, $q\geq q(\alpha)$, and $X\subseteq\mathbf{F}_{q}^{d}$ is a
spherical 4-point configuration spanning 2 dimensions, then

$\mathcal{N}_{X}(1_{A},1_{A},1_{A},1_{A})\geq c(\alpha).$

In other words, $A$ contains $c(\alpha)q^{3d-6}$ isometric copies of $X$.

###### Proof.

To set up, let $\epsilon>0$ be a parameter to be determined only in terms of
$\alpha$ and let $\varphi:(0,1]^{2}\rightarrow(0,1)$ be a function increasing
in both coordinates to be determined only in terms of $\epsilon$. We apply the
decomposition theorem to obtain a regular Bohr set $B=B(\Gamma,\rho)$ with
$\lvert\Gamma\rvert\leq C(\epsilon,\varphi)$ and $\rho\geq
c(\epsilon,\varphi)$ and functions
$f_{1},f_{2},f_{3}:\mathbf{F}_{q}^{d}\rightarrow[-2,2]$ such that

$\displaystyle 1_{A}$ $\displaystyle=f_{1}+f_{2}+f_{3}$ $\displaystyle f_{1}$
$\displaystyle=\nu_{B}*(1_{A}\sigma)$ $\displaystyle\lVert
f_{2}\rVert_{U^{2}_{\perp}(S)}$ $\displaystyle\leq\varphi(|\Gamma|^{-1},\rho)$
$\displaystyle\lVert f_{3}\rVert_{L^{2}(S)}$ $\displaystyle\leq\epsilon.$

We take
$\rho^{\prime}\in[\epsilon\rho/(400|\Gamma|),\epsilon\rho/(200|\Gamma|)]$ so
that $B_{1}:=B(\Gamma,\rho^{\prime})\prec_{\epsilon}B$. Throughout the
argument we will continue to take $q(\alpha)$ large enough so that, with
parameters other than $q$ fixed, we can absorb error terms that tend to zero
as $q\rightarrow\infty$ into a single $O(\epsilon)$ error. We now fix a
spherical 4-point configuration spanning 2 dimensions $X$, parameterize it as

$X=\\{0,v,w,av+bw\\}$

for $v,w$ linearly independent, and search for isometric copies of the form

$\\{x,x+y,x+z,x+ay+bz\\}\subseteq S$

where $|y|^{2}=|v|^{2}$, $|z|^{2}=|w|^{2}$, and $y\cdot z=v\cdot w$. To detect
these copies, we define the spherical measures

$\displaystyle\sigma_{x}(y)$
$\displaystyle:=\sigma_{x}^{-|v|^{2}/2,|v|^{2}}(y),\text{ and}$
$\displaystyle\sigma_{x,y}(z)$ $\displaystyle:=\sigma_{x,y}^{-|w|^{2}/2,v\cdot
w,|w|^{2}}(z)$

in which case we can parametrize our counting operator as

$\mathcal{N}_{X}(g_{0},g_{1},g_{2},g_{3})=\mathbf{E}_{x}g_{0}(x)\sigma(x)\mathbf{E}_{y,z}^{*}g_{1}(x+y)g_{2}(x+z)g_{3}(x+ay+bz)\sigma(x)\sigma_{x}(y)\sigma_{x,y}(z).$

Setting

$B_{2}:=B(\Gamma\cup\\{a\cdot\xi:\xi\in\Gamma\\}\cup\\{b\cdot\xi:\xi\in\Gamma\\},\rho^{\prime}/4),$

we note that restricting $y,z\in B_{2}+B_{2}$ ensures $y,z,ay+bz\in B_{1}$.
This is useful to us, since for our main term function $f_{1}$, for any
$x\in\mathbf{F}_{q}^{d}$ and any $x^{\prime}\in B_{1}$, 4 provides

(8) $f_{1}(x+x^{\prime})=f_{1}(x)+O(\epsilon).$

We will in fact work with the further restricted count

$\mathcal{N}_{X}^{B}(g_{0},g_{1},g_{2},g_{3}):=\mathbf{E}_{x}g_{0}\sigma(x)\mathbf{E}_{y,z}^{*}g_{1}(x+y)g_{2}(x+z)g_{3}(x+ay+bz)\sigma_{x}(y)\mu_{B_{2}}*\mu_{B_{2}}(y)\sigma_{x,y}(z)\mu_{B_{2}}*\mu_{B_{2}}(z).$

Setting $\beta:=|B_{2}|/q^{d}$, the size estimate (4) and the dependency of
our parameters guarantees

$\beta^{-1}\leq\left(\frac{C|\Gamma|}{\epsilon\rho}\right)^{3|\Gamma|}\leq
C(\alpha).$

We should establish that this restricted count is well normalized. That is, we
will show

(9)
$\mathcal{N}_{X}^{B}(\mathbf{1},\mathbf{1},\mathbf{1},\mathbf{1})=1+O(\epsilon),$

where here $\mathbf{1}$ stands for the constant 1 function. Applying Parseval
and extracting the $\xi=0$ term,

$\mathbf{E}_{z}\mu_{B_{2}}*\mu_{B_{2}}(z)\sigma_{x,y}(z)=1+\sum_{\xi\in\mathbf{F}_{q}^{d}\setminus\\{0\\}}\widehat{\mu}_{B_{2}}(\xi)^{2}\overline{\widehat{\sigma}_{x,y}(\xi)}+O(\epsilon),$

so we can apply 3 for

$\mathcal{N}_{X}^{B}(\mathbf{1},\mathbf{1},\mathbf{1},\mathbf{1})=\mathbf{E}_{x,y}\sigma(x)\sigma_{x}(y)\mu_{B_{2}}*\mu_{B_{2}}(y)+O\left(\beta^{-1}\sum_{\xi\in\mathbf{F}_{q}^{d}\setminus\\{0\\}}|\widehat{\mu}_{B_{2}}(\xi)|^{2}\mathbf{E}_{x,y}\sigma(x)\sigma_{x}(y)\delta_{x,y}(\xi)\right)+O(\epsilon).$

Without trying to be too careful, notice that uniformly in $\xi\neq 0$,

$\mathbf{E}_{x,y}\sigma(x)\sigma_{x}(y)\delta_{x,y}(\xi)\leq
q^{3}\mathbf{E}_{x,y}\delta_{x,y}(\xi)=O(q^{-1}),$

so we can apply Plancherel and ensure $q$ is sufficiently large to conclude

$\mathcal{N}_{X}^{B}(\mathbf{1},\mathbf{1},\mathbf{1},\mathbf{1})=\mathbf{E}_{x,y}\sigma(x)\sigma_{x}(y)\mu_{B_{2}}*\mu_{B_{2}}(y)+O(\epsilon).$

Arguing similarly to eliminate the average in $y$, we conclude (9). This
restricted count will be useful for us since we trivially have the lower bound

(10)
$\mathcal{N}_{X}(1_{A},1_{A},1_{A},1_{A})\geq\beta^{2}\mathcal{N}_{X}^{B}(1_{A},1_{A},1_{A},1_{A}),$

and we will spend the rest of the proof establishing the lower bound

$\mathcal{N}^{B}_{X}(1_{A},1_{A},1_{A},1_{A})\geq\alpha^{4}+O(\epsilon).$

From the decomposition $1_{A}=f_{1}+f_{2}+f_{3}$,

$\mathcal{N}^{B}_{X}(1_{A},1_{A},1_{A},1_{A})=\sum_{1\leq
i_{0},i_{1},i_{2},i_{3}\leq
3}\mathcal{N}^{B}_{X}(f_{i_{0}},f_{i_{1}},f_{i_{2}},f_{i_{3}}),$

and we handle separately the three cases (i) when each of
$i_{0},i_{1},i_{2},i_{3}$ equals 1, (ii) when one of $i_{0},i_{1},i_{2},i_{3}$
equals 2, and (iii) when one of $i_{0},i_{1},i_{2},i_{3}$ equals 3. Case (i)
will yield our main term of $\alpha^{4}$, and we will argue that cases (ii)
and (iii) contribute a negligible $O(\epsilon)$ error.

For the first case, applying (8) and (9) yields

$\mathcal{N}^{B}_{X}(f_{1},f_{1},f_{1},f_{1})=\mathbf{E}_{x}f_{1}(x)^{4}\sigma(x)\mathbf{E}_{y,z}\mu_{B_{2}}*\mu_{B_{2}}(y)\sigma_{x}(y)\mu_{B_{2}}*\mu_{B_{2}}(z)\sigma_{x,y}(z)+O(\epsilon).$

Arguing as we did to establish (9),

$\mathcal{N}^{B}_{X}(f_{1},f_{1},f_{1},f_{1})=\mathbf{E}_{x}f_{1}(x)^{4}\sigma(x)+O(\epsilon).$

Then from Hölder’s inequality,

$\mathcal{N}_{X}^{B}(f_{1},f_{1},f_{1},f_{1})\geq(\mathbf{E}_{x}f_{1}(x)\sigma(x))^{4}+O(\epsilon),$

and writing

$\displaystyle\mathbf{E}_{x}f_{1}(x)\sigma(x)$
$\displaystyle=\mathbf{E}_{x}\nu_{B}*1_{A}(x)\sigma(x)$
$\displaystyle=\mathbf{E}_{x}1_{A}(x)\nu_{B}*\sigma(x),$

we can apply (6) to conclude

$\mathcal{N}_{X}^{B}(f_{1},f_{1},f_{1},f_{1})\geq\alpha^{4}+O(\epsilon).$

For the second case, with one of $i_{0},i_{1},i_{2},i_{3}=2$, we apply Fourier
inversion to write

$\mu_{B_{2}}*\mu_{B_{2}}(y)\mu_{B_{2}}*\mu_{B_{2}}(z)=\sum_{\xi_{1},\xi_{2}\in\mathbf{F}_{q}^{d}}\widehat{\mu}_{B_{2}}(\xi_{1})^{2}\widehat{\mu}_{B_{2}}(\xi_{2})^{2}\chi(-\xi_{1}\cdot
y)\chi(-\xi_{2}\cdot z).$

Setting $\chi_{\xi}(x)=\chi(\xi\cdot x)$, this allows us to express

$\mathcal{N}_{X}^{B}(f_{i_{0}},f_{i_{1}},f_{i_{2}},f_{i_{3}})=\sum_{\xi_{1},\xi_{2}\in\mathbf{F}_{q}^{d}}\widehat{\mu}_{B_{2}}(\xi_{1})^{2}\widehat{\mu}_{B_{2}}(\xi_{2})^{2}\mathcal{T}_{X,S}(f_{i_{0}}\chi_{\xi_{1}+\xi_{2}},f_{i_{1}}\chi_{-\xi_{1}},f_{i_{2}}\chi_{-\xi_{2}},f_{i_{3}}).$

As $\lVert f\chi_{\xi}\rVert_{U^{2}_{\perp}(S)}=\lVert
f\rVert_{U^{2}_{\perp}(S)}$, we can apply the triangle inequality, 6 and
Plancherel to conclude

$\displaystyle|\mathcal{N}_{X}^{B}(f_{i_{0}},f_{i_{1}},f_{i_{2}},f_{i_{3}})|$
$\displaystyle\leq\beta^{-2}\lVert f_{2}\rVert_{U^{2}_{\perp}(S)}+O(\epsilon)$
$\displaystyle\leq\left(\frac{C|\Gamma|}{\epsilon\rho}\right)^{6|\Gamma|}\varphi(|\Gamma|^{-1},\rho)+O(\epsilon).$

We see that by taking

$\varphi(|\Gamma|^{-1},\rho)=\epsilon\left(\frac{c\epsilon\rho}{|\Gamma|}\right)^{6|\Gamma|},$

we can ensure case (ii) contributes at most $O(\epsilon)$.

For the third case, we will assume $i_{0}=3$, but each case is similar by
reindexing in $x_{0}$. Applying the triangle inequality and using that
$f_{i_{1}},f_{i_{2}},f_{i_{3}}$ are each at most 2 in absolute value,

$\mathcal{N}_{X}^{B}(f_{3},f_{i_{1}},f_{i_{2}},f_{i_{3}})\leq
8\mathbf{E}_{x}|f_{3}(x)|\sigma(x)\mathbf{E}_{y,z}^{*}\sigma_{x}(y)\mu_{B_{2}}*\mu_{B_{2}}(y)\sigma_{x,y}(z)\mu_{B_{2}}*\mu_{B_{2}}(z).$

Then arguing as we did to establish (9) we have

$\mathcal{N}_{X}^{B}(f_{3},f_{i_{1}},f_{i_{2}},f_{i_{3}})\leq
8\mathbf{E}_{x}|f_{3}(x)|\sigma(x)+O(\epsilon).$

Applying Cauchy-Schwarz and the fact that $\lVert
f_{3}\rVert_{L^{2}(S)}\leq\epsilon$, we conclude case (iii) again contributes
at most an $O(\epsilon)$ error term. In total we have shown

$\mathcal{N}^{B}_{X}(1_{A},1_{A},1_{A},1_{A})\geq\alpha^{4}+O(\epsilon)$

which we can insist is at least $\alpha^{4}/2$ by taking $\epsilon$
sufficiently small with respect to $\alpha$. Then since $\beta\geq c(\alpha)$,
we are done by recalling (10). ∎

###  5.3. Proof that Propositions 6 and 7 imply 2 in its full generality

It clearly suffices to establish the following

######  Theorem 9.

Let $\alpha\in(0,1)$ and $A\subseteq S$ with $|A|\geq\alpha q^{d-1}$. If
$d\geq 2k+6$, $q\geq q(\alpha,k)$, and $X\subseteq\mathbf{F}_{q}^{d}$ is a
spherical $(k+2)$-point configuration spanning $k$ dimensions, then

$\mathcal{N}_{X}(1_{A},\ldots,1_{A})\geq c(\alpha,k).$

In other words, $A$ contains $c(\alpha,k)q^{(k+1)d-(k+1)(k+2)/2}$ isometric
copies of $X$.

###### Proof.

The proof is essentially the same as for 4-point configurations spanning 2
dimensions, although the notation becomes more cumbersome. As before, to set
up, let $\epsilon>0$ be a parameter to be determined only in terms of $\alpha$
and $k$ and let $\varphi:(0,1]^{2}\rightarrow(0,1)$ be a function increasing
in both coordinates to be determined only in terms of $\epsilon$. We apply the
decomposition theorem to obtain a regular Bohr set $B=B(\Gamma,\rho)$ with
$\lvert\Gamma\rvert\leq C(\epsilon,\varphi)$ and $\rho\geq
c(\epsilon,\varphi)$ and functions
$f_{1},f_{2},f_{3}:\mathbf{F}_{q}^{d}\rightarrow[-2,2]$ such that

$\displaystyle 1_{A}$ $\displaystyle=f_{1}+f_{2}+f_{3}$ $\displaystyle f_{1}$
$\displaystyle=\nu_{B}*(1_{A}\sigma)$ $\displaystyle\lVert
f_{2}\rVert_{U^{2}_{\perp}(S)}$ $\displaystyle\leq\varphi(|\Gamma|^{-1},\rho)$
$\displaystyle\lVert f_{3}\rVert_{L^{2}(S)}$ $\displaystyle\leq\epsilon.$

We take
$\rho^{\prime}\in[\epsilon\rho/(400|\Gamma|),\epsilon\rho/(200|\Gamma|)]$ so
that $B_{1}:=B(\Gamma,\rho^{\prime})\prec_{\epsilon}B$. Throughout the
argument we will continue to take $q(\alpha,k)$ large enough so that, with
parameters other than $q$ fixed, we can absorb error terms that tend to zero
as $q\rightarrow\infty$ into a single $O(\epsilon)$ error. We now fix a
spherical $(k+2)$-point configuration spanning $k$ dimensions $X$
parameterized as

$X=\\{0,v_{1},\ldots,v_{k+1}\\}$

where $v_{k+1}=a_{1}v_{1}+\cdots+a_{k}v_{k}$ and search for isometric copies
of the form

$\\{x_{0},x_{0}+x_{1},\ldots,x_{0}+x_{k+1}\\}\subset S$

where $x_{k+1}=a_{1}x_{1}+\cdots+a_{k}x_{k}$ and $x_{i}\cdot x_{j}=v_{i}\cdot
v_{j}$ for $1\leq i,j\leq k$. To detect these copies, we define the spherical
measures

$\sigma_{x_{0},\ldots,x_{j-1}}(x_{j}):=(\sigma)^{-|v_{j}|^{2}/2,v_{1}\cdot
v_{j},\ldots,v_{j-1}\cdot v_{j},|v_{j}|^{2}}_{x_{0},\ldots,x_{j-1}}(x_{j})$

in which case we can parametrize our count as

$\mathcal{N}_{X}(g_{0},\ldots,g_{k+1})=\mathbf{E}_{x_{0}}g_{0}(x_{0})\sigma(x_{0})\mathbf{E}_{x_{1},\ldots,x_{k}}^{*}g(x_{0}+x_{k+1})\prod_{j=1}^{k}g(x_{0}+x_{j})\sigma_{x_{0},\ldots,x_{j-1}}(x_{j}).$

Setting

$B_{2}:=B\left(\Gamma\cup\bigcup_{j=1}^{k}\left\\{a_{j}\cdot\xi:\xi\in\Gamma\right\\},\dfrac{\rho^{\prime}}{2k}\right),$

ensures that if $x_{1},\ldots,x_{k}\in B_{2}+B_{2}$, then
$x_{1},\ldots,x_{k+1}\in B_{1}$. Again we will use that for
$x\in\mathbf{F}_{q}^{d}$ and $x^{\prime}\in B_{1}$,

(11) $f_{1}(x+x^{\prime})=f_{1}(x)+O(\epsilon).$

Our restricted count is given by

$\mathcal{N}_{X}^{B}(g_{0},\ldots,g_{k+1})=\mathbf{E}_{x_{0}}g_{0}(x_{0})\sigma(x_{0})\mathbf{E}_{x_{1},\ldots,x_{k}}^{*}g(x_{0}+x_{k+1})\prod_{j=1}^{k}g(x_{0}+x_{j})\mu_{B_{2}}*\mu_{B_{2}}(x_{j})\sigma^{(j+1)}(x_{j}).$

As before, setting $\beta:=|B_{2}|/q^{d}$, we have $\beta^{-1}\leq
C(\alpha,k)$ and a well-normalized restricted count

(12) $\mathcal{N}_{X}^{B}(\mathbf{1},\ldots,\mathbf{1})=1+O(\epsilon).$

Applying the straightforward lower bound

(13)
$\mathcal{N}_{X}(1_{A},\ldots,1_{A})\geq\beta^{k}\mathcal{N}_{X}^{B}(1_{A},\ldots,1_{A}),$

we will spend the rest of the proof establishing

$\mathcal{N}_{X}^{B}(1_{A},\ldots,1_{A})\geq\alpha^{k+2}+O(\epsilon).$

From the decomposition $1_{A}=f_{1}+f_{2}+f_{3}$,

$\mathcal{N}_{X}^{B}(1_{A},\ldots,1_{A})=\sum_{1\leq i_{0},\ldots,i_{k+1}\leq
3}\mathcal{N}_{X}^{B}(f_{i_{0}},\ldots,f_{i_{k+1}}),$

and we handle separately the three cases (i) when each of $i_{j}$ equals 1,
(ii) when one of $i_{j}$ equals 2, and (iii) when one of $i_{j}$ equals 3.
Case (i) will yield our main term of $\alpha^{k+2}$, and we will argue that
cases (ii) and (iii) contribute a negligible $O(\epsilon)$ error.

For the first case, applying (11) and (12) yields

$\mathcal{N}_{X}^{B}(f_{1},\ldots,f_{1})=\mathbf{E}_{x_{0}}f_{1}(x_{0})^{k+2}\sigma(x)\mathbf{E}_{x_{1},\ldots,x_{k}}\prod_{j=1}^{k}\mu_{B_{2}}*\mu_{B_{2}}(x_{j})\sigma^{(j+1)}(x_{j})+O(\epsilon).$

Arguing as we did to establish (9),

$\mathcal{N}_{X}^{B}(f_{1},f_{1},f_{1},f_{1})=\mathbf{E}_{x}f_{1}(x)^{k+2}\sigma(x)+O(\epsilon).$

Then from Hölder’s inequality,

$\displaystyle\mathcal{N}_{X}^{B}(f_{1},f_{1},f_{1},f_{1})$
$\displaystyle\geq(\mathbf{E}_{x}f_{1}(x)\sigma(x))^{k+2}+O(\epsilon)$
$\displaystyle=(\mathbf{E}_{x}1_{A}(x)\nu_{B}*\sigma(x))^{k+2}+O(\epsilon)$
$\displaystyle=\alpha^{k+2}+O(\epsilon).$

For the second case, we apply Fourier inversion to write

$\prod_{j=1}^{k}\mu_{B_{2}}*\mu_{B_{2}}(x_{j})=\sum_{\xi_{1},\ldots,\xi_{k}\in\mathbf{F}_{q}^{d}}\prod_{j=1}^{k}\widehat{\mu}_{B_{2}}(\xi_{j})^{2}\chi(-\xi_{j}\cdot
x_{j})$

in order to express

$\mathcal{N}_{X}^{B}(f_{i_{0}},\ldots,f_{i_{k+1}})=\sum_{\xi_{1},\ldots,\xi_{k}\in\mathbf{F}_{q}^{d}}\prod_{j=1}^{k}\widehat{\mu}_{B_{2}}(\xi_{j})^{2}\mathcal{N}_{X}(f_{i_{0}}\chi_{\xi_{1}+\cdots+\xi_{k}},f_{i_{1}}\chi_{-\xi_{1}},\ldots
f_{i_{k}}\chi_{-\xi_{k}},f_{i_{k+1}}).$

As before, $\lVert f\chi_{\xi}\rVert_{U^{2}_{\perp}(S)}=\lVert
f\rVert_{U^{2}_{\perp}(S)}$, so we apply the triangle inequality, 6 and
Plancherel to conclude

$\displaystyle|\mathcal{N}_{X}^{B}(f_{i_{0}},\ldots,f_{i_{k+1}})|$
$\displaystyle\leq\beta^{-k}\lVert f_{2}\rVert_{U^{2}_{\perp}(S)}+O(\epsilon)$
$\displaystyle\leq\left(\frac{C|\Gamma|}{\epsilon\rho}\right)^{C|\Gamma|}\varphi(|\Gamma|^{-1},\rho)+O(\epsilon).$

We see that by taking

$\varphi(|\Gamma|^{-1},\rho)=\epsilon\left(\frac{c\epsilon\rho}{|\Gamma|}\right)^{C|\Gamma|},$

we can ensure case (ii) contributes at most $O(\epsilon)$.

For the third case, we assume $i_{0}=3$ and apply the triangle inequality for

$\mathcal{N}_{X}^{B}(f_{3},f_{i_{1}},\ldots,f_{i_{k+1}})\leq
2^{k+1}\mathbf{E}_{x_{0}}|f_{3}(x_{0})|\sigma(x_{0})\mathbf{E}_{x_{1},\ldots,x_{k}}\prod_{j=1}^{k}\mu_{B_{2}}*\mu_{B_{2}}(x_{j})\sigma^{(j+1)}(x_{j})+O(\epsilon).$

Applying Cauchy-Schwarz and the fact that $\lVert
f_{3}\rVert_{L^{2}(S)}\leq\epsilon$, we conclude case (iii) again contributes
at most an $O(\epsilon)$ error term. In total we have shown

$\mathcal{N}^{B}_{X}(1_{A},\ldots,1_{A})\geq\alpha^{k+2}+O(\epsilon)$

which we can insist is at least $\alpha^{k+2}/2$ by taking $\epsilon$
sufficiently small with respect to $\alpha$ and $k$. Then since $\beta\geq
c(\alpha,k)$, we are done by recalling (13). ∎

It remains to prove both 6 and 7.

##  6\. Proof of 6

We establish 6 through two careful applications of Cauchy-Schwarz. In a sense,
each application of Cauchy-Schwarz replaces our configuration with a more
regular configuration, and we are left considering averages over geometric
rectangles rather than more complicated spherical configurations.

We fix a spherical $(k+2)$-point configuration $X$ spanning $k$ dimensions
parameterized by

$X=\\{v_{0},v_{1},\ldots,v_{k+1}\\}$

with $v_{1}-v_{0},\ldots,v_{k}-v_{0}$ linearly independent and

$v_{k+1}=\sum_{j=0}^{k}a_{j}v_{j}$

for coefficients $a_{j}\in\mathbf{F}_{q}$. It is enough by symmetry to show

$|\mathcal{N}_{X}(f_{0},\ldots,f_{k+1})|\leq\lVert
f_{k}\rVert_{U^{2}_{\perp}(S)}+O(q^{-1/8}).$

For $x_{0},\ldots,x_{k}\in\mathbf{F}_{q}^{d}$, we will understand that

$x_{k+1}=\sum_{j=0}^{k}a_{i}x_{i},$

and we will detect if $\\{x_{0},\ldots,x_{k+1}\\}\simeq X$ with the spherical
measures $\sigma(x_{0})$ and, for $1\leq j\leq k$,

$\sigma^{(j)}(x_{j}):=\sigma^{c_{0,j},\ldots,c_{j-1,j},\lambda}_{x_{0},\ldots,x_{j-1}}(x_{j})$

where we define $c_{i,j}=|v_{i}-v_{j}|^{2}/2-\lambda$, since if
$|x_{i}|^{2}=|x_{j}|^{2}=\lambda$, checking whether
$|x_{i}-x_{j}|^{2}=|v_{i}-v_{j}|^{2}$ amounts to checking whether $x_{i}\cdot
x_{j}=|v_{i}-v_{j}|^{2}/2-\lambda$. Then we can express our counting operator

$\mathcal{N}_{X}(f_{0},\ldots,f_{k+1})=\mathbf{E}_{x_{0},\ldots,x_{k}}\sigma(x_{0})\prod_{j=0}^{k+1}f_{j}(x_{j})\prod_{j=1}^{k}\sigma^{(j)}(x_{j})+O(q^{-1}),$

where we have included the negligible amount of linearly dependent vectors.
Rearranging and applying the triangle inequality,

$|\mathcal{N}_{X}(f_{0},\ldots,f_{k+1})|\leq\mathbf{E}_{x_{0},\ldots,x_{k-1}}\sigma(x_{0})\prod_{j=1}^{k-1}\sigma^{(j)}(x_{j})\left|\mathbf{E}_{x_{k}}f_{k}(x_{k})f_{k+1}(x_{k+1})\sigma^{(k)}(x_{k})\right|+O(q^{-1}).$

Introducing the differencing notation

$\Delta_{h}f(x):=f(x)\overline{f}(x+h),$

we square both sides and apply Cauchy-Schwarz for

$|\mathcal{N}_{X}(f_{0},\ldots,f_{k+1})|^{2}\leq\mathbf{E}_{x_{0},\ldots,x_{k-1}}\sigma(x_{0})\prod_{j=1}^{k-1}\sigma^{(j)}(x_{j})\mathbf{E}_{x_{k},h}\Delta_{h}f_{k}(x_{k})\Delta_{h}f_{k+1}(x_{k+1})\Delta_{h}\sigma^{(k)}(x_{k})+O(q^{-1}).$

We introduce new spherical measures with an additional condition involving $h$
defined by

$\displaystyle\sigma_{h}(x_{0})$ $\displaystyle:=\sigma_{h}^{0}(x_{0})$
$\displaystyle\sigma_{h}^{(j)}(x_{j})$
$\displaystyle:=\sigma_{h,x_{0},\ldots,x_{j-1}}^{0,c_{0,j},\ldots,c_{j-1,j},\lambda}(x_{j}).$

which allows us to rewrite

$\sigma(x_{0})\prod_{j=1}^{k-1}\sigma^{(j)}(x_{j})\Delta_{h}\sigma^{(k)}(x_{k})=\sigma_{h}(x_{0})\prod_{j=1}^{k-1}\sigma_{h}^{(j)}(x_{j})\sigma^{(k)}(x_{k})\sigma(x_{k}+h).$

Our bound from above can then be rearranged as

$|\mathcal{N}_{X}(f_{0},\ldots,f_{k+1})|^{2}\leq\mathbf{E}_{h,x_{0},\ldots,x_{k-1}}\sigma_{h}(x_{0})\prod_{j=1}^{k-1}\sigma_{h}^{(j)}(x_{j})\mathbf{E}_{x_{k}}\Delta_{h}f_{k}(x_{k})\Delta_{h}f_{k+1}(x_{k+1})\sigma^{(k)}(x_{k})\sigma(x_{k}+h)+O(q^{-1}).$

We claim that in the average above, the dependence of $x_{k+1}$ on $x_{k}$ is
superficial. That is, while

$x_{k+1}=\sum_{j=0}^{k}a_{j}x_{j},$

it must be the case that at least two coefficients $a_{j}$ are nonzero since
we are working with a $(k+2)$-point spherical configuration. In particular,
$a_{j}\neq 0$ for some $0\leq j<k$, and for this distinguished $j$, we can
reindex in $x_{j}$ in the average above to replace $x_{k+1}$ with

$x_{k+1}^{\prime}:=\sum_{j=0}^{k-1}a_{j}^{\prime}x_{j},$

allowing us to rearrange our bound above as

$|\mathcal{N}_{X}(f_{0},\ldots,f_{k+1})|^{2}\leq\mathbf{E}_{h,x_{0},\ldots,x_{k-1}}\Delta_{h}f_{k+1}(x_{k+1}^{\prime})\sigma_{h}(x_{0})\prod_{j=1}^{k-1}\sigma_{h}^{(j)}(x_{j})\mathbf{E}_{x_{k}}\Delta_{h}f_{k}(x_{k})\sigma^{(k)}(x_{k})\sigma(x_{k}+h)+O(q^{-1}),$

where we may have needed to adjust the implicit scalars in our spherical
measures. This allows us to proceed as before by applying the triangle
inequality for

$|\mathcal{N}_{X}(f_{0},\ldots,f_{k+1})|^{2}\leq\mathbf{E}_{h,x_{0},\ldots,x_{k-1}}\sigma_{h}(x_{0})\prod_{j=1}^{k-1}\sigma_{h}^{(j)}(x_{j})\left|\mathbf{E}_{x_{k}}\Delta_{h}f_{k}(x_{k})\sigma^{(k)}(x_{k})\sigma(x_{k}+h)\right|+O(q^{-1}),$

and Cauchy-Schwarz once more for

$|\mathcal{N}_{X}(f_{0},\ldots,f_{k+1})|^{4}\leq\mathbf{E}_{h,x_{0},\ldots,x_{k-1}}\sigma_{h}(x_{0})\prod_{j=1}^{k-1}\sigma_{h}^{(j)}(x_{j})\mathbf{E}_{x_{k},h^{\prime}}\Delta_{h^{\prime}}\Delta_{h}f_{k}(x_{k})\Delta_{h^{\prime}}\sigma^{(k)}(x_{k})\Delta_{h^{\prime}}\sigma(x_{k}+h)+O(q^{-1}).$

We can again reorganize our spherical measures, rewriting

$\sigma_{h}(x_{0})\prod_{j=1}^{k-1}\sigma_{h}^{(j)}(x_{j})\Delta_{h^{\prime}}\sigma^{(k)}(x_{k})\Delta_{h^{\prime}}\sigma(x_{k}+h)=\Delta_{h^{\prime}}\Delta_{h}\sigma(x_{k})\sigma_{x_{k},h,h^{\prime}}(x_{0})\prod_{j=1}^{k-1}\sigma_{x_{k},h,h^{\prime},x_{0},\ldots,x_{j-1}}(x_{j})$

for appropriate implicit scalars. Using this to rearrange our bound above, we
have

$|\mathcal{N}_{X}(f_{0},\ldots,f_{k+1})|^{4}\leq\mathbf{E}_{x_{k},h,h^{\prime}}\Delta_{h^{\prime}}\Delta_{h}f_{k}\sigma(x_{k})\mathbf{E}_{x_{0},\ldots,x_{k-1}}\sigma_{x_{k},h,h^{\prime}}(x_{0})\prod_{j=1}^{k-1}\sigma_{x_{k},h,h^{\prime},x_{0},\ldots,x_{j-1}}(x_{j})+O(q^{-1}).$

Applying 3, we have, uniformly in $x_{k},h,h^{\prime}$ when $d\geq 2k+6$,

$\mathbf{E}_{x_{0},\ldots,x_{k-1}}\sigma_{x_{k},h,h^{\prime}}(x_{0})\prod_{j=1}^{k-1}\sigma_{x_{k},h,h^{\prime},x_{0},\ldots,x_{j-1}}(x_{j})=1+O(q^{-1/2}),$

establishing

$|\mathcal{N}_{X}(f_{0},\ldots,f_{k+1})|^{4}\leq\mathbf{E}_{x_{k},h,h^{\prime}}\Delta_{h^{\prime}}\Delta_{h}f_{k}\sigma(x_{k})+O(q^{-1/2})$

from which we see $|\mathcal{N}_{X}(f_{0},\ldots,f_{k+1})|\leq\lVert
f_{k}\rVert_{U^{2}_{\perp}(S)}+O(q^{-1/8})$ as required. ∎

##  7\. An Inverse Theorem and Proof of 7

This section is dedicated to establishing 7. We begin in Section 7.1 by
establishing 10, an _inverse theorem_ which reveals that functions with large
$U^{2}_{\perp}(S)$ norm must exhibit Fourier bias.

###  7.1. An Inverse Theorem

One way to see that functions with large $U^{2}_{\perp}(S)$ norm exhibit
Fourier bias is to relate the $U^{2}_{\perp}(S)$ norm to the usual
$U^{2}(\mathbf{F}_{q}^{d})$ norm. In order to do so, define for
$v,w\in\mathbf{F}_{q}^{d}$ the normalized indicator function

$\ell_{v=w}=\begin{cases}q^{d}&\text{if }v=w\\\ 0&\text{otherwise}\end{cases}$

Then one can express

$\lVert
f\rVert_{U^{2}_{\perp}(S)}^{4}=\mathbf{E}_{x,y,z,w}\overline{f}\sigma(x)f\sigma(y)f\sigma(z)\overline{f}\sigma(w)\ell_{x+w=y+z}$

Expanding $\ell_{x+w=y+z}$ via orthogonality,

$\ell_{x+w=y+z}=\sum_{\xi\in\mathbf{F}_{q}^{d}}\chi(\xi(x+w-y-z)),$

providing the identity

$\lVert
f\rVert_{U^{2}_{\perp}(S)}^{4}=\sum_{\xi\in\mathbf{F}_{q}^{d}}|\widehat{f\sigma}(\xi)|^{4}.$

The right hand side is precisely $\lVert
f\sigma\rVert_{U^{2}(\mathbf{F}_{q}^{d})}$. It is tempting to here use
Plancherel to bound $\lVert f\rVert_{U^{2}_{\perp}(S)}^{4}$ above by
$\sup_{\xi\in\mathbf{F}_{q}^{d}}|\widehat{f\sigma}(\xi)|^{2}\lVert
f\sigma\rVert_{L^{2}}^{2}$, but this is not generally helpful since the
$L^{2}$ term may grow with $q$. By being a bit more careful, we establish the
following inverse theorem.

######  Theorem 10.

Let $f:\mathbf{F}_{q}^{d}\rightarrow\mathbf{C}$ with $|f|\leq 1$. If $d\geq
8$, then

$\lVert
f\rVert_{U^{2}_{\perp}(S)}\leq\sup_{\xi\in\mathbf{F}_{q}^{d}}\lvert\widehat{f\sigma}(\xi)\rvert^{1/4}+O(q^{-1/32}).$

###### Proof.

We apply absolute values and the triangle inequality for

$\lVert
f\rVert_{U^{2}_{\perp}(S)}^{4}\leq\mathbf{E}_{x}\sigma(x)\lvert\mathbf{E}_{y,z,w}\overline{f}\sigma(y)\overline{f}\sigma(z)f\sigma(w)\ell_{x=y+z-w}\rvert$

Applying Cauchy-Schwarz,

$\lVert
f\rVert_{U^{2}_{\perp}(S)}^{8}\leq\left(\mathbf{E}_{x}\sigma(x)\right)\left(\mathbf{E}_{x}\sigma(x)\lvert\mathbf{E}_{y,z,w}\overline{f}\sigma(y)\overline{f}\sigma(z)f\sigma(w)\ell_{x=y+z-w}\rvert^{2}\right).$

For any fixed $x$ and $y$, we have

$\mathbf{E}_{z,w}\sigma(z)\sigma(w)\ell_{x=y+z-w}=\mathbf{E}_{z}\sigma_{x+y}(z)$

for some measure $\sigma_{x+y}$. Then since $d\geq 4$, we can apply (1) and 3
for

$\lVert
f\rVert_{U^{2}_{\perp}(S)}^{8}\leq\mathbf{E}_{x}\sigma(x)\mathbf{E}_{\begin{subarray}{c}y_{1},z_{1},w_{1}\\\
y_{2},z_{2},w_{2}\end{subarray}}\overline{f}\sigma(y_{1})f\sigma(y_{2})\overline{f}\sigma(z_{1})f\sigma(z_{2})f\sigma(w_{1})\overline{f}\sigma(w_{2})\ell_{x=y_{1}+z_{1}-w_{1}}\ell_{x=y_{2}+z_{2}-w_{2}}+O(q^{-1/2})$

Moving the average in $x$ inside and rearranging a bit, this simplifies to

$\lVert
f\rVert_{U^{2}_{\perp}(S)}^{8}\leq\mathbf{E}_{\begin{subarray}{c}y_{1},z_{1},w_{1}\\\
y_{2},z_{2},w_{2}\end{subarray}}\overline{f}\sigma(y_{1})f\sigma(y_{2})\overline{f}\sigma(z_{1})f\sigma(z_{2})f\sigma(w_{1})\overline{f}\sigma(w_{2})\sigma(y_{1}+z_{1}-w_{1})\ell_{w_{2}-w_{1}=y_{2}-y_{1}+z_{2}-z_{1}}+O(q^{-1/2})$

Setting $f_{1}=f$ and $f_{2}=\overline{f}$, rearranging a bit more and
applying the triangle inequality,

$\lVert
f\rVert_{U^{2}_{\perp}(S)}^{8}\leq\mathbf{E}_{w_{1},w_{2}}\sigma(w_{1})\sigma(w_{2})\lvert\mathbf{E}_{y_{1},y_{2},z_{1},z_{2}}\sigma(y_{1}+z_{1}-w_{1})\ell_{w_{2}-w_{1}=y_{2}-y_{1}+z_{2}-z_{1}}\prod_{j=1,2}f_{j}\sigma(y_{j})f_{j}\sigma(z_{j})\rvert+O(q^{-1/2})$

Setting $f_{3}=f$ and $f_{4}=\overline{f}$ and again requiring $d\geq 4$, we
apply Cauchy-Schwarz as before and rearrange for

(14) $\lVert
f\rVert_{U^{2}_{\perp}(S)}^{16}\leq\mathbf{E}_{\begin{subarray}{c}y_{1},\ldots,y_{4}\\\
z_{1},\ldots,z_{4}\end{subarray}}\prod_{j=1}^{4}f_{j}\sigma(y_{j})f_{j}\sigma(z_{j})\ell_{y_{2}-y_{1}+z_{2}-z_{1}=y_{4}-y_{3}+z_{4}-z_{3}}W_{\begin{subarray}{c}y_{1},\ldots,y_{4}\\\
z_{1},\ldots,z_{4}\end{subarray}}+O(q^{-1/2})$

where

$W_{\begin{subarray}{c}y_{1},\ldots,y_{4}\\\
z_{1},\ldots,z_{4}\end{subarray}}=\mathbf{E}_{w}\sigma(w)\sigma(y_{2}-y_{1}+z_{2}-z_{1}+w)\sigma(y_{1}+z_{1}-w)\sigma(y_{3}+z_{3}-w)$

Since $d\geq 4$, we can restrict the sum in (14) to only consider the terms
when the vectors $y_{2}-y_{1}+z_{2}-z_{1},y_{1}+z_{1},y_{3}+z_{3}$ are
linearly independent at the cost of an error that can be absorbed in our
current error of $O(q^{-1/2})$. For these vectors, the function

$\sigma^{\prime}(w)=\sigma(w)\sigma(y_{2}-y_{1}+z_{2}-z_{1}+w)\sigma(y_{1}+z_{1}-w)\sigma(y_{3}+z_{3}-w)$

is a measure for which 3 applies, allowing us to conclude

$W_{\begin{subarray}{c}y_{1},\ldots,y_{4}\\\
z_{1},\ldots,z_{4}\end{subarray}}=\mathbf{E}_{w}\sigma^{\prime}(w)=1+O(q^{-1/2}),$

valid for $d\geq 8$. We have arrived at the estimate

$\lVert
f\rVert_{U^{2}_{\perp}(S)}^{16}\leq\mathbf{E}_{\begin{subarray}{c}y_{1},\ldots,y_{4}\\\
z_{1},\ldots,z_{4}\end{subarray}}\prod_{j=1}^{4}f_{j}\sigma(y_{j})f_{j}\sigma(z_{j})\ell_{y_{2}-y_{1}+z_{2}-z_{1}=y_{4}-y_{3}+z_{4}-z_{3}}+O(q^{-1/2})$

Expanding $\ell_{y_{2}-y_{1}+z_{2}-z_{1}=y_{4}-y_{3}+z_{4}-z_{3}}$ via
orthogonality, we have

$\ell_{y_{2}-y_{1}+z_{2}-z_{1}=y_{4}-y_{3}+z_{4}-z_{3}}=\sum_{\xi\in\mathbf{F}_{q}^{d}}\chi(\xi\cdot(y_{4}-y_{3}-y_{2}+y_{1}+z_{4}-z_{3}-z_{2}+z_{1})),$

from which we have the identity

$\mathbf{E}_{\begin{subarray}{c}y_{1},\ldots,y_{4}\\\
z_{1},\ldots,z_{4}\end{subarray}}\prod_{j=1}^{4}f_{j}\sigma(y_{j})f_{j}\sigma(z_{j})\ell_{y_{2}-y_{1}+z_{2}-z_{1}=y_{4}-y_{3}+z_{4}-z_{3}}=\sum_{\xi\in\mathbf{F}_{q}^{d}}|\widehat{f\sigma}(\xi)|^{8}$

It follows that we can conclude

$\displaystyle\lVert f\rVert_{U^{2}_{\perp}(S)}^{16}$
$\displaystyle\leq\sum_{\xi\in\mathbf{F}_{q}^{d}}|\widehat{f\sigma}(\xi)|^{8}+O(q^{-1/2})$
$\displaystyle\leq\sup_{\xi\in\mathbf{F}_{q}^{d}}|\widehat{f\sigma}(\xi)|^{4}\sum_{\xi\in\mathbf{F}_{q}^{d}}|\widehat{f\sigma}(\xi)|^{4}+O(q^{-1/2})$
$\displaystyle\leq\sup_{\xi\in\mathbf{F}_{q}^{d}}|\widehat{f\sigma}(\xi)|^{4}+O(q^{-1/2}),$

where we have used both $|\widehat{f\sigma}(\xi)|\leq 1+O(q^{-1/2})$ and
$\sum_{\xi\in\mathbf{F}_{q}^{d}}|\widehat{f\sigma}(\xi)|^{4}=\lVert
f\rVert_{U^{2}_{\perp}(S)}^{4}\leq 1+O(q^{-1/2})$. ∎

###  7.2. Proof of 7

In order to use the Fourier bias from 10 to prove 7, we will construct a
sequence of Bohr sets, refining at each stage, until we arrive at
$B^{\prime}\prec_{\epsilon}B$ for which $\nu_{B^{\prime}}*(f\sigma)$ and
$\nu_{B}*(f\sigma)$ are close in an $L^{2}$ sense. This will require a number
of technical lemmas, the first of which indicates that these two convolutions
are somewhat orthogonal.

######  Lemma 11.

Let $\epsilon\in(0,1)$, $B=B(\Gamma,\rho)$ a regular Bohr set with
$B^{\prime}\prec_{\epsilon}B$, and $f:\mathbf{F}_{q}^{d}\rightarrow\mathbf{C}$
with $|f|\leq 1$. If $d\geq 2$ and $q\geq q(|\Gamma|,\rho,\epsilon)$, then

$\lVert\nu_{B^{\prime}}*(f\sigma)\rVert_{2}^{2}-\lVert\nu_{B}*(f\sigma)\rVert_{2}^{2}\geq\lVert\nu_{B^{\prime}}*(f\sigma)-\nu_{B}*(f\sigma)\rVert_{2}^{2}+O(\epsilon).$

###### Proof.

By expanding the square,
$\lVert\nu_{B^{\prime}}*(f\sigma)-\nu_{B}*(f\sigma)\rVert_{2}^{2}$ is equal to

$\lVert\nu_{B^{\prime}}*(f\sigma)\rVert_{2}^{2}-2\mathbf{E}_{x}\nu_{B^{\prime}}*(f\sigma)(x)\nu_{B}*(f\sigma)(x)+\lVert\nu_{B}*(f\sigma)\rVert_{2}^{2},$

so it suffices to show that

$\lVert\nu_{B}*(f\sigma)\rVert_{2}^{2}\leq\mathbf{E}_{x}\nu_{B^{\prime}}*(f\sigma)(x)\nu_{B}*(f\sigma)(x)+O(\epsilon)$

Using the relationship $B^{\prime}\prec_{\epsilon}B$, we can apply 4 to see

$\displaystyle\lvert\widehat{\mu}_{B}(\xi)\rvert$
$\displaystyle=\lvert\mathbf{E}_{x\in B}\chi(x\cdot\xi)\rvert$
$\displaystyle\leq\lvert\mathbf{E}_{x\in B}\mathbf{E}_{y\in
B^{\prime}}\chi((x+y)\cdot\xi)\rvert+\epsilon$
$\displaystyle=\lvert\mathbf{E}_{x\in B}\chi(x\cdot\xi)\mathbf{E}_{y\in
B^{\prime}}\chi(y\cdot\xi)\rvert+\epsilon$
$\displaystyle\leq\lvert\widehat{\mu}_{B^{\prime}}(\xi)\rvert+\epsilon.$

in which case we apply Plancherel and this observation for

$\lVert\nu_{B}*(f\sigma)\rVert_{2}^{2}=\sum_{\xi\in\mathbf{F}_{q}^{d}}\lvert\widehat{\mu}_{B}(\xi)\rvert^{8}\lvert\widehat{f\sigma}(\xi)\rvert^{2}\leq\sum_{\xi\in\mathbf{F}_{q}^{d}}\lvert\widehat{\mu}_{B}(\xi)\rvert^{4}\left(\lvert\widehat{\mu}_{B^{\prime}}(\xi)\rvert+\epsilon\right)^{4}\lvert\widehat{f\sigma}(\xi)\rvert^{2}$

Writing $\beta=|B|/q^{d}$, we can apply Plancherel and (5) for

$\sum_{\xi\in\mathbf{F}_{q}^{d}}|\widehat{\mu_{B}}(\xi)|^{4}|\widehat{f\sigma}(\xi)|^{2}=\lVert\mu_{B}*\mu_{B}*f\sigma(x)\rVert_{2}^{2}\leq
1+O(\beta^{-1}q^{-1/2}).$

Using the bound (4), we see that we can take $q$ sufficiently large and apply
Plancherel again to conclude

$\displaystyle\lVert\nu_{B}*f\sigma\rVert_{2}^{2}$
$\displaystyle\leq\sum_{\xi\in\mathbf{F}_{q}^{d}}|\widehat{\mu}_{B}(\xi)|^{4}|\widehat{\mu}_{B^{\prime}}(\xi)|^{4}|\widehat{f\sigma}(\xi)|^{2}+O(\epsilon)$
$\displaystyle=\mathbf{E}_{x}\nu_{B^{\prime}}*(f\sigma)(x)\nu_{B}*(f\sigma)(x)+O(\epsilon).\qed$

With 11 in hand, we are ready to translate 10 into an energy increment,
showing that Fourier bias leads to a Bohr set refinement with increased
$L^{2}$ energy.

######  Proposition 12.

Let $\eta\in(0,1)$, $\epsilon\in(0,c\eta^{8})$ for $c>0$ sufficiently small,
$B=B(\Gamma,\rho)$ a regular Bohr set, and
$f:\mathbf{F}_{q}^{d}\rightarrow\mathbf{C}$ with $|f|\leq 1$. If $d\geq 2$,
$q\geq q(|\Gamma|,\rho,\eta,\epsilon)$, and

$\lVert f-\nu_{B}*(f\sigma)\rVert_{U^{2}_{\perp}(S)}\geq\eta,$

then there exists $B^{\prime}=B(\Gamma^{\prime},\rho^{\prime})$ with
$B^{\prime}\prec_{\epsilon}B$,
$\lvert\Gamma^{\prime}\rvert\leq\lvert\Gamma\rvert+1$, $\rho^{\prime}\geq
c(|\Gamma|,\epsilon)\rho$ and

$\lVert\nu_{B^{\prime}}*(f\sigma)\rVert_{L^{2}(S)}^{2}\geq\lVert\nu_{B}*(f\sigma)\rVert_{L^{2}(S)}^{2}+c\eta^{8}.$

###### Proof.

From the inverse theorem, $\lVert
f-\nu_{B}*(f\sigma)\rVert_{U^{2}_{\perp}(S)}\geq\eta$ implies the existence of
some $\gamma\in\mathbf{F}_{q}^{d}$ with

$|\widehat{f\sigma}(\gamma)-[(\nu_{B}*(f\sigma))\sigma]^{\wedge}(\gamma)|\geq\eta^{4}+O(q^{-1/32}).$

Expanding $[(\nu_{B}*(f\sigma))\sigma]^{\wedge}(\gamma)$, we can apply (1) for

$\displaystyle[(\nu_{B}*(f\sigma))\sigma]^{\wedge}(\gamma)$
$\displaystyle=\sum_{\xi\in\mathbf{F}_{q}^{d}}\widehat{\sigma}(\xi)\widehat{\mu}_{B}(\gamma-\xi)^{4}\widehat{f\sigma}(\gamma-\xi)$
$\displaystyle=\widehat{\mu}_{B}(\gamma)^{4}\widehat{f\sigma}(\gamma)+O\Big{(}q^{-1/2}\sum_{\gamma}|\widehat{\mu}_{B}(\xi)|^{4}|\widehat{f\sigma}(\xi)|\Big{)}$

Applying the uniform bounds $|\widehat{\mu}_{B}(\xi)|\leq 1$ and
$\lvert\widehat{f\sigma}(\xi)\rvert\leq 1+O(q^{-1/2})$ together with
Plancherel, we have

(15)
$[(\nu_{B}*(f\sigma))\sigma]^{\wedge}(\gamma)=\widehat{\mu}_{B}(\gamma)^{4}\widehat{f\sigma}(\gamma)+O(\beta^{-1}q^{-1/2}).$

in which case we can ensure $q$ is sufficiently large and refine our inverse
statement to

(16)
$|\widehat{f\sigma}(\gamma)||1-\widehat{\mu}_{B}(\gamma)^{4}|\geq\eta^{4}/2.$

Set $\Gamma^{\prime}=\Gamma\cup\\{\gamma\\}$. It follows from [26, Lemma 4.25]
that for any $\epsilon\in(0,1)$ we can choose a radius
$\rho^{\prime}\in[\epsilon\rho/(400|\Gamma|),\epsilon\rho/(200|\Gamma|)]$ such
that $B^{\prime}:=B(\Gamma^{\prime},\rho^{\prime})$ is regular, in which case
$B^{\prime}\prec_{\epsilon}B$. From 11,

$\lVert\nu_{B^{\prime}}*(f\sigma)\rVert_{2}^{2}-\lVert\nu_{B}*(f\sigma)\rVert_{2}^{2}\geq\lVert\nu_{B^{\prime}}*(f\sigma)-\nu_{B}*(f\sigma)\rVert_{2}^{2}+O(\epsilon),$

so we would like to provide a lower bound for

$\lVert\nu_{B^{\prime}}*f\sigma-\nu_{B}*f\sigma\rVert_{2}^{2}=\sum_{\xi\in\mathbf{F}_{q}^{d}}\lvert\widehat{\mu}_{B^{\prime}}(\xi)^{4}-\widehat{\mu}_{B}(\xi)^{4}\rvert^{2}\lvert\widehat{f\sigma}(\xi)\rvert^{2}\geq\lvert\widehat{\mu}_{B^{\prime}}(\gamma)^{4}-\widehat{\mu}_{B}(\gamma)^{4}\rvert^{2}\lvert\widehat{f\sigma}(\gamma)\rvert^{2}$

As $\gamma\in\Gamma^{\prime}$,
$\lvert\widehat{\mu}_{B^{\prime}}(\gamma)-1\rvert\leq\rho^{\prime}\leq\epsilon$.
Combining this with $|\widehat{f\sigma}(\gamma)|\leq 1+O(q^{-1/2})$ and (16),

$\lvert\widehat{\mu}_{B^{\prime}}(\gamma)^{4}-\widehat{\mu}_{B}(\gamma)^{4}\rvert^{2}\lvert\widehat{f\sigma}(\gamma)\rvert^{2}\geq|1-\widehat{\mu}_{B}(\gamma)^{4}|^{2}|\widehat{f\sigma}(\gamma)|^{2}+O(\epsilon)\geq\eta^{8}/4+O(\epsilon).$

Provided $\epsilon\leq c\eta^{8}$ for a sufficiently small absolute constant
$c>0$, we have managed to show

$\lVert\nu_{B^{\prime}}*(f\sigma)\rVert_{2}^{2}-\lVert\nu_{B}*(f\sigma)\rVert_{2}^{2}\geq
c\eta^{8}.$

For the same conclusion with $L^{2}(S)$ norms, we can apply (15) for

$\displaystyle\lVert\nu_{B}*(f\sigma)\rVert_{L^{2}(S)}^{2}$
$\displaystyle=\sum_{\xi\in\mathbf{F}_{q}^{d}}[\nu_{B}*(f\sigma)\sigma]^{\wedge}(\xi)\overline{\widehat{\mu}_{B}(\xi)^{4}\widehat{f\sigma}(\xi)}$
$\displaystyle=\lVert\nu_{B}*(f\sigma)\rVert_{2}^{2}+O(\beta^{-2}q^{-1/2}).$

and ensure $q$ is sufficiently large. ∎

Next we iterate 12 in order to find a Bohr set refinement for which
$\nu_{B^{\prime}}*(f\sigma)$ approximates $f$ well in a $U^{2}_{\perp}(S)$
sense.

######  Proposition 13.

Let $\eta\in(0,1)$, $\epsilon\in(0,c\eta^{8})$ for $c>0$ sufficiently small,
$B=B(\Gamma,\rho)$ a regular Bohr set, and
$f:\mathbf{F}_{q}^{d}\rightarrow\mathbf{C}$ with $|f|\leq 1$. If $d\geq 2$ and
$q\geq q(|\Gamma|,\rho,\eta,\epsilon)$, then there exists
$B^{\prime}=B(\Gamma^{\prime},\rho^{\prime})$ with
$B^{\prime}\prec_{\epsilon}B$, $\lvert\Gamma^{\prime}\rvert\leq
C(\lvert\Gamma\rvert,\eta)$, $\rho^{\prime}\geq
c(|\Gamma|,\eta,\epsilon)\rho$, and

$\lVert f-\nu_{B^{\prime}}*(f\sigma)\rVert_{U^{2}_{\perp}(S)}\leq\eta.$

###### Proof.

We will construct a sequence of Bohr sets

$B_{j+1}\prec_{\epsilon}B_{j}\prec_{\epsilon}\cdots\prec_{\epsilon}B_{1}\prec_{\epsilon}B_{0}$

where we write $B_{j}=B(\Gamma_{j},\rho_{j})$ and start with $B_{0}=B$. Our
goal is find $j\geq 1$ for which

(17) $\lVert f-\nu_{B_{j}}*(f\sigma)\rVert_{U^{2}_{\perp}(S)}\leq\eta.$

We set $\Gamma_{1}=\Gamma$ and choose
$\rho_{1}\in[\epsilon\rho/(400|\Gamma|),\epsilon\rho/(200|\Gamma|)]$ so that
$B_{1}\prec_{\epsilon}B_{0}$. Starting with $j=1$ and iterating, we are done
if (17) is satisfied. If (17) is false for $j\geq 1$, then by applying 12 with
$q$ sufficiently large, we obtain $B_{j+1}$ for which
$B_{j+1}\prec_{\epsilon}B_{j}$,
$\lvert\Gamma_{j+1}\rvert\leq\lvert\Gamma\rvert+j$, $\rho_{j+1}\geq
c(\lvert\Gamma\rvert,\eta,j)\rho$, and

$\lVert\nu_{B_{j+1}}*(f\sigma)\rVert_{L^{2}(S)}^{2}\geq\lVert\nu_{B_{j}}*(f\sigma)\rVert_{L^{2}(S)}^{2}+c\eta^{8}.$

From (6), we can ensure that that for every $j\geq 0$, we have

$\lVert\nu_{B_{j+1}}*(f\sigma)\rVert_{L^{2}(S)}^{2}\leq 2,$

in which case it is clear that we can iterate this process at most
$O(\eta^{-8})$ times before we arrive at some $j$ for which (17) is satisfied.
∎

Finally, we iterate 13 to find a Bohr set $B$ for which $\nu_{B}*(f\sigma)$
approximates $f$ arbitrarily well in the $U^{2}_{\perp}(S)$ norm, at the cost
of an $L^{2}$-error term. This is necessary since shrinking $\eta$ in 13 leads
to a loss of control of the size of the approximating Bohr set.

Letting $\epsilon>0$, we will construct a sequence of Bohr sets

$B_{J}\prec_{\epsilon}\cdots\prec_{\epsilon}B_{1}\prec_{\epsilon}B_{0}$

where we write $B_{j}=B(\Gamma_{j},\rho_{j})$. We start with
$\Gamma_{0}=\\{0\\}$, $\rho_{0}=1$, and $\eta_{0}=\varphi(1,1)$, in which case
$B_{0}=\mathbf{F}_{q}^{d}$. For $1\leq j\leq J$, we set
$\eta_{j}=\varphi(\lvert\Gamma_{j-1}\rvert^{-1},\rho_{j-1})$, ensure
$\epsilon\in(0,c\eta_{j}^{8})$, and apply 13 to obtain $B_{j}$ with
$B_{j}\prec_{\epsilon}B_{j-1}$, $\lvert\Gamma_{j}\rvert\leq
C(|\Gamma_{j-1}|,\eta_{j-1})$ and $\rho_{j}\geq c(|\Gamma_{j-1}|,\eta_{j-1})$
such that

$\lVert f-\nu_{B_{j}}*(f\sigma)\rVert_{U^{2}_{\perp}(S)}\leq\eta_{j}$

for sufficiently large $p$. We claim there exists some $j$ for which

(18)
$\lVert\nu_{B_{j+1}}*(f\sigma)-\nu_{B_{j}}*(f\sigma)\rVert_{L^{2}(S)}^{2}\leq\eta^{2}.$

Using 11, we can insist $\epsilon$ is sufficiently small with respect to
$\eta$ to guarantee

$\lVert\nu_{B_{j+1}}*(f\sigma)-\nu_{B_{j}}*(f\sigma)\rVert_{L^{2}(S)}^{2}\leq\lVert\nu_{B_{j+1}}*(f\sigma)\rVert_{L^{2}(S)}^{2}-\lVert\nu_{B_{j}}*(f\sigma)\rVert_{L^{2}(S)}^{2}+\eta^{2}/2.$

If
$\lVert\nu_{B_{j+1}}*(f\sigma)\rVert_{L^{2}(S)}^{2}\leq\lVert\nu_{B_{j}}*(f\sigma)\rVert_{L^{2}(S)}^{2}$,
then (18) would follow. Suppose instead that

$\lVert\nu_{B_{j+1}}*(f\sigma)\rVert_{L^{2}(S)}^{2}>\lVert\nu_{B_{j}}*(f\sigma)\rVert_{L^{2}(S)}^{2}$

for all $j$. From (6), $\lVert\nu_{B_{j}}*(f\sigma)\rVert_{L^{2}(S)}^{2}\leq
2$ for sufficiently large $q$, in which case the
$\lVert\nu_{B_{j}}*(f\sigma)\rVert_{L^{2}(S)}^{2}$ form a bounded increasing
sequence. It follows that we can take $J=O(\eta^{-2})$ and conclude (18) for
some now fixed $1\leq j\leq J$. We are then finished by setting $B=B_{j}$ and

$\displaystyle f_{1}$ $\displaystyle=\nu_{B_{j}}*(f\sigma)$ $\displaystyle
f_{2}$ $\displaystyle=f-\nu_{B_{j+1}}*(f\sigma)$ $\displaystyle f_{3}$
$\displaystyle=\nu_{B_{j+1}}*(f\sigma)-\nu_{B_{j}}*(f\sigma).$

This completes the proof of 7.∎

## Appendix: Necessity of the Spherical Condition

By an isometry of $\mathbf{F}_{q}^{d}$ we mean a linear map
$U:\mathbf{F}_{q}^{d}\to\mathbf{F}_{q}^{d}$ so that $|U(x)|^{2}=|x|^{2}$ for
all $x$. It is easy to see that $U(x)\cdot U(y)=x\cdot y$ for all $x$ and $y$,
hence $U^{T}U=UU^{T}=I$. We call a set $X=\\{x_{0},\ldots,x_{k}\\}$ _non-
degenerate_ if $V\cap V^{\bot}=\\{0\\}$ for the subspace
$V=\operatorname{Span}(X-X)$.

######  Lemma 14.

Suppose $X$ is a non-degenerate $k+1$-point configuration and
$Y=\\{y_{0},\ldots,y_{k}\\}$ is isometric to $X$. Then there exists a vector
$z$ and an isometry $U$ of $\mathbf{F}_{q}^{d}$ so that $Y=z+U(X)$.

###### Proof.

By performing a translation we can assume $x_{0}=y_{0}=0$. Assume that the
vertices are labeled so that $|y_{i}-y_{j}|^{2}=|x_{i}-x_{j}|^{2}$ for all
$0\leq i\leq j\leq k$. Let $V$ and $W$ be the subspace spanned by the sets $X$
and $Y$ respectively. Since $X$ and hence $Y$ are non-degenerate, we have that
the map $\phi:x_{i}\to y_{i},\ 1\leq i\leq k$ extends to an isometry
$\Phi:V\to W$. By Witt’s extension theorem [27] there is an orthogonal
transformation $U:\mathbf{F}_{q}^{d}\to\mathbf{F}_{q}^{d}$ extending the map
$\Phi$. Clearly $U(X)=Y$. ∎

Let us reformulate the property that a set $X$ is non-degenerate. Assume
$X=\\{0,x_{1},\dots x_{k}\\}$ and $X^{\prime}=\\{x_{1},\ldots,x_{k}\\}$ is a
linearly independent set such that
$\operatorname{Span}(X^{\prime})=\operatorname{Span}(X)$. Let $M$ be the
symmetric $l\times l$ matrix with entries $m_{ij}=x_{i}\cdot x_{j}$ for $1\leq
i,\,j\leq l$.

######  Lemma 15.

The set $X$ is non-degenerate if and only if the associated inner product
matrix $M$ has maximal rank.

###### Proof.

Clearly $X$ is non-degenerate if and only if $X^{\prime}$ is linearly
independent. Let $V=\operatorname{Span}(X^{\prime})$ and assume that $0\neq
v\in V$ such that $v\bot V$. Then $v=\sum_{i=1}^{l}b_{i}x_{i}$ and $v\cdot
x_{j}=\sum_{i=1}^{l}m_{ji}b_{i}=Mb_{j}=0$ for all $1\leq j\leq l$. Thus $Mb=0$
for a non-zero vector $b=(b_{j})$ and $M$ has rank less than $l$. Conversely
if $\operatorname{rank}(M)<l$ then $Mb=0$ for some $b=(b_{j})\neq 0$ and the
vector $v=\sum_{i}b_{i}x_{i}$ is orthogonal to
$V=\operatorname{Span}(X^{\prime})=\operatorname{Span}(X)$. ∎

Note that the inner product matrix drops rank if some non-trivial algebraic
relations between the distances $|x_{i}-x_{j}|^{2}$ of the points
$x_{i},\,x_{j}$ of $X$, hence generic configurations are non-degenerate. Next,
we show that the spherical condition is necessary over finite fields as well,
at least for non-degenerate sets. We follow the proof in [6] with some minor
modifications. First we give the following characterization of spherical sets.

######  Lemma 16.

Let $X=\\{x_{0},x_{1},\ldots,x_{k}\\}\subseteq\mathbf{F}_{q}^{d}$. Then $X$ is
spherical if and only if the following holds. For every
$c_{0},c_{1},\ldots,c_{k}\in\mathbf{F}_{q}$ if

$(I)\quad\sum_{i=0}^{k}c_{i}=0,\ \ \textit{ and}\ \ \ \
\sum_{i=0}^{k}c_{i}x_{i}=0,\ \ \textit{then also}$
$(II)\quad\sum_{i=0}^{k}c_{i}|x_{i}|^{2}=0.$

###### Proof.

Suppose $X$ is spherical, that is $|x_{i}-z|^{2}=r$ for $0\leq i\leq k$ for
some $z\in\mathbf{F}_{q}^{d}$ and $r\in\mathbf{F}_{q}$. Then
$|x_{i}|^{2}=r-|z|^{2}+2x_{i}\cdot z$ for all $i$, hence

$\sum_{i=0}^{k}c_{i}|x_{i}|^{2}=\sum_{i=0}^{k}c_{i}(|z|^{2}+r)+2\sum_{i=0}^{k}c_{i}x_{i}\cdot
z=0.$

Conversely, assume $X$ is not spherical. We show that there exists
$c_{0},\ldots,c_{k}$ satisfying (I) but not (II). One may assume that $X$ is
minimal that is $X^{\prime}$ is spherical for $X^{\prime}\subseteq X$,
$X^{\prime}\neq X$. Since a simplex is spherical there is a non-trivial linear
combination $\sum_{i=1}^{k}a_{i}(x_{i}-x_{0})=0$ and by reindexing the
vertices one may assume $a_{k}\neq 0$. Taking
$X^{\prime}=\\{x_{0},\ldots,x_{k-1}\\}$ one has $|x_{i}-z|^{2}=r$ for $1\leq
i<k$ but $|x_{k}-z|^{2}-r=b\neq 0$. Then
$|x_{i}|^{2}-|x_{0}|^{2}=2(x_{i}-x_{0})\cdot z$ for $1\leq i<k$ and
$|x_{k}|^{2}-|x_{0}|^{2}=2(x_{k}-x_{0})\cdot z\,+b$. Thus

$\sum_{i=1}^{k}a_{i}(|x_{i}|^{2}-|x_{0}|^{2})=2\sum_{i=1}^{k}a_{k}(x_{i}-x_{0})\cdot
z+a_{k}b=a_{k}b\neq 0.$

Taking $c_{0}=-\sum_{i=1}^{k}a_{i}$ and $c_{i}=a_{i}$ for $1\leq i\leq k$ the
claim follows. ∎

######  Lemma 17.

Let $X\subseteq\mathbf{F}_{q}^{d}$ be a non-spherical set of $k$ elements.
Then there exists a set $A\subseteq\mathbf{F}_{q}^{d}$ of size $|A|\geq
c_{k}q^{d}$ which does not contain any set $Y$ of the form $Y=z+U(X)$, where
$c_{k}>0$ is a constant depending only on $k$.

###### Proof.

Let $t\in\mathbf{F}_{q}^{\ast}$ such that $|\chi(t)-1|\geq 1/2$, $\chi$ being
a non-trivial character. Such a $t$ exists as $\sum_{t}\chi(t)=0$. If $X$ is
not spherical then, by scaling, there exists $c_{0},\ldots,c_{k}$ so that (I)
holds and $\sum_{i=0}c_{i}|x_{i}|^{2}=t$. If $Y=z+U(X)$ then
$\sum_{i=0}^{k}c_{i}y_{i}=0$ and moreover

$\sum_{i=0}^{k}c_{i}|y_{i}|^{2}=\sum_{i=0}^{k}c_{i}(|z|^{2}+2z\cdot
U(x_{i})+|x_{i}|^{2})=t.$

Let $B=B(\Gamma,\rho)$ be the Bohr set with $\Gamma=\\{c_{0},\ldots,c_{k}\\}$
and $\rho=\frac{1}{4(k+1)}$. By the estimate (4), $|B|\geq c_{k}q$, for some
constant $c_{k}>0$ depending only on $k$. Define
$A=\\{x\in\mathbf{F}_{q}^{d}:\ |x|^{2}\in B\\}$. Since the number of solutions
to $|x|^{2}=b$ is at least $\frac{1}{2}q^{d-1}$ uniformly for
$b\in\mathbf{F}_{q}$ for $q\geq q_{0}$, we have that
$|A|\geq\frac{c_{k}}{2}q^{d}$. We show that $Y=z+U(x)$ for some
$z\in\mathbf{F}_{q}^{d}$ and an orthogonal transformation $U$ then
$Y\subsetneq A$. Indeed if $Y\subseteq A$ then
$|\chi(c_{i}|y_{i}|^{2})-1|\leq\frac{1}{4(k+1)}$ for $0\leq i\leq k+1$. It
follows

$|\chi(\sum_{i=0}^{k}c_{i}|y_{i}|^{2})-1|=|\prod_{i=0}^{k}\chi(c_{i}|y_{i}|^{2})-1|\leq\sum_{i=0}^{k}|\chi(c_{i}|y_{i}|^{2})-1|\leq
1/4.$

This implies $|\chi(t)-1|\leq 1/4$ contradicting our choice of $t$. ∎

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