2 Definitions

Let ${\mathbf G}$ be a group with the identity $1$ . Given two sets $A,B\subset {\mathbf G}$ , define the product set of A and B as

$$ \begin{align*} AB:=\{ab ~:~ a\in{A},\,b\in{B}\}. \end{align*} $$

In a similar way, we define the higher product sets, e.g., $A^3$ is $AAA$ . Let $A^{-1} := \{a^{-1} ~:~ a\in A \}$ . As usual, having two subsets $A,B$ of a group ${\mathbf G}$ , denote by

$$ \begin{align*} {\mathsf{E}}(A,B) = |\{ (a,a_1,b,b_1) \in A^2 \times B^2 ~:~ a^{-1} b = a^{-1}_1 b_1 \}|, \end{align*} $$

the common energy of A and B. Clearly, ${\mathsf {E}}(A,B) = {\mathsf {E}}(B,A)$ , and by the Cauchy–Schwarz inequality,

(2.1) $$ \begin{align} {\mathsf{E}}(A,B) |A^{-1} B| \ge |A|^2 |B|^2 . \end{align} $$

In a little more general way, define

$$ \begin{align*} {\mathsf{E}}^L_k (A) = |\{ (a_1,\dots, a_k, b_1, \dots, b_k) \in A^{2k} ~:~ a^{-1}_1 b_1 = \dots = a^{-1}_k b_k \}| , \end{align*} $$

and, similarly,

$$ \begin{align*} {\mathsf{E}}^R_k (A) = |\{ (a_1,\dots, a_k, b_1, \dots, b_k) \in A^{2k} ~:~ a_1 b^{-1}_1 = \dots = a_k b^{-1}_k \}| . \end{align*} $$

For $k=2$ , we have ${\mathsf {E}}^L_k (A) = {\mathsf {E}}^R_k (A)$ , but for larger k, it is not the case. If there is no difference between ${\mathsf {E}}^L_k (A)$ and ${\mathsf {E}}^R_k (A)$ , then we write just ${\mathsf {E}}_k (A)$ . In this paper, we use the same letter to denote a set $A\subseteq {\mathbf G}$ and its characteristic function $A: {\mathbf G} \to \{0,1 \}$ .

First of all, we recall some notions and simple facts from the representation theory (see, e.g., [Reference Liebeck and Shalev23] or [Reference Serre27]). For a finite group ${\mathbf G}$ , let $\widehat {{\mathbf G}}$ be the set of all irreducible unitary representations of ${\mathbf G}$ . It is well known that size of $\widehat {{\mathbf G}}$ coincides with the number of all conjugate classes of ${\mathbf G}$ . For $\rho \in \widehat {{\mathbf G}}$ , denote by $d_\rho $ the dimension of this representation. Thus, ${\mathbf G}$ is a quasi-random group in the sense of Gowers (see [Reference Gowers7]) iff $d_{\rho } \ge |{\mathbf G}|^\varepsilon $ , where $\varepsilon>0$ and $\rho $ is any nontrivial irreducible unitary representation of ${\mathbf G}$ . We write $\langle \cdot , \cdot \rangle $ for the corresponding Hilbert–Schmidt scalar product $\langle A, B \rangle = \langle A, B \rangle _{HS}:= \mathrm {tr} (AB^*)$ , where $A,B$ are any two matrices of the same sizes. Put $| A\| = \sqrt {\langle A, A \rangle }$ . Finally, it is easy to check that for any matrices $A,B$ , one has $\| AB\| \le \| A\|_{o} \| B\|$ and $\| A\|_{o} \le \| A \|$ , where the operator $l^2$ -norm $\| A\|_{o}$ is just the maximal singular value of A.

For any function $f:{\mathbf G} \to \mathbb {C}$ and $\rho \in \widehat {{\mathbf G}}$ , define the matrix $\widehat {f} (\rho )$ , which is called the Fourier transform of f at $\rho $ by the formula

(2.2) $$ \begin{align} \widehat{f} (\rho) = \sum_{g\in {\mathbf G}} f(g) \rho (g). \end{align} $$

Then, the inverse formula takes place

(2.3) $$ \begin{align} f(g) = \frac{1}{|{\mathbf G}|} \sum_{\rho \in \widehat{{\mathbf G}}} d_\rho \langle \widehat{f} (\rho), \rho (g^{-1}) \rangle , \end{align} $$

and the Parseval identity is

(2.4) $$ \begin{align} \sum_{g\in {\mathbf G}} |f(g)|^2 = \frac{1}{|{\mathbf G}|} \sum_{\rho \in \widehat{{\mathbf G}}} d_\rho \| \widehat{f} (\rho) \|^2 . \end{align} $$

The main property of the Fourier transform is the convolution formula

(2.5) $$ \begin{align} \widehat{f*g} (\rho) = \widehat{f} (\rho) \widehat{g} (\rho) , \end{align} $$

where the convolution of two functions $f,g : {\mathbf G} \to \mathbb {C}$ is defined as

$$ \begin{align*} (f*g) (x) = \sum_{y\in {\mathbf G}} f(y) g(y^{-1}x) . \end{align*} $$

Given a function $f : {\mathbf G} \to \mathbb {C}$ and a positive integer k, we write $f^{(k)} = f^{(k-1)} * f$ for the kth convolution of f. Now, let $k\ge 2$ be an integer and $f_j : {\mathbf G} \to \mathbb {C}$ , $j\in [2k]$ be any functions. Denote by $\mathcal {C}$ the operator of convex conjugation. As in [Reference Shkredov31], define

(2.6) $$ \begin{align} {\mathsf{T}}_k (f_1,\dots, f_{2k}) = \frac{1}{|{\mathbf G}|} \sum_{\rho \in \widehat{{\mathbf G}}} d_\rho \langle \prod_{j=1}^k \mathcal{C}^j \widehat{f}_j (\rho), \prod_{j=k+1}^{2k} \mathcal{C}^j \widehat{f}_j (\rho) \rangle . \end{align} $$

Put ${\mathsf {T}}_k (f) = {\mathsf {T}}_k (f,\dots , f_{})$ . For example, we have, clearly, ${\mathsf {T}}_2 (A) = {\mathsf {E}} (A)$ . It is easy to see that ${\mathsf {T}}^{1/2k}_k (f)$ defines a norm of a function f (see [Reference Shkredov31]). This fact follows from the following inequality [Reference Shkredov31, Lemma 10]:

(2.7) $$ \begin{align} {\mathsf{T}}^{2k}_k (f_1,\dots, f_{2k}) \le \prod_{j=1}^{2k} {\mathsf{T}}_k (f_j) . \end{align} $$

In particular, ${\mathsf {E}}(A,A^{-1}) \le {\mathsf {E}}(A)$ .

Now, let us say a few words about varieties. Having a field ${\mathbb {F}}$ , define an (affine) variety in ${\mathbb {F}}^n$ to be the set of the form

$$ \begin{align*} V = \{ (x_1,\dots, x_n) \in {\mathbb{F}}^n ~:~ p_j (x_1,\dots,x_n) = 0\, \mbox{ for all } j \} , \end{align*} $$

where $p_j \in {\mathbb {F}}[x_1,\dots ,x_n]$ . Let us recall some basic properties of varieties. General theory of varieties and schemes can be found, e.g., in [Reference Hartshorne8]. The union of any finite number of varieties is, clearly, a variety, and the intersection of any number of varieties is a variety as well. Having a set X, we denote by $\mathrm {Zcl} (X)$ a minimal (by inclusion) variety, containing X. A variety over an algebraically closed field ${\mathbb {F}}$ is irreducible if it is not the union of two proper subvarieties. Every variety has a unique (up to inclusion) decomposition into finitely many irreducible components [Reference Hartshorne8]. The dimension of V is

$$ \begin{align*} \dim(V) = \max \{ n ~:~ V\supseteq X_n \supset X_{n-1} \supset \dots \supset X_0 \neq \emptyset \} , \end{align*} $$

where $X_j$ are irreducible subvarieties of V. We will frequently use the simple fact that if $V_1 \subseteq V_2$ are two varieties and $V_2$ is irreducible, then either $V_1=V_2$ , or $\dim (V_1) < \dim (V_2)$ . A variety is absolutely irreducible if it is irreducible over $\overline {{\mathbb {F}}}$ . In this paper, we consider just these varieties.

We define the degree of any irreducible variety V with $\dim (V) = d$ as in [Reference Heintz9], namely,

$$ \begin{align*} \deg (V) = \sup \{ |L \cap V| <\infty ~:~ L \mbox{ is } (n-d)\mbox{-dimensional affine subspace in } {\mathbb{F}}^n \}. \end{align*} $$

For an arbitrary variety V, we denote $\deg (V)$ to be the sum of the degrees of its irreducible components. Recall the generalized Bézout theorem (see [Reference Heintz9, Theorem 1]): for any varieties $U,V$ , one has

(2.8) $$ \begin{align} \deg (U\cap V) \le \deg(U) \deg(V) . \end{align} $$

The signs $\ll $ and $\gg $ are the usual Vinogradov symbols. If we want to underline the dependence on a parameter M, then we write $\ll _M$ and $\gg _M$ . All logarithms are to base $2$ . Sometimes, we allow ourselves to lose logarithmic powers of $|{\mathbb {F}}|$ . In this situation, we write $\lesssim $ and $\gtrsim $ instead of $\ll $ and $\gg $ .

3 On noncommutative Gowers norms

Let ${\mathbf G}$ be a group and $A\subseteq {\mathbf G}$ be a finite set. Let $ \| A \|_{\mathcal {U}^{k}} $ be the Gowers nonnormalized kth-norm [Reference Gowers6] of the characteristic function of A (in multiplicative form; see, say, [Reference Shkredov28]):

$$ \begin{align*} \| A \|_{\mathcal{U}^{k}} = \sum_{x_0,x_1,\dots,x_k \in {\mathbf G}}\, \prod_{\vec{\varepsilon} \in \{ 0,1 \}^k} A \left( x_0 x^{\varepsilon_1}_1 \dots x^{\varepsilon_k}_k \right)\kern-1.5pt, \end{align*} $$

where $\vec {\varepsilon } = (\varepsilon _1,\dots ,\varepsilon _k)$ . For example,

$$ \begin{align*}\| A \|_{\mathcal{U}^{2}} = \sum_{x_0,x_1,x_2\in {\mathbf G}} A(x_0) A(x_0 x_1 ) A(x_0 x_2) A(x_0 x_1 x_2) = {\mathsf{E}} (A) \end{align*} $$

is the energy of A, and $\| A \|_{\mathcal {U}^{1}} = |A|^2$ . For any $\vec {s} = (s_1,\dots ,s_k) \in {\mathbf G}^k$ , put

(3.1) $$ \begin{align} A_{\vec{s}} (x) = \prod_{\vec{\varepsilon} \in \{ 0,1 \}^k} A \left(x s^{\varepsilon_1}_1 \dots s^{\varepsilon_k}_k \right)\kern-1pt, \end{align} $$

and similar for an arbitrary function $f: {\mathbf G} \to {\mathbb {C}}$ , namely,

$$ \begin{align*}f_{\vec{s}} (x) = \prod_{\vec{\varepsilon} \in \{ 0,1 \}^k} \mathcal{C}^{\varepsilon _1+\dots+\varepsilon _k} f \left(x s^{\varepsilon_1}_1 \dots s^{\varepsilon_k}_k \right)\kern-1pt,\end{align*} $$

where $\mathcal {C}$ is the operator of the conjugation. For example, $A_s (x) = A(x) A(xs)$ or, in other words, $A_s = A\cap (As^{-1})$ . Then, obviously,

(3.2) $$ \begin{align} \| A \|_{\mathcal{U}^{k}} = \sum_{\vec{s}} |A_{\vec{s}}| . \end{align} $$

Also, note that

(3.3) $$ \begin{align} \| A \|_{\mathcal{U}^{k+1}} = \sum_{\vec{s}} |A_{\vec{s}}|^2 . \end{align} $$

Moreover, the induction property for Gowers norms holds (it follows from the definitions or see [Reference Gowers6])

(3.4) $$ \begin{align} \| A \|_{\mathcal{U}^{k+1}} = \sum_{s \in A^{-1}A} \| A_s \|_{\mathcal{U}^{k}} , \end{align} $$

e.g., in particular,

$$ \begin{align*}\| A \|_{\mathcal{U}^{3}} = \sum_{s \in A^{-1}A} {\mathsf{E}}(A_s) . \end{align*} $$

The Gowers norms enjoy the following weak commutativity property. Namely, let $k=n+m$ and $\vec {s} = (s_1,\dots ,s_k) = (\vec {u}, \vec {v})$ , where the vectors $\vec {u}, \vec {v}$ have lengths n and m, correspondingly. We have

$$ \begin{align*} \| A \|_{\mathcal{U}^k} = \sum_{\vec{s}} \sum_x \prod_{\vec{\varepsilon} \in \{ 0,1 \}^k} A \left( x s^{\varepsilon_1}_1 \dots s^{\varepsilon_k}_k \right) \end{align*} $$

(3.5) $$ \begin{align} = \sum_{\vec{u},\vec{v}}\, \sum_{x} \prod_{\vec{\eta} \in \{ 0,1 \}^m}\, \prod_{\vec{\omega } \in \{ 0,1 \}^n} A \left( u^{\eta_1}_1 \dots u^{\eta_m}_m x v^{\omega _1}_1 \dots v^{\omega _n}_n \right)\kern-1.5pt. \end{align} $$

In particular, $\| A^{-1} \|_{\mathcal {U}^k} = \| A \|_{\mathcal {U}^k}$ and $\| gA \|_{\mathcal {U}^k} = \| Ag \|_{\mathcal {U}^k} = \| A \|_{\mathcal {U}^k}$ for any $g\in {\mathbf G}$ . To obtain (3.5), just make the changing of variables $u_j x = x \tilde {u}_j$ for $j\in [m]$ .

It was proved in [Reference Gowers6] that ordinary Gowers kth-norms of the characteristic function of any subset of an abelian group ${\mathbf G}$ are connected to each other. In [Reference Shkredov28], the author shows that the connection for the nonnormalized norms does not depend on the size of the group ${\mathbf G}$ . Here, we formulate a particular case of Proposition 35 from [Reference Shkredov28], which relates $\| A \|_{\mathcal {U}^{k}}$ and $\| A \|_{\mathcal {U}^{2}}$ (see Remark 36 here).

Lemma 1 Let A be a finite subset of a commutative group ${\mathbf G}$ . Then, for any integer $k\ge 1$ , one has

$$ \begin{align*}\| A \|_{\mathcal{U}^{k+1}} \ge \frac{\| A \|^{(3k-2)/(k-1)}_{\mathcal{U}^{k}}}{\| A \|^{2k/(k-1)}_{\mathcal{U}^{k-1}}} . \end{align*} $$

In particular,

$$ \begin{align*}\| A \|_{\mathcal{U}^{k}} \ge {\mathsf{E}}(A)^{2^k-k-1} |A|^{-(3\cdot 2^k -4k -4)} . \end{align*} $$

Actually, one can derive Lemma 1 from Lemma 2, but to prove this more general result, we need an additional notation and arguments.

Given two functions $f,g: {\mathbf G} \to \mathbb {C}$ and an integer $k\ge 0$ , consider the “scalar product”

$$ \begin{align*} \langle f, g \rangle_k := \sum_{\vec{s}, t} \sum_{x} f_{\vec{s}} (x) \overline{g_{\vec{s}} (xt)} = \overline{\langle g, f \rangle_k} , \end{align*} $$

where $\vec {s} = (s_1,\dots , s_k)$ . For example, $\langle A, B \rangle _1 = {\mathsf {E}}(A,B)$ , $\langle A, B \rangle _0 = |A||B|$ , $\langle A, A \rangle _k = \|A\|_{\mathcal {U}^{k+1}}$ , and $\langle A, 1 \rangle _k = |{\mathbf G}| \|A\|_{\mathcal {U}^{k}}$ (for finite group ${\mathbf G}$ ). Clearly, $\langle f, g \rangle _0 = \left ( \sum _x f(x) \right ) \left ( \overline {\sum _x g(x)} \right )$ , but for $k\ge 1$ , it is easy to see that $\langle f, g \rangle _k \ge 0$ , because $\langle f, g \rangle _k = \sum _{\vec {s}_*, s_k} |(f_{\vec {s}_*} * \overline {\tilde {g}}_{\vec {s}_*}) (s_k)|^2 \ge 0$ , where $\vec {s}_* = (s_1,\dots , s_{k-1})$ , and $\tilde {g}(x) := g(x^{-1})$ . Also, note that

(3.6) $$ \begin{align} \langle A, B \rangle_k = \sum_{\vec{s}} |A_{\vec{s}}| |B_{\vec{s}}| = \sum_{\vec{s}_*} \sum_{s_k} |A_{\vec{s}_*} \cap A_{\vec{s}_*} s_k| |B_{\vec{s}_*} \cap B_{\vec{s}_*} s_k| . \end{align} $$

Lemma 2 Let ${\mathbf G}$ be a commutative group and $f,g : {\mathbf G} \to \mathbb {C}$ be functions. Then, for any integer $k\ge 1$ , one has

(3.7) $$ \begin{align} \langle f, g\rangle^{3+1/k}_k \le \langle f, g\rangle^{2}_{k-1} \langle f, g\rangle^{}_{k+1} \| f\|^{1/k}_{\mathcal{U}^k} \| g\|^{1/k}_{\mathcal{U}^k} , \end{align} $$

and hence for an arbitrary $k\ge 2$ and any sets $A,B \subseteq {\mathbf G}$ , the following holds:

(3.8) $$ \begin{align} {\mathsf{E}}(A,B) \le (|A||B|)^{\frac{3}{2} - \frac{\beta(k+2)}{2}} \langle A,B \rangle^{\beta}_k , \end{align} $$

where $\beta = \beta (k) \in [4^{-k}, 2^{-k+1}]$ .

For any (not necessary commutative) group ${\mathbf G}$ , if $\|A\|_{\mathcal {U}^k} \le |A|^{k+1-c}$ , where $c>0$ , then ${\mathsf {E}}(A) \le |A|^{3-c_*}$ with $c_* = c_*(c,k)>0$ .

Proof We have

(3.9) $$ \begin{align} \sigma:= \langle f, g \rangle_k = \sum_{\vec{s}, t} \sum_{x} f_{\vec{s}} (x) \overline{g_{\vec{s}} (xt)}, \end{align} $$

and our first task to estimate the size of the set of $(\vec {s},t)$ in the last formula. Basically, we consider two cases. If the summation in (3.9) is taken over the set

(3.10) $$ \begin{align} Q := \{ \vec{s} ~:~ \sum_t g_{\vec{s}} (t) \ge \sigma (2k \| f \|_{\mathcal{U}^{k}} )^{-1} \} , \end{align} $$

then it gives us $(1-1/2k)$ proportion of $\sigma $ . Cardinality of the set Q can be estimated as

$$ \begin{align*} |Q| \cdot \sigma (2k \| f \|_{\mathcal{U}^{k}} )^{-1} \le \|g \|_{\mathcal{U}^{k}}, \end{align*} $$

and hence $|Q| \le 2k \| f \|_{\mathcal {U}^{k}} \| g \|_{\mathcal {U}^{k}} \sigma ^{-1}$ . Now, we fix any $j\in [k]$ (without any loss of generality, we can assume that $j=1$ ), put $\vec {s}_* = (s_2,\dots ,s_{k})$ , and consider

(3.11) $$ \begin{align} Q_1 := \{ (\vec{s}_*, t) ~:~ \sum_{s_1} f_{\vec{s}_*} (s_1) \overline{g_{\vec{s}_*} (ts_1)} \ge \sigma (2k \langle f,g \rangle _{k-1} )^{-1} \}. \end{align} $$

Using the changing of the variables as in (3.5) (here we appeal to the commutativity of the group ${\mathbf G}$ ) and arguing as above, we have

$$ \begin{align*} |Q_1| \cdot \sigma (2k \langle f,g \rangle _{k-1} )^{-1} \le \sum_{\vec{s}_*, t} \sum_{s_1} f_{\vec{s}_*} (s_1) \overline{g_{\vec{s}_*} (ts_1)} \langle f,g \rangle _{k-1} . \end{align*} $$

Again, if the summation in (3.9) is taken over the set $Q_1$ , then it gives us $(1-1/2k)$ proportion of $\sigma $ . Hence, by the standard projection results (see, e.g., [Reference Bollobás and Thomason1]), we see that the summation in (3.9) is taken over a set $\mathcal {S}$ of vectors $(\vec {s},t)$ of size at most

(3.12) $$ \begin{align} |\mathcal{S}| \le ( (2k)^{k+1} \| f \|_{\mathcal{U}^{k}} \| g \|_{\mathcal{U}^{k}} \sigma^{-(k+1)} )^{1/k} \langle f,g \rangle^2_{k-1} . \end{align} $$

The proof below is rather technical in the commutative case. If the reader is interested in the general case, then it is possible to avoid all calculations before (3.17), where much simpler arguments are presented to estimate the size of the set $\mathcal {S}$ .

Returning to (3.12) and using the Cauchy–Schwarz inequality, we get

$$ \begin{align*} 2^{-4}\sigma^2 \le |\mathcal{S}|\kern-1pt \sum_{\vec{s}, t} \left| \sum_{x} f_{\vec{s}} (x) \overline{g_{\vec{s}} (xt)} \right|{}^2 \kern1pt{\le}\kern1pt ( (2k)^{k+1} \| f \|_{\mathcal{U}^{k}} \| g \|_{\mathcal{U}^{k}} \sigma^{-(k+1)} )^{1/k} \langle f,g \rangle^2_{k-1} \langle f, g\rangle^{}_{k+1}, \end{align*} $$

and we have (3.7) up to a constant depending on k. Using the tensor trick (see, e.g., [Reference Tao and Vu35]), we obtain the result with the constant one.

To prove inequality (3.8), we see by induction and formula (3.7) that one has, for $l\le k$ ,

$$ \begin{align*} \langle A,B \rangle_l \le\langle A,B \rangle^{\alpha _0 (l,k)}_0 \prod_{j=1}^{k-1} (\| A\|_{\mathcal{U}^j} \| B\|_{\mathcal{U}^j})^{\alpha _j(l,k)} \cdot \langle A,B \rangle^{\beta_k(l,k)}_k , \end{align*} $$

where $\alpha _j (l,k)$ , $\beta _j (l,k)$ are some nonnegative functions. In principle, in view of (3.7), these functions can be calculated via some recurrences, but we restrict ourselves giving just crude bounds for them. We are interested in $l=1$ and k is a fixed number, and hence we write

$$ \begin{align*} {\mathsf{E}}(A,B) = \langle A,B \rangle_1 \le \langle A,B \rangle^{\alpha _0}_0 \prod_{j=1}^{k-1} (\| A\|_{\mathcal{U}^j} \| B\|_{\mathcal{U}^j})^{\alpha _j} \cdot \langle A,B \rangle^{\beta}_k . \end{align*} $$

By homogeneity, we get

(3.13) $$ \begin{align} 2= \alpha _0 + \sum_{j=1}^{k-1} \alpha _j 2^j + 2^k \beta . \end{align} $$

In particular, $\beta \le 2^{-k+1}$ . Further taking $A=B$ equals a subgroup, we obtain one more equation

(3.14) $$ \begin{align} 3= 2\alpha _0 + 2\sum_{j=1}^{k-1} \alpha _j (j+1) + (k+2) \beta . \end{align} $$

Using trivial inequalities $\|A\|_{\mathcal {U}^j} \le |A|^{j+1}$ , $\|B\|_{\mathcal {U}^j} \le |B|^{j+1}$ and formula (3.14), we derive

$$ \begin{align*} {\mathsf{E}}(A,B) \le (|A| |B|)^{\alpha _0 + \sum_{j=1}^{k-1} \alpha _j (j+1)} \langle A,B \rangle^{\beta}_k = (|A||B|)^{\frac{3}{2} - \frac{\beta(k+2)}{2}} \langle A,B \rangle^{\beta}_k, \end{align*} $$

as required. Actually, if there is a nontrivial upper bound for $\|A\|_{\mathcal {U}^j}$ (and it will be so in the next section), then the last estimate can be improved in view of Lemma 1. Furthermore, our task is to obtain a good lower bound for $\beta $ . Put $\omega _j := 3+1/j> 3$ , $j\in [k-1]$ . Using (3.7), we get

(3.15) $$ \begin{align} \prod_{j=1}^{k-1} \langle A, B \rangle^{\omega _j x_j}_j \le S \prod_{j=1}^{k-1} \left( \langle A, B \rangle^2_{j-1} \langle A, B \rangle_{j+1} \right)^{x_j}, \end{align} $$

where S is a quantity depending on $\|A\|_{\mathcal {U}^j}$ , $\|B\|_{\mathcal {U}^j}$ , which we do not specify, and let $x_j$ be some positive numbers, which we will choose (indirectly) later. For $2 \le j \le k-2$ , put

(3.16) $$ \begin{align} x_{j-1} + 2 x_{j+1} = \omega _j x_{j} . \end{align} $$

Then, we obtain from (3.15)

$$ \begin{align*} \langle A, B \rangle^{4x_1}_1 \langle A, B \rangle^{\omega_{k-1} x_{k-1}}_{k-1} \le S \langle A, B \rangle^{2x_{2}}_1 \langle A, B \rangle^{x_{k-1}}_k \langle A, B \rangle^{x_{k-2}}_{k-1} . \end{align*} $$

Now, choosing $x_{k-2} = \omega _{k-1} x_{k-1}$ , we see that $\beta = x_{k-1}/(4x_1-2x_2)$ , and it remains to estimate $x_{k-1}$ in terms of $x_1, x_2$ . But for all j, one has $\omega _j \le 4$ , hence $x_{k-1} \ge 4^{-1} x_{k-2}$ , and similarly, from $x_{j-1} + 2 x_{j+1} = \omega _j x_{j}$ , $2 \le j \le k-2$ , we get $x_{j} \ge 4^{-1} x_{j-1}$ , and hence $x_{k-1} \ge 4^{-(k-1) }x_1$ . Furthermore, summing (3.16) over $2\le j \le k-2$ and putting $T=\sum _{j=1}^{k-1} x_j$ , we obtain

$$ \begin{align*} T-x_{k-1}-x_{k-2} + 2T - 2x_1 - 2x_2 = \sum_{j=2}^{k-2} \omega _j x_j \ge 3T - 3x_1 -3x_{k-1}, \end{align*} $$

and hence

(3.17) $$ \begin{align} x_1-2x_2 \ge x_{k-2} - 2x_{k-1} = (\omega _{k-1} - 2) x_{k-1}> 0 . \end{align} $$

In particular, it gives $x_j>0$ for all $j\in [k-1]$ , and thus indeed $\beta \ge 4^{-k}$ .

Now, suppose that ${\mathbf G}$ is an arbitrary group and $\|A\|_{\mathcal {U}^k} \le |A|^{k+1-c}$ , but ${\mathsf {E}}(A) \ge |A|^3/K$ , where $K\ge 1$ is a parameter. By the noncommutative Balog–Szemerédi–Gowers theorem (see [Reference Larsen and Pink22, Theorem 32] or [Reference Tao and Vu35, Proposition 2.43 and Corollary 2.46]), there is $a\in A$ and $A_* \subseteq a^{-1}A$ , $|A_*|\gg _K |A|$ such that $|A^3_*| \ll _K |A_*|$ . We can apply the previous argument to the set $A_*$ and obtain an estimate similar to Lemma 1

$$ \begin{align*}|A|^{k+1-c} \ge \| A \|_{\mathcal{U}^{k}} \ge \| A_* \|_{\mathcal{U}^{k}} \gg_K {\mathsf{E}}(A_*)^{2^{k-2}} |A_*|^{-(3\cdot 2^{k-2} -k -1)} \end{align*} $$

(3.18) $$ \begin{align} \gg_K {\mathsf{E}}(A)^{2^{k-2}} |A|^{-(3\cdot 2^{k-2} -k -1)} . \end{align} $$

Indeed, to bound ${\mathsf {E}}(A_*)$ via $\| A_*\|_{\mathcal {U}^{k+2}}$ using the argument as in the proof above, we need to estimate the size of the set $\mathcal {S}_k$ at each step k. But clearly, $|\mathcal {S}_k| \le |A_* A_*^{-1}|^{k+1} \ll _K |A_*|^{k+1}$ , and hence, by induction, we obtain ${\mathsf {E}}^{2^k}(A_*) \ll _K |A_*|^{3\cdot 2^k-k-3} \| A_* \|_{\mathcal {U}^{k+2}}$ , as required. Finally, from (3.18), it follows that $K^{C(k)}\gg |A|^{c}$ , where $C(k)$ is a constant depending on k only. This completes the proof.▪

A closer look to the proof (see, e.g., definition (3.11)) shows that for $k=1$ , estimate (3.7) of Lemma 1 takes place for any class functions f, g. Nevertheless, for larger k, this argument does not work.

4 The proof of the main result

Let ${\mathbf G}$ be an algebraic group in an affine or projective space of dimension n over the field ${\mathbb {F}}_q$ , and let $V\subseteq {\mathbf G}$ be a variety, $d=\dim (V)$ , and $D=\deg (V)$ . If V is absolutely irreducible, then by Lang and Weil [Reference Mockenhaupt and Tao16], we know that

(4.1) $$ \begin{align} \left| |V| - q^d \right| \le (d-1)(d-2) q^{d-1/2} + A(n,d,D) q^{d-1} , \end{align} $$

where $A(n,d,D)$ is a certain constant. By sufficiently large q, we mean that $q\ge q_0 (n,d,D,$ $\dim ({\mathbf G}),\deg ({\mathbf G}))$ and all constants below are assumed to depend on $n,d,D,\dim ({\mathbf G}),\deg ({\mathbf G})$ . In particular, for an absolutely irreducible variety V, one has $q^d \ll |V| \ll q^d$ . One can think about ${\mathbf G}$ and V as varieties defined over $\mathbb {Q}$ by absolutely irreducible polynomials. Then, by the Noether theorem [Reference Liebeck, Schul and Shalev24], we know that ${\mathbf G}$ and V reduce $\mathrm {mod~} p$ to some absolutely irreducible varieties defined over p, $p\notin S({\mathbf G},V)$ , where $S({\mathbf G},V)$ is a certain finite set of the primes.

Finally, for any set $W \subseteq {\mathbf G}$ , consider the quantity

(4.2) $$ \begin{align} t = t(W) := \max_{x\in {\mathbf G},\, \Gamma \le {\mathbf G}}\, \{ |\Gamma | ~:~ x\Gamma \subseteq W,\,\, \Gamma \mbox{ is an algebraic subgroup} \}. \end{align} $$

Now, we are ready to estimate different energies of varieties in terms of the quantity $t(V)$ . We are also able to give a nontrivial bound for any sufficiently large subset of V (see inequality (4.5)). The proof relies on Lemma 2, and this lemma works better in the abelian case allowing to find concrete $\delta = \delta (d,\varepsilon )$ in (1.1) for an abelian group ${\mathbf G}$ .

Theorem 1 Let ${\mathbf G}$ be an algebraic group, $V\subseteq {\mathbf G}$ be a variety, $d=\dim (V)$ , $D=\deg (V)$ , and $t=t(V)$ . Then, for any positive integer k and all sufficiently large q, one has

(4.3) $$ \begin{align} {\mathsf{E}}_k (V) \ll_{d,D} \frac{|V|^{k+1}}{q^{k-1}} + t |V|^k . \end{align} $$

In particular, if V is absolutely irreducible and V is not a coset of a subgroup, then ${\mathsf {E}}_k (V) \ll _{d,D} |V|^{k+1 - \frac {1}{d}}$ .

Similarly, one has

(4.4) $$ \begin{align} \| V\|_{\mathcal{U}^{k}} \ll_{d,D} |V|^{k+1} q^{-\frac{k(k-1)}{2}} + |V|^2 t^{k-1} + |V|^2 \sum_{j=1}^{k-2} |V|^{j} t^{k-1-j} q^{-\frac{j(1+j)}{2}} , \end{align} $$

and for any $A\subseteq V$ , the following holds:

(4.5) $$ \begin{align} \| A\|_{\mathcal{U}^{d+1}} \ll_{d,D} t|A|^{d+1} . \end{align} $$

Proof First of all, consider the case of an absolutely irreducible V. For any $g\in {\mathbf G}$ , we have either $\dim (V \cap gV) < \dim (V)$ , or g belongs to the stabilizer ${\textrm {Stab}}(V)$ of V. It is well known that any stabilizer under any action of an algebraic group is an algebraic subgroup (but not necessary irreducible). Clearly, ${\textrm {Stab}}(V) \subseteq v^{-1}V$ for any $v\in V$ , and hence either V is a coset of an (algebraic) subgroup, and hence there is nothing to prove, or $\dim ({\textrm {Stab}}(V)) < \dim (V)$ . The degree of the variety $gV \cap V$ is at most $\deg ^2 (V)$ by inequality (2.8). Similarly, because our topological space is a Noetherian one (see, e.g., [Reference Hartshorne8, p. 5]) and ${\textrm {Stab}}(V) = \bigcap _{v\in V} v^{-1}V$ , it follows that the cardinality of ${\textrm {Stab}}(V)$ can be estimated in terms of d and D thanks to (4.1). Hence, all parameters of all appeared varieties are controlled by $d,D$ , n (and, possibly, by $\dim ({\mathbf G}), \deg ({\mathbf G}))$ . Using the Lang–Weil formula (4.1), we obtain for sufficiently large q that $q^d/2 \le |V| \le 2q^d$ , say, further $|{\textrm {Stab}}(V)| \ll q^{d-1}$ , and, similarly, for any $g\notin {\textrm {Stab}}(V)$ , one has $|V \cap gV| \ll q^{d-1}$ . Hence,

(4.6) $$ \begin{align} {\mathsf{E}}_k (V) = \sum_{g\in {\mathbf G}} |V\cap gV|^k = \sum_{g\notin {\textrm{Stab}}(V)} |V\cap gV|^k + \sum_{g\in {\textrm{Stab}}(V)} |V\cap gV|^k \end{align} $$

(4.7) $$ \begin{align} \ll (q^{d-1})^{k-1} \sum_{g\in {\mathbf G}} |V\cap gV| + q^{d-1} |V|^k \ll (q^{d-1})^{k-1} |V|^2 + q^{d-1} |V|^k \ll |V|^{k+1 - \frac{1}{d}} . \end{align} $$

Now, to obtain (4.3), we apply the same argument but before we need to consider V as a union of its irreducible components $V=\bigcup _{j=1}^s V_j$ . Clearly, $s\le \deg (V)$ . Take any $g\in {\mathbf G}$ , and consider $V\cap gV = \bigcup _{i,j=1}^s (V_i \cap g V_j)$ . If, for all $i,j\in [s]$ , one has $\dim (V_i \cap g V_j) < \dim V$ , then for such g, we can repeat the previous calculations in (4.6)–(4.7). Consider the set of the remaining g, and denote this set by B. For any $g\in B$ , there is $i,j\in [s]$ such that $\dim (V_i \cap g V_j) = \dim (V)$ . In particular, $\dim (V_i) = \dim (V_j) = \dim (V)$ and $V_i \cap g V_j = V_i = gV_j$ by irreducibility of $V_i,V_j$ . Suppose that for the same pair $(i,j)$ , there is another $g_* = g_* (i,j) \in B$ such that $g_* V_j = V_i$ . Then, $g^{-1}_* g \in {\textrm {Stab}}(V_j)$ . It follows that $g\in g_* {\textrm {Stab}}(V_j)$ , and hence the set B belongs to $\bigcup _{i,j=1}^s g_* (i,j) {\textrm {Stab}}(V_j)$ plus at most $s^2 \le \deg ^2 (V)$ points. Hence, we need to add to the computations in (4.6)–(4.7) the term

$$ \begin{align*} s^2 (t+1) |V|^k \le \deg(V)^2 (t+1) |V|^k \ll t |V|^k, \end{align*} $$

as required.

To prove bound (4.4), let us obtain a generalization of (4.3). Put ${\mathsf {E}}^{(l)}_k = {\mathsf {E}}^{(l)}_k (V) := \sum _{\vec {s}} |V_{\vec {s}}|^k$ , where $\vec {s} = (s_1,\dots ,s_l)$ and $V_{\vec {s}}$ as in (3.1). Write $\vec {s} = (\vec {s}_*,s_l)$ . Because ${\mathsf {E}}^{(l)}_k = \sum _{\vec {s}_*} {\mathsf {E}}_k (V_{\vec {s}_*})$ , it follows that by the obtained estimate (4.3),

(4.8) $$ \begin{align} {\mathsf{E}}^{(l)}_k \ll \sum_{\vec{s}_*} \left( \frac{|V_{\vec{s}_*}|^{k+1}}{q^{k-1}} + t (V_{\vec{s}_*}) |V_{\vec{s}_*}|^{k} \right) \ll q^{-(k-1)} {\mathsf{E}}^{(l-1)}_{k+1} + t {\mathsf{E}}^{(l-1)}_{k} . \end{align} $$

Here, we have used the fact that the function t on a subset of V does not exceed $t(V)$ . Furthermore, inequality (4.8) gives by induction

(4.9) $$ \begin{align} {\mathsf{E}}^{(l)}_k \ll |V|^k \sum_{j=0}^l q^{-\frac{j(2k+j-3)}{2}} |V|^j t^{l-j} . \end{align} $$

Now, applying inequality (4.3) with the parameter $k=2$ and using the notation $\vec {s} = (\vec {s}_*,s_l)$ again, we get

$$ \begin{align*} \| V\|_{\mathcal{U}^{l+1}} = \sum_{\vec{s}} |V_{\vec{s}}|^2 = \sum_{\vec{s}_*} {\mathsf{E}} (V_{\vec{s}_*}) \ll \sum_{\vec{s}_*} \left( \frac{|V_{\vec{s}_*}|^3}{q} + t |V_{\vec{s}_*}|^2 \right) = q^{-1} {\mathsf{E}}^{(l-1)}_3 + t {\mathsf{E}}^{(l-1)}_2. \end{align*} $$

Hence, in view of (4.9), we derive

$$ \begin{align*} \| V\|_{\mathcal{U}^{l+1}} \ll |V|^2 \sum_{j=0}^{l-1} |V|^{j} t^{l-1-j} (|V| q^{-\frac{j(3+j)}{2}-1} + tq^{-\frac{j(1+j)}{2}}) \end{align*} $$

$$ \begin{align*} \ll |V|^2 t^l + |V|^{l+2} q^{-\frac{l^2+l}{2}} + |V|^2 \sum_{j=1}^{l-1} |V|^{j} t^{l-j} q^{-\frac{j(1+j)}{2}} . \end{align*} $$

Finally, take any $A\subseteq V$ . As above, put $\vec {s} = (s_1,\dots ,s_{d}) = (\vec {s}_*, s_d)$ . We have $A_{\vec {s}} \subseteq V_{\vec {s}}$ . Furthermore, by (3.6),

(4.10) $$ \begin{align} \| A \|_{\mathcal{U}^{d+1}} = \sum_{\vec{s}} |A_{\vec{s}}|^2 = \sum_{\vec{s}_*} {\mathsf{E}} (A_{\vec{s}_*}) = \sum_{\vec{s}_*} \sum_{s_d} |A_{\vec{s}_*} \cap A_{\vec{s}_*} s_d|^2 . \end{align} $$

Take any vector $\vec {z} = (z_1,\dots , z_{l})$ , $l< d$ , and consider the decomposition of the variable $V_{\vec {z}}$ onto irreducible components $V_{\vec {z}} (j)$ . As before, define the set $B(\vec {z})$ of all $g\in {\mathbf G}$ such that there are $V_{\vec {z}} (i), V_{\vec {z}} (j)$ with $V_{\vec {z}} (i) = gV_{\vec {z}} (j)$ . Then, arguing as above, we have $|B(\vec {z})| \ll t$ . Using formula (4.10), we get

(4.11) $$ \begin{align} \| A \|_{\mathcal{U}^{d+1}} \ll t \sum_{\vec{s}_*} |A_{\vec{s}_*}|^2 + \sigma = t \| A \|_{\mathcal{U}^{d}} + \sigma , \end{align} $$

where for all $\vec {s}$ in $\sigma $ , we have $\dim (V_{\vec {s}}) = 0$ . Hence,

(4.12) $$ \begin{align} \| A \|_{\mathcal{U}^{d+1}} \ll t \| A \|_{\mathcal{U}^{d}} + \sum_{\vec{s}} |A_{\vec{s}}| \ll t |A|^{d+1} + |A|^{d+1} \ll t |A|^{d+1} . \end{align} $$

This completes the proof.▪

Once again, the bounds above depend on $d,D$ , as well as on $\dim ({\mathbf G})$ , $\deg ({\mathbf G})$ and on the dimension of the ground affine space.

Corollary 3 Let ${\mathbf G}$ be an abelian algebraic group, $V\subseteq {\mathbf G}$ be a variety, $d=\dim (V)$ , and $D=\deg (V)$ . Then, for all sufficiently large q and any $A\subseteq V$ , one has

(4.13) $$ \begin{align} {\mathsf{E}}(A) \ll_{d,D} |A|^3 \left(\frac{t}{|A|} \right)^{(2^{d+1}-d-5)^{-1}} \mbox{ for } d\ge 2 \quad \mbox{and} \quad {\mathsf{E}}(A) \ll_{d,D} |A|^2 t \,,\mbox{ for } d=1. \end{align} $$

In particular, for any $A\subseteq V$ with $|A| \ge t^{1+c}$ , $c>0$ , there is $\delta =\delta (d,c)>0$ such that

(4.14) $$ \begin{align} {\mathsf{E}}(A) \ll_{d,D} |A|^{3-\delta} . \end{align} $$

Bound (4.14) takes place in any algebraic group. Moreover, let $B\subseteq {\mathbf G}$ be an arbitrary set. Then,

(4.15) $$ \begin{align} {\mathsf{E}}(A,B) \ll_{d,D} \left( \frac{t}{|A|} \right)^{\beta} \cdot |A|^{\frac{3}{2} - \frac{\beta d}{2}} |B|^{\frac{3}{2} + \frac{\beta d}{2}} , \end{align} $$

where $\beta = \beta (d) \in [4^{-d}, 2^{-d+1}]$ , and for any $k\ge 1$ , one has either ${\mathsf {T}}_k (A) \le |A|^{2k-1-c\beta /4}$ or

(4.16) $$ \begin{align} {\mathsf{T}}_{k+1} (A) \ll_{d,D} |A|^{2} {\mathsf{T}}_k (A) \cdot |A|^{-c\beta/4} . \end{align} $$

Proof Let $d\ge 2$ . By Theorem 1, we have $\| A\|_{\mathcal {U}^{d+1}} \ll t |A|^{d+1}$ . Using the second part of Lemma 1 with $k=d+1$ , we obtain

$$ \begin{align*} {\mathsf{E}}(A) \ll |A|^3 \left(\frac{t}{|A|} \right)^{(2^k-k-4)^{-1}} = |A|^3 \left(\frac{t}{|A|} \right)^{(2^{d+1}-d-5)^{-1}} . \end{align*} $$

If $d=1$ , then the arguments of the proof of Theorem 1 (see, e.g., (4.12)) give us

$$ \begin{align*} {\mathsf{E}}(A) \ll t |A|^2 + \sum_s |A_s| \ll t |A|^2 . \end{align*} $$

For an arbitrary algebraic group, use the last part of Lemma 2.

To derive (4.15), we can suppose that $|B| \ge |A|$ , because otherwise the required bound

$$ \begin{align*}{\mathsf{E}}(A,B)^2 \le {\mathsf{E}}(A) {\mathsf{E}}(B) \le \left( \frac{t}{|A|} \right)^{\beta} |A|^{3} |B|^3 \le \left( \frac{t}{|A|} \right)^{\beta} \cdot |A|^{3 - \beta d} |B|^{3 + \beta d} \end{align*} $$

takes place for $\beta = (2^{d+1}-d-5)^{-1}$ , $d\ge 2$ (and similar for $d=1$ ; see estimate (4.13)). Furthermore, let $|B|\ge |A|$ . Then, we use Lemma 2 with $k=d$ , combining with Theorem 1 (see formulae (3.6), (4.11), and (4.12)) and the assumption $|A| \le |B|$ to obtain

$$ \begin{align*} \!\!\!\!\!{\mathsf{E}}(A,B) \le (|A||B|)^{\frac{3}{2} - \frac{\beta(d+2)}{2}} \langle A,B \rangle^{\beta}_d \ll (|A||B|)^{\frac{3}{2} - \frac{\beta(d+2)}{2}} \left( \| B \|_{\mathcal{U}^d} + t \langle A,B \rangle_{d-1} \right)^{\beta} \end{align*} $$

$$ \begin{align*} &\ \quad\le (|A||B|)^{\frac{3}{2} - \frac{\beta(d+2)}{2}} \left( |B|^{d+1} + t |A| |B|^d \right)^{\beta} \ll (|A||B|)^{\frac{3}{2} - \frac{\beta(d+2)}{2}} t^{\beta} |B|^{\beta (d+1)}\\&\ \quad= t^{\beta} |A|^{\frac{3}{2} - \frac{\beta(d+2)}{2}} |B|^{\frac{3}{2} + \frac{\beta d}{2}}, \end{align*} $$

and (4.15) follows. Finally, to get (4.16), we use the dyadic pigeonhole principle and the fact that ${\mathsf {T}}^{1/2k} (f)$ defines a norm of f to find the number $\Delta>0$ and the set P such that $P = \{ x\in {\mathbf G} ~:~ \Delta < A^{(k)} (x) \le 2\Delta \}$ and ${\mathsf {T}}_{k+1} (A) \lesssim \Delta ^2 {\mathsf {E}}(A,P)$ . Thus (we assume that $c\le 1$ ),

$$ \begin{align*} {\mathsf{T}}_{k+1} (A) \kern1pt{\lesssim}\kern1pt (\kern-1pt t/|A|)^{\beta} |A|^{\frac{3}{2} - \frac{\beta d}{2}} (\Delta ^2 |P|)^{\frac{1}{2}-\frac{d\beta}{2}} (\Delta |P|)^{1+d\beta} \le |A|^{-\frac{c\beta}{2}} |A|^{\frac{3+2k}{2} - \frac{\beta d}{2}+\beta kd} \cdot {\mathsf{T}}^{\frac{1}{2}-\frac{d\beta}{2}}_k (\kern-1pt A) . \end{align*} $$

Suppose that ${\mathsf {T}}_k (A) \ge |A|^{2k-1-\varepsilon }$ , where $\varepsilon \le c\beta /4$ . In view of the last inequality and $\beta \le 2^{-d+1}$ , one has

$$ \begin{align*}|A|^{-\frac{c\beta}{2}} |A|^{\frac{3+2k}{2} - \frac{\beta d }{2}+\beta kd} \cdot {\mathsf{T}}^{\frac{1}{2}-\frac{d\beta}{2}}_k (A) \le |A|^{2-\varepsilon } {\mathsf{T}}_k (A) . \end{align*} $$

This completes the proof.▪

Theorem 1 and Corollary 3 imply the following criterion.

Corollary 5 Let ${\mathbf G}$ be a finite simple group, $V\subseteq {\mathbf G}$ be a variety, $d=\dim (V)$ , and $D=\deg (V)$ . Suppose that $t(V) = o(|V|)$ . Then, there is $\delta =\delta (d,n)>0$ such that for any $A\subseteq V$ , $|A| \gg |V|$ , the following holds:

(4.17) $$ \begin{align} {\mathsf{E}}(A) \ll_{d,D} |A|^{3-\delta} . \end{align} $$

Clearly, if $t(V) \gg |V|$ , then (4.17) does not hold, and hence Corollary 5 is indeed a criterion. Also, it gives a lower bound for $\delta $ of the form $\delta \gg 1/d$ . Recall that our current dependence on d in (4.17) has an exponential nature.

5 Applications

In [Reference Noether19, Reference Landazuri and Seitz20], the authors obtain the following results on growth of normal sets.

From Corollary 3, we obtain a result on the growth of an arbitrary subset of a conjugate class.

One can see that $t(C) \le |C|^{1-c_*}$ for a certain $c_*>0$ via the general bound on such intersections with generating sets (see [Reference Murphy17]) or, alternatively, from some modifications of Theorem 1 (see [Reference Noether19, Reference Landazuri and Seitz20]). Thus, Corollary 2 takes place for all large subsets of conjugate classes.

Question. Is it true that for any $A\subseteq C$ , where C is a conjugate class such that $|A|\ge |C|^{1-o(1)}$ , say, one has $A^n = {\mathbf G}$ , where n is a function on $\log |{\mathbf G}|/\log |A|$ ? For $n\ll \log |{\mathbf G}|/\log |A|$ ? For $C=C(x)$ , where x is a semisimple element?

Now, we are ready to obtain a nontrivial upper bound for any sufficiently large subset of a Chevalley group living in a variety differ from the maximal parabolic subgroup.

Theorem 3 Let ${\mathbf G}_r ({\mathbb {F}}_q)$ be a finite Chevalley group with rank r and odd q and $\Pi \le {\mathbf G}_r ({\mathbb {F}}_q)$ be its a maximal (by size) parabolic subgroup. Also, let $V \subset {\mathbf G}_r ({\mathbb {F}}_q)$ be a variety differ from all shifts of conjugates of $\Pi $ . Then, for any $A\subseteq V$ and $|A| \ge |\Pi | q^{-1+c}$ , $c>0$ , one has

(5.1) $$ \begin{align} \| \widehat{A} (\rho) \|_o \le |A|^{1-\delta } , \end{align} $$

where $\delta = \delta (c,r)>0$ and $\rho $ is any nontrivial representation of ${\mathbf G}_r ({\mathbb {F}}_q)$ .

Proof Because by the assumption $|A| \ge |\Pi | q^{-1+c}$ , it follows that $|V| \ge |\Pi | q^{-1+c}$ and hence V is rather large. Also, one can see that, trivially, $|A|\ge |\Pi | q^{-1+c} \gg q^{1+c}$ . Furthermore, by [Reference Shkredov30, Lemma 8], we know that $\Pi $ is the maximal (by size) subgroup of ${\mathbf G}({\mathbb {F}}_q)$ , and for all other subgroups $\Gamma \le {\mathbf G}({\mathbb {F}}_q)$ , one has $|\Gamma | \le q^{-1} |\Pi |$ ( $\Gamma $ is not conjugate to $\Pi $ , of course). In particular, in view of (4.1),

$$ \begin{align*}t(V) \le \max_{x,y \in {\mathbf G}_r ({\mathbb{F}}_q)} \{ q^{-1} |\Pi|, |V \cap x\Pi y| \} \ll |V| q^{-c_*} , \end{align*} $$

where $c_* = \min \{c,1\}$ . Take any subgroup H differ from all conjugates of $\Pi $ . Also, let $x\in {\mathbf G}_r ({\mathbb {F}}_q)$ be an arbitrary element. Our task is to estimate the above size of the intersection $A_* := A\cap xH \subseteq V$ . By Corollary 5 and estimate (2.1), we have

$$ \begin{align*}|A_*|^{1+\delta } \ll |A^{-1}_* A_*| \le |H| \le |\Pi| q^{-1} \le |A| q^{-c}, \end{align*} $$

and hence, in particular, $|A_*| \ll |A|^{(1+\delta )^{-1}}$ (actually, in this place of the proof, we can assume a weaker condition on size of A). By a similar argument and estimate (4.1), we derive that

$$ \begin{align*}|A\cap x \Pi y| \le |V\cap x\Pi y| \ll |\Pi|q^{-1} \le |A| q^{-c}, \end{align*} $$

and hence for any proper subgroup $\Gamma \subset {\mathbf G}_r ({\mathbb {F}}_q)$ and for all $x\in {\mathbf G}_r ({\mathbb {F}}_q)$ , one has $|A\cap x\Gamma | \ll |A|^{} q^{-c/2}$ (we use $|A| \gg q$ and assume that $\delta \le c$ ). In particular, A is a generating set of ${\mathbf G}_r ({\mathbb {F}}_q)$ . Combining this observation with the fact (see [Reference Lewko15]) that Chevalley groups are quasi-random in the sense of Gowers [Reference Gowers7], we obtain desired estimate (5.1) (see, e.g., [Reference Helfgott10, Reference Helfgott11], [Reference Shkredov29, Sections 8 and 10]). This completes the proof.▪

Now, we obtain an application of Corollary 3 to some questions about the restriction phenomenon. Recall that in this setting, our group ${\mathbf G}$ is ${\mathbf G} = {\mathbb {F}}^n$ , ${\mathbb {F}}$ is a finite field, $V\subseteq {\mathbb {F}}^n$ is a variety, and ${\mathbf G}$ acts on ${\mathbf G}$ via shifts. For any function $g:{\mathbb {F}}^n \to {\mathbb {C}}$ , consider the commutative analogue of (2.2)

$$ \begin{align*}\hat g(\xi) := \sum_{x\in {\mathbb{F}}^n} g(x) e(-x\cdot \xi), \end{align*} $$

as well as the inverse Fourier transform of a function $f:V \to {\mathbb {C}}$ ,

$$ \begin{align*}(fd\sigma)^\lor(x) := \frac{1}{|V|} \sum_{\xi\in V} f(\xi) e(x\cdot\xi), \end{align*} $$

where $e(x\cdot \xi ) = e^{2\pi i (x_1\xi _1 + \dots +x_n \xi _n)/\mathrm {char} ({\mathbb {F}})}$ for $x = (x_1,\dots ,x_n)$ , $\xi = (\xi _1,\dots ,\xi _n)$ . Thus, a “Lebesgue $L^q$ -norm” of f on V is defined as

$$ \begin{align*}\|f\|_{L^q(V,d\sigma)}:= \left( \frac{1}{|V|} \sum_{\xi\in V} |f(\xi) |^q \right)^{\frac{1}{q}}, \end{align*} $$

while for a function g, it is

$$ \begin{align*}\|g\|_{L^q({\mathbb{F}}^n)}:= \left( \sum_{x\in {\mathbb{F}}^n} |g(x) |^q \right)^{\frac{1}{q}}. \end{align*} $$

The finite-field restriction problem [Reference Lang and Weil21] for our variety V seeks exponent pairs $(q,r)$ such that one has the inequality

$$ \begin{align*}\left\|(f d \sigma)^{\vee}\right\|_{L^{r}\left({\mathbb{F}}^{n}\right)} \leq R^{*}(q \rightarrow r)\|f\|_{L^{q}(V, d \sigma)} \end{align*} $$

or, equivalently,

$$ \begin{align*}\|\widehat{g}\|_{L^{q^{\prime}}(V, d \sigma)} \leq R^{*}(q \rightarrow r)\|g\|_{L^{r^{\prime}}\left({\mathbb{F}}^{n}\right)} \end{align*} $$

takes place with a constant $R^{*}(q \rightarrow r)$ independent of the size of the finite field. As before, we use the notation $\lesssim $ and $\gtrsim $ instead of $\ll $ and $\gg $ , allowing ourselves to lose logarithmic powers of $|{\mathbb {F}}|$ .

Using the arguments of the proofs of [Reference Lang and Weil21, Lemma 5.1 and Proposition 5.2], we obtain the following result.

Proof According to our assumption that V does not contain any line, we see that the parameter $t(V)$ equals $1$ . Hence, by Corollary 3, we know that ${\mathsf {E}}(A) \ll |A|^{3-c} = |A|^\kappa $ , where $c=c(d)>0$ . Put $q=4/\kappa $ , and we want to obtain a good bound for $R^* (q \rightarrow 4)$ . We want to obtain an estimate of the form (see the proofs of [Reference Lang and Weil21, Lemma 5.1 and Proposition 5.2])

$$ \begin{align*} \sum_x (fV * fV )^2 (x) \lesssim \left( \sum_{x\in V} |f(x)|^q \right)^{4/q} , \end{align*} $$

where f is an arbitrary function (we can freely assume that f is positive). Using the dyadic pigeonhole principle, we need to prove the last bound for any $f=A$ with $A\subseteq V$ , and this is equivalent to

$$ \begin{align*} {\mathsf{E}}(A) \lesssim |A|^{4/q} = |A|^{3-c} . \end{align*} $$

This completes the proof.▪

Notice that if the variety V contains subspaces of positive dimension, then there is no any restriction-type result as in Theorem 4 in such generality (see, e.g., [Reference Lang and Weil21, Section 4]).