For fixed $c>0$ and prime $p$, if $A\subseteq\mathbb F_p^N$ has at least $c|A|^3$ <additive quadruple>[additive quadruples], then $2A-2A$ contains a vector subspace of size at least $c'(c,p)|A|$. Combine the <Balog-Szemerédi-Gowers theorem>, the <Freiman-Ruzsa theorem over a finite field>, and the <Finite-field Bogolyubov lemma>.

Additive energy produces a large subspace in a fourfold difference set

The hypothesis says that the <additive energy> satisfies $E(A)\geq c|A|^3$. The <Balog-Szemerédi-Gowers theorem> supplies $A'\subseteq A$ with
$$
|A'|\geq c_1(c)|A|,
\qquad
|A'+A'|\leq C_1(c)|A'|.
$$
By the <Freiman-Ruzsa theorem over a finite field>, $A'$ lies in a subspace $H$ with
$$
|H|\leq C_2(c,p)|A'|.
$$
Thus $A'$ has density at least $C_2^{-1}$ in $H$. Applying the <Finite-field Bogolyubov lemma> inside $H$ gives a subspace
$$
V\subseteq2A'-2A'\subseteq2A-2A
$$
of codimension bounded in terms of $c,p$ alone. Therefore
$$
|V|\geq c'(c,p)|H|\geq c'(c,p)|A|,
$$
which is the <additive energy produces a large subspace in a fourfold difference set> result.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 129 / 3 / ii / Solution

The finite-field <Bogolyubov lemma> states that if $A\subseteq G=\mathbb F_p^n$ has density $\alpha$, then $2A-2A$ contains a subspace of codimension at most $2\alpha^{-2}$.

Use normalized <Fourier analysis on a finite abelian group> and put $f=1_A$. Define
$$
S=\left\{\gamma\in\widehat G:|\widehat f(\gamma)|\geq\frac{\alpha^{3/2}}{\sqrt2}\right\}.
$$
By <Parseval identity>,
$$
|S|\frac{\alpha^3}{2}\leq\sum_\gamma|\widehat f(\gamma)|^2=\alpha,
$$
so $|S|\leq2\alpha^{-2}$. Let
$$
V=\{x\in G:\gamma(x)=1\text{ for every }\gamma\in S\}.
$$
Then $V$ is a subspace of codimension at most $|S|$.

The normalized representation function of $2A-2A$ is
$$
r(x)=(f*f*\widetilde f*\widetilde f)(x)
=\sum_{\gamma\in\widehat G}|\widehat f(\gamma)|^4\gamma(x),
$$
where $\widetilde f(x)=f(-x)$. For $x\in V$, all terms indexed by $S$ are nonnegative real numbers, while
$$
\sum_{\gamma\notin S}|\widehat f(\gamma)|^4
\leq\frac{\alpha^3}{2}\sum_\gamma|\widehat f(\gamma)|^2
=\frac{\alpha^4}{2}.
$$
The trivial character alone contributes $\alpha^4$, so $r(x)>0$. Hence $x\in2A-2A$, proving the <Finite-field Bogolyubov lemma>.

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 129 / 3 / i / Solution

{c}

If $A\subseteq\mathbb F_p^n$ has density $\alpha$, then $2A-2A$ contains a vector subspace of codimension at most $2\alpha^{-2}$. It is the annihilator of the Fourier spectrum on which $|\widehat{1_A}|$ is at least $\alpha^{3/2}/\sqrt2$.


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 Incoming links: Finite-field Bogolyubov lemma