2. Proof of Theorem 1·5

Notation. In this paper, we shall frequently consider subsets of products of two vectors spaces G and H. Given $X \subset G \times H$ and an element $x \in G$ , we write $X_{x \bullet} = \{y \in H \colon (x,y) \in X\}$ for the (vertical) slice of X in the column indexed by x. Likewise, for an element $y \in H$ , we write $X_{\bullet y} = \{x \in G \colon (x,y) \in X\}$ for the (horizontal) slice of X in the row indexed by y. Intuitively, we take a geometric perspective and informally use a Cartesian coordinate system, where G plays the role of the x-axis and H plays the role of the y-axis.

In order to study a transverse set A, we move to the set of orthogonal complements of its columns. Namely, for each $x \in G$ , we consider the subspace $V_x = A_{x \bullet}^\perp$ , where ${}^\perp$ is taken with respect to some fixed inner product on G. Observe that these subspaces satisfy $V_{x_1 + x_2} \subseteq V_{x_1} + V_{x_2}$ . Indeed, since A is a transverse set, its rows are subspaces so $A_{x_1 \bullet} \cap A_{x_2 \bullet} \subseteq A_{x_1 + x_2 \bullet}$ . Therefore

\[V_{x_1 + x_2} = A_{x_1 + x_2 \bullet}^\perp \subseteq (A_{x_1 \bullet} \cap A_{x_2 \bullet})^\perp = A_{x_1 \bullet}^\perp + A_{x_2 \bullet}^\perp = V_{x_1} + V_{x_2},\]

as claimed. This motivates the following definition.

Observe that in a transverse set A we have $A_{0 \bullet} = H$ , so $A_{0 \bullet}^\perp = \{0\}$ . Thus, every transverse set gives rise to a linear system of subspaces in a natural way. Likewise, it turns out that every linear system of subspaces produces a transverse set.

We remark also the following properties of a linear system of subspaces.

The following lemma allows us to pass from subspaces to linear maps. It is an exact version of [ Reference Milićević9 , lemma 34], obtained by taking $K = 1$ in that paper. To make the paper self-contained, we provide a proof.

It is important that we get further maps $\phi_1, \phi_2$ and $\phi_3$ for any given $\phi_4$ , rather than only being able to find a single quadruple of maps $(\phi_1, \dots, \phi_4)$ satisfying the properties in the conclusion of the lemma.

Furthermore, we need the related uniqueness result, which is an exact version of [ Reference Milićević9 , lemma 35], obtained by taking $K = 1$ and $r = 0$ in that paper. Like above, we offer a short proof.

We shall also make use of elementary fact that near-homomorphisms of vector spaces come from (exact) homomorphisms. The formulation below is a slight rephrasing of [ Reference Gowers and Milićević7 , corollary 26]. As with the two lemmas above, we give a proof for the sake of self-completeness.

Remark. Returning to the last remark from the introduction, in order to prove the multidimensional generalisation of Corollary 1·4 using the approach in this paper, we would need a higher-order generalisation of Lemma 2·6. Such a generalisation concerns extending multilinear maps defined on $1-o(1)$ proportion of multilinear varieties. Unlike the lemma above, such a higher-order result is highly non-trivial, and, in fact, is one of the main results from [ Reference Gowers and Milićević7 ].

Proof. For this proof, we need the notation for convolutions defined by differences instead of sums. Namely, if $f, g \;:\; G_1 \to \mathbb{R}$ are functions, then $f \,{\overline{\ast}}\, g$ is another function from $G_1$ to $\mathbb{R}$ , defined by $f { \,\overline{\ast}}\, g(x) = \sum_{y \in G_1} f(x + y) g(y)$ .

Let $\tilde{A} = \{(a, \phi(a)) \;:\; a \in A\} \subseteq G_1 \times G_2$ be the graph of the function $\phi$ . Write also $\tilde{G} = G_1 \times G_2$ . Our primary aim is to show that $\tilde{A}$ is almost a subgroup of $\tilde{G}$ , from which we shall deduce the functional conclusion fairly quickly. Note that $|\tilde{A}| = |A|$ and that $(1-\varepsilon)|G_1| \leq |A| \leq |G_1|$ . The assumption in the lemma means that $\tilde{A}$ has at least $(1-\varepsilon)|\tilde{A}|^3$ additive quadruples. Using the convolution notation, we have $\sum_{x \in \tilde{G}} {\unicode{x1D7D9}}_{\tilde{A}} { \,\overline{\ast}}\, {\unicode{x1D7D9}}_{\tilde{A}}(x)^2 \geq (1 - \varepsilon) |\tilde{A}|^3$ . Let S be the set of all $x \in \tilde{G}$ such that ${\unicode{x1D7D9}}_{\tilde{A}} { \,\overline{\ast}}\, {\unicode{x1D7D9}}_{\tilde{A}}(x) \geq (1- \sqrt{\varepsilon})|\tilde{A}|$ . Note that $\|{\unicode{x1D7D9}}_{\tilde{A}} { \,\overline{\ast}}\, {\unicode{x1D7D9}}_{\tilde{A}}\|_{\infty} \leq |\tilde{A}|$ . Observe that

\begin{align*}\sqrt{\varepsilon}|\tilde{A}|\sum_{x \in \tilde{G} \setminus S} {\unicode{x1D7D9}}_{\tilde{A}} { \,\overline{\ast}}\, {\unicode{x1D7D9}}_{\tilde{A}}(x) \leq &\sum_{x \in \tilde{G} \setminus S} (|\tilde{A}| - {\unicode{x1D7D9}}_{\tilde{A}} { \,\overline{\ast}}\, {\unicode{x1D7D9}}_{\tilde{A}}(x)){\unicode{x1D7D9}}_{\tilde{A}} { \,\overline{\ast}}\, {\unicode{x1D7D9}}_{\tilde{A}}(x)\\\leq &\sum_{x \in \tilde{G}} (|\tilde{A}| - {\unicode{x1D7D9}}_{\tilde{A}} { \,\overline{\ast}}\, {\unicode{x1D7D9}}_{\tilde{A}}(x)){\unicode{x1D7D9}}_{\tilde{A}} { \,\overline{\ast}}\, {\unicode{x1D7D9}}_{\tilde{A}}(x) \\= &|\tilde{A}|\Big(\sum_{x \in \tilde{G}}{\unicode{x1D7D9}}_{\tilde{A}} { \,\overline{\ast}}\, {\unicode{x1D7D9}}_{\tilde{A}}(x) \Big) - \Big(\sum_{x \in \tilde{G}}{\unicode{x1D7D9}}_{\tilde{A}} { \,\overline{\ast}}\, {\unicode{x1D7D9}}_{\tilde{A}}(x)^2\Big)\\\leq &\varepsilon |\tilde{A}|^3,\end{align*}

so $\sum_{x \in S} {\unicode{x1D7D9}}_{\tilde{A}} { \,\overline{\ast}}\, {\unicode{x1D7D9}}_{\tilde{A}}(x) \geq (1 - \sqrt{\varepsilon}) |\tilde{A}|^2$ , implying also $|S| \geq (1 - \sqrt{\varepsilon}) |\tilde{A}|$ .

We now show that $|S - S| \leq (1 + 7\sqrt{\varepsilon})|S|$ . Let $x, y \in S$ . Thus $|\tilde{A} \cap (\tilde{A} - x)| = {\unicode{x1D7D9}}_{\tilde{A}} { \,\overline{\ast}}\, {\unicode{x1D7D9}}_{\tilde{A}}(x) \geq (1-\sqrt{\varepsilon})|\tilde{A}| $ and $|\tilde{A} \cap (\tilde{A} - y)| = {\unicode{x1D7D9}}_{\tilde{A}} { \,\overline{\ast}}\, {\unicode{x1D7D9}}_{\tilde{A}}(y) \geq (1-\sqrt{\varepsilon})|\tilde{A}|$ . Hence,

\begin{align*}{\unicode{x1D7D9}}_{\tilde{A}} { \,\overline{\ast}}\, {\unicode{x1D7D9}}_{\tilde{A}}(x - y) = &|(\tilde{A} - x) \cap (\tilde{A} - y)| \geq |\tilde{A} \cap (\tilde{A} - x) \cap (\tilde{A} - y)| \\\geq& |\tilde{A} \cap (\tilde{A} - x)| + |\tilde{A} \cap (\tilde{A} - y)| - |\tilde{A}| \geq (1 - 2 \sqrt{\varepsilon})|\tilde{A}|.\end{align*}

However, using $\sum_{d \in \tilde{G}} {\unicode{x1D7D9}}_{\tilde{A}} { \,\overline{\ast}}\, {\unicode{x1D7D9}}_{\tilde{A}}(d)^2 \leq |\tilde{A}|^3$ and elementary inequality $(1 - 2 \sqrt{\varepsilon})^{-2} \leq 1 + 5\sqrt{\varepsilon}$ which holds when $\varepsilon \leq 2^{-8}$ , the inequality ${\unicode{x1D7D9}}_{\tilde{A}} { \,\overline{\ast}}\, {\unicode{x1D7D9}}_{\tilde{A}}(d) \geq (1 - 2 \sqrt{\varepsilon})|\tilde{A}|$ can hold for at most ${|\tilde{A}|^3}/{(1 - 2 \sqrt{\varepsilon})^2|\tilde{A}|^2} \leq (1 + 5\sqrt{\varepsilon})|\tilde{A}|$ elements $d \in \tilde{G}$ , showing that $|S - S| \leq (1 + 5\sqrt{\varepsilon})|\tilde{A}| \leq (1 + 7\sqrt{\varepsilon})|S|$ .

Since $\varepsilon \leq 28^{-2}$ , we have $|S - S| \leq ({5}/{4})|S|$ . A similar elementary argument to the one above, where one considers intersection of sets $S \cap S - a$ and $S \cap S - b$ for elements $a,b \in S - S$ , shows that $S - S$ is a subgroup of $\tilde{G}$ , which we denote by H. Fix arbitrary $s \in S$ . In particular, since $\varepsilon \leq 19^{-2}$ , we have

\[|H| = |S - S| \leq (1 + 7\sqrt{\varepsilon})|S| \leq (1 + 7\sqrt{\varepsilon})(1-\sqrt{\varepsilon})^2|\tilde{A}| \leq (1 + 10\sqrt{\varepsilon})|\tilde{A}|.\]

Since $S - s \subseteq H$ and thus $S \subseteq s + H$ , we have

\[(1 - \sqrt{\varepsilon}) |\tilde{A}|^2 \leq \sum_{d \in s + H} {\unicode{x1D7D9}}_{\tilde{A}} { \,\overline{\ast}}\, {\unicode{x1D7D9}}_{\tilde{A}}(d) = \sum_{x, y \in \tilde{G}} {\unicode{x1D7D9}}_{\tilde{A}}(x) {\unicode{x1D7D9}}_{\tilde{A}}(y) {\unicode{x1D7D9}}_{s + H}(x - y).\]

By averaging over $y \in \tilde{A}$ , we obtain an element y such that $|(s + y + H) \cap \tilde{A}| \geq (1 - \sqrt{\varepsilon}) |\tilde{A}|$ .

We now return to the initial functional perspective. Let $\pi_1 \;:\; G_1 \times G_2 \to G_1$ and $\pi_2 : G_1 \times G_2 \to G_2$ be the natural projections. Note that $\pi_1$ is injective on $\tilde{A}$ , which implies that

\[(1 - \sqrt{\varepsilon}) |\tilde{A}| \leq |\tilde{A} \cap (s + y + H)| = |\pi_1(\tilde{A} \cap (s + y + H))| \leq |\pi_1(H)| \leq |H| \leq (1 + 5\sqrt{\varepsilon})|\tilde{A}|.\]

Thus, $|\pi_1(H)| \gt ({1}/{2})|H|$ and $\pi_1(H)$ and H, being subgroups, both have sizes which are powers of p, so $|\pi_1(H)| = |H|$ , and thus $\pi_1$ is injective on H. Furthermore, since $|A| \geq (1-\varepsilon)|G_1|$ , we must have $\pi_1(H) = G_1$ . Therefore, writing $h \in s + y + H$ as $(\pi_1(h), \pi_2(h))$ and then defining a map $\Phi \;:\; \pi_1(h) \mapsto \pi_2(h)$ , we obtain an affine map from $G_1 = \pi_1(s + y + H)$ to $G_2$ . Note that H is then the graph of $\Phi$ . Since $|(s + y + H) \cap \tilde{A}| \geq (1 - \sqrt{\varepsilon}) |\tilde{A}|$ we have $\phi(a) = \Phi(a)$ for at least $(1 - \sqrt{\varepsilon}) |A| \geq (1 - 2\sqrt{\varepsilon})|G_1|$ elements $a \in A$ , as desired.

The heart of the proof of Theorem 1·5 is that a quasirandom linear system of subspaces is generated by a bilinear map of expected codimension, which is the content of the next proposition.

We remark that Proposition 2·7 is similar in spirit to [ Reference Milićević9 , theorem 33], and these two results can roughly be viewed as ‘99%’ and ‘1%’ cases of the same phenomenon. Given the additional structure in the setting of the present paper, we are able to obtain a more streamlined proof. An important difference from [ Reference Milićević9 ] is that throughout that paper one has sufficiently strong quasirandom properties, which is not the case in this paper. For that reason, we need an efficient regularity lemma for transverse sets, which we prove later.

Brief proof overview. As described in the introduction, our aim is to find parameterizing linear maps $\phi_x \;:\; \mathbb{F}_p^d \to V_x$ such that $\phi_{x} + \phi_{y} = \phi_{x + y - a} + \phi_a$ holds for most of choices of x, y and a. We shall first pick a such that most of pairs $(x,y) \in G$ are well behaved, in the sense that

\begin{align*}V_a + V_x + V_y = V_a + V_x + V_{x + y - a} = V_a + V_y + V_{x + y - a} =& V_x + V_y + V_{x + y - a} \\= &V_a + V_x + V_y + V_{x + y - a}\end{align*}

and subspaces $V_a, V_x, V_y$ and $V_{x + y - a}$ have dimension d. Fix an arbitrary linear isomorphism $\theta \colon \mathbb{F}_p^d \to V_a$ . Write $q = (x,y)$ . The displayed property allows us to apply Lemma 2·4 to obtain linear isomorphisms $\theta_q^1 \colon \mathbb{F}_p^d \to V_x$ , $\theta_q^2 \colon \mathbb{F}_p^d \to V_y$ and $\theta_q^3 \colon \mathbb{F}_p^d \to V_{x + y - a}$ , such that $\theta^1_q + \theta^2_q = \theta + \theta^3_q$ . In the long run, we want $\theta^3_q$ to depend only on x, and, moreover, in a linear fashion. To show that $\theta^3_q$ depends only on $x + y$ instead of both x and y, we rely on Lemma 2·5; this is achieved in Claim 2·9. To stress this sharper dependence, we write $\phi^3_{x+ y - a}$ instead of $\theta^3_q$ . In order to obtain linearity, we look at well-behaved pairs of the form (x, y), (x’, y), (x, y’) and (x’, y’). Algebraic manipulation shows that

\[\phi^3_{x + y - a} + \phi^3_{x' + y' - a} - \phi^3_{x' + y - a} - \phi^3_{x + y' - a} = 0.\]

Therefore, we may view $\phi^3$ as a map from a very dense subset of G to $\textrm{Hom}(\mathbb{F}_p^d, H)$ that respects most additive quadruples. An application of Lemma 2·6 completes the proof.

For the remainder of the proof, we fix an element $a \in G$ such that the set $\mathcal{T}$ of triples (x, y, z) such that the quadruple (a, x, y, z) satisfies properties in the conclusion of the claim above is of size at least $(1 - 8\varepsilon' )|G|^3$ . Let $\mathcal{P}$ be the set of all pairs $(x,y) \in G^2$ such that $|V_x| = |V_y| = |V_{x + y - a}| = p^d$ and the subspace $V_a + V_x + V_y$ has size $p^{3d}$ . By Lemma 2·3, for each $(x,y) \in \mathcal{P}$ , we have that

(2·1) \begin{align}V_a + V_x + V_y = V_a + V_x + V_{x + y - a} = V_a + V_y + V_{x + y - a} =& V_x + V_y + V_{x + y - a} \nonumber\\= &V_a + V_x + V_y + V_{x + y - a}.\end{align}

Since $|\mathcal{T}| \geq (1 - 8\varepsilon' )|G|^3$ , we also have that $|\mathcal{P}| \geq (1 - 8 \varepsilon')|G|^2$ .

Fix an arbitrary linear isomorphism $\theta \colon \mathbb{F}_p^d \to V_a$ . For each $q = (x, y) \in \mathcal{P}$ , equalities in (2·1) imply that the subspaces $V_a$ , $V_x$ , $V_y$ and $V_{x + y - a}$ satisfy the conditions of Lemma 2·4. Lemma 2·4 provides us with linear isomorphisms $\theta_q^1 \colon \mathbb{F}_p^d \to V_x$ , $\theta_q^2 \colon \mathbb{F}_p^d \to V_y$ and $\theta_q^3 \colon \mathbb{F}_p^d \to V_{x + y - a}$ , such that $\theta^1_q + \theta^2_q = \theta + \theta^3_q$ . Our next step is to show that these maps do not depend entirely on the pair q, but only on the relevant point in G.

Although $q = (x,y) \in \mathcal{P}$ is a pair, we misuse the notation and write $q_3 = x + y - a$ for the third element associated to q.

Proof. Let us focus first on the case $i = 1$ ; we shall comment on the other cases afterwards. Let us count the number N of pairs $(q, q') \in \mathcal{P}^2$ such that $q_1 = q'_1$ and $\theta^1_q = \theta^1_{q'}$ . Notice that whenever $q = (x, y), q' = (x, y') \in \mathcal{P}$ are two pairs with the same first element, then

\[\theta^1_q + \theta^2_q - \theta^3_q = \theta = \theta^1_{q'} + \theta^2_{q'} - \theta^3_{q'},\]

so $\theta^2_q - \theta^3_q - \theta^2_{q'} - \theta^3_{q'} + (\theta^1_q - \theta^1_{q'}) = 0$ . If the indexing elements satisfy $V_x \cap (V_y + V_{x + y - a} + V_{y'} + V_{x + y' - a}) = \{0\}$ , then by Lemma 2·5 we have $\theta^1_q = \theta^1_{q'}$ . On the other hand, Lemma 2·3 shows that $V_y + V_{x + y - a} + V_{y'} + V_{x + y' - a} = V_y + V_{x + y - a} + V_{y'}$ . By the choice of a, $V_x \cap (V_y + V_{x + y - a} + V_{y'}) = \{0\}$ holds for all but at most $8\varepsilon' |G|^3$ triples $(x,y, y') \in G^3$ . From this and Cauchy-Schwarz inequality, we conclude that

\[N \geq |G|^{-1} |\mathcal{P}|^2 - 8\varepsilon' |G|^3 \geq (1 - 24 \varepsilon') |G|^3.\]

Let $X_1$ be the set of all $x \in G$ such that the number of pairs (q, q ^′) above with x as the first element is at least $(1 - 10\sqrt{\varepsilon'}) |G|$ . Then $|X_1| \geq (1 - 10\sqrt{\varepsilon'}) |G|$ and we may take $\phi^1_x$ to be the most common map appearing as $\theta^1_q$ among triples with $q_1 = x$ .

The case $i = 2$ follows by symmetry. For the case $i = 3$ , consider $z \in G$ and $q = (x, z + a - x), q' = (x', z + a - x') \in \mathcal{P}$ agreeing in the third associated element, that is $q_3 = q'_3 = z$ . Observe that we need $V_z \cap (V_x + V_{z + a - x} + V_{x'} + V_{z + a - x'}) = \{0\}$ property to get the equality of the maps coming from different pairs. By Lemma 2·3, $V_x + V_{z + a - x} + V_{x'} + V_{z + a - x'} = V_x + V_{z + a - x} + V_{x'}$ , so we just need $V_z \cap (V_x + V_{z + a - x} + V_{x'}) = \{0\}$ . After a change of variables $y = z + a - x$ , this condition has the form guaranteed by Claim 2·8, so the same proof works in this case as well.

Let $\tilde{\mathcal{P}}$ be the set of all pairs $q \in \mathcal{P}$ for which the conclusion of the claim above holds. Let $\textrm{Hom}(\mathbb{F}_p^d, H)$ be the vector space of all linear homomorphisms from $\mathbb{F}_p^d$ to H. Write $\phi^3 \colon X_3 \to \textrm{Hom}(\mathbb{F}_p^d, H)$ for the map given by $\phi^3(x) = \phi^3_x$ . Let us observe that $\phi^3$ respects most additive quadruples. Consider the set $\mathcal{Q}$ of all quadruples $(x,x',y,y') \in G^4$ such that:

1. (i) $x, x' \in X_1, y, y' \in X_2, x + y - a, x' + y - a, x + y' - a, x' + y' - a \in X_3$ ;

2. (ii) $(x, y), (x', y), (x, y'), (x', y') \in \tilde{\mathcal{P}}$ .

By the work above, we have that $|\mathcal{Q}| \geq (1 - 300\sqrt{\varepsilon'})|G|^4$ . Each such quadruple gives rise to an additive quadruple respected by $\phi^3$ . Namely,

\begin{align*}&\phi^3(x + y - a) + \phi^3(x' + y' - a) - \phi^3(x' + y - a) - \phi^3(x + y' - a)\\&\hspace{2cm}=\Big(\phi^1_x+ \phi^2_y - \theta\!\Big) + \Big(\phi^1_{x'}+ \phi^2_{y'} - \theta\!\Big) - \Big(\phi^1_{x'}+ \phi^2_y - \theta\!\Big)- \Big(\phi^1_x+ \phi^2_{y'} - \theta\!\Big) = 0.\end{align*}

On the other hand, each additive quadruple in $X_3$ can arise at most $|G|$ times in this way. Hence, $\phi^3$ respects at least $(1 - 300\sqrt{\varepsilon'})|G|^3$ additive quadruples in $X_3$ .

We are now in position to apply Lemma 2·6. It provides us with an affine map $\Gamma \colon G \to \textrm{Hom}(\mathbb{F}_p^d, H)$ such that $\Gamma(x) = \phi^3(x)$ , and thus $\textrm{Im}\, \Gamma(x) = \textrm{Im}\, \phi^3_x = V_x$ , holds for at least $(1 - \varepsilon'') |G|$ of $x \in X_3$ , where $\varepsilon'' \leq 40\sqrt[4]{\varepsilon'}$ , provided $\varepsilon' \leq 10^{-15}$ . Finally, we define maps $\Phi \colon G \times \mathbb{F}_p^d \to H$ and $\Psi \colon \mathbb{F}_p^d \to H$ as $\Phi(x, \lambda) = \Gamma(x)(\lambda) - \Gamma(0)(\lambda)$ and $\Psi = \Gamma(0)$ , giving us the desired conclusion.

To prove Theorem 1·5 it thus remains to locate a sufficiently quasirandom piece. The following regularity lemma for transverse sets achieves this. The proof is based on dependent random choice and the algebraic structure is essential for ensuring the good bounds.

Proof. Let Y be the set of all $y \in H$ such that $|A_{\bullet y}| \geq ({\delta}/{2}) |G|$ . By averaging, we have $|Y| \geq ({\delta}/{2})|H|$ . Pick $y \in Y$ uniformly at random. Then $\mathbb{P}(x \in A_{\bullet y}) = ({|A_{x \bullet} \cap Y|}/{|Y|})$ . Let $X_{\text{bad}}$ be the set of elements $x \in A_{\bullet y}$ such that $|A_{x \bullet}| \leq ({\delta^2}/{100})|H|$ . Then

\begin{align*}\mathbb{E} \Big(|A_{\bullet y}| - 10 |X_{\text{bad}}|\Big) \geq &\frac{\delta}{2}|G| - 10\sum_{x \in G} \mathbb{P}(x \in X_{\text{bad}})\\= &\frac{\delta}{2}|G| - 10\sum_{x \in G}{\unicode{x1D7D9}}(|A_{x \bullet}| \leq \frac{\delta^2}{100}|H|) \mathbb{P}(x \in A_{\bullet y}) \geq \frac{\delta}{4}|G|.\end{align*}

Hence, there is a choice of $y_0$ such that $|A_{\bullet y_0}| \geq ({\delta}/{2})|G|$ and $|A_{x \bullet}| \geq ({\delta^2}/{100})|H|$ holds for at least $({9}/{10})|A_{\bullet y_0}|$ elements $x \in A_{\bullet y_0}$ . Hence, for an arbitrary $x \in A_{\bullet y_0}$ we may find $x' \in A_{\bullet y_0}$ such that $|A_{x' \bullet}| \geq ({\delta^2}/{100})|H|$ and $|A_{x + x' \bullet}| \geq ({\delta^2}/{100})|H|$ both hold. As A is transverse we have $({\delta^4}/{10^4}) |H| \leq |A_{x' \bullet} \cap A_{x + x' \bullet}| \leq |A_{x \bullet}|$ . Thus, $|A_{x \bullet}| \geq ({\delta^4}/{10^4}) |H|$ holds for all elements $x \in A_{\bullet y_0}$ .

Similarly, there exists an element $x_0 \in G$ such that $|A_{x_0 \bullet}| \geq ({\delta}/{2})|H|$ and $|A_{\bullet y}| \geq ({\delta^4}/{10^4}) |G|$ holds for all elements $y \in A_{x_0 \bullet}$ . Set $U_0 = A_{\bullet y_0}$ , $V_0 = A_{x_0 \bullet}$ and $d_0 = \lceil \log_p (10^4 \delta^{-4}) \rceil$ . Hence, $U_0$ and $V_0$ have codimension at most $d_0$ in G and H respectively, for each $x \in U_0$ we have $\textrm{codim}_{V_0} (A_{x \bullet} \cap V_0) \leq d_0$ and for each $y \in V_0$ we have $\textrm{codim}_{U_0} (A_{\bullet y} \cap U_0) \leq d_0$ .

We now perform an iterative procedure in which at step i we find further subspaces $U_i \leq U_0$ , $V_i \leq V_0$ , of codimension $\textrm{codim}_{U_0} U_i \leq O(i d_0 \log(2 \varepsilon^{-1}))$ and $\textrm{codim}_{V_0} V_i \leq O(i d_0)$ , such that the proportion of elements $x \in U_i$ such that

(2·2) \begin{equation}\textrm{codim}_{V_i} (A_{x \bullet} \cap V_i) \leq d_0 - i\end{equation}

is at least $1 - {1}/{2} \varepsilon$ . The elements x in (2·2) are those that have $A_{x \bullet} \cap V_i$ at least as large as we would like, thus we guarantee that most subspaces restricted to $V_i$ have at least the desired size, if not equal to the one required by property (i). The procedure terminates when we find subspaces $U_i$ and $V_i$ satisfying the claimed properties in the lemma, with d taken to be $d_0 - i$ . Let us emphasize that the property that is kept during the procedure, namely that (2·2) holds for most $x \in U_i$ , is weaker than the properties (i) and (ii) of the lemma. Hence, in each step of the argument, we shall either show that the current $U_i$ and $V_i$ have the properties (i) and (ii) or we shall find further $U_{i + 1}$ and $V_{i+1}$ for which the the properties described around (2·2) hold.

Suppose that we have subspaces $U_i, V_i$ for some $i \geq 0$ (when $i=0$ we know that $U_0$ and $V_0$ have the properties described around (2·2)). If both properties (i) and (ii) of the lemma hold for $U_i$ and $V_i$ and integer $d_0-i$ , we are done. We now distinguish between two cases, depending on which one of the two properties fails.

Property (i) fails. We thus have at least $\varepsilon|U_i|$ elements $x \in U_i$ such that $\textrm{codim}_{V_i} (A_{x \bullet} \cap V_i) \not= d_0-i$ . Since $\textrm{codim}_{V_i} (A_{x \bullet} \cap V_i) \leq d_0 - i$ holds for at least $(1 - \varepsilon/2)|U_i|$ elements $x \in U_i$ , we conclude that $\textrm{codim}_{V_i} (A_{x \bullet} \cap V_i) \leq d_0 - i - 1$ holds for at least $({\varepsilon}/{2})|U_i|$ elements $x \in U_i$ . Let $r = \lceil2\log_p(4 \varepsilon^{-1}) \rceil$ and take $y_1, \dots, y_{r}$ uniformly and independently at random from $V_i$ . Let $U_{i+1} = U_i \cap A_{\bullet y_1} \cap \dots\cap A_{\bullet y_{r}}$ and set $V_{i+1} = V_i$ . Let $X_{\text{bad}}$ be the set of all elements $x \in U_{i + 1}$ such that $\textrm{codim}_{V_i} (V_i \cap A_{x \bullet}) \geq d_0-i$ . Observe that, for each $x \in U_i$ , $\mathbb{P}(x \in U_{i+1}) = ({|A_{x \bullet} \cap V_i|}/{|V_i|})^r = p^{ - r \;\textrm{codim}_{V_i} (V_i \cap A_{x \bullet})}$ . Then

\begin{align*}\mathbb{E}\Big(|U_{i+1}| - 2\varepsilon^{-1}|X_{\text{bad}}|\Big) \geq p^{-r (d_0 - i - 1)}& \frac{\varepsilon}{2} |U_i| - 2 \varepsilon^{-1} p^{-r(d_0 - i)} |U_i| \\&= 2 \varepsilon^{-1}p^{-r(d_0 - i)} |U_i| \Big(p^r \frac{\varepsilon^2}{4} - 1\Big) \gt 0,\end{align*}

by our choice of r. Hence, there is a choice of $y_1, \dots, y_{r}$ producing the desired pair of subspaces $U_{i+1}$ and $V_{i+1}$ . Note that $\textrm{codim}_{U_i} U_{i+1} \leq r d_0 = O(d_0 \log_p(2 \varepsilon^{-1}))$ and $\textrm{codim}_{V_i} V_{i+1} = 0$ .

Property (ii) fails, but (i) holds. In this case have at least $3\varepsilon|U_i|^2$ pairs of elements $(x_1, x_2) \in U^2_i$ such that $\textrm{codim}_{V_i}( A_{x_1 \bullet} \cap A_{x_2 \bullet} \cap V_i) \not= 2(d_0-i)$ . We call such a pair $(x_1, x_2)$ irregular. Let R be the set of all $a \in U_i$ such that $\textrm{codim}_{V_i} (A_{a \bullet} \cap V_i) = d_0-i$ . Since property (i) holds, we have $|R| \geq (1 - \varepsilon) |U_i|$ . The number of irregular pairs $(x_1, x_2)$ such that $x_1 \notin R$ is at most $\varepsilon |U_i|^2$ . Hence, there at least $2\varepsilon|U_i|^2$ irregular pairs $(x_1, x_2) \in R \times U_i$ , so by averaging, there exists $a \in R$ such that (a, x) is irregular for at least $2\varepsilon|U_i|$ choices of $x \in U_i$ .

By assumption (2·2), at least $\varepsilon|U_i|$ of such elements $x \in U_i$ also have the property that $\textrm{codim}_{V_i} A_{x \bullet} \cap V_i \leq d_0-i$ . Note that

\begin{align*}&\textrm{codim}_{V_i}( A_{a \bullet} \cap A_{x \bullet} \cap V_i) = \dim V_i - \dim (A_{a \bullet} \cap A_{x \bullet} \cap V_i)\\ &\quad = \dim V_i - \dim (A_{a \bullet} \cap V_i) - \dim (A_{x \bullet} \cap V_i) + \dim \Big( (A_{a \bullet} \cap V_i) + (A_{x \bullet} \cap V_i) \Big)\\ &\quad = \textrm{codim}_{V_i}A_{a \bullet} + \textrm{codim}_{V_i}A_{x \bullet} - \dim V_i + \dim \Big( (A_{a \bullet} \cap V_i) + (A_{x \bullet} \cap V_i) \Big)\\ &\quad \leq \textrm{codim}_{V_i}A_{a \bullet} + \textrm{codim}_{V_i}A_{x \bullet} \leq 2(d_0 - i),\end{align*}

where we used $(A_{a \bullet} \cap V_i) + (A_{x \bullet} \cap V_i) \subseteq V_i$ in the inequality between the last two lines. Since (a, x) is irregular, we have $\textrm{codim}_{V_i}( A_{a \bullet} \cap A_{x \bullet} \cap V_i) \lt 2(d_0-i)$ . Setting $V_{i+1} = V_i \cap A_{a \bullet}$ , for such an x we have

\begin{align*}&\textrm{codim}_{V_{i+1}} (A_{x \bullet} \cap V_{i+1})\\ &\quad = \dim(V_{i+1}) - \dim(A_{x \bullet} \cap V_{i+1}) = \dim(V_i \cap A_{a \bullet}) - \dim( V_i \cap A_{x \bullet} \cap A_{a \bullet}) \\ &\quad= \Big(\!\dim V_i - \textrm{codim}_{V_i} (A_{a \bullet} \cap V_i)\Big) - \Big(\!\dim V_i - \textrm{codim}_{V_i}( A_{a \bullet} \cap A_{x \bullet} \cap V_i)\Big) \\ &\quad= \textrm{codim}_{V_i}( A_{a \bullet} \cap A_{x \bullet} \cap V_i) - \textrm{codim}_{V_i} (A_{a \bullet} \cap V_i) \leq d_0 - i - 1.\end{align*}

We may now repeat the same final step of the previous case, with the same choice of $r = \lceil2\log_p(4 \varepsilon^{-1}) \rceil$ . Take $y_1, \dots, y_{r}$ uniformly and independently at random from $V_{i+1}$ and let $U_{i+1} = U_i \cap A_{\bullet y_1} \cap \dots\cap A_{\bullet y_{r}}$ . By the calculations in the previous step, there exists a choice of $y_1, \dots, y_r$ such that $\textrm{codim}_{V_{i+1}} (A_{x \bullet} \cap V_{i+1}) \leq d_0 - i - 1$ for at least $\Big(1 - \frac{1}{2} \varepsilon\Big)|U_{i+1}|$ elements $x \in U_{i+1}$ . In this case, the codimension bounds are $\textrm{codim}_{U_i} U_{i+1} \leq r d_0 = O(d_0 \log_p(2 \varepsilon^{-1}))$ and $\textrm{codim}_{V_i} V_{i+1} \leq d_0$ .

We are now in position to prove Theorem 1·5.

Proof of Theorem 1·5. Let A be the given transverse set of density $\delta$ . Apply Lemma 2·10 with parameter $\varepsilon = 2^{-80} p^{-16 \lceil \log_p(10^4 \delta^{-4})\rceil}$ to get a positive integer $d \leq \lceil \log_p(10^4 \delta^{-4})\rceil$ and subspaces $U \leq G$ and $V \leq H$ such that:

1. (i) $\textrm{codim}_G U \leq O(\!\log^3_p(2 \delta^{-1}))$ and $\textrm{codim}_H V \leq O(\!\log^2_p(2 \delta^{-1}))$ ;

2. (ii) for at least $(1 - \varepsilon/3)|U|$ of elements $x \in U$ we have $|A_{x \bullet} \cap V| = p^{-d} |V|$ ; and

3. (iii) for at least $(1 - \varepsilon/3)|U|^2$ of pairs $(x_1, x_2) \in U^2$ we have $|A_{x_1 \bullet} \cap A_{x_2 \bullet} \cap V| = p^{-2d} |V|$ .

Note that $\varepsilon \leq 2^{-80} p^{-16 d} \lt 10^{-32} p^{-8d}$ . Fix an inner product $\cdot$ on V and, for each $x \in U$ , define subspace $V_x = (A_{x \bullet} \cap V)^\perp$ , where ${}^\perp$ indicates orthogonal complement in V with respect to $\cdot$ . As A is a transverse set, the collection $(V_x)_{x \in U}$ is a linear system of subspaces. By properties (ii) and (iii) we have that $|V_x| = p^d$ holds for at least $(1 - \varepsilon/3)|U|$ of elements $x \in U$ , and three equalities $|A_{x_1 \bullet} \cap V| = p^{-d} |V|$ , $|A_{x_2 \bullet} \cap V| = p^{-d} |V|$ and $|A_{x_1 \bullet} \cap A_{x_2 \bullet} \cap V| = p^{-2d} |V|$ hold simultaneously for at least $(1 - \varepsilon)|U|^2$ of pairs $(x_1, x_2) \in U^2$ . These three equalities together imply that

\begin{align*}|V_{x_1} \cap V_{x_2}| = \frac{|V_{x_1}| |V_{x_2}|}{|V_{x_1} + V_{x_2}|} = \frac{\frac{|V|}{|A_{x_1 \bullet} \cap V|} \frac{|V|}{|A_{x_2 \bullet} \cap V|}}{\frac{|V|}{|(V_{x_1} + V_{x_2})^\perp|}} = &\frac{|V| |(V_{x_1} + V_{x_2})^\perp|}{p^{-2d} |V|^2} \\= &\frac{|V_{x_1}^\perp \cap V_{x_2}^\perp|}{p^{-2d} |V|} = \frac{|A_{x_1 \bullet}\cap A_{x_2 \bullet} \cap V|}{p^{-2d} |V|} = 1,\end{align*}

hence $V_{x_1} \cap V_{x_2} = \{0\}$ .

We may apply Proposition 2·7 to linear system of subspaces $(V_x)_{x \in U}$ inside V to find a bilinear map $\Phi \colon U \times \mathbb{F}_p^d \to V$ and a linear map $\Psi \colon \mathbb{F}_p^d \to V$ such that $\{\Phi(x, \lambda) + \Psi(\lambda) \colon \lambda \in \mathbb{F}_p^d\} = V_x$ holds for at least $(1 - 200p^d \sqrt[8]{\varepsilon})|G|$ elements $x \in U$ . By our choice of $\varepsilon$ , we have $200p^d \sqrt[8]{\varepsilon} \leq \frac{1}{3} p^{-d}$ . Hence, there exists a set $X \subseteq U$ , of size $|X| \geq (1 - p^{-d}/3) |U|$ , such that $\{\Phi(x, \lambda) + \Psi(\lambda) \colon \lambda \in \mathbb{F}_p^d\} = (A_{x \bullet} \cap V)^\perp$ holds for all $x \in X$ .

Taking orthogonal complements in V, we get $\{\Phi(x, \lambda) + \Psi(\lambda) \colon \lambda \in \mathbb{F}_p^d\}^\perp = A_{x \bullet} \cap V$ . Let $V' = V \cap (\textrm{Im} \Psi)^\perp$ , and let $\beta_i(x,y) = \Phi(x, e_i) \cdot y$ for $i \in [d]$ , where $e_1, \dots, e_d$ is the standard basis of $\mathbb{F}_p^d$ . Then for each $x \in X$ , we have $\{y \in V' \colon \beta(x,y) = 0\} \subseteq A_{x \bullet}$ . We claim that in fact

\[\{(x,y) \in U \times V' \colon \beta(x,y) = 0\} \subseteq A.\]

To see this, take any $(x,y) \in U \times V'$ such that $\beta(x,y) = 0$ . The subspace $\{u \in U \colon \beta(u,y) = 0\}$ has codimension at most d inside U. As $|X| \geq (1 - p^{-d}/3) |U|$ , we conclude that $|X \cap \{u \in U \colon \beta(u,y) = 0\}| \geq ({2}/{3}) |\{u \in U \colon \beta(u,y) = 0\}|$ . Thus, as $\beta(x,y) = 0$ , we have that $(X \cap \{u \in U \colon \beta(u,y) = 0\}) + x \subseteq \{u \in U \colon \beta(u,y) = 0\}$ as well, so $(X \cap \{u \in U \colon \beta(u,y) = 0\}) + x$ and $(X \cap \{u \in U \colon \beta(u,y) = 0\})$ intersect, at some element u, say. Thus, $u, u -x \in X$ and $\beta(u, y) = \beta(u - x, y) = 0$ . By properties of elements in X, we have that $(u, y), (u - x, y) \in A$ . But A is a transverse set, so $(x,y) \in A$ , as desired.

Finally, let us remark that terms of the form $\log_p(2\delta^{-1})$ in the bounds above can be replaced by more esthetically pleasing $\log_p(\delta^{-1})$ . This stems from the fact that if the density $\delta$ is at least ${15}/{16}$ , then our transverse set A is in fact the full product $G \times H$ . We leave this simple observation as an exercise to the reader.