# Special Topics in Complexity Theory, Lectures 16-17

Special Topics in Complexity Theory, Fall 2017. Instructor: Emanuele Viola

### 1 Lectures 16-17, Scribe: Tanay Mehta

In these lectures we prove the corners theorem for pseudorandom groups, following Austin [Aus16]. Our exposition has several non-major differences with that in [Aus16], which may make it more computer-science friendly. The instructor suspects a proof can also be obtained via certain local modifications and simplifications of Green’s exposition [Gre05bGre05a] of an earlier proof for the abelian case. We focus on the case $G = \textit {SL}_2(q)$ for simplicity, but the proof immediately extends to other pseudorandom groups.

Theorem 1. Let $G = \textit {SL}_2(q)$. Every subset $A \subseteq G^2$ of density $\mu (A) \geq 1/\log ^a |G|$ contains a corner, i.e., a set of the form $\{(x, y), (xz, y), (x, zy) ~|~ z \neq 1\}$.

#### 1.1 Proof Overview

For intuition, suppose $A$ is a product set, i.e., $A = B \times C$ for $B, C \subseteq G$. Let’s look at the quantity

\begin{aligned}\mathbb {E}_{x, y, z \leftarrow G}[A(x, y) A(xz, y) A(x, zy)]\end{aligned}

where $A(x, y) = 1$ iff $(x, y) \in A$. Note that the random variable in the expectation is equal to $1$ exactly when $x, y, z$ form a corner in $A$. We’ll show that this quantity is greater than $1/|G|$, which implies that $A$ contains a corner (where $z \neq 1$). Since we are taking $A = B \times C$, we can rewrite the above quantity as

\begin{aligned} & \mathbb {E}_{x, y, z \leftarrow G}[B(x)C(y) B(xz)C(y) B(x)C(zy)] \\ & = \mathbb {E}_{x, y, z \leftarrow G}[B(x)C(y) B(xz)C(zy)] \\ & = \mathbb {E}_{x, y, z \leftarrow G}[B(x)C(y) B(z)C(x^{-1}zy)] \end{aligned}

where the last line follows by replacing $z$ with $x^{-1}z$ in the uniform distribution. If $\mu (A) \ge \delta$, then $\mu (B) \ge \delta$ and $\mu (C) \ge \delta$. Condition on $x \in B$, $y \in C$, $z \in B$. Then the distribution $x^{-1}zy$ is a product of three independent distributions, each uniform on a set of measure greater than $\delta$. By pseudorandomness $x^{-1}zy$ is $1/|G|^{\Omega (1)}$ close to uniform in statistical distance. This implies that the above quantity equals

\begin{aligned} & \mu (A) \cdot \mu (C) \cdot \mu (B) \cdot \left (\mu (C) \pm \frac {1}{|G|^{\Omega (1)}}\right )\\ & \geq \delta ^3 \left ( \delta - \frac {1}{|G|^{\Omega (1)}} \right ) \\ & \geq \delta ^4 /2 \\ & > 1/|G|. \end{aligned}

Given this, it is natural to try to write an arbitrary $A$ as a combination of product sets (with some error). We will make use of a more general result.

#### 1.2 Weak Regularity Lemma

Let $U$ be some universe (we will take $U = G^2$). Let $f:~U \rightarrow [-1,1]$ be a function (for us, $f = 1_A$). Let $D \subseteq \{d: U \rightarrow [-1,1]\}$ be some set of functions, which can be thought of as “easy functions” or “distinguishers.”

Theorem 2.[Weak Regularity Lemma] For all $\epsilon > 0$, there exists a function $g := \sum _{i \le s} c_i \cdot d_i$ where $d_i \in D$, $c_i \in \mathbb {R}$ and $s = 1/\epsilon ^2$ such that for all $d \in D$

\begin{aligned}\mathbb {E}_{x \leftarrow U}[f(x) \cdot d(x)] = \mathbb {E}_{x \leftarrow U}[g(x) \cdot d(x)] \pm \epsilon .\end{aligned}

The lemma is called ‘weak’ because it came after Szemerédi’s regularity lemma, which has a stronger distinguishing conclusion. However, the lemma is also ‘strong’ in the sense that Szemerédi’s regularity lemma has $s$ as a tower of $1/\epsilon$ whereas here we have $s$ polynomial in $1/\epsilon$. The weak regularity lemma is also simpler. There also exists a proof of Szemerédi’s theorem (on arithmetic progressions), which uses weak regularity as opposed to the full regularity lemma used initially.

Proof. We will construct the approximation $g$ through an iterative process producing functions $g_0, g_1, \dots , g$. We will show that $||f - g_i||_2^2$ decreases by $\ge \epsilon ^2$ each iteration.

1. Start: Define $g_0 = 0$ (which can be realized setting $c_0 = 0$).
2. Iterate: If not done, there exists $d \in D$ such that $|\mathbb {E}[(f - g) \cdot d]| > \epsilon$. Assume without loss of generality $\mathbb {E}[(f - g) \cdot d] > \epsilon$.
3. Update: $g' := g + \lambda d$ where $\lambda \in \mathbb {R}$ shall be picked later.

Let us analyze the progress made by the algorithm.

\begin{aligned} ||f - g'||_2^2 &~ = \mathbb {E}_x[(f - g')^2(x)] \\ &~ = \mathbb {E}_x[(f - g - \lambda d)^2(x)] \\ &~ = \mathbb {E}_x[(f - g)^2] + \mathbb {E}_x[\lambda ^2 d^2 (x)] - 2\mathbb {E}_x[(f - g)\cdot \lambda d(x)] \\ &~ \leq ||f - g||_2^2 + \lambda ^2 - 2\lambda \mathbb {E}_x[(f-g)d(x)] \\ &~ \leq ||f - g||_2^2 + \lambda ^2 - 2\lambda \epsilon \\ &~ \leq ||f-g||_2^2 - \epsilon ^2 \end{aligned}

where the last line follows by taking $\lambda = \epsilon$. Therefore, there can only be $1/\epsilon ^2$ iterations because $||f - g_0||_2^2 = ||f||_2^2 \leq 1$. $\square$

#### 1.3 Getting more for rectangles

Returning to the lower bound proof, we will use the weak regularity lemma to approximate the indicator function for arbitrary $A$ by rectangles. That is, we take $D$ to be the collection of indicator functions for all sets of the form $S \times T$ for $S, T \subseteq G$. The weak regularity lemma gives us $A$ as a linear combination of rectangles. These rectangles may overlap. However, we ideally want $A$ to be a linear combination of non-overlapping rectangles.

Claim 3. Given a decomposition of $A$ into rectangles from the weak regularity lemma with $s$ functions, there exists a decomposition with $2^{O(s)}$ rectangles which don’t overlap.

Proof. Exercise. $\square$

In the above decomposition, note that it is natural to take the coefficients of rectangles to be the density of points in $A$ that are in the rectangle. This gives rise to the following claim.

Claim 4. The weights of the rectangles in the above claim can be the average of $f$ in the rectangle, at the cost of doubling the distinguisher error.

Consequently, we have that $f = g + h$, where $g$ is the sum of $2^{O(s)}$ non-overlapping rectangles $S \times T$ with coefficients $\Pr _{(x, y) \in S \times T}[f(x, y) = 1]$.

Proof. Let $g$ be a partition decomposition with arbitrary weights. Let $g'$ be a partition decomposition with weights being the average of $f$. It is enough to show that for all rectangle distinguishers $d \in D$

\begin{aligned}|\mathbb {E}[(f-g')d]| \leq |\mathbb {E}[(f-g)d]|.\end{aligned}

By the triangle inequality, we have that

\begin{aligned}|\mathbb {E}[(f-g')d]| \leq |\mathbb {E}[(f-g)d]| + |\mathbb {E}[(g-g')d]|.\end{aligned}

To bound $\mathbb {E}[(g-g')d]|$, note that the error is maximized for a $d$ that respects the decomposition in non-overlapping rectangles, i.e., $d$ is the union of some non-overlapping rectangles from the decomposition. This can be argues using that, unlike $f$, the value of $g$ and $g'$ on a rectangle $S\times T$ from the decomposition is fixed. But, for such $d$, $g' = f$! More formally, $\mathbb {E}[(g-g')d] = \mathbb {E}[(g-f)d]$. $\square$

We need to get a little more from this decomposition. The conclusion of the regularity lemma holds with respect to distinguishers that can be written as $U(x) \cdot V(y)$ where $U$ and $V$ map $G \to \{0,1\}$. We need the same guarantee for $U$ and $V$ with range $[-1,1]$. This can be accomplished paying only a constant factor in the error, as follows. Let $U$ and $V$ have range $[-1,1]$. Write $U = U_+ - U_-$ where $U_+$ and $U_-$ have range $[0,1]$, and the same for $V$. The error for distinguisher $U \cdot V$ is at most the sum of the errors for distinguishers $U_+ \cdot V_+$, $U_+ \cdot V_-$, $U_- \cdot V_+$, and $U_- \cdot V_-$. So we can restrict our attention to distinguishers $U(x) \cdot V(y)$ where $U$ and $V$ have range $[0,1]$. In turn, a function $U(x)$ with range $[0,1]$ can be written as an expectation $\mathbb{E} _a U_a(x)$ for functions $U_a$ with range $\{0,1\}$, and the same for $V$. We conclude by observing that

\begin{aligned} \mathbb{E} _{x,y}[ (f-g)(x,y) \mathbb{E} _a U_a(x) \cdot \mathbb{E} _b V_b(y)] \le \max _{a,b} \mathbb{E} _{x,y}[ (f-g)(x,y) U_a(x) \cdot V_b(y)].\end{aligned}

#### 1.4 Proof

Let us now finish the proof by showing a corner exists for sufficiently dense sets $A \subseteq G^2$. We’ll use three types of decompositions for $f: G^2 \rightarrow \{0,1\}$, with respect to the following three types of distinguishers, where $U_i$ and $V_i$ have range $\{0,1\}$:

1. $U_1(x) \cdot V_1(y)$,
2. $U_2(xy) \cdot V_2(y)$,
3. $U_3(x) \cdot V_3(xy)$.

The last two distinguishers can be visualized as parallelograms with a 45-degree angle between two segments. The same extra properties we discussed for rectangles hold for them too.

Recall that we want to show

\begin{aligned}\mathbb {E}_{x, y, g}[f(x, y) f(xg, y) f(x, gy)] > \frac {1}{|G|}.\end{aligned}

We’ll decompose the $i$-th occurrence of $f$ via the $i$-th decomposition listed above. We’ll write this decomposition as $f = g_i + h_i$. We do this in the following order:

\begin{aligned} & ~f(x, y) \cdot f(xg, y) \cdot f(x, gy) \\ = & ~f(x, y) f(xg, y) g_3(x, gy) + f(x, y) f(xg, y) h_3(x, gy) \\ &~ \vdots \\ =&~ g_1 g_2 g_3 + h_1 g_2 g_3 + f h_2 g_3 + f f h_3 \end{aligned}

We first show that $\mathbb{E} [g_1 g_2 g_3]$ is big (i.e., inverse polylogarithmic in expectation) in the next two claims. Then we show that the expectations of the other terms are small.

Claim 5. For all $g \in G$, the values $\mathbb {E}_{x, y}[g_1(x, y) g_2(xg, y) g_3(x, gy)]$ are the same (over $g$) up to an error of $2^{O(s)} \cdot 1/|G|^{\Omega (1)}$.

Proof. We just need to get error $1/|G|^{\Omega (1)}$ for any product of three functions for the three decomposition types. By the standard pseudorandomness argument we saw in previous lectures,

\begin{aligned} \mathbb {E}_{x, y}[c_1 U_1(x)V_1(y) \cdot c_2 U_2(xgy)V_2(y) \cdot c_3 U_3(x)V_3(xgy)] \\ = c_1 c_2 c_3 \mathbb {E}_{x, y}[(U_1 \cdot U_3)(x) (V_1 \cdot V_2)(y) (U_2 \cdot V_3)(xgy)] \\ = c_1 c_2 c_3 \cdot \mu (U_1 \cdot U_3) \mu (V_1 \cdot V_2) \mu (U_2 \cdot V_3) \pm \frac {1}{|G|^{\Omega (1)}}. \end{aligned}

$\square$

Recall that we start with a set of density $\ge 1/\log ^{a} |G|$.

Claim 6. $\mathbb {E}_{g, x, y}[g_1 g_2 g_3] > \Omega (1/\log ^{4a} |G|)$.

Proof. By the previous claim, we can fix $g = 1_G$. We will relate the expectation over $x, y$ to $f$ by a trick using the Hölder inequality: For random variables $X_1, X_2, \ldots , X_k$,

\begin{aligned}\mathbb {E}[X_1 \dots X_k] \leq \prod _{i=1}^k \mathbb {E}[X_i^{c_i}]^{1/c_i} \text { such that } \sum 1/c_i = 1.\end{aligned}

To apply this inequality in our setting, write

\begin{aligned}\mathbb {E}[f] = \mathbb {E}\left [(f \cdot g_1 g_2 g_3)^{1/4} \cdot \left (\frac {f}{g_1}\right )^{1/4}\cdot \left (\frac {f}{g_2}\right )^{1/4}\cdot \left (\frac {f}{g_3}\right )^{1/4}\right ].\end{aligned}

By the Hölder inequality, we get that

\begin{aligned}\mathbb {E}[f] \leq \mathbb {E}[f \cdot g_1 g_2 g_3]^{1/4} \mathbb {E}\left [\frac {f}{g_1}\right ]^{1/4} \mathbb {E}\left [\frac {f}{g_2}\right ]^{1/4} \mathbb {E}\left [\frac {f}{g_3}\right ]^{1/4}.\end{aligned}

Note that

\begin{aligned} \mathbb {E}_{x, y} \frac {f(x,y)}{g_1(x, y)} & = \mathbb {E}_{x, y} \frac {f(x, y)}{\mathbb {E}_{x', y' \in \textit {Cell}(x,y)}[f(x', y')] } \\ & = \mathbb {E}_{x, y} \frac {\mathbb {E}_{x', y' \in \textit {Cell}(x, y)}[f(x',y')]}{\mathbb {E}_{x', y' \in \textit {Cell}(x,y)}[f(x', y')] }\\ & = 1 \end{aligned}

where $\textit {Cell}(x, y)$ is the set in the partition that contains $(x, y)$. Finally, by non-negativity of $f$, we have that $\mathbb {E}[f \cdot g_1 g_2 g_3]^{1/4} \leq \mathbb {E}[g_1 g_2 g_3]$. This concludes the proof. $\square$

We’ve shown that the $g_1 g_2 g_3$ term is big. It remains to show the other terms are small. Let $\epsilon$ be the error in the weak regularity lemma with respect to distinguishers with range $[-1,1]$.

Claim 7. $|\mathbb {E}[f f h_3]| \leq \epsilon ^{1/4}$.

Proof. Replace $g$ with $gy^{-1}$ in the uniform distribution to get

\begin{aligned} & \mathbb {E}^4_{x, y, g}[f(x,y) f(xg,y)h_3(x, gy)] \\ & = \mathbb {E}^4_{x, y, g}[f(x,y) f(xgy^{-1},y)h_3(x, g)] \\ & = \mathbb {E}^4_{x, y}[f(x,y) \mathbb {E}_g [f(xgy^{-1},y)h_3(x, g)]] \\ & \leq \mathbb {E}^2_{x, y} [f^2(x, y)] \mathbb {E}^2_{x, y} \mathbb {E}^2_g [f(xgy^{-1},y)h_3(x, g)]\\ & \leq \mathbb {E}^2_{x, y} \mathbb {E}^2_g [f(xgy^{-1},y)h_3(x, g)]\\ & = \mathbb {E}^2_{x, y, g, g'}[f(xgy^{-1}, y) h_3(x, g) f(xg'y^{-1}, y) h_3(x, g')], \end{aligned}

where the first inequality is by Cauchy-Schwarz.

Now replace $g \rightarrow x^{-1}g, g' \rightarrow x^{-1}g$ and reason in the same way:

\begin{aligned} & = \mathbb {E}^2_{x, y, g, g'}[f(gy^{-1}, y) h_3(x, x^{-1}g) f(g'y^{-1}, y) h_3(x, x^{-1}g')] \\ & = \mathbb {E}^2_{g, g', y}[f(gy^{-1}, y) \cdot f(g'y^{-1}, y) \mathbb {E}_x [h_3(x, x^{-1}g) \cdot h_3(x, x^{-1}g')]] \\ & \leq \mathbb {E}_{x,x',g,g'}[h_3(x, x^{-1}g) h_3(x, x^{-1}g') h_3(x', x'^{-1}g) h_3(x', x'^{-1}g')]. \end{aligned}

Replace $g \rightarrow xg$ to rewrite the expectation as

\begin{aligned} \mathbb {E}[h_3(x, g) h_3(x, x^{-1}g') h_3(x', x'^{-1}xg) h_3(x', x'^{-1}g')].\end{aligned}

We want to view the last three terms as a distinguisher $U(x) \cdot V(xg)$. First, note that $h_3$ has range $[-1,1]$. This is because $h_3(x,y) = f(x,y) - \mathbb{E} _{x', y' \in \textit {Cell}(x,y)} f(x',y')$ and $f$ has range $\{0,1\}$.

Fix $x', g'$. The last term in the expectation becomes a constant $c \in [-1,1]$. The second term only depends on $x$, and the third only on $xg$. Hence for appropriate functions $U$ and $V$ with range $[-1,1]$ this expectation can be rewritten as

\begin{aligned} \mathbb {E}[h_3(x, g) U(x) V(xg)], \end{aligned}

which concludes the proof. $\square$

There are similar proofs to show the remaining terms are small. For $fh_2g_3$, we can perform simple manipulations and then reduce to the above case. For $h_1 g_2 g_3$, we have a slightly easier proof than above.

##### 1.4.1 Parameters

Suppose our set has density $\delta \ge 1/\log ^a |G|$. We apply the weak regularity lemma for error $\epsilon = 1/\log ^c |G|$. This yields the number of functions $s = 2^{O(1/\epsilon ^2)} = 2^{O(\log ^{2c} |G|)}$. For say $c = 1/3$, we can bound $\mathbb{E} _{x,y,g}[g_1 g_2 g_3]$ from below by the same expectation with $g$ fixed to $1$, up to an error $1/|G|^{\Omega (1)}$. Then, $\mathbb {E}_{x,y,g=1}[g_1g_2g_3] \geq \mathbb {E}[f]^4 = 1/\log ^{4a}|G|$. The expectation of terms with $h$ is less than $1/\log ^{c/4} |G|$. So the proof can be completed for all sufficiently small $a$.

### References

[Aus16]    Tim Austin. Ajtai-Szemerédi theorems over quasirandom groups. In Recent trends in combinatorics, volume 159 of IMA Vol. Math. Appl., pages 453–484. Springer, [Cham], 2016.

[Gre05a]   Ben Green. An argument of shkredov in the finite field setting, 2005. Available at people.maths.ox.ac.uk/greenbj/papers/corners.pdf.

[Gre05b]   Ben Green. Finite field models in additive combinatorics. Surveys in Combinatorics, London Math. Soc. Lecture Notes 327, 1-27, 2005.