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}


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.


[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.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s