## 10 June 2017

### Cap Sets

Why it's difficult to avoid arithmetic progressions.

The following result was established in 2016:

Let $A$ be a subset of $\mathbb{F}_3^n$ containing no three-term arithmetic progression. Then, $|A|=o(2.8^n)$.

This has been described as a big breakthrough. Yet, the proof is simple enough to fit in 2 pages.

The elements of $\mathbb{F}_3^n$ are vectors with $n$ components, each component being an integer modulo 3. Vectors $a_1,a_2,a_3$ are said to form a three-term arithmetic progression when $a_2-a_1=a_3-a_2$. Because we are working modulo $3$, this is equivalent to $a_1+a_2+a_3=0$. Thus, we can rephrase the result as follows.

Let $A$ be a subset of $\mathbb{F}_3^n$ such that $a_1+a_2+a_3\ne0$ for all distinct $a_1,a_2,a_3\in A$. Then, $|A|=o(2.8^n)$.

If we were to not require $a_1,a_2,a_3$ to be distinct, then the hypothesis would essentially be just ‘false’, because any nonempty set contains a trivial three-term arithmetic progression if we are allowed to pick the same element three times. An equivalent formulation is to say ‘not all equal’ instead of ‘distinct’, because $a+a+b \ne 0$ if $a \ne b$.

Let $A$ be a subset of $\mathbb{F}_3^n$ such that $a_1+a_2+a_3\ne0$ for $a_1,a_2,a_3\in A$ not all equal. Then, $|A|=o(2.8^n)$.

The overall idea is to find lower and upper bounds for the dimension of some vector space of polynomials. We consider the set $M_n^d$ of monomials over $n$ variables that have total degree $\le d$ and all powers from $\{0,1,2\}$. The restriction on powers is not serious because $x^3=x$ by Fermat's little theorem. (One may think that $x^2=1$, but that is true only for nonzero $x$.) Let us write $S_n^d$ for the vector space of polynomials written with monomials from $M_n^d$. We have that $\dim S_n^d=|M_n^d|$; let us denote this quantity by $m_d$. Observe that $m_{\infty}=3^n$; indeed, if we allow any total degree, then polynomials can represent any function $\mathbb{F}_3^n\to\mathbb{F}_3$, by combining indicator polynomials $I_a(x)\stackrel{\text{def}}=\prod_{k=1}^n \bigl(1- (x_k-a_k)^2\bigr)$.

We define two subsets of $\mathbb{F}_3^n$: \begin{align} X &\stackrel{\text{def}}= \{\,a_1+a_2 \mid\text{a_1,a_2\in A distinct}\,\} \\ Y &\stackrel{\text{def}}= \{\,-a_3 \mid a_3\in A\,\} \end{align} These sets are disjoint, if $A$ satisfies our hypothesis. We will consider the subspace $V$ of the polynomials that vanish outside $Y$. We will derive a lower bound on $\dim V$, by simply counting points outside $Y$; and we will derive an upper bound on $\dim V$, using a lemma which says, roughly, that ‘any polynomial that is zero on all of $X$ is zero on much of $Y$’. The reasoning will work for an arbitrary $d$; when comparing the upper with the lower bound, we shall pick for $d$ a convenient value.

The bounds we aim to prove are the following: $$m_d - 3^n + |A| \le \dim V \le 2 m_{d/2}$$

Let us first see how we use these bounds, and we will soon return to how to prove them. By the correspondence $x^k \leftrightarrow x^{2-k}$, we see that monomials of total degree $\le d$ are in a one-to-one correspondence with monomials of total degree $\ge 2n-d$. Thus, $m_d$ equals $3^n-m_{2n-d-1}$. It follows that $|A| \le 2 m_{d/2} + m_{2n-d-1}$. If we pick $d$ such that $d/2=2n-d-1$, we get $|A| \le 3 m_{(2n-1)/3} \le 3m_{2n/3}$. Finally, one can show that $m_{2n/3}=o(2.8^n)$.

Why is $m_{2n/3}=o(2.8^n)$? [toggle answer]

We can use a Chernoff bound. Several useful inequalities involving probabilities have the form $$\Pr(X\in S) \le \mathbb{E}f(X)$$ where $X$ is a random variable, and $f$ is a function such that $f(x)\ge [x \in S]$. The notation $[\phi]$ stands for $1$ when $\phi$ holds, and $0$ otherwise. It is very easy to prove this: \begin{align} \Pr(X \in S) & = \sum_{x} [x \in S] \Pr(X=x) \\ &\le \sum_x f(x) \Pr(X=x) = \mathbb{E} f(X) \end{align} We pick $S \stackrel{\text{def}}= \{\,x\mid x\le 0\,\}$ and $f(x) \stackrel{\text{def}}= e^{-\alpha x}$. This tells us that $\Pr(X \le 0) \le \mathbb{E} e^{-\alpha X}$ for any $\alpha\ge 0$. Enough background – let us move back to $m_{2n/3}$.

The number $m_{2n/3}$ says in how many ways we can choose $n$ numbers from the set $\{0,1,2\}$ such that their sum is $\le 2n/3$. In other words, $m_{2n/3} = 3^n \cdot \Pr(Y_1+\cdots +Y_n\le 2n/3)$, where $Y_1,\ldots,Y_n$ are i.i.d. random variables taking values (uniformly) in $\{0,1,2\}$. If we define $X_k \stackrel{\text{def}}= Y_k-2/3$ and we use the inequality from above, we can calculate \begin{align} \Pr\biggl(\sum_{i=1}^n Y_i \le \frac{2n}{3}\biggr) & = \Pr\biggl(\sum_{i=1}^n X_i \le 0\biggr) \\&\le \mathbb{E} \biggl(e^{-\alpha\sum_{i=1}^n X_i}\biggr) = \mathbb{E} \biggl( \prod_{i=1}^n e^{-\alpha X_i}\biggr) = \prod_{i=1}^n \mathbb{E} e^{-\alpha X_i} \\&= \biggl( \frac{e^{-\frac{4}{3}\alpha}+e^{-\frac{1}{3}\alpha}+e^{\frac{2}{3}\alpha}}{3} \biggr)^n \end{align} Now we optimize for $\alpha\ge0$, and we get that $m_{2n/3} \lt 2.75510462^n$.

Now let us get back to proving the bounds on $\dim V$.

The lower bound is easy. The definition of $V$ is $\{\,P\in S_n^d \mid\text{$P(a)=0$for$a\in \mathbb{F}_3^n\setminus Y$}\,\}$. The dimension of $S_n^d$ is $m_d$, and each of the $|\mathbb{F}_3^n\setminus Y|$ constraints reduces the dimension by at most $1$. Done.

Why does adding a constraint $P(a)=0$ reduce the dimension by at most $1$? [toggle answer]

Consider a basis $P_1,\ldots,P_k$. Without loss of generality, we can assume that $P_i(a)\in\{0,1\}$ for all $i$. We partition $P_1,\ldots,P_k$ based on whether $P_i(a)$ is $0$ or $1$: for $Q_1,\ldots,Q_{k_1}$, we have $Q_i(a)=0$ for all $i$; for $R_1,\ldots,R_{k_2}$, we have $R_i(a)=1$ for all $i$; and $k=k_1+k_2$. If $k_2=0$, then the dimension is not reduced at all, as witnessed by the initial basis $P_1,\ldots,P_k$. If $k_2\gt 0$, then the dimension is reduced by $1$, as witnessed by the basis formed by $Q_1,\ldots,Q_{k_1}$ together with $R_2-R_1,R_3-R_1,\ldots,R_{k_2}-R_1$. [This is an instance of the rank-nullity theorem.]

The upper bound isn't quite so easy. My understanding is that the upper bound is the tear that led to the breakthrough. We'll do it in two steps, corresponding to these inequalities: $$\dim V \le |\Sigma| \le 2 m_{d/2}$$ where $\Sigma$ is a maximal support of a polynomial in $V$.

For the first inequality, we will show the contrapositive: if a polynomial $P \in V$ has support $\Sigma$ with $|\Sigma|\lt\dim V$, then $\Sigma$ is not maximal. We do this by finding another polynomial $Q \in V$ whose support is nonempty and disjoint from $\Sigma$. The support of $P+Q$ will be a strict superset of $\Sigma$.

Why does $Q$ exist? [toggle answer]

We reuse the argument from the previous gray box. We start with space $V$, and each additional constraint $Q(a)=0$ reduces the dimension by at most $1$. Thus, the space $\{\,Q\in V\mid\text{$Q(a)=0$for$a\in\Sigma$}\,\}$ has positive dimension.

For the second inequality, we show that any $P\in V$ has a support of size $\le 2 m_{d/2}$. I'll start with an example. Take $P({\bf x})=x_1 x_2 x_3+x_1^2$. This is an element of $S_3^3$ because we use $3$ variables and the total degree of each monomial is $\le 3$. The polynomial $P({\bf x}+{\bf y})$ will have monomials that use both $x$-variables and $y$-variables, but still have total degree $\le 3$: \begin{align} P({\bf x}+{\bf y}) = \left\{ \begin{aligned} &x_{1} x_{2} x_{3} + x_{2} x_{3} y_{1} + x_{1} x_{3} y_{2} + x_{3} y_{1} y_{2} \\ &+ x_{1} x_{2} y_{3} + x_{2} y_{1} y_{3} + x_{1} y_{2} y_{3} + y_{1} y_{2} y_{3} \\ &+ x_{1}^{2} + 2 \, x_{1} y_{1} + y_{1}^{2} \end{aligned} \right. \end{align}

For each monomial, in addition to the total degree (the sum of all powers), we can define an $x$-degree (the sum of all powers on $x$-variables), and a $y$-degree (the sum of all powers on $y$-variables). Since the total degree is $\le3$, it follows that the $x$-degree is $\le3/2$ or the $y$-degree is $\le3/2$ (or both). Let's put on the first line all those monomials with $x$-degree $\le3/2$, and on the second line the rest: P({\bf x}+{\bf y}) = \left\{ \begin{aligned} & x_{3} y_{1} y_{2} + x_{2} y_{1} y_{3} + x_{1} y_{2} y_{3} + y_{1} y_{2} y_{3} + 2 x_{1} y_{1} + y_{1}^{2} \\ & + x_{1} x_{2} x_{3} + x_{2} x_{3} y_{1} + x_{1} x_{3} y_{2} + x_{1} x_{2} y_{3} + x_{1}^{2} \end{aligned} \right.

Now we group monomials on the first line by $x$, and we group monomials on the second line by $y$. P({\bf x}+{\bf y}) = \left\{ \begin{aligned} & x_{3} y_{1} y_{2} + x_{2} y_{1} y_{3} + x_{1} (y_{2} y_{3} + 2 y_1) + (y_{1} y_{2} y_{3} + y_{1}^{2}) \\ & + (x_{1} x_{2} x_{3} + x_1^2) + x_{2} x_{3} y_{1} + x_{1} x_{3} y_{2} + x_{1} x_{2} y_{3} \end{aligned} \right.

Introducing some new notation, we can write the above as P({\bf x}+{\bf y}) = \left\{ \begin{aligned} & x_{3} F_{x_3}({\bf y}) + x_{2} F_{x_2}({\bf y}) + x_{1} F_{x_1}({\bf y}) + F_1({\bf y}) \\ & + G_1({\bf x}) + y_{1} G_{y_1}({\bf x}) + y_{2} G_{y_2}({\bf x}) + y_{3} G_{y_3}({\bf x}) \end{aligned} \right.

Finally, we instantiate this equality for all $({\bf x},{\bf y})\in A^2$. For an example, let's say $A=\{100,020\}$, where $100$ is a compact notation for the point $(1,0,0)\in\mathbb{F}_3^3$. Then, \begin{align} \begin{bmatrix} P(100+100) & P(100+020) \\ P(020+100) & P(020+020) \end{bmatrix} = &\begin{bmatrix} 0 \\ 0 \end{bmatrix}_{x_3} \begin{bmatrix} F_{x_3}(100) & F_{x_3}(020) \end{bmatrix}\\ &+ \begin{bmatrix} 0 \\ 2 \end{bmatrix}_{x_2} \begin{bmatrix} F_{x_2}(100) & F_{x_2}(020) \end{bmatrix} + \cdots \end{align}

The matrix corresponding to the term $x_3 F_{x_3}({\bf y})$ was factored into a column vector corresponding to $x_3$, and a row vector corresponding to $F_{x_3}({\bf y})$.

Of course, we can do the same manipulations for any polynomial $P\in S_n^d$, obtaining \begin{align} P({\bf x}+{\bf y}) = \sum_{m\in M_n^{d/2}} m({\bf x}) F_m({\bf y}) + \sum_{m\in M_n^{d/2}} m({\bf y}) G_m({\bf x}) \end{align}

and we can instantiate this equation for all $({\bf x},{\bf y})\in A^2$ to obtain a matrix identity. If $P\in V$, which implies that $P$ vanishes on $X$, then its matrix is diagonal. On the other hand, the matrix of each term $m({\bf x})F_m({\bf y})$ has rank 1, and similarly the matrix of each term $m({\bf y})G_m({\bf x})$. Because rank is subadditive, the matrix of a $P \in V$ has rank $\le 2 m_{d/2}$, which means that it has $\le2m_{d/2}$ nonzero elements on the diagonal. In other words, $P(-a) = P(a + a)\ne 0$ for $\le 2 m_{d/2}$ points $a\in A$.

This concludes the proof.

This post follows [Ellenberg, Gijswijt, On large subsets of $F_q^n$ with no three-term arithmetic progression], which presents a slightly more general result, for $\mathbb{F}_q^n$ rather than $\mathbb{F}_3^n$.

[Stefan Kiefer provided feedback on several drafts of this post. Thanks!]