Math Labs

Worked solution: The polynomial method resolves the cap set problem (Croot-Lev-Pach, Ellenberg-Gijswijt, 2016)

Step 1 of 7: The cap set problem: how big can a progression-free set be?
In plain words

Think of the game SET, played on cards described by nn ternary features: a cap set is a collection of cards with no three that form a legal "set" (the pattern that is forbidden is exactly a 3-term arithmetic progression x+y+z=0x+y+z=0 in F3n\mathbb{F}_3^n). For decades the best known upper and lower bounds on the largest possible cap set were painfully far apart, leaving open whether cap sets could occupy almost all of F3n\mathbb{F}_3^n or only an exponentially tiny sliver of it.

A⊆F3n, ∄ x,y,z∈A distinct with x+y+z=0  ⟹  ∣A∣≤O(cn), c<3A \subseteq \mathbb{F}_3^n,\ \nexists\, x,y,z \in A \text{ distinct with } x+y+z=0 \implies |A| \le O(c^n),\ c < 3
Detailed analysis

A cap set is a subset A⊂F3nA\subset\mathbb{F}_3^n containing no three distinct points x,y,z∈Ax,y,z\in A with x+y+z=0x+y+z=0; since −x=x+x-x=x+x holds only in characteristic 33, this is exactly the condition that no three distinct elements form an affine line (equivalently, a three-term arithmetic progression x,x+d,x+2dx,x+d,x+2d with d≠0d\ne0). Write r3(F3n)=max⁡∣A∣r_3(\mathbb{F}_3^n)=\max|A|.

The trivial bound ∣A∣≤3n|A|\le3^n was the only exponential statement available for decades: Meshulam (1995) proved ∣A∣=O(3n/n)|A|=O(3^n/n) by Fourier analysis on F3n\mathbb{F}_3^n, and Bateman–Katz (2012) pushed this to O(3n/n1+ϵ)O(3^n/n^{1+\epsilon}) — still within a polynomial factor of trivial. On the other side, the largest known constructions (Edel, and improvements by Tyrell) gave only about 2.2n2.2^n, so even the qualitative question 'is the growth rate strictly less than 33?' was open.

Ellenberg–Gijswijt (2017), building on the new polynomial-method breakthrough of Croot–Lev–Pach (2016) for (Z/4Z)n(\mathbb{Z}/4\mathbb{Z})^n, settle that question: ∣A∣≤O(cn)|A|\le O(c^n) with an explicit c=3e−I(2/3)≈2.756c=3e^{-I(2/3)}\approx2.756. The rest of this proof turns a cap set into a vanishing-polynomial problem and lets a dimension count do the work.

Terms in this step
Cap set
A subset AA of F3n\mathbb{F}_3^n containing no three distinct points x,y,zx,y,z with x+y+z=0x+y+z=0, i.e. no non-trivial three-term arithmetic progression.
Three-term arithmetic progression
Three elements x,y,zx,y,z with yy the average of xx and zz; over F3\mathbb{F}_3 this is exactly the equation x+y+z=0x+y+z=0.
Knowledge used in this step