Ellenberg–Gijswijt cap-set bound

There is a constant $C<3$ such that every cap set in ${\mathbb{F}}_3^n$ has cardinality less than $C^n$.

For the polynomial method in combinatorics, useOver ${\mathbb{F}}_3$, this is the indicator function of $x+y+z=0$. On $A^3$ it is therefore a diagonal tensor with $|A|$ nonzero diagonal entries. Every monomial in its expansion has individual exponents at most two and total degree at most $2n$, so one of its three variable blocks has degree at most $2n/3$. Grouping terms according to such a block and using the slice rank of a diagonal tensor giveswhere $m_n$ is the number of $\alpha \in \{0,1,2{\}}^n$ with ${\sum }_i{\alpha }_i\le 2n/3$. If $X_i=1-{\alpha }_i$ are independent random variables uniform on $\{-1,0,1\}$, thenAny exponential upper bound for this tail probability gives $m_n\le (3-ϵ{)}^n$ and hence the result after absorbing the factor three and finitely many small dimensions into $C<3$.

The number $m_n$ of monomials in $n$ variables with each exponent at most two and total polynomial degree at most $\lfloor 2n/3\rfloor$ satisfiesFor each exponent vector $a$ counted on the left, $1\le 2^{2n/3-{\sum }_i{a}_i}$. Summing and then allowing all $a\in \{0,1,2{\}}^n$ gives $m_n\le 2^{2n/3}(1+1/2+1/4{)}^n$. The base has cube $343/16<27$, so it is strictly less than three. This elementary weighted count gives the exponential saving in the polynomial method in combinatorics without an unstated probability estimate.