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Source. Theorem 8 with its proof, p. 4, of Eric Naslund and William F. Sawin, Upper bounds for sunflower-free sets, Forum Math. Sigma 5 (2017), Paper No. e15, doi:10.1017/fms.2017.12. Labels and pages here are those of arXiv:1606.09575v1, the edition named on the source card.

Statement

Definitions. $\mu_3^S$ is the Erdős-Szemerédi sunflower-free capacity of Theorem 3 (p. 1). A capset is a subset of $\mathbb F_3^n$ with no three-term arithmetic progression; with $A_n$ a largest capset in $\mathbb F_3^n$, the capset capacity is $C=\limsup_{n\to\infty}|A_n|^{1/n}$ (p. 4).

Theorem 8 (p. 4, quoted). "We have that $\mu_3^S\leq\sqrt{1+C}$ where $C$ is the capset capacity and $\mu_3^S$ is the Erdős-Szemeredi-sunflower-free capacity."

The paper presents this as a quantitative form of the result of Alon, Shpilka and Umans that the Ellenberg-Gijswijt bound implies $\mu_3^S<2$ (pp. 2 and 4). With the Ellenberg-Gijswijt bound $C\le2.7552$ it gives $\mu_3^S\le1.938$, which the paper notes is weaker than Theorem 3 (p. 4).

Proof pointer

P. 4. Pair the coordinates of $\{0,1\}^{2n}$ and code each pair by a symbol in $\{0,1,2,3\}$, with $3$ standing for $(1,1)$. For each pattern $x\in\{0,1\}^n$ of positions of the symbol $3$, the vectors with that pattern, read on the remaining coordinates as elements of $\mathbb F_3^{n-w(x)}$, form a capset, because the pairs $(0,0),(1,0),(0,1)$ form a sunflower. Summing the capset bound over $x$ gives $(1+C)^n$ for sets in $\{0,1\}^{2n}$.

Read depth

Claims checked: the definitions, the statement and the proof were read on the print. Nothing here is independently reviewed.

Dependencies

None in the corpus. The numerical consequence uses the Ellenberg-Gijswijt bound on the capset capacity (arXiv:1605.09223).

Bears on

• Problem 857: since $m(n,3)$ is $F_3(n)+1$, the theorem gives $m(n,3)\le(1.938+o(1))^n$, weaker than the bound of Theorem 3.