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Source. Theorem 5, p. 2, with its proof in Section 3, pp. 4--5, 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

Definition (p. 2, after Alon, Shpilka and Umans, Definition 2.5). For k≤Dk\le D, a kk-sunflower in (Z/DZ)n(\mathbb Z/D\mathbb Z)^n is a set of kk vectors that in each coordinate are either all different or all the same. A set A⊂(Z/DZ)nA\subset(\mathbb Z/D\mathbb Z)^n is sunflower-free when it contains no 3-sunflower; the abstract (p. 1) phrases this as: every triple of distinct x,y,z∈Ax,y,z\in A has a coordinate ii in which exactly two of xi,yi,zix_i,y_i,z_i are equal.

Theorem 5 (p. 2, quoted). "Let D≥3D\geq 3, and let A⊂(Z/DZ)nA\subset(\mathbb{Z}/D\mathbb{Z})^n be a sunflower-free set. Then

∣A∣≤cDn|A|\leq c_D^n

where cD=322/3(D−1)2/3c_D=\frac{3}{2^{2/3}}(D-1)^{2/3}."

Proof pointer

Section 3, pp. 4--5. The proof replaces polynomials by the characters of Z/DZ\mathbb Z/D\mathbb Z. By orthogonality, a product over coordinates of character sums gives a function T(x,y,z)T(x,y,z), displayed as (3.1) on p. 5, that is nonzero exactly when x,y,zx,y,z form a sunflower or are all equal; on a sunflower-free AA it is diagonal, so Lemma 6 (p. 2) bounds ∣A∣|A| by its slice rank. Each term of the expansion has at most 2n2n nontrivial characters, so one of the three variables carries at most 2n/32n/3 of them; grouping by that variable bounds the slice rank by 3∑k≤2n/3(nk)(D−1)k≤3cDn3\sum_{k\le2n/3}\binom nk(D-1)^k\le3c_D^n, using D≥3D\ge3. A tensor-power amplification removes the factor 33.

Read depth

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

Dependencies

Lemma 6 (p. 2), quoted by the paper from Tao. Nothing in the corpus.

Bears on

  • Problem 20: by Alon, Shpilka and Umans (Theorem 2.6), a bound CnC^n with CC independent of DD for 3-sunflower-free sets in (Z/DZ)n(\mathbb Z/D\mathbb Z)^n would give the problem's bound for k=3k=3 with c3=e⋅Cc_3=e\cdot C (p. 2). The paper calls Theorem 5 progress towards that conjecture, but its cDc_D grows like D2/3D^{2/3}, so it gives no bound independent of DD and proves no case of the problem.