2012
DOI: 10.1090/s0894-0347-2011-00725-x
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New bounds on cap sets

Abstract: We provide an improvement over Meshulam's bound on cap sets in F N 3 . We show that there exist universal ǫ > 0 and C > 0 so that any cap set in F N 3 has size at most C 3 N N 1+ǫ . We do this by obtaining quite strong information about the additive combinatorial properties of the large spectrum.

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Cited by 68 publications
(184 citation statements)
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“…Meshulam [M95], following the general lines of Roth's argument, has shown that if G is an abelian group of odd order, then r 3 (G) ≤ 2|G|/ rk(G) (where we use the standard notation rk(G) for the rank of G); in particular, r 3 (Z n m ) ≤ 2m n /n. Despite many efforts, no further progress was made for over 15 years, till Bateman and Katz in their ground-breaking paper [BK12] proved that r 3 (Z n 3 ) = O(3 n /n 1+ε ) with an absolute constant ε > 0.…”
Section: Background and Motivationmentioning
confidence: 99%
“…Meshulam [M95], following the general lines of Roth's argument, has shown that if G is an abelian group of odd order, then r 3 (G) ≤ 2|G|/ rk(G) (where we use the standard notation rk(G) for the rank of G); in particular, r 3 (Z n m ) ≤ 2m n /n. Despite many efforts, no further progress was made for over 15 years, till Bateman and Katz in their ground-breaking paper [BK12] proved that r 3 (Z n 3 ) = O(3 n /n 1+ε ) with an absolute constant ε > 0.…”
Section: Background and Motivationmentioning
confidence: 99%
“…Meshulam [12], following the general lines of Roth's argument, has shown that if G is an abelian group of odd order, then r 3 (G)≤2|G|/rk(G) (where we use the standard notation rk(G) for the rank of G). Despite many efforts, no further progress was made for over 15 years, till Bateman and Katz [163] in their ground-breaking paper proved that r 3 (Z 3 n ) = O(3 n / n 1+ε ) with an absolute constant ε>0. Abelian groups of even order were first considered by Lev in 2004, who, as a further elaboration on the Roth-Meshulam proof, has shown that r 3 (G)<2|G|/rk(2G) for any finite abelian group G; here 2G = {2g: g is in G}.…”
Section: Combinatorial Resultsmentioning
confidence: 99%
“…2 Note, however, that the problem of counting a 3-term arithmetic progressions becomes trivial in characteristic 2, as it amounts to computing the number of triples (x, y, z) ∈ A 3 satisfying x + y = 2z ≡ 0 mod 2. 3 Note that while any 3-term progression in F n 3 forms a complete line, the proof makes no use of this special geometric feature.…”
Section: The Classicsmentioning
confidence: 99%