aa r X i v : . [ m a t h . C O ] F e b LOWER BOUNDS FOR CORNER-FREE SETS
BEN GREEN
Abstract.
We show that for infinitely many N there is a set A ⊂ [ N ] of size 2 − ( c + o (1)) √ log N N not containing any configuration ( x, y ) , ( x + d, y ) , ( x, y + d ) with d = 0, where c = 2 q ≈ . . . . . Let q, d be large positive integers. For each x ∈ [ q d − π ( x ) =( x , . . . , x d − ) ∈ Z d for the vector of digits of its base q expansion, thus x = P d − i =0 x i q i , with 0 x i < q for all i .For each positive integer r , consider the set A r of all pairs ( x, y ) ∈ [ q d − for which k π ( x ) − π ( y ) k = r and q x i + y i < q for all i .We claim that A r is free of nontrivial corners. Suppose that ( x, y ) , ( x + d, y ) , ( x, y + d ) ∈ A r . Then k π ( x ) − π ( y ) k = k π ( x + d ) − π ( y ) k = k π ( x ) − π ( y + d ) k = r. (0.1)We claim that π ( x + d ) + π ( y ) = π ( x ) + π ( y + d ) . (0.2)To this end, we show that ( x + d ) i + y i = x i + ( y + d ) i for i = 0 , , . . . byinduction on i . A single argument works for both the base case i = 0 andthe inductive step. Suppose that, for some j >
0, we have the statementfor i < j . Write x > j := P i > j x i q i , and define ( x + d ) > j , y > j , ( y + d ) > j similarly. By the inductive hypothesis and the fact that x + ( y + d ) =( x + d ) + y , we see that x > j + ( y + d ) > j = ( x + d ) > j + y > j . Therefore x j + ( y + d ) j = ( x + d ) j + y j (mod q ). However by assumption we have q x j + ( y + d ) j , ( x + d ) j + y j < q , and so x j + ( y + d ) j = ( x + d ) j + y j .The induction goes through.With (0.2) established, let us return to (0.1). We now see that this state-ment implies that k a k = k a + b k = k a − b k = r , where a := π ( x ) − π ( y )and b := π ( x + d ) − π ( x ) = π ( y + d ) − π ( y ). By the parallelogram law2 k a k + 2 k b k = k a − b k + k a + b k , this immediately implies that b = 0.Since π is injective, it follows that d = 0 and so indeed A r is corner-free.The set of all pairs ( x, y ) with q x i + y i < q for all i has size ( q + O ( q )) d . Therefore by the pigeonhole principle there is some r such that A r > ( dq ) − ( q + O ( q )) d . Mathematics Subject Classification.
Primary .The author is supported by a Simons Investigator Grant.
Now for a given d set q := ⌊ (2 / √ d ⌋ and N := q d . Then A r ⊂ [ N ] × [ N ],and A r is free of corners. Moreover N − A r > ( dq ) − ( 34 + O ( 1 q )) d = 2 − ( c + o (1)) √ log N , where c = 2 q ≈ . . . . . Remark.
The construction came about by a careful study of the recentpreprint of Linial and Shraibman [1], where they used ideas from communica-tion complexity to obtain a bound with c = 2 p log e ≈ . . . . , improvingon the previously best known bound with c = 2 √ ≈ . . . . which comesfrom Behrend’s construction. By bypassing the language of communicationcomplexity one may simplify the construction, in particular avoiding the useof entropy methods. This yields a superior bound. References [1] N. Linial and A. Shraibman,
Larger corner-free sets from better NOF exactly- N pro-tocols, preprint, arXiv:2012.00421. Mathematical Institute, Radcliffe Observatory Quarter, Woodstock Road,Oxford OX2 6GG, England
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