Multiple recurrence for two commuting transformations

Abstract

This paper is devoted to a study of the multiple recurrence of two commuting transformations. We derive a result which is similar but not identical to that of one single transformation established by Bergelson, Host and Kra. We will use the machinery of "magic systems" established recently by B. Host for the proof.

Ergodic averages of commuting transformations with distinct degree polynomial iterates

“…Since by hypothesis E µ (f |Z k−1 ) = 0, by (10) we have |||f ||| k,µ = 0, and as a consequence |||f ||| k,µx = 0 for µ-almost every x ∈ X by (9). This completes the proof.…”

“…Since E(f −f|Z k−1 ) = 0, by (10) we have |||f −f||| k = 0, and by inequality (7), all these integrals are equal to zero. This establishes that g · D k f dµ is equal to g · D kf dµ, and the announced result follows.…”

“…where each of the functions f i is equal to eitherf or to f −f , and at least one of the functions f i is equal to f −f . Since E(f −f |Z k−1 ) = 0, by (10) we have |||f −f ||| k = 0, and by inequality (7), all these integrals are equal to zero. This establishes that g · D k f dμ is equal to g · D kf dμ, and the announced result follows.…”

“…For example, one can take T 2 = T 2 1 , and choose the (non-ergodic) transformation T 1 , and the set A, so that (3) fails with any power of µ(A) on the right hand side for every n ∈ N (Theorem 2.1 in [7]). If ℓ = 2, d 1 = d 2 = 1, and the joint action of the transformations T 1 , T 2 is ergodic, then the result remains true up to a change of the exponent on the right hand side [10]. But even under similar ergodicity assumptions, the result probably fails when 3 exponents agree no matter what exponent one uses on the right hand side (a conditional counterexample appears in Proposition 5.2 of [14]).…”

Multiple recurrence for non-commuting transformations along rationally independent polynomials

“…. , n ) without ergodicity assumptions [CFH11] and the family (n, n) if T 1 and T 2 are jointly ergodic [Chu11], although in the latter case the optimal bound is µ(A) 4 and not µ(A) 3 as one would expect from the weakly mixing case.…”