SYNCHRONOUS RECURRENCE KAMEL HADDAD, WILLIAM OTT, AND RONNIE PAVLOV Abstract. Auslander and Furstenberg asked the following question in [1]: If (x, y) is recurrent for all uniformly recurrent points y, is x necessarily a distal point? We call such a point x weakly product recurrent. Haddad and Ott [4] answered the Auslander/Furstenberg question in the negative by identifying a class of weakly product recurrent points that are not distal; however, the points in this class are not minimal. This suggests a refinement of the Auslander/Furstenberg question: If x is minimal and weakly product recurrent, is x necessarily distal? Here we introduce a new notion of product recurrence and we show that there exist minimal points that exhibit this new type of product recurrence and yet are not distal. Contents 1. A refinement of a question of Auslander and Furstenberg 1 2. A new type of product recurrence 3 2.1. Inductive construction 3 2.2. Properties of A 3 3. Discussion 6 References 6 1. A refinement of a question of Auslander and Furstenberg Recurrence has long been of central importance throughout ergodic theory and dynamical systems. Here we study a phenomenon which we call synchronous recurrence, where a recurrent point x X recurs in tandem with certain recurrent points from other topological dynamical systems. A topological dynamical system consists of a compact metric space X together with a continuous map f : X X. For x X and A X, let R f (x, A) = {i N : f i (x) A}. We say that x X is recurrent if R f (x, U) is infinite for every neighborhood U of x. We can strengthen the notion of recurrence by requiring that the sets R f (x, U) possess additional structure. A Furstenberg family F is a collection of subsets of N such that if F 1 F and F 1 F 2, then F 2 F. For a Furstenberg family F and x X, we say that x is F-recurrent if R f (x, U) F for every neighborhood U of x. Dynamicists have studied several important Furstenberg families. (a) (F ) Let F = {E N : E is infinite}. F -recurrence is simply recurrence. (b) (F s ) A subset E N is said to be syndetic if there exists M N such that E {k, k + 1,..., k + M 1} = for every k N (that is, E has bounded gaps). Let F s denote the family of syndetic subsets of N. F s -recurrence is often called uniform recurrence. (c) (F ps ) A subset T N is said to be thick if for every L N there exists k N such that {k, k + 1,..., k + L 1} T. A subset E N is said to be piecewise-syndetic if E is the intersection of a thick set and a syndetic set. Let F ps denote the family of piecewise-syndetic sets. (d) (F Banach ) A subset E N is said to have positive upper Banach density if lim max # (E {k, k + 1,..., k + M 1}) > 0. M k N M Let F Banach denote the family of subsets of N with positive upper Banach density. Note that (1) F s F ps F Banach F. Date: January 3, 2015. 2010 Mathematics Subject Classification. 37B05, 37B10, 37B20, 54H20. Key words and phrases. Auslander-Furstenberg question, distality, Furstenberg family, product recurrence, recurrence, topological dynamical system, van der Waerden theorem.
2 KAMEL HADDAD, WILLIAM OTT, AND RONNIE PAVLOV We now define various types of synchronous recurrence. Classically, a recurrent point x X is said to be product recurrent if given any recurrent point y in any topological dynamical system (Y, g), the pair (x, y) is recurrent for (X Y, f g). We generalize this notion as follows: For any Furstenberg family F, we say that a recurrent point x X is F-product recurrent (F-pr) if for every topological dynamical system (Y, g) and every F-recurrent point y Y, the pair (x, y) is recurrent for the product system (X Y, f g). In terms of Furstenberg families, product recurrence is exactly F -pr. The inclusions in (1) yield the implications (2) F -pr = F Banach -pr = F ps -pr = F s -pr. What about the reverse implications in (2)? While studying the relationship between product recurrence and distality for general semigroup actions, Auslander and Furstenberg asked the following question. Question 1.1. ([1]) If the pair (x, y) is recurrent for all uniformly recurrent (i.e. F s -recurrent) points y, is x necessarily a distal point? Since product recurrence and distality are equivalent in the context of topological dynamical systems [3, Theorem 9.11], Question 1.1 is equivalent to this: Does F s -product recurrence imply product recurrence? Haddad and Ott [4] answered this question in the negative by proving the following result. Theorem 1.2. ([4]) Let (X, f) be a topological dynamical system. The point x X is F s -pr if it has the following property: For every neighborhood V of x there exists n = n(v ) N such that if E Z + is any finite set satisfying i j n for all distinct i, j E, then there exists p Z + such that f p+i (x) V for all i E. The hypothesis of Theorem 1.2 is satisfied by any point with a dense orbit in a sufficiently mixing system. Corollary 1.3. ([4]) Let (X, f) be a topological dynamical system. The point x X is F s -pr if the following hold. (S1) The orbit of x is dense in X. (S2) For every neighborhood V of x, there exists N = N(V ) such that for any k N, if n i N for 1 i k, then the intersection is nonempty. V f n1 (V ) f (n1+ +n k) (V ) The mixing property (S2) holds for both the full one-sided shift on a finite number of symbols and mixing subshifts of finite type. Corollary 1.3 implies that every point with a dense orbit in such a system is F s -pr. Haddad and Ott used Corollary 1.3 to answer Question 1.1 in the negative: Any point with a dense orbit in the full shift on a finite number of symbols has a fixed point in its orbit closure and is therefore F s -pr but not distal. Subsequent work has shown that F s -pr points exist in other contexts as well (see e.g. [2, 6, 7]). For example, a closed subset A of a topological dynamical system (containing at least two points) is said to be weakly mixing of order two if for any open sets U 1, V 1, U 2, V 2 in X with A U i and A V i for i = 1, 2, there exists k > 0 such that f k (V i A) U i for i = 1, 2. Oprocha and Zhang [7, Corollary 10] prove that if A X is a weakly mixing set of order two for (X, f) and if distal points are dense in A, then A contains a residual subset of F s -pr points that are not product recurrent. While F s -pr points appear to be plentiful, F ps -product recurrence turns out to be restrictive: A F ps - pr point is not only minimal [2, Theorem 3.4], but also distal [7, Theorem 3]. We may now update the implications in (2): F s -pr = F ps -pr F Banach -pr F -pr. Any point with a dense orbit in the full shift on a finite number of symbols is not minimal. More generally, any point satisfying the hypothesis of Theorem 1.2 is necessarily not minimal [2, Theorem 4.11]. We arrive at the following question. Question 1.4. Let (X, f) be a topological dynamical system. If x X is F s -pr and uniformly recurrent (i.e. F s -recurrent), is x necessarily distal?
SYNCHRONOUS RECURRENCE 3 2. A new type of product recurrence Inspired by Question 1.4, we define a new type of product recurrence. Let L denote the collection of lattices in N: L = { an + b : a N, b Z +, 0 b a 1 }. We introduce the Furstenberg family F (s L ) that consists of syndetic subsets of N that have nonempty intersection with every lattice in L. This paper studies the associated notion of F (s L )-product recurrence. Clearly F (s L ) F s, and so F s -pr implies F (s L )-pr. Our main result partially answers Question 1.4: We show that if x is uniformly recurrent and F (s L )-pr, then x need not be distal. In particular, we construct a class of points that are both uniformly recurrent and F (s L )-pr and we show that this class contains points that are not distal. This result implies that F s -pr cannot be weakened to F (s L )-pr in Question 1.4. The inductive construction takes place in the full two-shift Σ + 2 = {0, 1}Z+ with shift map σ. 2.1. Inductive construction. For every k N, we inductively choose n k N, A k {0, 1} n k, and w k A k, and then create the desired points by taking limits of the words w k. Our techniques are quite similar to those used in e.g. [5, 8, 9]. We choose n k using the van der Waerden theorem. Theorem 2.1 (van der Waerden). For any m, p N, there exists k = k(m, p) such that for any partition of {1, 2,..., k} into m pieces C 1,..., C m, one of the C i contains a p-term arithmetic progression. Corollary 2.2. For any m, p N, there exists l = l(m, p) such that the following holds for any set B {1, 2,..., l}. If the elements b 1 < < b r of B satisfy b 1 1 m, l b r m, and b i+1 b i m for 1 i < r, then B contains a p-term arithmetic progression. Proof of Corollary 2.2. Given any m and p, choose l = k(m, p)+m+1, where k(m, p) is as in the statement of Theorem 2.1. For any B {1, 2,..., l} satisfying the assumption of Corollary 2.2, let B i denote the translate B + (i 1) for 1 i m and define C 1 = B 1, C 2 = B 2 \ B 1, and C i = B i \ i 1 j=1 B j for 2 < i m. The sets C i form a partition of {b 1, b 1 +1,..., b r +m 1} and this set contains b r b 1 +m (l m) (m+1)+m = k elements, so Theorem 2.1 implies that one of the sets in the partition, say C i, contains a p-term arithmetic progression. But C i B i and therefore B i (and hence B itself) contains a p-term arithmetic progression. We have shown that l satisfies the properties required of l(m, p). Returning to the inductive construction, start with n 0 = 0, n 1 = 1, A 1 = {0, 1}, and choose any w 1 A 1. Given k, select n k+1, A k+1, and w k+1 as follows. n k+1 : Any multiple of n k satisfying { k (3) n k+1 > max 2 n i, l ( } n k, (2 k+2 ) A k + 1)n k + nk 1. i=1 A k+1 : Set of all words of length n k+1 which are concatenations of words in A k such that every A k -word is used at least once and all but at most 2 k+1 A k of the concatenated words are w k. w k+1 : Any member of A k+1 that begins with w k. In this way, we may define n k, A k, and w k for all k, where we have many possible choices of n k and w k at each step. For instance, we could define n 1 = 1 and A 1 = {0, 1} as above and then choose w 1 = 1. Then n 2 need only be greater than max(1, l(1, 17)) = 17, so we could take n 2 = 18. Then A 2 consists of all concatentations of 18 words from A 1 (namely 0 and 1) such that both are used, and all but at most 8 are 1s. For instance, 010100000111100110 / A 2 since there are ten A 1 -words which are not equal to w 1 (namely the 0s). We may then choose w 2 to be any word in A 2 beginning with w 1 = 1, for instance w 2 = 110001111111011011. Then we choose n 3, and continue in this way. Fix any legal choices of n k, and let A denote the set of all sequences which can be obtained as limits of the form lim k w k for any legal choices of w k as above. 2.2. Properties of A. We show that every point in A is uniformly recurrent and F (s L )-recurrent. Further, we show that A contains points that are not distal (equivalently, not product recurrent). Proposition 2.3. Every x A is uniformly recurrent.
4 KAMEL HADDAD, WILLIAM OTT, AND RONNIE PAVLOV Proof of Proposition 2.3. Let x A; then x is the limit of some sequence (w k ). For every neighborhood U of x, there exists k so that the cylinder set [w k ] is contained in U and therefore R(x, [w k ]) R(x, U). But R(x, [w k ]) is precisely the set of locations at which w k appears within x; this location set is always syndetic because x is a concatenation of A k +1-words (since w m is such a concatenation for every m > k + 1) and every A k +1-word contains w k. Proposition 2.4. Every x A is F (s L )-pr. Proof of Proposition 2.4. In the sequel, for any γ Σ + 2 and n N we let U n(γ) denote the cylinder set [γ[0]γ[1] γ[n 1]]. Let x A. We must show that for every topological dynamical system (Y, g) and every F (s L )-recurrent point y Y, the pair (x, y) is recurrent for the product system (Σ + 2 Y, σ g). Because the full shift is universal in this context [4, Section 4], it suffices to consider Y = Σ + 2 and g = σ. Let z Σ + 2 be F (s L )-recurrent. Let p, b N. We show that R(x, U p (x)) R(z, U b (z)). The set R(z, U b (z)) is in F s L and is therefore syndetic; let q denote the maximum gap between any two consecutive elements of R(z, U b (z)). Choose k such that n k > max{p, b} and n k+1 > q. It suffices to show that R(x, U nk (x)) R(z, U b (z)), since clearly U nk (x) U p (x). Since w k+1 is an A k+1 -word, it begins with w k, and so w k+1 [0]w k+1 [1] w k+1 [n k 1] = w k. Let Q = n k+1 N. Choose any interval I = {c, c + 1,..., d} of natural numbers containing n k+2 elements. We claim that all but at most 2 k+3 A k+1 members of Q I are in R(x, U nk (x)). To see this, note that since x is a concatenation of A k+2 -words, x[c] x[d] is a subword of x[jn k+2 ] x[(j + 2)n k+2 1] for some j, a concatenation of a pair of A k+2 -words. Each of these A k+2 -words is a concatenation of A k+1 -words, all but at most 2 k+2 A k+1 of which are w k+1. This implies that all but at most 2 k+3 A k+1 elements of Q I are in R(x, U nk (x)). Turning to z: Since the maximum gap between consecutive elements of R(z, U b (z)) is less than n k+1 and n k+2 n k > l ( n k+1, (2 k+3 A k+1 + 1)n k+1 ), R(z, U b (z)) {1,..., n k+2 n k } contains an arithmetic progression P of length (2 k+3 A k+1 + 1)n k+1. Let s denote the smallest element of P. Since R(z, U nk+2 (z)) intersects every lattice in L, there exists r R(z, U nk+2 (z)) such that s + r 0 (mod n k+1 ). Note then that and for every s P, implying that z[r]z[r + 1] z[r + n k+2 1] = z[0]z[1] z[n k+2 1] z[s ]z[s + 1] z[s + b 1] = z[0]z[1] z[b 1] z[s + r]z[s + r + 1] z[s + r + b 1] = z[0]z[1] z[b 1] as well. Therefore, the translate P := P + r is contained in R(z, U b (z)), and its smallest element s + r is a multiple of n k+1. Let J = {s + r, s + r + 1,..., s + r + n k+2 1}; notice that J contains n k+2 elements and P J. We claim that P (Q J) > 2 k+3 A k+1. To see this, write Define P = { s + r + jt : 0 j < (2 k+3 A k+1 + 1)n k+1 }. E = { s + r + j(tn k+1 ) : 0 j < 2 k+3 A k+1 + 1 }. This verifies the claim since E P (Q J) and E = 2 k+3 A k+1 + 1. Since J contains n k+2 elements, all but at most 2 k+3 A k+1 elements of Q J lie in R(x, U nk (x)). It follows that P R(x, U nk (x)), and therefore R(x, U nk (x)) R(z, U b (z)) because P R(z, U b (z)). This completes the proof that x is F (s L )-pr. Proposition 2.5. There exists q A such that q[0] = 0 and q[ k F n k] = 1 for every nonempty finite set F N. The point q is not product recurrent.
SYNCHRONOUS RECURRENCE 5 Proof of Proposition 2.5. Such a point q cannot be product recurrent. Define the IP set { } P = n k : F N, F <. k F Now, define z Σ + 2 by z[0] = 0, z[p] = 0 for every p P, and z[i] = 1 for all other values of i. The point z is recurrent since for every k, n k R(z, U nk (z)). However, (q, z) is not recurrent; clearly R(z, U 1 (z)) = P, but R(q, U 1 (q)) P =. Therefore, q is not product recurrent. The construction of q proving Proposition 2.5 uses the following lemma. Lemma 2.6. There exist choices of w k such that for every k N and m Z +, there exists an A k -word v k,m = v k,m [0] v k,m [n k 1] with v k,m [i m] = 1 for all i P [m, m + n k 1]. In addition, v k,0 = w k for every k. Proof of Lemma 2.6. We induct on k. Assume that for a fixed value of k, w 1,..., w k have been chosen and that Lemma 2.6 holds for every m. We verify Lemma 2.6 for k + 1 and all values of m. List the elements of A k as a 1, a 2,..., a Ak with a 1 = w k, choose any legal value of n k+1, and define the auxiliary word u k+1 by u k+1 = a n k+1 n (2 k +1)( A k 1) k 1 a 2k +1 2 a 2k +1 (We note that n k+1 > l(n k, (2 k+2 A k + 1)n k ) (2 k+2 A k + 1)n k > (2 k + 1)( A k 1)n k, and so the above definition makes sense.) We modify u k+1 to define an A k+1 -word v k+1,m such that v k+1,m [i m] = 1 for all i P such that m i m + n k+1 1. For every A k -word u k+1 [j] u k+1 [j + n k 1] in u k+1, if P {m + j,..., m + j + n k 1}, then use the inductive hypothesis to replace u k+1 [j] u k+1 [j + n k 1] with v k,m+j. This procedure produces v k+1,m. We must show that this v k+1,m is in A k+1. Define A k. P k+1,m = P { i Z + : m i m + n k+1 1 }. The word v k+1,m has been created by modifying at most P k+1,m of the A k -words in u k+1. We note that P k+1,m cannot contain two elements of the form k F n k and k F n k where F {1,..., k} = F {1,..., k}, since any two such numbers would differ by at least n k+1 by (3). Therefore, (4) P k+1,m 2 k. Since every A k -word appears at least 2 k +1 times in u k+1, estimate (4) implies that every A k -word appears at least once in v k+1,m. To see that a 1 appears at least 2 k + 1 times in u k+1, notice that and n k 2 k (implied by (3)) imply Finally, all but at most n k+1 > l(n k, (2 k+2 A k + 1)n k ) (2 k+2 A k + 1)n 2 k n k+1 n k (2 k + 1)( A k 1) > (2 k+2 A k + 1)n k (2 k + 1)( A k 1) (2 k+2 A k + 1) 2 k (2 k + 1)( A k 1) = A k (2 2k+2 (2 k + 1)) + 2 k + (2 k + 1) > 2 k + 1. (2 k + 1)( A k 1) + P k+1,m (2 k + 1)( A k 1) + 2 k < 2 k+1 A k of the A k -words in v k+1,m are w k. We conclude that v k+1,m A k+1 for every m. When m = 0, the first word a 1 = w k of u k+1 need not be modified to create v k+1,0 since w k = v k,0 already contains 1 symbols at every P-indexed location. Therefore, v k+1,0 begins with w k, and so we may choose w k+1 = v k+1,0, completing the induction. Returning to the proof of Proposition 2.5, we realize q as a limit of the form q = lim k v k,0 = lim k w k, and so q A.
6 KAMEL HADDAD, WILLIAM OTT, AND RONNIE PAVLOV 3. Discussion Further progress on Question 1.4 might be made by working with idempotents in the enveloping semigroup. Auslander and Furstenberg adopt this point of view in [1]. In the context of general semigroup actions, they prove that a point is product recurrent if and only if it is fixed by all maximal idempotents in the semigroup [1, Theorem 2]. This result is then used to formulate sufficient conditions under which product recurrence implies distality. The work of Auslander and Furstenberg inspires the following. Question 3.1. What characterizes F s -product recurrence? In particular, does a characterization exist in terms of idempotents in the enveloping semigroup? References [1] J. Auslander and H. Furstenberg, Product recurrence and distal points, Transactions of the American Mathematical Society, 343 (1994), pp. 221 232. [2] P. Dong, S. Shao, and X. Ye, Product recurrent properties, disjointness and weak disjointness, Israel Journal of Mathematics, 188 (2012), pp. 463 507. [3] H. Furstenberg, Recurrence in ergodic theory and combinatorial number theory, Princeton University Press, 1981. [4] K. Haddad and W. Ott, Recurrence in pairs, Ergodic Theory and Dynamical Systems, 28 (2008), pp. 1135 1144. [5] F. Hahn and Y. Katznelson, On the entropy of uniquely ergodic transformations, Trans. Amer. Math. Soc., 126 (1967), pp. 335 360. [6] P. Oprocha, Weak mixing and product recurrence, Annales de l Institut Fourier, 60 (2010), pp. 1233 1257. [7] P. Oprocha and G. Zhang, On weak product recurrence and synchronization of return times, Advances in Mathematics, 244 (2013), pp. 395 412. [8] R. Pavlov, Some counterexamples in topological dynamics, Ergodic Theory Dynam. Systems, 28 (2008), pp. 1291 1322. [9] K. E. Petersen, A topologically strongly mixing symbolic minimal set, Trans. Amer. Math. Soc., 148 (1970), pp. 603 612. California State University San Marcos University of Houston URL, William Ott: www.math.uh.edu/ ott/ University of Denver URL, Ronnie Pavlov: www.math.du.edu/ rpavlov/