OPTIMAL LOWER BOUNDS FOR MULTIPLE RECURRENCE

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1 OPTIMAL LOWER BOUNDS FOR MULTIPLE RECURRENCE SEBASTIÁN DONOSO, ANH NGOC LE, JOEL MOREIRA, AND WENBO SUN Abstract. Let X, B, µ, T ) be an ergodic measure preserving system, A B and ɛ > 0. We study the largeness of sets of the form S = { n N: µa T fn) A T f2n) A... T fkn) A) > µa) k+ ɛ } for various families f,..., f k ) of functions f i : N Z. For k 3 and f i n) = ifn), we show that S has positive density if fn) = qp n) where q Z[x] satisfies q) = 0 and p n) is the sequence of primes; or when f is a Hardy field sequence. If T q is ergodic for some q N, then for all r Z, S is syndetic if fn) = qn + r. For f i n) = a i n, where a i are distinct integers, we show that S can be empty for k 4, and for k = 3 we found an interesting relation between the largeness of S and the existence and abundance) of solutions to certain linear equations in sparse sets of integers. We also provide several partial results when the f i are distinct polynomials.. Introduction.. Historical background. The classical Poincaré recurrence theorem states that for every measure preserving system X, B, µ, T ) and every set A B with µa) > 0, there exists some n N such that µa T n A) > 0. This result was improved by Khintchine in [7], who showed that under the same conditions, for every ɛ > 0, the set S := { n : µa T n A) > µa) 2 ɛ } is syndetic, meaning that it has bounded gaps. Taking a mixing system, one sees that the bound µa) 2 is optimal. In [4], Furstenberg established a multiple recurrence theorem, showing that for every measure preserving system X, B, µ, T ), every k N and set A B with µa) > 0, there exists a syndetic set S N such that for all n S, we have ) µa T n A T kn A) > 0. One could hope to improve Furstenberg s multiple recurrence theorem in the same way that Khintchine s theorem strengthens Poincaré s. Since for a system mixing of all orders, the left hand side of ) approaches µa) k+ as n, one could hope that under the same conditions as Furstenberg s multiple recurrence theorem, for every ɛ > 0, the set { 2) n : µa T n A T 2n A T kn A) > µa) k+ ɛ } is syndetic. This was showed to be true by Furstenberg when the system is weakly mixing, and the general case was finally settled by Bergelson Host and Kra in [3], who showed that if the system X, B, µ, T ) is ergodic, then the set in 2) is syndetic when k =, 2, 3 with the case k = following from Khintchine s theorem). However, the set in 2) may be empty if the system is not ergodic or if k 4:

2 2 SEBASTIÁN DONOSO, ANH NGOC LE, JOEL MOREIRA, AND WENBO SUN Theorem. See [3, Theorems 2. and.3]). There exist a non ergodic) measure preserving system X, B, µ, T ) and for each l N a set A B with µa) > 0 such that for every n N \ {0} µa T n A T 2n A) 2 µa)l There exist a totally ergodic measure preserving system X, B, µ, T ) and for each l N a set A B with µa) > 0 such that for every n N \ {0} µa T n A T 4n A) 2 µa)l The first part of this theorem explains why one needs to focus on ergodic systems when studying optimal recurrence. Furstenberg s multiple recurrence theorem has been extended in several different directions, each leading to the question of whether or under which conditions) can optimal recurrence be achieved. In this paper, we are mostly concerned with expressions of the form µa T fn) A T f2n) A T f kn) A) for various families f,..., f k ) of functions f i : N Z. In most cases where recurrence has been established, optimal recurrence can be obtained for weakly mixing systems cf. [] when the f i are polynomials and [2, 8, 9] for more general f i ), or when the functions are independent see [2, 3] for the case of linearly independent polynomials and [9] for more general f i with different growth). In the general case, besides the aforementioned paper [3], the main progress was obtained by Frantzikinakis in [6], where the case when k 3 and the f i are polynomials is studied in detail..2. Optimal recurrence along T fn), T 2fn),..., T kfn) ). Our first result concerns the sequence p n ) n N of primes and answers a question of Kra. Multiple recurrence along polynomials evaluated at primes was established by Frantzikinakis, Host and Kra in [0, ]. Our result states that one can also obtain optimal recurrence in this setting. Theorem.2. Let p n ) n N be the increasing) enumeration of the primes, let X, B, µ, T ) be an invertible ergodic measure preserving system and let f Z[x] be such that f) = 0. Then for every A B, ɛ > 0 and k {, 2, 3}, the set { 3) n N: µa T fp n) A T 2fpn) A T kfpn) A) > µa) k+ ɛ } has positive lower density. Theorem.2 follows from the stronger Theorem 3.2 below. syndetic in general. We remark that the set in 3) is not In fact, it follows from [22] that when fx) = x, for every non-trivial finite system, there exists A B such that the set in 3) has unbounded gaps. A similar result can be obtained if the sequence fp n ) is replaced with the sequence fn), where x is the largest integer not greater than x, and f is a function belonging to a Hardy field with polynomial growth and sufficiently far away from Q[x]. More precisely, denote by G the set of all equivalence classes of smooth functions R R, where f g if there exists a constant c > 0 such that fx) = gx) for all x > c. A Hardy field is a subfield of the ring G, +, ) which is closed under differentiation. Let H be the union of all Hardy fields. We say that a function ax) has polynomial growth if there exists d N such that ax)/x d 0. E {,...,N} The lower density de) of a set E N is the number de) = inf. N N

3 OPTIMAL LOWER BOUNDS FOR MULTIPLE RECURRENCE 3 Theorem.3. Let a H have polynomial growth and satisfy ax) cpx) / logx) for every c R, p Z[x]. Then for every invertible ergodic measure preserving system X, B, µ, T ), every A B, every ɛ > 0 and every k {, 2, 3}, the set { 4) n N: µa T an) A T 2 an) A T k an) A) > µa) k+ ɛ } has positive lower density. Theorem.3 is proved in Section 3.2. Examples of functions that satisfy the conditions in the previous theorem are ax) = x c where c > 0, c Z, ax) = x log x, ax) = x x 3, and ax) = x 3 + log x) 3. We point out that in Theorem.3 we cannot replace has positive density by is syndetic. This is easy to see for certain functions ax) growing slowly for instance ax) = x c when c < ). For such functions, an) is constant in arbitrarily long intervals and takes every value which is large enough. Therefore there are gaps of the set 4) which are arbitrarily long. On the other hand, we expect the set in 4) to be thick, i.e. contain arbitrarily long intervals, whenever a H has polynomial growth. Some evidence in this direction is given in [4], where the set { n N: µa T an) A T k an) A) > 0 } is shown to be thick, as well as the set in 4) when k =. Our third result concerns sequences of the form fn) = qn + r for fixed q, r Z and was suggested by Kra. Theorem.4. Let q, r Z, with q > 0, and X, B, µ, T ) be a measure preserving system with T q ergodic. Let A B, ɛ > 0 and k {, 2, 3}. Then the set { 5) n N: µa T qn+r) A T 2qn+r) A T kqn+r) A) > µa) k+ ɛ } is syndetic. Theorem.4 follows from Theorem 3.5 below, which deals with a more general situation involving Beatty sequences. Observe that the conclusion of Theorem.4 is equivalent to the statement that the intersection { n N: µa T n A T 2n A T kn A) > µa) k+ ɛ } qz + r ) is syndetic. If in 5) one replaces the optimal lower bound µa) k+ ɛ with 0, then the set is syndetic for any k N. This was proved in [5] for k = 2 and k = 3, and for larger k this is essentially the content of [5, Corollary 6.5]; see also [2]..3. Optimal recurrence along T an, T a2n,..., T adn ). Next we study obtaining optimal recurrence for the expression µt an A T a2n A T adn A) where a,..., a d are distinct integer numbers. In particular, if a i = i, then the results of Bergelson, Host and Kra tell us that we have optimal recurrence if and only if d 4. More generally, in [6] it is proved that if d 3, or d = 4 and a 2 + a 3 = a + a 4, then optimal recurrence holds, but any other case is not known. Expanding an argument of Ruzsa, presented in the appendix of [3], we prove that for d 5, one does not have optimal recurrence.

4 4 SEBASTIÁN DONOSO, ANH NGOC LE, JOEL MOREIRA, AND WENBO SUN Theorem.5. Let a <... < a 5 be pairwise distinct integers. There exists an ergodic system X, B, µ, T ) such that for every l > 0, there exists a set A B with µa) > 0 such that for every non-zero integer n. µt an A T a2n A... T a5n A) 2 µa)l Theorem.5 is proved in Section 4.. The cases not covered by the above results seem difficult to address. For instance, it is not known whether for every ergodic measure preserving system, every set A and every ɛ > 0 there exists a syndetic set of) n for which µa T 2n A T 3n A T 4n A) > µa) 4 ɛ. In [6], Frantzikinakis showed that a positive answer to this question would imply the existence of solutions to a certain linear equation in sparse sets. We obtain a converse result, showing a tight connection between optimal lower bounds for multiple recurrence and solutions to linear equations in sparse sets. In order to formulate our result, we need to introduce some notation. Definition.6. Let m, d, N N. Denote [N] := {0,,..., N }. Given a set E [N] m and a subspace V Q d m, denote D m,n V, E) = V Ed V [N] d m d m,n E) = E N m. Observe that a point x,..., x dm ) [N] d m belongs to V if and only if the coordinates x,..., x dm satisfy some system of linear equations. The reader should think of D m,n V, E) as the proportion of solutions to that system of equations with all variables in E. Definition.7. A subset S N is a Bohr 0 if there exist d N, ρ > 0 and α T d such that S = {n N : nα T d < ρ}, where T d denotes the distance to the identity in T d. Theorem.8. Let l > 4 and a,..., a 4 Z be distinct. Let V be the subspace of Q 4 spanned by a i,..., a i 4) for 0 i 2. 2 Let C > 0 and suppose that for every m N, every sufficiently large N and subset E [N] m, we have D m,n V m, E) Cd m,n E) l. Then for every invertible ergodic system X, B, µ, T ) and every A B with µa) > 0, there exists a Bohr 0 set S N such that sup S [M, N) Theorem.8 is proved in Section 4.3. n S [M,N) µt an A T a4n A) C 4 l ) l µa) l. Remark.9. It is easy to see that the conclusion of Theorem.8 implies that the set { 6) n N: µt an A T a4n A) C 4 ) } l µa) l ɛ l is syndetic for all ɛ > 0. Unfortunately, the condition D m,n V m, E) Cd m,n E) l seems difficult to verify in concrete instances, even for m =. We obtain a partial converse to Theorem.8 which shows that it is essentially as difficult as establishing optimal lower bounds for the corresponding multiple recurrence problem. 2 We adopt the convention that if some ai equals to 0, then a 0 i = 00 =.

5 OPTIMAL LOWER BOUNDS FOR MULTIPLE RECURRENCE 5 Theorem.0. Let l 4 and a,..., a 4 Z be distinct. Let V be the subspace of Q 4 spanned by a i,..., a i 4) for 0 i 2. Let C > 0 and suppose that for every ergodic system X, B, µ, T ) and every A B with µa) > 0, there exists a Bohr 0 set S N such that sup S [M, N) n S [M,N) µt an A T a4n A) CµA) l. Then for every m N, every sufficiently large N and every E [N] m, D m,n V m, E) Cβ m d m,n E) l, where β > 0 is an explicit constant depending only on a,..., a 4 and l. Theorem.0 is proved in Section 4.2. We provide some examples to illustrate Theorems.8 and.0. Example.. a, a 2, a 3, a 4 ) = 0, 2, 3, 4). 0, 4, 9, 6). It is not hard to show that In this case V is the Q-span of,,, ), 0, 2, 3, 4) and For convenience set m =. V = { x, y, z, w) Q 4 : x 6y + 8z 3w = 0 }. Then D,N V, E) is essentially the density of solutions of the equation x 6y + 8z 3w = 0 in E, i.e. the proportion of tuples x, y, z, w) [N] 4 satisfying x 6y + 8z 3w = 0 that belong to E 4. The condition in Theorem.8 can be rephrased informally as saying that this density can be bounded from below by the l-th power of the density d,n E) of the set E. Example.2. Suppose that a + a 2 = a 3 + a 4. In this case, an elementary computation shows that V = { x, y, z, w) Q 4 : sx y) + tz w) = 0 }, where s = a 3 a 4 and t = a 2 a. We can assume, without loss of generality, that both s and t are positive. Given N, m N and a set E [N] m, denote by P n) the number of pairs x, z) E 2 satisfying sx+tz = n for each n N m. Observe that if x, y, z, w) E 4 V m then sx+tz = sy +tw [ s+t)n ] m. We have P n) = E 2 and [ ] m n s+t)n It follows from the Cauchy-Schwarz inequality that 7) β m d m,n E) 4 D m,n V, E) N 3 V [N] 4 s+t) P n) 2 = E 4 V m. [ ] m n s+t)n E 4 N m s+t) E 4 V m, which in turn implies that m where 0 < β < it is easy to see that the it exists and is positive). N The conclusion 7) also follows from combining Theorem.0 with [6, Theorem C]..4. Optimal recurrence along polynomials. In [6], Frantzikinakis studied in detail the optimal recurrence for polynomial sequences with k 3 and dealt with most cases in that regime. However, some stubborn questions remain unanswered. For instance, it is not known if there exists l > 0 such that for every ergodic system X, B, µ, T ), every A B and every ɛ > 0, the set 8) { n N: µa T n A T 2n A T n2 A) > µa) l ɛ }

6 6 SEBASTIÁN DONOSO, ANH NGOC LE, JOEL MOREIRA, AND WENBO SUN is non-empty let alone syndetic). Intriguingly enough, if we replace n 2 by n 3 or by n d for any d > 2) and set l = 4 in 8), then by Theorem B and Section 4.2 in [6] we deduce that the set obtained is syndetic. We give a partial positive result for a situation where 8) is syndetic. Proposition.3. Let X, B, µ, T ) be an ergodic system and let Z 3 be the 3-step nilfactor of X see Section 2 for the definition). Assume that Z 3 is an inverse it of nilsystems that can be represented as G/Γ, with G is a connected. Then for every A B and ɛ > 0, the set { n N: µa T n A T 2n A T n2 A) > µa) 4 ɛ } is syndetic. Remark.4. In particular, the hypothesis of Proposition.3 is satisfied if X = G/Γ, B, µ, T ) is an ergodic nilsystem with G being a connected Lie group. We are unable to remove the connectedness assumption. Hence the general question regarding optimal recurrence for 0, n, 2n, n 2 ) remains open. However in next result, we provide an example of lack of optimal recurrence for this family in the case of two commuting transformations. Proposition.5. There exists a system X, B, µ, T, T 2 ), with T ergodic, T T 2 = T 2 T such that for every integer l > 0, there exists A B with µa) > 0 such that for every positive integer n. µa T n A T 2n A T n2 2 A) 2 µa)l Proposition.3 and Proposition.5 are proved in Section 5. Acknowledgements. We thank Bryna Kra for proposing to us the questions studied in this paper and for many helpful discussions. We thank the Mathematics Research Communities program of the AMS for giving us the optimal environment to start this project. The first author is grateful for the support of Fondecyt Iniciación Grant 6006, and the third author is supported by the NSF via grant DMS Background 2.. Nilmanifolds, nilsystems and nilsequences. Given a group G, we denote its lower central series by G = G G 2, where each term is defined by G i+ = [G i, G], i.e., G i+ is the subgroup of G generated by all the commutators [a, b] := aba b with a G i and b G. The group G is a k-step nilpotent group if G k+ is the trivial group. Let G be a k-step nilpotent Lie group and let Γ be a uniform i.e closed and cocompact) subgroup of G. The compact homogeneous space X := G/Γ is called a k-step nilmanifold. Let π : G X be the standard quotient map. We write X = π G ) where G is the identity element of G. Denote by G 0 the connected component of G containing the identity G. If X is connected, then X = πg 0 ). The space X is endowed with a unique probability measure that is invariant under translations by G. This measure is called the Haar measure for X, and denoted by µ X. For every τ G, the measure preserving system X, B, µ X, T ) given by T x = τ x, x X is called a k-step nilsystem, where B is the Borel σ-algebra of X. Let CX) denote the set of continuous functions on X. ψn) := ft n x) is called a basic k-step nilsequence. k-step nilsequences. For f CX) and x X, the sequence A k-step nilsequence is a uniform it of basic

7 OPTIMAL LOWER BOUNDS FOR MULTIPLE RECURRENCE 7 We say that a sequence x n ) n N is equidistributed on a nilmanifold X if for every F CX), we have N N N F x n ) = n= Similarly, we say that x n ) n N is well distributed on X if for all F CX). N X F x n ) = F dµ X. X F dµ X Nilfactors. Let X, B, µ, T ) be an ergodic measure preserving system. Suppose s j n)) n N is an integer valued sequence for j k. A factor Y, D, ν, S) of X is said to be characteristic for s n),..., s k n)) if for any bounded functions f,..., f k on X, we have N k T sjn) f j N k T sjn) E f j Y ) = 0, N N N n= j= n= j= where E f Y ) denotes the conditional expectation of f onto the factor Y and the it is taken in L 2 X, µ). Host and Kra [6] showed that there exists a characteristic factor for n, 2n,..., kn) which is an inverse it of k )-step nilsystems. We call this factor the k )-step nilfactor of X and denote it by Z k X) or Z k when there is no confusion) Limit formula for multiple averages on nilsystems. The following description of the iting distribution of multiple ergodic averages in nilsystems is essentially due to Ziegler [23]. Theorem 2.. Let a,..., a d Z be distinct, let X = G/Γ, B, µ, T ) be a k-step ergodic nilsystem and let f, f 2,..., f d L µ). For each i =,..., k, let Γ i = Γ G i and let µ i be the Haar measure of G i /Γ i. Then for µ-a.e. x = gγ X, we have 9) N f T an x)... f d T a dn x) =... G /Γ G k /Γ k i= d f i gg a i ) k ) k... g ai ) Γ dµ k g k Γ k )... dµ g Γ ). Remark 2.2. Theorem 2. in particular asserts that the right hand side of 9) does not depend on the choice of representative g i for the co-set g i Γ i. Remark 2.3. The statement in [23, Theorem.2] requires G to be connected and simply connected. These restrictions were removed in [3, Theorem 5.4], although in that paper the it is described in a different but equivalent) form; see also [20, Theorem 6.3]. Let X = G/Γ, B, µ, T ) be an ergodic nilsystem. Then its Kronecker factor Z is G/G 2 Γ), T ). Let π : X Z be the natural projection. Suppose that T x = τ x for all x X and some τ G. Let α be the projection of τ on Z. Define 0) S δ := {n N: α n Bδ)},

8 8 SEBASTIÁN DONOSO, ANH NGOC LE, JOEL MOREIRA, AND WENBO SUN where Bδ) is the ball in Z centered at 0 with radius δ. Observe that S δ is a Bohr 0 set and by ergodicity, the uniform density ds δ ) of S δ is ds δ ) := where µ Z is the Haar measure on Z. S δ [M, N) = µ Z Bδ)), We need the following proposition, whose proof for case d = 3 is sketched in [6, Page 35]. The proof for general d is similar and included here for completeness. Proposition 2.4. Let a,..., a d Z be distinct, let X = G/Γ, B, µ, T ) be a k-step ergodic nilsystem and let f, f 2,..., f d L µ). For each i =,..., k, let Γ i = Γ G i and let µ i be the Haar measure of G i /Γ i. Also, for each δ > 0 let S δ be defined by 0). Then for µ almost every x = gγ X, we have: δ 0 G 2/Γ 2 S δ [M, N) G 2/Γ 2 n S δ [M,N) i=... G 3/Γ 3 G k /Γ k i= d f i T ain x) = d f i gg a i ) k ) k... g ai ) Γ dµ k g k Γ k )... dµ 2 g 2 Γ )dµ 2 g Γ ). Proof. Let π : X Z be the natural projection. For any character χ of the compact abelian group Z = X/G 2, the composition χ π is in L µ), and χ πt n x) = χ nα + πx) ) for all n N and x X. On the other hand, χ πghγ) = χ πgγ) whenever h G 2. x = gγ X, we have ) N χ nα + πx) ) d f i T ain x) = i=... χ πgg Γ) ) d f i G /Γ G k /Γ k i= gg a i ) By Theorem 2., for µ-almost every k ) k... g ai ) Γ dµ k g k Γ k )... dµ g Γ ). As χ is a character of Z, we have χ nα+πx) ) = χnα)χ i πx) ), and χ πgg Γ) ) = χ πgγ) ) χ πg Γ) ). Note that x = gγ. After canceling χ πx) ) from both sides of ), we get: 2) N χnα) d f i T ain x) = i=... χ πg Γ) ) d f i G /Γ G k /Γ k i= gg a i ) k ) k... g ai ) Γ dµ k g k Γ k )... dµ g Γ ). We can approximate the Riemann integrable function Bδ) by finite linear combinations of characters, and so we can replace χ in 2) with Bδ) to get: 3) N Bδ) nα) d f i T ain x) = i=... Bδ) πg Γ) ) d f i G /Γ G k /Γ k The left hand side of 3) is equal to: m Z Bδ) ) i= S δ [M, N) gg a i ) k ) k... g ai n S δ [M,N) i= ) Γ dµ k g k Γ k )... dµ g Γ ). d f i T ain x).

9 OPTIMAL LOWER BOUNDS FOR MULTIPLE RECURRENCE 9 On the other hand, the right hand side of 3) is equal to: π Bδ))... G 2/Γ 2 G k /Γ k i= d f i gg a i ) Let µ δ be the probability measure on X defined by f dµ δ = m Z Bδ)) X k ) k... g ai π Bδ)) f dµ X ) Γ dµ k g k Γ k )... dµ g Γ ). f CX). Since µ X is invariant under the action of G and hence of G 2 ) and the set π Bδ)) is invariant under G 2, we have that µ δ is invariant under the action of G 2. Moreover, any it point of {µ δ : δ > 0} is supported on G 2 /Γ 2. This shows that µ δ µ G2/Γ 2 as δ 0, where µ G2/Γ 2 is the Haar measure on G 2 /Γ 2. Therefore, dividing both sides of 3) by µ Z Bδ)) and taking the it as δ 0, we obtain the desired conclusion. We also need the following proposition whose proof is sketched in [6, Page 34]. Proposition 2.5. Let X, B, µ, T ) be an ergodic system and define S δ as in 0). Let a, a 2, a 3 Z be distinct and f, f 2, f 3 L µ). Assume that Ef i Z 2 ) = 0 for some i 3. Then 4) f T an x)f 2 T a2n x)f 3 T a3n x) = 0, S δ [M, N) where the it is taken in L 2 µ). n S δ [M,N) Proof. Without loss of generality, we assume Ef Z 2 ) = 0. Let L be the it on the left hand side of 4) and ds δ ) be the Banach density of S δ. Then 5) ds δ )L = N Sδ n)f T an x)f 2 T a2n x)f 3 T a3n x) = N Bδ) nα)f T an x)f 2 T a2n x)f 3 T a3n x). Approximating the Riemann integrable function Bδ) by linear combinations of characters, it suffices to show 6) N χnα)f T an x)f 2 T a2n x)f 3 T a3n x) = 0 for all character χ of Z. Note that the it in the left hand side of 6) is equal to 7) χx) N χnα + x)f T an x)f 2 T a2n x)f 3 T a3n x). By [6, Theorem. and 2.], the above it exists in L 2 µ) and does not change if we replace f i by Ef i Z 3 ). Therefore, by approximation, we can assume that X, B, µ, T ) is a 3-step nilsystem. First suppose that X, B, µ, T ) is totally ergodic. Then we can assume that its Kronecker factor Z has the form G, G, m, α), where G is a connected compact abelian group, G is the Borel σ-algebra, m is the Haar measure and α is the rotation defined above. Since G is connected, there exists g G such that a 2 g = α. Let α/a 2 denote that element. Consider the system Y = X G, B G, µ m, T α/a 2 ). Since Ef Z 2 X)) = 0, for almost every ergodic component Y t of Y we have Ef Z 2 Y t )) = 0 one

10 0 SEBASTIÁN DONOSO, ANH NGOC LE, JOEL MOREIRA, AND WENBO SUN way to verify is to show f 3 = 0 where k is Host-Kra s seminorm defined in [6]). Hence by [6, Theorem 2.], 8) N T α/a 2 ) an f T α/a 2 ) a2n f 2 χ T α/a 2 ) a3n f 3 = 0, where the it is taken in L 2 µ m). Rewriting the left hand side of 8), we get 9) N χnα + y)f T an x)f 2 T a2n x)f 3 T a3n x) = χy) N for all y G. Since χy) 0 for all y G, 9) implies 6). χnα)f T an x)f 2 T a2n x)f 3 T a3n x) = 0 We now return to general situation without the total ergodicity assumption. Let k be the number of connected components of X. Since X, B, µ, T ) is ergodic, X, B, µ, T k ) is totally ergodic. For all 0 i k, applying the above argument with T k, T ai f, T a2i f 2, T a3i f 3 replacing T, f, f 2, f 3, respectively, we get N χkn + i)α)f T akn+i) x)f 2 T a2kn+i) x)f 3 T a3kn+i) x) = 0 for all character χ of Z. Taking the average over all 0 i k, we derive 6). This finishes the proof. 3. Optimal recurrence along T fn), T 2fn),..., T kfn) ) 3.. Optimal recurrence for the sequence of shifted primes. We begin this section by recalling the following classification of certain tuples Q n), Q 2 n), Q 3 n)) of polynomials, introduced in [6]. Definition 3.. A family of polynomials Q n), Q 2 n), Q 3 n) Z[n] is said to be of type e ), e 2 ) or e 3 ) if some permutation of them has the form e ) {lq, mq, rq} with 0 l < m < r and l + m r. e 2 ) {lq, mq, kq 2 + rq} e 3 ) {kq 2 + lq, kq 2 + mq, kq 2 + rq} for some q Q[n] and constants k, l, m, r Z with k 0. We prove a stronger version of Theorem.2. Theorem 3.2. Let p n ) n N be the increasing enumeration of the primes. Let X, B, µ, T ) be an ergodic measure preserving system, A B and ɛ > 0. Suppose Q, Q 2, Q 3 are integer polynomials with Q i 0) = 0 for i =, 2, 3. Then the sets and {n N : µa T Qpn ) A) > µa) 2 ɛ} {n N : µa T Qpn ) A T Q2pn ) A) > µa) 3 ɛ} have positive lower density. Moreover, the set 20) {n N : µa T Qpn ) A T Q2pn ) A T Q3pn ) A) > µa) 4 ɛ} also has positive lower density unless the polynomials are pairwise distinct and of type e ), e 2 ) or e 3 ).

11 OPTIMAL LOWER BOUNDS FOR MULTIPLE RECURRENCE Proof. We only prove the set in 20) has positive density the under given hypothesis, as the proofs for the other two sets are similar. Fix ɛ > 0 and assume that the family Q, Q 2, Q 3 is not of type e ), e 2 ) nor e 3 ). Denote for n N. cn) = µa T Qn) A T Q2n) A T Q3n) A) By [9, Theorem 4.], the sequence cn) can be decomposed as cn) = ψn) + δn), where ψn) is a nilsequence and 2) By [8, Theorem.], we also have 22) N N N δn) = 0. N δp n ) = 0. n= Since a nilsequence is a uniform it of basic nilsequences, there exists a basic nilsequence F b n Y ) such that ψn) F b n Y ) < ɛ/4 for all n N. Here F is a continuous function on a nilmanifold Y = G/Γ, b G acts ergodically on Y and Y = π G ) Y. Assume that Y has d connected components and Y 0 is the component containing Y. Observe that b dn Y Y 0 for all n N. Since the polynomial family Q n), Q 2 n), Q 3 n) is not of the types e ), e 2 ) nor e 3 ), the polynomial family P n) = Q dn), P 2 n) = Q 2 dn), P 3 n) = Q 3 dn) is also not of these types. Hence by [6, Theorem C], the set S = {n N : cdn) > µa) 4 ɛ/4} is syndetic. Together with 2), we get δdn) = 0, N [N] S which implies sup N [N] S n [N] S n [N] S cdn) F b dn Y ) < ɛ/4. We deduce that there exists an n such that F b dn Y ) > µa) 4 ɛ/2. Since b dn Y Y 0 and F is continuous, there is an open subset U of Y 0 such that F > µa) 4 3ɛ/4 on U. By [8, Corollary.4], the sequence b pn Y is equidistributed on Y 0 when restricted to p n mod d. Hence the set R := {n N : b pn Y U} has positive density, and for every n R we have F b pn Y ) > µa) 4 3ɛ/4. On the other hand, from 22) it follows that the set R := {n R : cp n ) < µa) 4 ɛ} has 0 density. Therefore the set R \ R has positive density and is contained in the set 20). This finishes the proof Optimal recurrence for Hardy sequences. We prove a slight generalization of Theorem.3. Theorem 3.3. Let a H have polynomial growth and satisfy ax) cpx) / logx) for every c R, p Z[x]. Let X, B, µ, T ) be an ergodic measure preserving system, A B and ɛ > 0. Let 0 l m r Z. Then the sets and {n N : µa T l an) A) > µa) 2 ɛ} {n N : µa T l an) A T m an) A) > µa) 3 ɛ} have positive lower density. If r = l + m then the set 23) {n N : µa T l an) A T m an) A T r an) A) > µa) 4 ɛ}

12 2 SEBASTIÁN DONOSO, ANH NGOC LE, JOEL MOREIRA, AND WENBO SUN also has positive lower density. Proof. The proof is similar to that of Theorem 3.2. Let cn) = µa T ln A T mn A T rn A) and write cn) = ψn) + δn), where ψn) is a nilsequence and δn) satisfies N δn) = 0. Let F b n Y ) be an approximation of ψn) as in proof of Theorem 3.2. By [8, Theorem.], This implies N N N δ an) ) = 0. n= 24) sup N N N c an) ) F b an) Y ) < ɛ/4. n= As in the proof of Theorem 3.2, there is an open subset U of Y such that F > µa) 4 3ɛ/4 on U. By [7, Theorem.2], the sequence b an) X ) is equidistributed on Y. Hence the set {n N : F b an) Y ) > µa) 4 3ɛ/4} has positive lower density. This fact combined with 24) implies the set {n N : c an) ) > µa) 4 ɛ} has positive lower density. Remark 3.4. In above proof, we do not utilize the fact that the open set U is inside the identity component Y 0. This is because the orbit along an) is equidistributed on the entire Y. On the other hand, the orbit along primes minus is only equidistributed on some connected components of Y Y 0 is one of them) Optimal recurrence for Beatty sequences. Let X, B, µ, T ) be a measure preserving system. The discrete spectrum σt ) is the set of eigenvalues θ T := R/Z for which there exists a non-zero eigenfunction f L 2 µ) satisfying ft x) = e 2πiθ fx) for µ-almost every x X. Given a measure preserving system X, B, µ, T ), the transformation T q is ergodic if and only if σt ) /q = {0}, where a denotes the abelian group generated by a, as we view /q as an element of the group T. Theorem.4 follows from the next result. Theorem 3.5. Let θ, γ R with θ > 0 and X, B, µ, T ) be an ergodic system whose discrete spectrum σt ) satisfies σt ) θ = {0}. Let 0 l m r Z. Then for any A B and ɛ > 0, the sets and are syndetic. If r = l + m then the set is also syndetic. {n N : µa T l θn+γ A) > µa) 2 ɛ} {n N : µa T l θn+γ A T m θn+γ A) > µa) 3 ɛ} {n N : µa T l θn+γ A T m θn+γ A T r θn+γ A) > µa) 4 ɛ}

13 OPTIMAL LOWER BOUNDS FOR MULTIPLE RECURRENCE 3 Proof. If 0 < θ, then the set S = { θn + γ : n N} is co-finite in N and the conclusion follows from [3, Theorem.2]. Assume θ >. Define cn), ψn), δn), Y = G/Γ, F b n Y ) as in the proof of Theorem 3.3. Then by [2, Theorem 2.], the discrete spectrum of Y, µ Y, b) is contained in the discrete spectrum of X, µ, T ). Hence σb) θ = {0}. Claim 3.6. The sequence b θn+γ Y ) n N is well distributed on Y. Proof. Let F CY ). It suffices to show 25) N F b θn+γ Y ) = Y F dµ Y. Let S = { θn + γ : n N}. Then an integer m belongs to S if m = θn + γ for some n N. This is equivalent to or θn + γ < m θn + γ n γ < mθ n + γ θ θ for some n N. This is equivalent to {mθ } J, where J = [0, γ/θ] γ)/θ, ). Let W = {nθ mod : n N}. Then W is a closed subgroup of T. Since σb) θ = {0}, for any F CY ) and G CW ), we have that N m=m F b m Y )Gmθ ) = Y F dµ Y W G dµ W. Approximating the Riemann integrable function J W by continuous functions, we then get or equivalently 26) µ W J W ) N m=m F b m Y ) J W mθ ) = µ W J W ) F dµ Y, Y N m=m F b m Y ) J W mθ ) = Y F dµ Y. Note that {mθ } J W if and only if m S, and the uniform density of S is exactly µ W J W ). Therefore the left hand side of 26) is the same as the left hand side of 25) This proves our claim. As pointed out in the proof of Theorem 3.2, there is an open set U of Y such that F > µa) 4 3ɛ/4 on U. Since b θn+γ X ) n N is well distributed on Y, the set S = {n N : F b θn+γ X ) > µa) 4 3ɛ/4} is syndetic. Since the sequence δn)) n N tends to zero in the uniform density, and the set { θn+γ : n S} has positive uniform density, we have Hence sup S [M, N) S [M, N) n S [M,N) n S [M,N) δ θn + γ ) = 0. c θn + γ ) F b θn+γ Y ) < ɛ/4. Since F b θn+γ Y ) > µa) 4 3ɛ/4 when n S, we get that the set of n S such that c θn + γ ) > µa) 4 ɛ is syndetic. This finishes the proof.

14 4 SEBASTIÁN DONOSO, ANH NGOC LE, JOEL MOREIRA, AND WENBO SUN 4. Optimal recurrence along T an, T a2n,..., T a kn ) 4.. Poor lower bound for k = 5. In this subsection, we prove Theorem.5. We adapt the proof of Theorem.3 in [3]. We use the measure preserving system X, B, µ, T ) where X = T 2 is the 2-dimensional torus, µ is the Haar measure, and T x, y) = x + α, y + 2x + α) for some irrational α R. It is well known that this system is totally ergodic. For every n Z and every point x, y) T 2 a quick computation shows that T n x, y) = x + nα, y + 2nx + n 2 α). Let l >. We take a suitably large L, C N and a set Λ {0,..., L } to be chosen later. Let B := [ b I b where I b := CL, b CL + ) C 2 L b Λ and let A = T B. For each n Z, in order for a point x, y) to belong to A T an A T a4n A we need y i := y + 2a i nx + a 2 i n2 α B for each i =,..., 5. Let b i Λ be such that y i I bi. We now need the following elementary lemma. Lemma 4.. Let a,..., a 4 Z be distinct and let M be the 4 3 matrix whose i, j) entry is a j i for i =,..., 4 and j = 0,, 2. For each i =,..., 4, let v i be ) i times the determinant of the matrix obtained from M by deleting the ith row. Then for every quadratic polynomial f R[x], v fa ) + v 2 fa 2 ) + v 3 fa 3 ) + v 4 fa 4 ) = 0. Proof. The claim amounts to the statement that the matrix fa ) a a 2 fa 2 ) a 2 a 2 2 fa 3 ) a 3 a 2 3 fa 4 ) a 4 a 2 4 has determinant 0. But this follows from the fact that any quadratic polynomial is a linear combination of the polynomials, x, x 2. In view of Lemma 4., there exist integers v,..., v 4 and ṽ 2,... ṽ 5 depending only on a,..., a 5 ) such that v y + + v 4 y 4 = 0 and ṽ 2 y ṽ 5 y 5 = 0. Therefore, if C is large enough, then also v b + + v 4 b 4 = ṽ 2 b ṽ 5 b 5 = 0, as it will be an integer which can be made smaller than when C is large enough. Suppose now that Λ does not contain any solution to v b + + v 4 b 4 = ṽ 2 b ṽ 5 b 5 = 0 except when b = = b 5. Then, if x, y) A T an A T a4n A, all the y i must belong to the same I b, which implies that x X n, where X n is the set of points x T satisfying 2na2 a )x T < /C 2 L. Since y B, the point y must belong to the set B 2a nx a 2 n 2 α, which being a shift of B has the same measure as B. We conclude that µa T an A T a4n A) µ T X n )µ T B) = 2 C 4 L 2 Λ. Since µa) = Λ C 2 L, a quick computation now shows that the proof will be complete once we construct a set Λ {0,..., L } with Λ > L /l and without non-constant solutions to v b + + v 4 b 4 = ṽ 2 b ṽ 5 b 5 = 0. The existence of such a set Λ is provided by the following lemma. Lemma 4.2. Let a,..., a 5 Z be pairwise distinct and let v i and ṽ i be described in the paragraph after Lemma 4.. For every ɛ > 0 and every large enough L N, there exists a set Λ {0,..., L } with

15 OPTIMAL LOWER BOUNDS FOR MULTIPLE RECURRENCE 5 Λ > L ɛ such that the only b,..., b 5 Λ satisfying v b + + v 4 b 4 = ṽ 2 b ṽ 5 b 5 = 0 also satisfy b = = b 5. Lemma 4.2 is a generalization of [3, Theorem 2.4], due to I. Ruzsa, corresponding to a i = i. The key to proving Lemma 4.2 is the following intermediate result. Lemma 4.3. Let a < < a 5 be integers and let v i and ṽ i be as described above. Let d N and let b,..., b 5 R d all have the same Euclidean norm. If v b + + v 4 b 4 = ṽ 2 b ṽ 5 b 5 = 0 then b = = b 5. Unfortunately Lemma 4.3 does not hold for arbitrary v i and ṽ i, as seen by the example v = v 3 = ṽ 3 = ṽ 5 = and v 2 = v 4 = ṽ 2 = ṽ 4 = which would provide a counterexample with d = and b = b 2 = b 5 = and b 3 = b 4 =. Indeed we will need to use the description of the v i and ṽ i given by Lemma 4. and this makes the proof somewhat cumbersome. Proof. The condition a < < a 5 implies that v, v 3 > 0 and v 2, v 4 < 0. Let S := v b + v 3 b 3 v + v 3, A := b S, B = b 2 S. Applying Lemma 4. to a constant polynomial, we get that v + v 2 + v 3 + v 4 = 0 and hence, together with v b + + v 4 b 4 = 0, that S = v 2 b 2 + v 4 b 4 )/v 2 + v 4 ). Then we have b = S + A, b 2 = S + B, b 3 = S v v 3 A, b 4 = S v 2 v 4 B. Our goal is to show that b = b 2 = b 3 = b 4 = S, and so it suffices to show that A = B = 0 the fact that also b 5 = S would then immediately follow from the equation ṽ 2 b ṽ 5 b 5 = 0). Since the quantity b i 2 S 2 does not depend on i, we find that the following 4 numbers are equal v 2 v3 2 27) A S, A, B S, B, A 2 2v S, A, B 2 2v 2 S, B. v 3 v 4 ) Equality between the first and third gives 2 S, A = A 2 v v 3 ; equality between the second and ) fourth gives 2 S, B = B 2 v 2 v 4 and then equality between the first two numbers implies 28) A 2 v v 4 = B 2 v 2 v 3. In order to show that A = B = 0, we first show that B is a positive scalar multiple of A. Once we do that, we have from 28) that B = vv 4 v 2v 3 A and hence, equality between the first and last quantities from ) 27) together with 2 S, A = A 2 v v 3 ) gives A 2 v = v ) ) v 2 A 2 v v 2 2 S, A A 2 v v v2 v v 2 + = 0. v 3 v 3 v 4 v 3 v 4 v 3 v 3 v 4 v 3 v 4 This implies that A = 0 unless v v 3 = v2 v 4. Using the description of each v i from Lemma 4. as a Vandermonde determinant, this is equivalent to a a 4 ) 2 = a 2 a 3 ) 2. Since we are assuming that a < a 2 < a 3 < a 4 this can not happen and hence A = 0. We have reduced the proof to showing that B is a positive scalar multiple of A. It is now the time to use the fact that also ṽ 2 b ṽ 5 b 5 = 0. From Lemma 4., we deduce that ṽ 5 = v, and so we can write b 5 in terms of S, A, B as b 5 = v ṽ 2 b 2 ṽ 3 b 3 ṽ 4 b 4 ) = S + ṽ3 v 3 A + v2 ṽ 4 ṽ 2 v 4 v 4 v v2 2 v4 2 ) B = S + αa + βb,

16 6 SEBASTIÁN DONOSO, ANH NGOC LE, JOEL MOREIRA, AND WENBO SUN where α := ṽ3 v 3 in terms of A 2 we compute and β := v2ṽ4 ṽ2v4 v 4v. Using the relations established above to write B 2, S, A and S, B b 5 2 S 2 = 2αβ A, B + A 2 [α 2 + α ) v + β 2 v2 + β v 3 v 4 )) v v 4 Since b 5 = b, we deduce that b 5 2 S 2 = A 2 v v 3. After a somewhat tedious computation, we eventually arrive at A, B = A 2 v v 2 v 3v 4 = A B. But this implies that B must be a positive scalar multiple of A as desired, finishing the proof. Proof of Lemma 4.2. Let C = v + + v 4 + ṽ ṽ 5, let d > 2/ɛ be a natural number and then let m N be large enough multiple of C depending only on C, d and ɛ in fact, we need that m dɛ 2 > C d 2 d). Set L = m d. We can express each number in {0,..., L } using d digits in base m expansion. Let { [ F := x 0 + x m x d m d : x i 0,..., m )}. C We have F = m/c) d. Let r : F N be the sum of the squares of the digits in base m, in other words, rx 0 + x m x d m d ) = x x 2 d. Then rf ) [0, dc2 /m 2 ). Therefore there exists r 0 [0, dc 2 /m 2 ) such that Λ := {x F : rx) = r 0 } has cardinality Λ m/c) d 2 /d. The choice of parameters above yields Λ > L ɛ. Finally, suppose that b,..., b 5 Λ satisfy v b + + v 4 b 4 = ṽ 2 b ṽ 5 b 5 = 0. We identify each b i with the vector in R d obtained from its digits in base m. Then b = = b 5. Since each digit in b i is at most m/c, there is no carryover when multiplying by v i or ṽ i and thus, the equations v b + + v 4 b 4 = ṽ 2 b ṽ 5 b 5 = 0 apply even when multiplication and addition is being performed in R d. Applying Lemma 4.3, we conclude that indeed b = = b 5 as desired Lack of solutions implies poor lower bounds for k = 4. In this subsection we prove Theorem.0. We need the following well known equidistribution result whose short proof we include for completeness. Lemma 4.4. For every Bohr 0 set S, every α R m whose coordinates are rationally independent, and every cube I T m, we have that { n S [M, N): n 2 α mod Z m I } S [M, N) = µ T mi). Proof. By assumption, we can write S = {n N: nx U}, where K is a compact abelian group, U K is a neighborhood of K such that U v 3 v 2 is Riemann integrable, and x U is a point such that {nx : n Z} = K and S = {n N: nx U}. Then it suffices to show that as, {n S [M, N): n 2 α I} = {n [M, N) : nx, n2 α) U I} converges to µ K U) µ T mi) = µ K T mu I), where µ K, µ T m and µ K T m are the Haar measures on K, T m and K T m, respectively. This follows once we show that the sequence nx, n 2 α) n N is well distributed on K T m. Since U is Riemann integrable, it suffices to show that for every character χ : K S C of K and every b Z m, if either χ is non-trivial or b 0, then 29) N e 2πib αn2 χx) n = 0. ].

17 OPTIMAL LOWER BOUNDS FOR MULTIPLE RECURRENCE 7 Let θ T be such that χx) = e 2πiθ. Since the group generated by x is dense in K, θ / Z unless χ is trivial. Then we can write e 2πib αn2 χx) n = e 2πin2b α+nθ). If b 0 then 29) follows from Weyl s equidistribution theorem, and if b = 0 then θ / Z and 29) follows quickly from evaluating the resulting geometric series. Proof of Theorem.0. Let l, a,..., a 4, C and V be as in the statement of the theorem. Since V has a basis of rational vectors, there exists a positive constant ɛ such that any point x Z 4 at a distance say in the l norm) less than ɛ from the closure V R 4 must in fact belong to V. Fix m, N 0 N and E [N 0 ] m. Let X = T 2m be the 2m dimensional torus endowed with the Lebesgue measure µ. Define the map T : X X via the formula T x, y) = x + α, y + 2x + α), x, y T m for some α T m whose coordinates are rationally independent. Then X, B, µ, T ) is ergodic, and for each n N, T n x, y) = x + nα, y + 2nx + n 2 α). For i = c,..., c m ) [N 0 ] m, denote B i = We can directly compute that µa) = [ c, c ) [ + ɛ cm, c ) m + ɛ, A = T m B i. N 0 N 0 N 0 N 0 i E ɛ N m 0 all j 4, T ajn x, y) A if and only if E = ɛ m d m,n0 E). Let x, y) T 2m and n N. Note that for 30) u j = u j x, y, n) := y + 2a j nx + a 2 jn 2 α B ij for some i j E. Fix such a point x, y, n) T 2m N. Then the vector u,..., u 4 ) = y,..., ) + 2nxa,..., a 4 ) + n 2 αa 2,..., a 2 4) B i B i4 belongs to the closure in R m 4 of V m. Since N 0 u j is at most ɛ away from the integer vector i j in the l [m]) distance), from the definition of ɛ we deduce that i,..., i 4 ) belongs to V m as well. Let W denote the collection of all tuples i,..., i 4 ) V m with i j E. By definition, D m,n0 V m, E) = By the discussion above, 3) µt an A T a4n A) = i,...,i 4) W X j= 4 Bij uj x, y, n) ) dµx, y). W V [N 0] 3 m. Fix i,..., i 4 ) W. If n N is such that 30) holds for all i 4, then considering 30) as a linear equation system with 4m equations and the coordinates of y, nx, n 2 α as unknowns i.e. 3m unknowns in total), we deduce that there exists 3 cubes I, I 2 and I 3 in T m with side length at most N 0 such that y I, nx I 2, n 2 α I 3. Lemma 4.4 implies that for any Bohr 0 set S, {n S [M, N): n 2 α I 3 } = µi 3 ) S [M, N) N0 m.

18 8 SEBASTIÁN DONOSO, ANH NGOC LE, JOEL MOREIRA, AND WENBO SUN Since µi ) N m 0 D m,n0 V m, E) = and µ{x T m : nx I 2 }) = µi 2 ) N0 m V [N 0 ] 3 m N0 3m V [N 0 ] 3 m i,...,i 4) W i,...,i 4) W sup S [M, N) N0 3m V [N 0 ] 3 sup m S [M, N) n S [M,N) N0 3m C V [N 0 ] 3 m µa)l = C ɛml N 3m 0 V [N 0 ] 3 m d m,n 0 E) l, n S [M,N) for any n 0, by 3) we conclude that X j= µt an A T a dn A) 4 Bij u j x, y, n)) dµx, y) ɛ which finishes the proof by taking β < l N 0 N V [N 0] Solutions to linear equations imply optimal lower bounds for k = 4. In this section we prove Theorem.8. We first need to reformulate the assumptions in terms of functions on a torus; this is the content of Lemma 4.6 below. We start with an estimate from harmonic analysis. Let A be a finite set, f : A R a function and p > 0. We denote by f L p with the normalized counting probability measure, i.e. /p f L p := fa) p). A We will also make use of the weak L p quasinorm: ) /p {a A: fa) > s} f L p w := sup s. s>0 A a A its usual L p quasinorm when A is endowed We remark that when p < these quasinorms do not satisfy the triangle inequality. We will only use these quasinorms with p < to invoke the following well known interpolation lemma. We include its short proof for completeness. Lemma 4.5. Let 0 < p < r < and let A be a finite set. For every function f : A R we have Proof. Combining the identity x r = x 0 rs r ds = r f r L r r r p f p L p w f r p L. with the definition of L r norm, we deduce the formula f r L = r r 0 s r A 0 s r [0,x] s) ds = r { fa) >s} ds = r a A Finally, using the definition of the weak L p norm we conclude s r {x>s} ds s r {a A : fa) > s} A f r L r s r p f p r L ds = r p w r p f p L p f r p w L. ds. The following lemma makes use of the quantities d m,n E) and D m,n V, E) introduced in Definition.6. Lemma 4.6 Equivalent inequalities). Let m, d, l N with l > d, let C > 0, V Q d be a subspace containing the vector,..., ) and V R d be its closure in R d. Then ) 2) 3) 4):

19 OPTIMAL LOWER BOUNDS FOR MULTIPLE RECURRENCE 9 ) For every large enough N and every subset E [N] m, we have D m,n V m, E) Cd m,n E) l. 2) For every large enough N and every function c: [N] m [0, ], we have V [N] d m ca )ca 2 )... ca d ) C c d. L d/l a i [N] m,a,...,a d ) V m w 3) For every large enough N and every function c: [N] m [0, ], we have V [N] d m ca )ca 2 )... ca d ) C d ) l c l L l. a i [N] m,a,...,a d ) V m 4) Let Y = V /Z d) m be a subtorus of T d m. For every measurable function f : T m [0, ], Y Remark 4.7. fy )fy 2 )... fy d ) dµ Y y,..., y d ) C d ) l ) l f dµ T m. l T m Whenever we have a point x in [N] dm or analogously for Q dm, T dm, etc.) we consider x = x i,j ) i=,...,m, j=,...,d with each x i,j [N]. We then write x = x,..., x d ) where each x i [N] m is the vector x i = x i,j ) m j=. Depending on the context, we may also write x = x,..., x m ), where now each x j [N] d is the vector x j = x i,j ) d i= vectors we are referring to). it should be clear at any point which For instance if v = v i,j ) i=,...,m, j=,...,d Q dm then v is in V m if for every i the vector v i,j ) d j= of Q d belongs to V ; and v is in E d if for every j the vector v i,j ) m i= is in E. Similarly in 2) and 3), the statement that a,..., a d ) V m should be interpreted by writing each a j as a i,j ) m i= and requiring that each vector a i,j ) d j= is in V. It might be true that 3) and 4) are also equivalent to ) and 2), but we don t have a proof and it is not needed in this paper. Proof. 2) ). Take ca) = E a). ) 2). Let p := d/l, observe that c L p w = N ml/d sup s 0 s {a [N] m : ca) > s} l/d and assume that the maximum is obtained at s = t. Let E = {a [N] m : ca) > t}. Since c t E, we have V [N] d m a i [N] m,a,...,a d ) V Invoking ), we get t d D m,n V m, E) C td E l = C c d N lm L. p w 2) 3). We only need to show 3) for c 0. By Lemma 4.5, 3) 4). Let c d L p w d l Y N = ca )ca 2 )... ca d ) t d D m,n V m, E). ) l c l L c l d L d a V [N] d ) m i= j= d l m [ ai,j N ) l c l L., a ) i,j T dm N and let µ N be the normalized probability measure supported on Y N. We claim that µ N! µ Y as N. Indeed, any it point of the sequence µ N ) N N must be supported on Y. Moreover, for any v V/Z d ) m, if N is larger than the denominators of all coordinates of v, then µ N! is invariant under v. We conclude that any it point of the sequence µ N! ) N N is supported on Y and invariant under Y, hence it must be µ Y.

20 20 SEBASTIÁN DONOSO, ANH NGOC LE, JOEL MOREIRA, AND WENBO SUN Now given f : T m [0, ], let c : [N] m [0, ] be the function ca) = N m [ /N,0) m fa/n + x)dx. When N is large enough, we have that Y N i= d fy i ) dµ N y,..., y d ) = N dm V [N] d m V [N] d m a V [N] d ) m [ N,0 ) dm d f i= a i [N] m,a,...,a d ) V m ca )ca 2 )... ca d ) C d l ) l c l L = C d l ) l T m f dµ T m) l. ai ) N + x i dx,..., x d ) We are now ready to prove Theorem.8. Proof of Theorem.8. Fix an ergodic system X, B, µ, T ) and a set A B with µa) > 0. Let Z 2 be the 2-step nilfactor of X, defined in Section 2.2. Using a standard approximation argument, we can assume that Z 2 is a 2-step nilsystem, so that Z 2 = G/Γ, B, µ, T ), where G is a 2-step nilpotent Lie group and Γ G is a uniform subgroup. Denote Z := G/Γ. In view of ergodicity, the topological system Z, T ) is minimal see, for instance, [3, Theorem 4..]). We can assume that G is generated by the connected component of the identity and τ. Indeed, the projection of the connected component of G onto Z = G/Γ is an open subset of Z as its pre-image under the natural map G Z is the union of all connected components of G having non-empty intersection with Γ and hence it is open) and by minimality of Z, τ) its orbit under τ is all of Z. Therefore, if we let G be the subgroup of G generated by the connected component of the identity and T, it follows that Z = G/Γ G). Since G is a 2-step nilpotent group, the commutator G 2 = [G, G] is inside the center of G, and hence the subgroup Γ 2 = G 2 Γ is normal in G. Therefore Z = G/Γ 2 )/Γ/Γ 2 ) and thus after modding out by Γ 2 we can assume that G 2 Γ = {e}, which implies that G 2 is a compact abelian Lie group. From [3, Theorem 4..4], it follows that G 2 is connected, and so G 2 must be a finite dimensional torus. Let K be the quotient K := Z/G 2 = G/ΓG 2 ) and note that it is also a compact abelian Lie group but it may be disconnected). Let π : G K be the natural projection, let a = πτ) and define S δ = {n N: a n Bδ)}, where Bδ) is the ball in K centered at the identity of K with radius δ. It suffices to show that 32) δ 0 S δ [M, N) n S δ [M,N) 4 X i= ft ain x) dµx) C 4 ) l ) l f dµ l X for all 0 f. By Proposition 2.5, the left hand side of 32) is 0 if we replace at least one of the four f s with f Ef Z 2 ). Since 0 Ef Z 2 ), it suffices to prove 32) under the assumption that X = Z 2 = G/Γ. 33) Since G is 2 step nilpotent, by Proposition 2.4, the left hand side of 32) equals to Z G 2 G 2 i= d f gg a i ) g ai 2 ) 2 Γ ) dµ G2 g 2 ) dµ G2 g ) dµ Z gγ),

21 OPTIMAL LOWER BOUNDS FOR MULTIPLE RECURRENCE 2 where µ X and µ G2 are the Haar measures on X and G 2. Recall that G 2 is a torus, say G 2 = T m. Consider the subgroup Y := { y,..., y 4 ) T m ) 4 : g, g 2 T m ) y i = ) ai g + where we now changed to the additive notation. Then we may rewrite 4 33) = fgy i Γ) dµ Y y y 2, y 3, y 4 ) dµ Z gγ), where µ Y Z Y i= } ai )g 2 T 4m, 2 is the Haar measure on Y. We can also describe Y in terms of V as Y = V /Z 4) m, where V is the closure V in R 4 or, equivalently, its R-span). For each g G let f g : G 2 R be the function defined by the formula f g g 2 Γ) = fgg 2 Γ) for all g 2 G 2. Then by Lemma 4.6, ) 4), and then Jensen s inequality, we conclude that 4 fgy i Γ) dµ Y y y 2, y 3, y 4 ) dµ Z gγ) C 4 ) l ) l f g dµ G2 dµ Z gγ) Z Y l i= Z G 2 C 4 ) l ) l f g dµ G2 dµ Z gγ) = C 4 ) l ) l f dµ Z. l Z G 2 l Z 5. Optimal recurrence along polynomials To state our results, we need to introduce a notion defined and studied in detail by Leibman in [20]. The C-complexity of a family of integer-valued polynomials {p,..., p d } is the minimum integer k for which the factor Z k is characteristic for this family in every ergodic nilsystem G/Γ, B, µ, T ) with G being connected. Note that the minimum value of k for general ergodic systems is an upper bound of the C-complexity and in some cases it is strictly larger. Proposition 5.. Let G/Γ, B, µ, T ) be an ergodic nilsystem where G is connected. Let Z be its Kronecker factor and let α Z be the rotation induced by T. Let q n), q 2 n) be two independent integer polynomials and set p = aq, p 2 = bq 2 and p 3 = cq + dq 2, a, b, c, d Z. Assume that the C-complexity of the family {p, p 2, p 3 } is equal to one. For δ > 0, let B δ be the ball in Z centered at 0 of radius δ and define S δ = {n N : q n)α, q 2 n)α) Bδ) Bδ)}. Let f, f 2, f 3 L µ) and assume Ef i Z ) = 0 for some i 3. Then 34) f T pn) x)f 2 T p2n) x)f 3 T p3n) x) = 0 S δ [M, N) where the it is taken in L 2 µ). n S δ [M,N) Proof. The proof is similar to that of Proposition 2.5, subject to some minor changes that we write explicitly. In this proof, all the its are taken in L 2 µ). Without loss of generality, we assume Ef Z ) = 0. Let L be the it on the left hand side of 34) and ds δ ) be the uniform density of S δ. Since {q n)α, q 2 n)α)} is well distributed on Z Z because q and q 2 are independent), we have 35) ds δ )L = N N Sδ n)f T pn) x)f 2 T p2n) x)f 3 T p3n) x) = Bδ) Bδ) q n)α, q 2 n)α)f T pn) x)f 2 T p2n) x)f 3 T p3n) x).