RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
Abstract. We consider for which sequences there exists a weakly mixing transformation
which is rigid along the sequence, and for which sequences there exists a weakly mixing
transformation which is not recurrent along the sequence.
1. Introduction
We consider here two properties of a dynamical system (X, β, p, T ), rigidity and recurrence. Ultimately we would like to characterize the times along which these properties
do, or do not occur, for different classes of transformations. But the main focus here is to
characterize which subsequences (nm ) in Z+ can be a sequence of times for rigidity, and
which can be a sequence of times for the failure of recurrence, for some weakly mixing
dynamical system.
These properties are in some sense opposites. Rigidity along the times (nm ) means that
the powers (T nm ) are converging in the strong operator topology to the identity; that is,
kf ◦ T nm − f k2 → 0 as m → ∞, for all f ∈ L2 (X). So, rigidity along (nm ) means that
p(T nm A ∩ A) → p(A) as m → ∞ for all A ∈ β. On the other hand, failure of recurrence
along the times (nm ) means that for some A ∈ β with p(A) > 0, we have p(T nm A ∩ A) = 0
for all m ≥ 1. Nonetheless, there are parallels between these two properties of the times
(nm ). For example, one basic fact is that neither property can occur for a transformation
T unless the times form a sparse subsequence. Also, these two properties cannot occur
without there being some combinatorial or algebraic structure (or the absence of it) on
the sequence of times (nm ). For example, in the case of non-recurrence, the sequence
cannot be the positive terms in an infinite set of differences because the positive terms in
D − D, for an infinite set D ⊂ Z+ , always forms a set of recurrence.
We first look at rigidity and also the more restrictive property of IP-rigidity. We see
that this property can be viewed as a spectral property and therefore connected with the
behavior of the values νb(nm ) of the Fourier transforms for positive Borel measures ν on T.
But also we see that while our sequences must be sparse, they also must not have certain
types of algebraic structure for rigidity to occur for weakly mixing transformations. We
then consider the failure of recurrence. This property cannot as easily be viewed as a
spectral property. However, we do know again that the sequences must be sparse and
cannot have certain types of algebraic structure for recurrence to fail for a weakly mixing
transformation.
2. Generalities on Rigidity and Weakly Mixing
Suppose we consider a dynamical system (X, β, p, T ). Unless it is noted otherwise, we
will be assuming that (X, β, p) is a standard Borel probability space. In particular it is
non-atomic and L2 (X, p) has a countable density subset (sometimes referred to as the
Date: February 15, 2010.
1
2
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
dynamical system being separable). We also assume that T is an invertible measurepreserving transformation (X, β, p). Notice that such a system must be aperiodic once it
is ergodic.
Given an increasing sequence (nm ) of integers we consider the family
A(nm ) = {A ∈ β : p T −nm A4A → 0}.
We now recall some basic and well-known facts about A(nm ). First, it is clear that
A(nm ) ⊂ β is a sub-σ-algebra which is also T -invariant. Moreover, if by d we denote the
pseudometric on β given by d(A1 , A2 ) = p(A1 4A2 ) then it is not hard to see that A(nm )
is the maximal σ-algebra A ⊂ β such that
T nm |A → Id|A as m → ∞.
Moreover
(2.1)
L2 (X, A(nm ), p) = {f ∈ L2 (X, β, p) : f ◦ T nm → f in L2 (X, β, p)}.
Indeed, the left hand inclusion is obtained from a standard approximation argument. For
the other direction, take [a, b] ⊂ R and ε > 0. With no loss of generality we can assume
that f : X → R and let f∗ denote the distribution of f . There exists δ > 0 such that
f∗ ([a − δ, b + δ] \ [a, b]) < ε. Set A = f −1 ([a, b]). Then
p T −nm A \ A ≤ p T −nt A \ f −1 ([a − δ, b + δ]) + p f −1 ([a − δ, b + δ]) \ f −1 ([a, b])
≤ p (f ◦ T nm )−1 ([a, b]) \ f −1 ([a − δ, b + δ]) + ε.
Since
2
δ p (f ◦ T
nm −1
) ([a, b]) \ f
−1
([a − δ, b + δ]) ≤
Z
|f ◦ T nm − f |2 dp,
our claim (2.1) follows.
If A(nm ) = β then one says that (nm ) is a rigidity time of (X, β, p, T ). Systems
possessing rigidity times are called rigid. It is the property (2.1) which shows that the
fact that (nm ) is a rigidity time of T (or that T is rigid) is a spectral property, that is it
is a spectral invariant of the associated Koopman operator UT on L2 (X, p) given by the
formula UT (f ) = f ◦ T . The following should make this clear.
First, note that f ◦ T nm → f in L2 (X, p) if and only if
Z
nm
hf ◦ T , f i :=
f ◦ T nm · f dp → kf k22 .
X
Now recall some basic notions of spectral theory (see e.g. [5], [20], [33]). For each f ∈
T
L2 (X, p) its spectral measure νfT is the Borel measure on T that is determined by νc
f (n) =
hf ◦ T n , f i for all n ∈ Z. Spectral measures are positive and have kνfT k1 = νfT (T) = kf k22 .
We also need the adjoint ν ∗ given by ν ∗ (E) = ν(E −1 ) for all Borel sets E ⊂ T. The adjoint
−1
has νb∗ (n) = νb(−n) for all n ∈ Z. It is clear that νfT = (νfT )∗ . If by Z(f ) we denote the
cyclic space of f , i.e. the smallest closed UT -invariant subspace containing f , then UT |Z(f )
is unitarily equivalent to the unitary operator V on L2 (T, νfT ), V (j)(z) = zj(z) for each
j ∈ L2 (T, νfT ). Each positive measure ν νfT is also a spectral measure, that is there is
f 0 ∈ Z(f ) such that νfT0 = ν. Among all spectral measures there exists a measure ν T such
that all other spectral measures are absolutely continuous with respect to ν T . The type of
the measure ν T is called the maximal spectral type of T . Note that if fi ∈ L2 (X, β), i ≥ 1,
RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
3
P
1
T
then the measure ν := ∞
i=1 2i kfi k22 νfi (which is a probability measure) satisfies ν νT
and hence it is a spectral measure:
(2.2)
∞
X
i−1
1
ν T = νFT for some F ∈ L2 (X, β).
2i kfi k22 fi
nm
T
, f i → kf k22 then for each f 0 ∈ Z(f ) we also have
Note that if νc
f (nm ) = hf ◦ T
0 2
T
model (in which f will correspond
νc
f 0 (nm ) → kf k2 . Indeed, by passing to the standard
R n
nm
to the constant function 1) we obtain hV 1, 1i = T z m dνfT (z) → 1 which is equivalent to
z nm converges to 1 in L2 (T, νfT ). Then for each j ∈ L2 (T, νfT ) which is bounded we clearly
have V nm (j)(z) = z nm j(z) converges to j in L2 (T, νfT ) and for a general j ∈ L2 (T, νfT ) we
use an approximation argument
kV nm j − jk2 ≤ kV nm j − V nm jε k2 + kV nm jε − jε k2 + kjε − jk2 .
It now easily follows that T is rigid if and only if for each function f ∈ L2 (X, p) there
exists (nm ) = (nm (f )) such that f ◦T nm → f in L2 (X, p). Indeed, we will obtain a rigidity
time for T by taking (nm (f )) for any f such that νfT is equivalent to ν T .
The maximal spectral type of a rigid transformation is a Dirichlet measure, that is for
T (n ) → kν T k . Also, a measure absolutely
some increasing sequence (nm ) we have νc
m
1
continuous with respect to a Dirichlet measure is also a Dirichlet measure. So by the
Riemann-Lebesgue Lemma, no spectral measure of a rigid transformation is absolutely
continuous and therefore ν T is always a singular measure. In particular, all rigid transformations have zero entropy.
Remark 2.1. J.-P. Thouvenot observed that T is rigid if and only if for each f ∈ L2 (X)
(or just for each characteristic function f = 1A , A ∈ β), there exists (nm ) such that
kf ◦ T nm − f k2 → 0 as m → ∞. That is, there is one sequence along which T is rigid
simultaneously for all functions if and only if for each L2 function (or each characteristic
function), there is a sequence along which T is rigid for that function. There are a
number of different ways to prove this, but here is an interesting approach via Krieger’s
Generator Theorem. It is sufficient to prove rigidity holds assuming that one has the
weaker condition of there being rigidity times for each characteristic function. In fact,
suppose that an automorphism T has the property that for each set A ∈ β there exists
(nm ) = (nm (A)) such that p (T −nm A4A) → 0. Then all spectral measures of functions of
the form χA , A ∈ β are singular, and since the family of such functions is linearly dense,
the maximal spectral type of T is singular. It follows that T has zero entropy and by
Krieger’s Generator Theorem, there exists a two element partition P = {A, Ac } which
is generating β. Now, let (nm ) = (nm (A)) and notice that for each k ≥ 1 and for each
W
i
−nm
B ∈ k−1
B4B) → 0. Hence by approximating the L2 functions by
i=0 T P we have p(T
simple functions, (nm ) is a rigidity time for T .
We can often use Baire category arguments to distinguish the behavior of transformations. For this we use the group Aut(X, β, p) of invertible measure-preserving transformations with the topology of strong operator convergence of the associated Koopman
operators. By a generic property, we mean that the property holds for a dense Gδ subset
of Aut(X, β, p). For example, it is well-known that the generic dynamical system is weakly
mixing. Also, for the generic transformation there exists a rigidity sequence. Hence, the
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V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
generic transformation is both weakly mixing and rigid. We will show this in a slightly
more general setting.
In order to see that a generic transformation is rigid we will show the following stronger
result which is a folklore.
Proposition 2.2. Given an increasing sequence (nm ) of natural numbers the set
R(nm ) = {S ∈ Aut(X, β, p) : S nmk → Id for some subsequence (nmk ) of (nm )}
is a residual subset of Aut(X, β, p).
Proof. Recall that a metric d compatible with the strong topology is given by
(2.3)
d(R, S) =
∞
X
1
−1
−1
(p
(RA
4SA
)
+
p
R
A
4S
A
)
i
i
i
i
i
2
i=1
where {Ai : i ≥ 1} is a dense subset in (β, p). It follows that given n ∈ Z and ε > 0 the
set
{S ∈ Aut(X, β, p) : d (S n , Id) < ε}
is open, and therefore the set
Ak,ε := {S ∈ Aut(X, β, p) : d (S nq , Id) < ε for some q ≥ k}
is open as well. It is also dense. Indeed, given A1 , . . . , Am and R ∈ Aut(X, β, p) we seek
S ∈ Aut(X, β, p) so that
(2.4)
p(SAi 4RAi ) is close to zero
and for some q ≥ k
(2.5)
p(S nq Ai 4Ai ) is close to zero
for i = 1, . . . , m (the argument below will show the same if we replace R and S by their
inverses). With no loss of generality we can assume that R is of rank one (as this family
is dense in Aut(X, β, p)), that is, we can see the sets A1 , . . . , Am as union of levels of high
Rokhlin towers for R. Now fix nq (we will see that the only condition on nq will be that
nq → ∞) and let hs be the height of a Rokhlin tower for R well approximating the sets
A1 , . . . , Am . We now divide this tower into consecutive subtowers (without changing the
levels) of height nq (first nq levels, then levels nq + 1 until 2nq , etc.). Then we define S
in the following way: inside each subtower of height nq the automorphism S acts as R
except of the top level on which we require that S sends this level into the bottom level
of the subtower (so for the first subtower the first level is sent into the second, ..., the
nq − 1st into the nq th, and the nq th into the first; for the second subtower the nq + first
level is sent into the nq + second, ..., the 2nq − 1st into the 2nq th, and the 2nq th into the
nq + first, etc.). Now, if nq /hs is sufficiently small than we can see that the hs /nq Rokhlin
towers (of height nq ) for S well approximate the sets Ai , and the error in (2.4), (2.5)
comes from the fact that these towers are cyclic. The total error is hence of order
hs 1
1
·
= ,
nq hs
nq
so we only need to know that nq → ∞.
RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
5
Now take 0 < εl → 0 and notice that the set
∞ \
∞
\
Ak,εl
l=1 k=1
is included in R(nm ) .
Remark 2.3. This should be compared with the beginning of the proof of Proposition 4.2.
Remark 2.4. We also notice that the metric d defined in (2.3) has the following properties:
d(R, S) = d(R−1 , S −1 ) and d(T R, T S) = d(R, S) once T commutes with R and S. Denote
kT k = d(T, Id). Then
kT n+m k = d(T n , T −m k ≤ d(T n , Id) + d(Id, T −m ) = kT n k + kT m k.
We recall now Wiener’s Lemma: Given a positive Borel measure ν on T we have
N
X
X
1
lim
|b
ν (n)|2 =
ν 2 ({x}).
N →∞ 2N + 1
n=−N
x∈T
N
P
1
2N
+1
N →∞
n=−N
It follows that ν is continuous if and only if lim
|b
ν (n)|2 = 0. The latter
condition is well known to be equivalent to the fact that νb(n) tends to zero along a
subsequence of density 1.
Now, a transformation is weakly mixing if and only if νf is continuous for each f ∈
L2 (X, p) which is mean-zero, that is (by Wiener’s Lemma) we have hf ◦ T n , f i tends to
zero along a sequence of density one. Denote a given such density one sequence by NfT .
As f changes, this sequence generally might need to change. However, it is classical that
we can choose a subsequence of density 1 which works for all L2 -functions. We give a
proof of this fact for reader’s convenience.
Proposition 2.5. Assume that T is weakly mixing. Then there is a sequence (nm ) in Z+
of density one set such that for all mean-zero f ∈ L2 (X), one has lim hf ◦ T nm , f i = 0.
m→∞
Proof. Let (fs ) be a sequence of mean-zero functions which is dense in the subspace
L2,0 (X, p) of L2 (X, p) consisting of the mean-zero functions. Because T is weakly mixing,
also T × T so is. In view of (2.2) there exists F ∈ L2 (X × X, p ⊗ p) such that
νFT ×T =
∞
X
s=1
∞
X
1
1
T ×T
ν T ∗ νfTs .
ν
=
4 fs
s
2 fs ⊗fs
s
2 kfs k2
2 kfs ⊗ fs k2
s=1
Since νFT ×T is continuous, there is a sequence (nm ) in Z+ of density one such that νb(nm ) →
T ×T
T
0 as m → ∞. But for every s ≥ 1, we have 2s kf k4 ν[
(n) ≥ |νc
(n)|2 for all n ∈ Z.
So it follows that for every s ≥ 1, we also have
s 2 F
T
νc
fs (nm )
fs
→ 0 as m → ∞. But then
T
for any mean-zero function f ∈ L2 (X), we have νc
f (nm ) → 0 as m → ∞ by the usual
approximation argument and Schwarz inequality.
We can rewrite the assertion of Proposition 2.5 as
(2.6)
UTnm → 0 weakly in the space L2,0 (X, p).
6
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
Each sequence (nt ) (not necessarily of density 1) of integers for which (2.6) holds is called
a mixing sequence for T . It is not hard to see that mixing sequences are not rigidity times
for T , in fact it is a complementary notion to rigidity sequences (indeed, (nt ) is a rigidity
sequence if and only if UTnt → Id). Any transformation possessing a mixing sequence is
weakly mixing. Finally notice that (nt ) is a mixing sequence of T if and only if for each
k∈Z
T (n + k) → 0 when t → ∞.
(2.7)
νc
t
The following result about mixing subsequences is also a folklore.
Proposition 2.6. Given an increasing sequence (nm ) of natural numbers the set
M(nm ) = {S ∈ Aut(X, β, p) : S nmk → 0
weakly, for some subsequence (nmk ) of (nm ) in L2,0 (X, p)}
is a residual subset of Aut(X, β, p).
Proof. Let {Ai : i ≥ 1} be a dense family in (β, d). Take ε > 0 and set
∞
X
1 −nk
< ε}.
M(k, ε) = {S ∈ Aut(X, β, p) :
p(S
A
∩
A
)
−
p(A
)p(A
)
i
j
i
j
i+j
2
i,j=1
Notice that M(k, ε) is open and, for 0 < εi → 0, consider the set
∞ [
∞
\
M(k, εi ).
M=
i=1 k=i
S∞
It is not hard to check that k=i M(k, εi ) is not empty and dense (each mixing transformation belongs to it), so M is Gδ and dense. If T ∈ M then for each i ≥ 1 there exists
ki ≥ i such that
∞
X
1 −nki
< εi .
p(T
A
∩
A
)
−
p(A
)p(A
)
r
s
r
s
r+s
2
r,s=1
Hence for each r, s ≥ 1
p(T −nki Ar ∩ As ) − p(Ar )p(As ) → 0 when i → ∞
and therefore (nki ) is a mixing sequence for T .
Let us recall that a positive finite Borel measure ν on T is called a Dirichlet measure if
there is an increasing sequence (nm ) of integers such that
(2.8)
νb(nm ) → kνk1 .
It is not hard to see that if ν1 ν and ν is Dirichlet measure, then so is ν1 .
Each ergodic transformation with discrete spectrum is rigid. In order to see this via the
Halmos-von Neumann Theorem, consider an ergodic rotation T x = x + x0 where X is a
compact metric monothetic group, x0 is its topological cyclic generator, and p stands for
Haar measure of X. Take any increasing sequence (nt ) of integers, and consider (nt · x0 ).
By passing to a subsequence if necessary, we can assume that nt x0 → y ∈ X. This is
equivalent to saying that T nt → S, where Sx = x + y. Because the convergence is taking
part in the strong operator topology, it is not hard to see that we will obtain
T ntk+1 −ntk → S ◦ S −1 = Id,
RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
7
and therefore T is rigid (indeed, nt+1 − nt → ∞ by Proposition 2.9 below). In fact, the
argument we have used shows that each purely atomic measure is a Dirichlet measure.
Moreover, we have also showed that in the discrete spectrum case the centralizer of T is
compact in the strong operator topology. The converse is also true (see, e.g. [24]). We
will see later that ergodic rotations are completely determined by their rigidity sequences
(see Corollary 3.14 below).
A positive finite Borel measure ν on T is called a Rajchman measure if its Fourier
transform vanishes at infinity, that is
(2.9)
νb(n) → 0 when |n| → 0.
Each absolutely continuous measure is Rajchman but there are Rajchman measures which
are singular. It is not hard to see that a measure absolutely continuous with respect to a
Rajchman measure is Rajchman, and that a Rajchman measure is singular with respect
to any Dirichlet measure. It easily follows that a dynamical system is strongly mixing if
and only if ν T is a Rajchman measure.
Remark 2.7. Proposition 2.5 certainly shows that rigidity sequences for weakly mixing
transformations are density zero. But Proposition 2.9 below shows also that more than
this is true without the assumption that T is weakly mixing.
Remark 2.8. It would be interesting to be able to characterize the sets N = {nm } of
density one that occur in Proposition 2.5. This of course inherently means characterizing
sets L of density zero that are the complements of such sets. Part of what characterizing rigidity sequences for weakly mixing sequences would do would at least characterize
certain types of sets L. However, the sets N could fail to have rigidity sequences in their
d ≥ δ for
complements and just have a complement that consists of all n such that |µ(n)|
some δ > 0 and some µ which is a continuous measure.
One general question that we would like to answer is this:
Question Which rigidity sequences of a transformation with discrete spectrum can be
realized using a weakly mixing transformation?
Lacunary sequences might seem to be good “candidates” to be rigidity sequences for
some weakly mixing transformations because they certainly have the necessary sparseness. However, even in this category of sequences the situation is not clear as there are
lacunary sequences which cannot even be rigidity sequences for any aperiodic and ergodic
transformations (e.g. (2n + 1); see Remark 3.4 c)), in particular, they are not rigidity
sequences for (non-trivial) ergodic rotations.
We do have at least one general principle in play here, but the argument has to be
different when there are eigenfunctions. If the system is not ergodic, or if some power
T n is not ergodic, then there can exist a non-zero, mean-zero function f ∈ L2 (X) and a
periodic sequence (nm ) such that f ◦ T nm = f for all m ≥ 1. Otherwise, the only way a
sequence can exhibit rigidity for a function, or for the whole dynamical system, is when
the sequence has gaps tending to ∞, and hence is certainly of density zero.
Proposition 2.9. Let (nm ) be an increasing sequence of integers.
a) Suppose T is totally ergodic. Let f0 ∈ L2 (X) be non-zero and mean-zero. If kf0 ◦T nm −
f0 k2 → 0 as m → ∞, then the sequence (nm ) has gaps tending to ∞ and hence has zero
density.
8
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
b) Suppose T is ergodic. If kf ◦ T nm − f k2 → 0 as m → ∞ for all f ∈ L2 (X), then (nm )
has gaps tending to ∞ and hence has zero density.
Proof. In a) we claim that nm+1 − nm → ∞ as m → ∞. Otherwise, there would be a
value d ≥ 1 such that d = nm+1 − nm infinitely often. It follows that f0 ◦ T d = f0 . This
is not possible since f0 is non-zero and mean-zero, and T is totally ergodic. To prove b),
one again argues that nm+1 − nm → ∞ as m → ∞ since otherwise there exists d ≥ 1
such that d = nm+1 − nm for infinitely many m, and hence f ◦ T d = f for all f ∈ L2 (X).
But this is impossible since our system is aperiodic. Indeed, for any D, using the Rokhlin
Lemma, there is a set B of positive measure such that T j B are pairwise disjoint for all
j, 0 ≤ j ≤ D. Take f0 supported on B that is non-zero and mean-zero. Then f0 ◦ T j 6= f0
for all j, 0 ≤ j ≤ D. Hence, once D > d, we cannot have f0 ◦ T d = f0 .
Remark 2.10. The proof shows certainly that being density zero is not enough for rigidity
because a sequence can be density zero and have infinitely many pairs of terms (nm , nm+1 )
with nm+1 − nm ≤ 10.
By using spectral measures and the Gaussian measure space construction (denoted
here by GMC), we can see that our basic question is equivalent to a fact about Fourier
transforms of measures. See Cornfeld, Fomin, and Sinai [5] for details about the GMC.
We will use the notation Gν for the transformation obtained by applying GMC to the
positive Borel measure ν.
First, if f ∈ L2 (X) is norm one, then rigidity along the times (nm ) for f means that
T
T
we will have lim νc
f (nm ) = 1. But when T is weakly mixing and f is mean-zero, then νf
m→∞
T
is continuous and so νc
f (n) → 0 along a sequence of values n which has density one. So,
phrased in a somewhat different fashion than above, this shows that for a weakly mixing,
T
rigid transformation, while νc
f (n) tends to 0 along a sequence of times n of density one,
T
there are also times (n ) along which lim νc
(n ) = 1.
m
m→∞ f
m
Conversely, if there is a continuous Borel probability measure ν on T such that νb(nm ) →
1 as m → ∞, then the GMC T = Gν gives us a weakly mixing dynamical system
(X, β, p, T ) and a mean-zero function f ∈ L2 (X) with kf k22 = 1 and νb(n) = hf ◦ T n , f i
for all n. Indeed, it is not hard to see that this construction gives us a weakly mixing
dynamical system such that T is rigid along the times (nm ); that is, one might say that
the GMC preserves rigidity. In particular, if ν = ν S for some S ∈ Aut(Y, βY , pY ) then S
and T := Gν have the same rigidity sequences.
Proposition 2.11. The sequence (nm ) is a rigidity sequence for some weakly mixing
dynamical system if and only if there is a continuous Borel probability measure ν on T
such that lim νb(nm ) = 1.
m→∞
Remark 2.12. The above of course means that ν is a Dirichlet measure. In what follows,
we will also say that the measure ν is rigid. Consequently, Dirichlet measures will be
also called rigid measures. In other words, a positive Borel measure ν is rigid when there
exists a sequence (nm ) such that lim νb(nm ) = ν(T).
m→∞
The same fact holds with appropriate changes if we ask for the stronger property that
we have rigidity along the times in the IP set generated by the sequence. Some notation
is useful for understanding this. The IP set generated by the sequence consists of all finite
RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
9
sums of distinct elements in the sequence. So the notation Σ = F S(nm ) for this IP set
makes sense. Here given m1 < · · · < mk , let σ = σ(nm1 , . . . , nmk ) = nm1 + · · · + nmk .
We say σ → ∞(IP ) if σ(1) = m1 tends to ∞. Also, given the dynamical system, we say
that T is IP-rigid on Σ if T σ → Id in the strong operator topology as σ → ∞(IP ). With
this understood, it is easy to see that there is a weakly mixing dynamical system that is
IP-rigid on Σ if and only if there is a continuous Borel probability measure ν on T such
that νb(σ) → 1 as σ ∈ F S(nm : m ≥ 1) tends to ∞ (IP).
There is another way of phrasing the Fourier transform condition above for rigidity, or
IP-rigidity, that is very useful. We leave the routine proof to the reader.
Proposition 2.13. Given a sequence (nm ) and a positive Borel measure on T, we have
νb(nm ) tends to kνk1 if and only if γ nm → 1 in measure with respect to ν as m → ∞.
Remark 2.14. The same type of result holds for νb(σ) tending to 1 as σ → ∞ (IP); it is
equivalent to γ σ converging to 1 in measure with respect to ν as σ → ∞ (IP).
The Fourier transform characterizations of rigidity sequences above suggests that we
ought to be able to take this further by finding the correct growth/sparsity condition on
a strictly increasing sequence (nm ) to be a rigidity sequence or IP-rigidity sequence for a
weakly mixing dynamical system. First, consider the property of IP-rigidity. It is easy to
see, from the spectral measure characterization of IP-rigidity above, that it is sufficient
to have a criterion that guarantees there is an uncountable Borel set of points K ⊂ T
such that for all γ ∈ K, we have γ σ → 1 as σ → ∞(IP ). Let us consider this in the
parametrization of T where γ = exp(2πix) for x ∈ [0, 1). We also denote by K the set of x
corresponding to exp(2πix) ∈ K. Our pointwise criterion then means that for all x ∈ K,
we have both cos(2πσx) → 1 and sin(2πσx) → 0 as σ → ∞ (IP ). It is enough to just
have sin(2πσx) → 0 as σ → ∞ (IP ). Indeed, with the usual notation that {z} = z − bzc
is the fractional part of a real number z, the {y = {2x} : x ∈ K} will give an uncountable
set of values y such that both cos(2πσy) → 1 and sin(2πσy) → 0 as σ → ∞ (IP )
Indeed, the pointwise spectral property above is equivalent to having rigidity along Σ for
all functions whose spectral measure in a given dynamical system is supported in K. This
is stronger than what is needed for rigidity along Σ for some weakly mixing dynamical
system. This weaker notion is equivalent to having a continuous positive measure ν
supported on K such that exp(2πσx) → 1 in measure with respect to ν as σ → ∞ (IP ).
3. Methods for Rigid Times for Weakly Mixing Transformations
We give a number of different approaches here for constructing weakly mixing transformations that have a specific type of sequence as a rigidity sequence. These methods
are sometimes overlapping, but the different approaches give us insights into the issues
nonetheless. Also, there are a variety of number theoretic and harmonic analysis connections with some of these methods; these are also explored in this section.
3.1. Diophantine approach. Denote by [x] the nearest integer to x ∈ R, choosing bxc
if {x} = 1/2. So |x − [x]| would be the distance of x to Z. We will denote this by kxk.
The distance kxk = {x} if {x} ≤ 21 and kxk = 1 − {x} if 12 ≤ {x}. We have the following
result.
10
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
Proposition 3.1. On some infinite perfect compact set K ⊂ [0, 1), sin(2πσx) → 0 as
σ → ∞(IP ) for all x ∈ K if and only if
∞
X
(3.1)
knm xk < ∞
m=1
for an uncountable set of x values.
Proof. We know that any uncountable Borel set in [0, 1) contains an infinite perfect compact subset. This makes the set theoretic aspects of this proposition work. By considering
separately the values of knm xk where it is {x} or it is 1 − {x}, it is not hard to see that
∞
P
Equation (3.1) implies that
sin(2πnm x) converges absolutely for an uncountable set.
m=1
Using the formula
sin(α + β) = sin(α) cos(β) + cos(α) sin(β)
k
P
repeatedly, we see that sin(2πxσ) =
fj sin(2πxnmj ) with coefficients fj that suitable
j=1
products of cosines. Here |fj | ≤ 1. So the convergence in Equation (3.1) tells us that for
an uncountable set we have sin(2πσx) → 0 as σ → ∞ (IP ).
Conversely, suppose we have for an uncountable set sin(2πσx) → 0 as σ → ∞ (IP ).
By doubling the x values, we see that this means that for an uncountable set K, if
x ∈ K, and > 0, we can choose M ≥ P
1 such that for all finite sets F of whole
numbers, all no smaller than M , we have k
nm xk ≤ . In particular, for all x ∈ K,
m∈F
we have knm xk → 0 as m → ∞. It follows that there is an uncountable set Ko ⊂ K
and some Mo such that for all
all finite sets F of whole numbers all no
P x ∈ Ko and
1
1
smaller than Mo , we have k
nm xk ≤ 100
. In particular, if m ≥ Mo , knm xo k ≤ 100
.
Now suppose
∞
P
m∈F
knm xo k = ∞ for some xo ∈ Ko . Then also
m=1
∞
P
knm xo k = ∞. Each
m=Mo
knm xo k is either {nm xo } or it is 1 − {nm xo }. Say I1 is the set of m ≥ Mo where the
first formula
the set of m ≥ Mo where the second formula holds. Then
P holds, and I2 is P
either
knm xo k = ∞ or
knm xo k = ∞. Assume it is the first case. Then by an
m∈I1
m∈I2
P
upcrossing argument, we can choose a finite set I ⊂ I1 such that
knm xo k ∈ ( 18 , 38 ). But
m∈I
P
P
then sin(2π
nm xo ) = sin(2π
knm xo k) ≥ √12 . This is not possible for x ∈ Ko because
m∈I P
m∈I
P
P
we know that k
nm xo k is small and so | sin(2π
nm xo )| = | sin(2πk
nm xo k)| ≤
m∈I
m∈I
m∈I
P
1
2πk
nm xo k ≤ 2π 100
. In the second case, again by an upcrossing argument, we can
m∈I
P
P
choose a finite set I ⊂ I2 such that
knm xo k ∈ ( 81 , 38 ). But then sin(2π
nm xo ) =
m∈I
m∈I
P
− sin(2π
knm xo k) ≤ − √12 . Again this is not possible for x ∈ Ko because we know that
m∈I
P
P
P
P
k
nm xo k is small and so | sin(2π
nm xo )| = | sin(2πk
nm xo k)| ≤ 2πk
n m xo k ≤
m∈I
1
2π 100
.
m∈I
m∈I
m∈I
It is not clear what growth property for (nm ) corresponds to Equation (3.1) holding for
an uncountable set of points. At least though this analysis shows that we can use results
RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
11
from Erdös and Taylor [9]. In particular, they show that a sufficient condition for the
∞
P
absolute convergence on an uncountable set that we need is that
nm /nm+1 converges.
m=1
See the beginning of the proof of Proposition 4.6. But sometimes much less of a growth
condition is needed. For example, if one just knows nm+1 /nm tends to infinity, but also
knows that the ratio is a whole number for large enough values of m, then we again have
∞
P
knm xk converging absolutely for an uncountable set of x values. However, they give
m=1
an example to show that the condition on the ratio being a whole number is necessary.
Erdös and Taylor [9] also observe that an earlier result of Eggleston [8] is relevant
here. Eggleston [8] showed that it is necessary to have something like nm+1 /nm tending
to infinity because if these ratios are bounded then one has exp(2πinm x) → 1 as m → ∞
for at most a countable set of values x. This fact is related to the weaker version of
the result above that is worth observing here, the pointwise criterion needed for rigidity
along the sequence itself. The question is: what growth condition on a strictly increasing
sequence (nm ) is needed for the sequence to admit a weakly mixing dynamical system for
which there is a non-trivial rigid function along (nm )? The same methods as above, using
the GMC, shows that it is sufficient to know when there is an uncountable set of points
K ⊂ [0, 1] such that for all x ∈ K, we have exp(2πinm x) → 1 as m → ∞. Hence, this
property can be characterized by having
lim knm xk = 0
(3.2)
j→∞
for an uncountable set of x values. As with convergence along an IP set, this property is
stronger than what is needed to produce just one weakly mixing dynamical system with
rigidity along (nm ). This property guarantees that any dynamical system, weakly mixing
or not, whose non-trivial spectral measures are supported in K, would have (nm ) as a
rigidity sequence.
These characterizations, Equation (3.2) and Equation (3.1), show the difference between
having rigidity along a sequence versus having rigidity along the IP set that the sequence
generates for all spectral measures supported in the set. Eggleston shows in the article
cited above that having lim mj+1 /mj = ∞ is sufficient for Equation (3.2) to hold for an
j→∞
uncountable set. But he also shows, as commented above, that it is necessary to know
that these ratios are not bounded. It is easy to construct examples of strictly increasing
sequences (mj ) with arbitrarily long blocks of terms on which the ratios mj+1 /mj = 2 on
the blocks while also Equation (3.2) holds for an uncountable set. So it is not necessary
to know that lim mj+1 /mj = ∞ in order to have Equation (3.2) for an uncountable set.
j→∞
Another way to look at the issue of when rigidity occurs is to ask this: given an
increasing sequence (nm ) of natural numbers is there a weakly mixing T such that for
a subsequence (nmk ) we have T nmk → Id in the strong operator topology? The answer
to this question is positive. The discussion above makes this clear; indeed, there is a
subsequence which is an IP rigidity sequence. The next (probably well-known) result also
addresses this question but with some additional Baire category aspects included. See
the end of Section 3.4.4 for another approach using continued fractions.
We want to make some general observations about the values of x such that knm xk → 0
as m → ∞. Alternatively, consider this set in its representation in T; we are then looking
for all γ ∈ T such that γ nm → 1 as m → ∞. The first important point is that this
12
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
is a subgroup of T, which we will denote by R(nm ). For example, it is clear that if
γ1 , γ2 ∈ R(nm ), then (γ1 γ2 )nm → 1 as m → ∞ also. So R(nm ) is a subgroup of T . It is
easy to see that it is a Borel set, indeed it is clearly an Fδσ because it is
∞ [
∞
∞
\
\
{γ : |γ nm − 1| ≤ 1/k}.
k=1 M =1 m=M
There is quite a bit of literature about such subgroups, and there is some interesting
descriptive set theory involved in the structure of this set too. First, consider this set
m
in the situation that the nm = qm are the denominators qm of the convergents pqm
of the
qm+1
continued fraction expansion of some fixed α. Sometimes the ratios qm are bounded. In
this case Larcher [27] showed that R(nm ) is just Zα + Z i.e. in T, we have R(nm ) just the
circle group generated by exp(2πiα). See also Kraaikamp and Liardet [22] who discuss
issues of speed of approach of kqm αk to 0. See also Host, Méla, and Parreau [17] where
this subgroup is considered extensively in the context of spectral analysis of dynamical
systems.
Also, one can reverse the question of the structure of subgroups R(nm ), by asking
which subgroups of T can be realized as such subgroups. It is generally known that
there are subgroups of the circle that are not even Lebesgue measurable, let along Borel
measurable. In communication with S. Solecki, we learned of references that are very
thorough in evaluating the descriptive set theoretic structure of subgroups of the circle.
For example, see the articles by Klee [21], Mauldin [31], Solecki [39], and Farah and
Solecki [10]. In addition, Solecki [40] has pointed out that there are even subgroups that
are in Fδσ which are not of the form R(nm ) for some (nm ). Moreover, subgroups like
R(nm ) are Polishable, see Farah and Solecki [10], and not all subgroups that are in Fδσ
are Polishable. In addition, he points out that there are Polishable Fδσ subgroups which
are not of the form R(nm ) for some (nm ). These results suggest that there is unlikely
to be a descriptive set theoretic characterization of the class of subgroups of the form
R(nm).
3.2. Symbolic approach. We can also deal with the rigidity question for certain types
of sequences when there are not necessarily pointwise results. In this section, we take
a symbolic approach. Later, in Section 3.3, we take an approach to this using Riesz
products, and then in Section 3.4, we will take more of an algebraic approach and give
examples using cocycle constructions.
We have seen that if nm+1 /nm is bounded then a pointwise approach to the rigidity
question will not work. So when this happens, the next result is giving us weakly mixing
transformations that are rigid along (nm ) even though R(nm ) is countable.
Proposition 3.2. Given an increasing sequence (nm ) such that nm+1 /nm is always a
whole number, there is a weakly mixing dynamical system for which there is rigidity along
(nm ).
Proof. let a1 = n1 and am+1 = nm+1 /nm for all m ≥ 1. So nm =
m
Q
k=1
ak and
nm
nM
m
Q
=
ak
k=M +1
∞
P
for all m ≥ M + 1 Consider the series representations for x ∈ [0, 1) of the form x =
m=1
bm
nm
where bm is a whole number such that 0 ≤ bm < am . Except for a set of Lebesgue measure
zero, such series are uniquely determined by x, and vice versa. We will write bm (x) to
RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
indicate the dependence of (bm ) on x. Let Π =
∞
Q
{0, . . . , am − 1} and let π =
m=1
13
∞
Q
πm
m=1
where πm is the uniform counting measure on {0, . . . , am − 1}. It is easy to see that
there is a one-to-one, onto Borel mapping Φ of [0, 1) to Π such that π ◦ Φ−1 is Lebesgue
measure..
Now we use a block construction to build a positive continuous Borel measure ν such
that knm xk → 0 in measure with respect to ν as m → ∞. Consider disjoint intervals
Ik = [Nk + 1, . . . , Nk+1 ], with 0 = No < N1 < N2 < . . . , and Nk+1 − Nk = |Ik |, the length
∞
P
of Ik , increasing to ∞. Fix (k ) with 0 < k ≤ 21 such that
k = ∞ and lim k = 0.
k→∞
k=1
Q
Define νk on Π as follows. Let 0k be the element in ( Ik {0, . . . , am − 1}) whose entries
are all 0. Let νk (0k ) = 1 − k and νk uniformly distributed over the points in \{0k } with
∞
Q
Q
the total mass νk (( Ik {0, . . . , am − 1})\{0k }) = k . Then let ν =
νk and consider the
k=1
measure ν ◦ Φ on [0, 1).
We know that ν and ν ◦ Φ are regular Borel probability measures because all finite
Borel measures are regular in this situation. Moreover, ν and ν ◦ Φ are continuous, i.e.
∞
∞
Q
P
they have no point masses. Indeed, ν({0}) =
(1 − k ) = 0 because
k = ∞, and by
k=1
k=1
the definition of ν for every other point x ∈ Π, we have ν({x}) ≤ ν({0}).
We claim that knm xk → 0 in measure with respect to ν ◦ Φ as M → ∞. We see that
∞
∞
P
P
bm
bm
knM xk =
. For any M ≥ 1, we have M ∈ Ik for a unique
=
nm /nM
aM +1 ...am
m=M +1
m=M +1
k = k(M ). Clearly, as M → ∞, we have k(M ) → ∞. Consider the set Dk of vectors
b = (bm ) ∈ Π such that bm = 0 for m ∈ Ik ∪ Ik+1 . We have ν(Dk ) = (1 − k )(1 − k+1 ) → 1
as k → ∞. But also for b ∈ Dk(M ) , we have bm = 0 for all m, M ≤ m ≤ Nk(M )+2 .
∞
∞
P
P
1
bm
≤
Hence, if Φ(x) ∈ Dk(M ) , then knM xk =
≤
aM +1 ...am
aM +1 ...am−1
m=Nk(M )+2 +1
m=Nk(M )+2 +1
P
1
1
= 2M +1 Nk(M
. But because we chose |Ik | increasing to ∞, we have
2m−M −1
)+2
m=Nk(M )+2 +1
1
2M +1 Nk(M
)+2
2
2
→ 0 as M → ∞. This means that, as M → ∞, we have ν ◦ Φ(Φ−1 Dk(M ) ) →
1 and for x ∈ Φ−1 Dk(M ) , we have knM xk → 0
This result leads to the following important special case.
Corollary 3.3. Given any whole number a ≥ 2, there is a weakly mixing dynamical
system for which there is rigidity along (am : m ≥ 1). However, there is never rigidity
along σ ∈ F S(am : m ≥ 1) as σ → ∞(IP ).
Proof. The first statement follows from Proposition 3.2 For the second part, first consider
the sequence (2j : j ≥ 1). When we take all finite sums 2j1 + · · · + 2jk with m ≤ j1 <
· · · < jk , then we get all whole numbers 2m s, s ≥ 1. Hence, to have rigidity for f ◦ T σ
m
as σ → ∞(IP ), with σ ∈ Σ = F S(2j : j ≥ 1) would imply that kf ◦ T 2 s − f k2 → 0 as
m → ∞, independently of s ≥ 1. Assume that f is mean-zero and not zero. Then we fix
m
m
m such that kf ◦ T 2 s − f k2 ≤ 21 for all s ≥ 1. But if T is weakly mixing, so is T 2 and
hence
m
lim hf ◦ T 2 s , f i → 0.
s→∞
14
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
This is not possible. The same argument works with any sequence (aj : j ≥ 1) in place
of (2j ) with a ∈ Z+ , a ≥ 2. The only difference is that we would need to use some fixed
number of finite sums of elements in Σ when a ≥ 3. For example, if we take σ1 + σ2
with each σi ∈ Σ, in the case that a = 3, then we would be able to represent all numbers
3m s if the smallest index in each σi is not smaller than m. But if kf ◦ T σi − f k2 ≤ for both i = 1 and i = 2, , then clearly kf ◦ T σ1 +σ2 − f k2 ≤ 2, and so using sums of
pairs of elements in Σ with all indices large allows us to represent any number of the form
2m s, s ≥ 1, with m large too.
Remark 3.4. a) See Proposition 3.31 for an improvement of the above negative result.
b) The contrast of this result with the main one in Section 3.1 is clear. For example, with
the sequence (nm ) given by nm = 2m as above, there is only a countable number (infinite
in this case) of values x such that knm xk tends to 0 as m → ∞.
c) If we assume that (nm ) is lacunary, but not necessarily as above a power of a fixed
whole number a ≥ 2, when is there still a continuous Borel probability measure on [0, 1)
with νb(nm ) → 1 as m → ∞? This seems to be a difficult problem because it is not just
about the growth rate inherent in lacunarity, but also about the algebraic nature of the
sequence (nm ). For example, if nm = 2m + 1 for all m, then γ nm tending to 1 implies that
m
m
m
m+1
m
γ 2 tends to γ, and so γ 2 = γ 2 /γ 2 tends to 1. But then γ nm /γ 2 = γ tends to 1.
This cannot happen in measure with respect to a non-zero continuous measure. Hence,
the lacunary sequence (2m + 1) does not admit rigidity in a weakly mixing dynamical
system. A similar type of calculation, carried out by a finite number of inductive steps of
taking differences, can be used to show that sequences like (am + p(m)) where the whole
number a ≥ 2 and p is a non-constant polynomial with integer coefficients, cannot be
rigidity sequences for a weakly mixing transformation. More generally, suppose we have a
K
P
linear function F (x1 , . . . , xK ) =
ck xk where c1 , . . . , ck ∈ Z. Suppose we know that for
k=1
all sufficiently large m, there are values mk ≥ m such that F (nm1 , . . . , nmK ) = c where c
is a fixed non-zero whole number. Then there cannot be a weakly mixing transformation
with rigidity along (nm ). We say that (nm ) is linearly separated if there are no such
functions F , values c, and values of nm . Not only is linear separation needed for rigidity,
but what is inherently also going on here is that if (nm ) is not linearly separated, then
the values of x such that lim exp(2πinm x) = 1 must force exp(2πix) to be M th roots of
m→∞
unity for some fixed value of M . Is something like this linear separation what one needs to
be able to construct a weakly mixing dynamical system that is rigid along the sequence?
d) On the other hand, it is not clear what happens for the lacunary sequence (2m +3m ), but
rigidity along this sequence is equivalent to rigidity along (2m ) and (3m ) simultaneously.
Can there be a continuous probability measure ν on T such that both νb(2m ) → 1 and
n
cu(3m ) → 1 as m → ∞?
e) Does there exist an increasing sequence (nm ) such that lim nnm+1
= 1, for which there
m
m→∞
is rigidity along (nm ) for some weakly mixing dynamical system? See Remark 3.24 for a
partial result in this direction. Concretely, recalling d) above, is there rigidity along (nm )
for some weakly mixing dynamical system if (nm ) is the sequence obtained by writing
{2k 3l : k, l ≥ 1} in increasing order? Note: results of Ajtai, Havas, and Komlos [2]
show that for any sequence (m ) decreasing to 0, no matter how slowly, there exists an
increasing sequence (nm ) such that nnm+1
≥ 1 + m but there is not rigidity along (nm ).
m
RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
1
M
Indeed, they give examples where
any positive Borel measure ν,
1
M
M
P
15
exp(2πinm x) = 0 for all x ∈ (0, 1), and hence for
m=1
M
P
m=1
νb(nm ) = 0.
3.3. Riesz Products. Here is another approach to defining continuous measures ν with
νb(nm ) → 1 as m → ∞. This method works in cases like (nm ) = (am ) for some a ≥ 2.
This method is like the classical method of Riesz products. There are many references
for Riesz products, but one that is closely related to the considerations in this article is
Host, Méla, and Parreau [17]. But unlike the classical approach, where one is taking Riesz
products in the frequency variable, we are taking Riesz products in the space variable,
using finitely supported discrete measures. Of course, this too has been used by others
to give constructions of measures. See Brown and Moran [3, 4] and Graham and McGehee [13]. This method may possibly give a way to handle cases that we have not been
able to handle before, but that is not clear. However, some other harmonic analysis issues
come into play here that lead to interesting conclusions in ergodic theory.
The basic approach is as follows. Choose (ak ) and (bk ) positive, with ak + bk = 1. We
∞
∞
∞
Q
Q
P
assume that lim bk = 0 and
ak =
(1 − bk ) = 0. So we assuming
bk = ∞. Take a
k→∞
k=1
k=1
k=1
sequence of points (xk ) and let ωk = ak δ1 + bk δexp(2πixk ) . Let νk = ωk ∗ ωk∗ = (a2k + b2k )δ1 +
ak bk δexp(2πixk ) + ak bk δexp(−2πixk ) . So we have νbk (j) = |c
ωk (j)|2 = 1 − 2ak bk (1 − cos(2πjxk )).
∞
Q
In particular, 0 ≤ νbk (j) ≤ 1 for all k and j. Consider the infinite product ν =
νk .
k=1
We take this as defined in the weak∗ sense; that is, we know that we have the Fourier
K
Q
transforms of the partial products ΠK =
νk converging and hence they give the Fourier
k=1
transform of some Borel probability ν in the limit. Then we evaluate νb(nm ) by looking
at what the infinite product does.
Here is an example of this approach. Take nm = 2m for all m ≥ 1 and let xk = 21k for
1
1
all k ≥ 1. We specifically choose bk = k+1
, so ak = 1 − k+1
. We defined ν by the spectral
condition
!
∞
1
Y
(1 − k+1
)
j
νb(j) =
1−2
(1 − cos(2π k )) .
k
+
1
2
k=1
Now consider the values νb(2m ). We always have
m
νb(2 ) =
∞
Y
k=m+1
!
1
(1 − k+1
)
2m
1−2
(1 − cos(2π k )) .
k+1
2
So we need to know what this does as m → ∞.
But we claim that νb(2m ) → 1 as m → ∞. That is, by taking the natural logarithm of
the expression, we need to show that
lim
m→∞
∞
X
k=m+1
2
1
(1 − k+1
)
1
(1 − cos(2π k−m )) = 0.
k+1
2
16
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
However, this is the same as showing
∞
1
X
)
(1 − k+1
1
lim
2
= 0.
2(k−m)
m→∞
k+1 2
k=m+1
∞
P
But
k=m+1
2
1
)
(1− k+1
1
k+1
22(k−m)
≤
22m+1
m
∞
P
k=m+1
1
22k
22m 1
m 22m
≤
=
1
.
m
So we have the estimate that
we needed.
We also need to make an argument that ν has no atoms. First, each of the partial
K
Q
products ΠK =
νk is a purely atomic measure. The choice of ak and bk guarantees that
k=1
for K ≥ 1, these partial products have a maximal mass at 1 with the value
K
Q
(a2k + b2k ) =
k=1
K
Q
(1−2ak bk ). Suppose that ν is suspected of having a non-trivial point mass at some point
k=1
< x < j+1
}
γ0 . For a fixed value of K, choose an open dyadic arc A = {exp(2πix) : j−1
2K
2K
which contains γo . Then there is a continuous function hK , 0 ≤ hK ≤ 1, with hK (γo ) = 1
and
R the support of hK , that is the closure of {hK > 0}, contained in A. We have for any M ,
hK dΠM ≤ ΠM (A). We can easily see that for M ≥ K, ΠM (A) ≤ K with lim K = 0.
K→∞
R
R
But then hK dΠM ≤ K too, so letting M tend to ∞ shows that hK dνR ≤ K . Then
letting K → ∞, this proves that ν({γo }) = 0 because for all K, ν({γo }) ≤ hK dν.
But the estimate we need is easy because, for M ≥ K, we can write the partial product
2M
2M
P
P
ΠM as a sum
ct (M )δexp(2πi t−1
with positive coefficients ct (M ) such that
ct (M ) = 1.
M )
2
t=1
t=1
The value of ΠM (A) can be seen to be no larger than the value that the measure µ given
by
M
Y
cj−1 (K)δexp(2πi j−1 ) + cj (K)δexp(2πi j ) + cj+1 (K)δexp(2πi j+1 )
νk
2K
2K
2K
k=K+1
, the measure δexp(2πi t−1
gives to A because for the other point masses δexp(2πi t−1
K )
K )
2
M
Q
has support disjoint from A. Since
2
M
Q
νk
k=K+1
νk itself is a probability measure, this shows
k=K+1
that ΠM (A) ≤ cj−1 (K) + cj (K) + cj+1 (K) ≤ 3
K
Q
(a2k + b2k ). We take K = 3
k=1
3
K
Q
K
Q
k=1
(1 − 2ak bk ). The choice of (ak ) shows that K → 0 as K → ∞. Indeed,
k=1
and lim ak = 1, so
k→∞
∞
Q
(a2k + b2k ) =
∞
P
bk = ∞
k=1
(1 − 2ak bk ) = 0.
k=1
Remark 3.5. a) The method above suggests that we might be able to choose (xk ) for
more general sequences, including perhaps ones that are not lacunary. But the method
implicitly needs inductively chosen good choices of x where exp(2πinm x) → 1 as m →
∞. It is not clear for a fixed (nm ) how to characterize when the set of x such that
lim exp(2πinm x) = 1 is empty, when it is finite, when it is countably infinite, or when
m→∞
it is uncountable!
RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
17
b) believe that what is needed generally for the construction as above of a continuous
Borel probability measure ν with lim νb(nm ) = 1 as m → ∞ is
m→∞
(a) lim bk = 0,
k→∞
∞
P
bk = ∞, and
(b)
k=1
(c) there is a sequence (xk ) such that lim
∞
P
m→∞ k=1
bk knm xk k2 = 0
For example, this method would work not only for the sequences above, but also for
sequences as in Proposition 3.2.
c) We have tried to prove directly that the examples above have a Fourier transform
converging to 0 at ∞ along a sequence of positive density. For example, one would want
to show that for some sequence of values j with positive density, we have
lim
j→∞
∞
X
bk (1 − cos(2π
k=1
j
) = ∞.
2k
But we cannot see how to prove this.
3.3.1. Disjoint Rigidity. The method above can perhaps give us some good examples to
look at for which we have two systems that are rigid but not simultaneously, indeed when
the convolution of their maximal spectral types is strongly mixing. For example, take the
measures ν(2) and ν(3) where
d
ν(2)(j)
=
∞ Y
j
1 − 2ak bk (1 − cos(2π k ))
2
k=1
d
ν(3)(j)
=
∞ Y
j
1 − 2ak bk (1 − cos(2π k )) .
3
k=1
and
d m ) → 1 and ν(3)(3
d m ) → 1 as m → ∞. But does
We have ν(2)(2
(3.3)
d ν(3)(j)
d
ν(2)(j)
→ 0 as j → ∞?
This is the same as asking if the series
∞
X
j
j
bk 1 − cos(2π k ) + 1 − cos(2π k )
2
3
k=1
tends to ∞ as j tends to ∞. This will not work with bk = k1 ; one can see this by taking
j = 6k for large values of k. But it may work with bk = √1k .
While we cannot answer the question 3.3 at this time, we can modify a construction of
LaFontaine [26] to obtain Proposition 3.10 below. We will need two procedural lemmas.
RB
First, given a finite Borel measure ω on R, we let F TB (ω)(n) = exp(−2πin Bx ) dω(x). We
0
R
use the notation ω
b for the Fourier transform on R given by ω
b (t) = exp(−2πitx) dω(x).
Let m = mR denote the Lebesgue measure on R, and let L2 (m) denote the Lebesgue
space L2 (R, βm , m).
18
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
Lemma 3.6. Suppose we have a positive regular Borel measure ω on R that has compact
∞
P
dω
support. Then dm
∈ L2 (m) if and only if
|b
ω (n)|2 < ∞.
n=−∞
Proof. By translating ω, we may assume without loss of generality that the support of ω
is a subset [0, B] for some whole number B. Then
Z
ω
b (n) =
exp(−2πinx) dω(x)
=
k+1
B−1
XZ
exp(−2πinx) dω(x)
k=0 k
=
1
B−1
XZ
exp(−2πin(x + k)) dω(x + k)
k=0 0
=
1
B−1
XZ
exp(−2πinx) dω(x + k)
k=0 0
Z1
=
exp(−2πinx) dΩ(x).
0
where dΩ(x) = 1[0,1) (x)
B−1
P
ω(x + k) is a positive Borel measure supported on [0, 1].
k=0
dΩ
That is, ω
b (n) = F T1 (Ω)(n) for all n. The usual classical argument shows that dm
∈
∞
P
L2 (m) if and only if
|F T1 (Ω)(n)|2 < ∞ because Ω is supported on [0, 1). Therefore,
n=−∞
∞
P
2
|b
ω (n)| < ∞ is equivalent to knowing that
n=−∞
dΩ
dm
∈ L2 ([0, 1), m).
dΩ
∈ L2 ([0, 1), m), then for each k = 0, . . . , B − 1, the positive
So we conclude that if dm
measure 1[0,1] dω(x + k) satisfies 0 ≤ 1[0,1] dω(x + k) ≤ dΩ(x), and so it also has a density
in L2 ([0, 1), m). Hence, by translating the terms back again and adding them together,
dω
dω
we also know that dm
is in L2 (R, m). Of course also conversely, if dm
is in L2 (R, m), then
∞
P
dΩ
dω
∈ L2 ([0, 1), m). This proves that dm
∈ L2 (m) if and only if
|b
ω (n)|2 < ∞.
dm
n=−∞
Remark 3.7. This remark and Lemma 3.6 are related to the ideas behind Shannon sampling and the Nyquist frequency for band-limited signals. In Lemma 3.6, if we replace ω
by a dilation of it, then one can see that more generally for a positive Borel measure ω
dω
on R with compact support, we have dm
∈ L2 (m) if and only if for some b > 0 (or for all
∞
P
b > 0), we have
|b
ω (bn)|2 < ∞.
n=−∞
The assumption that ω is positive is necessary here. Indeed, suppose we have a compactly supported complex-valued Borel measure ω on R. Suppose that the support of ω
dω
is a subset of [0, B]. Then by the usual classical argument, we know that dm
∈ L2 (m) if
∞
P
and only if
|F TB (ω)(n)|2 < ∞. But ω
b ( Bn ) = F TB (ω)(n) because ω is supported on
n=−∞
RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
[0, B]. So we have in this case,
dω
dm
∈ L2 (m) if and only if
∞
P
n=−∞
19
|b
ω ( Bn )|2 < ∞. Here, B
can be replaced by any larger value, but not necessarily by a smaller value. For example,
if we take ωo supported on [0, 1) and define dω(x) = dωo (x) − dωo (x − 1), then our value
of B = 2, but ω
b (n) = 0 for all n. However, the measure ωo could be singular to m and
hence ω might not have an L2 (R, m)-density.
We also want to make a few observations about the differences between convolving
measures on T and convolving their associated measures on R. Given a Borel measure ω
on T, let ωR denote the Borel measure on R given by ωR = ω ◦ E where E : [0, 1) → T by
E(x) = exp(2πix). Let mT denote the usual Lebesgue measure on R i.e. mT ◦E = 1[0,1) mR .
Lemma 3.8. If µ and ν are regular Borel measures on T, then µ ∗ ν is absolutely continuous with respect to mT if µR ∗ νR is absolutely continuous with respect to mR .
Proof. For f ∈ C(T), we have
Z
Z
f (γ)d(µ ∗ ν)(γ) =
Z
=
Z
=
Z
=
Z
f (γ1 γ2 )dµ(γ1 )dν(γ2 )
Z
f (E(x)E(y))dµR (x)dνR (y)
f ◦ E(x + y)dµR (x)dνR (y)
f ◦ E(z)d(µR ∗ νR )(z).
Hence, it is clear that if µR ∗ νR is absolutely continuous with respect to mR with density
F , then µ ∗ ν is absolutely continuous with respect to mT with density F ◦ E −1 .
Remark 3.9. The converse statement to Lemma 3.8 is not true without additional assumption, for example the assumption that both measures are positive. For example,
take a non-zero measure ωo on T supported in the arc E([0, 1/2). Let ω = ωo − ωo ∗ δ−1 .
Then let ν = δ1 + δ−1 , a positive discrete measure on T. We have
ω ∗ ν = ωo − ωo ∗ δ−1 + ωo ∗ δ−1 − ωo ∗ δ1
= ωo − ω ∗ δ1
= 0.
However, ωR = (ωo )R − (ωo )R ∗ δ1/2 and νR = δ0 + δ1/2 . So
ωR ∗ νR = (ωo )R − (ωo )R ∗ δ1/2 + (ωo )R ∗ δ1/2 − (ωo )R ∗ δ1
= (ωo )R − (ωo )R ∗ δ1 .
This measure is not zero as a measure on R. If also ωo is singular to mT , then we have
ω ∗ ν absolutely continuous with respect to mT , since it is 0, while ωR ∗ νR is not absolutely
continuous with respect to mR .
These two lemmas will help in proving the following.
Proposition 3.10. Assume that (X, βX , pX , T ) is a rigid, weakly mixing dynamical system. Then there is a weakly mixing, rigid dynamical system (Y, βY , pY , S) such that the
maximal spectral type of UT ×S in the orthocomplement of F := L2 (X, pX ) ⊗ 1Y ⊕ 1X ⊗
20
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
L2 (Y, pY ) is Rajchman. In other words, for all f ∈ L2 (X × Y, pX ⊗ pY ) that are orthogonal to both the X-measurable functions and the Y -measurable functions, we have
hf ◦ (T × S)n , f i → 0 as n → ∞. In fact, UT ×S has absolutely continuous spectrum on
F ⊥.
Proof. From the spectral point of view we want to show that given a continuous Dirichlet
measure µ, there is a continuous Dirichlet probability measure ν on T such that µ ∗ ν is
an absolutely continuous (hence a Rajchman measure). Indeed, we then can take µ = ν T
and we let
βY , pY , S) be given by S = Gν . As µ ∗ ν ∗k = (µ ∗ ν) ∗ ν ∗(k−1) we easily check
P(Y,
∞
that µ ∗ k=1 21k ν ∗k is still absolutely continuous. Given the references we are using, it
is better to carry out our construction in R. So consider first the measure µ ◦ E on R.
Lemma 3.8 shows that, to get our result, it will be enough to construct a suitable positive
Borel measure ν ◦ E on R, with support in [0, 1), such that (µ ◦ E) ∗ (ν ◦ E) is absolutely
continuous. For notational convenience we will denote µ ◦ E and ν ◦ E by µ and ν in the
rest of this proof.
Hence, suppose we have a continuous positive Borel measure µ on R, which is compactly
supported. Consider the
by µλ (E) = µ( λ1 E) for all Borel sets
R dilation µλ , λ > 0, given
R
E ⊂ R. Then µ
cλ (t) = exp(−2πitx) dµλ (x) = exp(−2πitλx) dµ(x) = µ
b(λt). We would
like to construct a suitable continuous positive Borel measure ν supported in [0, 1] such
∞
P
that J(λ) =
|c
µλ (n)|2 |b
ν (n)|2 is finite. But
n=−∞
Z1
J(λ) dm(λ) =
∞
X

|b
ν (n)|2 
n=−∞
0
Z1

|b
µ(λn)|2 dm(λ) .
0
Using Wiener’s lemma for measures on R (e.g. [19], Chapter VI.2) and the fact that µ is
continuous
Z1
Z
1 n
2
an := |b
µ(λn)| dm(λ) =
|b
µ(t)|2 dt → 0
n 0
0
as |n| → ∞. So, we are seeking a suitable ν such that
(3.4)
∞
X
|b
ν (n)|2 an < +∞.
n=−∞
Clearly, if ν were actually absolutely continuous with respect to m with a square integrable
density, then we would have this condition. But ν could not be rigid in this situation.
However, as LaFontaine points out, the two articles of Salem [37, 38] give a construction
of a Borel probability measure ν with support in [0, 1) with this property, and which
is also continuous and rigid. See LaFontaine [26] and Salem [37, 38] for the details. It
follows that there exists a continuous probability measure ν supported on [0, 1) such that
for some increasing sequence (nm ) of integers
(3.5)
and (3.4) holds.
νb(nm ) → 1
RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
21
It follows from (3.4) that for m-a.e. λ ∈ [0, 1]
∞
X
(3.6)
n=−∞
2
|µ\
λ ∗ ν(n)| =
∞
X
|c
µλ (n)|2 |b
ν (n)|2 = J(λ) < ∞.
n=−∞
The measure µλ ∗ν is supported in [0, 1+ λ1 ] and in view of Equation (3.6) and Lemma 3.6, it
λ ∗ν)
must be absolutely continuous with respect to m, with d(µdm
∈ L2 (m). Thus, if we choose
R∞
R∞
|b
µ(t)|2 |b
ν ( λt )|2 dm(t)
any value of λ satisfying (3.6), we have
|b
µ(λt)|2 |b
ν (t)|2 dm(t) = λ1
−∞
−∞
is finite. It follows that the measure µ ∗ ν1/λ is compactly supported and absolutely
continuous. In view of (3.5), νd
1/λ (λnm ) converges to 1 as m → ∞. Moreover, with respect
to m, for almost every λ, we would know that (nk λ : k ≥ 1) is uniformly distributed
modulo 1. Hence, for
some subsequence (nkj ) and some sequence of integers (mj ), we
have lim nkj λ − mj = 0. Then by the uniform continuity of νd
1/λ , we would also have
j→∞
νd
1/λ (mj ) → 1 as j → ∞. Hence, with respect to m, almost every choice of 0 < λ < 1 gives
ν1/λ supported on [0, λ] ⊂ [0, 1], which is a continuous rigid Borel probability measure,
such that µ ∗ ν1/λ is absolutely continuous.
Remark 3.11. The abstract argument used in the above proof may give some motivation
for finding some concrete examples of the phenomenon of continuous rigid measures whose
convolution has a Fourier transform vanishing at infinity, as we were considering above
with the Riesz product constructions for powers of 2 and powers of 3.
Remark 3.12. We can argue differently that if a continuous probability measure ρ supported on [0, 1] ⊂ R satisfies ρb(rk ) → 1 for some sequence of reals rk → ∞, then, as a
circle measure, ρ is Dirichlet. Indeed, consider the flow Vt (f )(x) = eitx f (x) on L2 (R, ρ).
Our assumption says that Vrk → Id. Consider then V{rk } , k ≥ 1. By passing to a subsequence if necessary and using the continuity of the unitary representation R 3 t 7→ Vt ,
we have V{rk } → Vs for some s ∈ [0, 1]. It follows that V[rk ] → Id ◦ (Vs )−1 = V−s . Then,
([rk+1 ] − [rk ]) is a rigidity sequence for ρ.
Corollary 3.13. Assume that (X, β, p, T ) is an ergodic dynamical system. Then T has
discrete spectrum if and only if for each weakly mixing rigid system (Y, βY , pY , S) the
Cartesian product system T × S remains rigid.
Proof. Assume that T is an ergodic rotation and let S be weakly mixing, S nm → Id. By
passing to a subsequence if necessary, T nm → R ∈ C(T ), whence T nm+1 −nm → Id and
still S nm+1 −nm → Id; thus T × S is rigid (see also Proposition 2.9).
To prove the converse suppose that T does not have discrete spectrum. but T × S
is rigid for each S which is rigid and weakly mixing. Then there is some continuous ν
T (n ) → kν T k then
with ν ν T . It follows that ν is a Dirichlet measure, and if νc
m
1
νb(nm ) → kνk1 . Consider the Gaussian system Gν given by ν. Then Gν is weakly mixing
and each sequence which is a rigid sequence for T is also a rigid sequence for Gν . It
follows that Gν × S is rigid for each weakly mixing rigid S which is in conflict with from
Proposition 3.10.
Corollary 3.14. Assume that (X, βX , pX , T ) is an ergodic rotation and (Y, βY , pY , S) is
ergodic and has the same rigidity sequences as T . Then S is isomorphic to T .
22
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
Proof. It follows directly from Corollary 3.13 that S has discrete spectrum. The result
follows from the proposition below.
The following is a folklore result.
Proposition 3.15. Assume that T and S are ergodic rotations. They are isomorphic if
and only if the have the same rigidity sequences.
Proof. Consider C(T ) and C(S) respectively. Both are given as weak closure of powers.
Take the map:
F : C(T ) → C(S), F (T n ) = S n , n ∈ Z.
We easily show that this extends to a homeomorphism equivariant with rotation by T
and S respectively.
Then we apply the fact (which is a consequence of the Halmos-von Neumann Theorem)
that T is isomorphic to the translation by T on C(T ) considered with Haar measure (recall
that a transformation has discrete spectrum if and only if its centralizer is compact). 3.4. Cocyle methods.
3.4.1. Tools. We will now describe tools using cocycles over rotations to produce weakly
mixing transformations with a prescribed sequence as rigidity times. We will start with
a transformation T having discrete spectrum and its sequence of rigidity. (In fact, for
applications, we will consider one dimensional irrational rotations by α and the sequence
given the denominators of α.) Then we will consider cocycles over T with values in
locally compact Abelian groups. We will then pass to the associated unitary operators
(weighted operators) and we will try to “lift” some rigidity sequences for the rotation to
the weighted operator. Once such an operator has continuous spectrum we apply the GMC
which preserves rigidity. Another option to obtain “good” weakly mixing transformations
will be to pass to Poissonian suspensions (in case we extend by a locally compact and not
compact group) - which in a sense will be even easier as ergodicity of Poissonian suspension
will be closely related to the fact that the cocycles are not coboundaries. See [5] (and also
recent [7], [35]) for details concerning ergodic properties of Poissonian suspensions.
3.4.2. Compact group extension, weighted operators. Assume that T is an ergodic automorphism acting on a standard probability Borel space (X, β, p). Let G be a compact
metric Abelian group with Haar measure λG . A measurable map ϕ : X → G is called
˙
a cocycle; in fact it generates a cocycle ϕ() (·) given by ϕ(n) (·) : X → G, n ∈ Z, by the
formula, for x ∈ X,

 ϕ(x) + ϕ(T x) + . . . + ϕ(T n−1 x) if n > 0,
(n)
0
if n = 0,
(3.7)
ϕ (x) =
 −(ϕ(T −n x) + . . . + ϕ(T −1 x)) if n < 0.
Using T and ϕ we define a compact group extension Tϕ of T which acts on the space
(X × G, B ⊗ B(G), p ⊗ λG ) by the formula
(3.8)
Tϕ (x, g) = (T x, ϕ(x) + g) for (x, g) ∈ X × G.
Notice that for each n ∈ Z and (x, g) ∈ X × G
(3.9)
(Tϕ )n (x, g) = (T n x, ϕ(n) (x) + g).
RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
23
b yields the decomThe natural decomposition of L2 (G, λG ) using the character group G
position
M
(3.10)
L2 (X × G, p ⊗ λG ) =
L2 (X, p) ⊗ (Cχ).
b
χ∈G
To understand ergodic and other mixing properties of Tϕ we need to study the associated
Koopman operator UTϕ ,
UTϕ F = F ◦ Tϕ , for F ∈ L2 (X × G, p ⊗ λG ).
As the (closed) subspaces L2 (X, p) ⊗ (Cχ) in (3.10) are UTϕ -invariant, we can examine
those mixing properties separately on all such subspaces (notice that for χ = 1 we consider
the original Koopman operator UT ). It is well-known and not hard to see that the map
T
acting
f ⊗ χ 7→ f provides a spectral equivalence of UTϕ |L2 (X,p)⊗(Cχ) and the operator Vχ◦ϕ
on L2 (X, β, p) by the formula
T
Vχ◦ϕ
f = χ ◦ ϕ · f ◦ T for f ∈ L2 (X, p).
(3.11)
Each such operator is an example of a weighted operator VξT over T , where ξ : X → S1 is
a measurable function with values in the (multiplicative) circle S1 and VξT f = ξ · f ◦ T .
Assume now that T is an ergodic transformation with discrete spectrum, i.e. without
loss of generality, we can assume that X is a compact monothetic metric group with
p = λX Haar measure on X, and T x = x + x0 where x0 and {nx0 : n ∈ Z} is dense in X.
Assume that ξ : X → S1 is measurable. By Helson’s analysis [14] (see also e.g. [18]):
Theorem 3.16 ([14]). For T as above, the maximal spectral type of VξT is either discrete
or continuous. If it is continuous then either it is singular or it is Lebesgue.
We will use only the first part of this theorem. An important practical point that
comes from this theorem is that once we find a function f ∈ L2 (X, β, p) such that σf is
continuous then VξT has purely continuous spectrum. Consider f = 1. Using (3.9), for
each n ∈ Z, we obtain
Z
T n
(3.12)
h Vξ
1, 1i =
ξ (n) (x) dp(x).
X
It follows that if there is a subsequence (nm )m≥1 such that
Z
(3.13)
ξ (nm ) (x) dp(x) → 0 ⇒ VξT has continuous spectrum.
X
It is also nice to note in passing that we have a bit stronger result:
Z
nm
(3.14)
ξ (nm ) (x) dp(x) → 0 ⇒ VξT
→ 0 weakly,
X
which follows directly from (3.13); indeed,
Z
ξ (nm ) (x)f (T nm x)f (x) dp(x) → 0
X
whenever f is a character of X.
It is well known and easy to check that if (nm ) is a rigidity time for T
nm
(3.15)
ξ (nm ) → 1 in measure ⇒ VξT
→ Id strongly.
24
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
3.4.3. R-extensions, weighted operators and Poissonian suspensions. We will now consider
the case G = R. We assume now that f : X → R is a cocycle for T acting ergodically on
a standard probability Borel space (X, β, p). We consider Tf
Tf (x, r) = (T x, f (x) + r)
acting on (X × R, p ⊗ λR ). Note that we are now on a standard Borel space with a σ-finite
(and not finite) measure. In particular, constants are not integrable.
To study spectrally Tf we will write it slightly differently, namely
Tf = Tf,τ
where τ is the natural action of R on itself by translations: τt (r) = r + t and
Tf,τ (x, r) = (T x, τf (x) (r)).
This transformation is a special case of so called Rokhlin extension (of T ), e.g. [28], and
the spectral analysis below is similar to the one in [28]. So let us just imagine a slightly
more general situation
Tf,S (x, y) = (T x, Sf (x) (y))
where S = (St ) is a flow acting on (Y, C, ν) (ν can be finite or infinite). We will denote
the spectral measure of a ∈ L2 (Y, C, ν) (for the Koopman representation t 7→ USt on
L2 (Y, C, ν)) by σa,S .
The space L2 (X × Y, p ⊗ ν) is nothing but a tensor product of two Hilbert subspaces,
so to understand spectral measures we only need to study spectral measures for tensors
a ⊗ b and we have
Z
(a ⊗ b) ◦ (Tf,S )n · a ⊗ b dp dν
X×Y
Z Z
a(T n x)a(x)b(Sf (n) (x) y)b(y) dp(x) dν(y)
=
X
Y
Z
=
n
Z
a(T x)a(x)
X
e
dσb,S (t) dp(x)
Y
Z Z
=
e
Y
2πitf (n) (x)
2πitf (n) (x)
n
a(T x)a(x) dp(x)
dσb,S (t).
X
Proposition 3.17. If T nk → Id and f (nk ) → 0 in measure then (nk ) is a rigidity sequence
for Tϕ,S .
Proof. Take a ∈ L∞ (X, p) and notice that by assumption for each t ∈ R
Z
Z
2πitf (nk ) (x)
nk
e
a(T x)a(x) dp(x) →
|a|2 dp.
X
X
By the Lebesgue Dominated theorem
Z Z
Z Z
2πitf (nk ) (x)
nk
2
e
a(T x)a(x) dp(x) dσb,S (t) →
|a| dp dσb,S = ka⊗bk2L2 (p⊗λR ) .
Y
X
Y
X
RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
25
We need more information about sequences of the form
Z
(n)
e2πitf (x) a(T n x)a(x) dp(x), n ∈ Z.
X
In fact, they turn out to be again Fourier coefficients of some spectral measures. Indeed,
consider Vt acting on L2 (X, β, p) by the formula
(Vt a)(x) = e2πitf (x) a(T x).
This is nothing but a weighted unitary operator and
Z
(n)
n
hVt a, ai =
e2πitf (x) a(T n x)a(x) dp(x).
X
(Notice that we came back to the finite measure-preserving case.)
Clearly, for a ∈ L2 (R) with compact support
Z
a ◦ St · a dr
σ
ba,S (t) =
R
Z
=
a(r + t)a(r) dr = (a ∗ a)(−t)
R
whence the Fourier transform of σa,S is square summable, and therefore this spectral
measure is absolutely continuous. In fact, the maximal spectral type of S is Lebesgue,
and we can see the maximal spectral type of UTf as an integral (against “Lebesgue”
measure) of the maximal spectral types of the family indexed by t ∈ R of weighted
operators.
Suppose now that UTf has an eigenvalue c, |c| = 1. Then we cannot have that all
spectral measures σa⊗b,Tf are continuous. In fact, we must have that c appears as an
eigenvalue for “many” Vt (on a set of t ∈ R of positive Lebesgue measure), and the
following result is well known (it is an exercise).
Lemma 3.18. c is an eigenvalue of Vt if and only if we can solve the following functional
equation:
e2πitf = c · ξ ◦ T /ξ
in measurable functions ξ : X → S1 .
It follows that having an eigenvalue for UTf means that we can solve the above multiplicative equations on a set of positive Lebesgue measure of t ∈ R. We are now in
the framework of the classical Helson’s problem (e.g. [32]) of passing from multiplicative
coboundaries to additive coboundaries. Using known results in this area ([32], and see
also the appendix in [29]) we obtain the following (remember that constant functions are
not elements of L2 (X × R, p ⊗ λR )).
Proposition 3.19. If UTf has an eigenvalue then f is an additive quasi-coboundary, that
is there exist a measurable g : X → R and r ∈ R such that f (x) = r + g(x) − g(T x) for
p-a.e. x ∈ X.
It follows that if f is a non-trivial cocycle then automatically UTf has continuous spectrum and classically the Poissonian suspension over Tf is ergodic (hence weakly mixing).
Recall that from spectral point of view Poissonian suspension over (X × R, p ⊗ λR , Tf )
will be the same as Gaussian functor over (L2 (X × R, p ⊗ λR ), UTf ). In particular, if
26
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
(Tf )nt → Id on L2 (X × R, p ⊗ λR ) then (nt ) will be a rigidity sequence for the suspension (in view of Proposition 3.17). From the above discussion it follows that to produce
“weakly mixing rigidity sequences” we need:
(i) f is not an additive coboundary,
(ii) T nt → Id,
(iii) f (nt ) → 0 in measure.
3.4.4. Denominators of α are “weakly mixing realizable”. We have already seen that the
sequence (2n ) is a rigidity sequence for a weakly mixing transformation. We can construct
some other explicit examples of rigidity sequences by using known results from the theory
of “smooth” cocycles over one-dimensional rotations. This will allow us to show that if α
is irrational, and (qn ) stands for its sequence of denominators then (qn ) is also a rigidity
sequence for a weakly mixing transformation. The most interesting case is of course the
bounded partial quotient case (for example for the Golden Mean).
So assume that ϕ : T → R is a smooth zero mean cocycle. We use the term “smooth”
here in a not very precise way; it may refer to a good speed of decaying of the Fourier
transform of ϕ.
We recall first that one of consequences of the Denjoy-Koksma Inequality for AC0
(absolutely continuous zero mean) cocycles is that
(3.16)
ϕ(qn ) → 0 uniformly
for every irrational rotation by α, see [15]. Another type of Denjoy-Koksma inequality
has been proved in [1] for functions ϕ whose Fourier transform is of order O(1/|n|) – as
its consequence we have the following:
(3.17)
1
If ϕ(n)
b
= o( |n|
), ϕ(0)
b = 0 then ϕ(qn ) → 0 in measure
for every irrational rotation by α.
We would like also to recall another (unpublished) result by M. Herman [16] (see also
[23] for generalizations).
Theorem 3.20. Assume that a zero-mean ϕ : T → R is in L2 (T, λT ) and its Fourier
transform is concentrated on a lacunary subset of Z. Suppose that
ϕ(x) = g(x) − g(x + α), λT − a.e.
for some irrational α ∈ [0, 1). Then g ∈ L2 (T, λT ).
Fix α ∈ [0, 1) irrational, and let α = [0 : a1 , a2 , . . .] stand for the continued fraction
expansion of α. Denote by (qn ) the sequence of denominators of α: q0 = 1, q1 = a1 and
qn+1 = an+1 qn + qn−1 for n ≥ 2. Then
an+2 qn+1 + qn
qn+2
=
≥ an+2 + 1 ≥ 2.
qn
qn
It follows that
(3.18)
(q2n ) is lacunary.
Moreover,
(3.19)
qn kqn αk ≤ 1 for each n ≥ 1.
RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
27
We define ϕ : T → R by
(3.20)
ϕ(x) =
∞
X
aq2n cos 2πiq2n x
n=0
where for n ≥ 1
1
aq2n = √ kq2n αk.
n
(3.21)
We then have ϕ : T → R, ϕ(n)
b
= o(1/|n|) and (aq2n ) ∈ l2 in view of (3.19). Now, suppose
that
ϕ(x) = g(x) − g(x + α)
(3.22)
for a measurable g : T → R. In view of Theorem 3.20 and (3.18), g ∈ L2 (T, λT ). Hence,
∞
X
g(x) =
bk e2πikx .
k=−∞
Furthermore, by comparing Fourier coefficients on both sides in (3.22),
bk = 0 if k 6= q2n and bq2n = aq2n /(1 − e2πiq2n α )
with (bq2n ) ∈ l2 . However
√
|bq2n | = aq2n /kq2n αk = 1/ n
which is a contradiction. We hence proved the following.
Proposition 3.21. For each irrational α ∈ [0, 1) there is a zero-mean ϕ : T → R such
that ϕ(n)
b
= o(1/|n|) and ϕ is not an additive coboundary.
Using (3.17) and Proposition 3.17 we hence obtain the following.
Proposition 3.22. For ϕ satisfying the assertion of Proposition 3.21 the sequence (qn )
of denominators of α is a rigidity sequence of Tϕ on L2 (T × R, λT ⊗ λR ) and UTϕ has
continuous spectrum.
By using GMC method or by passing to the relevant Poissonian suspension we obtain:
Corollary 3.23. For each sequence (qn ) of denominators there exists a weakly mixing
transformation R such that Rqn → Id.
Remark 3.24. (i) We would like to emphasize that in general the sequence (qn ) of denominators of α is not lacunary. Indeed, assume that α = [0 : a1 , a2 , . . .] stands for the
continued fraction expansion of α. Suppose that for a subsequence (nk ) we have ank = 1,
ank +1 → ∞. Then by the recurrence formula qm+1 = am+1 qm + qm−1 we obtain that
qn+1
lim inf
= 1.
n→∞
qn
It seems that the sequences of denominators are the only known non-lacunary sequences
which can be realized as rigidity sequences for weakly mixing transformations.
(ii) As the above shows {qn : n ≥ 1} is always a Sidon set (see [19],[36]); indeed,
{qn : n ≥ 1} = {q2n : n ≥ 1} ∪ {q2n+1 : n ≥ 0}.
It follows that the set of denominators is the union of two lacunary sets and hence is a
Sidon set ([19],[36]).
28
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
(iii) The assertion of Theorem 3.20 is true for functions whose Fourier transform is
concentrated on a Sidon set and when T is an arbitrary ergodic rotation on a compact
metric Abelian group (by the proof of the main result in [14] or by [23]).
(iv) It follows that to prove Proposition 3.21 we could have used all denominators, with
(for example) aqn = √1n kqn αk.
There are also more complicated constructions showing that for each irrational α there
is an absolutely continuous zero-mean ϕ : T → R which is not a coboundary – see [30].
We can then use such cocycles and (3.17) for another proof of the above corollary.
Corollary 3.25. There exists a weakly mixing transformation R such that Rfn → Id,
where (fn ) denotes the Fibonacci sequence.
Remark 3.26. It is not clear how to characterize rigidity sequences that cannot be IP
rigidity sequences. See Proposition 3.3, and its generalization Proposition 3.31 for examples of this phenomenon and. In reference to the above, it would be interesting to show
that F S((qn )) is NOT a rigidity net. When α is just the golden number this leads to the
following problem: Show that for a weakly mixing transformation R it is impossible to
have
for each ε > 0 there exists i0 such that for each i ≥ i0 and each m ≥ 1
d(Rmqi +k(m)qi−1 , Id) < ε
where k(m) = qi1 −1 + . . . + qis −1 whenever m = qi1 + . . . + qis (Ostrovski’s decomposition).
Remark 3.27. The above will be used to answer positively the following question: Given
an increasing sequence (nm ) of integers is there a weakly mixing transformation R such
that Rnmk → Id for some subsequence (nmk ) of (nm )?
In fact, we have already answered this question (see Proposition 2.2) but we will show
now a different approach. We start with the following. This is well-known, see e.g. [25].
Lemma 3.28. Given an increasing sequence (nm ) of natural numbers, consider the set
of α ∈ [0, 1) such that a subsequence of (nm ) is a subsequence of denominators of α. This
is a residual subset of [0, 1).
Now, given (nm ) choose any irrational α so that for some subsequence (nmk ) we have
all numbers nmk being denominators of α. Then use previous arguments to weakly mixing
“realization” of the whole sequence of denominators of α.
Remark 3.29. It was observed already above, but it is worth repeating here, that not
every lacunary sequence is a sequence of rigidity. Indeed, r2n = 22n , r2n+1 = 22n+1 + 1
cannot be a sequence of rigidity for an ergodic and aperiodic transformation.
3.5. Integer lacunarity case. We will now give an alternative proof of Proposition 3.2
using the cocycle methods that have been developed here. Assume that (nm )m≥0 is an
increasing sequence of positive integers such that n0 = 1, nm+1 /nm ∈ Z with
ρm := nm+1 /nm ≥ 2 for m ≥ 0.
(3.23)
Notice that in view of (3.23) there exists a constant C > 0 such that
(3.24)
1
n2m+1
+
1
n2m+2
+ ... ≤
C
for each m ≥ 0.
n2m
RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
29
Let X = Π∞
m=1 {0, 1, . . . , ρm − 1} which is a metrisable compact group when we consider the product topology and the addition is meant coordinatewise with carrying the
remainder to the right. On X we consider Haar measure pX which is the usual product
measure of uniform measures. Define T x = x + 1̂, where
1̂ = (1, 0, 0, . . .).
The resulting dynamical system is called the (nm )-odometer.
For each t ≥ 0 set
D0nm = {x ∈ X : x0 = x1 = . . . = xm−1 = 0}.
Note that {D0nm , T D0nm , . . . , T nm −1 D0nm } is a Rokhlin tower fulfilling the whole space X
and
1̂ ∈ T D0nm for each t ≥ 0.
(3.25)
b of X is discrete and is isomorphic to the (discrete) group of
The character group X
roots of unity of degree nm , m ≥ 0. More precisely, for m ≥ 0 set
χn, (x) = εjnm := e2πij/nm for x ∈ T j D0nm , j = 0, 1, . . . , nm − 1.
b = {χj : j = 0, 1, . . . , nm − 1, m ≥ 0}.
Then X
nm
From now on we will consider f ∈ L2,0 (X, pX ) whose Fourier transform is “concentrated” on {χnm : m ≥ 0}. We have
(3.26)
∞
X
f (x) =
anm χnm (x),
∞
X
|anm |2 < +∞.
m=1
m=1
A) Small divisors. Assume that f satisfies (3.26) and suppose that
f (x) = g(x) − g(x + 1̂) for pX -a.e. x ∈ X.
P
Suppose moreover that g ∈ L2 (X, pX ). Hence g(x) = χ∈Xb bχ χ(x) and by comparison of
Fourier coefficients on both sides in (3.26) we obtain
(3.27)
bχ = 0 whenever χ 6= χnm and anm = bnm (1 − χnm (1̂)) for m ≥ 1.
Using (3.25) we obtain that bnm =
|bnm |2 =
anm
,
1−εnm
so
|anm |2
= n2m |anm |2 for m ≥ 1.
|1 − εnm |2
We have proved the following
(3.27) has an L2 -solution if and only if (nm anm )m ∈ l2 .
(3.28)
B) Estimate of L2 -norms for the cocycle. Assume that n ∈ N then
f
(n)
(x) =
∞
X
anm (1 + χnm (1̂) + . . . + χnm ((n − 1)1̂))χnm (x)
m=1
=
∞
X
m=1
anm
n−1
X
j=0
!
εjnm
χnm (x).
30
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
Fix n = nm0 . We then have


nm0 −1
m
0
X
X
εjnm  χnm (x) +
anm 
f (nm0 ) (x) =
m=1
∞
X
anm 
X

εjnm  χnm (x)
j=0
n
0
1 − εnm
m
anm
χn (x).
1 − εnm m
+1
=
(3.29)
nm0 −1
m=m0 +1
j=0
Since |1 − εnm | = 1/nm and |1

∞
X
m=m0
n 0
− εnm
m |
≤ nm0 /nt ,
kf (nm0 ) k2L2 (X,pX ) ≤ n2m0
∞
X
|anm |2 .
m=m0 +1
C) Sidon sets and a “good” function. According to [36] (see Example 5.7.6 therein)
every infinite subset of a discrete group contains an infinite Sidon set. Hence we can choose
a subsequence (nmk ) of (nm ) so that
(3.30)
b
{χnmk : k ≥ 1} is a Sidon subset of X.
We set
(3.31)
f (x) =
∞
X
k=1
1
√
χnmk (x).
knmk
Suppose now that (3.27) has a measurable solution g (we should consider f real valued,
so in fact we should consider f + f below). In view of (3.30) and [23], g ∈ L2 (X, pX ). But
∞
X1
X
|nm anm |2 =
,
k
m=1
k≥1
whence by (3.28), we cannot obtain an L2 -solution. This means that f is not a measurable
coboundary. According to (3.29), the definition of f and (3.23) for each s ≥ 1
∞
∞
∞
X
X
X
kf (ns ) k2L2 (X,pX ) ≤ n2s
|anm |2 = n2s
|anmk |2 = n2s
|anmk |2
m=s+1
≤
k≥1: nmk ≥ns+1
k=ks
∞
n2s X 1
C
≤
→0
2
ks j=s+1 nj
ks
as clearly ks → ∞ when s → ∞.
Using our general method we hence proved the following.
Proposition 3.30. Assume that (nm ) is an increasing sequence of integers with nm+1 /nm
being an integer at least 2. Then there exists a weakly mixing transformation T such that
T nm → Id.
We will now discuss the problem of IP -rigidity along (nm ).PFirst of all notice that (nm )
is a sequence of IP rigidity for the (nm )-odometer (indeed, ∞
m=1 |χnm (1̂) − 1| < +∞).
nm
Let us also notice that if R is weakly mixing and R
→ Id then by passing to a
subsequence, we will get IP − Rnmk → Id. But if we then set mk = nmk then mk
divides mk+1 and (mk ) is a sequence of IP -rigidity for R. It means that if the sequence
(nm+1 /nm ) is unbounded then, at least in some cases, it is a sequence of IP -rigidity for a
RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
31
weakly mixing transformation. On the other hand we have already seen (Corollary 3.3)
that when ρt = a, t ≥ 1 then IP -rigidity does not take place. The proposition below
generalizes that result and shows that in the bounded case an IP -rigidity is excluded.
Proposition 3.31. Assume that nt+1 /nt ∈ N \ {0, 1}, t ≥ 0, and supt≥0 nt+1 /nt =: C <
+∞. Then for any weakly mixing transformation R for which Rnt → Id, the sequence
(nt ) is not a sequence of IP -rigidity.
Proof. Recall that T x = x + 1̂ where X stands for the (nt )-odometer. Each natural
number r ≥ 1 can be expressed in a unique manner as
N
X
r=
at nt , 0 ≤ at < ρt = nt+1 /nt .
t=0
Assume that T rm → IdX . Write
rm =
Nm
X
(m)
(m)
at n t , 0 ≤ at
< ρt
t=0
and set km = max{t ≥ 0 :
(m)
a0
(m)
= . . . = at
= 0} (so rm =
PNm
t=km
(m)
at nt ). We claim that
km → ∞ whenever m → ∞.
(3.32)
Indeed, suppose that the claim does not hold. Then without loss of generality we can
assume that there exists t0 ≥ 0 such that km = t0 for all m ≥ 1, that is
rm =
(m)
at0 nt0
+
Nm
X
(m)
(m)
at nt with 1 ≤ at0 < ρt0 .
t=t0 +1
{D0t0 +1 , . . . , Dnt0t+1
}
0 +1 −1
int0 +j
Consider the tower
and j ≥ 1 we have T
and let A = D0t0 +1 . Notice that for each i ≥ 0
A = A. It follows that
P Nm
(m)
(m)
T rm A = T at0 nt0 T t=t0 +1 at nt (A)
(m)
= T at0
nt0
(A) ∈ {D1t0 +1 , . . . , Dnt0t+1
},
0 +1 −1
(m)
where the latter follows from the fact that 1 ≤ at0 < ρt0 . So pX (T rm (A)4A) = 2/nt0 +1
and hence (rm ) is not a rigidity sequence for T , a contradiction. Thus (3.32) has been
shown.
Assume now that R is a weakly mixing transformation for which (nt ) is its IP -rigidity
sequence. It follows that we have a convergence along the net
(3.33)
R
PN
t=k
ηt nt
→ Id whenever k → ∞ and ηk = 1, 0 ≤ ηt ≤ 1, t ≥ k + 1.
We claim that also
(3.34) R
PN
t=k
at nt
→ Id whenever k → ∞ and 1 ≤ ak ≤ ρk −1, 0 ≤ at ≤ ρt −1, t ≥ k+1.
PN
Indeed, we write R t=k at nt as the
composition of at most S1 ◦. . .◦SD with D ≤ C = maxt ρt
PN
δt nt
t=k
automorphisms of the form R
with δt ∈ {0, 1} (to define the first automorphism
S1 we put δt = 1 as soon as at ≥ 1 and δt = 0 elsewhere, for the second automorphism
S2 we put δt = 1 as soon as at ≥ 2 and δt = 0 elsewhere, etc.). Notice that for each
i = 1, . . . , D
PN
Si = R t=ki nt with ki ≥ k.
32
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
PN
Therefore, if we assume that in (3.33), kR t=k nt k < ε for k ≥ K then kS1 ◦. . .◦SD k < Dε
(see Remark 2.4) and thus (3.34) follows.
Combining (3.34) and (3.32) we see that each rigidity sequence for T is also a rigidity
sequence for R. This is however contradictory to Corollary 3.14 (or rather to its proof).
Question. If (ρt ) is bounded, but not always a whole number, can it still (nt ) be a
sequence of IP -rigidity for some weakly mixing transformation? How fast does (ρt ) have
to grow for (nt ) to be a sequence of IP -rigidity for some weakly mixing transformation?
4. Non-recurrence
In the previous sections, we have seen that the characterization of which sequences (nk )
exhibit rigidity for some ergodic, or more specifically, weakly mixing dynamical system
will most certainly be difficult. The only aspect that is clear is that these sequences must
have density zero because their gaps tend to infinity. Lacunary sequences are always
candidates for consideration in such a situation. However, we have also observed that
more than lacunarity itself is needed to have rigidity in an ergodic dynamical system.
In a similar vein, we would like to characterize which increasing sequences (nm ) in Z+
have the property that for some ergodic system (X, β, m, T ) and such set A of positive
measure, the sets T nm A are disjoint from A for all m ≥ 1. That is, recurrence fails along
the sequence of powers T nm . For instance, is it sufficient that the sequence be lacunary
for this to happen for a weakly mixing transformation?
Remark 4.1. It is not hard to see that lacunary sequences do fail recurrence for some
ergodic dynamical system. Indeed, this happens even with ergodic rotations of T. See
Pollington [34] and de Mathan [6], and the later article in the same direction by Furstenberg [12]. They show that for any lacunary sequence (nm ) there is some γ ∈ T of infinite
order, and some δ > 0, such |γ nm − 1| ≥ δ for all m ≥ 1. The arguments there also
give information about the size of the set of rotations that work for a given lacunary
sequence. The constructions in these articles are made more difficult, as with a number
of other results about lacunary sequences, by not knowing the degree or nature of lacunarity. But if one just wants some ergodic dynamical system to exhibit non-recurrence,
then the construction is easier. Indeed, suppose (nm ) is lacunary, say nnk+1
≥ λ > 1 for
k
all m ≥ 1. Depending only on λ, we can choose K so that the subsequences (pm,j ) given
p
by pm,j = nj+Km , j = 0, . . . , K − 1 each have lacunary constant inf m+1,j
≥ 5. Then
pm,j
m≥1
for each j, a standard argument shows that there is a closed perfect set Cj such that for
1
all γ ∈ Cj , we have |γ pm,j − 1| ≥ 100
for all m ≥ 1. We can choose (γ0 , . . . , γK−1 ) with
each γj ∈ Cj and such that γ0 , . . . , γK−1 are independent. Then we would know that the
transformation T of the K-torus TK given by T (α0 , . . . , αK−1 ) = (γ0 α0 , . . . , γK−1 αK−1 )
is ergodic Also, for a sufficiently small , the arc of radius I around 1 in T will give a
set C = I × · · · × I ⊂ T K such that for all nm , we have T nm C and C disjoint. Indeed,
each nm is some pm,j and so T nm C and C are disjoint because in the jth coordinate T nm
p
corresponds to the rotation γj m,j I which is disjoint from I.
We can see that the non-recurrence phenomenon is both pervasive and not, depending
on how one chooses the quantifiers. As usual, the measure-preserving transformations of
a non-atomic separable probability space (X, β, m) can be given the weak topology, and
RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
33
become a complete pseudo-metric group G in this topology. By a generic transformation,
we mean the elements in some dense Gδ -set in G.
Proposition 4.2. The generic transformation is both weakly mixing and rigid, and moreover is not recurrent along some increasing sequence (ps ) in Z+ .
Proof. It is well-known that the generic transformation is weakly mixing and rigid. So for
some (nm ) one has kf ◦T nm −f k2 → 0 for all f ∈ L2 (X). But then it follows that for any set
A, lim m(T nm A∆A) = 0. Since T is ergodic, we can choose a set A with m(A) > 0 and
m→∞
T A and A disjoint. Let B = T A. We have B and A disjoint, and lim m(T nm −1 B∆A) =
m→∞
∞
P
1
0. Hence, we can pass to a subsequence (ms ) so that
m(T nms −1 B∆A) ≤ 100
m(A) =
s=1
1
m(B).
100
So
C = B\(
∞
[
T −(nms −1) (T nms −1 B∆A)
s=1
will have m(C) > 0. Also, C ⊂ B = T A is disjoint from A. But at the same time
T nms −1 C ⊂ T nms −1 B\(T nms −1 B∆A) for all s ≥ 1, and so T nms −1 C ⊂ A for all s ≥ 1.
Thus, for the generic transformation is weakly mixing, and for some sequence of powers
T ps and some set C of positive measure, we have non-recurrence because C is disjoint
from all T ps C.
Remark 4.3. In the result above, if one restricts (ps ) to be the same for the whole class,
then one has only a meager set of transformations.
We do have some specific, interesting examples of the failure of recurrence for a weakly
mixing dynamical system.
Proposition 4.4. The Chacon transformation is not recurrent for times
3m+1 −1
.
2
Proof. The transformation T that we are using here is constructed inductively as follows.
Take the current stack (single tower) Tm of interval and cut it in thirds Tm,1 , Tm,2 , and
Tm,3 . Add a spacer s of the size of levels above the middle third Tm,2 , and let the new
stack Tm+1 consist of Tm,1 , Tm,2 , s, and Tm,3 in that order from bottom to top. To see
the failure of recurrence, use the standard symbolic dynamics for T i.e. assign the symbol
1 to all the spacer levels and 0 to the rest of the levels. Then let Bm be the name of
length nm of a point in the base of the mth tower Tm . Then B0 = 0, B1 = 0010, and
Bm+1 = Bm Bm 1 Bm in general. It is a routine check that when one shifts Bk by nm for
k larger than m, and compares this with Bk , then they have no common occurrences of
1. So if A is the first added spacer level, then we have T nm A and A disjoint for all m.
m+1
The value of nm is 3 2 −1 .
Remark 4.5. Friedman and King [11] consider a class of weakly mixing, but not strongly
mixing, transformations constructed by Chacon; they prove that these, unlike the Chacon
transformation above, are lightly mixing and so there is always recurrence for these transformations along any increasing sequence. Hence, these transformations form a meager
set by the category result above. The obvious question here is what non-recurrence can
be shown for sequences via classical cutting and stacking? Is there an algebraic restriction
to these constructions as there is an algebraic restriction for rigidity?
34
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
If we have a sufficient growth rate assumed for (nm ), we can give a construction of
a weakly mixing transformation which exhibits non recurrence along the sequence. The
argument here starts like the constructions in Section 3.1
Proposition 4.6. Suppose we have an increasing sequence (nm ) such that
∞
P
nm /nm+1 <
k=1
∞. Then there is a weakly mixing transformation T for which (nm ) is an IP rigidity
sequence and such that T is not recurrent for (nm − 1).
Proof. Choose a non-decreasing sequence (hm ) such that
1/hm
nm /nm+1
→ ∞ and
∞
P
1/hm <
m=1
∞. Actually, we only need that 1/hm ≥ 10(nm /nm+1 ) for sufficiently large m, say m ≥ M .
Then we construct a Cantor set C with constituent intervals at each level that are arcs
of size 1/(nm hm ) around some of the nm -th roots of unity (determined as part of the
induction). Our conditions allow us to find in each such constituent interval many points
j/nm+1 because 1/nm+1 is sufficiently smaller than 1/nm hm , and then select in these
constituent intervals new ones of length 1/(nm+1 hm+1 ) around some of the nm+1 -th roots
of unity for m ≥ M . The resulting Cantor set C has the property that for all points
x in the set nm x is within 1/hm of an integer for m ≥ M . It follows that if we take a
continuous, positive measure νo on C, with νo ([0, 1)) = 1, then |1 − νbo (nm )| ≤ 1/hm for
m ≥ M.
Now we take the GMC construction corresponding to νo . This gives us a weakly mixing
dynamical system (X, β, p, T ) and a function f of norm one in L2 (X) such that kf ◦T nm −
f k22 ≤ 2/hm for m ≥ M . This function is not constant of course. We may assume it is
real valued by taking either its real part or its imaginary part, and rescaling. Then we
have a non-constant, real-valued function of norm one such that kf ◦ T nm − f k22 ≤ 4/hm
for m ≥ M .
Now we claim that both the positive part f + and the negative part f − of f satisfies
the same inequality. To see this, Write
Z
Z
nm
2
|f ◦ T − f | dp = |f + ◦ T nm − f − ◦ T nm − f + + f − |2 dp.
Expand this into the sixteen terms involved. Use that fact that the terms f + f − and
(f + ◦ T nm )(f − ◦ T nm ) are zero, and regroup terms to see that
Z
Z
nm
2
|f ◦ T − f | dp =
|f + ◦ T nm − f + |2 dp
Z
+
|f − ◦ T nm − f − |2 dp
Z
+
2(f + ◦ T nm )f − + 2(f − ◦ T nm )f + dp.
Because 2(f + ◦ T nm )f − + 2(f − ◦ T nm )f + is positive, we have
Z
Z
nm
2
|f ◦ T − f | dp ≥ |f + ◦ T nm − f + |2 dp
and
Z
|f ◦ T
nm
2
− f | dp ≥
Z
|f − ◦ T nm − f − |2 dp.
RIGIDITY AND NON-RECURRENCE ALONG SEQUENCES
35
In addition, the same argument above shows that for every constant L, the function
(f − L)+ also would satisfy this last estimate too. Hence, taking L = 1/2, we would have
F = f 1{f ≥L} satisfies this inequality too and not being the zero function. But let A =
{f ≥ L}. On T nk A\A, we would have |F ◦ T nm − F |2 ≥ 1/4. So m(T nm A\A) ≤ 64/hm .
But similarly, on A\T nm A, we would have |F ◦T nm −F |2 ≥ 1/4. So m(A\T nm A) ≤ 64/hm .
The result is that from our original GMC construction, we can infer the existence of a
proper set A of positive measure, which depends on the original function f and not on
∞
P
m, such that m(T nm A∆A) ≤ 64/hm for all m ≥ M . Hence,
m(T nm A∆A) < ∞.
m=1
We do have to also make certain that the set A here is a proper set i.e. m(A) < 1. But
since f is not constant, either both f + and f − are non-trivial, in which case the value of
L above will do, or we are in the situation where without loss of generality our function
f > 0 a.e. Then we can still choose an appropriate value of L, perhaps somewhat larger,
so that A is a proper set of positive measure.
∞
P
Now we can use the convergence of
m(T nm A∆A) to construct a set of positive
m=1
measure B for which T nm −1 B and B are disjoint for all m. The argument is a variation
on the one given in Proposition 4.2. There is the issue that T A and A are not necessarily
disjoint. But there is some subset Ao of A of positive measure such that T Ao and A are
∞
S
disjoint. So if we take B = T Ao \
T −(nm −1) (T nm A∆A), for suitably large M , then we
m=M
would have m(B) > 0, B ⊂ T Ao \A, and T nm −1 B ⊂ A for all m ≥ M . Hence T nm −1 B
and B disjoint for all m ≥ M .
Now we need to revise the result above so that we get the same disjointness for all
m. But here we know that T is weakly mixing, and consequently all of its powers are
ergodic. So one can inductively revise B as follows. One takes a subset B1 of B such that
T n1 −1 B1 and B1 are disjoint, then one takes a subset B2 of B1 such that T n2 −1 B2 and B2
are disjoint, and so on. After a finite number of steps one ends up with a subset BM −1 of
B such that T nm −1 BM −1 and BM −1 are disjoint for all m ≥ 1. Now, with BM −1 replacing
B, we have T nm −1 B and B disjoint for all m ≥ 1. So this construction gives a weakly
mixing transformation that is not only IP rigid along (nm ), but such that along (nm − 1)
it is not recurrent.
Remark 4.7. a) In the above, replacing our original sequence by (nm + 1), allows us to
∞
P
conclude this fact: whenever (nm ) is increasing and
nm /nm+1 < ∞, there exists a
m=1
weakly mixing transformation T and a set of positive measure B such that T nm B and B
are disjoint for all m. Of course now the transformation is IP rigid along (nm + 1).
b) The condition we are using of course will NOT hold for lacunary sequences like nm =
2m . But it is the case that if (km ) is lacunary, and δ > 1, then the subsequence (nm ) =
(kbmδ c ) will have our series property. So speeding up the exponent for a lacunary sequence
slightly will give us the type of non-recurrence that we want.
Remark 4.8. What can one say about having
∞
P
m(T nm A∆A) convergent for all A?
m=1
This is not so obviously impossible, although it probably is. But if one replaces this
by the corresponding functional version, which may be more general, than there would
36
V. BERGELSON, A. DEL JUNCO, M. LEMANCZYK, J. ROSENBLATT
be a problem. First, one can see that
∞
P
m(T nm A∆A) =
m=1
∞
P
kf ◦ T nm − f k22 if we
m=1
take f = 1A . So the question is, can we have a dynamical system in which the square
1/2
∞
P
2
nm
is always an L2 -function? But it is clear that
function Sf =
|f ◦ T − f |
m=1
this is not possible. If it were, then one can show there is a homogeneous inequality
kSf k2 ≤ Kkf k2 for some constant K. Then one takes again f = 1A , and sees that
∞
P
p(T nm A∆A) ≤ K 2 p(A). But this cannot be. Indeed, just take a very long Rokhlin
m=1
tower with A as the base and the left-hand side could exceed the right-hand side.
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V. Bergelson, Department of Mathematics
Ohio State University, Columbus, OH 43210, USA
E-mail: [email protected]
A. del Junco, Department of Mathematics
University of Toronto, Toronto, M5S 3G3, Canada
E-mail: [email protected]
M. Lemańczyk, Faculty of Mathematics and Computer Science
Nicolaus Copernicus University, Toruń, Poland, and
Institute of Mathematics, Polish Academy of Sciences, Warsaw, Poland
E-mail: [email protected]
J Rosenblatt, Department of Mathematics
University of Illinois at Urbana-Champaign, Urbana, IL 61801, USA
E-mail: [email protected]