1. Introduction
We study the question of whether knowing the exact times a point in an unknown dynamical system enters an open set is enough to determine the system. In this paper, a topological dynamical system is a pair
$(X,T)$
, where X is a compact metric space with metric d (all metrics will be referred to as d even for different spaces if no ambiguity arises) and
$T:X \to X$
is a homeomorphism. Recall that
$(X,T)$
is minimal if for all
$x_0 \in X$
the orbit
$T^{\mathbb {Z}}x_0 = \{T^n x_0 ~|~ n \in \mathbb {Z}\}$
is dense and is transitive if some orbit is dense. If
$x_0 \in X$
and
$U \subset X$
is an open set, then we can define the set of return times
$$ \begin{align*}\mathcal{R}_{(X,T)}(x_0, U) = \{ n \in \mathbb{Z} ~|~ T^nx_0 \in U \}.\end{align*} $$
Our question may now be stated as follows: given two minimal topological dynamical systems
$(X_1, T_1)$
and
$(X_2, T_2)$
, with
$x_1 \in X_1, x_2 \in X_2$
and
$U_1 \subset X_1, U_2 \subset X_2$
open such that
$\mathcal {R}_{(X_1,T_1)}(x_1, U_1) = \mathcal {R}_{(X_2,T_2)}(x_2, U_2)$
, is it necessarily true that
$(X_1, T_1)$
and
$(X_2, T_2)$
are isomorphic? If so, is there an isomorphism mapping
$x_1$
to
$x_2$
?
Recall that an isomorphism
$\phi :(X_1, T_1) \to (X_2, T_2)$
is a homeomorphism
$\phi :X_1 \to X_2$
such that
$\phi \circ T_1 = T_2 \circ \phi $
. One can contrast this to Taken’s reconstruction theorem [Reference Takens, Rand and Young6] as well as recent developments [Reference Gutman and Assani2], where one instead is given a system
$(X,T)$
where the dimension of X is d and asks whether the delay observation mapping
$X \to [0,1]^{2d+1}$
given by
$x \mapsto (h(x), h(Tx), \ldots , h(T^{2d+1}x))$
is injective for some generic continuous
$h:X \to [0,1]$
. Our question can instead be posed as asking to what extent the mapping
is injective up to isomorphism.
1.1. Reconstructing Kronecker systems
Let us now start with some motivating examples where no such isomorphism exists.
Example 1.1. Let
$X_1 = X_2 = \mathbb {T} = \mathbb {R}/\mathbb {Z}$
and suppose
$\alpha \in \mathbb {T}$
is irrational. Let
$$ \begin{align*}U_1=U_2 = (-\epsilon, \epsilon) \cup \big(-\epsilon + \tfrac{1}{2}, \epsilon + \tfrac{1}{2}\big) \subset \mathbb{T}.\end{align*} $$
Then for each
$n \in \mathbb {Z}$
, we have that
$n\alpha \in U_1$
if and only if
$n(\alpha + \tfrac {1}{2}) \in U_2$
. However, the minimal systems
$(\mathbb {T}, x \mapsto x+\alpha )$
and
$(\mathbb {T}, x \mapsto x + \alpha + \tfrac {1}{2})$
are not isomorphic.
Example 1.2. Let
$X_1 = \mathbb {T}$
,
$X_2 = \mathbb {T}^2$
,
$\alpha _1 = \sqrt {2} \in \mathbb {T}$
and
$\alpha _2 = (\sqrt {2}, \sqrt {3}) \in \mathbb {T}^2$
. Let
$U_1 \subset \mathbb {T}$
be any non-empty proper open subset and let
$U_2 = U_1 \times \mathbb {T}$
. Then clearly
$n \alpha _1 \in U_1$
if and only if
$n\alpha _2 \in U_2$
. Yet
$X_1$
and
$X_2$
are not even homeomorphic.
These two examples highlight that open sets with non-trivial translational symmetries are a source of issues. For a compact abelian metrizable group
$(K, +)$
and
$A \subset K$
, we can define the stabilizer
$$ \begin{align*}\operatorname{Stab}_K(A) = \{k \in K ~|~ A + k = A \}.\end{align*} $$
Note that in our examples above, the stabilizers are not trivial; the stabilizer of
$(-\epsilon , \epsilon ) \cup (-\epsilon + \tfrac {1}{2}, \epsilon + \tfrac {1}{2}) $
is
$\{0, \tfrac {1}{2}\}$
while the stabilizer of
$U_1 \times \mathbb {T}$
, for
$U_1 \subset \mathbb {T}$
non-empty and proper, is the vertical subgroup
$\{0\} \times \mathbb {T}$
.
Our first main result demonstrates that the only way two non-isomorphic Kronecker systems can yield the same set of return times is if the closure of one of the defining open sets has a non-trivial stabilizer.
Theorem 1.3. Let
$(K_1, +)$
and
$(K_2, +)$
be two compact metrizable abelian groups and suppose that
$\alpha _1 \in K_1$
and
$\alpha _2 \in K_2$
are such that
$\overline {\mathbb {Z}\alpha _i} = K_i$
for
$i=1,2$
. Suppose that
$U_1 \subset K_1$
and
$U_2 \subset K_2$
are open sets such that the stabilizers of their closures are trivial, that is,
$\operatorname {Stab}_{K_i}(\overline {U_i}) = \{0\}$
. Then if
$$ \begin{align*}\{n \in \mathbb{Z} ~|~ n\alpha_1 \in U_1\} = \{n \in \mathbb{Z} ~|~ n\alpha_2 \in U_2\},\end{align*} $$
then there exists an isomorphism of topological groups (so continuous)
$K_1 \to K_2$
mapping
$\alpha _1$
to
$\alpha _2$
.
Note that if one removes a point not on the trajectory
$\{n\alpha _i ~|~ n \in \mathbb {Z}\}$
from
$U_i$
, then
$U_i$
remains open, and the return times and the closures do not change, so it is indeed natural to study the stabilizer of the closure
$\overline {U_i}$
rather than
$U_i$
.
1.2. Detecting Kronecker factors of minimal systems
We now turn our attention to minimal topological systems. As demonstrated in Example 1.2, if a minimal system has a non-trivial factor, then it shares a set of return times with it. Of course, this is also the case if one replaces in that example the rotation
$(x,y) \mapsto (x+\sqrt {2},y+\sqrt {3})$
with an ergodic skew product such as
$\mathbb {T}^2 \to \mathbb {T}^2: (x,y) \mapsto (x+\sqrt {2}, x+y)$
. Our next result shows that essentially the only way a minimal system can share a set of return times with a Kronecker system is if the latter is a factor of the former.
Theorem 1.4. Let
$(X,T)$
be a minimal topological dynamical system and
$U \subset X$
be a non-empty open set and
$x_0 \in X$
. Let
$(K, +)$
be a compact metrizable abelian group and
$\alpha \in K$
such that
$\overline {\mathbb {Z} \alpha } = K$
and let
$U' \subset K$
be a non-empty open set. Suppose that
$$ \begin{align} \{ n \in \mathbb{Z} ~|~ T^nx_0 \in U \} = \{n \in \mathbb{Z} ~|~ n\alpha \in U'\} \end{align} $$
and that
$\operatorname {Stab}_K(\overline {U'}) = \{ 0 \}$
. Then the pointed Kronecker system
$(K,0 , k \mapsto k+\alpha )$
is a factor of the pointed system
$(X, x_0, T)$
, that is, there is a continuous map
$\phi : X \to K$
with
$\phi (x_0) = 0$
satisfying
$$ \begin{align*}\phi(Tx) = \phi(x) + \alpha\quad \text{ for all } x \in X.\end{align*} $$
Note that Theorem 1.4 immediately implies Theorem 1.3 (apply twice to both Kronecker systems to obtain two factor maps that are inverses), since an isomorphism of Kronecker systems that preserves the zero elements is necessarily a group isomorphism. The following example shows that we cannot replace the minimality assumption with transitivity.
Example 1.5. (Insufficient to assume transitivity instead of minimality) Let
$X \subset [-1, 1]$
be the set
$X = \{ x_n ~|~ n \in \mathbb {Z} \cup \{-\infty , \infty \} \}$
, where
$x_{\pm \infty } = \pm 1$
and for
$n\in \mathbb {Z}$
,
$$ \begin{align*}x_n = \begin{cases} -1 + \dfrac{1}{|n|}, & n < 0, \\ 0, & n = 0, \\ 1 - \dfrac{1}{n}, & n>0. \end{cases}\end{align*} $$
Thus X is closed and all
$x_n$
with
$n \in \mathbb {Z}$
are isolated points. So we have a homeomorphism
$T:X \to X$
mapping
$x_n$
to
$x_{n+1}$
and fixing
$x_{\pm \infty }$
. Hence, it is transitive (all
$x_n$
with
$n \in \mathbb {Z}$
have dense orbits) but not minimal. Now
$U = \{x_n ~|~ n \in 2\mathbb {Z}\}$
is open and the return times set
$2\mathbb {Z} = \{ n \in \mathbb {Z} ~|~ T^n x_0 \in U\}$
is also equal to
$\{n \in \mathbb {Z} ~|~ n\alpha \in U'\}$
, where
$U' = \{0\} \subset K = \mathbb {Z}/2\mathbb {Z}$
is open and
$\alpha = 1 \in K$
. However, there is no factor
$\pi $
from
$(X,T)$
to the Kronecker system
$(K, k \mapsto k+1)$
as it would yield the contradiction
$\pi (x_{\infty }) = \pi (Tx_{\infty }) = \pi (x_{\infty }) + 1$
.
1.3. Return times along a polynomial sequence
We recall the following result which follows immediately from the polynomial Weyl equidistribution [Reference Weyl7] and the fact that connected compact abelian groups have no non-trivial characters with finite image.
Proposition 1.6. Let K be a connected compact abelian group and suppose that
$P(x) \in \mathbb {Z}[x]$
is a non-constant polynomial. If
$\alpha \in K$
is such that
$\overline {\mathbb {Z} \alpha } = K$
, then
$P(\mathbb {Z})\alpha = \{P(n)\alpha ~|~ n \in \mathbb {Z}\}$
is dense in K.
Given this, it is natural to ask whether we can extend Theorem 1.3 to the case where we only know the set of return times inside a polynomial sequence, that is, the set
$\{ n ~|~ P(n)\alpha \in U\}$
. The next result shows that this is indeed the case.
Theorem 1.7. Let
$K_1$
and
$K_2$
be compact connected abelian groups, and suppose that
$\alpha _i \in K_i$
with
$\overline {\mathbb {Z}\alpha _i} = K_i$
. Let
$P(x) \in \mathbb {Z}[x]$
be a polynomial with
$P(0) = 0$
. Let
$U_i \subset K_i$
be non-empty open sets and let
$\mathcal {R}_i = \{n \in \mathbb {Z} ~|~ P(n)\alpha _i \in U_i\}$
. If
$\mathcal {R}_1 = \mathcal {R}_2$
and
$\operatorname {Stab}_{K_1}(\overline {U_1}) = \{0 \}$
and
$\operatorname {Stab}_{K_2}(\overline {U_2}) = \{0 \}$
, then there exists an isomorphism
$K_1 \to K_2$
(of topological groups, so continuous) mapping
$P(n)\alpha _1$
to
$P(n)\alpha _2$
for each
$n \in \mathbb {Z}$
. In particular, if
$P(\mathbb {Z})$
is not contained in any proper subgroup of
$\mathbb {Z}$
, then this isomorphism maps
$\alpha _1$
to
$\alpha _2$
.
Example 1.8. Consider the polynomial
$P(n) = n^5-n$
and let
$\alpha _1 = \sqrt {2} \in \mathbb {T}$
and
$\alpha _2 = \sqrt {2} + \tfrac {1}{5} \in \mathbb {T}$
. Then for any open set
$U \subset \mathbb {T}$
, we have that
$\{n \in \mathbb {Z} ~|~ P(n)\alpha _1 \in U\} = \{n \in \mathbb {Z} ~|~ P(n)\alpha _2 \in U\} $
and indeed there is an isomorphism (the identity map) mapping
$P(n)\alpha _1 $
to
$P(n)\alpha _2$
as they are equal, but there is no isomorphism mapping
$\alpha _1$
to
$\alpha _2$
.
We remark that it seems that there is no obvious way to extend Theorem 1.4 to polynomial times under the assumption that X is connected, since an example of Pavlov [Reference Pavlov4] shows that there exists a minimal connected system
$(X,T)$
such that
$\{T^{n^2} x_0 ~|~ n \in \mathbb {Z}\}$
is not dense for some
$x_0 \in X$
.
1.4. Return times sets of nilmanifolds
It is natural to try to extend Theorem 1.3 on the return times in Kronecker systems to other homogeneous spaces such as nilmanifolds. The following is a partial result on this, which establishes that if a Kronecker system
$(K,+)$
and a minimal nilsystem
$(G/\Gamma , T)$
share the same return times, with the closures of the defining open sets having trivial stabilizers under translations of K and G, respectively, then they must be isomorphic as dynamical systems.
Theorem 1.9. Let
$X=G/\Gamma $
be a nilmanifold, where G is a nilpotent Lie group and
$\Gamma \leq G$
is a cocompact discrete subgroup. Suppose that
$\tau \in G$
is such that
$(X,T)$
is minimal, where
$T:X \to X$
is given by
$Tx = \tau x$
for
$x \in X$
. Suppose that
$U \subset X$
is open and satisfies that
$\operatorname {Stab}_G(\overline {U}):= \{g \in G ~|~ g\overline {U} = \overline {U}\}$
consists of only those
$g \in G$
such that
$gx = x$
for all
$x \in X$
. Now suppose that
$(K,+)$
is a compact abelian group and
$\alpha \in K$
is such that
$\overline {\mathbb {Z} \alpha } = K$
, and
$V \subset K$
is open such that
$\operatorname {Stab}_K(\overline {V}) = \{0\}$
and
$$ \begin{align*}\{ n \in \mathbb{Z} ~|~ T^nx_0 \in U\} = \{n \in \mathbb{Z} ~|~ n\alpha \in V\}\end{align*} $$
for some
$x_0 \in X$
. Then the pointed system
$(X, x_0, T)$
is isomorphic to the pointed Kronecker system
$(K,0 , k \mapsto k+\alpha )$
, that is, there is a homeomorphism
$\phi : X \to K$
with
$\phi (x_0) = 0$
satisfying
$$ \begin{align*}\phi(Tx) = \alpha + \phi(x)\quad \text{ for all } x \in X.\end{align*} $$
Note that Theorem 1.4 shows that such a continuous map
$\phi $
exists but does not show that it is a homeomorphism (and indeed it may not be without the assumption on
$\operatorname {Stab}_G(\overline {U})$
). It suffices to show the existence of an inverse to
$\phi $
, which will be established in §4.
It is interesting to ask whether the same result holds more generally for two nilsystems rather than a nilsystem and a Kronecker system. Furthermore, one can extend this question to more general homogeneous spaces.
1.5. An explicit reconstruction for Jordan measurable sets
One way of stating Theorem 1.4 is that one can uniquely reconstruct a Kronecker system
$(K, x \mapsto x+\alpha )$
, where K is a compact group with
$\overline {\mathbb {Z}\alpha } = K$
, from the set
$\mathcal {R} = \{n ~|~ n\alpha \in U\}$
provided that U is some (unknown) open set such that
$\operatorname {Stab}_K(\overline {U})$
is trivial. If we further assume that U is Jordan measurable (recall that this means that
$m_K(\overline {U}) = m_K(U)$
, where
$m_K$
is the Haar measure) then the following result shows that there is in fact a rather explicit reconstruction of
$(K, \alpha )$
from
$\mathcal {R}$
. Let
$\mathbb {U} = \{ z \in \mathbb {C} ~|~ |z|=1\}$
denote the unit complex numbers.
Theorem 1.10. Let K be a compact abelian group and let
$\alpha \in K$
be such that
$\overline {\mathbb {Z}\alpha } = K$
and suppose
$U \subset K$
is a Jordan measurable open set such that
$\operatorname {Stab}_K(\overline {U})$
is trivial. Let
$\mathcal {R} = \{n ~|~ n\alpha \in U\}$
and let
Then
$\Lambda = \{ \unicode{x3bb} _1, \unicode{x3bb} _2, \ldots \}$
is countable and there is an injective continuous group homomorphism
$K \to \mathbb {U}^{\mathbb {N}}$
mapping
$\alpha $
to
$ \vec {\unicode{x3bb} } = (\unicode{x3bb} _1, \unicode{x3bb} _2, \ldots )$
. In particular, K is isomorphic to the closure of the subgroup generated by
$\vec {\unicode{x3bb} }$
.
Note that this implies Theorem 1.4 in the case where the open sets are Jordan measurable, but it is not difficult to construct examples with U not Jordan measurable where this reconstruction is invalid despite Theorem 1.4 still guaranteeing the uniqueness of the Kronecker system.
1.6. Further questions
We now gather some open questions. Our first question asks about a natural extension of Theorem 1.9 to two nilsystems and more general homogeneous spaces.
Question 1.11. Let
$X_1=G_1/\Gamma _1$
and
$X_2=G_2/\Gamma _2$
be homogeneous spaces, where
$G_i$
are Polish groups and
$\Gamma _i \leq G_i$
are cocompact discrete subgroups. For all
$i=1,2$
, suppose that
$U_i \subset X_i$
is an open set with closure having a trivial stabilizer in the sense that whenever
$g\in G$
is such that
$g\overline {U_i} = \overline {U_i}$
, then
$gx=x$
for all
$x \in X_i$
. Suppose that
$\tau _i \in G_i$
are such that
$(X_i, x \mapsto \tau _i x)$
are minimal systems and
$$ \begin{align*}\{n \in \mathbb{Z} ~|~ \tau_1^n x_1 \in U_1\} = \{n \in \mathbb{Z} ~|~ \tau_2^n x_2 \in U_2\}\end{align*} $$
for some
$x_1 \in X_1$
and
$x_2 \in X_2$
. Then, are the systems
$(X_1, x \mapsto \tau _1 x)$
and
$(X_2, x \mapsto \tau _2 x)$
isomorphic (via a map sending
$x_1$
to
$x_2$
)? If not true in general, is it true if
$G_i$
are nilpotent Lie groups?
The next question asks whether one can extend our results on minimal dynamical
$\mathbb {Z}$
systems to actions of other groups. It turns out that this seems difficult even without any topology, so we assume full orbits rather than just dense orbits as follows.
Question 1.12. Suppose that G is a group acting on sets
$X_1$
and
$X_2$
transitively. Thus there are
$x_1 \in X_1$
and
$x_2 \in X_2$
with
$Gx_1 = X_1$
and
$Gx_2 = X_2$
. Suppose that
$U_1 \subset X_1$
and
$U_2 \subset X_2$
are subsets with a trivial setwise stabilizer, that is,
$$ \begin{align*}\{g \in G ~|~ gU_1 = U_1\} = \{1\} = \{g \in G ~|~ gU_2 = U_2\}.\end{align*} $$
If
$\{g \in G ~|~ gx_1 \in U_1\} = \{g \in G ~|~ gx_2 \in U_2\}$
, then does it mean that the two actions are isomorphic? That is, does there exist a bijection
$\phi :X_1 \to X_2$
such that
$\phi (gx) = g\phi (x)$
for all
$g \in G$
and
$x \in X_1$
? Does there exist one mapping
$x_1$
to
$x_2$
?
Using the language of [Reference Albertson and Collins1] such group actions satisfying the trivial setwise stabilizer hypothesis are called
$2$
-distinguishable. This question is already interesting for G finite. In fact, we can simplify this to the following interesting question, which again is already interesting for finite sets (we will show this equivalence in §6).
Definition 1.13. Let G be a group acting on a set X. We say that
$U \subset X$
is simple (for this action) if whenever
$\pi :X \to X'$
is a factor (that is,
$X'$
is a G-set and
$\pi :X \to X'$
is a surjective map such that
$\pi (gx) = g\pi (x)$
for all
$g \in G$
,
$x \in X$
) such that
$U = \pi ^{-1}(U')$
for some
$U' \subset X'$
, then
$\pi $
is a bijection (isomorphism of G-sets). In other words, U is simple if it can not be realized as a preimage of a factor in a non-trivial way (that is, the factor is not an isomorphism).
Question 1.14. Let G be a group acting transitively on a set X and suppose that
$U \subset X$
has a trivial setwise stabilizer, that is,
$\{g \in G ~|~ gU = U\}$
consists of only
$1 \in G$
. Then, is it true that U is simple for this G-action? In other words, is it true that if
$\mathcal {B} \subset 2^X$
is a G-invariant partition of X such that U is a union of elements in
$\mathcal {B}$
, then
$\mathcal {B}$
consists of only singletons?
2. Bohr set return times of minimal systems (proof of Theorem 1.4)
Proof Proof of Theorem 1.4
Consider the product system
$(Z, S)$
, where
$Z = X \times K$
and
$S:Z \to Z$
is given by
$$ \begin{align*}S(x,k) = (Tx, k+\alpha).\end{align*} $$
Let
$\Theta = \overline {S^{\mathbb {Z}}z_0}$
be the orbit closure of
$z_0 = (x_0, 0) \in Z$
. Let
$\pi _X:\Theta \to X$
and
$\pi _K: \Theta \to K$
denote the projection maps, which are surjective by the minimality of
$(X,T)$
and
$(K, k \mapsto k+\alpha )$
. Let
$$ \begin{align*}H = \{ h \in K ~|~ (x_0, h) \in \Theta\}.\end{align*} $$
Claim 2.1. H is a closed subgroup of K.
Note that
$(x_0, 0) = z_0 \in \Theta $
so
$0 \in H$
. Now if
$h, h' \in H$
then that means
$(x_0, h), (x_0, h') \in \Theta $
and there exists a sequence
$n_1, n_2, \ldots \in \mathbb {Z}$
such that
$$ \begin{align*}S^{n_i}z_0 = (T^{n_i} x_0, n_i \alpha) \to (x_0, h)\end{align*} $$
as
$i \to \infty $
and another sequence
$n^{\prime }_1, n^{\prime }_2, \ldots \in \mathbb {Z}$
such that
$(T^{n^{\prime }_i} x_0, n^{\prime }_i \alpha ) \to (x_0, h')$
. Now fix
$\epsilon> 0.$
We may choose
$J = J(\epsilon ) \in \mathbb {N}$
such thatFootnote
†
$n_J\alpha \approx _{\epsilon } h$
and choose
$\delta>0$
such that
$$ \begin{align*}d(T^{n_J}x , x_0) < \epsilon \text{ whenever } d(x,x_0) < \delta.\end{align*} $$
Now take I large enough so that
$d(T^{n^{\prime }_I}x_0, x_0) < \delta $
and
$n I \alpha \approx _{\epsilon } h'$
. Thus
$d(T^{n_J + n^{\prime }_I}x_0, x_0) < \epsilon $
and
$(n_J + n^{\prime }_I)\alpha \approx _{2\epsilon } h + h'$
. As
$\epsilon>0$
is arbitrary, this means
$(x_0, h + h') \in \overline {S^{\mathbb {Z}}z_0} = \Theta $
, and thus
$h+h' \in H$
. Finally, notice that H is closed (as it is
$\pi _K(\Theta \cap (\{x_0\} \times K))$
) and so it is a closed non-empty sub-semigroup and a closed subgroup as K is compact.
Now for
$x \in X$
, let
$H_{x} = \{k \in K ~|~ (x, k) \in \Theta \}$
, and thus
$H = H_{x_0}$
.
Claim 2.2. For each
$x\in X$
, we have that
$H_{x}$
is a coset of H, that is,
$H_x$
and
$H_{x'}$
are translates of each other for all
$x, x' \in X$
.
To see this note that by minimality of
$(X,T)$
, there exists a sequence
$n_1, n_2, \ldots \in \mathbb {Z}$
such that
$T^{n_i}x \to x'$
, and by compactness we suppose that
$n_i\alpha \to \alpha _{x,x'} \in K$
. It follows that
$H_{x} + \alpha _{x,x'} \subset H_{x'}$
. In particular, this means that
$\alpha _{0, x} + \alpha _{x, 0} \in H$
, so
$$ \begin{align*}H + \alpha_{0,x} = H - (\alpha_{0, x} + \alpha_{x, 0}) + \alpha_{0,x} = H - \alpha_{x,0} \supset H_x\end{align*} $$
and thus
$H_x = H + \alpha _{0,x}$
is a coset of H as desired.
Claim 2.3.
$H \subset \operatorname {Stab}_K(\overline {U'})$
.
To see this, suppose that
$h \in H$
and
$u' \in U'$
. Then by the surjectivity of
$\pi _K:\Theta \to K$
, there exists
$x \in X$
such that
$(x, u') \in \Theta $
. Now we may find a sequence
$n_1, n_2, \ldots \in \mathbb {Z}$
such that
$n_i \alpha \to u'$
and
$T^{n_i}x_0 \to x$
. So for large enough i, we have
$n_i\alpha \in U'$
and by the assumption of the equality of return times (1), we have
$T^{n_i}x_0 \in U$
. Since
$(T^{n_i}x_0, n_i\alpha ) \in \Theta $
, we must have that
$(T^{n_i}x_0, n_i\alpha + h) \in \Theta $
as
$H_{T^{n_i}x_0}$
is a coset of H. This means that for each fixed i, we may find a sequence
$m_1, m_2, \ldots \in \mathbb {Z}$
such that
${S^{m_j}z_0 \to (T^{n_i}x_0, n_i\alpha + h)}$
. So for large enough j, we have that
$T^{m_j}x_0$
is so close to
$T^{n_i}x_0 \in U$
that
$T^{m_j}x_0 \in U$
and so again, by the equality of return times, we have
$m_j\alpha \in U'$
. But as
$m_j \alpha \to n_i\alpha + h$
, we must have that
$n_i\alpha + h \in \overline {U'}$
. Finally, as
$n_i\alpha \to u'$
, we must have that
$u' + h \in \overline {U'}$
. As
$u' \in U'$
is arbitrary, we must have that
$h + U' \subset \overline {U'}$
and by continuity of translation,
$h+\overline {U'} \subset \overline {U'}$
. Thus,
$h \in \operatorname {Stab}_K(\overline {U'})$
as claimed.
Now by the assumption that this stabilizer is trivial, we get that H is trivial and
$\pi _X:\Theta \to X$
is injective and so a homeomorphism. Thus our desired factor is
${\pi _K \circ \pi _X^{-1}:X \to K}$
.
3. Polynomial return times (proof of Theorem 1.7)
3.1. The case of compact connected abelian Lie groups
We now embark on proving Theorem 1.7. We first prove this for the case where
$K_1$
and
$K_2$
are compact connected abelian Lie groups, that is, subgroups of a finite dimensional torus. We leave the more subtle case of non-Lie groups to the next subsection.
Proposition 3.1. Let K be a compact abelian Lie group with
$\alpha \in K$
such that
$\mathbb {Z}\alpha $
is dense in K. Let
$K_0$
be the connected component of the identity, and thus
$d = |K / K_0|$
is finite (as K is a compact Lie group). Let
$P(x) \in \mathbb {Z}[x]$
be a non-constant polynomial with
$P(0) = 0$
. Then
$$ \begin{align*}\overline{P(d\mathbb{Z}) \alpha} = K_0\end{align*} $$
and
$$ \begin{align*}\overline{P(\mathbb{Z})\alpha} = \bigcup_{a \in A} (K_0 + a )\end{align*} $$
for some finite set
$A \subset K$
.
Proof Note that
$dk \in K_0$
for all
$k \in K$
. In particular,
$\mathbb {Z} d \alpha \subset K_0$
. Let
$K^{\prime }_0 = \overline {\mathbb {Z} d\alpha }$
. Note that
$K^{\prime }_0 \subset K_0$
and it is of finite index in K since
$$ \begin{align*}K = \overline{\mathbb{Z}\alpha} = \bigcup_{m = 0}^{d-1} \overline{(d\mathbb{Z} + m)\alpha} = \bigcup_{m = 0}^{d-1} (K^{\prime}_0 + m\alpha ).\end{align*} $$
Thus,
$K^{\prime }_0$
is of finite index in
$K_0$
as well. But as
$K_0$
is connected, we must have that
$K^{\prime }_0 = K_0$
. Now let
$Q(x) = ({1}/{d})P(dx) \in \mathbb {Z}[x]$
(as
$P(0) = 0$
). From Weyl equidistribution and connectedness of
$K_0$
, we see that the sequence
$Q(1)d\alpha , Q(2)d\alpha , Q(3) d\alpha , \ldots $
equidistributes in
$K_0$
. As
$Q(n)(d\alpha ) = P(dn)\alpha $
, this implies the first claim that
$$ \begin{align*}\overline{P(d\mathbb{Z}) \alpha} = K_0.\end{align*} $$
Now, for the second claim, we write
$$ \begin{align*}\overline{P(\mathbb{Z})\alpha} = \bigcup_{m=0}^{d-1} \overline{P(d\mathbb{Z} + m)\alpha} = \bigcup_{m=0}^{d-1} \overline{P_m(d\mathbb{Z})\alpha + a_m}\end{align*} $$
for some non-constant
$P_m(x) \in \mathbb {Z}[x]$
with
$P_m(0) = 0$
and
$a_m \in K$
. Applying the previous claim to the
$P_m$
, we get that
$\overline {P_m(d\mathbb {Z})\alpha + a_m}$
is a coset of
$K_0$
, as required.
Proposition 3.2. Let
$K_1$
and
$K_2$
be compact connected abelian Lie groups and suppose that
$\alpha _i \in K_i$
with
$\overline {\mathbb {Z}\alpha _i} = K_i$
. Let
$P(x) \in \mathbb {Z}[x]$
be a polynomial with
$P(0) = 0$
such that
$P(\mathbb {Z})$
is not contained in any proper subgroup of
$\mathbb {Z}$
. Let
$U_i \subset K_i$
be non-empty open sets and let
$\mathcal {R}_i = \{n \in \mathbb {Z} ~|~ P(n)\alpha _i \in U_i\}$
. If
$\mathcal {R}_1 = \mathcal {R}_2$
and
$\operatorname {Stab}_{K_1}(\overline {U_1}) = \{0 \}$
and
$\operatorname {Stab}_{K_2}(\overline {U_2}) = \{0 \}$
, then there exists an isomorphism
$K_1 \to K_2$
(of topological groups, so continuous) mapping
$\alpha _1$
to
$\alpha _2$
.
Proof Let
$\alpha = (\alpha _1, \alpha _2) \in K_1 \times K_2$
and let
$K = \overline {\mathbb {Z}\alpha } \subset K_1 \times K_2$
. Let
$\Theta = \overline {P(\mathbb {Z})\alpha }$
. We know that
$$ \begin{align*}\Theta = \bigcup_{a \in A} (K_0 + a)\end{align*} $$
for some finite
$A \subset K$
, where
$K_0$
is the identity connected component of K. Let
$\pi _i : K \to K_i$
be the projection maps. We know that
$\pi _i$
is surjective by the assumption that
$\overline {\mathbb {Z}\alpha _i} = K_i$
. In fact, since
$K_0$
is finite index in K, we have that
$\pi _i(K_0)$
of finite index in
$K_i$
and
${\pi _i(K_0) = K_i}$
by the assumption that
$K_i$
is connected. We now wish to show that
$\pi _i$
is injective on
$\Theta $
. To see this, suppose that
$\theta , \theta ' \in \Theta $
are such that
$\pi _1(\theta ) = \pi _1(\theta ')$
. We know that
$\theta \in K_0 + a$
and
$\theta ' \in K_0 + a'$
for some
$a,a' \in A$
. We now wish to show that
$\pi _2(\theta - \theta ') \in \operatorname {Stab}_{K_2}(\overline {U_2})$
. To do this, suppose that
$u_2 \in U_2$
. From
$\pi _2(K_0) = K_2$
, we get that
$\pi _2(K_0 + a') = K_2$
, and thus there exists
$x \in K_1$
such that
$(x, u_2) \in K_0 + a'$
. Now,
$$ \begin{align*}(x, u_2 + \pi_2(\theta - \theta')) = (x, u_2) + \theta - \theta' \in (K_0 + a') + (K_0 + a - a') = K_0 + a,\end{align*} $$
that is,
$(x, u_2 + \pi _2(\theta - \theta ')) \in \Theta $
. Now since
$(x, u_2) \in \Theta $
, this means there exists a sequence
$n_1, n_2, \ldots $
of integers such that
$P(n_j) \alpha \to (x, u_2)$
. In particular,
$P(n_j)\alpha _2 \to u_2$
, so
$P(n_j)\alpha _2 \in U_2$
for sufficiently large j. But since
$\mathcal {R}_1 = \mathcal {R}_2$
, this means that
$P(n_j)\alpha _1 \in U_1$
. Now, since
$(x, u_2) \in K_0 + a'$
and
$K_0 + a'$
is open in K, we must have that
$P(n_j)\alpha \in K_0 + a'$
for sufficiently large j. Thus
$P(n_j)\alpha + (\theta - \theta ') \in K_0 + a \subset \Theta $
for sufficiently large j. In particular, this means that we may choose an integer
$n^{\prime }_j$
so that
$P(n^{\prime }_j)\alpha $
is so close to
$P(n_j)\alpha + (\theta - \theta ')$
so that
$P(n^{\prime }_j)\alpha _1$
is sufficiently close to
$P(n_j)\alpha _1 \in U_1$
so that
$P(n^{\prime }_j)\alpha _1 \in U_1$
and so that
$P(n^{\prime }_j)\alpha \to (x, u_2 + \pi _2(\theta - \theta '))$
. However, as
$\mathcal {R}_1 = \mathcal {R}_2$
, this means that
$P(n^{\prime }_j)\alpha _2 \in U_2$
. However,
$P(n^{\prime }_j)\alpha _2 \to u_2 +\pi _2(\theta - \theta ')$
, which shows that
$u_2 +\pi _2(\theta - \theta ') \in \overline {U_2}$
. So we have shown that if
$u_2 \in U_2$
, then
$u_2 +\pi _2(\theta - \theta ') \in \overline {U_2}$
, and by continuity of addition this holds more generally for
$u_2 \in \overline {U_2}$
; thus
$\pi _2(\theta - \theta ') \in \operatorname {Stab}_{K_2}(\overline {U_2})$
. However, by the assumption that
$\operatorname {Stab}_{K_2}(\overline {U_2}) = \{0\}$
, we get that
$\pi _2(\theta ) = \pi _2(\theta ')$
, which means that
$\theta = \theta '$
. We have shown that
$\pi _1:\Theta \to K_1$
is injective and thus a homeomorphism. As
$K_1$
is connected, this means that
$\Theta $
is connected and so in fact
$ \Theta = K_0$
(as
$P(0)\alpha = (0,0)$
). So
$\pi _1: K_0 \to K_1$
is in fact an isomorphism of topological groups. By symmetry (using now the assumption that
$\operatorname {Stab}_{K_1}(\overline {U_1}) = \{0 \}$
), we get that
$\pi _2:K_0 \to K_2$
is an isomorphism of topological groups. So we have an isomorphism
$K_1 \to K_2$
mapping
$P(n)\alpha _1$
to
$P(n)\alpha _2$
for all
$n \in \mathbb {Z}$
. Finally, from the assumption that
$P(\mathbb {Z})$
is not contained in any proper subgroup, we have that
$1 = \sum _{i} a_i P(n_i)$
for some integers
$a_i$
and
$n_i$
. Thus
$\alpha _1 = \sum _{i} a_i P(n_i) \alpha _1$
is mapped to
$\sum _{i} a_i P(n_i)\alpha _2 = \alpha _2$
under this isomorphism.
3.2. Infinite-dimensional polynomial orbit
We now remove the assumption that our groups
$K_1$
and
$K_2$
are Lie groups (that is, embedded in a finite-dimensional torus). Let
$K_1$
and
$K_2$
be compact metrizable connected abelian groups. Note that this means they are closed subgroups of the countable-dimensional torus
$\mathbb {T}^{\mathbb {N}}$
and we may approximate them by compact Lie groups as follows.
Definition 3.3. Let
$(X, d_X)$
and
$(Y, d_Y)$
be metric spaces. For
$\delta>0$
, we say that a map
$\phi :X \to Y$
is a
$\delta $
-almost isometry if for all
$x_1,x_2 \in X$
we have that
$$ \begin{align*}|d_X(x_1, x_2) - d_Y(\phi(x_1), \phi(x_2))| < \delta.\end{align*} $$
We have a decreasing sequence
$\delta _1> \delta _2> \cdots $
of positive real numbers converging to
$0$
and, for each positive integer D, surjective continuous group homomorphisms
$\phi ^1_D: K_1 \to F^1_D$
and
$\phi ^2_D: K_2 \to F^2_D$
, where
$F^1_D$
and
$F^2_D$
are closed subgroups of
$\mathbb {T}^D$
and
$\phi ^1_D$
, and
$\phi ^2_D$
are
$\delta _D$
-almost isometries (we have fixed an invariant metric on each compact metrizable abelian group K, which we shall call
$d_K$
or d if clear from the context). For concreteness, we construct these maps conveniently as follows. Equip
$\mathbb {T}^{D}$
with the metric
$$ \begin{align*}d(x, y) = \sum_{i=1}^{D} 2^{-i}\| x_i - y_i\|,\end{align*} $$
where
$\| v\|$
is the shortest distance from
$v \in \mathbb {T}$
to
$0 \in \mathbb {T}$
, and
$\mathbb {T}^{\mathbb {N}}$
with the metric
$$ \begin{align*}d(x, y) = \sum_{i=1}^{\infty} 2^{-i}\| x_i - y_i\|,\end{align*} $$
that way,
$\phi _D^1$
and
$\phi _D^2$
are (restrictions to
$K_1$
and
$K_2$
, respectively) the projections onto the first D co-ordinates, which are clearly
$O(2^{-D})$
-almost isometries. The
$F_D^1$
and
$F_D^2$
are then defined to be the images of these, and thus
$\phi ^1_D: K_1 \to F^1_D$
and
$\phi ^2_D: K_2 \to F^2_D$
are surjective.
Now let
$P(x) \in \mathbb {Z}[x]$
be a non-constant polynomial such that
$P(0)=0$
. We now let
$$ \begin{align*}\Theta = \overline{P(\mathbb{Z}) \alpha} \subset K_1 \times K_2,\end{align*} $$
where
$\alpha = (\alpha _1, \alpha _2) \in K_1 \times K_2$
are such that
$\overline {\mathbb {Z}\alpha _i} = K_i$
for
$i=1,2$
. We let
$\phi _D:K_1 \times K_2 \to F_D^1 \times F_D^2$
be the map
$\phi _D = \phi _D^1 \times \phi _D^2$
. We also let
$\Theta _D = \phi _D(\Theta ) \subset F_D^1 \times F_D^2$
. We equip
$K_1 \times K_2$
and
$F_D^1 \times F_D^2$
with the metrics
$d((x_1, y_1), (x_2, y_2)) = \max \{d(x_1,x_2), d(y_1, y_2)\}$
, that way,
$\phi _D$
is also an
$O(2^{-D})$
-almost isometry. Observe that
$$ \begin{align*}\Theta_D = \overline{P(\mathbb{Z}) \phi_D(\alpha)}\end{align*} $$
and we may apply Proposition 3.1 to obtain that
$$ \begin{align} \Theta_D = \bigcup_{a \in A_D} (\Gamma_D + a) \end{align} $$
for some finite
$A_D \subset F_D^1 \times F_D^2$
and closed connected subgroup
$\Gamma _D \leq F_D^1 \times F_D^2$
. Note that
$\Gamma _D \subset \Theta _D$
as
$0 \in \Theta _D$
since
$P(0)=0$
.
Let
$$ \begin{align*}\Gamma = \bigcap_{D=1}^{\infty} \phi_D^{-1}(\Gamma_D)\end{align*} $$
and notice that this is a closed subgroup.
Lemma 3.4. We have
$\Gamma \subset \Theta $
.
Proof Since
$\phi _D$
is an
$O(2^{-D})$
-almost isometry, we have that if
$p \in \phi _D^{-1}(\Gamma _D)$
, then
$\phi _D(p) \in \Gamma _D \subset \phi _D(\Theta )$
and so
$d(p, \Theta ) < O(2^{-D})$
. However,
$\Theta $
is closed, so the proof is complete by letting
$D \to \infty $
.
Lemma 3.5. We have that
$\phi _D^{-1}(\Gamma _D) \supset \phi _{D+1}^{-1}(\Gamma _{D+1})$
.
Proof The natural projection
$F_{D+1}^1 \times F_{D+1}^2 \to F_{D}^1 \times F_{D}^2$
maps
$\Theta _{D+1}$
to
$\Theta _D$
so, in particular, maps the connected component
$\Gamma _{D+1}$
to the connected component
$\Gamma _D$
.
We let
$\pi _i: K_1 \times K_2 \to K_i$
and
$\pi _{i,D}:F_D^1 \times F_D^2 \to F_D^i$
denote the projections.
Lemma 3.6. For
$i=1,2$
, we have that
$\pi _i(\Gamma ) = K_i$
.
Proof Fix
$i \in \{1,2\}$
. We first show that
$\pi _{i,D}(\Gamma _D) = F_D^i$
. First notice that
$$ \begin{align*}\pi_{i,D}(\Theta_D) = \overline{P(\mathbb{Z})\phi_D(\alpha_i)} = F_D^i,\end{align*} $$
where the last equality follows from Weyl equidisitribution, connectendess of
$F_{D}^i$
, and
$\overline {\mathbb {Z} \phi _D(\alpha _i)} = F_D^i$
. However,
$$ \begin{align*}\pi_{i,D}(\Theta_D) = \bigcup_{a \in A_D} (\pi_{i,D}(\Gamma_D) + \pi_{i,D}(a))\end{align*} $$
is a disjoint union of closed sets, so by connectedness, they must all be the same and equal to
$\pi _{i,D}(\Gamma _D)$
, so indeed
$\pi _{i,D}(\Gamma _D) = F_D^i$
. Now let
$k \in K_i$
. Then by our first claim, we have that
$\phi _D^i(k) = \pi _{i,D}(\gamma _D)$
for some
$\gamma _D \in \Gamma _D$
. Now by surjectivity of
$\phi _D^i$
, we may take
$\beta _D \in \phi _D^{-1}(\Gamma _D)$
such that
$\phi _D(\beta _D) = \gamma _D$
. By compactness, we may pass to a subsequence such that
$\beta _{D_j} \to \beta $
for some
$\beta \in K_1 \times K_2$
. However, it now follows from Lemma 3.5 that
$\beta \in \Gamma $
. The proof will be complete if we can show that
$\pi _{i}(\beta ) = k$
. To see this, note that
$\phi _D^i (\pi _i(\beta )) = \pi _{i,D}(\phi _D(\beta ))$
but
$\phi _D(\beta )$
can be made arbitrarily close to
$\phi _D(\beta _D) = \gamma _D$
for large enough D. So
$\phi _D^i(\pi _i(\beta ))$
can be made arbitrarily close to
$\pi _{i,D}(\gamma _D) = \phi _D^i(k)$
. It now follows that
$\pi _i(\beta )$
is arbitrarily close to k for large enough D (as
$\phi _D^i$
is an
$O(2^{-D})$
-almost isometry), and thus they are equal.
Proposition 3.7. There exists
$A \subset K_1 \times K_2$
, with
$0 \in A$
, such that
$$ \begin{align*}\Theta = \bigcup_{a \in A} (\Gamma + a).\end{align*} $$
Proof We already know that
$\Gamma \subset \Theta $
by Lemma 3.4. It remains to show that if
$\gamma _0 + a \in \Theta $
for some
$\gamma _0 \in \Gamma $
and
$a \in K_1 \times K_2$
, then for every
$\gamma \in \Gamma $
, we have that
$\gamma + a \in \Theta $
. Let D be a positive integer, we thus have that
$\phi _D(\gamma _0 + a) \in \Theta _D$
and so Equation (2), and the fact that
$\Gamma \subset \phi _D^{-1}(\Gamma _D)$
(by definition) implies that
$\Gamma _D + \phi _D(a) \subset \Theta _D$
. In particular, this means that
$\phi _D(\gamma + a) \in \Theta _D$
. Thus, there exists a
$\beta _D \in \Theta $
such that
$\phi _D(\gamma +a) = \phi _D(\beta _D)$
. However, since
$\phi _D$
is an
$O(2^{-D})$
-almost isometry, we have that
$d(\gamma + a, \beta _D) < O(2^{-D})$
. Thus,
$\beta _D \to \gamma +a$
and since
$\Theta $
is closed, we must have
$\gamma +a \in \Theta $
.
Now let
$U_1 \subset K_1$
and
$U_2 \subset K_2$
be open sets. Let
$$ \begin{align*}\mathcal{R}_i = \{n \in \mathbb{Z} ~|~ P(n)\alpha_i \in U_i \}.\end{align*} $$
Proposition 3.8. If
$\mathcal {R}_1 = \mathcal {R}_2$
and
$\theta , \theta ' \in \Theta $
are such that
$\pi _1(\theta _1) = \pi _1(\theta _2)$
, then
$\pi _2(\theta - \theta ') \in \operatorname {Stab}_{K_2}(\overline {U_2})$
.
Proof We have
$\theta ' - \theta = (0,v)$
for some
$v \in K_2$
, so we wish to show
$v \in \operatorname {Stab}_{K_2}(\overline {U_2})$
. By Lemma 3.7, we have
$a,a' \in A$
such that
$\theta \in \Gamma + a$
and
$\theta ' \in \Gamma + a'$
. Now let
$u_2 \in U_2$
. Since
$\pi _2(\Gamma ) = K_2$
, there exists
$x \in K_1$
such that
$(x, u_2) \in \Gamma + a$
. Now fix
$\epsilon> 0$
such that the ball of radius
$\epsilon $
centered at
$u_2$
is contained in
$U_2$
. Since
$\pi _1 \vert _{\Gamma }:\Gamma \to K_1$
is a surjective homomorphism between compact groups, it is an open map and hence there exists a
$\delta>0$
such that if
$g_1 \in K_1$
with
$d(0, g_1) < \delta $
, then there exists a
$\gamma \in \Gamma $
with
$d(0, \gamma ) < \epsilon $
and
$\pi _1(\gamma ) = g_1$
. Now since
$(x, u_2) \in \Gamma +a \subset \Theta $
, there exists a positive integer
$n_1$
such that
$d(P(n_1)\alpha , (x,u_2)) < \max \{\delta , \epsilon \}$
. Writing
$P(n_1)\alpha = (x_1, y_1)$
with
$x_1 \in K_1, y_1 \in K_2$
, we see that
$d(y_1, u_2) < \epsilon $
and
$y_1 \in U_2$
. Additionally, since
$n_1 \in \mathcal {R}_2 = \mathcal {R}_1$
, we have
$x_1 \in U_1$
. We set
$g_1 = x_1 - x$
. Notice that
$d(0, g_1) < \delta $
and so there exists a
$\gamma \in \Gamma $
with
$d(0, \gamma ) < \epsilon $
such that
$\pi _1(\gamma ) = g_1$
. Writing
$\gamma = (g_1, g_2)$
for some
$g_2 \in K_2$
, we get that
$$ \begin{align*}(x_1, u_2 + g_2) = (x,u_2) + \gamma \in \Gamma + a + \gamma = \Gamma + a.\end{align*} $$
Thus,
$\theta " := (x_1, u_2 + g_2 + v) \in \Gamma + a' \subset \Theta $
. This means that there exists
$n_2 \in \mathbb {Z}$
such that
$P(n_2)\alpha $
is so close to
$\theta "$
that
$\pi _1(P(n_2)\alpha )$
is so close to
$x_1$
that
$\pi _1(P(n_2)\alpha ) \in U_1$
and
$d(\pi _2(P(n_2)\alpha ), u_2 + g_2 + v) < \epsilon $
. Thus
$n_2 \in \mathcal {R}_1 = \mathcal {R}_2$
and so
$\pi _2(P(n_2)\alpha ) \in U_2$
. This means that
$$ \begin{align*}d(u_2 + v, U_2) \leq d(u_2 + v + g_2, U_2) + \epsilon \leq d(u_2 + v +g_2, \pi_2(P(n_2)\alpha)) + \epsilon < 2\epsilon.\end{align*} $$
As
$\epsilon> 0$
was arbitrary and independent of v, we may take
$\epsilon \to 0$
to get that
$d(u_2 + v, U_2) = 0$
, and thus
$u_2 + v \in \overline {U_2}$
. Thus,
$U_2 + v \subset \overline {U_2}$
and by continuity of addition, we have that
$\overline {U_2} + v \subset \overline {U_2}$
. The reverse inclusion holds by swapping the roles of
$\theta $
and
$\theta '$
, and thus
$v \in \operatorname {Stab}_{K_2}(\overline {U_2})$
.
Proposition 3.9. If
$\mathcal {R}_1 = \mathcal {R}_2$
and
$\operatorname {Stab}_{K_1}(\overline {U_1})$
and
$\operatorname {Stab}_{K_2}(\overline {U_2})$
are trivial, then there exists an isomorphism of topological groups
$K_1 \to K_2$
mapping
$P(n)\alpha _1$
to
$P(n)\alpha _2$
for all
$n \in \mathbb {Z}$
.
Proof By Proposition 3.8 and the assumption
$\operatorname {Stab}_{K_2}(\overline {U_2}) = \{0\}$
, we have that the mapping
$\pi _1\vert _{\Theta }:\Theta \to K_1$
is injective and a homeomorphism. However, since
$\pi _1(\Gamma ) = K_1$
by Lemma 3.6, we must have that
$\Theta = \Gamma $
, so
$\pi _1\vert _{\Theta }:\Theta \to K_1$
is also a group homomorpishm. By symmetry (this time using
$\operatorname {Stab}_{K_1}(\overline {U_1}) = \{0\}$
), we get that
$\pi _2\vert _{\Theta }:\Theta \to K_2$
is also an isomorphism of topological groups. Thus
$\pi _2\vert _{\Theta } \circ (\pi _1\vert _{\Theta })^{-1}:K_1 \to K_2$
is the desired isomorphism.
4. Return times sets for nilsystems
We now prove Theorem 1.9. As noted, Theorem 1.4 already defines the map
$\phi :X \to K$
and so it remains to provide an inverse
$\psi :K \to X$
, which is established by the following result.
Proposition 4.1. Let
$X=G/\Gamma $
be a nilmanifold, where G is a nilpotent Lie group and
$\Gamma \leq G$
is a cocompact discrete subgroup. Suppose that
$\tau \in G$
is such that
$(X,T)$
is minimal, where
$T:X \to X$
is given by
$Tx = \tau x$
for
$x \in X$
. Suppose that
$U \subset X$
is open and satisfies that
$\operatorname {Stab}_G(\overline {U}):= \{g \in G ~|~ g\overline {U} = \overline {U}\}$
consists of only those
$g \in G$
such that
$gx = x$
for all
$x \in X$
. Now suppose that
$(K,+)$
is a compact Lie group and
$\alpha \in K$
is such that
$\overline {\mathbb {Z} \alpha } = K$
, and
$V \subset K$
is open such that
$$ \begin{align*}\{ n \in \mathbb{Z} ~|~ T^nx_0 \in U\} = \{n \in \mathbb{Z} ~|~ n\alpha \in V\}\end{align*} $$
for some
$x_0 \in X$
. Then the pointed system
$(X, x_0, T)$
is a factor of the pointed Kronecker system
$(K,0 , k \mapsto k+\alpha )$
, that is, there is a continuous map
$\psi : K \to X$
with
$\psi (0) = x_0$
satisfying
$$ \begin{align*}\psi(\alpha + k) = T\psi(x)\quad \text{ for all } k \in K.\end{align*} $$
Proof Let
$Z = K \times X$
and
$z_0 = (0, x_0) \in Z$
. Then
$Z = (K \oplus G) / (\{0\} \oplus \Gamma )$
is a nilmanifold and thus
$(Z,S)$
is a nilsystem where
$S:Z \to Z$
is given by
$T(k,x) = (k + \alpha , Tx)$
. Now let
$\Theta = \overline {S^{\mathbb {Z}}z_0}$
be the closure of the S-orbit of
$z_0$
. Then (see Theorem 9 in Chapter 11 of [Reference Host and Kra3]), we have that
$$ \begin{align*}\Theta = Hz_0\end{align*} $$
for some closed subgroup H of
$K \oplus G$
. Now let
$\pi _K:\Theta \to K$
and
$\pi _X:\Theta \to X$
denote the projection maps. We wish to show that
$\pi _K$
is injective as it would then be a homeomorphism and our desired factor map would be
$\pi _X \circ \pi _K^{-1}:K \to X$
. Suppose that
$\theta _1, \theta _2 \in \Theta $
are such that
$\pi _K(\theta _1) = \pi _K(\theta _2)$
. Thus we can write
$\theta _1 = h_1 z_0$
and
$\theta _2 = h_2 z_0$
for some
$h_1,h_2 \in H$
. For
$i=1,2$
, we may write
$h_i = (k_i, g_i)$
where
$k_i \in K$
and
$g_i \in G$
, and thus
$\theta _i = (k_i, g_i x_0)$
. However,
$k_1 = k_2$
and so we have
$h_0 := h_2h_1^{-1} = (0, g_2 g_1^{-1}) \in H$
and we have the relation
$h_0 \theta _1 = \theta _2$
. Now let
$u \in U$
be arbitrary. By surjectivity of
$\pi _X:\Theta \to X$
, we may find a
$h' \in H$
such that
$h'z_0 = (k', u) \in \Theta $
for some
$k' \in K$
. We may find a sequence of integers
$n_1, n_2, \ldots $
such that
$S^{n_i}z_0 \to h'z_0$
. In particular,
$T^{n_i}x_0 \to u$
and thus
$T^{n_i}x_0 \in U$
for large enough
$n_i$
. This must mean that
$n_i \alpha \in V$
. Now
$h_0 S^{n_i}z_0 = (n_i\alpha , g_2g_1^{-1}T^{n_i}z_0) \in \Theta $
and so, for each fixed i, we may find a sequence of integers
$m_1, m_2, \ldots $
such that
$\lim _{j \to \infty } S^{m_j}z_0 = (n_i\alpha , g_2g_1^{-1}T^{n_i}x_0)$
. In particular, this means that
$m_j\alpha \to n_i\alpha \in V$
and so, for large enough j, we have
$m_j \alpha \in V$
and so
$T^{m_j}x_0 \in U$
. However,
$T^{m_j}x_0 \to g_2g_1^{-1}T^{n_i}x_0$
and so
$g_2g_1^{-1}T^{n_i}x_0 \in \overline {U}$
. Finally, taking the limit as
$i \to \infty $
and using
$T^{n_i}x_0 \to u$
, we have that
$g_2g_1^{-1} u\in \overline {U}$
. As
$u \in U$
was arbitrary, this shows that
$g_2g_1^{-1} U \subset \overline {U}$
and
$g_2g_1^{-1} \overline {U} \subset \overline {U}$
. The reverse inclusion follows from swapping the roles of
$\theta _1$
and
$\theta _2$
, and thus
$g_2g_1^{-1} \in \operatorname {Stab}_G(\overline {U})$
. Thus,
$g_2g_1^{-1}$
stabilizes every
$x \in X$
thus,
$g_2x_0 = g_2g_1^{-1}g_1x_0 = g_1x_0$
, so in fact
$\theta _2 = \theta _1$
. Thus
$\pi _K$
is injective as desired.
5. Spectral construction for Jordan measurable sets (proof of Theorem 1.10)
Let
$U \subset K$
be a Jordan measurable open subset of a compact abelian group K. Let
$\alpha \in K$
be an element such that
$\mathbb {Z} \alpha $
is dense in K. Let
$\Lambda $
denote the set of unit complex numbers
$\unicode{x3bb} $
such that
does not converge to
$0$
as
$N \to \infty $
.
Lemma 5.1. We have
$\Lambda \subset \{ \chi (\alpha ) ~|~ \chi \in \widehat {K} \}.$
In fact, if
is the Fourier decomposition in
$L^2(K)$
, then
$\Lambda $
is the set of those
$\chi (\alpha )$
for which
$c_{\chi } \neq 0$
.
Proof Let
$\mathbb {U}$
be the unit complex numbers. Suppose that
$\unicode{x3bb} \in \mathbb {U}$
but
$\unicode{x3bb} \notin \{ \chi (\alpha ) ~|~ \chi \in \widehat {K} \}$
. Note that for any character
$\chi $
on K, we have that
$$ \begin{align*}\frac{1}{N}\sum_{n=1}^N \unicode{x3bb}^n \chi(n\alpha) = \frac{1}{N}\sum_{n=1}^N(\unicode{x3bb}\chi(\alpha))^n \to 0.\end{align*} $$
As any continuous function
$f:K \to \mathbb {C}$
can be uniformly approximated by a linear combination of characters, we have that
$$ \begin{align*}\frac{1}{N}\sum_{n=1}^N \unicode{x3bb}^n f(n\alpha) \to 0.\end{align*} $$
Now observe that there exist continuous functions
$f_{\epsilon }^+: K \to [0,1]$
such that
In particular, we choose these functions so that
$$ \begin{align*}m_K(\operatorname{supp}f_{\epsilon}^+ \setminus U) = m_K(\operatorname{supp}f_{\epsilon}^+ \setminus \overline{U}) < \epsilon,\end{align*} $$
where
$m_K$
denotes the Haar measure on K and, in the first equality, we used that
$\overline {U}$
and U are
$m_K$
almost the same (as U is Jordan measurable). Now, as
$\operatorname {supp}f_{\epsilon }^+ \setminus U$
is a closed set, its indicator function can be written as a pointwise decreasing limit of continuous functions, and thus we have that
$$ \begin{align*}\limsup_{N \to \infty} \frac{1}{N} |\{ n \in [1,N] ~|~ n\alpha \in \operatorname{supp}f_{\epsilon}^+ \setminus U \}| \leq m_K(\operatorname{supp}f_{\epsilon}^+ \setminus U) < \epsilon.\end{align*} $$
This means that
Thus
and so we have shown the first claim that
$$ \begin{align*}\Lambda \subset \{ \chi(\alpha) ~|~ \chi \in \widehat{K} \}.\end{align*} $$
Now suppose that
$\unicode{x3bb} \in \{ \chi (\alpha ) ~|~ \chi \in \widehat {K} \}$
. Thus
$\unicode{x3bb} = \overline {\chi (\alpha )}$
for some unique (by density of
$\mathbb {Z}\alpha $
) character
$\chi $
on K. However,
is Riemann-integrable, and thus we have that
Now let
$E = \{ \chi (\alpha ) ~|~ \chi \in \widehat {K} \}$
. Note that the mapping
$\widehat {K} \to E$
mapping
$\chi $
to
$\chi (\alpha )$
is injective (by density of
$\mathbb {Z} \alpha $
) and thus bijective. Consequently, let
$\Gamma \subset \widehat {K}$
be the set corresponding to
$\Lambda $
, that is,
$\Lambda = \{ \chi (\alpha ) ~|~ \chi \in \Gamma \}$
.
Now observe that the mapping
$\iota :K \to \mathbb {U} ^{\widehat {K}}$
given by
$\iota (k) = ( \chi (k) )_{\chi \in \widehat {K}}$
is injective and continuous, and thus a homeomorphism onto its image. Now let
$$ \begin{align*}\pi: K \to \mathbb{U}^{\Gamma}\end{align*} $$
be the projection given by
$$ \begin{align*}k \mapsto ( \chi(k) )_{\chi \in \Gamma}.\end{align*} $$
Proposition 5.2. We have
$\operatorname {ker}(\pi ) \subset \operatorname {Stab}_K(\overline {U})$
.
Proof Suppose that
$k \in \operatorname {ker} \pi $
. Thus,
$\chi (k) = 1$
for all
$\chi \in \Gamma $
. However, since
we have that
for almost all
$x \in K$
. As U and
$\overline {U}$
are
$m_K$
almost equal, this means that the open set
$(U - k) \setminus \overline {U}$
has zero measure and is thus empty. This means that
$u - k \in \overline {U}$
for all
$u \in U$
and
$k \in \operatorname {ker}(\pi )$
. Thus
$\operatorname {ker(\pi )} \subset \operatorname {Stab}_K(\overline {U})$
.
Proof Proof of Theorem 1.10
The assumption that
$\operatorname {Stab}_K(\overline {U})$
is trivial, together with Proposition 5.2, implies that
$\pi $
is injective and an isomorphism onto its image. However, the image of
$\pi $
is the closed subgroup of
$\mathbb {U}^{\Gamma }$
generated by
$$ \begin{align*}\pi(\alpha) = ( \chi(\alpha) )_{\chi \in \Gamma}.\end{align*} $$
Finally, by applying the natural isomorphism
$\mathbb {U}^{\Gamma } \cong \mathbb {U}^{\Lambda }$
(that is, the one induced by the bijection
$\chi \mapsto \chi (\alpha )$
from
$\Gamma $
to
$\Lambda $
), we get that K is isomorphic to the closed subgroup of
$\mathbb {U}^{\Lambda }$
generated by
$ ( \unicode{x3bb} )_{\unicode{x3bb} \in \Lambda }. $
6. Equivalence of Question 1.12 and Question 1.14
We now show that Question 1.12 and Question 1.14 are equivalent. Recall that a block system of a group action
$G\curvearrowright X$
is a partition
$\mathcal {B} \subset 2^X$
of X (collection of non-empty disjoint sets whose union is X) that is G-invariant (that is, if
$g \in G$
and
$B \in \mathcal {B}$
, then
$gB \in \mathcal {B}$
). Let us define the block system generated by
$U \subset X$
(for this action) to be the smallest (with respect to inclusion) block system such that U is a union of elements of
$\mathcal {B}$
. More concretely, the block system generated by U consists of those minimal non-empty sets that can be written as intersections of elements in
$\{ gU ~|~ g \in G\} \cup \{ X \setminus gU ~|~ g \in G\}$
.
Proposition 6.1. Let G be a group acting transitively on a set
$X_1$
and also acting transitively on a set
$X_2$
, and suppose that
$U_1 \subset X_1$
and
$U_2 \subset X_2$
are simple for the respective actions. Suppose that
$x_1 \in X_1$
and
$x_2 \in X_2$
are such that
$$ \begin{align*}\{g \in G ~|~ gx_1 \in U_1\} = \{g \in G ~|~ gx_2 \in U_2\}.\end{align*} $$
Then there is an isomorphism
$\phi :X_1 \to X_2$
of G-sets mapping
$x_1$
to
$x_2$
, that is,
$\phi (gx) = g\phi (x)$
for all
$g \in G$
and
$x \in X_1$
.
Proof Let
$\mathcal {B}$
denote the block system generated by
$U_1$
. There is a well-defined map
$\pi : X_1 \to \mathcal {B}$
, where for
$x \in X_1$
, we define
$\pi (x)$
to be the unique element of
$\mathcal {B}$
containing
$x \in X_1$
. Note that
$\pi $
is a morphism of G-actions, that is,
$g \pi (x) = \pi (gx)$
for all
$g \in G$
and
$x \in X_1$
. Note that
$$ \begin{align*}U_1 = \pi^{-1}\bigg(\bigcup_{\{B \in \mathcal{B} ~|~B \subset U_1\}} B \bigg)\end{align*} $$
since, by definition,
$U_1$
is a union of elements of
$\mathcal {B}$
. Since
$U_1$
is simple, we must have that
$\pi $
is an isomorphism. This means
$\{x\} \in \mathcal {B}$
for all
$x \in X_1$
. Now let
$$ \begin{align*}R(V) = \{g \in G ~|~ gx_1 \in V\}\end{align*} $$
for any
$V \subset X_1$
. Notice the properties
$R(gV)=gR(V)$
and
$R(\bigcap _{i \in \mathcal {I}} V_i)= \bigcap _{i \in \mathcal {I}} R(V_i)$
and
$R(X_1 \setminus V) = G \setminus R(V)$
. Now let
$\mathcal {B}_G$
denote the block system generated by
$R(U_1)$
for the action of
$G \curvearrowright G$
by left multiplication. By the aforementioned properties, we have that
$\mathcal {B}_G$
consists of those sets of the form
$R(B)$
for
$B \in \mathcal {B}$
and so we have that
$\mathcal {B}_G$
consists of exactly the sets of the form
$R(\{x\})$
for
$x \in X_1$
. Note that
$R(\{x_1\}) = \{g \in G ~|~ gx_1 = x_1\} = \operatorname {Stab}(x_1)$
and any other
$R(\{x\})$
must be a coset of
$\operatorname {Stab}(x_1)$
(by transitivity
$x = gx_1$
for some
$g \in G$
and so
$R(\{x\}) = gR(\{x_1\})$
). In fact
$\operatorname {Stab}(x_1)$
may be described as member of
$\mathcal {B}_G$
that contains
$1 \in G$
. Since
$X_1 \cong G/\operatorname {Stab}(x_1)$
, this means that we have an isomorphism of G-actions
$X_1 \to \mathcal {B}_G$
that maps
$x_1$
to the unique element of
$\mathcal {B}_G$
containing
$1 \in G$
(it maps x to
$R(\{x\})$
). Notice that
$\mathcal {B}_G$
is the same if we replace
$X_1,x_1,U_1$
with
$X_2, x_2, U_2$
, respectively, by the assumption that
$$ \begin{align*}\{g \in G ~|~ gx_1 \in U_1\} = \{g \in G ~|~ gx_2 \in U_2\}.\end{align*} $$
Indeed we have an isomorphism
$X_1 \cong X_2$
of G-actions (they are both isomorphic to G acting on
$\mathcal {B}_G$
).
This demonstrates the equivalence of the two questions as follows. If the answer to Question 1.14 is affirmative, then the two sets in Question 1.12 are simple and the Proposition 6.1 provides an affirmative answer. Conversely, if
$U \subset X$
is a subset in a transitive G-set X with a trivial setwise stabilizer, then we have a factor
$\pi :X \to \mathcal {B}$
, where
$\mathcal {B}$
is the block system generated by U, mapping
$x \in X$
to the unique element of
$\mathcal {B}$
containing x (as constructed in the proof of the Proposition 6.1). However,
$\pi (U)$
also has a trivial setwise stabilizer and
$\{g \in G ~|~ gx_0 \in U\} = \{g \in G ~|~ g\pi (x_0) \in \pi (U)\}$
, so if the answer to Question 1.12 is affirmative, then
$\pi $
must be an isomorphism and so
$\mathcal {B}$
consists of singletons.
Acknowledgements
The authors were partially supported by the Australian Research Council grant DP210100162. They are grateful to Sean Gasiorek and Robert Marangell for interesting discussions.
