@
Appl. Gen. Topol. 23, no. 1 (2022), 179-187

doi:10.4995/agt.2022.15187

© AGT, UPV, 2022

Topologically mixing extensions of

endomorphisms on Polish groups

John Burke and Leonardo Pinheiro

Department of Mathematical Sciences, Rhode Island College, 600 Mt Pleasant Ave, Providence,

RI 02908 (jburke@ric.edu, lpinheiro@ric.edu)

Communicated by J. Galindo

Abstract

In this paper we study the dynamics of continuous endomorphisms
on Polish groups. We offer necessary and sufficient conditions for a
continuous endomorphism on a Polish group to be weakly mixing. We
prove that any continuous endomorphism of an abelian Polish group can
be extended in a natural way to a topologically mixing endomorphism
on the countable infinite product of said group.

2020 MSC: 37B99.

Keywords: weak mixing; Polish group; hypercyclicity criterion.

1. Introduction

The theory of discrete dynamical systems is concerned with the behavior of
the iterates of a continuous map on a (usually compact) metric space. The
most interesting and studied examples include maps that, in some sense, ‘mix’
the space. For a nice survey on the subject see the article by Kolyada and
Snoha [7].

Formally, let X be a topological space and f : X → X a continuous map.
Let

fn(x) = f ◦f ◦ · · · ◦f︸ ︷︷ ︸
n−fold

denote the n− th iteration of the map f.

Received 28 February 2021 – Accepted 08 January 2022

http://dx.doi.org/10.4995/agt.2022.15187
https://orcid.org/0000-0002-1633-3606


J. Burke and L.Pinheiro

We say f is topologically transitive if given any two non-empty open subsets
U and V of X there exists a natural number n ≥ 1 such that fn(U) ∩V 6= ∅.

A continuous map f : X → X is said to be topologically mixing if given any
two non-empty open subsets U and V of X there exists a natural number N
such that fn(U)∩V 6= ∅, whenever n > N. A map f such that f×f is transi-
tive in X ×X is called a weakly mixing map. Topological mixing is a stronger
condition than weak mixing which is generally stronger than topological tran-
sitivity. For example, the irrational rotation of the circle is a topologically
transitive map that is not weakly mixing.

In the setting of linear operators acting on a Banach space, a very cele-
brated result is the set of sufficient conditions for topological mixing known
as the Hypercyclicity Criterion. The result first appeared in Kitai [6] and was
later independently rediscovered by Gethner and Shapiro [5]. In [1], Bes and
Peris showed that satisfying slightly laxer hypothesis than those in the Hyper-
cyclicity Criterion is equivalent to the weak mixing of the operator in question.
In [8] Reed and De la Rosa constructed a topologically transitive operator
on a Banach Space that is not weak mixing and hence does not satisfy the
Hypercyclicity Criterion, settling the question of whether the conditions were
necessary.

Notice that the general theory of discrete dynamical systems is usually not
concerned with any underlying algebraic structure of the space X. Operator
theorists studying linear dynamics, on the other hand, only consider maps pre-
serving the linear structure of the underlying space. In this note we will revisit
this theme in a more general setting; we will study the dynamics of continu-
ous endomorphisms on Polish groups. In particular, we conclude that every
topologically mixing and weakly mixing continuous endomorphism on a Polish
group satisfies conditions analogous to those in the Hypercylicity Criterion.

2. Main Results

In what follows G will denote a metric, complete, separable topological
group: a Polish group for short.

If G has no isolated points, then a result by G. D. Birkhoff [2] tells us that a
continuous map ϕ : G → G is topologically transitive if and only if there exists
an element x0 such that the orbit orb(x0,ϕ) = {x0,ϕ(x0),ϕ2(x0),ϕ3(x0), · · ·}
is dense in X. In other words, any element of G can be arbitrarily approximated
by a sequence of elements in the orbit of x0. In this setting, ϕ is weakly mixing
if there exists (h,f) in G×G whose orbit under ϕ×ϕ is dense.

Recall that for a group G, with the group operation written as multiplication,
an endomorphism is a map ϕ : G → G such that ϕ(gh) = ϕ(g)ϕ(h) for all g
and h in G.

Before we continue with our discussion, we introduce the following definition:

Definition 2.1 (Weak Mixing Criterion). Let G be a Polish semigroup with
identity e. We say that a continuous endomorphism ϕ : G → G satisfies the
Weak Mixing Criterion if there exists an increasing sequence {nk} of natural

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Topologically mixing extensions of endomorphisms on Polish groups

numbers, dense sets F,H ⊂ G, and maps ψnk : F → G such that for any f ∈ F
and h ∈ H:

(i) ϕnk (h) → e as k →∞.
(ii) ψnk (f) → e as k →∞.
(iii) ϕnkψnk (f) → f as k →∞.

Chan [3] and Moothatu [9] independently remarked that if ϕ is a continuous
endomorphism on a Polish group satisfying the Weak Mixing Criterion in the
particular case nk = k,k ≥ 0, then ϕ is topologically mixing. In the same
paper Moothatu showed that if G is a compact Hausdorff group and ϕ is a
topologically transitive endomorphism, then ϕ is weakly mixing.

In view of Chan and Moothatu’s observation we ask if every weak mixing
continuous endomorphism on a Polish group must satisfy the Weak Mixing
Criterion. We answer the question in the affirmative for compact Polish groups
in contrast to de la Rosa and Reed result for Banach spaces presented in [8]. The
connection between the two settings comes from the fact that continuous linear
operators are endomorphisms on the additive abelian group of a topological
vector space. Indeed, we have:

Theorem 2.2. Let G be a Polish group with identity e and f : G → G a
continuous endomorphism. Then ϕ satisfies the Weak Mixing Criterion if and
only if ϕ is weakly mixing.

Proof. Assume ϕ satisfies the the Weak Mixing Criterion for a sequence {nk}
and dense sets F and H. Let U1,U2,V1 and V2 be nonempty open subsets of
G. In order to show that ϕ is weakly mixing, it suffices to exhibit n ∈ N such
that (ϕn ×ϕn) (U1 ×U2) ∩ (V1 ×V2) 6= ∅. Since F and H are dense in G, we
can choose f1 ∈ V1 ∩F, f2 ∈ V2 ∩F , h1 ∈ U1 ∩H, and h2 ∈ U2 ∩H. Then

h1ψnk (f1) → h1e = h1 ∈ U1 as k →∞
Similarly,

h2ψnk (f2) → h2e = h2 ∈ U2 as k →∞,
so for sufficiently large k

(h1ψnk (f1),h2ψnk (f2)) ∈ U1 ×U2.
Now, notice that

ϕnk (h1ψnk (f1)) = ϕ
nk (h1)ϕ

nk (ψnk (f1)) → ef1 = f1 ∈ V1 as k →∞

and also

ϕnk (h2ψnk (f2)) = ϕ
nk (h2)ϕ

nk (ψnk (f2)) → ef2 = f2 ∈ V2 as k →∞.
Again, for k large enough,

(ϕnk (U1) ×ϕnk (U2)) ∩ (V1 ×V2) 6= ∅,
and we can conclude ϕ is weakly mixing.

Now, we assume that is ϕ weakly mixing with (h,f) an element whose orbit
is dense in G×G. Let id be the identity map on G. Since id×ϕn commutes

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J. Burke and L.Pinheiro

with ϕ×ϕ and its image is dense in G×G, we have that for any n ∈ N, the
orbit of (h,ϕn(f)) is also dense. This is implies that for all U ⊂ G open, there
is u ∈ U such that (h,u) has dense orbit under ϕ×ϕ. We will now exhibit the
sets H and F and construct the sequence {nk} in the statement of the Weak
Mixing Criterion.

For each natural number k > 1, consider Bk be the open ball in G centered
at e of radius 1/k. Since Bk is open in G, and left multiplication is continuous,
the set Bk × hBk is open in G × G. By the observation above, we can pick
wk ∈ Bk such that, the orbit of (h,wk) under ϕ × ϕ is dense in G × G. It
follows that there exists an integer nk > k such that

(ϕnk ×ϕnk )(h,wk) ∈ (Bk ×hBk).
That is, we can inductively construct an increasing sequence of positive integer
{nk} such that:

(i) ϕnk (h) ∈ Bk and
(ii) ϕnk (wk) ∈ hBk.

Letting k →∞, we have that:
wk → e because wk ∈ Bk, ϕnk (h) → e by (i), and ϕnk (wk) → h by (ii) .

Now, let F = H = {h,ϕ(h),ϕ2(h), · · ·} and define ψnk : F → G by
ψnk (ϕ

j(h)) = ϕj(wk).

We have:

(i) ϕnk (ϕj(h)) = ϕj(ϕnk (h)) → ϕj(e) = e,
(ii) ψnk (ϕ

j(h)) = ϕj(wk) → ϕj(e) = e, and
(iii) ϕnk (ψnk (ϕ

j(h))) = ϕnk (ϕj(wk)) = ϕ
j(ϕnk (wk)) → ϕj(h),

which is what we needed to show. �

It is worth noting again, that if the subsequence {nk} is given by nk = k
then we can conclude that ϕ is topologically mixing. We can now state:

Corollary 2.3. Let G be a compact Polish group and let ϕ be a continuous
homomorphism on G. Then ϕ is topologically transitive if and only if it satisfies
the Weak Mixing Criterion.

Proof. If ϕ is topologically transitive, then by a result of Moothatu [9] ϕ must
be weak mixing and by Theorem 2.2 it satisfies the Weak Mixing Criterion.
Conversely, if ϕ satisfies that Weak Mixing Criterion then its transitivity follows
immediately from Theorem 2.2. �

A natural question is whether any Polish group supports a topologically
mixing endomorphism. We show that if the group in question is an abelian
group that can be written as a countable infinite product of isomorphic copies
of one of its closed subgroups, then the answer is affirmative. More precisely,
we show that any continuous endomorphism on an abelian Polish group can be
extended in a natural way to a topologically mixing continuous endomorphism
of the infinite direct product of said group. The result is analogous to that of
Chan [4].

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Topologically mixing extensions of endomorphisms on Polish groups

Theorem 2.4. Let G be an abelian Polish group, and ϕ : G → G a contin-
uous endomorphism. There exists a continuous endomorphism Φ :

∏∞
i=1 G →∏∞

i=1 G such that:
(i) Φ is topologically mixing, and
(ii) j ◦ϕ = Φ ◦ j where j : G →

∏∞
i=1 G is defined by j(g) = (g,e,e, ...)

Proof. We will construct the endomorphism Φ and check it is topologically
mixing by verifying it satisfies the Weak Mixing Criterion.

Consider a bounded metric d on G, we can define a metric ρ on
∏∞
i=1 G by

ρ(g,h) =

∞∑
i=1

d(gi,hi)

2i
, forf,g ∈ G

which induces the product topology on
∏∞
i=1 G.

For each g ∈
∏∞
i=1 G , write

g = (g1,g2,g3, · · ·), gi ∈ G,
and define the maps Φ, Ψ :

∏∞
i=1 G →

∏∞
i=1 G by

Φ(g) = (ϕ(g1)g2,g3,g4, · · ·)
and

Ψ(g) = (e,g1,g2,g3, · · ·)
where e is the identity in G.

First we verify that Φ is an endomorphism in
∏∞
i=1 G. Indeed, let f,g,∈∏∞

i=1 G and we have

Φ(fg) = Φ((f1,f2,f3, ...)(g1,g2,g3, ...))

= Φ(f1g1,f2g2,f3g3, ...)

= (ϕ(f1g1)f2g2,f3g3, ...)

= (ϕ(f1)f2ϕ(g1)g2,f3g3, ...)

= (ϕ(f1)f2,f3, ...)(ϕ(g1)g2,g3, ...)

= Φ(f)Φ(g).

Notice that we have used the fact that G is abelian on the third equality from
the bottom.

The continuity of Φ follows from the fact that the group operation is con-
tinuous. Now, we will check that j ◦ϕ = Φ ◦ j. Indeed, for any g ∈ G,

(j ◦ϕ)(g) = j(ϕ(g))
= (ϕ(g),e,e, ...)

= (ϕ(g)e,e, ...)

= Φ((ϕ(g),e,e, ...))

= Φ(j(g)

= (Φ ◦ j)(g)

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J. Burke and L.Pinheiro

Now, note that for all g ∈
∏∞
i=1 G, Φ(Ψ(g)) = g and Ψ

ng → ẽ as n → ∞
where ẽ = (e,e,e, · · ·) is the identity element of

∏∞
i=1 G. We will now verify

that Φ is mixing. Let H be a dense set in G and consider the subgroup D̃ in∏∞
i=1 G whose elements are of the form (h1,h2,h3, · · · ,hk,e,e,e, . . . ) for some

natural k and hi ∈ H. This is clearly dense in
∏∞
i=1 G and for any element

h = (h1,h2,h3, · · · ,hk,e,e,e, . . . ) ∈ D̃, we have

Φk(h) =
(
ϕk(h1)ϕ

k−1(h2) · · ·ϕ(hk−1)hk,e,e, · · ·
)

and

Ψk(Φk(h)) = (e,e, · · · ,e︸ ︷︷ ︸
k positions

,ϕk(h1)ϕ
k−1(h2) · · ·ϕ(hk−1)hk,e,e, · · ·).

Notice that:

ρ(Ψk(Φk(h)), ẽ) =
d(ϕik(h0)ϕ

k−1(h1) · · ·ϕ(hk−1)hk,e)
2k

≤
1

2k+1

So we conclude that as n →∞, ρ(Ψk(Φk(h)), ẽ) → 0.
Now, consider the set:

F̃ = {g(Ψn(Φn(g))−1 : g ∈ D̃,n ∈ N},

and note that

lim
n→∞

g(Ψn(Φn(g))−1) = g lim
n→∞

(Ψn(Φn(g))−1)

= g( lim
n→∞

(Ψn(Φn(g−1)))

= gẽ

= g

which shows that F̃ is dense in
∏∞
i=1 G.

Also, let f ∈ F̃, so f = g(Ψn(Φn(g))−1 for some g ∈ D̃ and we have

Φn(f) = Φn(g(Ψn(Φn(g))−1))

= Φn(g)Φn(Ψn(Φn(g))−1)

= Φn(g)Φn(Ψn(Φn(g−1)))

= Φn(g)(Φn(g))−1

= ẽ.

The third equality follows from the fact that Ψ is a left inverse for Φ. We can
then conclude that Φn → ẽ on F̃ and by the results of Chan [3] and Moothatu
[9], we conclude Φ is mixing. �

Now, we define a weakly mixing continuous endomorphism on the infinite
product in an analogous fashion to Theorem 2.4.

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Topologically mixing extensions of endomorphisms on Polish groups

Theorem 2.5. Let G be an abelian Polish group and let ϕ : G → G be a
continuous endomorphism. Then the map Φ :

∏∞
i=1 G →

∏∞
i=1 G defined by:

Φ(a1,a2,a3, . . . ) = (ϕ(a1
−1)a−12 ,a3

−1,a4
−1, . . . )

is weakly mixing.

Proof. This proof will be a slight modification of the proof of Theorem 2.4. We
follow the notation in the proof of Theorem 2.4 and let Ψ and D̃ be defined in
the same manner. It is not hard to verify that Φ is a continuous endomorphism.

As in the proof of Theorem 2.4, we have Ψk(g) → ẽ as k → ∞ and thus
Ψ2k(g) → ẽ as k → ∞. Also observe that Φ2k(Ψ2k(g)) = g. Next note that
since G is an abelian group we have,

Φ2(a1,a2,a3, . . . ) = (ϕ
2(a1)ϕ(a2)a3,a4,a5 . . . ).

One can thus verify that

Ψ2k(Φ2k(h)) = (e,e, · · · ,e︸ ︷︷ ︸
2k positions

,ϕ2k(h0)ϕ
2k−1(h1) · · ·ϕ(h2k−1)h2k,e,e, · · ·).

and

ρ(Ψ2k(Φ2k(h)), ẽ) =
d̂(ϕ2k(h0)ϕ

2k−1(h1) · · ·ϕ(h2k−1)h2k,e)
22k

≤
1

22k+1

We conclude that as n →∞, ρ(Ψ2k(Φ2k(h)), ẽ) → 0.
Now, let D be a dense set in G and define F̃ as follows,

F̃ = {g(Ψ2n(Φ2n(g))−1 : g ∈ D,n ∈ N}.
Notice that

lim
n→∞

g(Ψ2n(Φ2n(g))−1 = g( lim
n→∞

(Ψ2n(Φ2n(g))−1)

= g( lim
n→∞

(Ψ2n(Φ2n(g))−1

= gẽ

= g

which shows that F̃ is dense in
∏∞
i=1 G.

Also, let f ∈ F̃, so f = g(Ψ2n(Φ2n(g))−1 for some g ∈
∏∞
i=1 G and we have

Φ2n(f) = Φ2n(g(Ψ2n(Φ2n(g))−1)

= Φ2n(g)Φ2n(Ψ2n(Φ2n(g))−1)

= Φ2n(g)(Φ2n(g))−1

= ẽ.

The third equality follows from the fact that Ψ is a left inverse for Φ. Hence,
Φ2n(f) → ẽ on F̃ and by Theorem 2.2, we can conclude Φ is weakly mixing.

�

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J. Burke and L.Pinheiro

3. Some General Results

The first result in this section is concerned with the size of the orbit of
an endomorphism; the second result gives an idea of how common are group
elements with a dense orbit.

Before we do so, we will fix some notation. For a topological space X and
A a subset of X, we denote the closure of A by Ā and its interior by A◦. We

say that a subset of a topological space is somewhere dense when A
◦
6= ∅.

Proposition 3.1. Let T be a continuous endomorphism on a Polish group. If
T has an element g with a somewhere dense orbit then the subgroup generated
by g is clopen in G.

Proof. It is a well-known fact that if a subgroup of a topological group has
non-empty interior then the subgroup is open. Since

orb(T,g)
◦
⊂〈orb(T,x)〉 6= ∅

we have that 〈orb(T,x)〉 is open but being the closure of the orbit, it is closed.
�

We immediately get the following.

Corollary 3.2. If G is a connected Polish group, then the subgroup generated
by an element with a somewhere dense orbit is dense.

Proof. The result follows from the basic fact that the only clopen subsets of a
connected space are the empty set and the space itself. �

Now, we turn our attention to the algebraic structure of the set of group
elements whose orbit is dense. We begin with the observation made by Kitai
in [6] that a continuous self-map f on a complete metrizable space X without
isolated points is topologically transitive if and only if the set of transitive
elements tr(f) = {x ∈ X; orb(x,f) is dense} is a dense Gδ set.

We then have :

Proposition 3.3. Let G be a Polish group that admits a topologically transitive
endomorphism, then every element x ∈ G is the product of two elements whose
orbits are dense under T .

Proof. Let T be a topologically transitive endomorphism on G and let x ∈ G.
We have that both tr(T) and tr(T)x−1 are dense Gδ sets, by Baire’s Theorem
their intersection is not empty. Let h be an element in the intersection, so
h = gx−1 with g and h with dense orbit. �

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Topologically mixing extensions of endomorphisms on Polish groups

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