1 Introduction and main results
Let G be a locally compact topological group and
$\mathcal {L}(G)$
the space of all its closed subgroups endowed with the Vietoris topology. An open prebase of the Vietoris topology on
$\mathcal {L}(G)$
is formed by the sets
$$ \begin{align*} \mathrm{D}_{1}(U) &= \{L\in \mathcal{L}(G)\mid L\subseteq U\}, \\ \mathrm{D}_{2}(V) &= \{L\in \mathcal{L}(G)\mid L\cap V\ne \varnothing\}, \end{align*} $$
where
$U, V$
range over all open subsets of G.
A locally compact group G is said to be compactly ruled if any compact subset of G is contained in a compact subgroup. The subspace of all compact subgroups, open subgroups, compact open subgroups and compactly ruled subgroups of G are denoted respectively by
$\mathcal {L}_c(G)$
,
$\mathcal {L}_o(G)$
,
$\mathcal {L}_{oc}(G)$
and
$\mathcal {L}_{cr}(G)$
. Clearly,
$$ \begin{align*}\mathcal{L}_{oc}(G)\subseteq \mathcal{L}_c(G)\subseteq \mathcal{L}_{cr}(G).\end{align*} $$
We denote the closure of a subset A of a topological space X by
$\overline {A}$
. Our first result gives a characterisation of compactly ruled groups in terms of the Vietoris space
$\mathcal {L}(G)$
.
Proposition (Proposition 2.6 below).
Let G be a locally compact group. Then, the following conditions are equivalent:
-
(1) G is compactly ruled;
-
(2)
$\overline {\mathcal {L}_c(G)}=\mathcal {L}(G)$
; -
(3)
$\overline {\mathcal {L}_{oc}(G)}=\{L\in \mathcal {L}(G)\mid G_0\subseteq L\}$
; -
(4)
$G\in \overline {\mathcal {L}_{oc}(G)}$
; -
(5)
$G\in \overline {\mathcal {L}_c(G)}$
.
Remark 1.1. The Vietoris topology
$\tau _v$
on the set of closed subgroups
$\mathrm {Sub}(G)$
of a locally compact group G is finer than the Chabauty topology
$\tau _c$
on
$\mathrm {Sub}(G)$
. Moreover, if F is a subset of
$\mathrm {Sub}(G)$
, then the group G belongs to the closure of F in
$(\mathrm {Sub}(G), \tau _v)$
if and only if G belongs to the closure of F in
$(\mathrm {Sub}(G), \tau _c)$
. However, the above proposition remains true if the Vietoris topology is replaced with the Chabauty topology on
$\mathrm {Sub}(G)$
(compare with [Reference Kadri11, Theorem 3.18]).
Notation 1.2. The identity component of a topological group G, denoted by
$G_0$
, is the connected component of the identity element in G.
Let G and H be two locally compact groups. For every topological morphism
$\varphi \colon G\to H$
, let us denote by
$\varphi _*$
the induced function
$\varphi _*\colon \mathcal {L}_c(G)\to \mathcal {L}_c(H)$
defined by
$L\mapsto \varphi (L)$
. Clearly,
$\varphi _*$
is continuous [Reference Donjuán, Jonard-Pérez and López-Poo4]. We are interested in the relationship between the openness of
$\varphi $
and the openness of
$\varphi _*$
(see [Reference Csikós, Kátay, Kocsis and Pálfy2, Question 1.3] and [Reference Fisher and Gartside5, page 859]). The main results are the following.
Theorem A (Theorem 3.10 below).
Let G be a locally compact group and H a locally compact group with compact identity component (for example, a totally disconnected group or a compactly ruled group). Let
$\varphi \colon G\to H$
be a continuous morphism. Then,
$\varphi $
is open if the following conditions are satisfied:
-
(1)
$\varphi _*\colon \mathcal {L}_c(G)\to \mathcal {L}_c(H)$
is open; -
(2) there is a compact open subgroup A of G such that
$H_0\subseteq \varphi (A)$
.
Theorem B (Theorem 3.11 below).
Let G, H be locally compact groups with compact identity components and
$\varphi \colon G\to H$
a continuous morphism. Then, the following conditions are equivalent:
-
(1)
$\varphi $
is open; -
(2)
$\varphi _*$
is open and
$H_0\subseteq \bigcap _{L\in \mathcal {L}_{oc}(G)} \varphi (L)$
; -
(3)
$\varphi _*$
is open and there is
$L\in \mathcal {L}_{oc}(G)$
such that
$H_0\subseteq \varphi (L)$
.
In particular, the following corollary holds.
Corollary. Let G be a locally compact group with compact identity component, H a locally compact totally disconnected group and
$\varphi \colon G\to H$
a continuous morphism. Then,
$\varphi $
is open if and only if
$\varphi _*$
is open.
2 Space of compact subgroups
If A is a closed subgroup of a topological group G, then
$[A, G]$
denotes the lattice of all closed subgroups of G containing A:
$[A, G]=\{L\in \mathcal {L}(G)\mid A\subseteq L\}$
. We begin with the following elementary (and probably well-known) lemma which will be very useful.
Lemma 2.1. Let G be a locally compact group and N a closed normal subgroup of G. Then, the canonical projection
$\pi \colon G\to G/N$
induces a bijective continuous map
$\pi _\bullet \colon [N, G]\to \mathcal {L}(G/N)$
defined by
$H\mapsto \pi (H)$
. Moreover, if N is compact, then
$\pi _\bullet $
is a homeomorphism with continuous inverse
$\pi ^\bullet \colon \mathcal {L}(G/N) \to [N, G]$
given by
$L\mapsto \pi ^{-1}(L)$
.
Proof. The continuity of
$\pi _\bullet $
follows from the fact that, for every open subset V of
$G/N$
, we observe that
$\pi _\bullet ^{-1}(\mathrm {D}_{1}(V))= \mathrm {D}_{1}(\pi ^{-1}(V))\cap [N, G]$
and
$\pi _\bullet ^{-1}(\mathrm {D}_{2}(V))= \mathrm {D}_{2}(\pi ^{-1}(V))\cap [N, G]$
. However, if N is compact, then
$\pi $
is a closed map and the continuity of
$\pi ^\bullet $
follows from the fact that
$(\pi ^\bullet )^{-1}(\mathrm {D}_{1}(U))= \mathrm {D}_{1}(G/N\setminus \pi (G\setminus U))$
and
$(\pi ^\bullet )^{-1}(\mathrm {D}_{2}(U))= \mathrm {D}_{2}(\pi (U))$
for every open subset U of G.
Remark 2.2. In general, the map
$\pi ^\bullet $
is not continuous (see [Reference Protasov12, Example 1]).
Proposition 2.3 [Reference Herfort, Hofmann and Russo8, Proposition 1.3].
For a locally compact group G, the following properties are equivalent:
-
(1) G is compactly ruled;
-
(2) every finite subset of G is contained in a compact subgroup;
-
(3) every compact subset of G is contained in a compact open subgroup;
-
(4) every open compactly generated subgroup of G is compact;
-
(5) G is a directed union of compact open subgroups;
-
(6)
$G_0$
is compact and
$G/G_0$
is compactly ruled.
Proposition 2.4 (Protasov, [Reference Protasov13]; see also [Reference Protasov15, Theorem 15]).
For every locally compact group G,
$\mathcal {L}_{cr}(G)$
is the closure of
$\mathcal {L}_c(G)$
in
$\mathcal {L}(G)$
.
Combining [Reference Djocovic3, Theorem 3] and the above proposition gives the next result.
Corollary 2.5. For any connected locally compact group G,
$\mathcal {L}_c(G)$
is closed in
$\mathcal {L}(G)$
.
Let G be a compactly ruled group and
$(G_\alpha )_{\alpha \in I}$
a directed family of compact open subgroups of G covering G. Then,
$$ \begin{align*} \mathcal{L}_c(G) &= \bigcup_{\alpha\in I} \mathcal{L}(G_\alpha),\\ \mathcal{L}_{oc}(G) &= \bigcup_{\alpha\in I} \mathcal{L}_o(G_\alpha), \\ \mathcal{L}(G) = \overline{\mathcal{L}_c(G)}&=\overline{\bigcup_{\alpha\in I} \mathcal{L}(G_\alpha)}. \end{align*} $$
Compactly ruled groups admit the following characterisation in terms of the Vietoris space
$\mathcal {L}(G)$
.
Proposition 2.6. Let G be a locally compact group. Then, the following conditions are equivalent:
-
(1) G is compactly ruled;
-
(2)
$\overline {\mathcal {L}_c(G)}=\mathcal {L}(G)$
; -
(3)
$\overline {\mathcal {L}_{oc}(G)}=[G_0, G]$
; -
(4)
$G\in \overline {\mathcal {L}_{oc}(G)}$
; -
(5)
$G\in \overline {\mathcal {L}_c(G)}$
.
Proof. (1)
$\Rightarrow $
(2). As G is compactly ruled, any closed subgroup of G is also compactly ruled [Reference Herfort, Hofmann and Russo8, Proposition 1.9(i)] and therefore,
$\mathcal {L}_{cr}(G)=\mathcal {L}(G)$
. Then,
$\overline {\mathcal {L}_c(G)}=\mathcal {L}_{cr}(G)=\mathcal {L}(G)$
.
(2)
$\Rightarrow $
(1). This follows from Proposition 2.4.
(1)
$\Rightarrow $
(3). Assume that G is compactly ruled and consider first the case when G is totally disconnected. Let
$(G_\alpha )_{\alpha \in I}$
be a directed family of compact open subgroups of G covering G. Then,
$\bigcup _{\alpha \in I} \mathcal {L}(G_\alpha )$
is dense in
$\mathcal {L}(G)$
and, for every
$\alpha \in I$
,
$\mathcal {L}_{oc}(G_\alpha )$
is dense in
$\mathcal {L}(G_\alpha )$
(see [Reference Gartside and Smith6, Lemma 3.2] or [Reference Hamrouni and Jlali7, Theorem 3.12]) and therefore,
$\bigcup _{\alpha \in I} \mathcal {L}_{oc}(G_\alpha )$
is dense in
$\mathcal {L}(G)$
. Thus,
$\mathcal {L}_{oc}(G)$
is dense in
$\mathcal {L}(G)$
. In the general case, consider the canonical projection
$\pi \colon G\to G/G_0$
. Since
$G_0$
is compact, by Lemma 2.1,
$\pi ^\bullet $
is a homeomorphism of
$\mathcal {L}(G/G_0)$
onto
$[G_0, G]$
. However, also since
$G_0$
is compact,
$\pi $
is perfect, that is, for every compact subset
$C\subseteq G/G_0$
, the preimage
$\pi ^{-1}(C)$
is also compact [Reference Stroppel16, Lemma 32.8] and, therefore,
$\pi ^\bullet (\mathcal {L}_{oc}(G/G_0))=\mathcal {L}_{oc}(G)$
. Hence,
$\overline {\mathcal {L}_{oc}(G)}=[G_0, G]$
.
The implications (3)
$\Rightarrow $
(4)
$\Rightarrow $
(5) are trivial.
(5)
$\Rightarrow $
(1). Let B be an open compactly generated subgroup of G. Let
$(H_i)_{i\in I}$
be a net of compact subgroups with limit G. As B is open, the net
$(H_i\cap B)_{i\in I}$
converges to B. Since the Vietoris topology is finer than the Chabauty topology, [Reference Wang17, Lemma 5.7] implies that B is compact. Consequently, any open compactly generated subgroup of G is compact, which proves that G is compactly ruled.
3 The relationship between openness of
$\varphi $
and
$\varphi _*$
The open mapping theorem for topological groups states that a surjective continuous morphism of a locally compact group G onto a locally compact group H is an open mapping if G is
$\sigma $
-compact [Reference Hofmann and Morris10, Theorem 1.6]. Recently, Cowling, Hofmann and Morris proved the following theorem [Reference Cowling, Hofmann and Morris1, Theorem 2, Corollaries 7, 4].
Theorem 3.1. Let
$\varphi \colon G\to H$
be a morphism of locally compact groups. Then, the following conditions are equivalent:
-
(1)
$\varphi $
is open; -
(2)
$\varphi (L)$
is open in H for all open subgroups L of G; -
(3)
$\varphi (L)$
is open in H for all compactly generated open subgroups L of G; -
(4) there is a compactly generated open subgroup L of G such that
$\varphi (L)$
is open in H; -
(5)
$\gamma \colon G/G_0\to H/H_0$
, given by
$gG_0\mapsto \varphi (g)H_0$
, is open and
$H_0=\bigcap _{L\in \mathcal {L}_o(G)} \varphi (L)$
.
Proposition 3.2 [Reference Protasov14, Lemma 2].
Let
$\varphi $
be a continuous morphism of a compact group G onto a compact group H. Then, the induced map
$\varphi _*\colon \mathcal {L}(G)\to \mathcal {L}(H)$
is open.
Remark 3.3. Another proof of this result in the case of profinite groups can be found in [Reference Fisher and Gartside5, Proposition 5].
In what follows, we often use the next obvious statement.
Lemma 3.4. Let
$\varphi \colon G\to H$
be a topological morphism of locally compact groups and L a compact subgroup of G. Then,
$\varphi _*(\mathcal {L}(L))= \mathcal {L}(\varphi (L))$
.
Proof. Let
$B{\kern-1pt}\in{\kern-1pt} \varphi _*(\mathcal {L}(L))$
and let
$A{\kern-1pt}\in{\kern-1pt} \mathcal {L}(L)$
such that
$\varphi _*(A){\kern-1pt}={\kern-1pt}B$
. As
$A{\kern-1pt}\subseteq{\kern-1pt} L$
,
$\varphi (A){\kern-1pt}\subseteq{\kern-1pt} \varphi (L)$
and so
$B\in \mathcal {L}(\varphi (L))$
. Hence,
$\varphi _*(\mathcal {L}(L))\subseteq \mathcal {L}(\varphi (L))$
. Conversely, let
$B\in \mathcal {L}(\varphi (L))$
. We have
$B=B\cap \varphi (L)=\varphi (\varphi ^{-1}(B)\cap L)=\varphi _*(A)$
, where
$A=\varphi ^{-1}(B)\cap L\in \mathcal {L}(L)$
. Hence,
$B\in \varphi _*(\mathcal {L}(L))$
and, therefore,
$\mathcal {L}(\varphi (L)) \subseteq \varphi _*(\mathcal {L}(L))$
.
A locally compact group G is periodic if any element belongs to a compact subgroup.
Proposition 3.5 (Subgroup with open Vietoris subspace).
Let G be a locally compact group and H a closed subgroup of G.
-
(1)
$\mathcal {L}(H)$
is homeomorphic to the closed subspace
$\{L\in \mathcal {L}(G)\mid L\subseteq H\}$
of
$\mathcal {L}(G)$
. -
(2) If
$\mathcal {L}(H)$
is open in
$\mathcal {L}(G)$
, then
$\mathcal {L}(H\cap A)$
is open in
$\mathcal {L}(A)$
for every closed subgroup A of G. -
(3) If H is open in G, then
$\mathcal {L}(H)$
is open in
$\mathcal {L}(G)$
. -
(4) If G is totally disconnected, then H is open in G if and only if
$\mathcal {L}(H)$
is open in
$\mathcal {L}(G)$
. -
(5) If G is periodic and if
$\mathcal {L}(H)$
is open in
$\mathcal {L}(G)$
, then
$HG_0$
is open in G.
Proof. (1) See [Reference Protasov12, Lemma 1].
(2) This follows from the continuity of the map
$\mathcal {L}(A)\to \mathcal {L}(G)$
defined by
$L\mapsto L$
.
(3) If H is open in G, then
$\mathrm {D}_{1}(H)=\mathcal {L}(H)$
and so
$\mathcal {L}(H)$
is open in
$\mathcal {L}(G)$
.
(4) Suppose
$\mathcal {L}(H)$
is open in
$\mathcal {L}(G)$
. As the trivial subgroup
$\{e\}\in \mathcal {L}(H)$
, there exists a compact subgroup U of G such that
$\{e\}\in \mathrm {D}_{1}(U)\subseteq \mathcal {L}(H)$
. Then,
$U\subseteq H$
and so H is open in G.
(5) We first consider the situation when G is compact. Let
$\pi \colon G\to G/G_0$
be the canonical projection. Since the induced map
$\pi _*$
is open,
$\pi _*(\mathcal {L}(H))=\mathcal {L}(\pi (H))$
is open in
$\mathcal {L}(G/G_0)$
and therefore, by condition (4),
$\pi (H)$
is open in
$G/G_0$
. Hence,
$HG_0$
is open in G. Consider now the case when G is periodic. Let
$(G_i)_{i\in I}$
be a family of compact open subgroups of G covering G. For any
$i\in I$
, by condition (2),
$\mathcal {L}(H\cap G_i)$
is open in
$\mathcal {L}(G_i)$
and hence,
$(H\cap G_i)G_0$
is open in
$G_i$
and so in G. Then,
$HG_0= \bigcup _{i\in I} (H\cap G_i)G_0$
is open in G.
Remark 3.6. As is well known, the Lie groups are precisely the locally compact groups without arbitrary small subgroups, and they admit the following characterisation in terms of the Vietoris space
$\mathcal {L}(G)$
: G is a Lie group if and only if the trivial subgroup
$\{e\}=\mathcal {L}(\{e\})$
is open in
$\mathcal {L}(G)$
. In general,
$\mathcal {L}(H)$
open in
$\mathcal {L}(G)$
does not imply that H is open in G.
Proposition 3.2 can be improved slightly as follows.
Proposition 3.7. Let
$\varphi $
be an open continuous morphism of a compact group G into a locally compact group H. Then, the induced map
$\varphi _*\colon \mathcal {L}(G)\to \mathcal {L}_c(H)$
is open.
Proof. Since
$\varphi (G)$
is open in H, by Proposition 3.5(3),
$\varphi _*(\mathcal {L}(G))=\mathcal {L}(\varphi (G))$
is open in
$\mathcal {L}(H)$
and so in
$\mathcal {L}_c(H)$
. As
$\varphi _*\colon \mathcal {L}(G)\to \varphi _*(\mathcal {L}(G))$
is open,
$\varphi _*\colon \mathcal {L}(G)\to \mathcal {L}_c(H)$
is open.
Proposition 3.8. In a totally disconnected locally compact group, every compact subgroup is contained in a compact open subgroup.
Proof. See [Reference Hofmann9, proof of Lemma 1.16] or [Reference Willis18, Lemma 7].
Proposition 3.9. Let G be a locally compact group with compact identity component
$G_0$
. Then, the following hold.
-
(1) Every compact subgroup of G is contained in a compact open subgroup.
-
(2)
$\mathcal {L}_c(G)=\bigcup _{H\in \mathcal {L}_{oc}(G)} \mathcal {L}(H)$
. -
(3)
$\mathcal {L}_c(G)$
is open in
$\mathcal {L}(G)$
. -
(4) Let H be a compact subgroup of G. If
$\mathcal {L}(H)$
is open in
$\mathcal {L}(G)$
, then
$HG_0$
is open in G.
Proof. (1) As
$G_0$
is compact, the canonical projection
$\pi \colon G\rightarrow G/G_0$
is perfect [Reference Stroppel16, Lemma 32.8]. The result follows from the fact that
$G/G_0$
is a totally disconnected locally compact group and Proposition 3.8.
(2) Trivial.
(3) Follows from condition (2) and Proposition 3.5(3).
(4) Let O be a compact open subgroup of G containing the compact subgroup
$HG_0$
. By condition (2),
$\mathcal {L}(H\cap O)$
is open in
$\mathcal {L}(O)$
and hence,
$(H\cap O)G_0$
is open in O and so in G. As
$(H\cap O)G_0= HG_0\cap O =HG_0$
, then
$HG_0$
is open in G.
Theorem 3.10. Let G be a locally compact group and H a locally compact group with compact identity component. Let
$\varphi \colon G\to H$
be a continuous morphism. Then,
$\varphi $
is open if the following conditions are satisfied:
-
(1)
$\varphi _*\colon \mathcal {L}_c(G)\to \mathcal {L}_c(H)$
is open; -
(2) there is a compact open subgroup A of G such that
$H_0\subseteq \varphi (A)$
.
Proof. By Proposition 3.5(3),
$\mathcal {L}(A)$
is open in
$\mathcal {L}(G)$
and so
$\varphi _*(\mathcal {L}(A)) =\mathcal {L}(\varphi (A))$
is open in
$\mathcal {L}_c(H)$
. By Proposition 3.9(3), we deduce that
$\mathcal {L}(\varphi (A))$
is open in
$\mathcal {L}(H)$
. As
$H_0\subseteq \varphi (A)$
, by Proposition 3.9(4),
$\varphi (A)$
is open in H and therefore, by Theorem 3.1,
$\varphi $
is open.
Theorem 3.11. Let G, H be locally compact groups with compact identity components and
$\varphi \colon G\to H$
a continuous morphism. Then, the following conditions are equivalent:
-
(1)
$\varphi $
is open; -
(2)
$\varphi _*$
is open and
$H_0\subseteq \bigcap _{L\in \mathcal {L}_{oc}(G)} \varphi (L)$
; -
(3)
$\varphi _*$
is open and there is
$L\in \mathcal {L}_{oc}(G)$
such that
$H_0\subseteq \varphi (L)$
.
Proof. (1)
$\Rightarrow $
(2). For every
$L\in \mathcal {L}_{oc}(G)$
, the subgroup
$\varphi (L)$
is open in H and so
$H_0\subseteq \varphi (L)$
. Then,
$H_0\subseteq \bigcap _{L\in \mathcal {L}_{oc}(G)} \varphi (L)$
. Let now
$\Omega $
be an open subset of
$\mathcal {L}_c(G)$
. For
${L\in \mathcal {L}_{oc}(G)}$
, in view of Proposition 3.2, the map
$\varphi _*\colon \mathcal {L}(L)\to \mathcal {L}(\varphi (L))=\varphi _*(\mathcal {L}(L))$
is open and therefore,
$\varphi _*(\mathcal {L}(L)\cap \Omega )$
is open in
$\mathcal {L}(\varphi (L))$
and so, by Proposition 3.5(3), in
$\mathcal {L}(H)$
since
$\varphi (L)$
is open in H. Thus,
$\mathcal {L}(\varphi (L))$
is open in
$\mathcal {L}_c(H)$
. As
${\Omega =\bigcup _{L\in \mathcal {L}_{oc}(G)} (\mathcal {L}(L)\cap \Omega )}$
, it follows that
$\varphi _*(\Omega )= \bigcup _{L\in \mathcal {L}_{oc}(G)} \varphi _*(\mathcal {L}(L)\cap \Omega )$
is open in
$\mathcal {L}_c(H)$
, which proves the openness of
$\varphi _*$
.
Since
$\mathcal {L}_{oc}(G)\ne \varnothing $
, the implication (2)
$\Rightarrow $
(3) is trivial.
(3)
$\Rightarrow $
(1). Since
$\mathcal {L}(L)$
is open in
$\mathcal {L}_c(G)$
,
$\varphi _*(\mathcal {L}(L))=\mathcal {L}(\varphi (L))$
is open in
$\mathcal {L}_c(H)$
and so in
$\mathcal {L}(H)$
(Proposition 3.9(3)). By Proposition 3.9(4),
$\varphi (L)$
is open in H and the result follows from Theorem 3.1.
In particular, we deduce the following corollary.
Corollary 3.12. Let G be a locally compact group with compact identity component, H a locally compact totally disconnected group and
$\varphi \colon G\to H$
a continuous morphism. Then,
$\varphi $
is open if and only if
$\varphi _*$
is open.
Remark 3.13. This result was proved by Cowling et al. for the case of profinite groups [Reference Cowling, Hofmann and Morris1, Corollary 9].
