1 Introduction
Let G be a locally compact group. For
$x\in G$
, let
$C_x = \{ gxg^{-1} \mid g\in G\}$
be the conjugacy class containing x and let
$B(G) = \{ x\in G \mid \overline {C_x} ~\mathrm {is ~~ compact} \}$
. It is easily seen that
$B(G)$
is a characteristic subgroup containing the centre
$Z(G) = \{ x\in G \mid gx =xg\ \mathrm{for~all} g\in G \}$
. In general,
$B(G)$
is not closed. There are locally compact groups G with
$B(G)$
as a proper dense subgroup (see [Reference Tits9, Proposition 3]), but if G is a totally disconnected, locally compact (tdlc) group that is generated by a compact set, then [Reference Möller7, Theorem 2] shows that
$B(G)$
is closed. The subgroup
$B(G)$
plays a crucial role in the location of finite central (positive) measures. A measure is called central if it is invariant under the conjugate action of the group G (see [Reference Greenleaf, Moskowitz and Preiss-Rothschild4, Theorem 1.5]).
A conjugacy class supporting a central measure is in
$B(G)$
for connected Lie groups [Reference Greenleaf, Moskowitz and Preiss-Rothschild5, Theorem 1′]. We prove this result for a certain extension of abelian groups (see Theorem 1.1) and obtain interesting applications (see Corollary 3.1) as in [Reference Greenleaf, Moskowitz and Preiss-Rothschild5, Theorem 3]. We also obtain some general results in the case of tdlc groups (see Section 4).
There is a close connection with finite covolume subgroups (see [Reference Greenleaf, Moskowitz and Preiss-Rothschild4, Reference Greenleaf, Moskowitz and Preiss-Rothschild5]). In this context, we say that a locally compact group has property (M) if, for any closed subgroup H of G with
$G/H$
admitting a finite G-invariant measure,
$G/H$
is compact (see [Reference Bader, Caprace, Gelander and Mozes1, Reference Mostow8] for further details on property (M)). It can be easily seen that abelian groups and compact groups have property (M). We obtain the following result.
Theorem 1.1. Let G be a locally compact group and let
$x\in G$
be such that
${\mu (C_x)=1}$
for some conjugate-invariant finite measure
$\mu $
. If G has a closed abelian normal subgroup A such that
$G/A$
has property (M), then
$C_x$
has compact closure: that is,
$x\in B(G)$
.
In particular, if G is a metabelian locally compact group, then
$C_x$
has compact closure: that is,
$x\in B(G)$
.
Remark 1.2. An example of a metabelian group G having a finite covolume subgroup that is not cocompact is provided in [Reference Bader, Caprace, Gelander and Mozes1]. Hence, groups under consideration in Theorem 1.1 need not have property (M).
2 Automorphisms of bounded displacement
Following [Reference Tits9], a (bicontinuous) automorphism
${\alpha }$
of a locally compact group G is called an automorphism of bounded displacement if
$\{ {\alpha }(x)x^{-1} \mid x\in G \}$
has compact closure. If
${\alpha }$
is an inner automorphism defined by
$x\in G$
(that is,
${\alpha }(g) = xgx^{-1}$
), then
${\alpha }$
is an automorphism of bounded displacement if and only if
$x\in B(G)$
. Thus, in order to identify elements in
$B(G)$
, we study automorphisms of bounded displacement. We obtain the following result.
Proposition 2.1. Let G be a locally compact group containing a closed abelian normal subgroup A such that
$G/A$
has property (M). Suppose that
${\alpha }$
is an automorphism of G such that
$G/H$
has a finite G-invariant measure, where
$H= \{x \in G \mid {\alpha } (x) =x \}$
and
${\alpha } (A)=A$
. Then
${\alpha }$
is an automorphism of bounded displacement.
Our proof relies on certain shift-invariant properties in convolutions of probability measures on locally compact groups. For a locally compact group X, let
${\cal M} (X)$
be the space of all regular Borel probability measures (that is, positive measures with total measure one) on X equipped with the weak* topology: that is,
$\rho _n \to \rho \in {\cal M} (X)$
if
$\int f(x) \,d\rho _n(x) \to \int f(x) \,d\rho (x)$
for all continuous bounded functions f on X. For
$x\in X$
and
$\rho $
in
${\cal M} (X)$
,
$x\rho $
and
$ \rho x \in {\cal M} (X)$
are defined by
$x\rho (E) = \rho (x^{-1}E)$
and
$\rho x(E) = \rho (Ex^{-1})$
for any Borel set E in X. The convolution of two measures
$\mu , {\lambda } \in {\cal M} (X)$
is denoted by
$\mu *{\lambda } $
and is defined by
$\mu *{\lambda } (E) = \int \mu (Ex^{-1}) \,d{\lambda } (x)$
for any Borel set E in X. For any automorphism
${\alpha } $
of X and
$\mu \in {\cal M} (X)$
, define
${\alpha } (\mu ) $
by
${\alpha } (\mu ) (E) = \mu ({\alpha } ^{-1}(E)) $
for any Borel set E in X (see [Reference Heyer6] for more details on probability measures on groups).
Lemma 2.2. Let G be a locally compact group and let A be a closed abelian normal subgroup of G. Suppose that
${\beta } \colon G \to A$
is a continuous map such that
${\beta } (xy) = {\beta } (x) x{\beta } (y) x^{-1}$
for all
$x, y \in G$
. Then
$H = \{ x\in G \mid {\beta } (x) =e\}$
is a closed subgroup. If
$G/H$
has a finite G-invariant measure, then
${\beta } (A)$
is contained in a compact subgroup of A.
Proof. Since
${\beta }$
is continuous, H is a closed subgroup. Assume that
$G/H$
carries a finite G-invariant measure
$\mu $
, say. Normalising
$\mu $
, we may assume that
$\mu $
is a probability measure. For
$x\in G$
and
$h \in H$
, we have
${\beta } (xh) = {\beta } (x)$
. Hence,
${\beta } $
induces a continuous map
$\tilde {\beta } \colon G/H \to A$
defined by
$\tilde {\beta } (xH) = {\beta } (x)$
for all
$x\in G$
. Take
${\lambda } = \tilde {\beta } (\mu )$
. Then
${\lambda }$
is a Borel probability measure on A such that
${\beta } (x) x{\lambda } x^{-1} = {\lambda }$
for all
$x\in G$
. Since A is an abelian group,
${\lambda } ={\beta } (a) a {\lambda } a^{-1}= {\beta } (a) {\lambda }$
. Let
$M=\{ a \in A \mid a {\lambda } = {\lambda } \}$
. Then, by [Reference Heyer6, Theorems 1.2.4 and 1.2.7], M is a compact subgroup of A and
${\beta } (A) \subset M$
.
Proof of Proposition 2.1.
Let
$N= \overline {AH}$
. Then N is a closed
${\alpha }$
-invariant subgroup of G. Since
$H\subset N$
,
$G/N$
has a finite G-invariant measure. Since
$G/N$
is a quotient of
$G/A$
, the assumption that
$G/A$
has property (M) implies that
$G/N$
is compact.
Define
${\beta }$
on
$N $
by
${\beta } (x) = {\alpha } (x) x^{-1}$
for all
$x\in N$
. Since
${\alpha } (x) =x$
for all
$x\in H$
, we have
${\beta } (x) \in A$
. Thus,
${\beta } \colon N \to A$
is a well-defined continuous map. Also,
${\beta } (xy) = {\beta } (x) x{\beta } (y)x^{-1}$
for all
$x, y \in N$
and
$H= \{ x\in N \mid {\beta } (x) =e \}$
. Since
$G/H$
has a finite G-invariant measure and
$H\subset N$
, it follows that
$N/H$
has a finite N-invariant measure. By Lemma 2.2, there is a compact subgroup M of A such that
${\beta } (A)\subset M$
. Since
${\beta } (ah) = {\beta } (a)$
, we have
${\beta } (AH)\subset M$
and, since
$AH$
is dense in N,
${\beta } (N) \subset M$
. Thus,
${\alpha } (x) \subset Mx$
for all
$x\in N$
and M is a compact subgroup of A.
Since
$G/N$
is compact, there is a compact subset C of G such that
$G= CN$
. For
$g\in G$
, there are
$x\in N$
and
$y \in C$
such that
$g = yx$
. This implies that
${\alpha } (g) g^{-1}={\alpha } (y) {\alpha } (x) x^{-1} y^{-1} \subset {\alpha } (C)MC^{-1}$
. Thus,
${\alpha }$
is an automorphism of bounded displacement.
3 Theorem 1.1 and applications
Proof of Theorem 1.1.
Let
${\alpha }$
be the inner automorphism defined by
$x^{-1}$
and define
$H= \{ g \in G \mid gx= xg \}$
. The group G has a canonical action on
$G/H$
on the left and G acts on
$C_x$
by conjugation. The map
$\theta \colon C_x \to G/H$
defined by
$\theta (gxg^{-1}) = gH$
is a well-defined G-equivariant Borel isomorphism and
$\theta (\mu )$
is a finite G-invariant measure on
$G/H$
. It follows from Proposition 2.1 that
${\alpha }$
is an automorphism of bounded displacement. Thus, there is a compact set C such that
${\alpha } (g) \subset Cg$
for all
$g\in G$
. This implies that
$x^{-1}gxg^{-1}\subset C$
for all
$g\in G$
and hence
$gxg^{-1} \subset xC$
for all
$g\in G$
. Thus, the conjugacy class
$C_x$
containing x has compact closure.
The following result on the centraliser of finite covolume subgroups is a kind of density theorem in the sense that any element commuting with a finite covolume subgroup commutes with all elements up to a compact set.
Corollary 3.1. Let G be a locally compact group containing a closed abelian normal subgroup A such that
$G/A$
has property (M) and H is a finite covolume subgroup of G. Then
$Z(H) \subset B(G)$
.
Proof. Let
$x\in Z(H)$
. Then
$Z(x)$
contains H, and hence
$G/Z(x)$
has a finite G-invariant measure. Let
$\eta \colon G/Z(x) \to C_x$
be
$\eta (g) = gxg^{-1}$
. Then
$\eta $
is a well-defined continuous map preserving the G-action. Hence, the conjugacy class of x supports a finite G-invariant measure. Therefore, by Theorem 1.1, the conjugacy class of x has compact closure. Thus,
$Z(H) \subset B(G)$
.
We now look at the case when A, as above, has no compact subgroups. As examples,
${\mathbb R} ^n$
and
${\mathbb Z} ^n$
have no compact subgroups.
Corollary 3.2. Let
$G,H, A$
and
${\alpha }$
be as in Proposition 2.1. Suppose that A has no compact subgroups. Then
$ G/H$
is compact. In particular, if
$x \in G$
is such that
${\mu (C_x)>0} $
for some conjugate-invariant finite measure
$\mu $
, then the corresponding conjugacy class
$C_x$
is compact.
Proof. Let
${\beta } \colon A \to A$
be
${\beta } (a) = {\alpha } (a)a^{-1}$
for all
$a \in A$
. Then
${\beta }$
is a continuous map and
${\beta } (xy) = {\beta }(x)x{\beta }(y)x^{-1} = {\beta }(x){\beta }(y)$
as A is abelian. By Lemma 2.2,
${\beta }(A)$
is contained in a compact subgroup of A. Since A has no compact subgroup,
${\beta }$
is trivial. Thus,
$A \subset H$
and
$G/H$
is a quotient of
$G/A$
. By the assumption that
$G/A$
has property (M),
$G/H$
is compact.
4 Tdlc groups
We now consider tdlc groups. For an automorphism
${\alpha }$
of a tdlc group G, define the
${\alpha }$
-invariant subgroups
$U_{\alpha } = \{ x\in G \mid \lim _{n \to \infty } {\alpha } ^n(x) = e \}$
and
$M_{\alpha } = \{ x\in G \mid \overline {\{ {\alpha } ^n(x) \mid n \in {\mathbb Z} \} } \text { is compact}\}$
, where
$U_{\alpha }$
is called the contraction group of the automorphism
${\alpha }$
(see [Reference Baumgartner and Willis2] for various results on
$U_{\alpha }$
and
$M_{\alpha } $
). In the tdlc case, we obtain the following result.
Proposition 4.1. Let G be a tdlc group and let
${\alpha }$
be an automorphism of G. Suppose that
$G/H$
has a finite G-invariant measure, where
$H = \{x \in G \mid {\alpha } (x) =x \}$
. Then
${\alpha }$
fixes a compact open subgroup of G and
$M_{\alpha }$
is a subgroup of finite index in G.
Proof. Let
$N= \overline {HU_{\alpha }}$
. Since
${\alpha } (x) =x$
for
$x\in H$
, H normalises
$U_{\alpha }$
. Then N is a tdlc group invariant under
${\alpha }$
. Let
${\beta } \colon N/H \to N$
be
${\beta } (xH) = {\alpha } (x) x^{-1}$
for all
$x\in N$
. Since
${\alpha } (x) = x$
for all
$x\in H$
,
${\beta }$
is a well-defined continuous map on
$N/H$
. Since
$N= \overline {HU_{\alpha } }$
and
${\alpha } (x) =x $
for all
$x\in H$
, we have
${\beta } (N)\subset \overline {U_{\alpha }}$
. By [Reference Baumgartner and Willis2, Corollaries 3.27 and 3.30],
$\overline {U_{\alpha }} = U_0U_{\alpha }$
, where
$U_0$
is an
${\alpha }$
-invariant compact subgroup of G. In fact, we have
${U_0 = \overline {U_{\alpha } }\cap \overline {U_{{\alpha } ^{-1}}}}$
. This implies that
${\alpha } ^n(x)U_0 \to U_0$
for all
$x\in \overline {U_{\alpha } }$
.
Let
$\bar {\alpha } \colon N/H \to N/H$
be
$\bar {\alpha } (xH) = {\alpha } (x)H$
for all
$x\in N$
. Since
${\alpha } (H) =H$
,
$\bar {\alpha }$
is a continuous map (in fact,
$\bar {\alpha }$
is an N-equivariant homeomorphism).
For
$x\in N$
,
${\beta } (\bar {\alpha } (xH)) = {\alpha } ^2 (x) {\alpha } (x^{-1}) = {\alpha } ({\beta } (x))$
. Thus,
${\beta } \bar {\alpha } = {\alpha } {\beta }$
.
Since
$H\subset N$
,
$N/H$
has an N-invariant probability measure
$\mu $
, say. Let
${\lambda } = {\beta } (\mu )$
. Since
${\beta }$
is a continuous map,
${\lambda }$
is a probability measure on N. Since
$\mu $
is an N-invariant probability measure and
$\bar {\alpha } (gxH) = {\alpha } (g) \bar {\alpha } (xH)$
for all
$g, x\in G$
, it follows that
$\mu $
is
$\bar {\alpha }$
-invariant. Since
${\beta } \bar {\alpha } = {\alpha } {\beta }$
,
${\lambda }$
is
${\alpha }$
-invariant, and hence
${\lambda } = {\alpha } ^n ({\lambda } ) $
for all n.
Let
$\rho $
be the normalised Haar measure on the compact subgroup
$U_0$
. Since
${\alpha } (U_0) = U_0$
,
$\rho $
is
${\alpha }$
-invariant. Since
${\alpha } ^n(x)U_0 \to U_0$
for all
$x\in \overline {U_{\alpha } }$
, we have
${\alpha } ^n({\lambda } * \rho )\to \rho $
. But
${\alpha } ^n ({\lambda } *\rho ) = {\alpha } ^n ({\lambda } ) *{\alpha } ^n (\rho ) = {\lambda } *\rho $
, so
${\lambda } * \rho =\rho $
. By considering the supports of the measures, we see that
${\lambda }$
is supported on
$U_0$
.
Since
$\mu $
is an N-invariant measure on
$N/H$
, the support of
$\mu $
is N. Since
${\beta } (\mu ) = {\lambda }$
,
$\overline {{\beta } (N)}$
is the support of
${\lambda }$
, and hence
${\beta } (N) \subset U_0$
. This implies that
${\alpha } (x) \subset U_0x$
for all
$x\in U_{\alpha }$
. Let
$x\in U_{\alpha }$
. Then, for each
$n \geq 1$
, we have
${\alpha } ^n (x) = g_nx$
for some
$g_n \in U_0$
. Since
$U_0$
is compact and
${\alpha } ^n(x) \to e$
as
$n \to \infty $
, we have
$x\in U_0$
. Thus,
$U_{\alpha }\subset U_0 \subset \overline {U_{\alpha }}$
, and hence
$\overline {U_{\alpha } }= U_0$
. Similarly, we may show that
$\overline {U_{{\alpha } ^{-1}} }= U_0$
. By [Reference Baumgartner and Willis2, Proposition 3.24],
${\alpha }$
fixes a compact open subgroup K, say.
Since
$M_{\alpha } = \{ x\in G \mid \overline {\{ {\alpha } ^n(x) \mid n \in {\mathbb Z} \} } \text { is compact} \}$
,
$K\subset M_{\alpha }$
, and so
$M_{\alpha }$
is open. Since
${\alpha } (x) = x$
for all
$x\in H$
,
$H\subset M_{\alpha }$
. This implies that
$G/M_{\alpha }$
is a discrete G-space with a finite G-invariant measure and hence
$G/M_{\alpha }$
is finite.
As a consequence, we have the following result on property (M) for tdlc groups.
Corollary 4.2. Let G be a tdlc group and let H be a closed subgroup of G such that
$G/H$
has a finite G-invariant measure. If H is a compactly generated abelian group, then
$G/H$
is compact.
Remark 4.3. For the metabelian counter-example provided in [Reference Bader, Caprace, Gelander and Mozes1], the finite covolume subgroup H, although abelian, is not compactly generated. It may be noted that any compactly generated abelian tdlc group is a direct product of
${\mathbb Z} ^n$
and a compact group.
Proof of Corollary 4.2.
Let H be a compactly generated abelian subgroup of G with finite covolume. For
$x\in H$
, let
${\alpha } _x$
be the inner automorphism defined by x on G: that is,
${\alpha } _x(g) =xgx^{-1}$
for all
$g\in G$
. Since H is abelian,
$H\subset Z(x)$
. Thus,
$G/Z(x)$
has a finite G-invariant measure. By Proposition 4.1, each
$x\in H$
fixes a compact open subgroup in G. Since H is a compactly generated abelian group, [Reference Willis10, Theorem 5.9] implies that H normalises a compact open subgroup K of G. Then
$HK$
is a open subgroup G and hence
$G/HK$
is a discrete G-space with a finite G-invariant measure. Therefore,
$G/HK$
is finite. Thus,
$G/H$
is compact.
We also have the following result for expansive automorphisms. An automorphism
${\alpha }$
of a tdlc group G is called expansive if
$\bigcap {\alpha } ^n(U) = \{ e \}$
for some compact open subgroup U of G (see [Reference Glöckner and Raja3] for more details on expansive automorphisms on tdlc groups).
Corollary 4.4. Let G be a tdlc group and let
${\alpha }$
be an expansive automorphism of G. Suppose that
$G/H$
has a finite G-invariant measure, where
$H = \{x \in G \mid {\alpha } (x) =x \}$
. Then
${\alpha }$
fixes a compact open subgroup of G that is normalised by H and
$G/H$
is compact.
Proof. By Proposition 4.1,
$\overline {U_{\alpha }}= \overline {U_{{\alpha } ^{-1}}} =U_0 = \overline {U_{\alpha } }\cap \overline {U_{{\alpha } ^{-1}}}$
. Since
${\alpha }$
is expansive on G, Lemma 1.1 of [Reference Glöckner and Raja3] implies that
$U_{\alpha } U_{{\alpha } ^{-1}}$
is open, and hence
$U_0$
is open. Since
${\alpha } (x) =x$
for all
$x\in H$
and
$U_0 = \overline {U_{\alpha }}$
, it follows that H normalises
$U_0$
. Since
$G/H$
has a finite G-invariant measure,
$G/HU_0$
is a discrete G-space with a finite G-invariant measure and hence
$G/HU_0$
is finite. Thus,
$G/H$
is compact.
Acknowledgement
I thank the referee for suggestions that improved the exposition and led to the correct version of Corollary 3.2.
