A sufficient condition for a locally compact almost simple group to have open monolith

TL;DR

This paper provides a sufficient condition for a totally disconnected, locally compact group with a topologically simple monolith to have that monolith be open and abstractly simple, advancing understanding of the structure of such groups.

Contribution

It introduces a new criterion that guarantees the openness and simplicity of the monolith in totally disconnected, locally compact groups.

Findings

The monolith is open under the given condition.

The monolith is shown to be abstractly simple.

The paper advances structural understanding of locally compact groups.

Abstract

We obtain a sufficient condition, given a totally disconnected, locally compact group $G$ with a topologically simple monolith $S$, to ensure that $S$ is open in $G$ and abstractly simple.

Equations

$$\mathrm{LC}(G)=\{{\mathrm{C}}_G(K)\mid K\le G,{\mathrm{N}}_G(K)\text{ is open in }G\},$$

$${\mathrm{con}}_G(h):=\{g\in G\mid h^{n}g{h}^{-n}\to 1\text{ as }n\to +\infty \}.$$

Full text

Acknowledgement

The impetus for writing this note was a question of Waltraud Lederle, as well as some questions arising from [1] and from an ongoing project with Alejandra Garrido and David Robertson. I thank Waltraud Lederle and my collaborators for their insightful questions and comments.

We recall some definitions from [1].

Definition**.**

Let $G$ be a totally disconnected, locally compact (t.d.l.c.) group. We say $G$ is expansive if there is a neighbourhood $U$ of the identity in $G$ such that $\bigcap_{g\in G}gUg^{-1}=\{1\}$. The group $G$ is regionally expansive if there is a compactly generated open subgroup $O$ of $G$ such that $O$ is expansive; equivalently, every open subgroup containing $O$ is expansive.

A topological group $G$ is monolithic if there is a unique smallest nontrivial closed normal subgroup of $G$, called the monolith $\mathrm{Mon}(G)$ of $G$. A t.d.l.c. group $G$ is robustly monolithic if it is monolithic and the monolith is nondiscrete, regionally expansive, and topologically simple.

Note that every topologically simple t.d.l.c. group $S$ is expansive, and hence if $S$ is compactly generated, then it is regionally expansive. Thus in the context of topologically simple groups, ‘regionally expansive’ should be considered a generalization of ‘compactly generated’. The definition of ‘robustly monolithic’ allows us to consider a more general situation, where the regionally expansive topologically simple group $S$ is embedded as a closed normal subgroup in some larger t.d.l.c. group $G$, such that $\mathrm{C}_{G}(S)=\{1\}$. It is then natural to ask how complex the quotient $G/S$ can be as a topological group.

Here is a sufficient condition for $G/S$ to be discrete, in other words, for $S$ to be open in $G$; in the situation described, in fact $S$ is abstractly simple.

Theorem 1**.**

Let $G$ be a robustly monolithic t.d.l.c. group. Suppose that $G$ has an open subgroup of the form $K\times L$ where $K$ and $L$ are nontrivial closed subgroups of $G$. Then for every nontrivial subgroup $H$ of $G$ such that $\mathrm{Mon}(G)\leq\mathrm{N}_{G}(H)$, without assuming that $H$ is closed, it follows that $H$ is open in $G$ and contains $\mathrm{Mon}(G)$. In particular, $\mathrm{Mon}(G)$ itself is abstractly simple and open in $G$.

The proof is based on the local structure theory developed in [2], [3] and [1]; we briefly recall the necessary background.

Definition**.**

We define the quasi-centralizer $\mathrm{QC}_{G}(H)$ of a subgroup $H$ of a topological group $G$ to be the set of elements $g\in G$ such that $g$ commutes with some open subgroup of $H$, and write $\mathrm{QZ}(G):=\mathrm{QC}_{G}(G)$. We say $G$ is [A]-semisimple if $\mathrm{QZ}(G)=\{1\}$, and whenever $A$ is an abelian subgroup of $G$ with open normalizer, then $A=\{1\}$.

Given an [A]-semisimple t.d.l.c. group $G$, the (globally defined) centralizer lattice of $G$ is the set

[TABLE]

equipped with the partial order of inclusion of subsets of $G$. Within $\mathrm{LC}(G)$, the (globally defined) decomposition lattice $\mathrm{LD}(G)$ consists of those $K\in\mathrm{LC}(G)$ such that $K\mathrm{C}_{G}(K)$ is open in $G$.

By [1, Proposition 5.1.2], every robustly monolithic t.d.l.c. group $G$ is [A]-semisimple, so the definitions above apply. Note that if $G$ is [A]-semisimple, then so is every open subgroup of $G$.

By construction, given $K\in\mathrm{LC}(G)$, then $K$ is closed in $G$ and $\mathrm{N}_{G}(K)$ is open in $G$. It is shown in [2] that $\mathrm{LC}(G)$ is a Boolean algebra, on which the map $K\mapsto\mathrm{C}_{G}(K)$ is the complementation map. The centralizer lattice is a local invariant of $G$, in the sense that if $O$ is any open subgroup of $G$, then $\mathrm{LC}(G)$ is $O$-equivariantly isomorphic to $\mathrm{LC}(O)$ via the map $K\mapsto K\cap O$. The decomposition lattice is a local invariant of $G$ in the same manner; in particular, it accounts for all direct factors of open subgroups of $G$. The decomposition lattice has the following additional property:

$(*)$ Given $A_{1},\dots,A_{n}\in\mathrm{LD}(G)$ with least upper bound $A$ in $\mathrm{LD}(G)$, and given open subgroups $B_{i}$ of $A_{i}$, then as a subset of $A$, the product $B_{1}B_{2}\dots B_{n}$ is a neighbourhood of the identity.

(To see why $(*)$ holds, note that we can choose compact open subgroups $B_{i}$ of $A_{i}$ that normalize each other, so that $B=B_{1}B_{2}\dots B_{n}$ is a compact subgroup of $G$; the fact that $B$ is an open subgroup of an element of $\mathrm{LD}(G)$ then follows by [2, Theorem 4.5].)

There is a natural action of $G$ on $\mathrm{LC}(G)$ by conjugation, which preserves the partial order and hence the Boolean algebra structure; note also that the stabilizers of this action are open. There is then a corresponding continuous action of $G$ by homeomorphisms on the Stone space $\mathfrak{S}(\mathrm{LC}(G))$ of $\mathrm{LC}(G)$, which is a compact zero-dimensional Hausdorff space. The latter action has useful dynamical properties. Given a group $G$ acting on a topological space $X$, we say the action is minimal if every orbit is dense, and compressible if there is a nonempty open subset $Y$ such that for every nonempty open subset $Z$ of $X$, there is $g\in G$ such that $gY\subseteq Z$.

Lemma 2**.**

Let $G$ be a robustly monolithic t.d.l.c. group and let $\mathcal{A}$ be a $G$-invariant subalgebra of $\mathrm{LC}(G)$. Then the $G$-action on $\mathfrak{S}(\mathcal{A})$ is continuous and the $\mathrm{Mon}(G)$-action is minimal and compressible.

Proof.

The action is continuous because stabilizers of elements of $\mathcal{A}$ are open. By [1, Theorem 7.3.3], the action of $\mathrm{Mon}(G)$ on $\mathfrak{S}(\mathrm{LC}(G))$ is minimal and compressible; these properties pass to any quotient $\mathrm{Mon}(G)$-space. ∎

We can now prove the theorem.

Proof of Theorem 1.

As noted above, $G$ is [A]-semisimple, and hence the centralizer lattice $\mathrm{LC}(G)$ is a Boolean algebra, with $G$-invariant subalgebra $\mathrm{LD}(G)$. The condition that $G$ has an open subgroup that splits nontrivially as a direct product then amounts to the condition that $|\mathrm{LD}(G)|>2$. Applying Stone duality, the $G$-space $X_{C}:=\mathfrak{S}(\mathrm{LC}(G))$ admits a $G$-equivariant quotient space $X_{D}:=\mathfrak{S}(\mathrm{LD}(G))$ with $|X_{D}|>1$.

Let $S=\mathrm{Mon}(G)$. By Lemma 2, the action of $G$ on $X_{D}$ is continuous and the action of $S$ is minimal and compressible. In particular, $S$ acts nontrivially on $X_{D}$. Since $S$ is topologically simple and the action is continuous, in fact $S$ acts faithfully on $X_{D}$, so the action of $G$ on $X_{D}$ is also faithful.

Claim: Let $H$ be a nontrivial subgroup of $G$, such that $\mathrm{N}_{G}(H)$ is open in $G$ and contains $S$ (do not assume that $H$ is closed). Then $H$ is open in $G$ and $S\leq H$.

We begin the proof of the claim with some reductions. By [1, Lemma 5.1.4], the normalizer $\mathrm{N}_{G}(H)$ is robustly monolithic with monolith $S$. Since $\mathrm{N}_{G}(H)$ is open in $G$, it also has an open subgroup that splits nontrivially as a direct product. Thus we may assume without loss of generality that $G=\mathrm{N}_{G}(H)$. Since $H$ and $S$ are normal, we have $[H,S]\leq H\cap S$. Since $\mathrm{C}_{G}(S)=\{1\}$, it follows that $H$ and $S$ have nontrivial intersection; thus we may replace $H$ with $H\cap S$ and assume that $H$ is a normal subgroup of $S$. Since $S$ is topologically simple, it follows that $H$ is dense in $S$. By continuity, we then see that the action of $H$ on $X_{D}$ is minimal and compressible.

Since $|X_{D}|>1$, $X_{D}$ is zero-dimensional, and the action is compressible, there is a nonempty clopen subset $\alpha$ of $X$ and $h\in H$ such that $h\alpha$ is properly contained in $\alpha$. Correspondingly, there is $K\in\mathrm{LD}(G)$ and $h\in H$, such that $hKh^{-1}$ is a subgroup of $K$ that is closed but not open in $K$. It then follows that $K$ has an open subgroup of the form $hKh^{-1}\times L$ where $L=\mathrm{C}_{K}(hKh^{-1})$; in turn, $L$ is a nontrivial element of $\mathrm{LD}(G)$. By [4, Proposition 5.1], we have $\mathrm{con}_{G}(h)\leq H$, where

[TABLE]

By [3, Proposition 6.14], the intersection $L^{*}=\mathrm{con}_{G}(h)\cap L$ is open in $L$. Let $\beta$ be the clopen subset of $X_{D}$ corresponding to $L\in\mathrm{LD}(G)$. Since $X_{D}$ is compact and the action of $H$ on $X_{D}$ is minimal, there are $h_{1},\dots,h_{n}\in H$ such that $X_{D}=\bigcup^{n}_{i=1}h_{i}\beta$. By $(*)$ it follows that $\langle h_{i}L^{*}h^{-1}_{i}\mid 1\leq i\leq n\rangle$ is open in $G$; thus $H$ is open in $G$. In particular, $H$ is closed in $S$; since $H$ is dense in $S$, it follows that $H=S$. This completes the proof of the claim.

The claim applies in particular when $S=H$, so $S$ is open in $G$. Thus the assumption that $\mathrm{N}_{G}(H)$ is open follows automatically from assuming that $\mathrm{N}_{G}(H)$ contains $S$. We have therefore shown that given a nontrivial subgroup $H$ of $G$ such that $S\leq\mathrm{N}_{G}(H)$, then $H$ is open in $G$ and $S\leq H$. As this conclusion applies in particular to any nontrivial normal subgroup of $S$, we see that $S$ is abstractly simple. ∎