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I am looking for non-trivial examples of the following:

  • $G$ is a locally compact group
  • $H\subset G$ a closed subgroup
  • Both are unimodular and non-discrete
  • The quotient space $G/H$ is compact, but $G$ is not compact

Trivial cases would be $G=G_1\times G_2$, $H=H_1\times H_2$ with $H_1=G_1$ and $H_2$ being discrete in $G_2$, or $G_2$ being compact. The same goes for semi-direct products instead of direct products.

Examples with $G$ being topologically simple would be nice.

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  • $\begingroup$ To really remove most trivial examples, you'd better assume that $H$ has a trivial core (and that $H$ is nondiscrete). The core of $H$ is the kernel of the $G$-action on $G/H$; this is the largest $G$-normal subgroup of $H$. $\endgroup$
    – YCor
    Commented Oct 10, 2023 at 7:43
  • $\begingroup$ Yes right, normal $H$ was not on my mind, but, out of curiosity, which examples with normal $H$ would you know? $\endgroup$
    – user473423
    Commented Oct 10, 2023 at 8:11
  • $\begingroup$ This is not what I said in my comment. $\endgroup$
    – YCor
    Commented Oct 10, 2023 at 9:06
  • $\begingroup$ No, but it is a special case. But still, which trivial examples did you have in mind? $\endgroup$
    – user473423
    Commented Oct 10, 2023 at 13:47
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    $\begingroup$ The general recipe to get examples with nontrivial core: start from an example $(G,H)$ (possibly $H$ is discrete). Find $G'$ (unimodular) with a quotient homomorphism $p:G'\to G$. Then get the pair $(G',p^{-1}(H))$. $\endgroup$
    – YCor
    Commented Oct 10, 2023 at 13:50

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You can find many such examples among groups acting on trees. Let $T$ be a $k$-regular tree and let $G$ be the subgroup of $\operatorname{Aut}(T)$ of automorphisms stabilizing each of the 2 parts of the bipartition of the vertex set $V(T)$. (This is a simple subgroup of $\operatorname{Aut}(T)$ of index $2$). Note that $G$ is a totally disconnected locally compact group w.r.t. the permutation topology. Fix a coloring $\lambda \colon E(T) \to \{ 1, \dots, k \}$ of the edges of the tree. We use the notation $T_v$ for the star around a vertex $v$, i.e., the collection of $k$ edges having $v$ as one of its end points.

Now fix a finite permutation group $F \leq \operatorname{Sym}(k)$. Then we can consider the Burger-Mozes universal group $$ U(F) = \{ g \in G \mid \lambda_{|T_{g.v}} \circ g \circ \lambda_{|T_v}^{-1} \in F \text{ for all } v \in V(T) \} ; $$ intuitively, this is the largest subgroup of $G$ with prescribed local action equal to $F$. Clearly, it is a closed subgroup.

Then $U(F)$ always acts transitively on each of the two partitions of the vertex set $V(T)$, so it is cocompact. Moreover, it is non-discrete as soon as $F$ does not act freely on $\{ 1,\dots,k \}$, and in addition, it is simple if (and only if) $F$ is transitive and generated by its point stabilizers. In particular, it is unimodular in this case.

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    $\begingroup$ Your definition of $G$ is strange because many elements of the coset $\mathrm{Aut}(G)\smallsetminus T$ have no inversion (e.g., a translation of odd length). $G$ is just the group of elements mapping one/every vertex to even distance. $\endgroup$
    – YCor
    Commented Oct 10, 2023 at 9:10
  • $\begingroup$ @YCor Thanks, I have modified the definition of $G$ (this is indeed what I meant to say). $\endgroup$
    – Tom De Medts
    Commented Oct 10, 2023 at 9:19
  • $\begingroup$ To be explicit, the simplest example seems to be the inclusion $U(F_1)\subseteq U(F_2)$ with $F_2$ the symmetric group on 3 elements, and $F_1$ cyclic of order 2, i.e., generated by a transposition. $\endgroup$
    – YCor
    Commented Oct 10, 2023 at 11:03

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