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In the article cited below, I. Castellano gives a proof for the following result (Proposition 4.1).

Let $G$ be a compactly generated totally disconnected locally compact group. Suppose that $G$ acts discretely on a tree $\mathcal T$ such that

  1. the group $G$ is acting without edge inversions;
  2. the quotient graph $G\backslash\mathcal T$ is finite;
  3. the edge stabilisers $G_e$ are compact open subgroups of $G$.

Then the vertex stabilizers $G_v$ are compactly generated.

Here acting discretely means that the stabilizers are open subgroups of $G$.

The proof as cited is rather involved and uses cohomology arguments, while the proposition looks rather innocent. Is there a more straightforward, elementary proof for this result? Note that the proposition should also hold for trees that are not locally finite (and in this case, the topology becomes a lot more subtle).

[1] Castellano, I., Rational discrete first degree cohomology for totally disconnected locally compact groups. arXiv:1506.02310 [math.GR].

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  • 3
    $\begingroup$ Yes I think it can be proved using plain Bass-Serre theory. I have to think about details... $\endgroup$ – YCor Sep 10 '19 at 20:14
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As YCor suggests, proceed by Bass–Serre theory. We can write $G$ in the form

$\frac{G_{v_1} \ast \dots \ast G_{v_m} \ast F(E)}{\langle \langle \overline{e}\alpha_e(g)e\alpha_{\overline{e}}(g)^{-1} \; (g \in G_e), \; e\overline{e}, \; e \; (e \in E') \rangle \rangle}$

where $E$ is a set of representatives for the edges in the quotient graph, $E' \subseteq E$ is a set of representatives for the edges in a spanning tree of the quotient graph, and $\alpha_e: G_e \rightarrow G_{o(e)}$ is the natural embedding of an edge group into the vertex group of the origin of the edge. Note that $E$ is finite and the Bass–Serre relations are gluing together vertex groups along compact open subgroups $G_e$. (This also ensures that everything is fine in terms of the group topology: you can just start with the topology of one of the edge groups, and then as a group topology, it extends uniquely to the whole of $G$.) Write $F$ for the image of $F(E)$ in $G$.

Let's look at $G_{v_1}$. We have a compactly generated open subgroup $H_0$ of $G_{v_1}$ generated by $\alpha_e(G_e)$ for all $e \in E$ such that $o(e)=v_1$. In other words, all the amalgamation of $G_{v_1}$ with the other vertex groups happens inside $H_0$.

Now write $G_{v_1}$ as a directed union $\bigcup_{i \in I}H_i$, where each $H_i$ is compactly generated and $0$ is the least element of $I$. Let $K_i$ be the subgroup of $G$ generated by $H_i \cup G_{v_2} \cup G_{v_3}\cup\dots\cup G_{v_n} \cup F$. Then $G = \bigcup_{i \in I}K_i$; since $G$ is compactly generated, in fact $G = K_i$ for some $i$. In particular, every $g \in G_{v_1}$ can be expressed as a product of elements of $H_i \cup G_{v_2} \cup G_{v_3}\cup\dots\cup G_{v_n} \cup F$. Using the normal form theorem for graphs of groups, we conclude that $g \in H_i$. Thus $G_{v_1}$ is compactly generated.

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  • 1
    $\begingroup$ Thank you! This is indeed more natural. I will try to generalise this method to groups acting on polyhedral complexes (using complexes of groups in the sense of Bridson and Haefliger). $\endgroup$ – Jens Bossaert Oct 10 '19 at 12:47

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