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I am looking for examples of totally disconnected, locally compact groups, which are not sigma-compact. For a start any such an example would do, so that I can a feeling for those groups and how to find them. (In fact I was not very successful so far in finding appropriate examples. Discrete uncountable groups do not count...)

In particular I am looking for such groups $G$, which are algebraically a (semi)direct product of subgroups $N$ and $H$, with the topology being different to the product topology.

Note that sigma-compactness is an obstruction to such an example of a group topology, which is the reason for this particular restriction.

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You should look at my paper with Mikhail Ershov and Thomas Weigel about commensurators of profinite groups http://arxiv.org/PS_cache/arxiv/pdf/0810/0810.2060v1.pdf.

Edit: Let $G$ be a profinite group. There are two natural ways to associate a topology with $\rm{Comm}(G)$. With the Aut-topology you take the biggest topology such that the homomorphisms of the automorphism group an open subgroups of $G$ to $\rm{Comm}(G)$ are continues for all open subgroups. You do not always get a topological group and if you do, its properties are not always nice. But many times it does coincide with a natural topolgoy that exists. The other topology is the strong-topolgy in which you force the image of $G$ in $\rm{Comm}(G)$ to have the quotient topolgy and to be an open subgroup in $\rm{Comm}(G)$. Now, you always do get a locally compact topological groups, but it is not always $\sigma$-compact. (Look at sections 7 and 8 of the paper.)

Let me give two examples: 1. Take $G=\mathbb{Z}_p$, then $\rm{Comm}(G)=\mathbb{Q}^{*}_p$. The Aut-topology yields the natural topology while the strong topolgy yields the discrete topology. 2. Take $G=\rm{PSL}_n(\mathbb{F}_p[[t]])$, then $\rm{Comm}(G)=\rm{PSL}_n(\mathbb{F}_p((t))) \rtimes \rm{Aut}(\mathbb{F}_p((t)))$. In the Aut-topology $\rm{PSL}_n(\mathbb{F}_p[[t]]) \rtimes \rm{Aut}(\mathbb{F}_p((t)))$ is an open subgroup. However, in the strong-topology $\rm{PSL}_n(\mathbb{F}_p[[t]])$ is an open subgroup and as Colin mentioned below $\rm{Comm}(G)$ is not $\sigma$-compact.

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This looks interesting. Thank you for the hint. –  Abel Stolz Mar 30 '11 at 11:19
    
@Yiftach: Could you elaborate on how you would show that some of the classes of groups occuring in your paper are not $\sigma$-compact? –  Tom De Medts Mar 30 '11 at 11:39
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Take a profinite group $G$ with $VZ(G)=1$ such that $Out(G)$ is uncountable. Then $Comm(G)_S$ is not $\sigma$-compact because the embedded copy of $Aut(G)$ is not $\sigma$-compact (topologically it's a disconnected union of uncountably many copies of $G$). Conversely if $L$ is a non-$\sigma$-compact t.d.l.c. group with a countably based open compact subgroup $U$, then some open subgroup of $U$ must have non-$\sigma$-compact normaliser, because there are only countably many pairs of open subgroups of $U$. –  Colin Reid Mar 30 '11 at 17:48
    
BTW, the paper should soon appear in TAMS. So we have just fixed quite a lot of typos that are still in the current version. –  Yiftach Barnea Mar 30 '11 at 21:07
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Unless I am missing something, a very large class of examples is furnished by uncountable discrete groups.

Edit: I missed that the OP does not want such examples. You could of course modify such examples to get other examples, e.g. by considering products $G \times H$ where $G$ is uncountable discrete and $H$ is compact and totally disconnected. Probably these examples are not very exciting either...

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The OP says he does not want to consider such examples in the end of the first paragraph. –  Qiaochu Yuan Mar 30 '11 at 16:29
    
Ah, that's what I was missing... –  Pete L. Clark Mar 30 '11 at 17:18
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