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If $X$ is a locally finite graph, (i.e. each vertex has finite index), is it true that the automorphism group Aut($X$) of the graph X is locally compact?

Here, Aut($X$) has compact open topology; and topology of $X$ is the weak topology when we consider $X$ as a CW-complex (see. Hatcher, Algebraic Topology - Graphs and Free Groups).

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    The question refers to automorphisms which are linear in the edges? Since you say "we consider X as a CW-complex", by Aut(X) I would understand the group of self-CW-homeomorphisms of X — but even for the connected graph with two vertices, also known as [0,1], the group of self-CW-homeomorphisms is not locally compact.2011-07-04
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    @Mariano: presumably the OP means graph automorphisms since, as you say, the result is not otherwise true.2011-07-04

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It is true under the additional hypothesis that the locally finite graph is connected. Here the key observation is that the stabilizer of each vertex is compact. For a proof of this, see e.g. Lemma 1 in this paper.

However, local compactness does not hold in general. For instance, if the vertex set is countably infinite and there are no edges, then the automorphism group is just the symmetric group on a countable set and one can show directly from the definition of the compact open topology that the identity does not admit a compact neighborhood.

More generally, local compactness holds if there are finitely many connected components. Perhaps the necessary and sufficient condition is that there be only finitely many connected components $X_i$ such that $\operatorname{Aut} X_i$ is noncompact?

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    Can any of the connected components be a finite graph? (I don't know any algebraic topology to grasp the construction of $\mathrm{Aut}(X)$'s topology so I can't visualize this.)2011-08-25