I'm studying the Creutz-Shalom paper on the normal subgroup theorem for dense commensurators of lattices. The authors state their results for locally compact and compactly generated groups. I do not have a good intuitive idea of these objects. My naive thinking would be that in compactly generated groups there is a compact neighbourhood of the identity. Are there easy examples that help distinguish the two concepts? Thank you very much for your time.
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3$\begingroup$ An example (compactly generated but not locally compact) is a dense free subgroup in a connected Lie group, with the induced topology. I think the concept is of limited interest (esp. in the context the authors you mention). Some authors, however, implicitly mean locally compact when they say "compactly generated". $\endgroup$– YCorCommented May 13, 2014 at 18:47
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2 Answers
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Functional analysis abounds with infinite dimensional (and so not locally compact) locally convex spaces which (as topological groups under addition) are compactly generated. For example the dual of an infinite dimensional Banach space with the weak $\ast$ topology. The unit ball is compact.
Edit in view of the comment below. It seems that there is some ambiguity in the phrase "compactly generated" in the OP. I assumed from the context that it was used in the sense of topological group theory, i. e., that there is a compact subset which generates the group algebraically. If it is intended to mean compactly generated as a topological space, then the same example works---we just take the topology of uniform convergence on the compact subsets of the Banach space rather than the weak $\ast$ topology. This is compactly generated à la Kelley (Banach-Dieudonné theorem).
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2$\begingroup$ So, as an example: an infinite-dimensional Hilbert space in its weak topology is compactly generated, but not locally compact (and of course not completely metrizable). I do not think this deserves a downvote, since this is an "easy example" as requested. $\endgroup$ Commented May 13, 2014 at 15:28
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$\begingroup$ According to Kelley's definition, a compactly generated Hausdorff topological space is one where a set is closed if and only if its intersection with every compact set is closed. Particular cases are locally compact Hausdorff spaces and metrizable spaces. So, in particular all Banach and Fréchet spaces are compactly generated. Why would the weak* dual of a Banach space be compactly generated? It is of course $\sigma$ compact but I do not see the argument for being compactly generated. $\endgroup$– TaQCommented May 14, 2014 at 6:19
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1$\begingroup$ The phrase "compactly generated" in the OP was clearly intended in the sense of topological group theory, i.e., that there is a compact subset which generates the whole group algebraically. $\endgroup$– couperinCommented May 14, 2014 at 7:27
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Edit: $(\mathbb{Q},+)$ with the order topology is compactly generated (by the set $\{1/n \mid n \in \mathbb{N}\} \cup \{0\}$) but not locally compact. I would say this is the 'easiest' example that answers the OP's question. (It's also a good example of an infinite totally disconnected group with no proper open subgroups, which is something that cannot occur for locally compact groups.) The groups mentioned by YCor in his comment have similar topological properties.
Old answer: If 'compactly generated' is used in the loose sense that there is a compact subset that generates a dense subgroup, then all bets are off. However, if there is a generating set in the strict sense that is compact, then we can at least say 'completely metrisable $\Rightarrow$ locally compact':
We may assume that the compact generating set $X$ contains inverses (since the set of inverses of $X$ is also compact), and we also note that $X^n$ is compact for all $n$. Thus the group $G$ is a union of a countable ascending chain of compact sets. One of these has non-empty interior, by the Baire category theorem.
