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We know that if $G$ is a topological group that contains a torsion element and $G$ satisfies additional conditions such as $G$ discrete or $G$ finite-dimensional, then the classifying space $BG$ is infinite-dimensional.

My question: can we remove the additional condition such that the statement still holds? Namely, is there any counterexample for $G$ infinite-dimensional?

Edit: Sorry, G contains a torsion element should be G is not torsion-free. Thank you for your thoughtful inputs.

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  • $\begingroup$ I think that $1_G$ is a torsion element. However "$G$ is not torsion-free" would be a very standard way to say that $G$ has no torsion element other than $1$. $\endgroup$
    – YCor
    Commented Nov 17, 2021 at 6:10
  • $\begingroup$ If there is a normal subgroup consisting of torsion elements, then maybe you can still conclude some infiniteness via the corresponding bundle. Consider the situation for $S^1$ and any of its cyclic subgroups. I'm not sure how necessary normal is here. $\endgroup$
    – ThorbenK
    Commented Nov 17, 2021 at 6:42
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    $\begingroup$ Can $G$ be infinite dimensional but $BG$ be finite dimensional? $\endgroup$
    – markvs
    Commented Nov 17, 2021 at 6:45
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    $\begingroup$ @markvs Every space is weakly equivalent to the classifying space of some group and the corresponding group is weakly equivalent to the loop space of that space. So start with $S^n$ which is finite dimensional and then since the loop space of the sphere has homology in infinitely many degrees the corresponding group will be infinite dimensional. $\endgroup$
    – ThorbenK
    Commented Nov 17, 2021 at 7:01
  • $\begingroup$ @ThorbenK Sorry for my ignorance. It seems that you invoked some proposition that any $\mathbb E_1$-group is given by some topological group. Do you have a reference for this? $\endgroup$
    – Z. M
    Commented Nov 17, 2021 at 8:03

2 Answers 2

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Here is sort of a canonical example.

Consider $GL(\mathbb H)$ the group of invertible operators on a Hilbert space. By Kuipers theorem it is contractible. But $GL(\mathbb H)$ acts freely and properly on itself. Hence the classifying space is $BGL(\mathbb H)=GL(\mathbb H)/GL(\mathbb H)$ a point. Consequences are that Hilbert bundles are trivial over paracompact spaces, K-theory is represented by Fredholm operators etc.

Of course $GL(\mathbb H)$ contains a lot of torsion. Any finite group is a subgroup of $GL(\mathbb H)$ for example.

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Let $A$ be a discrete group with a torsion element. Let $EA$ be the geometric realization of the action groupoid of $A$ acting on itself by left multiplication. $EA$ is the geometric realization of a simplicial group, so it is a topological group. It is well-known that $EA$ is a contractible space, and therefore the classifying space of $EA$ is contractible too. Because $A$ has torsion elements, $EA$ has torsion elements too.

If you want an example where the group itself is not contractible, let $G$ be any group such that $BG$ is homotopy equivalent to a finite space. Then $G\times EA$ is a topological group with torsion elements, and its classifying space is still homotopy finite.

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