1
$\begingroup$

Let $G$ be a compact Lie group acting effectively on a simply connected space $X$. Let Homeo$(X)$ be the group of all homeomorphisms of $X$ with itself given the compact open topology.

Is the topology on $G$ the same as the subspace topology it inherits from Homeo$(X)$? If not then is one topology finer than the other?

I'm not quite sure where to begin here. If I can somehow show that $G$ is closed in Homeo$(X)$ then I will be done. But I am unsure how to proceed (or even if it is true).

My ultimate goal is to see if $G$ is compact as a subspace of Homeo$(X)$. So even if I can show that the subspace topology on $G$ is contained in the topology of $G$ then it is enough (I'm not sure how to prove this either). However I was wondering if the two topologies are equal?

$\endgroup$
1
$\begingroup$

Thanks to the answer by Eric Wofsey to another similar question of mine, I figured out the answer to this question.

The two topologies agree if $X$ is locally compact and Hausdorff.

If $X$ is locally compact then by the proposition A.14 here we have that $G\hookrightarrow\operatorname{Homeo}(X)$ is continuous. Now since a continuous, injective map from a compact space to a Hausdorff space is a homeomorphism onto its image, we are done.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.