Let $G$ be a connected compact Lie group, consider the left/right action on itself. For any finite $A\subset G$, consider the centralizer $Z_G(A):=\{g\in G| a g= g a\}$.
Q: is $Z_G(A)$ a connected subgroup?
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What is $A$? Is it a subset? Is it a subgroup? Is it an abelian subgroup?
In the first two cases, the answer is no: the centralizer of the binary icosahedral group inside $SU(2)$ is $Z(SU(2))=\mathbb Z/2$. If $A$ is required to be abelian, then it again doesn't work: consider a subgroup of order two inside $SO(3)$. Its centraliser is $ \mathbb Z/2 \ltimes S^1$.
If $A$ is abelian and $G$ is also assumed to be simply connected, then it might be the case that the centralizer is always connected, but I'm not sure.
answered Sep 26, 2014 at 16:22
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$\begingroup$ Concerning the "simply connected" question, this only makes sense if $G$ is assumed semisimple. Though I'm not sure how far the compact Lie group literature goes, I also suspect that centralizers in that case are connected (in the standard topology). This is certainly true in the often-parallel case of a semisimple algebraic group over an algebraically closed field when you are given a semisimple element (automatic for a compact Lie group), by Steinberg and Springer. There too the (Zariski) connectedness may fail when the group is not simply connected. $\endgroup$ Sep 27, 2014 at 17:25
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$\begingroup$ For the case where $A$ is Abelian and $G$ is simply connected, if $\alpha$ is a short root of $\mathrm{Spin}_7$ (or whatever the compact form is called), then I think that the centraliser of $\alpha^\vee(-1)$ is $\mathrm{SU}_2 \times \mathrm{SU}_2 \times \mathrm{PU}_2$. If that is true, and if $\beta$ is "a root in $\mathrm{PU}_2$", then the centraliser of $A = \langle\alpha^\vee(-1), ((1/2)\beta^\vee)(-1)\rangle$ is $\mathrm{SU}_2 \times \mathrm{SU}_2 \times \mathbb Z/2\mathbb Z$. (I'm being a bit vague because, as @JimHumphreys points out, I'm actually thinking of algebraic groups.) $\endgroup$– LSpiceOct 11, 2018 at 18:47