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Exercise 16.2 from Daniel Bump - Lie Groups.

Let $G$ be a compact connected Lie group and let $g\in G$. Show that the centralizer $C_{G(g)}$ of $g$ is connected.

I have some problems verifying this, i have tried to use that in that case the exponential map is surjective, and things of maximal torus, but i have not achieved it, i would appreciate some answer.

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Observe that $C(g)=C(<g>)$ i.e the centralizer of an element is the centralizer of the subgroup generated by it. Next observe that $C(H)=C(\overline{H})$ for any subgroup of $H\subset G$. This means that the centralizer of $g$ is equal to the centralizer of $\overline{<g>}$.

$\overline{<g>}$ is a compact abelian subgroup of $G$. If it's connected then its a torus. For a torus we have the follwing

Theorem 16.6( Daniel Bump - Lie Groups.). Let $G$ be a compact connected Lie group and $S \subset G$ a torus (not necessarily maximal). Then the centralizer $C_{G}(S)$ is a closed connected Lie subgroup of $G$.

Every element of a compact connected lie group is contained in a maximal torus. And for a torus we have:

Corollary 15.1( Daniel Bump - Lie Groups.). Each compact torus $T$ has a generator. Indeed, generators are dense in $T$ .

This means that the assertion is true for a dense set in $G$.

In general I dont think its true. A counter example will need to be a an element with a discrete closure subgroup. There are no counter examples in the unitary group as elements commute if they have the same eigenspaces. We can diagonelize a matrix and degenerate it’s eagenvalues to 1 while preserving the eigenspaces. This means the assertion is true in the unitary group.

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  • $\begingroup$ In "Theodore Frankel - The geometry of physics, chapter 20.4b. Averaging over a Compact Group " we have that every compact Lie group can be consider as a subgroup of the unitary group, in that case we could make a proof. Thank you. $\endgroup$ – Casko Bain May 24 at 14:52
  • $\begingroup$ @CaskoBain I was thinking about this. The problem is,that our path to the identity might not stay in our subgroup $\endgroup$ – Elad May 24 at 14:57
  • $\begingroup$ @CaskoBain I have also seen somewhere that the assertion is true for simply connected compact Lie group. Page 101 in A.Borel Semi simple groups and riemannian symmetric space. I didn’t have the background to understand the proof so I don’t know where it uses the fact that it’s simply connected. $\endgroup$ – Elad May 24 at 15:11

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