Centralizer of one element on a compact connected Lie group

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.

Observe that $$C(g)=C()$$ 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 $$\overline{}$$.

$$\overline{}$$ 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.

• 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. – Casko Bain May 24 at 14:52