Let $G$ be a Lie group. Let $Z_G(g)$ be a centralizer of $g \in G$. Prove that $Z_G(g)$ is an closed Lie subgroup of $G$ and prove that Lie algebra of $Z_G(g)$ is $\mathfrak z = \{x \in \mathfrak g \mid \operatorname{Ad}_g(x) = x\}$.
The first part is easy since we can show that $Z_G(g)$ is a stabilizer of $g$ of action of $G$ on itself by conjugations ($h \mapsto hgh^1$).
The second part is trickier. Let $x \in \mathfrak g$ be such that $\operatorname{Ad}_g (x) = x$. If $\phi : G \rightarrow G$ ($h \mapsto ghg^-1$), then we can use functoriality of exponential map
$\require{AMScd}$
$$\begin{CD}
\mathfrak g @>\operatorname{Ad}_g>> \mathfrak g \\
@V\operatorname{exp}VV @VV\operatorname{exp}V \\
G @>>\phi> G
\end{CD}$$
to deduce that $\operatorname{exp}(x) \in Z_G(g)$. But how can I see that $\mathfrak z$ is exactly fixed points of $\operatorname{Ad}_g$?
Best Answer
Suppose that $x\in \mathfrak{z}$ for every real $t$, $g\exp(tx)g^{-1} = \exp(tx)$. This implies that $\left.\frac{d}{dt}\right|_{t=0} g\exp(tx)g^{-1} = \left.\frac{d}{dt}\right|_{t=0}\exp(tx)$. This is equivalent to saying that $\operatorname{Ad}_g(x) = x$.