# Lie algebra of groups' element centralizer

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 $$\operatorname{Ad}_g$$?

## 1 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 $$\operatorname{Ad}_g(x) = x$$.