Kernel of the group adjoint representation - Imported Knowlpedia Content

Let $G$ be a Lie group with Lie algebra $\mathfrak g$ , and let

\mathrm{Ad}:G\to \mathrm{Aut}(\mathfrak g)

be the adjoint representation, obtained by differentiating conjugation.

Lemma (Kernel of Ad).

For any Lie group $G$ , the center satisfies $Z(G)\subseteq \ker(\mathrm{Ad})$ .
If $G$ is connected, then $\ker(\mathrm{Ad})=Z(G)$ .

More generally, for arbitrary (not necessarily connected) $G$ , the kernel of $\mathrm{Ad}$ equals the centralizer of the identity component $G^\circ$ .

Remarks

Proof idea (connected case). If $\mathrm{Ad}_g=\mathrm{id}$ , then conjugation by $g$ has derivative equal to the identity at $e$ , hence acts trivially on $\mathfrak g$ . This forces conjugation by $g$ to fix a neighborhood of $e$ (via the exponential chart from the exponential map), and since $G$ is generated by any neighborhood of $e$ when connected, $g$ commutes with all of $G$ .

Context. This lemma relates the kernel of the adjoint representation to the center. In particular, for connected $G$ , $\mathrm{Ad}$ is faithful exactly when $Z(G)$ is trivial, while it has discrete kernel exactly when $Z(G)$ is discrete; compare the discrete-kernel criterion. The infinitesimal analogue is $\ker(\mathrm{ad})=Z(\mathfrak g)$ .