Kernel of ad and the center - Imported Knowlpedia Content

Let $\mathfrak g$ be a Lie algebra. The adjoint representation is the linear map

\mathrm{ad}:\mathfrak g\to \mathfrak{gl}(\mathfrak g),\qquad \mathrm{ad}_x(y)=[x,y].

Lemma.

\ker(\mathrm{ad}) \;=\; Z(\mathfrak g),

where $Z(\mathfrak g)$ is the center of the Lie algebra.

Remarks

Proof. By definition, $x\in\ker(\mathrm{ad})$ iff $\mathrm{ad}_x(y)=0$ for all $y\in\mathfrak g$ , i.e. $[x,y]=0$ for all $y$ . This is exactly the defining condition for $x\in Z(\mathfrak g)$ .

Context. This identifies the failure of $\mathrm{ad}$ to be injective with central directions and clarifies the meaning of inner derivations: $\mathrm{ad}_x$ depends only on the class of $x$ modulo the center. At the group level, a related statement is ker(Ad) equals the group center (connected case).