Let g\mathfrak g be a . Its is

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

Lemma.

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

where Z(g)Z(\mathfrak g) is the of g\mathfrak g.

Remarks

Proof. By definition, xker(ad)x\in\ker(\operatorname{ad}) if and only if [x,y]=0[x,y]=0 for every ygy\in\mathfrak g, which is precisely the condition xZ(g)x\in Z(\mathfrak g).

Thus adx\operatorname{ad}_x depends only on the class of xx modulo the center.