Center of a simple Lie algebra is trivial

Let $\mathfrak g$ be a simple Lie algebra . Then its center is trivial:

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

Reason. The center $Z(\mathfrak g)=\{z\in\mathfrak g:[z,x]=0\ \forall x\in\mathfrak g\}$ is always an ideal: if $z$ commutes with everything, then so does $[x,z]=0$ for any $x$, and invariance under brackets is automatic (this is a special case of the general “invariance under adjoint action” viewpoint). Since $\mathfrak g$ is simple, $Z(\mathfrak g)$ must be either $0$ or all of $\mathfrak g$. If $Z(\mathfrak g)=\mathfrak g$, then $\mathfrak g$ is abelian, contradicting the definition of simple.

This fact is often used when comparing simplicity to semisimplicity (see semisimple Lie algebra ) and when analyzing the kernel of adjoint representations (compare kernel of ad is the center ).