= Normalizer-centralizer equality for a toral abelian subalgebra
{title2=$N_{\mathfrak g}(\mathfrak a)=C_{\mathfrak g}(\mathfrak a)$}
If $\mathfrak a$ is a <toral subalgebra> and an <abelian Lie algebra>, its <normalizer of a Lie subalgebra> equals its <centralizer of a Lie subalgebra>. For $x$ in the former and $a\in\mathfrak a$, one has $[a,x]\in\mathfrak a$, hence $[a,[a,x]]=0$. A <diagonalizable> <linear map> has the same <kernel> as its square, by expansion in an <eigenbasis>. Applying this to $\operatorname{ad}a$ gives $[a,x]=0$ for every $a$. The reverse containment follows directly from the definitions. No assumption that $\mathfrak g$ is a <semisimple Lie algebra> is needed.
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