Cartan subalgebras are self-normalizing

Statement

Let $\mathfrak g$ be a finite-dimensional complex semisimple Lie algebra, and let $\mathfrak h\subseteq\mathfrak g$ be a Cartan subalgebra, understood here as a maximal toral subalgebra. Then

N_{\mathfrak{g}}(\mathfrak{h})=\{X\in\mathfrak{g}:[X,\mathfrak{h}]\subset \mathfrak{h}\}=\mathfrak{h}.

Proof from the root-space decomposition

Use the root-space decomposition

\mathfrak g=\mathfrak h\oplus\bigoplus_{\alpha\in\Phi}\mathfrak g_\alpha.

Write $X=X_0+\sum_{\alpha\in\Phi}X_\alpha$ , where $X_0\in\mathfrak h$ and $X_\alpha\in\mathfrak g_\alpha$ . If $X$ normalizes $\mathfrak h$ , then for every $H\in\mathfrak h$ ,

[H,X]=\sum_{\alpha\in\Phi}\alpha(H)X_\alpha\in\mathfrak h.

The sum is direct, so $\alpha(H)X_\alpha=0$ for every $H$ . For each root $\alpha$ , some $H$ has $\alpha(H)\ne0$ ; hence $X_\alpha=0$ . Therefore $X=X_0\in\mathfrak h$ .

Convention outside the semisimple case

For a general finite-dimensional Lie algebra over an algebraically closed field of characteristic $0$ , a common definition says that a Cartan subalgebra is a nilpotent, self-normalizing subalgebra. Under that convention, the displayed equality is part of the definition rather than a separate lemma. The substantive semisimple statement above is that the alternative “maximal toral” characterization implies self-normalization.

References

James E. Humphreys, Introduction to Lie Algebras and Representation Theory, Springer, 1972, §8. Publisher record.
Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed., Birkhäuser, 2002, Chapter II. Publisher record.