Past exam of the lie algebras course of Part C of the mathematics course of the University of Oxford / 2023 / 3 / a

The normalizer of a Lie subalgebra and the centralizer of a Lie subalgebra areThe second is always contained in the first. Conversely, let $x\in N_{\mathfrak{g}}(\mathfrak{a})$ and $a\in \mathfrak{a}$. Then $[a,x]\in \mathfrak{a}$, and, because $\mathfrak{a}$ is an abelian Lie algebra,By the definition of a toral subalgebra, $\mathrm{ad}a$ is diagonalizable. A diagonalizable linear map $T$ satisfies $\ker T^2=\ker T$: in an eigenbasis, both kernels consist of exactly the eigenvectors whose eigenvalue is zero. Thus $[a,x]=0$. This holds for every $a\in \mathfrak{a}$, so $x\in C_{\mathfrak{g}}(\mathfrak{a})$ andThis is the normalizer-centralizer equality for a toral abelian subalgebra; it does not require $\mathfrak{g}$ to be a semisimple Lie algebra.

Use the general definition of a Cartan subalgebra: a nilpotent Lie algebra that is a Lie subalgebra and equals its normalizer of a Lie subalgebra. In a semisimple Lie algebra, the stated structural fact says every Cartan subalgebra $\mathfrak{h}$ is a toral subalgebra and an abelian Lie algebra. Part (i) therefore givesConversely, suppose $\mathfrak{h}$ is a toral subalgebra and an abelian Lie algebra with $C_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$. An abelian Lie algebra is a nilpotent Lie algebra, because its lower central series of a Lie algebra vanishes at the second term. Part (i) gives $N_{\mathfrak{g}}(\mathfrak{h})=C_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$. Thus $\mathfrak{h}$ satisfies the defining conditions of a Cartan subalgebra, proving