Lie subalgebra

A Lie subalgebra is a vector subspace closed under the Lie bracket.

A maximal Lie subalgebra $\mathfrak{h}$ of $\mathfrak{g}$ is a proper Lie subalgebra with no intermediate proper Lie subalgebra. A useful sufficient criterion is that the quotient representation $\mathfrak{g}/\mathfrak{h}$ be an irreducible representation under the adjoint action of $\mathfrak{h}$: any intermediate Lie subalgebra gives an invariant subspace of this quotient, hence must be either $\mathfrak{h}$ or $\mathfrak{g}$. For example, $\mathfrak{so}(p)\oplus \mathfrak{so}(q)\subset \mathfrak{so}(p+q)$ has quotient given by the external tensor product of group representations of the two defining modules. For $p,q\ge 3$ these are complex irreducible representations, making the inclusion maximal.

The centralizer of a Lie subalgebra $\mathfrak{a}$ is the intersection of the centralizer of an element of a Lie algebra as that element ranges over $\mathfrak{a}$. It is a vector subspace and a Lie subalgebra: for $x,y$ commuting with $\mathfrak{a}$, the Jacobi identity gives $[[x,y],a]=[x,[y,a]]-[y,[x,a]]=0$. It is contained in the normalizer of a Lie subalgebra. If $\mathfrak{a}$ is an abelian Lie algebra, it is contained in its own centralizer.

A toral subalgebra of a Lie algebra $\mathfrak{g}$ is a Lie subalgebra $\mathfrak{a}$ such that $\mathrm{ad}a$ is diagonalizable for every $a\in \mathfrak{a}$. Here diagonalizability is over the stated ground field, and refers to the Adjoint representation. In arguments that use a simultaneous weight decomposition, commutativity of $\mathfrak{a}$ is stated as well. For a semisimple Lie algebra over an algebraically closed field of characteristic zero, Cartan subalgebras are examples. Semisimplicity on the defining representation of a symplectic Lie algebra follows from adjoint semisimplicity in the symplectic Lie algebra.

The normalizer of a Lie subalgebra $H\subseteq L$ isIt is the largest Lie subalgebra of $L$ in which $H$ is an ideal of a Lie algebra.

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 $\mathrm{ad}a$ 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.