Toral subalgebra explained

In mathematics, a toral subalgebra is a Lie subalgebra of a general linear Lie algebra all of whose elements are semisimple (or diagonalizable over an algebraically closed field). Equivalently, a Lie algebra is toral if it contains no nonzero nilpotent elements. Over an algebraically closed field, every toral Lie algebra is abelian;^[1] thus, its elements are simultaneously diagonalizable.

In semisimple and reductive Lie algebras

A subalgebra

ak h of a semisimple Lie algebra ak g is called toral if the adjoint representation of ak h on ak g , \operatorname{ad}( ak h) \subset ak{gl}( ak g) is a toral subalgebra. A maximal toral Lie subalgebra of a finite-dimensional semisimple Lie algebra, or more generally of a finite-dimensional reductive Lie algebra, over an algebraically closed field of characteristic 0 is a Cartan subalgebra and vice versa. In particular, a maximal toral Lie subalgebra in this setting is self-normalizing, coincides with its centralizer, and the Killing form of ak g restricted to ak h is nondegenerate.

For more general Lie algebras, a Cartan subalgebra may differ from a maximal toral subalgebra.

In a finite-dimensional semisimple Lie algebra

ak g over an algebraically closed field of a characteristic zero, a toral subalgebra exists. In fact, if ak g has only nilpotent elements, then it is nilpotent (Engel's theorem), but then its Killing form is identically zero, contradicting semisimplicity. Hence, ak g must have a nonzero semisimple element, say x; the linear span of x is then a toral subalgebra.

See also

• Maximal torus, in the theory of Lie groups

Notes and References

1. Proof (from Humphreys): Let x \in ak{h} . Since \operatorname{ad}(x) is diagonalizable, it is enough to show the eigenvalues of \operatorname{ad} _{ak{h} }(x) are all zero. Let y \in ak{h} be an eigenvector of \operatorname{ad} _{ak{h} }(x) with eigenvalue λ . Then x is a sum of eigenvectors of \operatorname{ad} _{ak{h} }(y) and then -λ y = \operatorname{ad} _{ak{h} }(y)x is a linear combination of eigenvectors of \operatorname{ad} _{ak{h} }(y) with nonzero eigenvalues. But, unless λ = 0 , we have that -λ y is an eigenvector of \operatorname{ad} _{ak{h} }(y) with eigenvalue zero, a contradiction. Thus, λ = 0 . \square