Toral subalgebra

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).[1] 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][2] thus, its elements are simultaneously diagonalizable.

In semisimple and reductive Lie algebras

A subalgebra ${\displaystyle {\mathfrak {h}}}$ of a semisimple Lie algebra ${\displaystyle {\mathfrak {g}}}$ is called toral if the adjoint representation of ${\displaystyle {\mathfrak {h}}}$ on ${\displaystyle {\mathfrak {g}}}$, ${\displaystyle \operatorname {ad} ({\mathfrak {h}})\subset {\mathfrak {gl}}({\mathfrak {g}})}$ 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,^[citation needed] over an algebraically closed field of characteristic 0 is a Cartan subalgebra and vice versa.[3] In particular, a maximal toral Lie subalgebra in this setting is self-normalizing, coincides with its centralizer, and the Killing form of ${\displaystyle {\mathfrak {g}}}$ restricted to ${\displaystyle {\mathfrak {h}}}$ is nondegenerate.

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

In a finite-dimensional semisimple Lie algebra ${\displaystyle {\mathfrak {g}}}$ over an algebraically closed field of a characteristic zero, a toral subalgebra exists.[1] In fact, if ${\displaystyle {\mathfrak {g}}}$ has only nilpotent elements, then it is nilpotent (Engel's theorem), but then its Killing form is identically zero, contradicting semisimplicity. Hence, ${\displaystyle {\mathfrak {g}}}$ 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

References

1. Humphreys 1972, Ch. II, § 8.1.
2. Proof (from Humphreys): Let ${\displaystyle x\in {\mathfrak {h}}}$. Since ${\displaystyle \operatorname {ad} (x)}$ is diagonalizable, it is enough to show the eigenvalues of ${\displaystyle \operatorname {ad} _{\mathfrak {h}}(x)}$ are all zero. Let ${\displaystyle y\in {\mathfrak {h}}}$ be an eigenvector of ${\displaystyle \operatorname {ad} _{\mathfrak {h}}(x)}$ with eigenvalue ${\displaystyle \lambda }$. Then ${\displaystyle x}$ is a sum of eigenvectors of ${\displaystyle \operatorname {ad} _{\mathfrak {h}}(y)}$ and then ${\displaystyle -\lambda y=\operatorname {ad} _{\mathfrak {h}}(y)x}$ is a linear combination of eigenvectors of ${\displaystyle \operatorname {ad} _{\mathfrak {h}}(y)}$ with nonzero eigenvalues. But, unless ${\displaystyle \lambda =0}$, we have that ${\displaystyle -\lambda y}$ is an eigenvector of ${\displaystyle \operatorname {ad} _{\mathfrak {h}}(y)}$ with eigenvalue zero, a contradiction. Thus, ${\displaystyle \lambda =0}$. ${\displaystyle \square }$
3. Humphreys 1972, Ch. IV, § 15.3. Corollary

• Borel, Armand (1991), Linear algebraic groups, Graduate Texts in Mathematics, vol. 126 (2nd ed.), Berlin, New York: Springer-Verlag, ISBN 978-0-387-97370-8, MR 1102012
• Humphreys, James E. (1972), Introduction to Lie Algebras and Representation Theory, Berlin, New York: Springer-Verlag, ISBN 978-0-387-90053-7