Toral_subalgebra Knowpia

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 ${\mathfrak {h}}$ of a semisimple Lie algebra ${\mathfrak {g}}$ is called toral if the adjoint representation of ${\mathfrak {h}}$ on ${\mathfrak {g}}$ , $\operatorname {ad} ({\mathfrak {h}})\subset {\mathfrak {gl}}({\mathfrak {g}})$ 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 ${\mathfrak {g}}$ restricted to ${\mathfrak {h}}$ is nondegenerate.

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

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

References

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

[Hum-1] Humphreys 1972, Ch. II, § 8.1.

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

[3] Humphreys 1972, Ch. IV, § 15.3. Corollary

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Summary

In semisimple and reductive Lie algebras

See also

References