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 [math]\displaystyle{ \mathfrak h }[/math] of a semisimple Lie algebra [math]\displaystyle{ \mathfrak g }[/math] is called toral if the adjoint representation of [math]\displaystyle{ \mathfrak h }[/math] on [math]\displaystyle{ \mathfrak g }[/math], [math]\displaystyle{ \operatorname{ad}(\mathfrak h) \subset \mathfrak{gl}(\mathfrak g) }[/math] 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 [math]\displaystyle{ \mathfrak g }[/math] restricted to [math]\displaystyle{ \mathfrak h }[/math] is nondegenerate.
For more general Lie algebras, a Cartan subalgebra may differ from a maximal toral subalgebra.
In a finite-dimensional semisimple Lie algebra [math]\displaystyle{ \mathfrak g }[/math] over an algebraically closed field of a characteristic zero, a toral subalgebra exists.[1] In fact, if [math]\displaystyle{ \mathfrak g }[/math] has only nilpotent elements, then it is nilpotent (Engel's theorem), but then its Killing form is identically zero, contradicting semisimplicity. Hence, [math]\displaystyle{ \mathfrak g }[/math] 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
- ↑ 1.0 1.1 1.2 Humphreys 1972, Ch. II, § 8.1.
- ↑ Proof (from Humphreys): Let [math]\displaystyle{ x \in \mathfrak{h} }[/math]. Since [math]\displaystyle{ \operatorname{ad}(x) }[/math] is diagonalizable, it is enough to show the eigenvalues of [math]\displaystyle{ \operatorname{ad}_{\mathfrak{h}}(x) }[/math] are all zero. Let [math]\displaystyle{ y \in \mathfrak{h} }[/math] be an eigenvector of [math]\displaystyle{ \operatorname{ad}_{\mathfrak{h}}(x) }[/math] with eigenvalue [math]\displaystyle{ \lambda }[/math]. Then [math]\displaystyle{ x }[/math] is a sum of eigenvectors of [math]\displaystyle{ \operatorname{ad}_{\mathfrak{h}}(y) }[/math] and then [math]\displaystyle{ -\lambda y = \operatorname{ad}_{\mathfrak{h}}(y)x }[/math] is a linear combination of eigenvectors of [math]\displaystyle{ \operatorname{ad}_{\mathfrak{h}}(y) }[/math] with nonzero eigenvalues. But, unless [math]\displaystyle{ \lambda = 0 }[/math], we have that [math]\displaystyle{ -\lambda y }[/math] is an eigenvector of [math]\displaystyle{ \operatorname{ad}_{\mathfrak{h}}(y) }[/math] with eigenvalue zero, a contradiction. Thus, [math]\displaystyle{ \lambda = 0 }[/math]. [math]\displaystyle{ \square }[/math]
- ↑ Humphreys 1972, Ch. IV, § 15.3. Corollary
- Borel, Armand (1991), Linear algebraic groups, Graduate Texts in Mathematics, 126 (2nd ed.), Berlin, New York: Springer-Verlag, ISBN 978-0-387-97370-8
- Humphreys, James E. (1972), Introduction to Lie Algebras and Representation Theory, Berlin, New York: Springer-Verlag, ISBN 978-0-387-90053-7, https://archive.org/details/introductiontoli00jame
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