Toral subalgebra

Lie algebra all of which elements are semisimple

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 h {\displaystyle {\mathfrak {h}}} of a semisimple Lie algebra g {\displaystyle {\mathfrak {g}}} is called toral if the adjoint representation of h {\displaystyle {\mathfrak {h}}} on g {\displaystyle {\mathfrak {g}}} , ad ⁡ ( h ) ⊂ g l ( 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 g {\displaystyle {\mathfrak {g}}} restricted to h {\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 g {\displaystyle {\mathfrak {g}}} over an algebraically closed field of a characteristic zero, a toral subalgebra exists.[1] In fact, if g {\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, g {\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. ^ a b c Humphreys 1972, Ch. II, § 8.1.
  2. ^ Proof (from Humphreys): Let x ∈ h {\displaystyle x\in {\mathfrak {h}}} . Since ad ⁡ ( x ) {\displaystyle \operatorname {ad} (x)} is diagonalizable, it is enough to show the eigenvalues of ad h ⁡ ( x ) {\displaystyle \operatorname {ad} _{\mathfrak {h}}(x)} are all zero. Let y ∈ h {\displaystyle y\in {\mathfrak {h}}} be an eigenvector of ad h ⁡ ( x ) {\displaystyle \operatorname {ad} _{\mathfrak {h}}(x)} with eigenvalue λ {\displaystyle \lambda } . Then x {\displaystyle x} is a sum of eigenvectors of ad h ⁡ ( y ) {\displaystyle \operatorname {ad} _{\mathfrak {h}}(y)} and then − λ y = ad h ⁡ ( y ) x {\displaystyle -\lambda y=\operatorname {ad} _{\mathfrak {h}}(y)x} is a linear combination of eigenvectors of ad h ⁡ ( y ) {\displaystyle \operatorname {ad} _{\mathfrak {h}}(y)} with nonzero eigenvalues. But, unless λ = 0 {\displaystyle \lambda =0} , we have that − λ y {\displaystyle -\lambda y} is an eigenvector of ad h ⁡ ( y ) {\displaystyle \operatorname {ad} _{\mathfrak {h}}(y)} with eigenvalue zero, a contradiction. Thus, λ = 0 {\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