# Difference between revisions of "Cartan subalgebra"

of a finite-dimensional Lie algebra $\mathfrak g$ over a field $k$

A nilpotent subalgebra of $\mathfrak g$ which is equal to its normalizer in $\mathfrak g$ . For example, if $\mathfrak g$ is the Lie algebra of all complex square matrices of a fixed order, then the subalgebra of all diagonal matrices is a Cartan subalgebra in $\mathfrak g$ . A Cartan subalgebra can also be defined as a nilpotent subalgebra $\mathfrak t$ in $\mathfrak g$ which is equal to its Fitting null-component (cf. Weight of a representation of a Lie algebra) $$\mathfrak g _{0} = \{ {X \in \mathfrak g} : { \forall H \in \mathfrak t \ \exists n _{X},H \in \mathbf Z \ ( ( \mathop{\rm ad}\nolimits \ H ) ^ {n _{X},H} (X) = 0 )} \} ,$$ where $\mathop{\rm ad}\nolimits$ denotes the adjoint representation (cf. Lie algebra) of $\mathfrak g$ .

Suppose further that $k$ is of characteristic zero. Then for any regular element $x \in \mathfrak g$ , the set $\mathfrak n ( X ,\ \mathfrak g )$ of all elements of $\mathfrak g$ which are annihilated by powers of $\mathop{\rm ad}\nolimits \ X$ is a Cartan subalgebra of $\mathfrak g$ , and every Cartan subalgebra of $\mathfrak g$ has the form $\mathfrak n ( X ,\ g )$ for some suitable regular element $X$ . Each regular element belongs to one and only one Cartan subalgebra. The dimension of all the Cartan subalgebras of $\mathfrak g$ are the same and are equal to the rank of $\mathfrak g$ . The image of a Cartan subalgebra under a surjective homomorphism of Lie algebras is a Cartan subalgebra. If $k$ is algebraically closed, then all Cartan subalgebras of $\mathfrak g$ are conjugate; more precisely, they can be transformed into another by operators of the algebraic group $D$ of automorphisms of $\mathfrak g$ whose Lie algebra is the commutator subalgebra of $\mathop{\rm ad}\nolimits \ \mathfrak g$ . If $\mathfrak g$ is solvable, then the above assertion holds without the hypothesis that $k$ be algebraically closed.

Let $G$ be either a connected linear algebraic group over an algebraically closed field $k$ of characteristic zero, or a connected Lie group, and let $\mathfrak g$ be its Lie algebra. Then a subalgebra $\mathfrak t$ of $\mathfrak g$ is a Cartan subalgebra if and only if it is the Lie algebra of a Cartan subgroup of $G$ .

Let $\mathfrak g$ be a subalgebra of the Lie algebra $\mathfrak g \mathfrak l (V)$ of all endomorphisms of a finite-dimensional vector space $V$ over $k$ , and let $\overline{\mathfrak g}$ be the smallest algebraic Lie algebra in $\mathfrak g \mathfrak l (V)$ containing $\mathfrak g$ (cf. Lie algebra, algebraic). If $\overline{\mathfrak t}$ is a Cartan subalgebra of $\overline{\mathfrak g}$ , then $\overline{\mathfrak t} \cap \mathfrak g$ is a Cartan subalgebra of $\mathfrak g$ , and if $\mathfrak t$ is a Cartan subalgebra of $\mathfrak g$ and $\overline{\mathfrak t}$ is the smallest algebraic subalgebra of $\mathfrak g \mathfrak l (V)$ containing $\mathfrak t$ , then $\overline{\mathfrak t}$ is a Cartan subalgebra of $\overline{\mathfrak g}$ and $\mathfrak t =\overline{\mathfrak t} \cap \mathfrak g$ .

Let $k \subset K$ be a field extension. A subalgebra $\mathfrak t$ of $\mathfrak g$ is a Cartan subalgebra if and only if $\mathfrak t \otimes _{k} K$ is a Cartan subalgebra of $\mathfrak g \otimes _{k} K$ .

Cartan subalgebras play an especially important role when $\mathfrak g$ is a semi-simple Lie algebra (this was used by E. Cartan ). In this case, every Cartan subalgebra $\mathfrak t$ of $\mathfrak g$ is Abelian and consists of semi-simple elements (see Jordan decomposition), and the restriction of the Killing form to $\mathfrak t$ is non-singular.

How to Cite This Entry:
Cartan subalgebra. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Cartan_subalgebra&oldid=44226
This article was adapted from an original article by V.L. Popov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article