# User:Maximilian Janisch/latexlist/Algebraic Groups/Cartan subalgebra

of a finite-dimensional Lie algebra $8$ over a field $k$

A nilpotent subalgebra of $8$ which is equal to its normalizer in $8$. For example, if $8$ 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 $8$. A Cartan subalgebra can also be defined as a nilpotent subalgebra $1$ in $8$ which is equal to its Fitting null-component (cf. Weight of a representation of a Lie algebra)

$$\mathfrak { g } 0 = \{ X \in \mathfrak { g } : \forall H \in \mathfrak { t } \exists \mathfrak { n } X , H \in Z ( ( \text { ad } H ) ^ { n } X , H ( X ) = 0 ) \}$$

where $a d$ denotes the adjoint representation (cf. Lie algebra) of $8$.

Suppose further that $k$ is of characteristic zero. Then for any regular element $x \in \mathfrak { Q }$, the set $n ( X , g )$ of all elements of $8$ which are annihilated by powers of $\operatorname { ad } X$ is a Cartan subalgebra of $8$, and every Cartan subalgebra of $8$ has the form $n ( X , g )$ 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 $8$ are the same and are equal to the rank of $8$. 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 $8$ are conjugate; more precisely, they can be transformed into another by operators of the algebraic group $\Omega$ of automorphisms of $8$ whose Lie algebra is the commutator subalgebra of $g$. If $8$ is solvable, then the above assertion holds without the hypothesis that $k$ be algebraically closed.

Let $k$ be either a connected linear algebraic group over an algebraically closed field $k$ of characteristic zero, or a connected Lie group, and let $8$ be its Lie algebra. Then a subalgebra $1$ of $8$ is a Cartan subalgebra if and only if it is the Lie algebra of a Cartan subgroup of $k$.

Let $8$ be a subalgebra of the Lie algebra $\mathfrak { d } ( V )$ of all endomorphisms of a finite-dimensional vector space $V$ over $k$, and let $8$ be the smallest algebraic Lie algebra in $\mathfrak { d } ( V )$ containing $8$ (cf. Lie algebra, algebraic). If $1$ is a Cartan subalgebra of $8$, then is a Cartan subalgebra of $8$, and if $1$ is a Cartan subalgebra of $8$ and $1$ is the smallest algebraic subalgebra of $\mathfrak { d } ( V )$ containing $1$, then $1$ is a Cartan subalgebra of $8$ and $t = \overline { t } \cap g$.

Let $k \subset K$ be a field extension. A subalgebra $1$ of $8$ is a Cartan subalgebra if and only if $t \otimes _ { k } K$ is a Cartan subalgebra of $g \otimes k ^ { K }$.

Cartan subalgebras play an especially important role when $8$ is a semi-simple Lie algebra (this was used by E. Cartan [1]). In this case, every Cartan subalgebra $1$ of $8$ is Abelian and consists of semi-simple elements (see Jordan decomposition), and the restriction of the Killing form to $1$ is non-singular.

#### References

 [1] E. Cartan, "Sur la structure des groupes de transformations finis et continus" , Paris (1894) Zbl 25.0638.02 [2] N. Jacobson, "Lie algebras" , Interscience (1962) ((also: Dover, reprint, 1979)) MR0148716 MR0143793 Zbl 0121.27504 Zbl 0109.26201 [3] C. Chevalley, "Theory of Lie groups" , 1 , Princeton Univ. Press (1946) MR0082628 MR0015396 Zbl 0063.00842 [4] , Theórie des algèbres de Lie. Topologie des groupes de Lie , Sem. S. Lie , Ie année 1954–1955 , Ecole Norm. Sup. (1955) Zbl 0068.02102

An element $h \in \mathfrak { q }$ is called regular if the dimension of the Fitting null-component of the endomorphism $h$ of $8$ is minimal. "Almost-all" elements of $8$ are regular in the sense that the condition of being regular defines a Zariski-open subset. The result that the Fitting null-component of $h$ for $h$ regular is a Cartan subalgebra holds for finite-dimensional Lie algebras over any infinite field [a4], p. 59.