§ 2. Cartan subalgebras and regular elements of a Lie algebra

From no. 2 onwards, the field $k$ is assumed to be infinite.

1. Cartan subalgebras

Definition 1 00TJ

Let $\mathfrak{g}$ be a Lie algebra. A Cartan subalgebra of $\mathfrak{g}$ is a nilpotent subalgebra of $\mathfrak{g}$ equal to its own normalizer.

Later we shall obtain the following results:

1) if $k$ is infinite, $\mathfrak{g}$ has Cartan subalgebras (no. 3, Cor. 1 of Th. 1);

2) if $k$ is of characteristic zero, all Cartan subalgebras of $\mathfrak{g}$ have the same dimension (§3, no. 3, Th. 2);

3) if $k$ is algebraically closed and of characteristic 0, all Cartan subalgebras of $\mathfrak{g}$ are conjugate under the group of elementary automorphisms of $\mathfrak{g}($ §3, no. 2, Th. 1).

Example 1 00TK

If $\mathfrak{g}$ is nilpotent, the only Cartan subalgebra of $\mathfrak{g}$ is $\mathfrak{g}$ itself (Chap. I, §4, no. 1, Prop. 3).

Example 2 00TL

Let $\mathfrak{g}=\mathfrak{g}\mathfrak{l}(n, k)$ , and let $\mathfrak{h}$ be the set of diagonal matrices belonging to $\mathfrak{g}$ . We show that $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$ . First, $\mathfrak{h}$ is commutative, hence nilpotent. Let $(E_{ij})$ be the canonical basis of $\mathfrak{g}\mathfrak{l}(n, k)$ , and let $x=\sum\mu_{ij}E_{ij}$ be an element of the normalizer of $\mathfrak{h}$ in $\mathfrak{g}$ . If $i\not=j$ , formulas (5) of Chap. I, §1, no. 2 show that the coefficient of $E_{ij}$ in $[E_{ii}, x]$ is $\mu_{ij}$ . Since $E_{ii}\in \mathfrak{h}$ , $[E_{ii}, x]\in \mathfrak{h}$ , and the coefficient in question is zero. Thus $\mu_{ij}= 0$ for $i\not=j$ , so $x\in \mathfrak{h}$ , which shows that $\mathfrak{h}$ is indeed a Cartan subalgebra of $\mathfrak{g}$ .

Example 3 00TM

Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$ and let $\mathfrak{g}_1$ be a subalgebra of $\mathfrak{g}$ containing $\mathfrak{h}$ . Then $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}_1$ ; this follows immediately from Def. 1.

Proposition 1 00TN

Let $\mathfrak{g}$ be a Lie algebra and let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$ . Then $\mathfrak{h}$ is a maximal nilpotent subalgebra of $\mathfrak{g}$ .

Let $\mathfrak{h}'$ be a nilpotent subalgebra of $\mathfrak{g}$ containing $\mathfrak{h}$ . Then $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{h}'$ (Example 3), so $\mathfrak{h}=\mathfrak{h}'$ (Example 1).

There exist maximal nilpotent subalgebras that are not Cartan subalgebras (Exerc. 2).

Proposition 2 00TO

Let $(\mathfrak{g}_i)_{i\in I}$ be a finite family of Lie algebras and $\mathfrak{g}=\prod_{i\in I}\mathfrak{g}_i$ .

The Cartan subalgebras of $\mathfrak{g}$ are the subalgebras of the form $\prod_{i\in I}\mathfrak{h}_i$ , where $\mathfrak{h}_i$

is a Cartan subalgebra of $\mathfrak{g}_i$ .

If $\mathfrak{h}_i$ is a subalgebra of $\mathfrak{g}_i$ with normalizer $\mathfrak{n}_i$ , then $\prod\mathfrak{h}_i$ is a subalgebra of $\mathfrak{g}$ with normalizer $\prod\mathfrak{n}_i$ ; if the $\mathfrak{h}_i$ are nilpotent, $\prod\mathfrak{h}_i$ is nilpotent; thus, if $\mathfrak{h}_i$ is a Cartan subalgebra of $\mathfrak{g}_i$ for all $i,\prod\mathfrak{h}_i$ is a Cartan subalgebra of $\mathfrak{g}$ . Conversely, let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$ ; the projection $\mathfrak{h}_i$ of $\mathfrak{h}$ onto $\mathfrak{g}_i$ is a nilpotent subalgebra of $\mathfrak{g}_i$ , and $\prod\mathfrak{h}_i$ is a nilpotent subalgebra of $\mathfrak{g}$ containing $\mathfrak{h}$ ; hence $\mathfrak{h}=\prod\mathfrak{h}_i$ (Prop. 1); thus, for all $i,\mathfrak{h}_i$ is its own normalizer in $\mathfrak{g}_i$ , and so is a Cartan subalgebra of $\mathfrak{g}_i$ .

Example 4 00TP

If $k$ is of characteristic $0,\mathfrak{g}\mathfrak{l}(n, k)$ is the product of the ideals $\mathfrak{s}\mathfrak{l}(n, k)$ and $k.1$ . It follows from Example 2 and Prop. 2 that the set of diagonal matrices of trace 0 in $\mathfrak{s}\mathfrak{l}(n, k)$ is a Cartan subalgebra of $\mathfrak{s}\mathfrak{l}(n, k)$ .

Proposition 3 00TQ

Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{h}$ a subalgebra of $\mathfrak{g}$ , and $k'$ an extension of $k$ . Then $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$ if and only if $\mathfrak{h}\otimes_kk'$ is a Cartan subalgebra of $\mathfrak{g}\otimes_kk'$ .

Indeed, $\mathfrak{h}$ is nilpotent if and only if $\mathfrak{h}\otimes_kk'$ is (Chap. I, §4, no. 5). On the other hand, if $\mathfrak{n}$ is the normalizer of $\mathfrak{h}$ in $\mathfrak{g}$ , the normalizer of $\mathfrak{h}\otimes_kk'$ in $\mathfrak{g}\otimes_kk'$ is $\mathfrak{n}\otimes_kk'$ (Chap. I, §3, no. 8).

Proposition 4 00TR

Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{h}$ a nilpotent subalgebra of $\mathfrak{g}$ . Then $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$ if and only if $\mathfrak{g}^0(\mathfrak{h}) =\mathfrak{h}$ .

If $\mathfrak{g}^0(\mathfrak{h}) =\mathfrak{h},\mathfrak{h}$ is its own normalizer (§1, Prop. 10 (i)), so $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$ . Assume that $\mathfrak{g}^0(\mathfrak{h})\not=\mathfrak{h}$ . Consider the representation of $\mathfrak{h}$ on $\mathfrak{g}^0(\mathfrak{h})/\mathfrak{h}$ obtained from the adjoint representation by passage to the quotient. By applying Engel’s theorem (Chap. I, §4, no. 2, Th. 1), we see that there exists $x\in \mathfrak{g}^0(\mathfrak{h})$ such that $x \notin \mathfrak{h}$ and $[\mathfrak{h}, x]\subset \mathfrak{h}$ ; then $x$ belongs to the normalizer of $\mathfrak{h}$ in $\mathfrak{g}$ , so $\mathfrak{h}$ is not a Cartan subalgebra of $\mathfrak{g}$ .

Corollary 1 00TS

Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{h}$ a Cartan subalgebra of $\mathfrak{g}$ . If $k$ is infinite, there exists $x\in \mathfrak{h}$ such that $\mathfrak{h}=\mathfrak{g}^0(x)$ .

Indeed, $\mathfrak{h}=\mathfrak{g}^0(\mathfrak{h})$ and we can apply Prop. 9 (ii) of §1.

Corollary 2 00TT

Let $f:\mathfrak{g}\rightarrow \mathfrak{g}'$ be a surjective homomorphism of Lie algebras. If $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g},f(\mathfrak{h})$ is a Cartan subalgebra of $\mathfrak{g}'$ .

Indeed, $f(\mathfrak{h})$ is a nilpotent subalgebra of $\mathfrak{g}'$ . On the other hand, consider the representation $x \rightarrow$ ad $f(x)$ of $\mathfrak{h}$ on $\mathfrak{g}'$ . By Prop. 9 (iv) of §1, no. 3, $f(\mathfrak{g}^0(\mathfrak{h})) ={\mathfrak{g}'}^0(\mathfrak{h})$ . Now $\mathfrak{g}^0(\mathfrak{h}) =\mathfrak{h}$ , and on the other hand it is clear that ${\mathfrak{g}'}^0(\mathfrak{h}) ={\mathfrak{g}'}^0(f(\mathfrak{h}))$ . Hence, $f(\mathfrak{h}) ={\mathfrak{g}'}^0(f(\mathfrak{h}))$ and it suffices to apply Prop. 4.

Corollary 3 00TU

Let $\mathfrak{h}$ be a Cartan subalgebra of a Lie algebra $\mathfrak{g}$ , and let $\mathscr{C}^n\mathfrak{g}(n\geq 1)$ be a term of the descending central series of $\mathfrak{g}$ (Chap. I, §1, no. 5). Then $\mathfrak{g}=\mathfrak{h}+\mathscr{C}^n\mathfrak{g}$ .

Indeed, Corollary 2 shows that the image of $\mathfrak{h}$ in $\mathfrak{g}/\mathscr{C}^n\mathfrak{g}$ is a Cartan subalgebra of $\mathfrak{g}/\mathscr{C}^n\mathfrak{g}$ , hence is equal to $\mathfrak{g}/\mathscr{C}^n\mathfrak{g}$ since $\mathfrak{g}/\mathscr{C}^n\mathfrak{g}$ is nilpotent (Example 1).

Corollary 4 00TV

Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{h}$ a Cartan subalgebra of $\mathfrak{g}$ , and $\mathfrak{a}$ a subalgebra of $\mathfrak{g}$ containing $\mathfrak{h}$ .

(i) $\mathfrak{a}$ is equal to its own normalizer in $\mathfrak{g}$ .

(ii) Assume that $k=\mathbf{R}$ or $\mathbf{C}$ ; let G be a Lie group with Lie algebra $\mathfrak{g}$ , A the integral subgroup of G with Lie algebra $\mathfrak{a}$ . Then A is a Lie subgroup of G, and it is the identity component of the normalizer of A in G.

Let $\mathfrak{n}$ be the normalizer of $\mathfrak{a}$ in $\mathfrak{g}$ . Since $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{n}$ (Example $3$ ) $,\{0\}$ is a Cartan subalgebra of $\mathfrak{n}/\mathfrak{a}$ (Cor. 2), hence is equal to its normalizer in $\mathfrak{n}/\mathfrak{a}$ ; in other words, $\mathfrak{n}=\mathfrak{a}$ . Assertion (ii) follows from (i) and Chap. III, §9, no. 4, Cor. of Prop. 11.

Corollary 5 00TW

Let $\mathfrak{g}$ be a Lie algebra, E a subset of $\mathfrak{g}$ . Let E operate on $\mathfrak{g}$ by the adjoint representation. Then E is a Cartan subalgebra of $\mathfrak{g}$ if and only if $E =\mathfrak{g}^0(E)$ .

The condition is necessary (Prop. 4). Assume now that $E =\mathfrak{g}^0(E)$ . By Prop. 2 (ii) of §1, no. 1, E is then a subalgebra of $\mathfrak{g}$ . If $x\in E$ , ad $_Ex$ is nilpotent since $E\subset \mathfrak{g}^0(E)$ ; hence the algebra E is nilpotent. But then E is a Cartan subalgebra by Prop. 4.

Corollary 6 00TX

Let $\mathfrak{g}$ be a Lie algebra, let $k_0$ be a subfield of $k$ such that $[k:k_0]<+\infty$ , and let $\mathfrak{g}_0$ be the Lie algebra obtained from $\mathfrak{g}$ by restricting the field of scalars to $k_0$ . Let $\mathfrak{h}$ be a subset of $\mathfrak{g}$ . Then $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$ if and only if $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}_0$ .

This follows from Cor. 5, since the condition $\mathfrak{h}=\mathfrak{g}^0(\mathfrak{h})$ does not involve the base field.

Proposition 5 00TY

Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{c}$ its centre, $\mathfrak{h}$ a vector subspace of $\mathfrak{g}$ . Then $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$ if and only if $\mathfrak{h}$ contains $\mathfrak{c}$ and $\mathfrak{h}/\mathfrak{c}$ is a Cartan subalgebra of $\mathfrak{g}/\mathfrak{c}$ .

Assume that $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$ . Since $[\mathfrak{c},\mathfrak{g}]\subset \mathfrak{h}$ , we have $\mathfrak{c}\subset \mathfrak{h}$ . On the other hand, $\mathfrak{h}/\mathfrak{c}$ is a Cartan subalgebra of $\mathfrak{g}/\mathfrak{c}$ by Cor. 2 of Prop. 4.

Assume that $\mathfrak{h}\supset \mathfrak{c}$ and that $\mathfrak{h}/\mathfrak{c}$ is a Cartan subalgebra of $\mathfrak{g}/\mathfrak{c}$ . Let $f$ be the canonical morphism from $\mathfrak{g}$ to $\mathfrak{g}/\mathfrak{c}$ . The algebra $\mathfrak{h}$ , which is a central extension of $\mathfrak{h}/\mathfrak{c}$ , is nilpotent. Let $\mathfrak{n}$ be the normalizer of $\mathfrak{h}$ in $\mathfrak{g}$ . If $x\in \mathfrak{n}$ , $[f(x),\mathfrak{h}/\mathfrak{c}]\subset \mathfrak{h}/\mathfrak{c}$ , hence $f(x)\in \mathfrak{h}/\mathfrak{c}$ , and so $x\in \mathfrak{h}$ . This proves that $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$ .

Corollary 00TZ

Let $\mathscr{C}_{\infty}\mathfrak{g}$ be the union of the ascending central series of the Lie algebra $\mathfrak{g}$ (Chap. I, §1, no. 6). The Cartan subalgebras of $\mathfrak{g}$ are the inverse images of the Cartan subalgebras of $\mathfrak{g}/\mathscr{C}_{\infty}\mathfrak{g}$ .

Indeed, the centre of $\mathfrak{g}/\mathscr{C}_i\mathfrak{g}$ is $\mathscr{C}_{i+1}\mathfrak{g}/\mathscr{C}_i\mathfrak{g}$ , and the corollary follows immediately from Prop. 5 by induction.

Remark 00U0

$\mathscr{C}_{\infty}\mathfrak{g}$ is the smallest ideal $\mathfrak{n}$ of $\mathfrak{g}$ such that the centre of $\mathfrak{g}/\mathfrak{n}$ is zero; it is a characteristic and nilpotent ideal of $\mathfrak{g}$ .

2. Regular elements of a Lie algebra

[Recall that $k$ is assumed to be infinite from now on.]

Let $\mathfrak{g}$ be a Lie algebra of dimension $n$ . If $x\in \mathfrak{g}$ , write the characteristic polynomial of ad $x$ in the form

det(T $-$ ad $x$ ) $=\sum_{i=0}^na_i(x)T^i$ , with $a_i(x)\in k$ .

We have $a_i(x) = (-1)^{n-i}$ Tr $(\bigwedge^{n-i}$ ad x), cf. Algebra, Chap. III, §8, no. 11. This shows that $x \rightarrow a_i(x)$ is a homogeneous polynomial map of degree $n-i$ from $\mathfrak{g}$ to $k($ Algebra, Chap. IV, §5, no. 9).

Remark 1 00U1

If $\mathfrak{g}\not=\{0\},a_0= 0$ since (ad $x$ ) $(x) = 0$ for all $x\in \mathfrak{g}$ .

Remark 2 00U2

Let $k'$ be an extension of $k$ . Write det(T $-$ ad $x'$ ) $=\sum_{i=0}^na'_i(x')T^i$ for $x'\in \mathfrak{g}\otimes_kk'$ . Then $a'_i|\mathfrak{g}=a_i$ for all $i$ .

Definition 2 00U3

The rank of $\mathfrak{g}$ , denoted by rk( $\mathfrak{g}$ ), is the smallest integer $l$ such that $a_l\not= 0$ . An element $x$ of $\mathfrak{g}$ is called regular if $a_l(x)\not= 0$ .

For all $x\in \mathfrak{g}$ , rk( $\mathfrak{g}$ ) $\leq$ dim $\mathfrak{g}^0(x)$ , and equality holds if and only if $x$ is regular.

The set of regular elements is dense and open in $\mathfrak{g}$ for the Zariski topology (App. I).

Example 1 00U4

If $\mathfrak{g}$ is nilpotent, rk( $\mathfrak{g}$ ) $=$ dim $\mathfrak{g}$ and all elements of $\mathfrak{g}$ are regular.

Example 2 00U5

Let $\mathfrak{g}=\mathfrak{s}\mathfrak{l}(2, k)$ . If $x=(\gamma -\alpha )\in \mathfrak{g}$ , an easy calculation gives

\beta \gamma

det(T $-$ ad $x$ ) $= T^3-4(\alpha \beta +\gamma^2)T$ .

If the characteristic of $k$ is $\not= 2$ , then rk( $\mathfrak{g}$ ) $= 1$ and the regular elements are those $x$ such that $\alpha \beta +\gamma^2\not= 0$ .

Example 3 00U6

Let V be a vector space of finite dimension $n$ , and $\mathfrak{g}=\mathfrak{g}\mathfrak{l}(V)$ . Let $\overline{k}$ be an algebraic closure of $k$ . Let $x\in \mathfrak{g}$ , and let $\lambda_1, . . . , \lambda_n$ be the roots in $\overline{k}$ of the characteristic polynomial of $x$ (each root being written a number of times equal to its multiplicity). The canonical isomorphism from $V^*\otimes V$ to $\mathfrak{g}$ is compatible with the $\mathfrak{g}$ -module structures of these two spaces, in other words it takes $1\otimes x-^tx\otimes 1$ to ad $x$ (Chap. I, §3, no. 3, Prop. 4). In view of §1, Prop. 4 (i), it follows that the roots of the characteristic polynomial of ad $x$ are the $\lambda_i-\lambda_j$ for $1\leq i\leq n,1\leq j\leq n$ (each root being written a number of times equal to its multiplicity). Thus, the rank of $\mathfrak{g}$ is $n$ , and $x$ is regular if and only if each $\lambda_i$ is a simple root of the characteristic polynomial of $x$ .

Proposition 6 00U7

Let $\mathfrak{g}$ be a Lie algebra, $k'$ an extension of $k$ , and $\mathfrak{g}'=$ $\mathfrak{g}\otimes_kk'$ .

(i) An element $x$ of $\mathfrak{g}$ is regular in $\mathfrak{g}$ if and only if $x\otimes 1$ is regular in $\mathfrak{g}'$ .

(ii) rk( $\mathfrak{g}$ ) $=$ rk( $\mathfrak{g}'$ ).

This follows from Remark 2.

Proposition 7 01IY

Let $(\mathfrak{g}_i)_{i\in I}$ be a finite family of Lie algebras, and let $\mathfrak{g}=$ $\prod_{i\in I}\mathfrak{g}_i$ .

(i) An element $(x_i)_{i\in I}$ of $\mathfrak{g}$ is regular in $\mathfrak{g}$ if and only if, for all $i\in I,x_i$ is regular in $\mathfrak{g}_i$ .

(ii) rk( $\mathfrak{g}$ ) $=\sum_{i\in I}$ rk( $\mathfrak{g}_i$ ).

Indeed, for any $x= (x_i)_{i\in I}\in \mathfrak{g}$ , the characteristic polynomial of ad $_{\mathfrak{g}}x$ is the product of the characteristic polynomials of the ad $_{\mathfrak{g}_i}x_i$ .

Proposition 8 00U8

Let $f:\mathfrak{g}\rightarrow \mathfrak{g}'$ be a surjective homomorphism of Lie algebras.

(i) If $x$ is a regular element of $\mathfrak{g},f(x)$ is regular in $\mathfrak{g}'$ . The converse is true if Ker $f$ is contained in the centre of $\mathfrak{g}$ .

(ii) rk( $\mathfrak{g}$ ) $\geq$ rk( $\mathfrak{g}'$ ).

Put rk( $\mathfrak{g}$ ) $=r$ , rk( $\mathfrak{g}'$ ) $=r'$ . Let $x\in \mathfrak{g}$ . The characteristic polynomials of ad $x$ , ad $f(x)$ and ad $x|$ Ker $f$ are of the form

P(T) = T^n+a_{n-1}(x)T^{n-1}+\cdots +a_r(x)T^r

Q(T) = T^{n'}+b_{n'-1}(x)T^{n'-1}+\cdots +b_{r'}(x)T^{r'}

R(T) = T^{n''}+c_{n''-1}(x)T^{n''-1}+\cdots +c_{r''}(x)T^{r''}

where the $a_i, b_i, c_i$ are polynomial functions on $\mathfrak{g}$ , with $a_r\not= 0, b_{r'}\not= 0, c_{r''}\not= 0$ . We have P = QR, so $r=r'+r''$ and $a_r(x) =b_{r'}(x)c_{r''}(x)$ , which proves (ii) and the first assertion of (i). If Ker $f$ is contained in the centre of $\mathfrak{g}$ , $R(T) = T^{n''}$ and so $a_r(x) =b_{r'}(x)$ , hence the second assertion of (i).

Corollary 00U9

Let $\mathscr{C}_n\mathfrak{g}(n\geq 0)$ be a term of the ascending central series of $\mathfrak{g}$ (Chap. I, §1, no. 6). The regular elements of $\mathfrak{g}$ are those whose image in $\mathfrak{g}/\mathscr{C}_n\mathfrak{g}$ is regular.

Proposition 9 00UA

Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{g}'$ a subalgebra of $\mathfrak{g}$ . Every element of $\mathfrak{g}'$ regular in $\mathfrak{g}$ is regular in $\mathfrak{g}'$ .

For $x\in \mathfrak{g}'$ , the restriction of ad $_{\mathfrak{g}}x$ to $\mathfrak{g}'$ is ad $_{\mathfrak{g}'}x$ , and so defines an endomorphism $u(x)$ of the vector space $\mathfrak{g}/\mathfrak{g}'$ by passage to the quotient. Let $d_0(x)$ (resp. $d_1(x)$ ) be the dimension of the nilspace of ad $_{\mathfrak{g}'}(x)$ (resp. of $u(x)$ ), and let $c_0$ (resp. $c_1$ ) be the minimum of $d_0(x)$ (resp. $d_1(x)$ ) when $x$ belongs to $\mathfrak{g}'$ . There exist non-zero polynomial maps $p_0, p_1$ from $\mathfrak{g}'$ to $k$ such that

d_0(x) =c_0\Leftarrow \Rightarrow p_0(x)\not= 0,d_1(x) =c_1\Leftarrow \Rightarrow p_1(x)\not= 0

Since $k$ is infinite, the set S of $x\in \mathfrak{g}'$ such that $d_0(x) =c_0$ and $d_1(x) =c_1$ is non-empty. Every element of S is regular in $\mathfrak{g}'$ . On the other hand, S is the set of elements of $\mathfrak{g}'$ such that the nilspace of ad $_{\mathfrak{g}}x$ has minimum dimension, and thus contains every element of $\mathfrak{g}'$ regular in $\mathfrak{g}$ .

Remark 3 00UB

Elements of $\mathfrak{g}'$ regular in $\mathfrak{g}$ do not necessarily exist. If at least one does exist, the set of these elements is precisely the set denoted by S in the above proof.

3. Cartan subalgebras and regular elements

Theorem 1 00UC

Let $\mathfrak{g}$ be a Lie algebra.

(i) If $x$ is a regular element of $\mathfrak{g},\mathfrak{g}^0(x)$ is a Cartan subalgebra of $\mathfrak{g}$ .

(ii) If $\mathfrak{h}$ is a maximal nilpotent subalgebra of $\mathfrak{g}$ , and if $x\in \mathfrak{h}$ is regular in $\mathfrak{g}$ , then $\mathfrak{h}=\mathfrak{g}^0(x)$ .

(iii) If $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$ , then dim( $\mathfrak{h}$ ) $\geq$ rk( $\mathfrak{g}$ ).

(iv) The Cartan subalgebras of $\mathfrak{g}$ of dimension rk( $\mathfrak{g}$ ) are the $\mathfrak{g}^0(x)$ where $x$ is a regular element.

Let $x$ be a regular element of $\mathfrak{g}$ and let $\mathfrak{h}=\mathfrak{g}^0(x)$ . Clearly $\mathfrak{h}^0(x) =\mathfrak{h}$ . Since $x$ is regular in $\mathfrak{h}$ (Prop. 9), rk( $\mathfrak{h}$ ) $=$ dim( $\mathfrak{h}$ ), so $\mathfrak{h}$ is nilpotent. On the other hand, $\mathfrak{h}=\mathfrak{g}^0(x)\supset \mathfrak{g}^0(\mathfrak{h})\supset \mathfrak{h}$ , so $\mathfrak{h}=\mathfrak{g}^0(\mathfrak{h})$ is a Cartan subalgebra of $\mathfrak{g}$ (Prop. 4). This proves (i).

If $\mathfrak{h}$ is a maximal nilpotent subalgebra of $\mathfrak{g}$ , and if $x\in \mathfrak{h}$ is regular in $\mathfrak{g}$ , then $\mathfrak{h}\subset \mathfrak{g}^0(x)$ and $\mathfrak{g}^0(x)$ is nilpotent by (i), so $\mathfrak{h}=\mathfrak{g}^0(x)$ , which proves (ii).

If $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$ , there exists $x\in \mathfrak{h}$ such that $\mathfrak{h}=\mathfrak{g}^0(x)$ (Cor. 1 of Prop. 4), so dim( $\mathfrak{h}$ ) $\geq$ rk( $\mathfrak{g}$ ), which proves (iii). If in addition dim( $\mathfrak{h}$ ) $=$ rk( $\mathfrak{g}$ ) $,x$ is regular. Finally, if $x'$ is regular in $\mathfrak{g},\mathfrak{g}^0(x')$ is a Cartan subalgebra by (i), and is obviously of dimension rk( $\mathfrak{g}$ ). This proves (iv).

We shall see in §3, Th. 2 that, when $k$ is of characteristic zero, all the Cartan subalgebras of $\mathfrak{g}$ have dimension rk( $\mathfrak{g}$ ).

Corollary 1 00UD

Every Lie algebra $\mathfrak{g}$ has Cartan subalgebras, and the rank of $\mathfrak{g}$ is the minimum dimension of a Cartan subalgebra.

Corollary 2 00UE

Let $f:\mathfrak{g}\rightarrow \mathfrak{g}'$ be a surjective homomorphism of Lie algebras. If $\mathfrak{h}'$ is a Cartan subalgebra of $\mathfrak{g}'$ , there exists a Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$ such that $\mathfrak{h}'=f(\mathfrak{h})$ .

Let $\mathfrak{a}=f^{-1}(\mathfrak{h}')$ . By Cor. $1,\mathfrak{a}$ has a Cartan subalgebra $\mathfrak{h}$ . By Cor. 2 of Prop. $4,f(\mathfrak{h}) =\mathfrak{h}'$ . We show that $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$ . Let $\mathfrak{n}$ be the normalizer of $\mathfrak{h}$ in $\mathfrak{g}$ . It is enough to prove that $\mathfrak{h}=\mathfrak{n}$ . If $x\in \mathfrak{n},f(x)$ belongs to the normalizer of $\mathfrak{h}'$ in $\mathfrak{g}'$ , so $f(x)\in \mathfrak{h}'$ and $x\in \mathfrak{a}$ ; but $\mathfrak{h}$ is its own normalizer in $\mathfrak{a}$ , so $x\in \mathfrak{h}$ .

Corollary 3 00UF

Every Lie algebra $\mathfrak{g}$ is the sum of its Cartan subalgebras.

The sum $\mathfrak{s}$ of the Cartan subalgebras of $\mathfrak{g}$ contains the set of regular elements of $\mathfrak{g}$ (Th. 1 (i)). Since this set is dense in $\mathfrak{g}$ for the Zariski topology, $\mathfrak{s}=\mathfrak{g}$ .

Proposition 10 00UG

Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{a}$ a commutative subalgebra of $\mathfrak{g}$ and $\mathfrak{c}$ the commutant of $\mathfrak{a}$ in $\mathfrak{g}$ . Assume that ad $_{\mathfrak{g}}x$ is semi-simple for all $x\in \mathfrak{a}$ . Then the Cartan subalgebras of $\mathfrak{c}$ are the Cartan subalgebras of $\mathfrak{g}$ containing $\mathfrak{a}$ .

Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{c}$ . Since $\mathfrak{a}$ is contained in the centre $\mathfrak{z}$ of $\mathfrak{c},\mathfrak{a}\subset \mathfrak{z}\subset \mathfrak{h}$ (Prop. 5). Let $\mathfrak{n}$ be the normalizer of $\mathfrak{h}$ in $\mathfrak{g}$ . Then

[\mathfrak{a},\mathfrak{n}]\subset [\mathfrak{h},\mathfrak{n}]\subset \mathfrak{h}

Since the ad $_{\mathfrak{g}}x,x\in \mathfrak{a}$ , are semi-simple and commute with each other, it follows from Algebra, Chap. VIII, §5, no. 1, that there exists a vector subspace $\mathfrak{d}$ of $\mathfrak{n}$ stable under ad $_{\mathfrak{g}}\mathfrak{a}$ and such that $\mathfrak{n}=\mathfrak{h}\oplus \mathfrak{d}$ . Then $[\mathfrak{a},\mathfrak{d}]\subset \mathfrak{h}\cap \mathfrak{d}= 0$ , so $\mathfrak{d}\subset \mathfrak{c}$ . Thus, $\mathfrak{n}$ is the normalizer of $\mathfrak{h}$ in $\mathfrak{c}$ , and hence $\mathfrak{n}=\mathfrak{h}$ , so $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$ containing $\mathfrak{a}$ .

Conversely, let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$ containing $\mathfrak{a}$ . Then $\mathfrak{h}=$ $\mathfrak{g}^0(\mathfrak{h})\subset \mathfrak{g}^0(\mathfrak{a})$ , and by hypothesis $\mathfrak{g}_0(\mathfrak{a}) =\mathfrak{g}^0(\mathfrak{a}) =\mathfrak{c}$ . Hence $\mathfrak{a}\subset \mathfrak{h}\subset \mathfrak{c}$ and $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{c}$ (for it is equal to its own normalizer in $\mathfrak{g}$ , and so a fortiori in $\mathfrak{c}$ ).

Proposition 11 00UH

Let $\mathfrak{n}$ be a nilpotent subalgebra of a Lie algebra $\mathfrak{g}$ . There exists a Cartan subalgebra of $\mathfrak{g}$ contained in $\mathfrak{g}^0(\mathfrak{n})$ .

Put $\mathfrak{a}=\mathfrak{g}^0(\mathfrak{n})$ . Then $\mathfrak{n}\subset \mathfrak{a}$ since $\mathfrak{n}$ is nilpotent. If $x\in \mathfrak{a}$ , let $P(x)$ be the determinant of the endomorphism of $\mathfrak{g}/\mathfrak{a}$ defined by ad $x$ . Denote by $\mathfrak{a}'$ the set of $x\in \mathfrak{a}$ such that $P(x)\not= 0$ , which is an open subset of $\mathfrak{a}$ in the Zariski topology; the relations $x\in \mathfrak{a}'$ and $\mathfrak{g}^0(x)\subset \mathfrak{a}$ are equivalent. By Prop. 7 (ii) of §1, no. 2, there exists $y\in \mathfrak{n}$ such that $\mathfrak{g}^0(y) =\mathfrak{a}$ , and $y\in \mathfrak{a}'$ so $\mathfrak{a}'$ is non-empty. Since $\mathfrak{a}'$ is open, its intersection with the set of regular elements of $\mathfrak{a}$ is non-empty. Let $x$ be an element of this intersection. Then $\mathfrak{g}^0(x)\subset \mathfrak{a}$ and $\mathfrak{g}^0(x)$ is a Cartan subalgebra of $\mathfrak{a}$ , hence is nilpotent. On the other hand, Prop. 10 (i) of §1, no. 3, shows that $\mathfrak{g}^0(x)$ is its own normalizer in $\mathfrak{g}$ ; it is therefore a Cartan subalgebra of $\mathfrak{g}$ , which completes the proof.

4. Cartan subalgebras of semi-simple Lie algebras

Theorem 2 00UI

Assume that $k$ is of characteristic 0. Let $\mathfrak{g}$ be a semi-simple Lie algebra, $\mathfrak{h}$ a Cartan subalgebra of $\mathfrak{g}$ . Then $\mathfrak{h}$ is commutative, and all of its elements are semi-simple in $\mathfrak{g}$ (Chap. I, §6, no. 3, Def. 3).

Since $\mathfrak{h}=\mathfrak{g}^0(\mathfrak{h}),\mathfrak{h}$ is reductive (§1, Prop. 11), hence commutative since it is nilpotent. On the other hand, the restriction of the adjoint representation of $\mathfrak{g}$ to $\mathfrak{h}$ is semi-simple (loc. cit.), so the elements of $\mathfrak{h}$ are semi-simple in $\mathfrak{g}$ (Algebra, Chap. VIII, §5, no. 1).

Corollary 1 00UJ

If $x\in \mathfrak{h}$ and $y\in \mathfrak{g}^{\lambda}(\mathfrak{h})$ , we have $[x, y] =\lambda (x)y$ .

Indeed, $\mathfrak{g}^{\lambda(x)}(x) =\mathfrak{g}_{\lambda(x)}(x)$ since ad $x$ is semi-simple.

Corollary 2 00UK

Every regular element of $\mathfrak{g}$ is semi-simple.

Indeed, such an element belongs to a Cartan subalgebra (no. 3, Th. 1 (i)).

Corollary 3 00UL

Let $\mathfrak{h}$ be a Cartan subalgebra of a reductive Lie algebra $\mathfrak{g}$ .

a $)\mathfrak{h}$ is commutative.

b) If $\rho$ is a finite dimensional semi-simple representation of $\mathfrak{g}$ , the elements of $\rho (\mathfrak{h})$ are semi-simple.

Let $\mathfrak{c}$ be the centre of $\mathfrak{g}$ , and $\mathfrak{s}$ its derived algebra. Then $\mathfrak{g}=\mathfrak{c}\times \mathfrak{s}$ , so $\mathfrak{h}=\mathfrak{c}\times \mathfrak{h}'$ , where $\mathfrak{h}'$ is a Cartan subalgebra of $\mathfrak{s}$ (Prop. 2). In view of Th. $2,\mathfrak{h}'$ is commutative, hence so is $\mathfrak{h}$ . Moreover, $\rho (\mathfrak{h}')$ consists of semi-simple elements and so does $\rho (\mathfrak{c})$ (Chap. I, §6, no. 5, Th. 4); assertion b) follows.

Exercises

See the exercises for § 2.