§ 2. Root system of a split semi-simple Lie algebra

1. Split semi-simple Lie algebras

Definition 1 00YK

Let $\mathfrak{g}$ be a semi-simple Lie algebra. A Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$ is called splitting if, for all $x\in \mathfrak{h}$ , ad $_{\mathfrak{g}}x$ is triangularizable. A semi-simple Lie algebra is called splittable if it has a splitting Cartan subalgebra. A split semi-simple Lie algebra is a pair $(\mathfrak{g},\mathfrak{h})$ where $\mathfrak{g}$ is a semi-simple Lie algebra and $\mathfrak{h}$ is a splitting Cartan subalgebra of $\mathfrak{g}$ .

Remark 1 00YL

Let $\mathfrak{g}$ be a semi-simple Lie algebra, $\mathfrak{h}$ a Cartan subalgebra of $\mathfrak{g}$ . For all $x\in \mathfrak{h}$ , ad $_{\mathfrak{g}}x$ is semi-simple (Chap. VII, §2, no. 4, Th. 2). Thus, to say that $\mathfrak{h}$ is splitting means that ad $_{\mathfrak{g}}x$ is diagonalizable for all $x\in \mathfrak{h}$ .

Remark 2 00YM

If $k$ is algebraically closed, every semi-simple Lie algebra $\mathfrak{g}$ is splittable, and every Cartan subalgebra of $\mathfrak{g}$ is splitting. When $k$ is not algebraically closed, there exist non-splittable semi-simple Lie algebras (Exerc. $2a$ ) $)$ ; moreover, if $\mathfrak{g}$ is splittable, there may exist Cartan subalgebras of $\mathfrak{g}$ that are not splitting (Exerc. $2b$ ) $)$ .

Remark 3 00YN

Let $\mathfrak{g}$ be a semi-simple Lie algebra, $\mathfrak{h}$ a Cartan subalgebra of $\mathfrak{g}$ , and $\rho$ a finite dimensional injective representation of $\mathfrak{g}$ such that $\rho (\mathfrak{h})$ is diagonalizable. Then ad $_{\mathfrak{g}}x$ is diagonalizable for all $x\in \mathfrak{h}$ (Chap. VII, §2, no. 1, Example 2), so $\mathfrak{h}$ is splitting.

Remark 4 00YO

We shall see (§3, no. 3, Cor. of Prop. 10) that if $\mathfrak{h},\mathfrak{h}'$ are splitting Cartan subalgebras of $\mathfrak{g}$ , there exists an elementary automorphism of $\mathfrak{g}$ transforming $\mathfrak{h}$ into $\mathfrak{h}'$ .

Remark 5 00YP

Let $\mathfrak{g}$ be a reductive Lie algebra. Then $\mathfrak{g}=\mathfrak{c}\times \mathfrak{s}$ where $\mathfrak{c}$ is the centre of $\mathfrak{g}$ and $\mathfrak{s}=\mathscr{D}\mathfrak{g}$ is semi-simple. The Cartan subalgebras of $\mathfrak{g}$ are the subalgebras of the form $\mathfrak{h}=\mathfrak{c}\times \mathfrak{h}'$ where $\mathfrak{h}'$ is a Cartan subalgebra of $\mathfrak{s}$ (Chap. VII, §2, no. 1, Prop. 2). Then $\mathfrak{h}$ is called splitting if $\mathfrak{h}'$ is splitting relative to $\mathfrak{s}$ . This leads in an obvious way to the definition of splittable or split reductive algebras.

2. Roots of a split semi-simple Lie algebra

In this number, $(\mathfrak{g},\mathfrak{h})$ denotes a split semi-simple Lie algebra.

For any $\lambda \in \mathfrak{h}^*$ , denote by $\mathfrak{g}^{\lambda}(\mathfrak{h})$ , or simply by $\mathfrak{g}^{\lambda}$ , the primary subspace of $\mathfrak{g}$ relative to $\lambda$ (cf. Chap. VII, §1, no. 3). Recall that $\mathfrak{g}^0=\mathfrak{h}$ (Chap. VII, §2, no. 1, Prop. 4), that $\mathfrak{g}$ is the direct sum of the $\mathfrak{g}^{\lambda}$ (Chap. VII, §1, no. 3, Prop. 8 and 9), that $\mathfrak{g}^{\lambda}$ is the set of $x\in \mathfrak{g}$ such that $[h, x] =\lambda (h)x$ for all $h\in \mathfrak{h}$ (Chap. VII, §2, no. 4, Cor. 1 of Th. 2), and that the weights of $\mathfrak{h}$ on $\mathfrak{g}$ are the linear forms $\lambda$ on $\mathfrak{h}$ such that $\mathfrak{g}^{\lambda}\not= 0$ (Chap. VII, §1, no. 1).

Definition 2 00YQ

A root of $(\mathfrak{g},\mathfrak{h})$ is a non-zero weight of $\mathfrak{h}$ on $\mathfrak{g}$ .

Denote by $R(\mathfrak{g},\mathfrak{h})$ , or simply by R, the set of roots of $(\mathfrak{g},\mathfrak{h})$ . We have

\mathfrak{g}=\mathfrak{h}\oplus \bigoplus_{\alpha\in R}\mathfrak{g}^{\alpha}

Proposition 1 00YR

Let $\alpha ,\beta$ be roots of $(\mathfrak{g},\mathfrak{h})$ and let $\langle \cdot ,\cdot \rangle$ be a non-degenerate invariant symmetric bilinear form on $\mathfrak{g}($ for example the Killing form of $\mathfrak{g})$ .

(i) If $\alpha +\beta \not= 0,\mathfrak{g}^{\alpha}$ and $\mathfrak{g}^{\beta}$ are orthogonal. The restriction of $\langle \cdot ,\cdot \rangle$ to $\mathfrak{g}^{\alpha}\times \mathfrak{g}^{-\alpha}$ is non-degenerate. The restriction of $\langle \cdot ,\cdot \rangle$ to $\mathfrak{h}$ is non-degenerate.

(ii) Let $x\in \mathfrak{g}^{\alpha},y\in \mathfrak{g}^{-\alpha}$ and $h\in \mathfrak{h}$ . Then $[x, y]\in \mathfrak{h}$ and

\langle h,[x, y]\rangle =\alpha (h)\langle x, y\rangle

Assertion (i) is a particular case of Prop. 10 (iii) of Chap. VII, §1, no. 3. If $x\in \mathfrak{g}^{\alpha},y\in \mathfrak{g}^{-\alpha}$ and $h\in \mathfrak{h}$ , we have $[x, y]\in \mathfrak{g}^{\alpha-\alpha}=\mathfrak{h}$ , and

\langle h,[x, y]\rangle =\langle [h, x], y\rangle =\langle \alpha (h)x, y\rangle =\alpha (h)\langle x, y\rangle

Theorem 1 00YS

Let $\alpha$ be a root of $(\mathfrak{g},\mathfrak{h})$ .

(i) The vector space $\mathfrak{g}^{\alpha}$ is of dimension 1.

(ii) The vector subspace $\mathfrak{h}_{\alpha}= [\mathfrak{g}^{\alpha},\mathfrak{g}^{-\alpha}]$ of $\mathfrak{h}$ is of dimension 1. It contains a unique element $H_{\alpha}$ such that $\alpha (H_{\alpha}) = 2$ .

(iii) The vector subspace $\mathfrak{s}_{\alpha}=\mathfrak{h}_{\alpha}+\mathfrak{g}^{\alpha}+\mathfrak{g}^{-\alpha}$ is a Lie subalgebra of $\mathfrak{g}$ .

(iv) If $X_{\alpha}$ is a non-zero element of $\mathfrak{g}^{\alpha}$ , there exists a unique $X_{-\alpha}\in \mathfrak{g}^{-\alpha}$ such that $[X_{\alpha}, X_{-\alpha}] =-H_{\alpha}$ . Let $\varphi$ be the linear map from $\mathfrak{s}\mathfrak{l}(2, k)$ to $\mathfrak{g}$ that takes $X_+$ to $X_{\alpha},X_-$ to $X_{-\alpha}$ , and $H$ to $H_{\alpha}$ ; then $\varphi$ is an isomorphism from the Lie algebra $\mathfrak{s}\mathfrak{l}(2, k)$ to the Lie algebra $\mathfrak{s}_{\alpha}$ .

a) Let $h_{\alpha}$ be the unique element of $\mathfrak{h}$ such that $\alpha (h) =\langle h_{\alpha}, h\rangle$ for all $h\in \mathfrak{h}$ . By Prop. $1, [x, y] =\langle x, y\rangle h_{\alpha}$ for all $x\in \mathfrak{g}^{\alpha},y\in \mathfrak{g}^{-\alpha}$ ; on the other hand $\langle \mathfrak{g}^{\alpha},\mathfrak{g}^{-\alpha}\rangle \not= 0$ . Hence $\mathfrak{h}_{\alpha}= [\mathfrak{g}^{\alpha},\mathfrak{g}^{-\alpha}] =kh_{\alpha}$ .

b) Choose $x\in \mathfrak{g}^{\alpha},y\in \mathfrak{g}^{-\alpha}$ such that $\langle x, y\rangle = 1$ , so $[x, y] =h_{\alpha}$ . Recall that $[h_{\alpha}, x] =\alpha (h_{\alpha})x, [h_{\alpha}, y] =-\alpha (h_{\alpha})y$ . If $\alpha (h_{\alpha}) = 0$ , it follows that $kx+ky+kh_{\alpha}$ is a nilpotent subalgebra $\mathfrak{t}$ of $\mathfrak{g}$ ; since $h_{\alpha}\in [\mathfrak{t},\mathfrak{t}]$ , ad $_{\mathfrak{g}}h_{\alpha}$ is nilpotent (Chap. I, §5, no. 3, Th. 1), which is absurd since ad $_{\mathfrak{g}}h_{\alpha}$ is non-zero semi-simple. So $\alpha (h_{\alpha})\not= 0$ . Hence there exists a unique $H_{\alpha}\in \mathfrak{h}_{\alpha}$ such that $\alpha (H_{\alpha}) = 2$ , which proves (ii).

c) Choose a non-zero element $X_{\alpha}$ of $\mathfrak{g}^{\alpha}$ . There exists $X_{-\alpha}\in \mathfrak{g}^{-\alpha}$ such that $[X_{\alpha}, X_{-\alpha}] =-H_{\alpha}$ (since $[X_{\alpha},\mathfrak{g}^{-\alpha}] =\mathfrak{h}_{\alpha}$ by $b$ ) $)$ . Then

[H_{\alpha}, X_{\alpha}] =\alpha (H_{\alpha})X_{\alpha}= 2X_{\alpha},[H_{\alpha}, X_{-\alpha}] =-\alpha (H_{\alpha})X_{-\alpha}=-2X_{-\alpha}

[X_{\alpha}, X_{-\alpha}] =-H_{\alpha}

hence $kX_{\alpha}+kX_{-\alpha}+kH_{\alpha}$ is a subalgebra of $\mathfrak{g}$ and the linear map $\varphi$ from $\mathfrak{s}\mathfrak{l}(2, k)$ to $kX_{\alpha}+kX_{-\alpha}+kH_{\alpha}$ such that $\varphi (X_+) =X_{\alpha},\varphi (X_-) =X_{-\alpha}$ , $\varphi (H) =H_{\alpha}$ is an isomorphism of Lie algebras.

d) Assume that dim $\mathfrak{g}^{\alpha}>1$ . Let $y$ be a non-zero element of $\mathfrak{g}^{-\alpha}$ . There exists a non-zero element $X_{\alpha}$ of $\mathfrak{g}_{\alpha}$ such that $\langle y, X_{\alpha}\rangle = 0$ . Choose $X_{-\alpha}$ as in c), and consider the representation $\rho :u \rightarrow$ ad $_{\mathfrak{g}}\varphi (u)$ from $\mathfrak{s}\mathfrak{l}(2, k)$ to $\mathfrak{g}$ . We have

\rho (H)y= [\varphi (H), y] = [H_{\alpha}, y] =-2y

\rho (X_+)y= [\varphi (X_+), y] = [X_{\alpha}, y] =\langle X_{\alpha}, y\rangle h_{\alpha}= 0

Thus, $y$ is primitive for $\rho$ , of weight $-2$ , which contradicts Prop. 2 of §1, no. 2. This proves (i).

e) Assertion (iii) is now a consequence of c). On the other hand, if $X_{\alpha}$ is a non-zero element of $\mathfrak{g}^{\alpha}$ , the element $X_{-\alpha}$ constructed in c) is the unique element of $\mathfrak{g}^{-\alpha}$ such that $[X_{\alpha}, X_{-\alpha}] =-H_{\alpha}$ since dim $\mathfrak{g}^{-\alpha}= 1$ . The last assertion of (iv) is a consequence of c). Q.E.D.

The notations $h_{\alpha},H_{\alpha},\mathfrak{s}_{\alpha}$ will be retained in what follows. (To define $h_{\alpha}$ , we take $\langle \cdot ,\cdot \rangle$ equal to the Killing form.) If $X_{\alpha}$ is a non-zero element of $\mathfrak{g}^{\alpha}$ , the isomorphism $\varphi$ of Th. 1 and the representation $u \rightarrow$ ad $_{\mathfrak{g}}\varphi (u)$ of $\mathfrak{s}\mathfrak{l}(2, k)$ on $\mathfrak{g}$ will be said to be associated to $X_{\alpha}$ .

Corollary 00YT

Let $\Phi$ be the Killing form of $\mathfrak{g}$ . For all $a, b\in \mathfrak{h}$ ,

\Phi (a, b) =\sum_{\gamma\in R}\gamma (a)\gamma (b)

Indeed, ad $a$ .ad $b$ leaves each $\mathfrak{g}^{\gamma}$ stable, and its restriction to $\mathfrak{g}^{\gamma}$ is the homothety with ratio $\gamma (a)\gamma (b)$ ; if $\gamma \not= 0$ , dim $\mathfrak{g}^{\gamma}= 1$ .

Proposition 2 00YU

Let $\alpha , \beta \in R$ .

(i) $\beta (H_{\alpha})\in \mathbf{Z}$ .

(ii) If $\Phi$ denotes the Killing form of $\mathfrak{g},\Phi (H_{\alpha}, H_{\beta})\in \mathbf{Z}$ .

Let $X_{\alpha}$ be a non-zero element of $\mathfrak{g}^{\alpha}$ , and let $\rho$ be the representation of $\mathfrak{s}\mathfrak{l}(2, k)$ on $\mathfrak{g}$ associated to $X_{\alpha}$ . The eigenvalues of $\rho (H)$ are 0 and the $\beta (H_{\alpha})$ for $\beta \in R$ . Hence (i) follows from §1, no. 2, Cor. of Prop. 2. Assertion (ii) follows from (i) and the Cor. of Th. 1. Q.E.D.

Let $\alpha \in R,X_{\alpha}$ a non-zero element of $\mathfrak{g}^{\alpha},X_{-\alpha}$ the element of $\mathfrak{g}^{-\alpha}$ such that $[X_{\alpha}, X_{-\alpha}] =-H_{\alpha}$ , and $\rho$ the representation of $\mathfrak{s}\mathfrak{l}(2, k)$ on $\mathfrak{g}$ associated to $X_{\alpha}$ . Let $\pi$ be the representation of $\mathbf{S}\mathbf{L}(2, k)$ on $\mathfrak{g}$ compatible with $\rho ($ §1, no. 4, Th. 2). Since ad $X_{\alpha}$ is nilpotent (Chap. VII, §1, no. 3, Prop. 10 (iv)), $\pi (e^{X_+}) =e^{adX_{\alpha}}$ is an elementary automorphism of $\mathfrak{g}$ . Similarly, $\pi (e^{X_-}) =$ $e^{adX_{-\alpha}}$ is an elementary automorphism of $\mathfrak{g}$ . Hence $\pi (\mathbf{S}\mathbf{L}(2, k))\subset$ Aut $_e(\mathfrak{g})$ . We make use of the notation $\theta (t)$ of §1, no. 5. For $t\in k^*$ , put

\theta_{\alpha}(t) =\pi (\theta (t)) =e^{adtX_{\alpha}}e^{adt^{-1}X_{-\alpha}}e^{adtX_{\alpha}} \tag{1}

Lemma 1 00YV

(i) For all $h\in \mathfrak{h},\theta_{\alpha}(t).h=h-\alpha (h)H_{\alpha}$ .

(ii) For all $\beta \in R,\theta_{\alpha}(t)(\mathfrak{g}^{\beta}) =\mathfrak{g}^{\beta-\beta(H_{\alpha})\alpha}$ .

(iii) If $\alpha , \beta \in R,\beta -\beta (H_{\alpha})\alpha \in R$ .

Let $h\in \mathfrak{h}$ . If $\alpha (h) = 0, [X_{\alpha}, h] = [X_{-\alpha}, h] = 0$ , so $\theta_{\alpha}(t).h=h$ . On the other hand, the formulas (5) of §1, no. 5 show that $\theta_{\alpha}(t).H_{\alpha}=-H_{\alpha}$ . This proves assertion (i). It follows that $\theta_{\alpha}(t)^2|\mathfrak{h}=$ Id. If $x\in \mathfrak{g}^{\beta}$ and $h\in \mathfrak{h}$ ,

[h, \theta_{\alpha}(t)x] =\theta_{\alpha}(t).[\theta_{\alpha}(t)h, x]-\beta (\theta_{\alpha}(t)h).\theta_{\alpha}(t)x

= (\beta (h)-\alpha (h)\beta (H_{\alpha})).\theta_{\alpha}(t)x

= (\beta -\beta (H_{\alpha})\alpha )(h).\theta_{\alpha}(t)x

so $\theta_{\alpha}(t)x\in \mathfrak{g}^{\beta-\beta(H_{\alpha})\alpha}$ . This proves (ii). Assertion (iii) follows from (ii).

Theorem 2 00YW

(i) The set $R = R(\mathfrak{g},\mathfrak{h})$ is a reduced root system in $\mathfrak{h}^*$ .

(ii) Let $\alpha \in R$ . The map $s_{\alpha ,H_{\alpha}}:\lambda \rightarrow \lambda -\lambda (H_{\alpha})\alpha$ from $\mathfrak{h}^*$ to $\mathfrak{h}^*$ is the unique reflection $s$ of $\mathfrak{h}^*$ such that $s(\alpha ) =-\alpha$ and $s(R) = R$ . For all $t\in k^*$ , $s$ is the transpose of $\theta_{\alpha}(t)|\mathfrak{h}$ .

First, R generates $\mathfrak{h}^*$ , for if $h\in \mathfrak{h}$ is such that $\alpha (h) = 0$ for all $\alpha \in R$ , then ad $h= 0$ and hence $h= 0$ since the centre of $\mathfrak{g}$ is zero. By definition, $0\notin R$ . Let $\alpha \in R$ . Since $\alpha (H_{\alpha}) = 2,s=s_{\alpha ,H_{\alpha}}$ is a reflection such that $s(\alpha ) =-\alpha$ . Then $s(R) = R$ by Lemma 1 (iii), and $\beta (H_{\alpha})\in \mathbf{Z}$ for all $\beta \in R$ (Prop. 2 (i)). This shows that R is a root system in $\mathfrak{h}^*$ . For all $h\in \mathfrak{h}$ and $\lambda \in \mathfrak{h}^*$ ,

\langle s(\lambda ), h\rangle =\langle \lambda -\lambda (H_{\alpha})\alpha , h\rangle =\langle \lambda , h-\alpha (h)H_{\alpha}\rangle =\langle \lambda , \theta_{\alpha}(t)h\rangle

so $s$ is the transpose of $\theta_{\alpha}(t)|\mathfrak{h}$ . Finally, we show that the root system R is reduced. Let $\alpha \in R$ and $y\in \mathfrak{g}^{2\alpha}$ . Since $3\alpha \notin R$ (Chap. VI, §1, no. 3, Prop. $8$ ) $, [X_{\alpha}, y] = 0$ ; on the other hand, $[X_{-\alpha}, y]\in \mathfrak{g}^{-\alpha+2\alpha}=\mathfrak{g}^{\alpha}=kX_{\alpha}$ , so $[X_{\alpha},[X_{-\alpha}, y]] = 0$ ; thus

4y= 2\alpha (H_{\alpha})y= [H_{\alpha}, y] =-[[X_{\alpha}, X_{-\alpha}], y] = 0

so $y= 0$ and $\mathfrak{g}^{2\alpha}= 0$ . In other words, $2\alpha$ is not a root. Q.E.D.

Identify $\mathfrak{h}$ canonically with $\mathfrak{h}^{**}$ . With the notations of Chap. VI, §1, no. 1, we then have, by Th. 2 (ii),

$H_{\alpha}=\alpha^{\vee}$ for all $\alpha \in R$ . (2)

The $H_{\alpha}$ thus form the root system $R^{\vee}$ in $\mathfrak{h}$ inverse to R.

We shall call $R(\mathfrak{g},\mathfrak{h})$ the root system of $(\mathfrak{g},\mathfrak{h})$ . The reflections $s_{\alpha ,H_{\alpha}}$ will be denoted simply by $s_{\alpha}$ . The Weyl group, group of weights, Coxeter number $. .$ . of $R(\mathfrak{g},\mathfrak{h})$ are called the Weyl group, group of weights, Coxeter number $. .$ . of $(\mathfrak{g},\mathfrak{h})$ . As in Chap. VI, §1, no. 1, we consider the Weyl group as operating not only on $\mathfrak{h}^*$ , but also on $\mathfrak{h}$ by transport of structure, so that $s_{\alpha}=\theta_{\alpha}(t)|\mathfrak{h}$ . Since the $\theta_{\alpha}(t)$ are elementary automorphisms of $\mathfrak{g}$ , we have:

Corollary 00YX

Every element of the Weyl group of $(\mathfrak{g},\mathfrak{h})$ , operating on $\mathfrak{h}$ , is the restriction to $\mathfrak{h}$ of an elementary automorphism of $\mathfrak{g}$ .

For a converse of this result, see §5, no. 2, Prop. 4.

Remark 1 00YY

If $\mathfrak{h}_{\mathbf{Q}}$ (resp. $\mathfrak{h}^*_{\mathbf{Q}}$ ) denotes the $\mathbf{Q}$ -vector subspace of $\mathfrak{h}$ (resp. $\mathfrak{h}^*$ ) generated by the $H_{\alpha}$ (resp. the $\alpha$ ), where $\alpha \in R$ , then $\mathfrak{h}$ (resp. $\mathfrak{h}^*$ ) can be identified canonically with $\mathfrak{h}_{\mathbf{Q}}\otimes_{\mathbf{Q}}k$ (resp. with $\mathfrak{h}^*_{\mathbf{Q}}\otimes_{\mathbf{Q}}k$ ) and $\mathfrak{h}^*_{\mathbf{Q}}$ can be identified with the dual of $\mathfrak{h}_{\mathbf{Q}}$ (Chap. VI, §1, no. 1, Prop. 1). We call $\mathfrak{h}_{\mathbf{Q}}$ and $\mathfrak{h}^*_{\mathbf{Q}}$ the canonical $\mathbf{Q}$ -structures on $\mathfrak{h}$ and $\mathfrak{h}^*($ Algebra, Chap. II, §8, no. 1, Def. 1). When we mention $\mathbf{Q}$ -rationality for a vector subspace of $\mathfrak{h}$ , for a linear form on $\mathfrak{h}$ , etc., we shall mean these structures, unless we indicate otherwise. When we mention Weyl chambers, or facets, of $R(\mathfrak{g},\mathfrak{h})$ , we shall work in $\mathfrak{h}_{\mathbf{Q}}\otimes_{\mathbf{Q}}\mathbf{R}$ or $\mathfrak{h}^*_{\mathbf{Q}}\otimes_{\mathbf{Q}}\mathbf{R}$ , that we shall denote by $\mathfrak{h}_{\mathbf{R}}$ and $\mathfrak{h}^*_{\mathbf{R}}$ .

Remark 2 00YZ

The root system $R^{\vee}$ in $\mathfrak{h}$ defines a non-degenerate symmetric bilinear form $\beta$ on $\mathfrak{h}$ (Chap. VI, §1, no. 1, Prop. 3), namely the form $(a, b) \rightarrow$ $\sum_{\alpha\in R}\langle \alpha , a\rangle \langle \alpha , b\rangle$ . By the Cor. to Th. 1, this form is just the restriction of the

Killing form to $\mathfrak{h}$ . The extension of $\beta |\mathfrak{h}_{\mathbf{Q}}\times \mathfrak{h}_{\mathbf{Q}}$ to $\mathfrak{h}_{\mathbf{Q}}\otimes_{\mathbf{Q}}\mathbf{R}$ is positive non-degenerate (Chap. VI, §1, no. 1, Prop. 3). On the other hand, we see that the inverse form on $\mathfrak{h}^*$ of the restriction to $\mathfrak{h}$ of the Killing form on $\mathfrak{g}$ is the canonical bilinear form $\Phi_R$ of R (Chap. VI, §1, no. 12).

Let $(\mathfrak{g}_1,\mathfrak{h}_1), (\mathfrak{g}_2,\mathfrak{h}_2)$ be split semi-simple Lie algebras, $\varphi$ an isomorphism from $\mathfrak{g}_1$ to $\mathfrak{g}_2$ such that $\varphi (\mathfrak{h}_1) =\mathfrak{h}_2$ . By transport of structure, the transpose of the map $\varphi |\mathfrak{h}_1$ takes $R(\mathfrak{g}_2,\mathfrak{h}_2)$ to $R(\mathfrak{g}_1,\mathfrak{h}_1)$ .

Proposition 3 00Z0

Let $\mathfrak{g}$ be a semi-simple Lie algebra, $\mathfrak{h}_1$ and $\mathfrak{h}_2$ splitting Cartan subalgebras of $\mathfrak{g}$ . There exists an isomorphism from $\mathfrak{h}^*_1$ to $\mathfrak{h}^*_2$ that takes $R(\mathfrak{g},\mathfrak{h}_1)$ to $R(\mathfrak{g},\mathfrak{h}_2)$ .

(For more precise results, see §3, no. 3, Cor. of Prop. 10, and §5, no. 3, Prop. 5).

Let $k'$ be an algebraic closure of $k,\mathfrak{g}'=\mathfrak{g}\otimes_kk',\mathfrak{h}'_i=\mathfrak{h}_i\otimes_kk'$ . Then $R(\mathfrak{g}',\mathfrak{h}'_i)$ is the image of $R(\mathfrak{g},\mathfrak{h}_i)$ under the map $\lambda \rightarrow \lambda \otimes 1$ from $\mathfrak{h}^*_i$ to $\mathfrak{h}^*_i\otimes_kk'={\mathfrak{h}'_i}^*$ . By Chap. VII, §3, no. 2, Th. 1, there exists an automorphism of $\mathfrak{g}'$ taking $\mathfrak{h}'_1$ to $\mathfrak{h}'_2$ , hence an isomorphism $\varphi$ from ${\mathfrak{h}'_1}^*$ to ${\mathfrak{h}'_2}^*$ that takes $R(\mathfrak{g}',\mathfrak{h}'_1)$ to $R(\mathfrak{g}',\mathfrak{h}'_2)$ . Then $\varphi |\mathfrak{h}^*_1$ takes $R(\mathfrak{g},\mathfrak{h}_1)$ to $R(\mathfrak{g},\mathfrak{h}_2)$ , and hence $\mathfrak{h}^*_1$ to $\mathfrak{h}^*_2$ . Q.E.D.

In view of Prop. 3, the root system of $(\mathfrak{g},\mathfrak{h})$ depends, up to isomorphism, only on $\mathfrak{g}$ and not on $\mathfrak{h}$ . In the same way, the Weyl group, group of weights $. .$ . of $(\mathfrak{g},\mathfrak{h})$ are simply called, by abuse of language, the Weyl group, group of weights $. .$ . of $\mathfrak{g}$ (cf. also §5, no. 3, Remark 2). If the Dynkin graph of $\mathfrak{g}$ is of type $A_l$ , or $B_l,. .$ . (cf. Chap. VI, §4, no. 2, Th. 3), we say that $\mathfrak{g}$ is of type $A_l$ , or $B_l,. .$ ..

Recall that, if $\alpha$ and $\beta$ are linearly independent roots, the set of $j\in \mathbf{Z}$ such that $\beta +j\alpha \in R$ is an interval $[-q, p]$ of $\mathbf{Z}$ containing 0, with $p-q=$ $-\langle \beta , \alpha^{\vee}\rangle =-\beta (H_{\alpha})$ (Chap. VI, §1, no. 3, Prop. 9).

Proposition 4 00Z1

Let $\alpha$ and $\beta$ be linearly independent roots. Let $p$ (resp. $q$ ) be the largest integer $j$ such that $\beta +j\alpha$ (resp. $\beta -j\alpha$ ) is a root.

(i) The vector subspace $\sum_{-q\leq j\leq p}\mathfrak{g}^{\beta+j\alpha}$ of $\mathfrak{g}$ is a simple $\mathfrak{s}_{\alpha}$ -module of dimension $p+q+ 1$ .

(ii) If $\alpha +\beta$ is a root, then $[\mathfrak{g}^{\alpha},\mathfrak{g}^{\beta}] =\mathfrak{g}^{\alpha+\beta}$ .

Let $X_{\alpha}$ (resp. $x$ ) be a non-zero element of $\mathfrak{g}^{\alpha}$ (resp. $\mathfrak{g}^{\beta+p\alpha}$ ). Then

[X_{\alpha}, x]\in \mathfrak{g}^{\beta+(p+1)\alpha}= 0

[H_{\alpha}, x] = (\beta (H_{\alpha}) +p\alpha (H_{\alpha}))x= (-p+q+ 2p)x= (p+q)x

Thus, $x$ is primitive of weight $p+q$ for the representation of $\mathfrak{s}\mathfrak{l}(2, k)$ on $\mathfrak{g}$ associated to $X_{\alpha}$ ; but the $\mathfrak{s}\mathfrak{l}(2, k$ )-module $\sum_{-q\leq j\leq p}\mathfrak{g}^{\beta+j\alpha}$ is of dimension $p+q+ 1$ ; hence it is simple (§1, no. 2, Prop. 2). If $\alpha +\beta \in R$ , then $p\geq 1$ , so the elements of $\mathfrak{g}^{\beta}$ are not primitive, and hence $[X_{\alpha},\mathfrak{g}^{\beta}]\not= 0$ . Since $[\mathfrak{g}^{\alpha},\mathfrak{g}^{\beta}]\subset$ $\mathfrak{g}^{\alpha+\beta}$ , we see finally that $[\mathfrak{g}^{\alpha},\mathfrak{g}^{\beta}] =\mathfrak{g}^{\alpha+\beta}$ .

Remark 3 00Z2

Recall that, by Chap. VI, §1, no. 3, Cor. of Prop. 9, the integer $p+q+ 1$ can only take the values $1,2,3,4$ .

Remark 4 00Z3

Let $(\mathfrak{g},\mathfrak{h})$ be a split reductive Lie algebra, $\mathfrak{c}$ the centre of $\mathfrak{g},\mathfrak{g}'=\mathscr{D}\mathfrak{g}$ , $\mathfrak{h}'=\mathfrak{h}\cap \mathfrak{g}'$ . Then $\mathfrak{h}=\mathfrak{c}\times \mathfrak{h}'$ , and we identify $\mathfrak{h}^{'*}$ with a vector subspace of $\mathfrak{h}^*$ . For any $\lambda \in \mathfrak{h}^*$ such that $\lambda \not= 0$ , the primary subspace $\mathfrak{g}^{\lambda}$ relative to $\lambda$ is equal to ${\mathfrak{g}'}^{\lambda|\mathfrak{h}'}$ . A non-zero weight of $\mathfrak{h}$ on $\mathfrak{g}$ is called a root of $(\mathfrak{g},\mathfrak{h})$ ; every root vanishes on $\mathfrak{c}$ . Denote by $R(\mathfrak{g},\mathfrak{h})$ the set of roots of $(\mathfrak{g},\mathfrak{h})$ ; it can be identified canonically with $R(\mathfrak{g}',\mathfrak{h}')$ . Let $\alpha \in R(\mathfrak{g},\mathfrak{h})$ . We define $h_{\alpha},H_{\alpha}$ , $\mathfrak{s}_{\alpha}$ , the isomorphisms $\mathfrak{s}\mathfrak{l}(2, k)\rightarrow \mathfrak{s}_{\alpha}$ , and the representations of $\mathfrak{s}\mathfrak{l}(2, k)$ on $\mathfrak{g}$ associated to $\alpha$ , as in the semi-simple case.

3. Invariant bilinear forms

Proposition 5 00Z4

Let $(\mathfrak{g},\mathfrak{h})$ be a split semi-simple Lie algebra, $\Phi$ an invariant symmetric bilinear form on $\mathfrak{g}$ , and W the Weyl group of $(\mathfrak{g},\mathfrak{h})$ . Then the restriction $\Phi '$ of $\Phi$ to $\mathfrak{h}$ is invariant under W. Moreover, if $\Phi$ is non-degenerate, so is $\Phi '$ .

Let $\alpha \in R$ , let $X_{\alpha}$ be a non-zero element of $\mathfrak{g}^{\alpha},\rho$ the associated representation of $\mathfrak{s}\mathfrak{l}(2, k)$ on $\mathfrak{g}$ , and $\pi$ the representation of $\mathbf{S}\mathbf{L}(2, k)$ on $\mathfrak{g}$ compatible with $\rho$ . Then $\Phi$ is invariant under $\rho$ , and hence under $\pi ($ §1, no. 4). In particular, $\Phi '$ is invariant under $\theta_{\alpha}(t)|\mathfrak{h}$ (no. 2), and hence under W. The last assertion follows from Prop. 1 (i).

Proposition 6 00Z5

Let $(\mathfrak{g},\mathfrak{h})$ be a split semi-simple Lie algebra, $\Phi$ a non-degenerate invariant symmetric bilinear form on $\mathfrak{g}$ . For all $\alpha \in R$ , let $X_{\alpha}$ be a non-zero element of $\mathfrak{g}^{\alpha}$ . Let $(H_i)_{i\in I}$ be a basis of $\mathfrak{h}$ , and $(H'_i)_{i\in I}$ the basis of $\mathfrak{h}$ such that $\Phi (H_i, H'_j) =\delta_{ij}$ . The Casimir element associated to $\Phi$ in the enveloping algebra of $\mathfrak{g}$ (Chap. I, §3, no. 7) is then

\sum\frac{1}{\Phi(X_{\alpha}, X_{-\alpha})}X_{\alpha}X_{-\alpha}+\sum H_iH'_i

\alpha \in Ri\in I

Indeed, by Prop. $1,\Phi (H_i, X_{\alpha}) =\Phi (H'_i, X_{\alpha}) = 0$ for all $i\in I,\alpha \in R$ , and $\Phi (\frac{1}{\Phi(X_{\alpha},X_{-\alpha})}X_{\alpha}, X_{-\beta})=\delta_{\alpha \beta}$ for all $\alpha , \beta \in R$ .

4. The coefficients n $_{\boldsymbol{\alpha \beta }}$

In this number, we again denote by $(\mathfrak{g},\mathfrak{h})$ a split semi-simple Lie algebra.

Lemma 2 00Z6

There exists a family $(X_{\alpha})_{\alpha\in R}$ such that, for all $\alpha \in R$ ,

$X_{\alpha}\in \mathfrak{g}^{\alpha}$ and $[X_{\alpha}, X_{-\alpha}] =-H_{\alpha}$ .

Let $R_1$ be a subset of R such that $R = R_1\cup (-R_1)$ and $R_1\cap (-R_1) =\emptyset$ . For $\alpha \in R_1$ , choose an arbitrary non-zero element $X_{\alpha}$ of $\mathfrak{g}^{\alpha}$ . There exists a unique $X_{-\alpha}\in \mathfrak{g}^{-\alpha}$ such that $[X_{\alpha}, X_{-\alpha}] =-H_{\alpha}$ (Th. 1 (iv)). Then

[X_{-\alpha}, X_{\alpha}] =H_{\alpha}=-H_{-\alpha}.Q.E.D

If $(X_{\alpha})_{\alpha\in R}$ is one family satisfying the conditions of Lemma 2, the most general family satisfying these conditions is $(t_{\alpha}X_{\alpha})_{\alpha\in R}$ where $t_{\alpha}\in k^*$ and $t_{\alpha}t_{-\alpha}= 1$ for all $\alpha \in R$ .

In the remainder of this number, we denote by $(X_{\alpha})_{\alpha\in R}$ a family satisfying the conditions of Lemma 2. We denote by $\langle \cdot ,\cdot \rangle$ a non-degenerate invariant symmetric bilinear form on $\mathfrak{g}$ .

Every $x\in \mathfrak{g}$ can be written uniquely in the form

x=h+\sum_{\alpha\in R}\mu_{\alpha}X_{\alpha}(h\in \mathfrak{h}, \mu_{\alpha}\in k)

The bracket of two such elements can be calculated by means of the following formulas:

[h, X_{\alpha}] =\alpha (h)X_{\alpha}

0 if $\alpha +\beta \notin R\cup \{0\}$

$[X_{\alpha}, X_{\beta}] =-H_{\alpha}$ if $\alpha +\beta = 0$

$N_{\alpha \beta}X_{\alpha+\beta}$ if $\alpha +\beta \in R$

the $N_{\alpha \beta}$ being non-zero elements of $k$ .

Lemma 3 00Z7

For all $\alpha \in R$ ,

\langle X_{\alpha}, X_{-\alpha}\rangle =-\frac{1}{2}\langle H_{\alpha}, H_{\alpha}\rangle

Indeed,

2\langle X_{\alpha}, X_{-\alpha}\rangle =\langle \alpha (H_{\alpha})X_{\alpha}, X_{-\alpha}\rangle =\langle [H_{\alpha}, X_{\alpha}], X_{-\alpha}\rangle

=\langle H_{\alpha},[X_{\alpha}, X_{-\alpha}]\rangle =-\langle H_{\alpha}, H_{\alpha}\rangle

Lemma 4 00Z8

Let $\alpha , \beta \in R$ be such that $\alpha +\beta \in R$ . Let $p$ (resp. $q$ ) be the largest integer $j$ such that $\beta +j\alpha \in R$ (resp. $\beta -j\alpha \in R$ ). Then,

N_{\alpha ,\beta}N_{-\alpha ,\alpha+\beta}=-p(q+ 1) \tag{3}

N_{-\alpha ,\alpha+\beta}\langle H_{\beta}, H_{\beta}\rangle =-N_{-\alpha ,-\beta}\langle H_{\alpha+\beta}, H_{\alpha+\beta}\rangle \tag{4}

N_{\alpha ,\beta}N_{-\alpha ,-\beta}= (q+ 1)^2 \tag{5}

Let $\rho$ be the representation of $\mathfrak{s}\mathfrak{l}(2, k)$ on $\mathfrak{g}$ defined by $X_{\alpha}$ . The element $e=X_{\beta+p\alpha}$ is primitive of weight $p+q$ (Prop. 4 (i)). Put

$e_n=(-n1)$ ! $^n\rho (X_-)^ne$ for $n\geq 0$ .

By Prop. 1 of §1,

(ad $X_{\alpha}$ ) $e_p= (q+ 1)e_{p-1}$

(ad $X_{-\alpha}$ )(ad $X_{\alpha}$ ) $e_p=-p(q+ 1)e_p$ .

This proves (3) since $e_p$ is a non-zero element of $\mathfrak{g}^{\beta}$ .

The form $\langle \cdot ,\cdot \rangle$ being invariant, we have

\langle [X_{-\alpha}, X_{\alpha+\beta}], X_{-\beta}\rangle =-\langle X_{\alpha+\beta},[X_{-\alpha}, X_{-\beta}]\rangle

N_{-\alpha ,\alpha+\beta}\langle X_{\beta}, X_{-\beta}\rangle =-N_{-\alpha ,-\beta}\langle X_{\alpha+\beta}, X_{-\alpha-\beta}\rangle

which, in view of Lemma 3, proves (4).

The restriction of $\langle \cdot ,\cdot \rangle$ to $\mathfrak{h}$ is non-degenerate and invariant under the Weyl group (Prop. 5). Identify $\mathfrak{h}$ and $\mathfrak{h}^*$ by means of this restriction. If $\gamma \in R,H_{\gamma}$ is identified with $2\gamma /\langle \gamma , \gamma \rangle$ (Chap. VI, §1, no. 1, Lemma 2); hence, for all $\gamma , \delta \in R$ ,

\langle \gamma , \gamma \rangle \langle H_{\delta}, H_{\delta}\rangle

\overline{\langle\delta , \delta\rangle}=\overline{\langle H_{\gamma}, H_{\gamma}\rangle} \tag{6}

Now, by Chap. VI, §1, no. 3, Prop. 10,

\frac{\langle\alpha + \beta , \alpha + \beta\rangle}{\langle\beta , \beta\rangle}=\frac{q + 1}{p} \tag{7}

so, by (3), (4), (6), (7),

N_{\alpha ,\beta}N_{-\alpha ,-\beta}=-N_{\alpha ,\beta}N_{-\alpha ,\alpha+\beta}\frac{\langle H_{\beta}, H_{\beta}\rangle}{\langle H_{\alpha+\beta}, H_{\alpha+\beta}\rangle}

=-N_{\alpha ,\beta}N_{-\alpha ,\alpha+\beta}\frac{q + 1}{p}= (q+ 1)^2

Definition 3 00Z9

A Chevalley system for $(\mathfrak{g},\mathfrak{h})$ is a family $(X_{\alpha})_{\alpha\in R}$ such that

(i) $X_{\alpha}\in \mathfrak{g}^{\alpha}$ for all $\alpha \in R$ ;

(ii) $[X_{\alpha}, X_{-\alpha}] =-H_{\alpha}$ for all $\alpha \in R$ ;

(iii) the linear map from $\mathfrak{g}$ to $\mathfrak{g}$ which is equal to $-1$ on $\mathfrak{h}$ and which takes $X_{\alpha}$ to $X_{-\alpha}$ for all $\alpha \in R$ is an automorphism of $\mathfrak{g}$ .

The extension of this definition to the case where $(\mathfrak{g},\mathfrak{h})$ is split reductive is immediate.

We shall show (§4, no. 4, Cor. of Prop. 5) that Chevalley systems for $(\mathfrak{g},\mathfrak{h})$ exist.

Proposition 7 00ZA

Let $(X_{\alpha})_{\alpha\in R}$ be a Chevalley system for $(\mathfrak{g},\mathfrak{h})$ . We retain the notation of Lemma 4. Then, $N_{-\alpha ,-\beta}= N_{\alpha ,\beta}$ and $N_{\alpha ,\beta}=\pm (q+ 1)$ for $\alpha , \beta , \alpha +\beta \in R$ .

Let $\varphi$ be the automorphism of $\mathfrak{g}$ considered in Def. 3 (iii). Then

N_{-\alpha ,-\beta}X_{-\alpha-\beta}= [X_{-\alpha}, X_{-\beta}] = [\varphi (X_{\alpha}), \varphi (X_{\beta})] =\varphi ([X_{\alpha}, X_{\beta}])

=\varphi (N_{\alpha ,\beta}X_{\alpha+\beta}) = N_{\alpha ,\beta}X_{-\alpha-\beta}

so $N_{-\alpha ,-\beta}= N_{\alpha ,\beta}$ . Now $N_{\alpha ,\beta}=\pm (q+ 1)$ by (5).

Proposition 8 00ZB

Let $(X_{\alpha})_{\alpha\in R}$ be a Chevalley system for $(\mathfrak{g},\mathfrak{h})$ . Let M be a $\mathbf{Z}$ -submodule of $\mathfrak{h}$ containing the $H_{\alpha}$ and contained in the group of weights of $R^{\vee}$ . Let $\mathfrak{g}_{\mathbf{Z}}$ be the $\mathbf{Z}$ -submodule of $\mathfrak{g}$ generated by M and the $X_{\alpha}$ . Then $\mathfrak{g}_{\mathbf{Z}}$ is a $\mathbf{Z}$ -Lie subalgebra of $\mathfrak{g}$ , and the canonical map from $\mathfrak{g}_{\mathbf{Z}}\otimes_{\mathbf{Z}}k$ to $\mathfrak{g}$ is an isomorphism.

If $\alpha , \beta \in R$ are such that $\alpha +\beta \in R$ , then $N_{\alpha ,\beta}\in \mathbf{Z}$ (Prop. 7). On the other hand, if $\alpha \in R$ and $h\in M$ , then $\alpha (h)\in \mathbf{Z}$ (Chap. VI, §1, no. 9). This proves that $\mathfrak{g}_{\mathbf{Z}}$ is a $\mathbf{Z}$ -Lie subalgebra of $\mathfrak{g}$ . On the other hand, M is a free abelian group of rank dim $\mathfrak{h}($ Algebra, Chap. VII, §3, Th. 1), so $\mathfrak{g}_{\mathbf{Z}}$ is a free abelian group of rank dim $\mathfrak{g}$ ; this implies the last assertion.

Exercises

See the exercises for § 2.