$[X_{-\alpha}, X_{\beta}] = 0$ if $\alpha \not=\beta$ (4)

(ad $X_{\alpha}$)$^{1-n(\beta ,\alpha)}X_{\beta}= 0$ if $\alpha \not=\beta$ (5)

(ad $X_{-\alpha}$)$^{1-n(\beta ,\alpha)}X_{-\beta}= 0$ if $\alpha \not=\beta$. (6)

The family $(H_{\alpha})_{\alpha\in B}$ is a basis of $\mathfrak{h}$. If $X_{\alpha}\not= 0$ and $X_{-\alpha}\not= 0$ for all $\alpha \in B$, the Lie algebra $\mathfrak{g}$ is generated by the $X_{\alpha}$ and the $X_{-\alpha}(\alpha \in B)$.

(Recall that, if $\alpha , \beta \in B$ and $\alpha \not=\beta ,n(\beta , \alpha )$ is an integer $\leq 0$, so formulas (5) and (6) make sense.)

Formulas (1), (2) and (3) are clear. If $\alpha \not=\beta ,\beta -\alpha$ is not a root since every element of R is a linear combination of elements of B with integer coefficients all of the same sign (Chap. VI, §1, no. 6, Th. 3). This proves (4). In view of Chap. VI, §1, no. 3, Prop. 9, this also proves that the $\alpha$-chain defined by $\beta$ is

hence $\beta +(1-n(\beta , \alpha ))\alpha \notin R$, which proves (5). The equality (6) is established in a similar way. The family $(H_{\alpha})_{\alpha\in B}$ is a basis of $R^{\vee}$, and hence of $\mathfrak{h}$. If $X_{\alpha}\not= 0$ and $X_{-\alpha}\not= 0$ for all $\alpha \in B$, then $[X_{\alpha}, X_{-\alpha}] =\lambda_{\alpha}H_{\alpha}$ with $\lambda_{\alpha}\not= 0$, so the last assertion follows from §3, no. 3, Prop. 9 (iii).

$[X_{\alpha}^0, X_{-\beta}^0] = 0$ if $\alpha \not=\beta$. (14)

Indeed, relation (9) can be written

which proves (10) and (14). Relation (11) is clear. Next

hence (13). Finally,

$0 = [H_{\alpha}^0,[X_{\beta}^0, X_{-\gamma}^0]]$ by $(10),(11),(14)$ (15)

$= [[H_{\alpha}^0, X_{\beta}^0]-n(\gamma , \alpha )X_{\beta}^0, X_{-\gamma}^0]$ by (13)

$= [[H_{\alpha}^0, X_{\beta}^0]-n(\beta , \alpha )X_{\beta}^0, X_{-\gamma}^0]$ by (14);

now, considering the empty word immediately gives

so (15) implies that

for all $\gamma_1, . . . , \gamma_n\in B$; this proves (12).

Since $X_{-\alpha}^0(\emptyset ) =\alpha$, it is clear that the $X_{-\alpha}^0$ are linearly independent.

Assume that $\sum_{\alpha}a_{\alpha}H_{\alpha}^0= 0$; then, for all $\beta \in B$,

since det($n(\beta , \alpha )$)$\not= 0$, it follows that $a_{\alpha}$ = 0 for all $\alpha$. Assume that $\sum_{\alpha}a_{\alpha}X_{\alpha}^0= 0$. In view of formulas (7), (8), (9),

for all $\beta \in B$. It follows that $a_{\beta}= 0$ for all $\beta$. Since $X_{\alpha}^0, H_{\alpha}^0, X_{-\alpha}^0$ are of degree $-1,0,1$, respectively, the lemma follows from what has gone before.

Let I be the set $B\times \{-1,0,1\}$. Put $x_{\alpha}= (\alpha ,-1),h_{\alpha}= (\alpha ,0)$, and $x_{-\alpha}= (\alpha ,1)$. Let $\mathfrak{a}$ be the Lie algebra defined by the generating family I and the following set $\mathscr{R}$ of relators:

$[x_{\alpha}, x_{-\beta}]$ if $\alpha \not=\beta$

(cf. Chap. II, §2, no. 3). By Lemma 1, there exists a unique linear representation $\rho$ of $\mathfrak{a}$ on E such that

In view of Lemma 2, this proves the following result:

In the following, we identify $x_{\alpha}, h_{\alpha}, x_{-\alpha}$ with their canonical images in $\mathfrak{a}$.

Indeed, there exists an involutive automorphism of the free Lie algebra L(I) satisfying these conditions. It leaves $\mathscr{R}\cup (-\mathscr{R})$ stable, and hence defines by passage to the quotient an involutive automorphism of $\mathfrak{a}$ satisfying the conditions of the lemma. The uniqueness follows from the fact that $\mathfrak{a}$ is generated by the elements $x_{\alpha}, h_{\alpha}, x_{-\alpha}(\alpha \in B)$.

This automorphism is called the canonical involutive automorphism of $\mathfrak{a}$.

Let Q be the set of radical weights of R; this is a free $\mathbf{Z}$-module with basis B (Chap. VI, §1, no. 9). There exists a graduation of type Q on the free Lie algebra L(I) such that $x_{\alpha}, h_{\alpha}, x_{-\alpha}$ are of degrees $\alpha ,0,-\alpha$, respectively (Chap. II, §2, no. 6). Now the elements of $\mathscr{R}$ are homogeneous. Hence there exists a unique graduation of type Q on $\mathfrak{a}$ compatible with the Lie algebra structure of $\mathfrak{a}$ and such that $x_{\alpha}, h_{\alpha}, x_{-\alpha}$ are of degrees $\alpha ,0,-\alpha$, respectively. For any $\mu\in Q$, denote by $\mathfrak{a}^\mu$ the set of elements of $\mathfrak{a}$ homogeneous of degree $\mu$.

For $\mu\in Q$, let $\mathfrak{a}^{(\mu)}$ be the set of $x\in \mathfrak{a}$ such that $[h_{\alpha}, x] =\langle \mu, \alpha^{\vee}\rangle x$ for all $\alpha \in B$. The sum of the $\mathfrak{a}^{(\mu)}$ is direct. To prove the lemma, it therefore suffices to show that $\mathfrak{a}^\mu\subset \mathfrak{a}^{(\mu)}$. Let $\alpha \in B$. The endomorphism $u$ of the vector space $\mathfrak{a}$ such that $u|\mathfrak{a}^\mu=\langle \mu, \alpha^{\vee}\rangle .1$ is a derivation of $\mathfrak{a}$ such that $ux=$ (ad $h_{\alpha}$)$.x$ for $x=x_{\beta}, x=h_{\beta}, x=x_{-\beta}$; hence $u=$ ad $h_{\alpha}$, which proves our assertion.

(ii) The Lie algebra $\mathfrak{a}_-$ is generated by the family $(x_{-\alpha})_{\alpha\in B}$.

(iii) The family $(h_{\alpha})_{\alpha\in B}$ is a basis of the vector space $\mathfrak{a}^0$.

(iv) The vector space $\mathfrak{a}$ is the direct sum of $\mathfrak{a}_+,\mathfrak{a}^0,\mathfrak{a}_-$.

Let $\mathfrak{r}$ (resp. $\mathfrak{n}$) be the Lie subalgebra of $\mathfrak{a}$ generated by $(x_{\alpha})_{\alpha\in B}$ (resp. $(x_{-\alpha})_{\alpha\in B}$), and $\mathfrak{h}$ the vector subspace of $\mathfrak{a}$ generated by $(h_{\alpha})_{\alpha\in B}$. Since the $x_{\alpha}$ are homogeneous elements of $\mathfrak{a}_+,\mathfrak{r}$ is a graded subalgebra of $\mathfrak{a}_+$; hence, $[\mathfrak{h},\mathfrak{r}]\subset \mathfrak{r}$, so $\mathfrak{h}+\mathfrak{r}$ is a subalgebra of $\mathfrak{a}$; since

$[x_{-\alpha},\mathfrak{r}]\subset \mathfrak{h}+\mathfrak{r}$ for all $\alpha \in B$. Similarly, $\mathfrak{n}$ is a graded subalgebra of $\mathfrak{a}_-$, one has $[\mathfrak{h},\mathfrak{n}]\subset \mathfrak{n},\mathfrak{h}+\mathfrak{n}$ is a subalgebra of $\mathfrak{n}$, and $[x_{\alpha},\mathfrak{n}]\subset \mathfrak{h}+\mathfrak{n}$ for all $\alpha \in B$. Put $\mathfrak{a}'=\mathfrak{r}+\mathfrak{h}+\mathfrak{n}$. The preceding shows that $\mathfrak{a}'$ is stable under ad $x_{\alpha}$, ad $h_{\alpha}$ and ad $x_{-\alpha}$ for all $\alpha \in B$, and hence is an ideal of $\mathfrak{a}$. Since $\mathfrak{a}'$ contains $x_{\alpha}, h_{\alpha}, x_{-\alpha}$ for all $\alpha \in B,\mathfrak{a}'=\mathfrak{a}$. It follows from this that the inclusions $\mathfrak{r}\subset \mathfrak{a}_+,\mathfrak{h}\subset \mathfrak{a}^0$, $\mathfrak{n}\subset \mathfrak{a}_-$ are equalities, which proves the proposition.

Let L be the Lie subalgebra of E generated by B. By Chap. II, §3, Th. 1, L can be identified with the free Lie algebra generated by B. The left regular representation of E on itself is clearly injective, and defines by restriction to L an injective representation $\rho '$ of the Lie algebra L on E. Let $\varphi$ be the unique homomorphism from L to $\mathfrak{a}_-$ which takes $\alpha$ to $x_{-\alpha}$ for all $\alpha \in B$. Then, for all $\alpha \in B,\rho (\varphi (\alpha ))$ is the endomorphism of left multiplication by $\alpha$ on E, so $\rho \circ \varphi =\rho '$, which proves that $\varphi$ is injective. Thus, $(x_{-\alpha})_{\alpha\in B}$ is a basic family for $\mathfrak{a}_-$. Since $\theta (x_{-\alpha}) =x_{\alpha}$ for all $\alpha$ (cf. Lemma $4$)$, (x_{\alpha})_{\alpha\in B}$ is a basic family for $\mathfrak{a}_+$.

The second formula follows from the first by using the automorphism $\theta$. To prove the first, it suffices to show that $[x_{\gamma}, y_{\alpha \beta}] = 0$ for all $\gamma \in B$. We distinguish three cases.

Case 1$:\gamma \not=\alpha$ and $\gamma \not=\beta$. In this case, $x_{\gamma}$ commutes with $x_{-\alpha}$ and $x_{-\beta}$, and hence with $y_{\alpha \beta}$.

Case 2$:\gamma =\beta$. In this case, $x_{\gamma}$ commutes with $x_{-\alpha}$, so

$[x_{\gamma}, y_{\alpha \beta}] =$ (ad $x_{-\alpha}$)$^{1-n(\beta ,\alpha)}[x_{\gamma}, x_{-\beta}]$

$=-$(ad $x_{-\alpha}$)$^{1-n(\beta ,\alpha)}h_{\beta}=-n(\alpha , \beta$)(ad $x_{-\alpha}$)$^{-n(\beta ,\alpha)}x_{-\alpha}$.

If $n(\beta , \alpha )<0$, this expression is zero since (ad $x_{-\alpha}$)$.x_{-\alpha}= 0$. If $n(\beta , \alpha ) = 0$, then $n(\alpha , \beta ) = 0$. In both cases, $[x_{\gamma}, y_{\alpha \beta}] = 0$.

Case 3$:\gamma =\alpha$. In the algebra of endomorphisms of $\mathfrak{a}$, $[$-ad $h_{\alpha}$, ad $x_{-\alpha}] = 2$ ad $x_{-\alpha}$

and [ad $x_{\alpha}$, ad $x_{-\alpha}] =$-ad $h_{\alpha}$; thus, by §1, Lemma 1,

[ad $x_{\alpha}$, (ad $x_{-\alpha}$)$^{1-n(\beta ,\alpha)}] = (1-n(\beta , \alpha$))(ad $x_{-\alpha}$)$^{-n(\beta ,\alpha)}($-ad $h_{\alpha}-n(\beta , \alpha ))$.

Consequently,

$[x_{\gamma}, y_{\alpha \beta}] =$ [ad $x_{\alpha}$, (ad $x_{-\alpha}$)$^{1-n(\beta ,\alpha)}]x_{-\beta}+$ (ad $x_{-\alpha}$)$^{1-n(\beta ,\alpha)}$(ad $x_{\alpha}$)$x_{-\beta}$

$=-(1-n(\beta , \alpha$))(ad $x_{-\alpha}$)$^{-n(\beta ,\alpha)}$(ad $h_{\alpha}+n(\beta , \alpha )$)$x_{-\beta}$

+ (ad $x_{-\alpha}$)$^{1-n(\beta ,\alpha)}$(ad $x_{\alpha}$)$x_{-\beta}$.

Now $[h_{\alpha}, x_{-\beta}] +n(\beta , \alpha )x_{-\beta}= 0$ and $[x_{\alpha}, x_{-\beta}] = 0$, so $[x_{\gamma}, y_{\alpha \beta}] = 0$.

Let $\mathfrak{n}'=\sum_{\alpha ,\beta\in B,\alpha\not=\beta}kx_{\alpha \beta}$. Since each $x_{\alpha \beta}$ is homogeneous in $\mathfrak{a}, [\mathfrak{a}^0,\mathfrak{n}']\subset \mathfrak{n}'$

(Lemma 5 and Prop. 2). Let U (resp. V) be the enveloping algebra of $\mathfrak{a}$ (resp. $\mathfrak{a}_+$), and $\sigma$ the representation of U on $\mathfrak{a}$ defined by the adjoint representation of $\mathfrak{a}$. The ideal of $\mathfrak{a}$ generated by $\mathfrak{n}'$ is $\sigma (U)\mathfrak{n}'$. Now $\mathfrak{a}=\mathfrak{a}_++\mathfrak{a}^0+\mathfrak{a}_-$ (Prop. $2$)$,\sigma (\mathfrak{a}_-)\mathfrak{n}'= 0$ (Lemma 6), and $\sigma (\mathfrak{a}^0)\mathfrak{n}'\subset \mathfrak{n}'$ by the preceding. By the Poincaré-Birkhoff-Witt theorem, $\sigma (U)\mathfrak{n}'=\sigma (V)\mathfrak{n}'$, which proves the first assertion of the lemma. It follows that the ideal of $\theta (\mathfrak{a}_+) =\mathfrak{a}_-$ generated by the $\theta (x_{\alpha \beta}) =y_{\alpha \beta}(\alpha , \beta \in B, \alpha \not=\beta )$ is the ideal $\theta (\mathfrak{n})$ of $\mathfrak{a}$. Q.E.D.

The ideal $\mathfrak{n}+\theta (\mathfrak{n})$ of $\mathfrak{a}$ is graded since it is generated by homogeneous elements. Consequently, the Lie algebra $\mathfrak{a}/(\mathfrak{n}+\theta (\mathfrak{n}))$ is a Q-graded Lie algebra; in the remainder of this paragraph, it is denoted by $\mathfrak{g}_B$, or simply by $\mathfrak{g}$. By Prop. 2, if $\mathfrak{g}^\mu\not= 0$ then $\mu\in Q_+$, or $\mu\in Q_-$, or $\mu= 0$. Denote by $X_{\alpha}$ (resp. $H_{\alpha}, X_{-\alpha}$) the canonical image of $x_{\alpha}$ (resp. $h_{\alpha}, x_{-\alpha}$) in $\mathfrak{g}$. In view of the definition of $\mathfrak{a},\mathfrak{n}$ and $\theta (\mathfrak{n})$, it follows that $\mathfrak{g}$ is the Lie algebra defined by the generating family $((X_{\alpha}, H_{\alpha}, X_{-\alpha}))_{\alpha\in B}$ and the relations

(ad $X_{\alpha}$)$^{1-n(\beta ,\alpha)}X_{\beta}= 0(\alpha \not=\beta )$ (21)

(ad $X_{-\alpha}$)$^{1-n(\beta ,\alpha)}X_{-\beta}= 0(\alpha \not=\beta )$. (22)

Let $z\in \mathfrak{g}$ and $\mu\in Q$. Then $z\in \mathfrak{g}^\mu$ if and only if $[H_{\alpha}, z] =\langle \mu, \alpha^{\vee}\rangle z$ for all $\alpha \in B$. This follows from Lemma 5.

Since $\mathfrak{a}^0\cap (\mathfrak{n}+\theta (\mathfrak{n})) = 0$, the canonical map from $\mathfrak{a}^0$ to $\mathfrak{g}^0$ is an isomorphism. Consequently, $(H_{\alpha})_{\alpha\in B}$ is a basis of the vector space $\mathfrak{g}^0$. The commutative subalgebra $\mathfrak{g}^0$ of $\mathfrak{g}_B$ will be denoted by $\mathfrak{h}_B$ or simply by $\mathfrak{h}$. There exists a unique isomorphism $\mu \rightarrow \mu_B$ from V to $\mathfrak{h}^*$ such that $\langle \mu_B, H_{\alpha}\rangle =\langle \mu, \alpha^{\vee}\rangle$ for all $\mu\in V$ and all $\alpha \in B$.

The involutive automorphism $\theta$ of $\mathfrak{a}$ defines by passage to the quotient an involutive automorphism of $\mathfrak{g}$ that will also be denoted by $\theta$. We have $\theta (X_{\alpha}) =X_{-\alpha}$ for $\alpha \in B\cup (-B)$, and $\theta (H_{\alpha}) =-H_{\alpha}$.

The proof follows that of Lemmas 8, 9, 10, 11.

Assume that $\alpha \in B$. Let $\mathfrak{g}'$ be the set of $z\in \mathfrak{g}$ such that (ad $X_{\alpha}$)$^pz= 0$ for sufficiently large $p$. Since ad $X_{\alpha}$ is a derivation of $\mathfrak{g},\mathfrak{g}'$ is a subalgebra of $\mathfrak{g}$. By (21), $X_{\beta}\in \mathfrak{g}'$ for all $\beta \in B$. By (17), (19), (20), $H_{\beta}\in \mathfrak{g}'$ and $X_{-\beta}\in \mathfrak{g}'$ for all $\beta \in B$. Hence $\mathfrak{g}'=\mathfrak{g}$ and ad $X_{\alpha}$ is locally nilpotent. Since ad $X_{-\alpha}=\theta$(ad $X_{\alpha}$)$\theta^{-1}$, we see that ad $X_{-\alpha}$ is locally nilpotent.

We shall see that $\mathfrak{g}$ is finite dimensional, so that ad $X_{\alpha}$ is actually nilpotent.

For all $\alpha \in B$, let $s_{\alpha}$ be the reflection in V defined by $\alpha$. Since W(R) is generated by the $s_{\alpha}$ (Chap. VI, §1, no. 5, Remark 1), it suffices to prove the lemma when $w=s_{\alpha}$. In view of Lemma 8, we can define

It is verified as in Chap. I, §6, no. 8, that $\theta_{\alpha}$ is an automorphism of $\mathfrak{g}$. We have

$^1$ An endomorphism $u$ of a vector space V is called locally nilpotent (or almost

nilpotent) if, for every $v\in V$, there exists a positive integer $n$ such that $u^n(v) = 0$

(cf. Commutative Algebra, Chap. IV, §1, no. 4, Def. 2). Then exp($u$), or $e^u$, is

defined by the formula $e^u(v) =\sum_{n\geq 0}(1/n$!)$u^n(v)$ for all $v\in V$. $\theta_{\alpha}(H_{\alpha}) =e^{adX_{\alpha}}e^{adX_{-\alpha}}(H_{\beta}-n(\alpha , \beta )X_{\alpha})$

If $z\in \mathfrak{g}^\mu$,

so $\theta_{\alpha}^{-1}z\in \mathfrak{g}^{s_{\alpha}\mu}$. This shows that $\theta^{-1}_{\alpha}\mathfrak{g}^\mu\subset \mathfrak{g}^{s_{\alpha}\mu}$. Since $\theta_{\alpha}$ is an automorphism and since this inclusion holds for all $\mu\in Q$, we see that $\theta_{\alpha}^{-1}\mathfrak{g}^\mu=\mathfrak{g}^{s_{\alpha}\mu}$, which proves the lemma.