If $s\in G$, it is clear that $\varepsilon (s)\in$ Aut(R$,B$). On the other hand, the map

$\varepsilon |G : G\rightarrow$ Aut(R$,B$)

is bijective by Th. 2 of §4, no. 4.

1 $\longrightarrow T_Q\longrightarrow^f$ Aut($\mathfrak{g},\mathfrak{h}$)$\longrightarrow^{\varepsilon}$ A(R) $\longrightarrow$ 1

is exact.

a) Let $\varphi \in$ Ker $f$. Then $\varphi (\alpha ) = 1$ for all $\alpha \in R$. Since R generates the group $Q(R),\varphi$ is the identity element of $T_Q$.

b) Let $\varphi \in T_Q$. The restriction of $f(\varphi )$ to $\mathfrak{h}=\mathfrak{g}^0$ is the identity, so

Im $f\subset$ Ker$\varepsilon$.

c) Let $s\in$ Ker$\varepsilon$. Then $s|\mathfrak{h}=$ Id$_{\mathfrak{h}}$. For all $\alpha \in R$, we have $s(\mathfrak{g}^{\alpha}) =\mathfrak{g}^{\alpha}$, and there exists a $t_{\alpha}\in k^*$ such that $sx=t_{\alpha}x$ for all $x\in \mathfrak{g}^{\alpha}$. Writing down the condition that $s\in$ Aut($\mathfrak{g}$), we obtain the relations

$t_{\alpha}t_{-\alpha}= 1$ for all $\alpha \in R$

$t_{\alpha}t_{\beta}=t_{\alpha+\beta}$ when $\alpha , \beta , \alpha +\beta \in R$.

Under these conditions, there exists $\varphi \in T_Q$ such that $\varphi (\alpha ) =t_{\alpha}$ for all $\alpha \in R$ (Chap. VI, §1, no. 6, Cor. 2 of Prop. 19). Then $s=f(\varphi )$. Hence, Ker $\varepsilon \subset$ Im $f$.

d) The image of Aut($\mathfrak{g},\mathfrak{h}$) under $\varepsilon$ contains W(R) by §2, no. 2, Cor. of Th. 2, and contains Aut(R$,B$) by Prop. 1. Hence this image is equal to A(R).

Indeed, $G\cap \varepsilon^{-1}(W(R)) =\{1\}$ by Prop. 1, and

Aut($\mathfrak{g},\mathfrak{h}$) $= G.\varepsilon^{-1}(W(R))$

since $\varepsilon$ is surjective (Prop. 2).

Indeed, Aut($\mathfrak{g},\mathfrak{h}$) operates transitively on the set of framings of $(\mathfrak{g},\mathfrak{h})$ by §4, no. 4, Th. 2. Cor. 2 now follows from Cor. 1.

This follows immediately from Prop. 2.

Let $\alpha \in R,X_{\alpha}\in \mathfrak{g}^{\alpha},X_{-\alpha}\in \mathfrak{g}^{-\alpha}$ be such that $[X_{\alpha}, X_{-\alpha}] =-H_{\alpha}$. We have seen (§2, no. 2, Th. 2) that, for all $t\in k^*$, the restriction of the elementary automorphism

to $\mathfrak{h}$ is the transpose of $s_{\alpha}$; so $\varepsilon (\theta_{\alpha}(t)) =s_{\alpha}$ and consequently $\theta_{\alpha}(t)\theta_{\alpha}(-1)\in$ Ker $\varepsilon$.

Let $\rho$ be the representation of $\mathfrak{s}\mathfrak{l}(2, k)$ on $\mathfrak{g}$ associated to $X_{\alpha}$. Let $\pi$ be the representation of $\mathbf{S}\mathbf{L}(2, k)$ compatible with $\rho$. Introduce the notations $\theta (t), h(t)$ of §1, no. 5. Since $\rho (H) =$ ad $H_{\alpha}$, the elements of $\mathfrak{g}^{\lambda}$ are of weight $\lambda (H_{\alpha})$ for $\rho$. By §2, no. $2,\theta_{\alpha}(t)\theta_{\alpha}(-1) =\pi (\theta (t)\theta (-1)) =\pi (h(t))$. Hence the restriction of $\theta_{\alpha}(t)\theta_{\alpha}(-1)$ to $\mathfrak{g}^{\lambda}$ is the homothety of ratio $t^{\lambda(H_{\alpha})}($§1, no. 5, Prop. 6), hence the lemma.

$T_P\longrightarrow^qT_Q\longrightarrow^f$ Aut($\mathfrak{g},\mathfrak{h}$)

is contained in Aut$_e(\mathfrak{g})$.

Let B be a basis of R. Then $(H_{\alpha})_{\alpha\in B}$ is a basis of $R^{\vee}$, and the dual basis of $(H_{\alpha})_{\alpha\in B}$ in $\mathfrak{h}^*$ is a basis of the group P(R). Hence the group $T_P$ is generated by the homomorphisms $\lambda \rightarrow t^{\lambda(H_{\alpha})}(t\in k^*, \alpha \in B)$. If $\varphi$ is the restriction of such a homomorphism to Q(R), Lemma 1 proves that $f(\varphi )\in$ Aut$_e(\mathfrak{g})$, hence the proposition.

Let $\overline{k}$ be an algebraic closure of $k$. The map which associates to any automorphism $s$ of $\mathfrak{g}$ the automorphism $s\otimes 1$ of $\mathfrak{g}\otimes_k\overline{k}$ is an injective homomorphism from Aut($\mathfrak{g}$) to Aut($\mathfrak{g}\otimes_k\overline{k}$). We denote by Aut$_0(\mathfrak{g})$ the normal subgroup of Aut($\mathfrak{g}$) which is the inverse image of Aut$_e(\mathfrak{g}\otimes_k\overline{k})$ under this homomorphism; this is the set of automorphisms of $\mathfrak{g}$ that become elementary on extending the base field from $k$ to $\overline{k}$. It is clear that Aut$_e(\mathfrak{g})$ is independent of the choice of $\overline{k}$, and that Aut$_e(\mathfrak{g})\subset$ Aut$_0(\mathfrak{g})$. The groups Aut$_0(\mathfrak{g})$ and Aut$_e(\mathfrak{g})$ can be distinct (Chap. VII, §13, no. 1). If $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$, put

Aut$_e(\mathfrak{g},\mathfrak{h}) =$ Aut$_e(\mathfrak{g})\cap$ Aut($\mathfrak{g},\mathfrak{h}$), Aut$_0(\mathfrak{g},\mathfrak{h}) =$ Aut$_0(\mathfrak{g})\cap$ Aut($\mathfrak{g},\mathfrak{h}$).

Then $\varepsilon (s) = 1$.

By extension of $k$, we are reduced to the case where $s\in$ Aut$_e(\mathfrak{g},\mathfrak{h})$. The dimension of the nilspace of $s-1$ is at least dim $\mathfrak{h}$ (Chap. VII, §4, no. 4, Prop. 9). Hence $(s-1)|\mathfrak{h}$ is nilpotent. Since $s|\mathfrak{h}\in A(R^{\vee}),s|\mathfrak{h}$ is of finite order, and hence semi-simple (Chap. V, Appendix, Prop. 2). Consequently, $(s-1)|\mathfrak{h}= 0$, which proves that $\varepsilon (s) = 1$.

(ii) If $k$ is algebraically closed, $q(T_P) = T_Q$.

There exist a basis $(\lambda_1, . . . , \lambda_l)$ of P(R) and integers $n_1\geq 1, . . . , n_l\geq 1$ such that $(n_1\lambda_1, . . . , n_l\lambda_l)$ is a basis of Q(R). We have $m=n_1. . . n_l$. Let $\psi \in T_Q$ and put $\psi (n_1\lambda_1) =t_1, . . . , \psi (n_l\lambda_l) =t_l$. For $i= 1, . . . , l$, put $m_i=$ $\prod_{j\not=i}n_j$. Let $\chi$ be the element of $T_P$ such that $\chi (\lambda_1) =t^{m_1}_1, . . . , \chi (\lambda_l) =t^{m_l}_l$. Then

so $\chi |Q(R) =\psi^m$. This proves (i). If $k$ is algebraically closed, every element of $k^*$ is the $m$th power of an element of $k^*$, so every element of $T_Q$ is the $m$th power of an element of $T_Q$; hence, (ii) follows from (i).

a) Let $\varphi \in T_Q$ and let $\overline{k}$ be an algebraic closure of $k$. By Lemma $3,\varphi$ extends to an element of Hom(P(R)$, k^*$). By Prop. 3,

$f(\varphi )\otimes 1\in$ Aut$_e(\mathfrak{g}\otimes_k\overline{k},\mathfrak{h}\otimes_k\overline{k})$.

Hence $f(\varphi )\in$ Aut$_0(\mathfrak{g},\mathfrak{h})$, and Ker $\varepsilon \subset$ Aut$_0(\mathfrak{g},\mathfrak{h})$.

b) The image of Aut$_e(\mathfrak{g},\mathfrak{h})$ under $\varepsilon$ contains W(R) (§2, no. 2, Cor. of Th. 2). In view of a), we see that $\varepsilon^{-1}(W(R))\subset$ Aut$_0(\mathfrak{g},\mathfrak{h})$.

c) It remains to prove that Aut$_0(\mathfrak{g},\mathfrak{h})\subset \varepsilon^{-1}(W(R))$. In view of b), it suffices to prove that $\varepsilon$(Aut$_0(\mathfrak{g},\mathfrak{h})$)$\cap$ Aut(R$,B$), where B denotes a basis of R, reduces to $\{1\}$.

Let $s\in$ Aut$_0(\mathfrak{g},\mathfrak{h})$ be such that $\varepsilon (s)\in$ Aut(R$,B$). The subgroup of A(R) generated by $\varepsilon (s)$ has a finite number of orbits on R. Let U be such an orbit, of cardinal $r$, and $\mathfrak{g}^U=\sum_{\beta\in U}\mathfrak{g}^{\beta}$. Let $\beta_1\in U$, and put $\beta_i=\varepsilon (s)^{i-1}\beta_1$ for

$1\leq i\leq r$, so that $U =\{\beta_1, . . . , \beta_r\}$. Let $X_{\beta_1}$ be a non-zero element of $\mathfrak{g}^{\beta_1}$, and put $X_{\beta_i}=s^{i-1}X_{\beta_1}$ for $1\leq i\leq r$. There exists $c_U\in k^*$ such that $s^rX_{\beta_1}=$ $c_UX_{\beta_1}$, hence $s^rX_{\beta_i}=c_UX_{\beta_i}$ for all $i$, and consequently $s^r|\mathfrak{g}^U=c_U.1$. Let $\varphi \in T_Q$, and $s'=s\circ f(\varphi )$, which by a) is an element of Aut$_0(\mathfrak{g},\mathfrak{h})$. We have ${s'}^r|\mathfrak{g}^U=c'_U.1$, where

Put $B =\{\alpha_1, . . . , \alpha_l\}$ and $\sum_{i=1}^r\beta_i=\sum_{j=1}^lm^U_j\alpha_j$. Since $\varepsilon (s)\in$ Aut(R$,B$), the

$m^U_j$ are integers of the same sign and not all zero. We have

Now $\varphi$ can be chosen so that $c'_U\not= 1$ for every orbit U; indeed, this reduces to choosing elements $\varphi (\alpha_1) =t_1, . . . , \varphi (\alpha_l) =t_l$ of $k^*$ which are not annihilated by a finite number of polynomials in $t_1, . . . , t_l$, not identically zero. For such a choice of $\varphi ,\varepsilon (s') = 1$ by Lemma 2, so

This follows from Prop. 1, Cor. 1 of Prop. 2, and Prop. 4.

Let $e_1=(\mathfrak g,\mathfrak h_1,B_1,(X^1_\alpha)_{\alpha\in B_1})$, $e_2=(\mathfrak g,\mathfrak h_2,B_2,(X^2_\alpha)_{\alpha\in B_2})$ be two framings of $\mathfrak g$. There exists at least one element of $\operatorname{Aut}_0(\mathfrak g)$ that transforms $e_1$ into $e_2$ (Prop. 1 and Prop. 4). Let $\overline{k}$ be an algebraic closure of $k$. There exists an element of $\operatorname{Aut}_e(\mathfrak g\otimes_k\overline{k})$ that transforms $\mathfrak h_1\otimes_k\overline{k}$ into $\mathfrak h_2\otimes_k\overline{k}$ (Chap. VII, §3, no. 2, Th. 1). Hence, by Prop. 4 and Cor. 2 of Prop. 2, there exists an element $\varphi$ of $\operatorname{Aut}_e(\mathfrak g\otimes_k\overline{k})$ that transforms the framing $(\mathfrak g\otimes_k\overline{k},\mathfrak h_1\otimes_k\overline{k},B_1,(X^1_\alpha)_{\alpha\in B_1})$ of $\mathfrak g\otimes_k\overline{k}$ into the framing $(\mathfrak g\otimes_k\overline{k},\mathfrak h_2\otimes_k\overline{k},B_2,(X^2_\alpha)_{\alpha\in B_2})$. Since $\mathfrak h_1$ and the $X^1_\alpha$ (resp. $\mathfrak h_2$ and the $X^2_\alpha$) generate $\mathfrak g_1$ (resp. $\mathfrak g_2$), we have $\varphi(\mathfrak g_1)=\mathfrak g_2$, so $\varphi$ is of the form $\psi\otimes 1$ where $\psi\in\operatorname{Aut}_0(\mathfrak g)$, and $\psi$ transforms $e_1$ into $e_2$.

Indeed, every element of $\operatorname{Aut}(\mathfrak g)$ transforms $(\mathfrak g,\mathfrak h,B,(X_\alpha)_{\alpha\in B})$ into a framing of $\mathfrak g$. By Prop. 5, every coset of $\operatorname{Aut}(\mathfrak g)$ modulo $\operatorname{Aut}_0(\mathfrak g)$ meets $G$ in exactly one point.

Q.E.D.

It follows from Cor. 1 that the group Aut($\mathfrak{g}$)$/$Aut$_0(\mathfrak{g})$ can be identified with Aut(R$,B$), and is isomorphic to the group of automorphisms of the Dynkin graph of R.

This follows from Cor. 1 and Chap. VI, Plates I to IX.

By §3, no. 3, Cor. of Prop. 10, Aut$_0(\mathfrak{g}) =$ Aut$_e(\mathfrak{g})$.Aut$_0(\mathfrak{g},\mathfrak{h})$. On the other hand, $\varepsilon$(Aut$_e(\mathfrak{g},\mathfrak{h})$)$\supset W(R)$ by §2, no. 2, Cor. of Th. 2, so Aut$_0(\mathfrak{g},\mathfrak{h}) =$ Aut$_e(\mathfrak{g},\mathfrak{h})$.Ker$\varepsilon$.

We can assume that $k$ is algebraically closed. Let $s\in$ Aut$_0(\mathfrak{g})$ be such that $s(\mathfrak{p}_1) =\mathfrak{p}_2$. Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$ contained in $\mathfrak{b}\cap s(\mathfrak{b}) ($§3, no. 3, Prop. 10). Since $\mathfrak{h}$ and $s(\mathfrak{h})$ are Cartan subalgebras of $s(\mathfrak{b})$, there exists $u\in [\mathfrak{b},\mathfrak{b}]$ such that $e^{adu}(\mathfrak{h}) =s(\mathfrak{h})$ (Chap. VII, §3, no. 4, Th. 3). Replacing $s$ by $e^{-adu}s$, we are reduced to the case in which $s(\mathfrak{h}) =\mathfrak{h}$, and $s$ then induces on $\mathfrak{h}$ an element $\sigma$ of the Weyl group W of $(\mathfrak{g},\mathfrak{h})$ (Prop. 4). Let C be the Weyl chamber corresponding to $\mathfrak{b}$. Then $\mathfrak{p}_1$ and $\mathfrak{p}_2$ correspond to facets $F_1$ and $F_2$ of $\mathfrak{h}_{\mathbf{R}}$ contained in the closure of C. We have $\sigma (F_1) = F_2$. Since $\sigma \in W$, this implies that $F_1= F_2$ (Chap. V, §3, no. 3, Th. 2) so $\mathfrak{p}_1=\mathfrak{p}_2$.

Let K be the Killing form of $\mathfrak{g}$. If $s\in$ Aut($\mathfrak{g}$),

for all $x, y\in \mathfrak{g}$. Conversely, let $s$ be an element of V satisfying (2) and (3) for all $x, y\in \mathfrak{g}$. Then Ker($s$) $= 0$, so $s$ is bijective and $s\in$ Aut($\mathfrak{g}$). But, for all $x, y\in \mathfrak{g}$, the maps $s \rightarrow [sx, sy]$ and $s \rightarrow K(sx, sy)$ from V to $\mathfrak{g}$ and $k$ are polynomial.

(i) The group $f(T_Q)$ is closed in Aut($\mathfrak{g}$) in the Zariski topology.

(ii) The group $f(q(T_P))$ is dense in $f(T_Q)$ in the Zariski topology.

Assertion (i) follows from the equality $f(T_Q) =$ Aut($\mathfrak{g},\mathfrak{h}$)$\cap$Ker$\varepsilon$ (Prop. 2). Put $m= (P(R) : Q(R))$. Let F be a polynomial function on V; we assume that F vanishes on the $m$th power of every element of $f(T_Q)$, and show that $F|f(T_Q) = 0$; in view of Lemma 3, this will prove (ii).

The set $V'$ of elements of V inducing the identity on $\mathfrak{h}$ and leaving each $\mathfrak{g}^{\alpha}$ stable can be identified with $k^R$. Let $F'$ be the restriction of F to $V'=k^R$; this is a polynomial function. We have $f(T_Q)\subset V'$. Let $B = (\alpha_1, . . . , \alpha_l)$ be a basis of R. For all $t= (t_1, . . . , t_l)\in k^{*B}$, let $\varphi (t)$ be the homomorphism from Q(R) to the group $k^*$ that extends $t$. Then $F'(f(\varphi (t)))$ can be written as a finite sum