Indeed, $y-y'\in$ Ker(ad$_{\mathfrak{g}}x$) and (ad$_{\mathfrak{g}}h$)$(y-y') =-2(y-y')$. But ad$_{\mathfrak{g}}x$ is injective on Ker($p+$ ad$_{\mathfrak{g}}h$) for every integer $p >0 ($§1, no. 2, Cor. of Prop. 2).

It is clear that $e^{ad_{\mathfrak{g}}(\mathfrak{n})}.h\subset h+\mathfrak{n}$. We shall prove that, if $v\in \mathfrak{n}$, then $h+v\in e^{ad_{\mathfrak{g}}(\mathfrak{n})}.h$. It suffices to prove that $h+v\in e^{ad_{\mathfrak{g}}(\mathfrak{n})}.h+\mathscr{C}0\mathfrak{n}$ for all $p\geq 1$ (since $\mathscr{C}^p\mathfrak{n}= 0$ for sufficiently large $p$). This is clear for $p= 1$ since $\mathscr{C}^1\mathfrak{n}=\mathfrak{n}$. Assume now that we have proved the existence of $y_p\in \mathfrak{n}$ and $z_p\in \mathscr{C}^p\mathfrak{n}$ such that $h+v=e^{ad_{\mathfrak{g}}y_p}.h+z_p$. Since (ad$_{\mathfrak{g}}h$)$(\mathfrak{n}) =\mathfrak{n}$, (ad$_{\mathfrak{g}}h$)$|\mathfrak{n}$ is a bijection from $\mathfrak{n}$ to $\mathfrak{n}$, hence its restriction to $\mathscr{C}^p\mathfrak{n}$, which leaves $\mathscr{C}^p\mathfrak{n}$ stable, is also bijective; consequently, there exists $z\in \mathscr{C}^p\mathfrak{n}$ such that $z_p= [z, h]$. Then

so

which establishes our assertion by induction on $p$.

If $u\in \mathfrak{p}$ and $v\in \mathfrak{q}$, there exists $w\in \mathfrak{g}$ such that $v= [x, w]$, so

On the other hand, $\mathfrak{p}$ is a subalgebra of $\mathfrak{g}$, so $[\mathfrak{p}\cap \mathfrak{q},\mathfrak{p}\cap \mathfrak{q}]\subset \mathfrak{p}\cap \mathfrak{q}$.

Let $\mathfrak{n}=$ Ker(ad $x$)$\cap$ Im(ad $x$). For all $p\in \mathbf{Z}$, let $\mathfrak{g}_p=$ Ker(ad $h-p$). By §1, no. 3 (applied to the adjoint representation of $kx+ky+kh$ on $\mathfrak{g}$), we have that $\mathfrak{n}\subset \sum_{p>0}\mathfrak{g}_p$, so ad$_{\mathfrak{g}}n$ is nilpotent for all $n\in \mathfrak{n}$, and $[h,\mathfrak{n}] =\mathfrak{n}$. We have $[x, h'-h] = 0$ and $[x, y-y'] =h'-h$, so $h'-h\in \mathfrak{n}$. By Lemmas 2 and 3, there exists $z\in \mathfrak{n}$ such that $e^{ad_{\mathfrak{g}}z}h=h'$. Since $z\in$ Ker ad$_{\mathfrak{g}}x$, we have $e^{ad_{\mathfrak{g}}z}x=x$. Lemma 1 now proves that $e^{ad_{\mathfrak{g}}z}y=y'$. Q.E.D.

Let G be a group of automorphisms of $\mathfrak{g}$. Then two $\mathfrak{s}\mathfrak{l}_2$-triplets $(x, h, y)$, $(x', h', y')$ are said to be G-conjugate if there exists $g\in G$ such that $gx=x'$, $gh=h', gy=y'$.

Consider the following conditions:

(i) $x$ and $x'$ are G-conjugate;

(ii) $(x, h, y)$ and $(x', h', y')$ are G-conjugate;

(iii) $\mathfrak{t}$ and $\mathfrak{t}'$ are G-conjugate.

We have (i) $\Leftarrow \Rightarrow$ (ii) $=\Rightarrow$ (iii). If $k$ is algebraically closed, the three conditions are equivalent.

(i) $\Leftarrow \Rightarrow$ (ii): This follows from Lemma 4.

(ii) $=\Rightarrow$ (iii): This is clear.

We assume that $k$ is algebraically closed and prove that (iii) $=\Rightarrow$ (i). We treat first the case in which $\mathfrak{t}=\mathfrak{t}'=\mathfrak{g}=\mathfrak{s}\mathfrak{l}(2, k)$. Since ad$_{\mathfrak{g}}x$ is nilpotent, the endomorphism $x$ of $k^2$ is nilpotent (Chap. I, §6, Th. 3), so there exists a matrix $A\in \mathbf{G}\mathbf{L}(2, k)$ such that $AxA^{-1}=X$, and consequently an automorphism $\alpha$ of $\mathfrak{s}\mathfrak{l}(2, k)$ such that $\alpha (x) =x'$; now $\alpha \in$ Aut$_e(\mathfrak{g}) ($§5, no. 3, Cor. 2 of Prop. 5). We now pass to the general case; we assume that $\mathfrak{t}$ and $\mathfrak{t}'$ are G-conjugate and prove that $x$ and $x'$ are G-conjugate. We can assume that $\mathfrak{t}=\mathfrak{t}'$. By the preceding, there exists $\beta \in$ Aut$_e(\mathfrak{t})$ such that $\beta x=x'$. Now, if $t\in \mathfrak{t}$ is such that ad$_{\mathfrak{t}}t$ is nilpotent, then ad$_{\mathfrak{g}}t$ is nilpotent; so $\beta$ extends to an element of Aut$_e(\mathfrak{g})$.

Put $C = [A,B]$. Since $[A,C] = 0$,

$[A$, BC$^p] = [A,B]C^p= C^{p+1}$

for every integer $p\geq 0$. Consequently, Tr(C$^p$) $= 0$ for $p\geq 1$, which proves that C is nilpotent (Algebra, Chap. VII, §3, no. 5, Cor. 4 of Prop. 13). Now let $\overline{k}$ be an algebraic closure of $k$, and let $\lambda \in \overline{k},x\in V\otimes_k\overline{k}$ be such that AB$x=\lambda x$, $x\not= 0$. The relation $[[B,A],A] = 0$ shows that $[B,A^p] =p[B,A]A^{p-1}$ for every integer $p\geq 0$. Let $r$ be the smallest integer such that $A^rx= 0$. Then

$\lambda A^{r-1}x= A^{r-1}$AB$x= A^rBx=$ BA$^rx-[B,A^r]x=-r[B,A]A^{r-1}x$.

Since $[B,A]$ is nilpotent and since $A^{r-1}x\not= 0$, this proves that $\lambda = 0$. Thus, all the eigenvalues of AB are zero, hence the lemma.

Let $\mathfrak{g}'$ be the solvable Lie algebra $kh+kx$. Since $x\in [\mathfrak{g}',\mathfrak{g}']$, ad$_{\mathfrak{g}}x$ is nilpotent (Chap. I, §5, no. 3, Th. 1); let $\mathfrak{n}$ be its kernel. Since [ad $h$, ad $x] =$ 2 ad $x$, we have (ad $h$)$\mathfrak{n}\subset \mathfrak{n}$. Let $z\in \mathfrak{g}$ be such that $h=-[x, z]$. For any integer $n\geq 0$, put $M_n=$ (ad $x$)$^n\mathfrak{g}$. If $n >0$, we have (§1, no. 1, Lemma 1)

[ad $z$,(ad $x$)$^n] =n$((ad $h$)$-n+$ 1)(ad $x$)$^{n-1}$

so, if $u\in M_{n-1}$,

$n$((ad $h$)$-n+ 1$)$u\in$ (ad $z$)(ad $x$)$u+ M_n$.

Since (ad $h$)$\mathfrak{n}\subset \mathfrak{n}$, it follows that

((ad $h$)$-n+ 1$)$(\mathfrak{n}\cap M_{n-1})\subset \mathfrak{n}\cap M_n$.

Since ad $x$ is nilpotent, $M_n= 0$ for sufficiently large $n$. Consequently, the eigenvalues of ad $h|\mathfrak{n}$ are integers $\geq 0$. Thus, the restriction of ad $h+ 2$ to $\mathfrak{n}$ is invertible.

Now $[h, z] + 2z\in \mathfrak{n}$ since

Hence there exists $z'\in \mathfrak{n}$ such that $[h, z'] + 2z'= [h, z] + 2z$, that is, $[h, y] =-2y$, putting $y=z-z'$. Since $[x, y] = [x, z] =-h$, this completes the proof.

Let $\mathfrak{n}=$ Ker(ad $x$)$^2$. If $z\in \mathfrak{n}$, then [ad $x$,[ad $x$, ad $z]] =$ ad([$x,[x, z]]$) $= 0$. By Lemma 5, ad $x\circ$ ad $z$ is nilpotent, so Tr(ad $x\circ$ ad $z$) $= 0$. This shows that $x$ is orthogonal to $\mathfrak{n}$ with respect to the Killing form $\Phi$ of $\mathfrak{g}$. Since

$\Phi$((ad $x$)$^2y, y'$) $=\Phi (y$, (ad $x$)$^2y')$

for all $y, y'\in \mathfrak{g}$, and since $\Phi$ is non-degenerate, the orthogonal complement of $\mathfrak{n}$ is the image of (ad $x$)$^2$. Hence there exists $y'\in \mathfrak{g}$ such that $x=$ (ad $x$)$^2y'$. Put

we have $[h, x] = 2x$ and $h\in$ (ad $x$)$(\mathfrak{g})$. It now suffices to apply Lemma 6.

This follows from Prop. 1 and 2.

of such matrices such that $\alpha = 1$. Then $G'= (G,G)$.

If $\alpha , \alpha '\in K^*$ and $\beta , \beta '\in K$,

0 $\alpha^{-1}$ 0 $\alpha^{'-1}$ 0 $\alpha^{-1}$ 0 $\alpha^{'-1}$

0 1

In particular,

$(1\beta )((\alpha '$ 0 $)((1\beta )^{-1}(\alpha '$ 0 $)^{-1}=(1\beta (1-{\alpha '}^2))$.

0 1 0 $\alpha^{'-1}$ 0 1 0 $\alpha^{'-1}$ 0 1

But there exists $\alpha '_0\in K^*$ such that $\alpha '_0\not= 1$ and $\alpha '_0\not=-1$, and then $k.(1-{\alpha '_0}^2)$ $=k$, hence the lemma.

Let $x$ be a non-zero nilpotent element of $\mathfrak{g}$. Choose $h, y\in \mathfrak{g}$ be such that $(x, h, y)$ is an $\mathfrak{s}\mathfrak{l}_2$-triplet (Prop. 2). The subalgebra $\mathfrak{s}$ of $\mathfrak{g}$ generated by $(x, h, y)$ can be identified with $\mathfrak{s}\mathfrak{l}(2, k)$. Let $\rho$ be the representation $z \rightarrow$ ad$_{\mathfrak{g}}z$ of $\mathfrak{s}=\mathfrak{s}\mathfrak{l}(2, k)$ on $\mathfrak{g}$, and let $\pi$ be the representation of $\mathbf{S}\mathbf{L}(2, k)$ compatible with $\rho ($§1, no. 4). The image of $\pi$ is generated by the exp($t$ ad$_{\mathfrak{g}}x$) and the exp($t$ ad$_{\mathfrak{g}}y$) with $t\in k($Algebra, Chap. III, §8, no. 9, Prop. 17), hence is contained in Aut$_e(\mathfrak{g})$. Since $\mathbf{S}\mathbf{L}(2, k)$ is equal to its derived group (Lemma 7 and loc. cit.), exp(ad$_{\mathfrak{g}}x$) belongs to the derived group G of Aut$_e(\mathfrak{g})$. Hence Aut$_e(\mathfrak{g})$ is equal to G. Assume now that $\mathfrak{g}$ is splittable. Since Aut$_0(\mathfrak{g})/$Aut$_e(\mathfrak{g})$ is commutative (§5, no. 3, Remark 3), the preceding proves that the derived group of Aut$_0(\mathfrak{g})$ is Aut$_e(\mathfrak{g})$.

The condition is clearly necessary. It is sufficient by Lemma 6.

The eigenvalues of ad$_{\mathfrak{g}}h$ belong to $\mathbf{Z}($§1, no. 2, Cor. of Prop. 2). Hence $h\in \mathfrak{h}_{\mathbf{Q}}$. There exists an element $w$ of the Weyl group of $(\mathfrak{g},\mathfrak{h})$ such that $\alpha (wh)\geq 0$ for all $\alpha \in B$ (Chap. VI, §1, no. 5, Th. 2 (i)). In view of §2, no. 2, Cor. of Th. 2, we are reduced to the case in which $\alpha (h)\in \mathbf{N}$ for all $\alpha \in B$. Let $R_+$ be the set of positive roots relative to B, and $R_-=-R_+$. There exists an $\mathfrak{s}\mathfrak{l}_2$-triplet in $\mathfrak{g}$ of the form $(x, h, y)$. Let T be the set of roots $\beta$ such that $\beta (h) =-2$. Then $T\subset R_-$ and $y\in \sum_{\beta\in T}\mathfrak{g}^{\beta}$. Assume that there

exists $\alpha \in B$ such that $\alpha (h)>2$. For all $\beta \in T$, we have $(\alpha +\beta )(h)>0$, so $\alpha +\beta \notin R_-$ and $\alpha +\beta \not= 0$; on the other hand, since $\beta \in R_-$ and $\alpha \in B$, we have $\alpha +\beta \notin R_+$; hence $\alpha +\beta \notin R\cup \{0\}$, so $[\mathfrak{g}^{\alpha},\mathfrak{g}^{\beta}] = 0$. Thus, $[y,\mathfrak{g}^{\alpha}] = 0$. But ad$_{\mathfrak{g}}y|\mathfrak{g}^{\alpha}$ is injective since $\alpha (h)>0 ($§1, no. 2, Cor. of Prop. 2). This contradiction proves that $\alpha (h)\leq 2$ for all $\alpha \in B$.

Indeed, every semi-simple element of $\mathfrak{g}$ is conjugate under Aut$_e(\mathfrak{g})$ to an element of $\mathfrak{h}$.

Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}^0$. This is a Cartan subalgebra of $\mathfrak{g}$ containing $h$ (Chap. VII, §2, no. 3, Prop. 10). We have $h\in \mathfrak{h}_{\mathbf{Q}}$. Let R be the root system of $(\mathfrak{g},\mathfrak{h}), Q$ the group of radical weights. There exists a basis B of R such that $\alpha (h)\geq 0$ for all $\alpha \in B$.

Let U be the set of $z\in \mathfrak{h}$ such that $\alpha (z)\not= 0$ for all $\alpha \in B$. Let $(H'_{\alpha})_{\alpha\in B}$ be the basis of $\mathfrak{h}$ dual to B. If $z\in U$, there exists a homomorphism from Q to $k^*$ that takes any $\gamma \in Q$ to $\prod_{\alpha\in B}\alpha (z)^{\gamma(H'_{\alpha})}$. By §5, Prop. 2 and 4, the

endomorphism $\varphi (z)$ of the vector space $\mathfrak{g}$ which induces on $\mathfrak{g}^{\gamma}$ the homothety with ratio $\prod_{\alpha\in B}\alpha (z)^{\gamma(H'_{\alpha})}$ is an elementary automorphism of $\mathfrak{g}$, which clearly

belongs to $G_h$.

Let $s\in \mathfrak{h}$. If $\gamma \in R$ is such that $\mathfrak{g}^{\gamma}\cap \mathfrak{g}^2\not= 0$,

since $\alpha (h)\geq 0$ for all $\alpha \in B$, and since the $\gamma (H'_{\alpha})$ are integers either all $\geq 0$ or all $\leq 0$, we have $\gamma (H'_{\alpha})\in \mathbf{N}$ for all $\alpha \in B$. Thus, we can consider (for $z\in \mathfrak{h}$) the endomorphism $\psi (z)$ of the vector space $\mathfrak{g}^2$ that induces on $\mathfrak{g}^{\gamma}\cap \mathfrak{g}^2$ the homothety with ratio $\prod_{\alpha\in B}\alpha (z)^{\gamma(H'_{\alpha})}$. The map $z \rightarrow \psi (z)$ from $\mathfrak{h}$ to End($\mathfrak{g}^2$) is polynomial. For $z\in U$, we have $\psi (z) =\varphi (z)|\mathfrak{g}^2$.

Let $\gamma_1, . . . , \gamma_r$ be the distinct roots of $(\mathfrak{g},\mathfrak{h})$ vanishing on $h$. If $y_1\in \mathfrak{g}^{\gamma_1}, . . .$, $y_r\in \mathfrak{g}^{\gamma_r}$, we have $e^{ady_1}. . . e^{ady_r}\in G_h$. We can thus define a map $\rho$ from $\mathfrak{h}\times \mathfrak{g}^{\gamma_1}\times \cdots \times \mathfrak{g}^{\gamma_r}$ to $\mathfrak{g}^2$ by putting

for $z\in \mathfrak{h}, y_1\in \mathfrak{g}^{\gamma_1}, . . . , y_r\in \mathfrak{g}^{\gamma_r}$. This map is polynomial, and $\rho (U,\mathfrak{g}^{\gamma_1}, . . . ,\mathfrak{g}^{\gamma_r})$ $\subset G_hx$. By Chap. VII, App. I, Prop. 3 and 4, it suffices to prove that the tangent linear map of $\rho$ is surjective at some point.

Now let T be the tangent linear map of $z \rightarrow \psi (z)$ at $h_0=\sum_{\alpha\in B}H'_{\alpha}$. Then $T(z)$ is the endomorphism of $\mathfrak{g}^2$ that induces on $\mathfrak{g}^{\gamma}\cap \mathfrak{g}^2$ the homothety with ratio

Thus, the tangent linear map of $z \rightarrow \rho (z,0, . . . ,0)$ at $h_0$ is the map $z \rightarrow [z, x]$; its image is $[x,\mathfrak{h}]$. The tangent linear map at 0 of the map $y_1 \rightarrow \rho (h_0, y_1,0, . . . ,0)$ is the map $y_1 \rightarrow \psi (h_0)[y_1, x]$; this last map has image $\psi (h_0)[x,\mathfrak{g}^{\gamma_1}] = [x,\mathfrak{g}^{\gamma_1}]$. Similarly, the tangent linear map at 0 of the map $y_i \rightarrow \rho (h_0,0, . . . ,0, y_i,0, . . . ,0)$ has image $[x,\mathfrak{g}^{\gamma_i}]$. Finally, the tangent linear map of $\rho$ at $(h_0,0, . . . ,0)$ has image

\*The group $G_h$ is an algebraic group with Lie algebra ad $\mathfrak{g}^0$.\*

(i) $h$ and $h'$ are G-conjugate;

(ii) $(x, h, y)$ and $(x', h', y')$ are G-conjugate.

We only have to prove the implication (i) $=\Rightarrow$ (ii), and we are reduced immediately to the case in which $h=h'$. Introduce $\mathfrak{g}^2$ and $G_h$ as in Lemma 8. We have $x\in \mathfrak{g}^2$, and $[x,\mathfrak{g}^0] =\mathfrak{g}^2$ by §1, no. 2, Cor. of Prop. 2. Hence $G_hx$ contains a subset of $\mathfrak{g}^2$ that is dense and open in the Zariski topology, and so does $G_hx'$. So there exists $a\in G_h$ such that $a(x) =x'$. We have $a(h) =h$, and consequently $a(y) =y'$ (no. 1, Lemma 1).

This follows from Cor. 1, the Cor. of Prop. 2, and the Cor. of Prop. 5.

This follows from Cor. 1, Prop. 1, and the Cor. of Prop. 5.

(i) $x$ is principal;

(ii) $h$ is principal;

(iii) $(x, h, y)$ is principal.

For $p\in \mathbf{Z}$, let $\mathfrak{g}^p=$ Ker(ad $h-p$). Let $\mathfrak{g}'=\sum_{p\in\mathbf{Z}}\mathfrak{g}^{2p}$. If $\mathfrak{g}$ is considered as

a module over $\mathfrak{a}=kx+kh+ky,\mathfrak{g}'$ is the sum of the simple submodules of odd dimension (§1, no. 2, Cor. of Prop. 2). Let $l$ (resp. $l'$) be the length of $\mathfrak{g}$ (resp. $\mathfrak{g}'$) considered as an $\mathfrak{a}$-module. By §1, no. 2,

dim(Ker ad $x$) $=l\geq l'=$ dim(Ker ad $h$)$\geq$ rk($\mathfrak{g}$).

The equivalence of (i) and (iii) follows immediately. On the other hand, condition (ii) means that dim(Ker ad $h$) $=$ rk($\mathfrak{g}$) and $\mathfrak{g}'=\mathfrak{g}$, in other words that

dim(Ker ad $h$) $=$ rk($\mathfrak{g}$) and $l=l'$. The equivalence of (ii) and the other conditions follows.