Proposition 10

Assume that $\mathfrak{g}$ is splittable.

(i) The group Aut$_e(\mathfrak{g})$ is dense in Aut$_0(\mathfrak{g})$ in the Zariski topology.

(ii) The groups Aut$_e(\mathfrak{g})$ and Aut$_0(\mathfrak{g})$ are connected in the Zariski topology.

By Prop. $3,f(q(T_P))\subset$ Aut$_e(\mathfrak{g})$. For all $s\in$ Aut$_e(\mathfrak{g})$, the closure of $s.f(q(T_P))$ in the Zariski topology contains $s.f(T_Q)$ by Prop. 9. Hence the closure of Aut$_e(\mathfrak{g})$ contains Aut$_e(\mathfrak{g}).f(T_Q) =$ Aut$_0(\mathfrak{g})$ (Prop. 6). This proves (i).

Let Aut$_e(\mathfrak{g}) =\Omega \cup \Omega '$ be a partition of Aut$_e(\mathfrak{g})$ formed by relatively open subsets in the Zariski topology, and with $\Omega \not=\emptyset$. If $\omega \in \Omega$ and if $x$ is a nilpotent element of $\mathfrak{g}$, the map $\tau :t \rightarrow \omega$ exp($t$ ad $x$) from $k$ to Aut$_e(\mathfrak{g})$ is polynomial, hence continuous in the Zariski topology; consequently, $\tau (k)$ is connected; since $\omega \in \tau (k)$, we have $\tau (k)\subset \Omega$. Thus, $\Omega$.(exp ad $kx$)$\subset \Omega$, so $\Omega$.Aut$_e(\mathfrak{g})\subset \Omega$ and $\Omega =$ Aut$_e(\mathfrak{g})$. This proves that Aut$_e(\mathfrak{g})$ is connected. It follows, by (i), that Aut$_0(\mathfrak{g})$ is connected. Q.E.D.

We shall see (§8, no. 4, Cor. of Prop. 6) that Aut$_0(\mathfrak{g})$ is closed in V in the Zariski topology, and that it is the connected component of the identity element of Aut($\mathfrak{g}$). On the other hand, Aut$_e(\mathfrak{g})$ is not in general closed in the Zariski topology.

\*Assume that $(\mathfrak{g},\mathfrak{h})$ is split. The group Aut$_0(\mathfrak{g})$ is the group $G(k)$ of $k$-points of a connected semi-simple algebraic group G with trivial centre (adjoint group). The group $f(T_Q)$ is equal to $H(k)$, where H is the Cartan subgroup of G with Lie algebra $\mathfrak{h}$. The inverse image $\widetilde{H}$ of H in the universal covering $\widetilde{G}$ of G (in the algebraic sense) has $T_P$ as its group of $k$-points. The image of $\widetilde{G}(k)$ in $G(k) =$ Aut$_0(\mathfrak{g})$ is the group Aut$_e(\mathfrak{g})$.\*