Lemma 3

Assume that G is simply-connected. Let C be a chamber of $\mathfrak{t}$, and $u$ an automorphism of G such that T and C are stable under $u$. Then the set of points of T fixed by $u$ is connected.

Since G is simply-connected, $\Gamma (T)$ is generated by the nodal vectors $K_{\alpha}$ $(\alpha \in R(G,T))$, and hence has a basis consisting of the family of the $K_{\alpha}$ for which $\alpha$ belongs to the basis B(C) defined by C (Chap. VI, §1, no. 10). Thus, it suffices to prove that, if $\varphi$ is an automorphism of the torus T leaving stable a basis of $\Gamma$(T), the set of fixed points of $\varphi$ is connected. Decomposing this basis into the disjoint union of the orbits of the group generated by $\varphi$, we are reduced to the case in which $T =\mathbf{U}^n$ and $\varphi$ is the automorphism $(z_1, . . . , z_n) \rightarrow (z_2, . . . , z_n, z_1)$; in this case the fixed points of $\varphi$ are the points $(z, z, . . . , z)$ for $z\in \mathbf{U}$, which form a connected subgroup of T.