Proposition 11

The homomorphism $f(G,T) :\Gamma (T)\rightarrow \pi_1(G)$ is surjective. Its kernel is the subgroup $N(G,T)$ of $\Gamma (T)$ generated by the family of nodal vectors $(K_{\alpha})_{\alpha\in R(G,T)}$.

The homomorphism $f(G,T)$ is surjective by Prop. 3 (§2, no. 4). We denote by $A(G,T)$ the assertion: “the kernel of $f(G,T)$ is generated by the $K_{\alpha}$” which it remains to prove, and distinguish several cases:

$a) G$ is simply-connected. Let $\rho :\mathfrak{g}_{\mathbf{C}}\rightarrow \mathfrak{g}\mathfrak{l}(V)$ be a linear representation of $\mathfrak{g}_{\mathbf{C}}$ on a finite dimensional complex vector space V. Restricting to $\mathfrak{g}$, we obtain a representation of $\mathfrak{g}$ on the real vector space $V_{(\mathbf{R})}$; since G is simply-connected, there exists an analytic linear representation $\pi$ of G on $V_{(\mathbf{R})}$ such that $\rho = L(\pi )$. It follows from Prop. 7 of no. 3 that the image $\delta (X(T))$ of X(T) in $\mathfrak{t}^*_{\mathbf{C}}$ contains all the weights of $\rho$ on V. This being true for every representation $\rho$ of $\mathfrak{g}_{\mathbf{C}}$, it follows from Chap. VIII, §7, no. 2, Th. 1 that $\delta (X(T))$ contains the group of weights of $\delta$(R), which is by definition the set of $\lambda \in \mathfrak{t}^*_{\mathbf{C}}$ such that $\lambda (H_{\delta(\alpha)})\in \mathbf{Z}$ for all $\alpha \in R$, in other words, $\lambda (K_{\alpha})\in 2\pi i\mathbf{Z}$ for all $\alpha \in R$. Thus, the group X(T) contains all the elements $\lambda$ of $X(T)\otimes \mathbf{Q}$ such that $\langle \lambda , K_{\alpha}\rangle \in \mathbf{Z}$ for all $\alpha \in R$, which implies by duality that $\Gamma (T)$ is generated by the $K_{\alpha}$, hence the assertion $A(G,T)$.

$b) G$ is the direct product of a simply-connected group $G'$ and a torus S. Then T is the direct product of a maximal torus $T'$ of $G'$ with $S,\Gamma (T)$ can be identified with $\Gamma (T')\times \Gamma$(S), $\pi_1(G)$ with $\pi_1(G')\times \pi_1$(S), and $f(G,T)$ with the homomorphisms with components $f(G',T')$ and $f(S,S)$. Since $f(S,S)$ is bijective, the canonical map $\Gamma (T')\rightarrow \Gamma (T)$ maps Ker $f(G',T')$ bijectively onto Ker $f(G,T)$. Moreover, the $K_{\alpha}$ belong to the Lie algebra of the derived group $G'$ of G, hence to the image of $\Gamma (T')$, so it is immediate that $A(G',T')$ implies $A(G,T)$, hence assertion $A(G,T)$, in view of a).

c) General case. There exists a surjective morphism $p: G'\rightarrow G$ with finite kernel, where $G'$ is the direct product of a simply-connected group by a torus (§1, no. 4, Prop. 4). If $T'$ is the inverse image of T in $G'$ (this is a maximal torus of $G'$ by §2, no. 3, Prop. 1), and N the kernel of $p$, we have exact sequences $0\rightarrow N\rightarrow G'\rightarrow G\rightarrow 0$ and $0\rightarrow N\rightarrow T'\rightarrow T\rightarrow 0$, hence a commutative diagram with exact rows (no. 2, Remark 1 and General Topology, Chap. XI, in preparation)

0 $\longrightarrow \Gamma (T')\longrightarrow \Gamma (T)\longrightarrow$ N $\longrightarrow$ 0

$f(G',T')f(G,T)$ Id$_N$

0 $\longrightarrow \pi_1(G')\longrightarrow \pi_1(G)\longrightarrow$ N $\longrightarrow 0$.

It follows immediately from the snake diagram (Algebra, Chap. X, p. 4, Prop. 2) that $A(G',T')$ implies $A(G,T)$, hence the proposition, in view of b).