Corollary

— There exists a morphism of Lie groups $\nu:\mathbf{SU}(2,\mathbf C)\to G$ with the following properties:

a) The image of $\nu$ commutes with the kernel of $\alpha$.

b) For all $a\in\mathbf U$, we have

If $\nu_1$ and $\nu_2$ are two morphisms from $\mathbf{SU}(2,\mathbf C)$ to $G$ with the preceding properties, there exists $a\in\mathbf U$ such that

By Th. 1 and Prop. 6 of §3, no. 6, there exists a morphism of Lie groups

that is surjective with discrete kernel. Then $\nu^{-1}(\mathbf T\cap S_\alpha)$ is a maximal torus of $\mathbf{SU}(2,\mathbf C)$ (§2, no. 3, Prop. 1). Since the maximal tori of $\mathbf{SU}(2,\mathbf C)$ are conjugate (§2, no. 2, Th. 2), we can assume, replacing $\nu$ by $\nu\circ\operatorname{Int}s$ (with $s\in\mathbf{SU}(2,\mathbf C)$) if necessary, that $\nu^{-1}(\mathbf T\cap S_\alpha)$ is the group of diagonal matrices in $\mathbf{SU}(2,\mathbf C)$. Then for all $a\in\mathbf U$, and the map is a root of $\mathbf{SU}(2,\mathbf C)$, and hence is equal to one of the two maps or (§3, no. 6, formulas (19)). In the first case, the homomorphism $\nu$ has the required properties; in the second case, the homomorphism $\nu\circ\operatorname{Int}\theta$ has them (loc. cit., formulas (18)).

If $\nu_1$ and $\nu_2$ are two morphisms from $\mathbf{SU}(2,\mathbf C)$ to $G$ satisfying the stated conditions, they both map $\mathbf{SU}(2,\mathbf C)$ into $S_\alpha$ (condition a)), hence are both universal coverings of $S_\alpha$. Hence, there exists an automorphism $\varphi$ of $\mathbf{SU}(2,\mathbf C)$ such that $\nu_2=\nu_1\circ\varphi$, and we conclude by using Prop. 9 of no. 4.

It follows from the preceding corollary that the homomorphism $\nu_{\mathbf T}$ from $\mathbf U$ to $\mathbf T$, defined by

for $a\in\mathbf U$, is independent of the choice of $\nu$. Denote by $K_\alpha\in\Gamma(\mathbf T)$ the image under $\Gamma(\nu_{\mathbf T})$ of the element $2\pi i$ of $\Gamma(\mathbf U)=2\pi i\mathbf Z$; it is called the nodal vector associated to the root $\alpha$. We have $\langle\alpha,K_\alpha\rangle=2$, in other words (no. 2, formula (2)) $\delta(\alpha)(K_\alpha)=4\pi i$; since $K_\alpha$ belongs to the intersection of $\mathfrak t$ and the $\mathscr L(S_\alpha)(\mathbf C)$, we have where $H_{\delta(\alpha)}$ is the inverse root associated to the root $\delta(\alpha)$ of $(\mathfrak g_{\mathbf C},\mathfrak t_{\mathbf C})$ (Chap. VIII, §2, no. 2). In other words, when $\Gamma(\mathbf T)\otimes\mathbf R$ is identified with the dual of $X(T)\otimes \mathbf{R}$ via the pairing $\langle ,\rangle ,K_{\alpha}$ is identified with the inverse root $\alpha^{\vee}\in (X(T)\otimes \mathbf{R})^*$.