Assume first that H is connected; every element of $X(H)$ vanishes on the derived group $D(H)$ of H, hence we have a homomorphism $X(H/D(H))\longrightarrow X(H)$. But $H/D(H)$ is connected and commutative, hence is a torus, and $X(H/D(H))$ is a free $\mathbf Z$-module of finite type (Spectral Theories, Chap. II, §2, no. 1, Cor. 2 of Prop. 1). In the general case, it follows from the exactness of the sequence

where $X(H_0)$ is free of finite type and $X(H/H_0)$ is finite, that $X(H)$ is of finite type.

By Spectral Theories, Chap. II, §1, no. 7, Th. 4, the orthogonal complement of the kernel of $a_i$ is the subgroup $A_i$ of $X(H)$ generated by $a_i$; by loc. cit., Cor. 2 of Th. 4, the orthogonal complement of $\bigcap\operatorname{Ker}a_i$ is the subgroup of $X(H)$ generated by the $A_i$, hence the proposition.

Recall (Algebra, Chap. IX) that, by definition, this means that the linear maps $X(S)\rightarrow$ Hom$_{\mathbf{Z}}(\Gamma (S),\mathbf{Z})$ and $\Gamma (S)\rightarrow$ Hom$_{\mathbf{Z}}(X(S),\mathbf{Z})$ associated to this bilinear form are bijective.

It is immediate that if the conclusion of the proposition is true for two tori, it is also true for their product. Thus, since every torus of dimension $n$ is isomorphic to $\mathbf{U}^n$, we are reduced to the case in which $S =\mathbf{U}$. In this particular case, the assertion is immediate.

Let $f: S\rightarrow S'$ be a morphism of tori. Then, each of the linear maps $X(f) : X(S')\rightarrow X(S)$ and $\Gamma (f) :\Gamma (S)\rightarrow \Gamma (S')$ is the transpose of the other: for all $a'\in X(S')$ and all $X\in \Gamma$(S),

If $f$ is a morphism of Lie groups from S to $S'$, the homomorphism $X(f)$ is simply the dual of $f$ in the sense of Spectral Theories, Chap. II, §1, no. 7. The map $\varphi \rightarrow \widehat{\varphi}$ from Hom$_{\mathbf{Z}}(X(S'),X(S))$ to $M(S,S')$ defined in loc. cit. is the inverse of the map $f \rightarrow X(f)$ from $M(S,S')$ to Hom$_{\mathbf{Z}}(X(S'),X(S))$; the latter is thus bijective. If we identify $\Gamma (S)$ (resp. $\Gamma (S')$) with the dual $\mathbf{Z}$-module of X(S) (resp. $X(S')$) (Prop. $3$)$,\Gamma (f)$ coincides with the transpose of the homomorphism $X(f)$, hence the proposition.

Since $\rho$ is semi-simple (§1, no. 1), it suffices to prove the lemma in the case in which $\rho$ is simple. In that case, the commutant Z of $\rho (G)$ in End( $\widetilde{V}$) reduces to homotheties (Algebra, Chap. VIII, §3, no. 2, Th. 1); thus, the image of the homomorphism $\widetilde{\rho}$ is contained in the subgroup $\mathbf{C}^*.1_V$ of GL( $\widetilde{V}$), and there exists $\lambda \in X(G)$ such that $\widetilde{V} = \widetilde{V}_{\lambda}(G)$.

Formula (9), with $T'= T, g=n$, implies that $P(\rho ,T)$ is stable under $w$; further, $\widetilde{\rho}(n)$ induces an isomorphism from $\widetilde{V}_{\lambda}(T)$ to $\widetilde{V}_{w\lambda}(T)$ (formula (8)), hence the proposition.

The homomorphism $\rho$ is injective if and only if its restriction to T is injective (§2, no. 6, Prop. 9). Further, since the canonical homomorphism $\mathbf{G}\mathbf{L}(V)\rightarrow \mathbf{G}\mathbf{L}( \widetilde{V})$ is injective, we can replace $\rho$ by $\widetilde{\rho}$. It then follows from (7) that the kernel of the restriction of $\rho$ to T is the intersection of the kernels of the elements of $P(\rho ,T)$. Thus, the conclusion follows from Prop. 2 of no. 1.

The linear representation $L(\rho )$ of $\mathfrak{t}$ in $\mathfrak{g}\mathfrak{l}( \widetilde{V})$ extends to a homomorphism of $\mathbf{C}$-Lie algebras

Moreover, recall (no. 1) that we have associated to every element $\lambda$ of X(T) a linear form $\delta (\lambda )$ on $\mathfrak{t}_{\mathbf{C}}$ such that

(exp$_Tx$)$^{\lambda}=e^{\delta(\lambda)(x)},x\in \mathfrak{t}$. (10)

Recall finally (Chap. VII, §1, no. 1) that, for any map $\mu:\mathfrak{t}_{\mathbf{C}}\rightarrow \mathbf{C}$, we denote by $\widetilde{V}_\mu(\mathfrak{t}_{\mathbf{C}})$ the vector subspace of $\widetilde{V}$ consisting of the $v$ such that $(\widetilde{L}(\rho )(u))(v) =\mu(u).v$ for all $u\in \mathfrak{t}_{\mathbf{C}}$.

We now deduce from (7) and loc. cit., Prop. 3:

b) The map $\delta : X(T)\rightarrow$ Hom$_{\mathbf{C}}(\mathfrak{t}_{\mathbf{C}},\mathbf{C})$ induces a bijection from $P(\rho ,T)$ to the set of weights of $\mathfrak{t}_{\mathbf{C}}$ on $\widetilde{V}$.

Note first that, if W operates on $\mathfrak{t}_{\mathbf{C}}$ by associating to any element $w$ of W the endomorphism $L(w)_{(\mathbf{C})}$ of $\mathfrak{t}_{\mathbf{C}}$, the map $\delta$ is compatible with the operation of W on X(T) and Hom$_{\mathbf{C}}(\mathfrak{t}_{\mathbf{C}},\mathbf{C})$.

Assume now that $k=\mathbf{R}$. Denote by $\sigma$ the conjugation of $\widetilde{V}$ relative to V, defined by $\sigma (x+iy) =x-iy$ for $x, y$ in V; for every complex vector subspace E of $\widetilde{V}$, the smallest subspace of $\widetilde{V}$ rational over $\mathbf{R}$ and containing E is $E +\sigma (E)$. In particular, for all $\lambda \in X(T)$, there exists a real vector subspace $V(\lambda )$ of V such that the subspace $V(\lambda )_{(\mathbf{C})}$ of $\widetilde{V}$ is $\widetilde{V}_{\lambda}(T) + \widetilde{V}_{-\lambda}(T)$ (note that $\sigma ( \widetilde{V}_{\lambda}(T)) = \widetilde{V}_{-\lambda}(T)$). We have $V(\lambda ) = V(-\lambda )$, and the $V(\lambda )$ are the isotypical components of the representation of T on V induced by $\rho$.

a) The centre C(G) of G is a closed subgroup of T, equal to the intersection of the kernels of the roots. The canonical map $X(T/C(G))\rightarrow X(T)$ is injective with image Q(R).

b) The compact group C(G) is isomorphic to the dual of the discrete group $X(T)/Q(R) ($Spectral Theories, Chap. II, §1, no. 1, Def. 2).

c) C(G) reduces to the identity element if and only if Q(R) is equal to X(T).

By §2, no. 2, Cor. 2 of Th. 2, C(G) is contained in T. Since this is the kernel of the adjoint representation, it is the intersection of the kernels of the roots, in other words the orthogonal complement of the subgroup Q(R) of X(T). Thus, the proposition follows from Spectral Theories, Chap. II, §1, no. 7, Th. 4 and no. 5, Th. 2.

Assume first of all that C(G) reduces to the identity element, in other words that X(T) = Q(R) (Prop. 8). Let $f$ be an automorphism of G inducing the identity on T, and $\varphi = L(f)_{(\mathbf{C})}$; then $\varphi$ is an automorphism of $\mathfrak{g}_{\mathbf{C}}$ inducing the identity on $\mathfrak{t}_{\mathbf{C}}$. By Chap. VIII, §5, no. 2, Prop. 2, there exists a unique homomorphism $\theta : Q(R)\rightarrow \mathbf{C}^*$ such that $\varphi$ induces on each $\mathfrak{g}^{\alpha}$ the homothety with ratio $\theta (\alpha )$. Since $\varphi$ leaves stable the real form $\mathfrak{g}$ of $\mathfrak{g}_{\mathbf{C}}$, it commutes with the conjugation $\sigma$ of $\mathfrak{g}_{\mathbf{C}}$ with respect to $\mathfrak{g}$; but $\sigma (\mathfrak{g}^{\alpha}) =\mathfrak{g}^{-\alpha}$, so $\theta (-\alpha ) =\overline{\theta(\alpha)}$ for all $\alpha \in R$. This implies that $\theta (\alpha )\theta (\alpha ) =\theta (\alpha )\theta (-\alpha ) = 1$. It follows that $\theta$ takes values in $\mathbf{U}$, and hence corresponds by duality to an element $t$ of T such that (Ad $t$)$_{(\mathbf{C})}=\varphi$, so Int $t=f$.

In the general case, the preceding applies to the group $G/C(G)$, whose centre reduces to the identity element, and to its maximal torus $T/C(G)$. It follows that, if $f$ is an automorphism of G inducing the identity on T, there exists an element $t$ of T such that $f$ and Int $t$ induce by passage to the quotient the same automorphism of $G/C(G)$. But, since the canonical morphism $D(G)\rightarrow G/C(G)$ is a finite covering (§1, no. 4, Cor. 1 of Prop. 4), $f$ and Int $t$ induce the same automorphism of D(G), hence of $D(G)\times C(G)$, and hence also of G (loc. cit.).

If $u$ is equal to Int $g$, with $g\in G$, the subgroup $H_0= Z(g)_0$ is of maximal rank (§2, no. 2, Cor. 3). Conversely, if H contains a maximal torus S, the automorphism $u$ is of the form Int $s$ with $s\in S$ (Prop. 9).

(i) $R(Z(S)_0,T)$ is the set of $\alpha \in R(G,T)$ such that $\alpha (S) =\{1\}$.

(ii) The centre of $Z(S)_0$ is the intersection of the Ker$\alpha$ for $\alpha \in R(Z(S)_0,T)$.

(iii) If S is connected, Z(S) is connected.

The Lie algebra $L(Z(S))_{(\mathbf{C})}$ consists of the invariants of S on $\mathfrak{g}_{\mathbf{C}}$ (Chap. III, §9, no. 3, Prop. 8), and hence is the direct sum of $\mathfrak{t}_{\mathbf{C}}$ and the $\mathfrak{g}^{\alpha}$ for which $\alpha (S) =\{1\}$, hence (i). Assertion (ii) follows from Prop. 8 (no. 4), and assertion (iii) has already been proved (§2, no. 2, Cor. 5 of Th. 2).

Let $Z'_{\alpha}$ be the centralizer of (Ker $\alpha$ )$_0$. By Lemma 2, this is a connected closed subgroup of G, and $R(Z'_{\alpha},T)$ is the set of $\beta \in R(G,T)$ such that $\beta$((Ker$\alpha$ )$_0$) $=\{1\}$. Clearly, $\{\alpha ,-\alpha \} \subset R(Z'_{\alpha},T)$. Conversely, let $\beta \in R(Z'_{\alpha},T)$; since (Ker $\alpha$ )$_0$ is of finite index in Ker $\alpha$, there exists an integer $r\not= 0$ such that $t^{r\beta}= 1$ for $t\in$ Ker $\alpha$. From the exactness of the sequence

$0\longrightarrow \mathbf{Z}\longrightarrow X(T)\longrightarrow$ X(Ker $\alpha$ )$\longrightarrow 0$

corresponding by duality to the exact sequence

0 $\longrightarrow$ Ker$\alpha \longrightarrow$ T $\longrightarrow^{\alpha}\mathbf{U}\longrightarrow 0$,

it follows that $r\beta$ is a multiple of $\alpha$; by Chap. VIII, §2, no. 2, Th. 2 (i), this implies that $\beta \in \{\alpha ,-\alpha \}$. Thus, $R(Z'_{\alpha},T) =\{\alpha ,-\alpha \}$. It follows (Lemma 2) that the centre of $Z'_{\alpha}$ is Ker$\alpha$, so $Z'_{\alpha}= Z_{\alpha}$. Finally, by Cor. 1 of Prop. 4 (§1, no. 4), $D(Z_\alpha)$ is a connected closed semi-simple subgroup of $G$; it is of rank $1$ because $\mathscr{L}(Z_\alpha)(\mathbf C)=\mathfrak{g}^\alpha+\mathfrak{g}^{-\alpha}+[\mathfrak{g}^\alpha,\mathfrak{g}^{-\alpha}]$.

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 $\alpha^{\vee}\in (X(T)\otimes \mathbf{R})^*$.