Proposition 8

Let G be a connected compact Lie group, T a maximal torus of G, H a Lie group and $\varphi : L(G)\rightarrow L(H)$ a homomorphism of Lie algebras. There exists a morphism of Lie groups $f: G\rightarrow H$ such that $L(f) =\varphi$ if and only if there exists a morphism of Lie groups $f_T: T\rightarrow H$ such that $L(f_T) =\varphi |L(T)$; then $f_T=f|T$.

If $f: G\rightarrow H$ is a morphism of Lie groups such that $L(f) =\varphi$, the restriction $f_T$ of $f$ to T is the unique morphism from T to H such that $L(f_T) =\varphi |L(T)$. Conversely, let $f_T: T\rightarrow H$ be a morphism of Lie groups such that $L(f_T) =\varphi |L(T)$. Let $\widetilde{D}(G)$ and $p$ be as above; the map $L(p)$ induces an isomorphism from $L( \widetilde{D}(G))$ to the derived algebra $\mathfrak{b}$ of L(G). There exists a morphism of Lie groups $\widetilde{f}: \widetilde{D}(G)\rightarrow H$ such that $L( \widetilde{f}) = (\varphi |\mathfrak{b})\circ L(p)$ (Chap. III, §6, no. 1, Th. 1). The morphisms $t \rightarrow \widetilde{f}(t)$ and $t \rightarrow f_T(p(t))$ from $\widetilde{T}$ to H induce the same homomorphism of Lie algebras, and hence coincide. Applying Lemma 2, we deduce the existence of a morphism $f: G\rightarrow H$ such that $L(f)$ and $\varphi$ coincide on L(T) and $\mathfrak{b}$. Since $L(G) =\mathfrak{b}+ L(T)$, we have $L(f) =\varphi$.