Let $G$ be a connected simple complex Lie group and $\mathfrak{g}$ be its Lie algebra. Let us fix a root decomposition, let $\mathfrak{b}_\pm$, $\mathfrak{n}_+$ and $\mathfrak{h}$ be the corresponding choices of Borel subalgebras, nilpotent subalgebras and a Cartan subalgebra. Let $\mathcal{B}_\pm$, $\mathcal{N}_\pm$, $\mathcal{H}$ be the corresponding connected subgroups of $G$.
I have a morphism of Lie algebras $f: \mathfrak{b}_+\rightarrow \mathfrak{b}_+$ constructed from the following procedure: first, start with an isometry $f : \Gamma_1 \rightarrow \Gamma_2$ for some subsets of the chosen set of simple roots $\Gamma_1,\Gamma_2\subseteq \Pi$; then, extend it to $f : \mathbb{Z}\Gamma_1 \rightarrow \mathbb{Z}\Gamma_2$, and then define $f: \mathfrak{b}_+(\Gamma_1) \rightarrow \mathfrak{b}_+(\Gamma_2)$ (Borels generated by $\Gamma_1$ and $\Gamma_2$) via $f(h_\alpha) = h_{f(\alpha)}$ and $f(e_\alpha) = e_{f(\alpha)}$, $\alpha > 0$ (Chevalley basis, let's say). Lastly, extend $f$ by zero to $f : \mathfrak{b}_+ \rightarrow \mathfrak{b}_+$. The question is: can it be integrated to a morphism $F : \mathcal{B}_+ \rightarrow \mathcal{B}_+$?
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Since $\mathcal{N}_+\subseteq \mathcal{B}_+$ is always simply connected, it is possible to integrate the morphism at least to $F|_{\mathcal{N}_+} : \mathcal{N}_+ \rightarrow \mathcal{N}_+$.
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We know that $\mathcal{B}_+ = \mathcal{N}_+ \rtimes \mathcal{H}$, so if it were possible, under some conditions, to define $F|_{\mathcal{H}} : \mathcal{H} \rightarrow \mathcal{H}$, it would be possible, I guess, to extended it to $F:\mathcal{B}_+ \rightarrow \mathcal{B}_+$. Under what conditions would it be possible?
Best Answer
I have found an answer to this problem in a book by Serre, Semisimple complex Lie algebras. Let $\mathfrak{g}$ be a semisimple complex Lie algebra with a fixed form, $\mathfrak{h}$ be a Cartan subalgebra and $\mathcal{H}$ be the corresponding Cartan subgroup (a set of roots $\Phi$ is fixed). Let $e: \mathfrak{h}\rightarrow\mathcal{H}$ be a group homomorphism defined as $e(h) = \exp(2\pi i h)$. Define abelian subgroups $\Gamma_1,\Gamma \subseteq \mathfrak{h}$ via $\Gamma_1:=\langle h_{\alpha^{\vee}} \ | \ \alpha \in \Phi\rangle$ and $\Gamma := \{h \in \mathfrak{h} \ | \ \langle h,h_{\alpha} \rangle \in \mathbb{Z} \ \text{for all} \ \alpha \in\Phi\}$. Then:
This characterization completely answers my question. If one wants to integrate something from a Cartan subalgebra to the Cartan subgroup, one simple checks what happens with $\Gamma$ under the given morphism.