simple $L$-module of highest weight $\pi$. Then $${V({\nu})} \subset {V({{\lambda}})} \otimes {V({\mu})} \iff {V_L({\nu})} \subset {V_L({{\lambda}})} \otimes {V_L({\mu})}.$$

If $\mathfrak a$ is any Lie algebra, denote ${\mathfrak U}(\mathfrak a)$ the corresponding universal enveloping algebra.

Suppose that ${V_L({\nu})} \subset {V_L({{\lambda}})} \otimes {V_L({\mu})}$; fix maximal vectors $v_{\lambda}\in {V_L({{\lambda}})}$ and $v_\mu \in {V_L({\mu})}$ for the Borel subgroup $B\cap L \subset L$ and fix $p \in {\mathfrak U}(\mathfrak
l\cap\mathfrak u^-) \otimes {\mathfrak U}(\mathfrak l\cap\mathfrak u^-)$ such that $p\,(v_{\lambda}\otimes v_\mu) \in {V_L({{\lambda}})} \otimes {V_L({\mu})}$ is a maximal vector of weight $\nu$. Since ${V_L({{\lambda}})} \otimes {V_L({\mu})}
\subset {V({{\lambda}})} \otimes {V({\mu})}$, we only need to prove that $p\,(v_{\lambda}\otimes v_\mu)$ is a maximal vector for $B$ too. If ${\alpha}\in {\Delta}'$ then we have $e_{\alpha}p\,(v_{\lambda}\otimes v_\mu) = 0$ by hypothesis. On the other hand, if ${\alpha}\in {\Delta}{\smallsetminus}{\Delta}'$, notice that $e_{\alpha}$ commutes with $p$, since by its definition $p$ is supported only on the $f_{\alpha}$’s with ${\alpha}\in {\Delta}'$. Since $v_{\lambda}\otimes
v_\mu$ is a maximal vector for $B$, then we get $$e_{\alpha}p\,(v_{\lambda}\otimes v_\mu) = p\, e_{\alpha}(v_{\lambda}\otimes v_\mu) = 0;$$ thus $p\,(v_{\lambda}\otimes v_\mu)$ generates a simple $G$-module of highest weight $\nu$.

Assume conversely that ${V({\nu})} \subset {V({{\lambda}})} \otimes {V({\mu})}$ and fix $p
\in {\mathfrak U}(\mathfrak u^-) \otimes {\mathfrak U}(\mathfrak u^-)$ such that 