$Hom_G(\pi,\sigma)$ = $Hom_{\mathfrak{g}}(d\pi,d\sigma)$? Let $G$ be a Lie group. Let $\mathfrak{g}$ be the corresponding Lie algebra. Let $(\pi,V)$ and $(\sigma, W)$ be representations of $G$, with corresponding differentials $d\pi$ and $d\sigma$, which are $\mathfrak{g}$ representations. I want to know about the relationship between $Hom_G(\pi,\sigma)$ and $Hom_{\mathfrak{g}}(d\pi,d\sigma)$. Are they equal? Any help or reference is very much appreciated.

Considering the $G$-module $\text{Hom}_{\mathbb R}(V,W)$, we have $\text{Hom}_G(V,W)=\text{Hom}_{\mathbb R}(V,W)^G$ and $\text{Hom}_{\mathfrak g}(V,W) = \text{Hom}_{\mathbb R}(V,W)^{\mathfrak g}$, so it suffices to study the relation between $V^G$ and $V^{\mathfrak g}$ for a $G$-module $V$ (and that this also special case of your formulation when taking $V$ to be the trivial $G$-module). Now show that for any $v\in V$, the closed subgroup $\text{Stab}_G(v) := \{g\in G\ |\ g.v=v\}$ has Lie algebra $\text{Lie}(\text{Stab}_G(v))=\text{stab}_{\mathfrak g}(v) := \{X\in{\mathfrak g}\ |\ X.v = 0\}$, and use that for a connected Lie group $G$ and a closed subgroup $H\subseteq G$ we have $H=G$ if and only if $\text{Lie}(H) = \text{Lie}(G)$. 
For $G$ not connected, you only have the inclusion $V^G \subset V^{\mathfrak g}$, and the other inclusion is false in general as you can see by looking at the case where $G$ is finite.
