Notation of subscript $Hom_{\mathfrak{g}}(A,B)=Hom_{\mathfrak{b}}(C,B)?$ Let $\mathfrak{g}$ be a Lie algebra, and let $\mathfrak{b}$ be a Borel subalgebra of $\mathfrak{g}$.
Let $A,B$ be $\mathfrak{g}$-modules, and $C$ be a $\mathfrak{b}$-module.
What does the subscript mean here: 
$$Hom_{\mathfrak{g}}(A,B)=Hom_{\mathfrak{b}}(C,B)?$$
I guess this means that $\psi:A\to B$ satisfies $\mathfrak{g}\cdot \psi = \psi \cdot \mathfrak{g}$? and the same but for $\mathfrak{b}$ in the image? So $\mathfrak{g}$ and $\mathfrak{b}$ equivariance?
Source, end of page 1: http://www.math.harvard.edu/~gaitsgde/267y/catO.pdf
 A: They are introducing a Verma module implicitly via an adjunction. To be concrete, suppose $M$ and $N$ are $\mathfrak g$- and $\mathfrak b$-modules, respectively. Because $\mathfrak b$ is a subalgebra of $\mathfrak g$, $M$ can be made into a $\mathfrak b$-module, denote it by $UM$. It then makes sense to consider the hom-set $\hom_{\mathfrak b}(N,UM)$. 
Similarly, we can consider the extended $\mathfrak g$-module obtained from $N$ as $U(\mathfrak g)\otimes_{U(\mathfrak b)} N$, denote this by $EN$, and we can consider the hom-set $\hom_{\mathfrak g}(EN,M)$. There is a natural isomorphism
$\hom_{\mathfrak g}(EN,M) \to \hom_{\mathfrak b}(N,UM)$
which the author is considering to be an identification. 
Explicitly, this maps a $\mathfrak g$-linear map $f : EN\to M$ to the $\mathfrak b$-linear map $g : N\to UM$ such that $g(n) = f(1\otimes n)$. The inverse assigns a map $g$ to the map $f$ such that $f(a\otimes n) = a g(n)$, and it is readily checked this defines the desired natural isomorphisms.
A: $\operatorname{Hom}_\mathfrak{g}(A,B)$ simply denotes the set of $\mathfrak{g}$-module homomorphisms from $A$ to $B$, and $\operatorname{Hom}_\mathfrak{b}(C,B)$ denotes the set of $\mathfrak{b}$-module homomorphisms from $C$ to $B$. That is, an element $f\in\operatorname{Hom}_\mathfrak{g}(A,B)$ is a linear map $f : A\to B$ such that $f(x\cdot a) = x\cdot f(a)$ for all $x\in\mathfrak{g}$, $a\in A$.
The subscript tells you which Lie algebra you are considering the arguments as modules over: if $\mathfrak{b}$ and $\mathfrak{g}$ are both relevant in some application or topic, then writing simply $\operatorname{Hom}(A,B)$ (for $\mathfrak{g}$-modules $A$ and $B$) might be ambiguous. After all, $A$ and $B$ can be thought of as either $\mathfrak{g}$-modules OR as $\mathfrak{b}$-modules via restriction of scalars.
