Understanding how ${\rm Hom}_k(V,W)$ is a representation of $G$. During classes, the lecturer defines what is a $G$-linear map and $\operatorname{Hom}_G(V,W)$ and he said:

Recall that $\operatorname{Hom}_k(V, W)$ is a $G$-rep via $(g \phi)(v) = g(\phi(g^{-1}v))$ for $\phi \in \operatorname{Hom}_k(V, W)$, $g \in G$, and $v \in V$. Then $\phi \in \operatorname{Hom}_G(V, W)$ precisely if $g \phi = \phi$ for all $g \in G$.

My questions are:

*

*Is $\operatorname{Hom}_G(V,W)$ define given these two representations of $G$, $(\rho,V)$ and $(\rho',W)$, right? Or is it for any representations from $V$ and $W$ of $G$?

*He said that $\operatorname{Hom}_k(V,W)$ is a $G$-rep, then which means that for every $g\in G$ we are going to associate it with an element from $\operatorname{Hom}_k(V,W)$, or it is associated with an element that is a transformation from $\alpha$ to $\beta$ with $\alpha,\beta \in \operatorname{Hom}_k(V,W)$? I also don't understand what representation he is associating with $g$. I think he is sometimes writing just $g$ instead of $\rho(g)$ or $\rho'(g)$ but that's confusing me. So in that case that would be $\rho'(g)(\phi(\rho(g)^{-1}v))$ which is in $W$?

Sorry for asking many questions, I hope someone can clear my doubts. Thank you very much in advance!
 A: Given two $G$-representations on $V$ and $W$, we can define a representation on the space of linear maps by the formula you wrote. This acts on a linear map $a:V\to W$ and gives a new linear map $ga:V\to W$. The linear maps that satisfy $ga=a$ are called $G$-invariant ( the elements of your $Hom_G(V,W)$).
It is a common practice to omit the letter $\rho$ and just right $gv$ instead of $\rho(g)v$ in order to make the notation lighter.
A: You have the right idea. But, it might be useful to see it spelled out with all of its gory details.

Remember that to say that a $k$-vector space $V$ is a $G$-rep means that there is a group homomorphism $\rho: G \to \operatorname{GL}(V)$, the target being the group of invertible $k$-linear transformations on $V$.
In order to turn $\operatorname{Hom}_k(V, W)$ into a $G$-rep, we need a group homomorphism
$$
G \to \operatorname{GL}\bigl( \operatorname{Hom}_k(V, W) \bigr).
$$
Exactly how this is a $G$-rep depends on how $V$ and $W$ are $G$-reps. I think this answers your first question. In other words, to spell out the formula more explicitly, suppose that
$\rho: G \to \operatorname{GL}(V)$
and
$\rho': G \to \operatorname{GL}(W)$.
Then we can define
$\operatorname{Hom}(\rho, \rho'): 
G \to \operatorname{GL}\bigl( \operatorname{Hom}_k(V, W) \bigr)$
via
$$
\operatorname{Hom}(\rho, \rho')(g)(\phi) 
= \rho'(g) \circ \phi \circ \rho(g)^{-1} 
$$
for all $g \in G$ and all $\phi \in \operatorname{Hom}_k(V, W)$.
This map is easier to picture like this:
$$
\biggl( V \xrightarrow{\;\;\phi\;\;} W \biggr) 
\quad\longmapsto\quad
\biggl( V \xrightarrow{\rho(g)^{-1}} 
V \xrightarrow{\;\;\phi\;\;} 
W \xrightarrow{\rho'(g)\;} W \biggr)
$$
As a composition of $k$-linear maps, it's pretty clear that the image lands back in $\operatorname{Hom}_k(V, W)$, as required.
We have to check that this construction respects the group structure on $G$. And this is where the inverse comes in. Since for any $g, h \in G$ and any $G$-rep $\rho$,
$$
\rho(gh)^{-1} = \rho\bigl( (gh)^{-1} \bigr) 
= \rho\bigl( h^{-1} g^{-1} \bigr)
= \rho\bigl( h^{-1} \bigr) \circ \rho\bigl( g^{-1} \bigr)
= \rho(h)^{-1} \circ \rho(g)^{-1}, 
$$
we have
\begin{align} 
\operatorname{Hom}(\rho, \rho')(gh)(\phi) 
&= \rho'(gh) \circ \phi \circ \rho(gh)^{-1} \\
&= \rho'(g) \circ \rho'(h) 
\circ \phi \circ \rho(h)^{-1} \circ \rho(g)^{-1} \\
&= \rho'(g) \circ 
\operatorname{Hom}(\rho, \rho')(h)(\phi) 
\circ \rho(g)^{-1} \\
&= \operatorname{Hom}(\rho, \rho')(g) 
\bigl( \operatorname{Hom}(\rho, \rho')(h)(\phi) \bigr) \\
&= \bigl( \operatorname{Hom}(\rho, \rho')(g) \circ
\operatorname{Hom}(\rho, \rho')(h) \bigr) (\phi)
\end{align}
for all $\phi \in \operatorname{Hom}_k(V, W)$.
Thus,
$$
\operatorname{Hom}(\rho, \rho')(gh) 
= \operatorname{Hom}(\rho, \rho')(g) 
\circ \operatorname{Hom}(\rho, \rho')(h), 
$$
in $\operatorname{GL}\bigl( \operatorname{Hom}_k(V, W) \bigr)$ for all $g, h \in G$, and together with the inverse property,
$$
\operatorname{Hom}(\rho, \rho')(g^{-1}) 
= \bigl( \operatorname{Hom}(\rho, \rho')(g) \bigr)^{-1}  
$$
which I leave to you to verify, this shows that $\operatorname{Hom}(\rho, \rho')$ is a $G$-rep, as desired.
