Confusion about the definiton of $\mathrm{Hom}_G(V_1,V_2)$

Question

I have a bit of trouble understanding the wording in sone lecture notes.

Let $(\rho_1, V_1)$ and $(\rho_2, V_2)$ be two complex, finite dimensional representations of a group $G$. We want to investigate the space $\operatorname{Hom}_G(V_1,V_2)$ that is, the space of linear maps $\varphi: V_1 \to V_2$ such that $\varphi \rho_1(g) = \rho_2(g) \varphi$.

I always thought of $\operatorname{Hom}$ as the space of all homomorphisms, i.e. structure preserving maps. In the above case I'd just read this as: For any $v,w\in V_1$ we have $\varphi(v+w) = \varphi(v)+ \varphi(w)$, etc. Now the group is incorporated via the condition $\varphi \rho_1(g) = \rho_2(g) \varphi$, but I don't really understand what this has to do with "structure preserving"..

Can somebody maybe elaborate?

Answer 1

You can rephrase the condition as the fact that for all $g\in G$ and all $v\in V_1$, $$\varphi(g\cdot v) = g\cdot \varphi(v)$$ if you write $g\cdot v$ for the action of $g$ through $\rho_1$ on $V_1$ (and through $\rho_2$ on $V_2$). Thus you preserve the structure of $G$-module.

Answer 2

You need to think of $V_1$ and $V_2$ as $KG$-modules, where $K$ is the field of the representations, and the module action for $V_1$ is defined by $g\cdot v = \rho_1(g)(v)$ for $g \in G$ and $v \in V_1$, and similarly for $V_2$.

Then ${\rm Hom}_G(V_1,V_2)$ (which is a vector space over $K$) consists of the $KG$-homomorphisms from $V_1$ to $V_2$.