# Tensor product group representations and spaces of intertwiners.

Let $V_{1}$, $V_{2}$, $W_{1}$, and $W_{2}$ be the carrier spaces of representations of some finite group $G$. Suppose also that $G$ acts trivially on $V_{1}$ and $V_{2}$. I would like to prove the following isomorphism between spaces of intertwining maps: $$\text{Hom}_{G}(V_{1}\otimes W_{1}, V_{2}\otimes W_{2})\cong \text{Hom}(V_{1},V_{2})\otimes \text{Hom}_{G}(W_{1},W_{2}).$$ I think this is probably very easy, but I'm getting a little confused with the notation.

Edit: I've attempted a proof below. Any comments would be appreciated.

• It is not true. Take for example some suitable $1$-dimensional representations such that the tensor products are isomorphic but $V_1\not\cong V_2$ and $W_1\not\cong W_2$. – Tobias Kildetoft Mar 16 '16 at 20:34
• If $G$ acts trivially on $V$ then $V\otimes W$ is just the direct sum of a number of copies of $W$ equal to the dimension of $V$. – Tobias Kildetoft Mar 16 '16 at 20:58
• @TobiasKildetoft Ah yes of course, thank you. I wrote down the wrong statement accidentally. I've now edited the question. – zander89 Mar 16 '16 at 21:00
• Yes, with the new addition, it becomes very easy as both sides are just a number of copies of $Hom(W_1,W_2)$. – Tobias Kildetoft Mar 16 '16 at 21:01
• @TobiasKildetoft Ok I see what you mean. Thankyou! I've attempted a proof below, but it seems a bit longwinded. Do you have any suggestions for simplifications? – zander89 Mar 17 '16 at 21:13

## 1 Answer

First note the following construction. Given two representations $(\pi,V)$ and $(\eta,W)$ of a group $G$, we can form the the homomorphism representation $(\rho_{V,W},\text{Hom}(V,W))$, defined by $$\rho_{V,W}(g)T:=\eta(g)\circ T\circ\pi(g^{-1}).$$ The space $\text{Hom}_{G}(V,W)$ of intertwiners can be interpreted as the fixed points of $\text{Hom}(V,W)$ under this group action.

Now consider the isomorphism $$\psi:\text{Hom}(V_{1},V_{2})\otimes \text{Hom}(W_{1},W_{2})\rightarrow \text{Hom}(V_{1}\otimes W_{1}, V_{2}\otimes W_{2})$$ obtained by sending each pure tensor $T\otimes S$ to the tensor product $T\otimes S$ and extending by linearity. The spaces $\text{Hom}(V_{1},V_{2})\otimes \text{Hom}(W_{1},W_{2})$ and $\text{Hom}(V_{1}\otimes W_{1},V_{2}\otimes W_{2})$ carry the $G$-representations $\rho_{V_{1},V_{2}}\otimes\rho_{W_{1},W_{2}}$ and $\rho_{V_{1}\otimes W_{1},V_{2}\otimes W_{2}}$ respectively. The map $\psi$ intertwines these two representations, and so they are equivalent. As a consequence, $\psi$ must restrict to an isomorphism between the $G$-fixed space of its domain and the $G$-fixed space of its image. The latter is just $\text{Hom}_{G}(V_{1}\otimes W_{1},V_{2}\otimes W_{2})$, and the former is equal to $\text{Hom}(V_{1},V_{2})\otimes\text{Hom}_{G}(W_{1},W_{2})$, since $\rho_{V_{1},V_{2}}$ acts trivially on $\text{Hom}(V_{1},V_{2})$.