Schur's Lemma says:

Let $\rho_1:G\to GL(V_1)$ and $\rho_2:G\to GL(V_2)$ be two irreductible representations over $\mathbb{C}$ of a group $G$. If $f:V_1\to V_2$ is an equivariant linear transformation, then:

1) Either $f$ is an isomorphism or $f=0$ and

2) If $V_1=V_2$ and $\rho_1=\rho_2$, then $f=\lambda\text{ id}$ for some $\lambda\in\mathbb{C}$.

Surely, this implies $\dim \hom(V,V)^G=1$. What I don't understand is why it implies that $\dim \hom(V_1,V_2)^G=1$ if $V_1\cong V_2$.

If $V_1$ and $V_2$ are isomorphic but not equal, we cannot prove that $f$ has an eigenvalue as it doesn't even make sense. So it seems to me that Schur's Lemma does not apply here.


If $(V,\rho_1)$ is $G$-isomorphic to $(W,\rho_2)$ then $\hom_G(V,W)$ is linearly isomorphic to $\hom_G(V,V)$. To prove this you only need to know that $\hom_G(V,-)$ is a functor. Functors map isomorphisms to isomorphisms.

For a direct proof: if $f:V\to W$ is a $G$-isomorphism then the induced map

$$\hom_G(f):\hom_G(V,V)\to\hom_G(V,W)$$ $$\hom_G(f)(h)=f\circ h$$

is a linear isomorphism. The inverse is given by $\hom_G(f^{-1})$.


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