I've been looking for a proof that if $V$ is an irreducible representation of a finite group $G$ $\dim(\mathrm{Hom}(V,W)^G)$ is equal to the multiplicity of $V$ in the decomposition of $W$, where $ \mathrm{Hom}(V,W)^G$ is the invariant subspace under $G$ of $ \mathrm{Hom}(V,W)$. Everything is clear except when everyone states one form or another of $$\mathrm{dim(Hom}(V,W_1\oplus W_2)^G)=\mathrm{dim(Hom(}V,W_1)^G)+\mathrm{dim(Hom(}V,W_2)^G)$$

this fact is not obvious to me, even though it might be under a layer of rust over my linear algebra.

I know that $\dim(\mathrm{Hom}(V,W_1\oplus W_2))=\dim(\mathrm{Hom}(V,W_1))+ \dim(\mathrm{Hom}(V,W_2))$ from $\mathrm{Hom}(V,W)\cong V^*\otimes W$, but how can I see that this statement translates to the restriction over the invariant subspace under $G$?


The idea is that $\oplus$ is also a direct sum as $G$-modules, and $\hom (V_1,V_2)^G$ is the same as $\hom_G(V_1,V_2)$, that is, morphisms of $G$-modules (also called morphisms of representations or intertwiners)

Therefore a morphism of $G$-modules $V\to W\oplus Z$ is the same as morphisms of $G$-modules $V\to W, V\to Z$

  • $\begingroup$ I think I get the idea morally, but is there a more explicit way to see it? What relation is there between the space of morphisms between two vector spaces and G-module morphisms between two G-modules? Why should the direct sum obey the same rule? $\endgroup$ – user438666 Jun 26 '19 at 12:05
  • $\begingroup$ Well the relation is that the space of $G$-morphisms is the space of invariants. Then as to why direct sum of $G$-modules obeys the same rules with respect to $G$-morphisms, that's just basic module theory : if you have any ring $R$ and $R$-modules $A,B,C$ then $\hom_R(A,B\oplus C) \simeq \hom_R(A,B)\oplus \hom_R(A,C)$. That's because for finitely many factors, direct sum and direct product coincide, then that isomorphism is just the statement of the universal property of a product $\endgroup$ – Max Jun 26 '19 at 12:39

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