# Dimension of the space of equivariant homomorphisms.

For a complex representation $$\rho$$ on $$V$$, I am interested in the dimension of the space of equivariant homomorphisms, i.e. $$\varphi :V\to V$$, s.t. $$\varphi \circ \rho =\rho \circ \varphi$$. I know that if $$V$$ decomposes into pairwise inequivalent, irreducible representations $$V=V_1\oplus \cdots \oplus V_n$$, according to Schur's Lemma the dimension must be $$n$$ (because on each $$V_i$$, $$\varphi$$ acts as a multiple of the identity).

But what if the representations aren't inequivalent but each occurs $$\lambda_i$$ times: $$V=V_1\otimes \mathbb{C}^{\lambda _1}\oplus \cdots \oplus V_n\otimes \mathbb{C}^{\lambda _n}$$, this should give us more freedom in "connecting" each of the $$V_i$$.

Intuitively I would think that for each of the identical $$\lambda _i$$ spaces I can select another of the $$\lambda _i$$ spaces, which would make $$\lambda _i!$$ options, and the dimension would turn out to be: $$\displaystyle \sum \limits _i\lambda _i!$$
I'd be happy if someone could quickly tell me whether this is true :).

There are no factorials involved in the dimension of $$\mathrm{Hom}_{\Bbb C[G]}(V,V)$$. Let's just use Schur's lemma a bunch of times and the fact that $$\mathrm{Hom}(V,U\oplus W)=\mathrm{Hom}(V,U)\oplus\mathrm{Hom}(V,W)$$ (and similarly in the first argument). We have $$\mathrm{Hom}_{\Bbb C[G]}(V,V)=\mathrm{Hom}_{\Bbb C[G]}(\oplus_i V_i^{\lambda_i},\oplus_i V_i^{\lambda_i})=\bigoplus_i \mathrm{Hom}_{\Bbb C[G]}(V_i^{\lambda_i},V_i^{\lambda_i})=\bigoplus_i\mathrm{Hom}_{\Bbb C[G]}(V_i,V_i^{\lambda_i})^{\lambda_i}=\bigoplus_i\mathrm{Hom}_{\Bbb C[G]}(V_i,V_i)^{\lambda_i^2}$$ We know by Schur that $$\mathrm{Hom}_{\Bbb C[G]}(V_i,V_i)$$ has dimension $$1$$. Thus the total dimension is $$\sum_i \lambda_i^2$$