Dimension of the space of equivariant homomorphisms.

Question

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 :).

Answer 1

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 $\sum_i \lambda_i^2$