Choices of decompositions of a representation into irreducible components (Serre, Ex. 2.8)

The following is exercise 2.8 in Serre's Linear Representations of Finite Groups. If $V$ is a representation of a group $G$, recall it has a canonical decomposition $V=\oplus_1^n V_i$, where each $V_i$ is the sum of some number of copies of irreducible representations isomorphic to the same irreducible representation $W_i$. The exercise concerns the number of ways a $V_i$ may be written as a sum of irreducible representations.

Let $H$ be the vector space of linear maps $h:W_i\rightarrow V$ such that $\rho_s h = h \rho_s$ for all $s\in G$. Schur's lemma implies that in fact $h\in H_i$ maps $W_i$ into $V_i$.

(a) Show that the dimension of $H_i$ is equal to the number of times $W_i$ appears in $V_i$, i.e., to $\dim V_i/\dim W_i$. [Reduce to the case $V=W_i$ and use Schur's lemma.]

(b) Let $G$ act on $H_i\otimes W_i$ through the tensor product of the trivial representation of $G$ on $H_i$ and the given representation on $W_i$. Show that the map $$F: H_i\otimes W_i\rightarrow V_i$$ defined by the formula $$F(\sum h_\alpha \cdot w_\alpha)=\sum h_\alpha(w_\alpha)$$ is an isomorphism of $H_i\otimes W_i$ onto $V_i$. [Same method.]

(c) Let $(h_1,...h_k)$ be a basis of $H_i$ and form the direct sum $\oplus_1^k W_i$ of $k$ copies of $W_i$. The system $(h_1,\dots, h_k)$ defines in an obvious way a linear mapping $h$ of $\oplus_1^k W_i$ into $V_i$. Show that it is an isomorphism of representations and that each representation is thus obtainable. [Apply (b), or argue directly.]

Part (a) I am comfortable with. Pick one decomposition of $V_i$ as a sum of irreducible representations. By Schur's lemma, we see that $h$ is determined by where the identity element is sent to on each copy of $W_i$. Further, it must be sent to some multiple of the identity. So the map is determined by the choice of the scale factors for each copy of $W_i$ that appears in $V_i$.

Part (b) works the same way. Decompose $V_i$ into a sum of copies of $W_i$, and let $h_j$ be a basis for $H_i$ such that $h_j$ maps $W_i$ isomorphically into one of the copes and is zero on the rest. Then we can decompose any element $v\in H_i\otimes W_i$ as $\sum h_j w_j$, which gets sent to $$w_1\oplus w_2 \oplus \cdots \oplus w_n$$ in $V$. (I realize I'm being a little sloppy with the isomorphisms here, but that idea seems correct.)

However, I am a little hesitant about part (c). First, let's proceed directly. Clearly each choice of basis ($h_i$) gives an isomorphism of representations. Further, each isomorphism is determined by its action on each copy of $W_i$. Call these maps $g_i$. The $g_i$ must be independent, or the map is not an isomorphism, so they form a basis of $H_i$. Hence we have a one to one correspondence of bases and isomorphisms.

Deriving (c) from (b) seems slightly tricker. I suppose (b) allows us to reduce the problem to classifying maps $\oplus_1^k W_i\rightarrow H_i\otimes W_i$. Can we then argue as in part (b)? We can decompose any element $v\in H_i\otimes W_i$ as $\sum h_j w_j$, where $h_j$ is a basis for $H_i$. So each choice of basis $(h_j)$ gives a different isomorphism with $H_i\otimes W_i$. But it seems tricky to justify that these are all of the isomorphism without resorting to getting your hands dirty as in the direct approach. It would be nice if there was a "coordinate free" way to do this. Hence my question.

Is there a slick way to see that (b) implies (c)?

In other words, can we prove that (b) implies (c) in a way that isn't just a rewrite of the direct approach I sketched?

Given a basis $(h_1, \dots, h_k)$, we can define $h: W_i \oplus \dots \oplus W_i \to V_i$ by $(w_1, \dots, w_k) \mapsto h_1(w_1) + \dots + h_k(w_k)$. To see that $h$ is an isomorphism of vector spaces, it is enough to show that it is surjective. But this follows from (b) because any element of $V_i$ is of the form $\sum H_\alpha(w_\alpha)$, and $(h_1, \dots, h_k)$ is a basis, so this sum can be expressed as a linear combination of these basis vectors. The proof that $h$ is an isomorphism of representations is similar to the one given for $F$. Conversely, given an isomorphism $h: W_i \oplus \dots \oplus W_i \to V_i$ of representations, we can obtain linear maps $h_j: W_i \to V_i$ by letting $h_j = h \circ \eta_j$ where $\eta_j: W_i \hookrightarrow W_i \oplus \dots \oplus W_i$ is inclusion into the $j$th summand. And because $h$ is an isomorphism of representations, we will have $h_j$ commuting with $\rho$. Since there are $k$ maps, they form a basis if they are linearly independent. If they were linearly dependent, then this would contradict that $h$ is an isomorphism of vector spaces because we could construct a nontrivial vector in the kernel. Thus, every isomorphism arises in the way described in (c).
• I see. So essentially knowing (b) means you don't need to fix a decomposition for $V_i$ in the first part of the argument. – Potato Apr 27 '15 at 7:36